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# Physical Mathematics

## on Mar 03, 2014

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Pythagoras, Kinematics, Kepler Orbit, Pendulum.

Pythagoras, Kinematics, Kepler Orbit, Pendulum.

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## Physical MathematicsPresentation Transcript

• From First Principles PART 0 – PHYSICAL MATHEMATICS July 2014 – R1.5 Maurice R. TREMBLAY
• 2014 MRT θo )(tancos2 ),,( oo 2oo Lh L|g| hLv − = θθ θ vo A project in business and science is a collaborative enterprise, frequently involving research or design, that is carefully planned to achieve a particular aim.1 A project is a temporary endeavor to create a ‘unique’ SERVICE.2 1 Oxford English Dictionary 2 PMBOK 4th Edition – Project Management Institute (PMI® ) A Bα r(t) dr(t)/dt=v(t) ZAB L h AB [ ] ∫≡ B A i ABttZ e)()](),([ vr  D ACTION ACTION PLAN MILESTONE SCOPE ACTION S The time t-integral of the difference between Kinetic K (e.g. when a mass is in motion) and Potential V (e.g. due to gravity and height) energies: ACTION RISK A Project Manager and a Basketball Shot DN EXECUTION 2014 PMONETWORK ENGINEERING =S ∫ t o K−V )( dtt to t g otan 2 tan θα −= L h Basketball weight: 1.31 lb diameter: 9.47 in. Fair=−kv ≅ 0
• Contents PART 0 – PHYSICAL MATHEMATICS Pythagorean ‘3-4-5’ Theorem The Right Triangle Definitions & Algebra Special Relativity Scalars and Vectors Trigonometry Vector Rules Direction Cosines Non-uniform Acceleration Problem Solution Kinematics of a Basketball Shot Problem Coordinates Solution Visualizing Newton’s 2nd Law using Kinetics Trajectory of a Projectile with Air 2014 MRT The Simple Pendulum – Diagram of Forces Diagram of Forces & Force Components Newton’s Law & Equation of Motion Frequency & Period Mathematical Models Advanced Models I & II Calculus of Variations General Path Rules Vector Calculus The Line Integral Vector Theorems Complex Functions Matrix Operations Derivative of f (z) Rotation Transformations Contour Integrals Cauchy’s Integral Formula Fourier Series and Transforms Electromagnetism and Relativity “The enormous usefulness of mathematics in the natural sciences is something bordering on the mysterious and that there is no rational explanation for it.” Eugene Wigner, ‘The unreasonable effectiveness of mathematics in the natural sciences.’ Richard Courant lecture in mathematical sciences delivered at New York University, May 11, 1959 (1960).
• You will notice three primary colors in the slides: Dark Red, Blue and Green. They are either are curves, vectors or variables in Figures, variables or terminology within the text, bold italic terms (e.g., how something is called – either mathematical or physical) or even to help in focusing on ‘results’ such as equations boxed with one of these colors: •Dark Red is good for skimming through and paying attention to the key stuff; •Blue is the ‘from first principles’ part (and pretty much the reason for doing this so anything blue you need to read through to understand the whole thing!) and; •Green is the physical part where the subject matter (i.e., the physics) is highlighted. Otherwize, these colors just makes for a nicer presentation! Italics are also used to spell out definitions (e.g., of variables and/or constants) or key physical statements. Since mathematics is a language you will be able to peruse a slide by homing onto the mathematics which is set in black. Scalars are typically black italic (e.g., the Cartesian coordinates x, y or z) while vectors are typically black bold (e.g., the position vector r). This applies to constants, variables, &c. This presentation is key in developing a sense of ‘reading’ the physics through ‘reading’ equations… It was the reason why I did this – to be able to peruse the slides and take on the whole whopping content in order to assimilate a result, equation, complex term or mathematical or physical conclusion. Notation “Now one may ask, ‘What is mathematics doing in a physics lecture?’ We have several possible excuses: first, of course, mathematics is an important tool, but that would only excuse us for giving the formula in two minutes. On the other hand, in theoretical physics we discover that all our laws can be written in mathematical form; and that this has a certain simplicity and beauty about it. So, ultimately, in order to understand nature it may be necessary to have a deeper understanding of mathematical relationships.” Richard Feynman, Feynman Lectures on Physics, Vol. I (1964). 2014 MRT
• Pythagorean† ‘3-4-5’ Theorem In any ‘right’ triangle*, the area of the square whose side is the hypotenuse c is equal to the sum of the areas of the squares whose sides are the two legs a and b. In vector c (or the 4-vector u ) notation we write: (direction) _ In scalar c2 ≡ |c|2 (or the invariant ds2 ) notation we write: a2 ⊕ b2 = c2 (magnitude) _ N.B., The Theorem only works in 2-dimensions: a flat table, a piece of 8.5 × 11 paper, &c. In 3 & 4-dimensions, We have to add depth and interval of light travel… † Pythagoras, about 572-497 B.C. *The right angle is indicated by the □. c is the opposite side of the right-angle and a and b are the adjacent sides to this 90° angle (or π/2 radians). Furthermore, the sum of all the angles (only 3 in a triangle or even 3∠) within the ‘right’ triangle is also equal to 180o . This slide shows that paying attention to detail in physics is crucial: each letter or symbol has a meaning – it is a language and it has a meaning which at times is far reaching! ≡ ⊕ = ba c cba  =⊕c  2014 MRT HYPOTENUSE ADJACENT OPPOSITE c2 b2a2 O  3 45 32 42 52 a bc
• cba =+ 22 :Algebra “The square on the hypotenuse is equal to the sum of the squares on the other two sides – one adjacent to the right angle and the other opposite.” If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the algebraic (latin: algebra) equation: a2 + b2 = c2 or, solved for c (using the square-root symbol – i.e., “√ ” which is the inverse “…2 ” function of squaring things – i.e., Einstein’s Mass-Energy formula: E = m‘c2 ’): or, using a = 3 and b = 4, we get: 32 + 42 = (3×3) + (4×4) = 9 + 16 = 25 32 + 42 = 52 ⇒ √(9 + 16) = √(25) = 5  With the Pythagorean theorem, x2 + y2 = 12 = 1×1 = 1 meter-stick long (a Metric reference frame), we have cos2 θ +sin2 θ =1 or with exp(Im(t)) = eiθ = cosθ + i sinθ we get eiπ +1=0 when θ = π ≡ PI()≅ 3.14… = 180° and i = √(−1) is the complex number and i = eiπ/2 . 222 cba =+:Algebra a bc HYPOTENUSE ADJACENT OPPOSITE 2014 MRT
• D V This is a galaxy located 1.5 Mpc away receding from our galaxy with a velocity of 110 km/s giving Ho = V /D = 73 km/s/Mpc.              Ho = 73 km/s/Mpc Hubble’s Law: Velocity = [Slope] × Distance Again, in the language of math (i.e., in an algebraic equation): y = m × x + 0 where y is the observed velocity of the galaxy away from us, usually in km/s. m, the slope, is a “Constant” – the slope of the graph – here in km/s/Mpc. x is the distance to the galaxy in Mpc. Finally, given any m, with x = 0 we have y = 0, then b = 0 (see Graph). Ho = 50 km/s/Mpc Velocity(Km/s) Distance (Mpc / 1×106 parsec) 2014 MRT       ×=      galaxythe of galaxythe to VelocityDistance oH = D V Ho
• The dominant motion in the universe is the smooth expansion known as Hubble’s Law. Recessional Velocity of a Galaxy = Hubble’s Constant × Distance In 1929, Hubble estimated the value of the expansion factor, now called the Hubble constant, to be about Ho ~ 50 km/s/Mpc. Today the value is still rather uncertain, but is generally believed to be in the range of Ho ~ 45-90 km/s/Mpc. In the language of Physics (i.e., a mathematical law representing a true observable phenomenon): V = Ho × D V is the observed velocity of the galaxy away from us, usually in km/s. Ho is Hubble’s “Constant”, in km/s/Mpc. D is the distance to the galaxy in Mpc (1 Mpc ~ 3×1022 m) Velocity(Km/s) 2014 MRT Distance (Mpc / 1×106 parsec)
• where the symbol × (cross product) represents the vector product. In the Figure, the variable θ is the angle between the vectors a and c. Vectors, on the other hand, are ‘oriented’ objects and the vector ‘addition’ rule is: Examples of vectors are: position, displacement, velocity, acceleration, force and torque. a c=+ b bc a θ In a plane (e.g., a piece of paper on a large table) d c×ad = Vector d comes out of front of page.  Adding the vectors a and b together produce a resultant vector c. The product of the vectors a and c, respectively, produces a new       Scalars are quantities represented by ‘magnitude’ only. For example: mass, volume, density, energy and temperature. The magnitude is represented by the ‘absolute value’ symbol |…| which means |a|=|±a|=a, the magnitude of the vector a.   One of the key properties of vectors is the capability of generating a new vector from the product of 2 vectors on a plane – this new vector will be perpendicular (out of page): × = ˆ k j i ˆ ˆ ˆ k j i ˆ ˆ ˆ k j i ˆ ˆ ˆi ˆj ˆk ˆi ˆj ˆk × =ˆj ˆk ˆi × =ˆk ˆi ˆj and × =ˆi ˆi 0 × =ˆj ˆj 0 × =ˆk ˆk 0 2014 MRT O   With this definition, we can create an orthogonal basis for a vectors space in three dimensions consisting in mutually perpendicular unit vectors , and :
• θθ coscos ⋅=⇒=== ca c a eHypothenus sideAdjacent Cosine θθ sinsin ⋅=⇒=== cb c b eHypothenus sideOpposite Sine a b a b c a c b =⋅===== c cθ θ θ cos sin tan sideAdjacent sideOpposite Tangent 1sincos 22 =+ θθ Now, the equation: a2 + b2 = c2 will give with the above: Here’s a rough proof: Dividing each term by c gives us cos2 θ +sin2 θ = 1. √[(c⋅cosθ )2 + (c⋅sinθ )2 ] = √[(c)2 ⋅(cosθ )2 + (c)2 ⋅(sinθ )2 ] = √[c2 ⋅(cos2 θ + sin2 θ )] = √[c2 ⋅(1)] = √(c2 ) = = c  With emphasis to show that “ √ ” is the inverse function of “ …2 ” because in they end, they nullify each other. Definitions: Sym. Name Ratio of… (i.e., side a devided by side c) 2 c 2014 MRT a bc HYPOTENUSE ADJACENT OPPOSITE θ c⋅sinθ c⋅cosθ Hence: If: and
• a c θ b In Space (e.g. look around!) c× = a sinθad = c The vector product is given by: The scalar product is given by: The vector commutativity addition rule is given by (i.e., the ‘sum’ order is irrelevant!): d c = a + b a+b= c×ad = c × a= − Vector d comes out of bottom.  Adding arbitrarily the vectors a and b together produce a resultant vector c. The product of the vectors a and c, respectively, produces a vector d and the angle θ is shown.       while the vector associativity addition rule – introducing a new vector e (e.g., a unit normal vector) – is: e( b+a b+a=)+ e + ( ) where the scalar product symbol ‘ • ’ (dot product) of a vector a with a vector b.   The magnitude of the vector product of two vectors can be constructed by taking the product of the magnitudes of the vectors times the sine of the angle (<180 degrees) between them. b• = ba cosθ b a•=a where the symbol ‘ × ’ (cross product) represents the vector product whereas the magnitude of this new vector is: ˆ 2014 MRT O e Unit vector e comes out from the top. ˆ ˆ ˆ ˆ
• The position vector, r, of an object located at a Point P(r) =P(x,y,z) is then given in a ‘Cartesian’ coordinate system as a function of Cartesian unit vectors: , and : The angles α, β, and γ are the angles that OP makes with the three coordinate axes. Here we have x = r , y = r m, and z = r n where  = cosα, m = cosβ, and n = cosγ (with 1 = 2 + m2 + n2 ). The quantities , m, and n are called direction ˆ k j i ˆ ˆ y x z rα β γ • In the language of line segments (i.e., OA, AB, BP, OP, &c – see Figure below), an object located at P from an origin O is represented by the relation: P(x,y,z)B A BPABOAOP ++= The position of a physical object is completely specified by its position vector (some- times called radius vector). To start with, the position vector is a vector drawn (pretty much) from the origin of a (say Cartesian) coordinate system to the object in question. ˆi ˆj ˆk The vectors , and are three components of r; they are the vector representations of the projections of r on the three coordinate axes, respectively. The quantities rx, ry, and rz are the magnitudes of the vector components in the three respective directions (e.g., in Cartesian: x, y, and z.) ˆirx ˆjry ˆkrz Using the Pythagorean theorem, we find that: 1ˆ ˆcosˆcosˆcosˆ = ++=≡ r kji r r r and γβα ˆr 2014 MRT ! )()()()( 222 22222 zyx BPABOAOP ++= ++== r r kji kjir ˆ )ˆcosˆcosˆ(cos ˆˆˆ r r rrrOP zyx = ++= ++== γβα O • •
• gs is the value of g at Earth’s surface (9.807 m/s/s or m/s2 ), Rs is the radius of the Earth (6.38×106 m), h is the altitude above Earth (m), gh is the value of g at altitude h (m/s/s). The percentage difference in radii is 1.0025%! These different values for the Earth’s radius is a characteristic of the Earth’s oblateness (e.g., Earth kinda looks like an egg!) 2 s s sh       + = hR R gg gh Rs gs h • The Earth’s equatorial radius of curvature in the ‘meridian’ is b2 /a or 6335.437 km. The Earth’s ‘polar’ radius of curvature is a2 /b or 6399.592 km. Difference is ∆ ≅ 1 %! The Earth’s radius at a geodetic latitude, ϕ, is given by the radius of the Earth formula: The plot of gh =gs*(Rs/(Rs+h))^2 in Excel shows that gh will decrease linearly with altitude. e.g., The space shuttle reaches an altitude of 300 km. h R(ϕ) Space Shuttle Orbit ti tf = ti + T • = gsgh C e l l 22 2222 S )sin()cos( )sin()cos( )( ϕϕ ϕϕ ϕ ba ba RR + + == 2014 MRT •
• The unit vector in the direction of r is thus identified as: The dot product of two vectors a and b with angle ϕ between them is a scalar quantity defined by the equation: The cross product of two vectors a and b is a vector defined by the equation (the product is noncommutative!): where r is a unit vector in the direction perpendicular to both a and b. From the Figure it is seen that the magnitude is equal to the product of the length of one of the two vectors and the projection of the other on a line perpendicular to the first vector, which is equal to the area of the parallelogram formed with a and b as sides. The result is not dependent on the order of multiplication and, hence, the dot product is commutative (i.e., independent of the order of summation): kji r r r r ˆcosˆcosˆcosˆ γβα ++==≡ r ϕcosab=•ba abbaabba •=•⇒=•−• 0 The equation suggests a convenient procedure for determining the component of a vector r along any chosen direction r, the result being (see Figure): ϕcosˆ r=•rr abrba ×× −== ˆ)sin( ϕba ˆ Component of r along r by dot product and cross product of two vectors is normal to the plane of the two vectors. r •P ˆr 2014 MRT ϕ r • rˆ ˆ ˆr ϕ a b a× b ˆ  |r|cosϕ
• When resolved into rectangular (i.e., Cartesian) components, the dot product is: since i× j=k (j× i=−k), j× k=i (k× j=−i) & k× i=j (i× k=−j) and i× i=0, j× j=0 & k× k=0. The result of the cross product can be conveniently expressed by the determinant: 2014 MRT since i• j=j• i=0, j• k=k• j=0 & k• i=i• k=0 and i• i=1, j• j=1 & k• k=1 while the cross product becomes: ˆ ˆ ˆ ijk kjkiji 0kjkikj0jikiji0 kkjkikkjjjijkijiii kkjkikkjjjijkijiii kjikjiba ˆ)(ˆ)(ˆ)( )ˆˆ)(()ˆˆ)(()ˆˆ)(( )ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ( )ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ( ˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆ )ˆˆˆ()ˆˆˆ( yzzyzxxzxyyx yzzyxzzxxyyx yzxzzyxyzxyx zzyzxzzyyyxyzxyxxx zzyzxzzyyyxyzxyxxx zyxzyx babababababa babababababa babababababa bababababababababa bababababababababa bbbaaa −−−= −−−= += = = = ++ ×+×+× ×+×−×++×−×+×+ ×+×+×+×+×+×+×+×+× ×+×+×+×+×+×+×+×+× ++×++× ↑ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ zzyyxx zzyzxzzyyyxyzxyxxx zzyzxzzyyyxyzxyxxx zyxzyx bababa bababababababababa bababababababababa bbbaaa ++= •••••••••= •••••••••= •=• )ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ( ˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆ )ˆˆˆ()ˆˆˆ( kkjkikkjjjijkijiii kkjkikkjjjijkijiii kjikjiba ++++++++ ++++++++ ++++ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ijkkji kji ba ˆ)(ˆ)(ˆ)(ˆˆˆ ˆˆˆ yzzyxzzxxyyx yx yx zx zx zy zy zyx zyx babababababa bb aa bb aa bb aa bbb aaa −−−=+== +−−× ↑↑
• Certain multiple products of vectors are occasionally encountered and we list two of the most common ones in the following: 2014 MRT cbabcacba )()()( ••= −×× and )()()( bacacbcba ××× •=•=•
• Let’s get right into the physics of the cross product: the Moment of a Vector! We consider a vector F and any point O in 3-D space. If we draw a vector r from O to any point on F or on the line of action of F (see Figure) and form the cross product with r as the first vector, then the result will be a moment M about the axis through O in a direction perpendicular to the plane containing r and F, the direction of which is indicated by the unit vector n: The moment M will be independent of where r terminates of F or on its line of action. nFrM ˆ)sin( ϕrF== × M nˆ r ϕ In a plane (e.g., 8½×11 paper) The vector M comes out of front of page. 2014 MRT The Moment vector M normal to plane of position vector r and Force vector F is r× F. F O ˆ
• As an application of a moment of force (here torque τ is used in a context of an applied weight w=−mg)letus consider the scenario depicted in the Figure which shows a diagram of an arm and the biceps muscle. The muscle is attached at a point about 4 cm from the socket which acts as a pivot point. If a weight of 50 N is held in the hand, what is the tension in the muscle? (Assume that the forearm is a horizontal uniform rod of weight 15 N and lengthL=30cm andthattheforce exertedby the muscle acts at 10° to the vertical.) The Figure (τ out of page) shows the forces acting on the forearm, where FH and FV are the horizontal and vertical components of the force exerted by the pivot of the forearm. 2014 MRT Since the muscle is close to the pivot point, it must exert a force much larger than the weight of the ball (i.e., using wBall =mg: 50 N =m ⋅ 9.8 m/s2 ⇒ m≅ 5 kg). L=30 cm Biceps (Tension T) 4 cm θ = 10° d +τ wArm wBall FH FV By taking torques about the pivot we avoid the force exer- ted by the pivot. Thus, using the components of the torque τ =r× F (i.e., with its magnitude being |τ |= |F||r|sinθ for d, L and T instead) explicitely under static equilibrium conditions: 0)1()3.0()50()1( 2 3.0 )15()1()04.0()10cos( 090sin90sin 2 1 90sincos =⋅⋅−⋅      ⋅−⋅⋅= =⋅⋅−⋅⋅−⋅⋅+=∑ mN m Nm BallArm   T LwLwdT θτ T Thus, T = |T|=438 N, which is considerably greater than the weight wBall (i.e., 50 N or ∼10×). As for the forces FH and FV : V y H x wwTF TF F F ==−−= −−= === = ∑ ∑ NNNN NN BallArm 3665015)10)(cos438( cos 76)10)(sin438( sin   θ θ +g Humerus Radius Ulna 0wLwLTdτ ==∑ BallArm ×−×−×+ 2 1 τ d L
• Finite angular rotations of a rigid body do not obey the law of composition (which states that the sum of two vectors is represented by the diagonal of a parallelogram formed by the two vectors as sides.) It has magnitude equal to the angles of rotation, which can be directed along the axis of rotation. However, two such rotations along different axes are not commutative and will not add up to the diagonal of a parallelogram (e.g., Pick up an iPhone and lay it flat in your right hand so that your fingers can wrap around it. Rotate it 90° about the 1- and 3-axes with your left hand. You need to hold it up sideways! Then repeat the procedure in the reverse order. You will need to hold it up facing you! Thus finite angular rotations are shown to be non-commutative: 1⋅3≠3⋅1) Infinitesimal rotations however can be shown to be commutative (i.e., R⋅R−R⋅R=0 so here we have 1⋅3=3⋅1⇒1⋅3−3⋅1=0) and to possess all the properties of a vector. ê1 r ω1 2014 MRT The angular velocity represented by a vector. r1 O ê2 To show this, consider the displacement of a point P(r) due to two infinitesimal rotations ω1 dt and ω2 dt about any two axes, where ω1 and ω2 are their respective rotational speeds. Let the direction of each of the axes of rotation be indicated by unit vectors ê1 and ê2 as shown in the Figure, and we will perform the rotations in the order 1 and 2, then repeat in the reverse order to examine the final result. Because of the infinitesimal rotation (ω1 dt) ê1, the end of the displacement vector r defining the position P will be displaced by an amount: re ×11 ˆ)ω( td and the new position is defined by the vector: rerr ×+ 111 ˆ)ω( td= ê3 ω2 ê1 r ê1× r P P (ω1 dt)ê1× r
• Next allow the second infinitesimal rotation (ω2dt)ê2, in which case the final position of P(r2) is defined by the vector r2: reererer rerererrerr ××+×+×+ ×+×+×+×+ 11222211 11221112212 ˆ)ω(ˆ)ω(ˆ)ω(ˆ)ω( ]ˆ)ω([ˆ)ω(]ˆ)ω([ˆ)ω( tdtdtdtd tdtdtdtd = == 2014 MRT r1 ω2 r2 P Neglecting the second-order term (i.e., ω1ω2(e2× e1 r)dt2 ) above, we arrive at the result: tdtd rrreerr ×ω+ω+×++ )()ˆωˆω( 2122112 == since ωn=ωnên. ê1 ê2 O ê3 ω1 r ê1 r If we repeat the operation in the reverse order, we will find the equation for r2 to be identical to the case above, indicating that infinitesimal rotations are commutative (at least to first-order since we neglected the second-order – or quadratic – term.)
• Time [s] Speed,v[mi/h] Useful unit conversions: • 1 mile [mi] = 1.61 kilometers [km] • 1 foot [ft] = 0.3048 meter [m]. • 1 Km = 1000 m = 103 m. • 1 m = 3.281 ft. • 1 mi = 5280 ft. For example, we have: • 5 mi/h = 8 km/h = 2.2 m/s.* • 10 mi/h = 16.09 km/h = 4.2 m/s. • 50 mi/h = 80.47 km/h = 37.8 m/s. • 100 mi/h = 160.93 km/h = 71.5 m/s. * For example, to convert mi/h into m/s: . s m 2.2 s3600 h1 km1 m10 mi1 km1.6 h1 mi5.0 h mi 0.5 3 =                        = The graph below represents the speed, v, as a function of time, t, of a 1966 Alpha Romeo GT. (Reprinted from Road and Track, November 1986.) a) Determine the average acceleration in mi/h/s for each of the three speeds. b) Suppose that the acceleration is constant (Uniform Acceleration) from rest until the shifting from 3rd to 4th speed. What would be the distance traveled? c) Determine the real distance traveled up to the shifting from the 3rd to the 4th speeds. Non-uniform Acceleration 2014 MRT
• Time [s] Speed,v[mi/h] (4s,37mi/h) (0 s,0 mi/h) mi/h/s8.4 s5.4s5.9 mi/h371mi/h6 32 = − − =→a mi/h/s2.3 s107s1 mi/h614mi/h8 43 = − − =→a (9.5s,61mi/h) mi/h/s2.9 s0s4 mi/h0mi/h37 21 = − − =→a (17s,84mi/h) a) Using a = ∆v/∆t = (vf − vi)/(tf − ti) (see Plot below), we get • up to 1st →2nd : 9.2 mi/h/s (~15 km/h/s) • up to 2nd →3rd : 4.8 mi/h/s, (~8 km/h/s) • up to 3rd →4th : 3.2 mi/h/s (~5 km/h/s) b) Using P1: (0s,0mi/h) and P2 : (17s,84mi/h), we get a = 4.94 (mi/h)/s (~8 km/h/s). Then, by checking units and using the kinematics formula and solving for s = (vf 2 − vo 2 )/2a, we get the distance 2000 ft. c) 1200 ft. ∆v t [s] v [mi/h] 9.5 s4.5 s 37 mi/h 61 mi/h if if tt vv t v a − − = ∆ ∆ = The slope of a v(t) plot: )( 2 tfv = ∆t This acceleration is considered to be constant from rest until the shifting from 3rd to 4th speed: a = (vf – vo)/(tf – to) = (84–0)/(17–0) a ≅ 5 mi/h/s        P1 P2 2014 MRT is the acceleration a(t).
• A player throws a basketball with an initial velocity vo along an initial angle θo [theta] towards a hoop situated at a horizontal distance L and at a height h above the throw stance. Assume no air-resistance nor jump. α Show that the scalar initial velocity required to reach the hoop is given by the relation: Finally, g is the constant acceleration due to Earth’s gravity (∼10 meters per second, per second or 10 m/s2 † ) and is given by g = (GM⊕)/R⊕ 2 , where G is Newton’s constant of universal attraction (6.673×10−11 m3 /kg/s2 ), M⊕ is the mass of the Earth (5.9742×1024 kg) and R⊕ is the equatorial radius of the Earth (6,378,100 m). † Basketball players in the United States of America would understand g as: The speed of an object in free- fall will increase by 32 feet per second with each second it falls or 32 ft/s2 , which is always directed towards the center of the Earth. )(tancos2 oo 2o Lh Lg v − = θθ * The time (t-) dependent parabolic trajectory of a basketball is given by (g/2)t2 +(vo)t+(so−s)=0which is of the form k1t 2 +k2t +k3 = 0 where k1=|g|/2, k2=|vo| and k3=|so−s|. Also, the absolute value means |x|=|±x|= x. A basketball follows a path in xy-tSpace-Time – showing a parabolic trajectory.* 201 4 PM O Allstrea m θo Kinematics of a Basketball Shot 2014 MRT Efficiency in service delivery: otan 2 tan θα −= L h
• z F y F x F zyx ∂ ∂ + ∂ ∂ + ∂ ∂ =•F∇ Point P at the tip of the distance vector r is given in Rectangular (Cartesian) Coordinates by the intersection of constant x, constant y and constant z planes (i.e. a box in 3-D!) – relative to the frame of the basketball. zyx ˆˆˆ z f y f x f f ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ zy x zyx F ˆˆ ˆ/// ˆˆˆ       ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ + +      ∂ ∂ − ∂ ∂ =∂∂∂∂∂∂= y F x F x F z F z F y F FFF zyx xyzx yz zyx ×∇ 2 2 2 2 2 2 z f y f x f ff ∂ ∂ + ∂ ∂ + ∂ ∂ =•=∇2 ∇∇ y x r P ˆ k j i ˆ ˆ k i ˆ Constant y plane j Constant x plane kjirr ˆˆˆ),,( zyxzyx ++== Constant z plane y x z r O zyxF ˆˆˆ zyx FFF 2222 ∇+∇+∇=∇ y x z O d d y dx d z dzdydxdV = z O kji ˆˆˆ dzdydxd ++= d is an infinitesimal differential increment of length. dV is an infinitesimal differential increment of volume. 2222 dzdydxd ++= 2014 MRT The Laplacian of a vector function F= F(x,y,z): The Laplacian of a scalar function f = f (x,y,z): As a vector product of a vector function F=F(x,y,z): As a scalar product of a vector function F= F(x,y,z): The gradient of a scalar function f = f (x,y,z):We can make the following geometric objects into physical realities if we substitute the scalar initial velocity vo for f = f (x,y,z) and gravity g =− = − for the vector F = F (x,y,z). so g Fans P ˆ x y ˆ 2 2 1 oo tt gvss ++=− As a function of the distance of the ball from the fans, the displacement s= s(x,y,z) is given by the quadratic equation represented as a function of time variable t: where vo is the initial velocity vector and g is gravity. Basketball Path r s ss −− ssoo Plane of free throw vo yx ˆˆ − ˆ ˆ ˆj ˆy • • ˆx yˆ
• x )(tancos2 ),,( cos2 )(tan coscos sin )sin( cos )cos( oo 2oo o 22 o 2 o 2 oo 2 1 oo oo 2 2 1 oo 2 2 1 oo oo oo 2 2 1 oo Lh Lg hLv v Lg Lh v L g v L vh tgtvhy tatvyy v L ttvLx tatvxx yy xx − = −=       −      = −== ++= =⇒== ++= θθ θ θ θ θθ θ θ θ θ )algebra...12(Grade whereas, along the y-axis, we have: The initial velocity vo is obtained by using the Kinematic Equation of Motion of a projectile: This is the required initial velocity (speed) of the basketball. The acceleration components are given by |ax|=0 and |ay|=g. The kinematics along the x-axis are: The required velocity vector vo will depend on gravity g andonthe scalar parameters θo,Landh. ŷ [m/s] According to Einstein’s Theory of Gravity: Space acts on matter, telling it how to move; in turn, matter reacts back on space, telling it how to curve. (g) vo= vox + voy vox=vocosθo voy=vosinθo vo θo Vector Addition hmax Gravity ° yxyx tt yxtt aa v avv v vyxsavss +++++ = ∆ ∆ == ∆ ∆ === ,, andwhere oo o o 2 2 1 oo ˆˆ so 2014 MRT ˆ (since ay = −g) Note: vy= 0 at hmax (θ = 0) and since there is no air-resistance, vx = vox= cte.
• Imagine that a mass m of 4 Kg (4⋅1000 g) is being pushed by a constant force F of 10 Newtons [N]. a) Calculate the uniform acceleration a of the mass m over a distance of 45 meters [m]. b) Represent the motion using a strobe graph in 1-second [s] time intervals and show the various distances s(t) and velocities v(t) corresponding to these time t intervals. Assume that there is no friction between the block and the surface (µ=0): Imagine a 4 Kg block of ice sliding on ice! tav = 0 1 s 2 s 3 s 4 s 5 s 6 s 0 5 m/s 7.5 m/s 10 m/s 12.5 m/s Speed v(t): Time t: 2.5 Visualizing Newton’s 2nd Law using Kinetics 2 2 m/s5.2 Kg4 m/sKg10 Kg4 Newtons10 = ⋅ ==⇒== a m a F a In this case, Newton’s 2nd law is simply F=ma. Dividing this equation by m on both side of the equality sign, we get:      F 2 2 1 tas= 0 5 m 11.25 m 20 m 31.25 m 45 m Distance s(t): 1.25 When a constant force F is applied to a mass m it will move with constant acceleration a. Every 1s, a strobe ‘’ captures the motion of a 4 Kg mass under an applied 10 N ‘constant’ force… 0 2.5 m/s2 Acceleration a=|a|: 2 m/s2.5=a2.5 m/s2 2.5 m/s2 2.5 m/s2 2.5 fkBut if a friction force fk (=µN) is present (i.e., µ ≠ 0) it will oppose the acceleration.        N =+mg Nµ µ 40≅ = gmfk 1 s 2 s 3 s 4 s 5 s … s 2014 MRT The accelleration increases by 2.5 m/s (constantly) every second – 2.5 m/s /s! The acceleration due to gravity g is 9.8 m/s /s!
• As an application of a accelerated motion consider Atwood’s machine. It was an apparatus that allowed a direct measurement of Newton’s second law. It could also be used to measure the gravitational constant g (we’ll use m/s2 ). Two equal blocks of mass M hang on either side of a pulley (see Figure). A small square rider of mass m is placed on one block. When the block is released, it accelerates for a distance h until the rider is caught by a ring that allows the block to pass. From then on the system moves at a constant speed which is measured by timing the fall through a distance d. Show that: 2014 MRT Decomposing the vertical components of the net force: 2 2 2 )2( thm dmM g + = maMamgMgTF MaMgTMaMgTF y y −−=−−+= +=⇒+=−+= ∑ ∑ )2( )1( and finally for h to d, d =vt which implies v=d/t so we get g! From 0 to h, v2 =+2ah which implies that a =v2 /2h and: g h v m mM =      + 2 2 2 ga m mM maMamgMgMaMg =      + ∴ −−+=+ 2 With T from  we put it into  to obtain after some algebra: An Atwood machine. M M m h d T Mg + where t is the time moved at constant speed.   T   +
• As an application of a friction force let us consider the scenario depicted in the Figure. A car travels in a horizontal circle around a curve of radius R banked at an angle θ to the horizontal. If µs is the coefficient of static friction, whow that the maximum speed possible without sliding sideways is (do consider that aR =(v2 /R)ên and assume that µs tanθ <1): 2014 MRT A car is traveling on a banked road at a constant speed v. We start by decomposing the components of the net force: 0sincos 0cossin =−−+= =−−−= ∑ ∑ θθ θθ cy csx FgmNF FgmfF From equation  above, we get since fs =µs N: θµ θµ tan1 )tan( s sgR v − + =max mg y Ncosθ N θ θ Nsinθ x aR fs   θ θµ θ θ cos sin cos sin gmNgmf F ss c + = + = whereas from equation  above, we obtain: θθ sincos cFgmN += Thus: θθµµ θ θµθµ tantan cos sincos gmFgm Fgm F css css c ++= + = and finally with Fc = maR =m vmax 2 /R and we get vmax above.         − + =⇒+=− θµ θµ θµθµ tan1 tan )tan()tan1( s s cssc gmFgmF and then:
• vxi vyj −kvyj −kvxi Gravity Velocity v i j Air Resistance −mg Fair=−kv Below is a free body diagram (not to scale) for a projectile that experiences air resistance and the effects of gravity (the dashed vectors are the x and y components of velocity and air resistance). Here we assume that the air resistance is in the direction opposite of the projectile’s (e.g. baseball) velocity. Trajectory of a Projectile with Air Resistance When the velocity of the projectile is 4 m/s, the air resistance is 7 N. When we double the velocity to 8 m/s, the air resistance doubles to 14 N accordingly. In this case, k = 7/4 = 1.75 N•s/m – i.e. 75% more resisitance to travel than without air resistance! We can write that the air resistance is an inverse linear function in the velocityv with k as the drag coefficient (Atmospheric drag is highlighted below but not considering uncertainties in atmospheric fluctuations, Earth’s oblateness, &c. If we consider only the cross-sectional area A perpendicular to the direction of motion, the dimensionless drag coefficient CD assiciated with A, and other parameters such as the speed of the mass m relative to the atmosphere va and the atmospheric density ρ at the mass’ altitude): 2014 MRT tCoefficien Ballistic 1 2 1 air =   −=−= mg ACD aD vACvkF ρ
• Neglecting the Magnus effect (i.e., basketball rotation via (16/3)π2 rbasketball 2 ρ ω × v) the formula for calculating the drag force of an object moving through a fluid is thus: 2 2 1 aDD vACF ρ= where ρ is the density of the fluid and A is the area of the ball. For air at sea level, ρ=1.2 Kg/m3 . The coefficient of drag, CD ~0.5 for most spheres for most sports conditions. For basketball with v=20 m/s and A=0.046 m2 . FD =½⋅1.2⋅400⋅0.5⋅0.046=5.5 N. Weight w=mg =4.5⋅1.31=6 N. So this is comparable to drag force. CD is not a fixed number but depends on Reynolds number and surface smoothness. This calculation of drag force is close enough. Note that this calculation almost replicates the terminal velocity data from the table above. Range R of a ball launched at 45 degrees. From kinematics we have R= vo 2 /g where vo is the launch velocity assuming the ball is launched from the height of the basket... tall man with jump shot. For a 3 point shot, R=24 feet=7.3 m. Given g=10 m/s2 , vo =√(Rg)=√(7.3⋅10)=8.5 m/s. For 3 point shot, launch velocity in vacuum is 8.5 m/s so drag force FD =(8.5/45)2 =0.036≥3.6% of weight. Since the range varies as √K, and the drag force is proportional to the loss of energy. The range is reduced by about 1.8% or 0.13 m. The basket has an allowance of 0.15/2 m so the loss of range doesn’t cause a clean miss. The players obviously practice shooting at the actual target hoop to get used to the way it looks and to get the feel for the launch velocity from experience... the only way. For shorter shots, the change in range is considerably smaller because the launch
• x x x vk td vd mam −== If g is the acceleration due to gravity – usually taken to be 9.81 m/s2 near the Earth’s surface; θ the angle at which the projectile is launched; v the velocity at which the projectile is launched; yo the initial height of the projectile; d the total horizontal distance travelled by the projectile (e.g. baseball). (83.5m, 0.0m) @ 5.17s (52.0m, 31.7m) @ 2.26s Values for the mass and terminal velocity for a baseball: m = 0.145 kg (5.1 oz) vo = 44.7 m/s (100 mi/h) g = −9.81 m/s² (−32.2 ft/s²) vt = −33.0 m/s (−73.8 mi/h) k mv x θcoso max = (203.6m, 0.0m) @ 6.44s g gyvv v d t o 2 2)sin(sin cos ++ == θθ θ The total horizontal distance d traveled by the projectile. (k = 0):       = − 2 1 sin 2 1 v gd θ [ ]o 2 2)sin(sin cos gyvv g v d ++= θθ θ With Air Resistance Without The time of flight t is the time it takes for the projectile to finish its trajectory. (k = 0) is: The red path is the path taken by our projectile modelled by the equation above, and the green path is taken by an idealized projectile, one that ignores air resistance altogether. Ignoring air resistance isn’t a very good idea. In some cases it's more accurate to assume Fair ∝ v2 meaning when air resistance increases by a factor of p the resistance increases by p2 . To go back to the first example of proportionality, when we doubled the velocity to 8 m/s, the air resistance would instead be quadrupled to 28 N – this only adds to the large amount of error in neglecting air resistance.  45Kg/s,0431.0 m/s0.33 )m/s81.9)(Kg145.0( 2 == − − == θ tv mg k (101.7m, 50.9m) @ 3.22s x = 250 Components along x Components along y gmvk td vd mam y y y +−== Maximum value that can be reached by x: Turns out ignoring air resistance isn’t a very good idea (in this case at least): without it a pitcher could throw a home run with 270 ft to spare! (The mechanics of pitching at 45 degrees notwithstanding.) )e1( sin )( 2 o kt k g k v k gt ty − −      ++−= θ is the angle θ at which a projectile must be launched in order to go a distance d, given the initial velocity v (k = 0).       −+      += θθ θ cos 1ln cos sin)( o 2 o o v kx k g v x k g vxy y 2014 MRT
• L m “Weight of Bob” Due to Gravity Snapshot The Simple Pendulum The key force acting on the bob is that of gravity. The acceleration due to gravity is 9.807 m/s/s. You accelerate by increasing your speed by 9.807 [m/s] with each [s] of fall. Another ‘Force’ is that of the tension on the string exerted by the weight of the bob. It is not important to our present analysis of finding the period of the pendulum. g = 9.807 (m/s)/s 2014 MRT
• θ L m F − mgsinθ Center of Mass −Wtg Bob’s ‘Weight’ Vector (Directed towards the center of the Earth) −Wθ sinθ = (Opposite side / Hypotenuse) = Wtg / W sinθ = Wtg / (m × g) ⇒ Wtg = m × g × sinθ cosθ = (Adjacent side / Hypotenuse) = Wθ / W cosθ = Wθ / m × g ⇒ Wθ = m × g × cosθ ‘Tangential’ component of the bob’s “weight” ‘Azimutal’ component of the bob’s weight tanθ = (Opposite side / Adjacent side) tanθ = sinθ / cosθ = Wtg / Wθ The tangent of both components is thus: θ θ 2014 MRT Bob W= mg
• θ θ L W= mg mbob F=ma s −Wtg With movement (displacement s) there is variation of displacement, ∆s, within a given period of time, ∆t, expressed by: ∆s / ∆t = (s2 − s1) / (t2 − t1) = vtg . This is called the ‘tangent’ velocity (i.e. speed). So, v = s / t when the velocity is constant – there is no acceleration. If there is a variation of velocity with time, there is acceleration (i.e change of speed with time – faster/slower): a = v / t hence the use of ΣF = ma instead of ΣF = 0 (stationary). ΣF = − Wtg = m a The Sum of all Forces is equal to m times a Newton’s 2nd Law: Velocity Acceleration (m = mbob) 2014 MRT
• θ L W= mg m F s − mgsinθ θ − mgsinθ = m a ΣF = m a Newton’s 2nd Law: This is the equation that predicts the position of the bob of length L given g and the angular frequency ω: s(t) = Lθ (t) = A sin(ωt) − Wtg = m a where A is the amplitude and t is the time (in seconds). The displacement s of the bob of mass m is a function of the amplitude A and time t. − mgθ = m a − mgθ = m[d2 (Lθ )/dt2 ] m[d2 θ /dt2 ] − m(g/L)θ = 0 d2 θ /dt2 − (g/L)θ = 0 For θ small sinθ ≈θ (an infinitesimal approximation) If a = dv/dt ⇒ a = d(ds/dt)/dt = d2 s/dt2 where s = L θ : (i.e. divide by L ) (i.e. divide by m ) ( )Lg /ω = 2014 MRT
• θ = θ ' θ θ L W=mg m F s − mgsinθ θ θ L W=mg m F s − mgsinθ where the ‘Angular’ frequency is given by: ω2 = g/L ⇒ and by definition we have ω = 2π / T . The period T is given by: gLT /π2= Lg /ω = θ θ L W=mg m F s − mgsinθ This is the Harmonic Oscillator Eq.: d2 θ /dt2 − ω2 θ = 0 Newton’s 2nd Law: ΣF = m a − mgθ = m a − mgsinθ = m a − Wtg = m a W= mg F For θ small sinθ ≈θ (an infinitesimal approximation)  − mgθ = m[d2 (Lθ )/dt2 ] m[d2 θ /dt2 ] − m(g/L)θ = 0 d2 θ /dt2 − (g/L)θ = 0 If a = dv/dt ⇒ a = d(ds/dt)/dt = d2 s/dt2 where s = L θ : 2014 MRT
• For the pendulum system these forces have two components, one tangential to the arc described by the weight's motion, and the other in the radial direction (perpendicular to the arc described by the weight's motion). Since the weight doesn't move in the radial direction (the length of the rod is constant), we can concentrate on the tangential form of Newton's second law, which states that the tangential force is proportional to the tangential acceleration: F = ma. The mathematical and physical conclusions we have obtained is that: Pendulum Period 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 0.00 0.50 1.00 1.50 2.00 2.50 3.00 Length [m] Period[s-1] • The period T is independent of the bob weight. Mass, m, is not present: T = f (L). • The period is independent of the amplitude. • The square of the period varies directly with the length L. Mathematically, this means: Plot of T =2*PI()*SQRT(L) in Excel L g L T ×≅= 4 π4 2 2 on Earth g L T π2= or, by taking the square root on both sides: [s]Period For a 0.25 m (25 cm) long plastic rod to which is attached a mass m, the period of the pendulum will be 1 s. 0.25 (0.25 m,1.00 s) C e l l 2014 MRT
• L mg k =with If we include ‘damping’ and the correct representation of ‘angular dependence’, as well as a ‘forcing function’, A × f (t) (of amplitude A coupled to its influence over a function of time over the period f (t) ), the ‘general’ equation of motion of the pendulum is: If we assume a linearized version of the first, this case is called the small angle approximation (for we have for any small θ that sinθ (t) ~ θ (t) ). When the pendulum is linear, it is missing the damping (friction) term in its motion which is apparent since there is no forcing function on the right hand side of the equation. The motion of the pendulum can be described as a simple, ideal, linear harmonic oscillator: )()(sin )()( 2 2 tfAtk dt td dt td m =++ θ θ η θ )()( )()( 2 2 tfAtk dt td dt td m =++ θ θ η θ 0)( )( 2 2 =+ tk dt td m θ θEquation of motion of a harmonic oscillator such as the pendulum. 2014 MRT
• In classical mechanics, a simple harmonic oscillator is a system which, when displaced from its equilibrium position, experiences a restoring force F proportional to the ‘vector’ displacement x according to Hooke's law (the ‘ − ’ sign is key and indicates that the acts to oppose the displacement) : where k is a positive constant dependent on the material used to communicate the force. and x(t) is the displacement as a function of time t. The solution to this equation is: xF k−= )ωsin( ω )ωcos()( o o t v txtx       ⊕= Period (T ))(tx ox ω/π2 )time(t y 10.10155.500.500.000.300.3 5.3 +<<×++<<+<<− − zyx z 2014 MRT
• )()()()()( :]),(),([ )]([ )]([ :]),([ )]([ )]([ )()]([ )]([ o1 1 o 1 o 1 o tδtδtδdttδdttδ δ F δ F δFtttFFδ d tdF tδF δ F δFttFFδ d tdF tδF td td δ td tδd tδ td d t t t t t t nnn ξξξξξ ξ ξ ξ ξ ξξξ ξ ξ ξ ξ ξ ξξ ξ ξ ξ ξξ ξ −=== ∂ ∂ + ∂ ∂ === ∂ ∂ ===         =≡ ∫∫       if if Calculus of Variations Since the points (ξo,to)Initial and (ξo,t1)Final are anchored, we have δξ(to) = δξ (t1) = 0. Then the variation of the function ξ(t) is, δξ (t): The fundamental problem of the calculus of variations is that of finding a function ξ(t) such that the functional S is stationary (an extremum) for small variations in ξ(t). This integral corresponds to calculating the sum of infinitesimal products between functionals (functions of functions) and the differentials of time dt calculated on an interval of time going from t =to to t =t1. In short, the functional evolves according to the condition that the initial function ξ(t) and its temporal derivative in time t. The integral also allows us to calculate the progression of the functional F[ξ(t),δξ(t)/dt,t] in time as long as the calculation is done within the limits t =to to t =t1, the observation time. Then, if ξ(t) is a function of time t, δξ(t) will be the variation of the function ξ(t). Two of the infinite number of paths Γ which can be drawn between (ξo,to)Initial and (ξo,t1)Final are indicated – one being the reference [ ] path Γξ and the other the varied [ ] path Γξ . ξ t t1 to Final Initial t )()()( tttδ ξξξ −= and the rules associated to these variations are: )()()( tttδ ξξξ −= ξΓξΓ 2014 MRT Now, we still need to review yet another fundamental concept – that of the Lagrangian. ξ (t) ξ (t)δξ (t) dt
• If the functional is defined by a (classical) Lagrangian L (i.e., the difference between the kinetic energy T due to the state of motion and the potential energy V due to the positioning of the particle) we find that the Euler-Lagrange equation are given by the variation of the action S=∫Ldt (i.e., the action is the time integral of the Lagrangian): or td d δ L δ L Lδ ξ ξξ ξ ξ ξ = ∂ ∂ + ∂ ∂ =   and Then we get by applying the rules of variations: dtδ L δ L dttttLδSδ t t t t ∫∫       ∂ ∂ + ∂ ∂ == 1 o 1 o ]),(),([ ξ ξ ξ ξ ξξ    dtδ dt dL δ Lt t∫       ∂ ∂ + ∂ ∂ = 1 o )( ξ ξ ξ ξ  0 1 o 1 o =               ∂ ∂ − ∂ ∂ + ∂ ∂ = ∫ dtδ L dt dL δ L t t t t ξ ξξ ξ ξ  With the Initial and Final anchored points (i.e., δξ(to)=δξ(t1)=0), we have the extremum: 0 1 o 1 o =               ∂ ∂ − ∂ ∂ == ∫∫ t t t t dtδ L dt dL dtLδSδ ξ ξξ  tdtttLδtdtttLδSδ t t t t ∫∫ == 1 o 1 o ]}),(),([{]),(),([ ξξξξ  0 2014 MRT
• With δξ =εη(t) where ε is a scaling parameter and η(t) is a continuous function of time, we get the Euler-Lagrange equation: 0=       ∂ ∂ − ∂ ∂ ξξ  L td dL ( ) )( 2 1 ),,( 2 rrrr VmtL VTL −= −=  energypotentialminusenergyKinetic The Euler-Lagrange equation then provides us with a remarkable conclusion: (Force)0 )]([)]([ 2 2 12 2 1 F r rr rr r rr rr =−⇒=− ∂ ∂ =         ∂ −∂ − ∂ −∂ =      ∂ ∂ − ∂ ∂ ⇒      ∂ ∂ − ∂ ∂ V td d m VVm td dVmL td dLL td dL ∇    ξξ where we used the total differential dV=∇V •dr to identify the gradient of the gravita- tional potential ∇V, and furthermore – given the definition of the force F as the variation (indicated by the derivative d) of the momentum p with respect to time t – we have: ar rrr Fa vvp F mm td d m td d td d m td d mm td d m td md td d ===      ======  )( 2 2 or If we consider a particle with a position vector r andvelocityvector ,we have:dtd /rr = So the Lagrangian equation provides naturally the relation ‘mass times acceleration’ – Newton’s Second Law of motion! We can also now define the generalized momentum: i i q L ∂ ∂ =p 2014 MRT
• For example, by using the Euler-Lagrange equation, let us determine the minimum (shortest) distance between two points [x1, y1] and [x2, y2] in Cartesian coodinates. Comparing the above integrand with the functional integrand ∫F(y,y′,x)dx(with y′= dy/dx), we find that: The element of distance is: 21 2 22 1               +=+= xd yd ydxdsd The distance between the two points is given by: 2014 MRT where ∫∫               +== 11 11 11 11 , , 21 2 , , 1 yx yx yx yx xd xd yd sds 21 2 1),,(               +=′ xd yd xyyF 0=      ∂ ∂ − ∂ ∂ y F xd d y F
• Since or, after integrating: 212 ])(1[ 0 y y y F y F ′+ ′ = ′∂ ∂ = ∂ ∂ and the Euler-Lagrange equation, ∂F/∂y–d(∂F/∂y′)/dx=0, becomes: 2014 MRT or Constant== A xd yd xdAyd = 0 ])(1[ 212 =        ′+ ′ − y y xd d Constant== ′+ C y xdyd 212 ])(1[ This equation may be written as: where A=(C/1–C)½ . Integrating (i.e., y =∫ Adx) the above differential equation, we obtain: BxAy += which is the equation of a straight line for the constant B.
• L nˆ r θ In a plane (e.g., 8½×11 paper) The vector L comes out of front of page. Scalarsarequantitiesrepresented by‘magnitude’ only,e.g.,coordinates (x,y,z), mass (m), speed (v),volume(V),density(ρ),energy (E),temperature (T),surface (S)&inertia (I ). Vectors, on the other hand, are ‘oriented’ objects that can also depend on other vectors: 2014 MRT General Path Rules )( MomentumLinearvp m= )()(ˆˆsin component-andMomentumAngular zvmrLz ⊥==== nLnprprL θ× where m is theconstantmass and v the velocity vector. Examples of vectors:position (r), displacement (d), velocity (v), acceleration (a), force (F), torque (τ ) and surface (S). 0jjjin 0iiinj 0nnnji == == == ˆˆˆˆˆ ˆˆˆˆˆ ˆˆˆˆˆ ×× ×× ×× and and and m The position vector r and momentum vector p. The vector product of the position r and momentum p produces angular momentum L= |r||p|sinθ n.ˆ r⊥ p=mv We can now create an orthogonal basis (i.e., perpendicular to each other) for a vector space in three dimensions con- sisting of mutually perpendicular unit vectors , and :î jˆ kˆ p px î ˆpy j Lz = xpy−ypx where the symbol × (cross product) represents the vector product, n is a unit vector and wherethemagnitude (orsize)isrepresented bythe‘absolute value’symbol |…| (i.e.,|n|=1). ˆ ˆ î kˆ jˆ jik ˆˆˆ =×ikj ˆˆˆ =×kji ˆˆˆ =× î kˆ jˆ î kˆ jˆ Setting n for the unit normal k (see Figure) we get the algebra: ˆ ˆ One of the key properties of vectors is their capability of generating a new vector from the product of two vectors on a plane–thisnew vector will be perpendicular(out of page):
• Dividing both sides of the above equation by the scalar ∆s allows us to obtain ∆F/∆s= PQ /∆s which is a vector along PQ. If ∆F/∆s approaches a ‘limit’ (i.e., mathematically lim) as ∆s approaches zero, this ‘limit’ is called the derivative of F with respect to s, that is: F(s) Consider a vector F(s) where s is a scalar variable. Here F is a function of s; for each value of s, there is a corresponding value of F (see Figure). The increment in F(s) when s changes to s +∆s is given by: where d/dx is the symbol for the derivative of a function f (in this case – along x only). The uncertainty principle limits ∆x. 2014 MRT PQFFF =∆+=∆ )()( sss − Illustration of the physical meaning of the variation, ∆F, of a vector F(s) by an amount ∆s (i.e., F(s +∆s)) inin the direction of the segment PQ. O P x afxaf x y afxafy xafxaf ∆ −∆+ = ∆ ∆ −∆+=∆ ∆+=∆+ )()( )()( )()( hence havewe Since x xfxxf xd xfd ∆ −∆+ = → )()( lim )( 0Δx :SlopemalInfinitesi Geometrically, the d y/dx number represents the infinitesimal slope at P: y xO ∆x ∆y )( xaf ∆+ )(af a xa ∆+ Q P )(xfy = Secant Tangent Given f a continuous function defined by y= f (x) we have: Q F(s+∆s) ∆F=F(s +∆s) – F(s) kji FFFF ˆˆˆ)()( limlim 00 sd Fd sd Fd sd Fd s sss ssd d zyx ss ++ − =      ∆ ∆+ =      ∆ ∆ = →∆→∆ d x P xp h x ∆ ≈∆e.g.d y F(s)+∆F=F(s +∆s)
• ∆C=∆A+∆B where C=A+B and C+∆A=A+∆A+B+∆A. The derivative of the vector sum A+B (this also applies to the vector difference: A−B=−(B−A)) is obtained in the following manner: Now consider the derivative of the scalar product where C=A•B (same as C=B•A). Here we obtain: BAC ∆∆=∆ + sd d sd d sd d sssd d ss BA BA BACC ++ + ==      ∆ ∆∆ =      ∆ ∆ = →∆→∆ )(limlim 00 where 2014 MRT B AB ABA •+•=•=      ∆ ∆ = →∆ sd d sd d sd d s C sd Cd s )(lim 0 and BABABA ∆•∆+•∆+∆•=∆C 0limlimlim 000 = ∆ ∆ •∆=      ∆ ∆ •∆ →∆→∆→∆ ss sss B A B A since 0)]()([limlim 00 =∆+=∆ →∆→∆ sss ss AAA − On dividing both sides of ∆C=∆A+∆B by ∆s and taking the ‘limit’ as ∆s→0, we obtain: Derivative of a vector C =A+B Derivative of a scalar C=A•B
• The following relations can also be proved (represents the ‘calculus’ of vector analysis): The physical (or vector) quantities with F, the Newtonian force, being the outcome: ( ) ( ) ( ) ( ))()()( )( )( )( sUUs sd Ud sd d UU sd d tss td sd sd d td d a sd d aa sd d sd d sd d sd d sd d === == = = = and where- scalarconstantais vectorconstantais VVV V V AA A A C0 C B AB ABA + ×+×× RuleChain 2014 MRT v v aa vvp Fr r vvp  ≡=⇔====≡=⇔= td d m td d m td md td d td d m )( and where t is time(parameter)andvand a are calledvelocity andacceleration, respectively. delˆˆˆ ≡ ∂ ∂ ∂ ∂ ∂ ∂ = kji zyx ++∇ The del (ornabla) operator, ∇, is a differential operator and is of immense importance in physics. It is defined by:
• In differentiating a vector, the usual rules of the limiting process apply: tttd d tt ∆ ∆ = ∆ ∆ = →∆→∆ rrrrr 00 lim )( lim −+ If the vector r is referenced to a fixed coordinate system, the ∆r is the vector change relative to the coordinates which is also the total change, and the equation above is the total derivative of r. êt r ω Differentiation of a vector r referenced to rotating coordinates. O φ ∆θ x z y If the vector r is referenced to a fixed coordinate system (e.g., a Cartesian one such as Oxyz in the Figure), the ∆r is the vector change relative to the coordinates which is also the total change, and dr/dt=lim∆t→0 (∆r/∆t) is the total derivative of r. If the vector r is referenced to a rotating coordinate system such as the one shown in the Figure, the vector r remaining stationary relative to the rotating axes will undergo a change: φθ sinr∆ where êt is a unit vector along the tangent. rω× along the tangent to the dotted circle, and its rate of change is established by the limit: t t rr t eˆ)sinω(sinlim 0 φφ θ =      ∆ ∆ →∆ Since this expression is equal to the cross product of ω and r, we conclude that due to the rotation of the vector ω of the coordinates the vector undergoes a rate of change of: 2014 MRT ∆ r
• This ω × r term occurs in addition to the vector change relative to the coordinate system, so that the total derivative relative to inertial axes is: rω rr ×+ scoordinate toRelative Inertial       =      td d td d This equation applies to any vector quantity and is of fundamental importance to dynamics where body-fixed axes are often used.
• The subject of kinematics is the study of motion. It is concerned with space and time, and with the time rate of change of vector quantities relating to the geometry of motion td d tt tt rrrrr v = ∆ ∆ = ∆ ∆ = →∆→∆ 00 lim )( lim −+ We consider first the motion of a point in a fixed coordinate system xyz. The position of a point P(r) which is in continuous motion along a curve such as s in the Figure is specified by its position vector r, the magnitude and direction of which are functions of time. In time ∆t, r changes to r+∆r, and its velocity is given by the time derivative: 2014 MRT Time rate of change of a vector r. y x z O r P r+∆r The direction of v can be shown to coincide with the limiting direction of ∆r as it approches zero, or the tangent to the curve s at P(r). By rewriting v in the form: s ∆r êt the limiting value of ∆r/∆s is a unit vector along the tangent to the curve, êt, so that the velocity can also be written as: t s st ∆ ∆ ∆ ∆ = →∆ r v 0 lim tt s td sd eev ˆˆ == Acceleration is the time rate of change of velocity v, and by observing the vector change from v to v+∆v in time ∆t, we obtain: td d tt tt vvvvv a = ∆ ∆ = ∆ ∆ = →∆→∆ 00 lim )( lim −+
• Consider a particle P moving along a curve s fixed in a stationary Oxy plane, as shown in the Figure. The position of P(r) is established by the position vector r=rêr where êr is a unit vector which is always oriented along r. To determine the velocity v of P, we differentiate r (i.e., d(uv)/dx=udv/dx+vdu/dx) recognizing that êr changes direction: θθ eeθ e ˆˆ ˆ  == r r td d × êθ 2014 MRT O êr r P td d rr td d r td rd td rd td d r r r r r e e e e er v ˆ ˆ ˆ ˆ )ˆ( ++ ===≡ The unit vector êr is one of fixed magnitude which is rotating with angular velocity θ about an axis through the origin perpendicular to the xy plane. Its derivative (or its velo- city) is the cross product of the vectors θ and êr, which is a vector perpendicular to r, or: The trajectory s. x y θ ∆θ s ⋅ ⋅ where êθ is a unit vector in the direction perpendicular to r. θ ⋅
• Our equation dr/dt=(dr/dt)êr +r(dêr /dt) may now be written with dêr /dt=θ × êr as: td d rrr td d r td rrd td d td d rr θ θ θ θθθ θ e e eeevr a ˆ ˆ)( ˆ)ˆˆ( 2 2   +++ + ==≡≡ θθ eerθe r v ˆˆˆ  rrr td d rr +×+ ==≡ since r=rêr which expresses the velocity in terms of the radial and transverse components (or parallel and perpendicular components.) The acceleration of P can be determined by differentiating dr/dt=rêr +rθ êθ above: 2014 MRT We can view this problem as that of a point P moving along a set of rotating axes with direction êr which êθ . The point P always moves along the êr axis, and its relative velocity along it is r which corresponds to the first term in dr/dt=rêr +θ × r above. The second term, θ × r, is the velocity of the coincident point P due to rotation θ . ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ As before, the derivative of a unit vector rotates it by 90° and multiplies it with its angular rate θ, so that: ⋅ r td d eeθ e ˆˆ ˆ θθ θ  −== × Our equation above for the acceleration reduces to radial and transverse components: Incidentally, the term 2rθ is the Coriolis (1792-1883) acceleration. θθθθ eea ˆ)2(ˆ)( 2  rrrr r ++−= ⋅ ⋅
• êt 2014 MRT O ên r P Unit vectors ên and êt moving along curve s, in the xy plane. x y θ ∆ϕϕ s ∆s ρ To resolve the acceleration into tangential and normal components, we start with: tsesv ˆ =≡ Differentiating and noting that: n t td d e e ˆ ˆ ϕ= where ên is a unit vector along the radius of curvature ρ which is normal to the tangent to the curve s at P, and ϕ is the angular rate of the radius of curvature as shown in the Figure, we obtain: ⋅ nt ss eea ˆˆ ϕ += Since the length ∆s is related to the radius of curvature ρ and to the angle ∆ϕ swept out by ρ: ϕρ ϕρ  = ∆=∆ s s the acceleration a= s êt + sϕên can be expressed in the following alternate forms: ⋅⋅⋅⋅ se ee eea     ×ϕ+ + + t nt nt s s s s ˆ ˆˆ ˆˆ 2 2 = = = ϕρ ρ s
• ϕϕ ϕ ee e ˆωˆ ˆ == td d td d r ω êr Average Earth orbital speed: 〈v⊕ 〉 ~ 29.783 km/s (107,218 km/h) and the average distance of the Sun from the Earth: 〈R 〉 ~ 150,000,000 km (8.31 minutes at the speed of light). ω is the Earth’s angular speed around the Sun. v⊕ ϕϕ eeee e ee r v ˆωˆ)ˆω(ˆ ˆ ˆ)ˆ( r td rd r td rd td d r td rd r td d td d rr r rr +=+=      +      ===⊕ 17.5 18.0 18.5 19.0 v⊕ êr dêϕ /dt dêr /dt êϕ 2014 MRT Variation of the radial unit vector with time. R′ R′′ t = 17.0 td dϕ = ϕϕ ϕ d d d d r r r == ê ê Variation of the azimutal unit vector with time. rr td d td d ee e ˆωˆ ˆ −=−= ϕϕ ϕ ϕrs = r = R
• ω Average Earth orbital speed: 〈v⊕ 〉 ~ 29.783 km/s (107,218 km/h) and the average distance of the Sun from the Earth: 〈R 〉 ~ 150,000,000 km (8.31 minutes at the speed of light). ω is the Earth’s angular speed around the Sun. td d r td d r td rd td d td rd td rd td d td d td d r r ϕ ϕ e e e e vr a ˆ )ω(ˆ ω ω ˆ ˆ 2 2 +      +++      === ⊕ ⊕ 17.5 18.0 18.5 19.0 a⊕ 2014 MRT R′ R′′ t = 17.0 0 ω 0 ==⇒== td d td d α ϕ ϕϕ α eeee v a ˆω2ˆωˆωˆ 2 2 2       ++         −=      +== ⊕ ⊕ td rd rr td rd r td rd td d td d rr r = R ϕα ee ˆ)ω2(ˆ)ω( 2 r r r vrra ++−=
• The vector ∇ f is called the gradient of f (r) (and is often written grad f ). Differentiating both sides of the above equation with respect to r, we obtain the directional derivative: dr r • Hence the gradient of f is a vector whose component in any direction, dr, is the deriva- tive of f with respect to r. Note that when ∇ f is parallel to dr, d f has its maximum value. 2014 MRT ( )rr rrr ∇∇∇ ≡•= ∂ ∂ + ∂ ∂ + ∂ ∂ = dfzd z f yd y f xd x f fd )()()( Illustration of the physical meaning of the gradient,∇f , of a field f (r) and ∇f •dr/dr is just the scalar components of ∇ f in the direction of dr. Let f (x,y,z) be a single-valued scalar function with continuous first derivatives in a certain region of space. If r is the position vector of an object located at P(x,y,z), we have r=xi+yj+zk and dr=dxi+dyj+dzk (see Figure). The total derivative of f (r) is:ˆ ˆ ˆ ˆ ˆ ˆ rd d f rd fd rr •= ∇ )( O P Field f (e.g., scalar field ) Γ The scalar operator ∇ •∇ ≡∇2 ≡ is also useful in physics; it is called the Laplacian (Laplace, 1749-1827) operator. In equation form, it is (e.g., in Cartesian coordinates): Multiplication involving the ∇ operator is extremely useful in physics. The two products denoted by ∇ •F≡divF and ∇ × F ≡curlF are called the divergence of F and the curl of F, respectively. 2 2 2 2 2 2 2 zyx ∂ ∂ + ∂ ∂ + ∂ ∂ =•≡∇≡ ∇∇ since i•i =j•j=k•k=1 and i•j =i•k= j•k= 0.ˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆ ∇ f θ dr rd fd  Vector Calculus
• whereas the curl of the same vector F is (ibid): The divergence of a vector F is given by (e.g., for Cartesian coordinates): If at some point P: Fdiv  ≡ ∂ ∂ + ∂ ∂ + ∂ ∂ =• z F y F x F zyx F∇ Fˆˆˆ ˆˆˆ ˆˆˆ  curlkji kji kji F y ≡        ∂ ∂ − ∂ ∂       ∂ ∂ − ∂ ∂         ∂ ∂ − ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ = y F x F z F x F z F y F FF yx FF zx FF zy FFF zyx xxzyz yxzxzy zyx +− +−×∇ The physical interpretation of the curl of a vector is connected with the rotation (or circulation) of a vector field such as V. 2014 MRT 0 0 0 0 alirrotation solenoidal sink source asclassifiedisthen betosaidisthen atahasthen atahasthen VV V V V V =      = < > • ×∇ ∇ P P
• To find a possible interpretation of the curl of a vector, let us consider a body rotation with uniform angular speed ω about an axis z. Let us define the vector angular velocity ω to be the vector of length ω =|ω | extending along z in the direction in which a right- handed screw would advance if subject to the same rotation as the body. Finally, let r be the vector drawn from any point P(r) on the axis z to an arbitrary (i.e., of unspecified value) point P in the body (see Figure – from which we see that the radius at which P rotates is |r|⋅|sinϕ|.) The linear speed of P is thus v=|v|=ω|r|⋅|sinϕ|=|ω |⋅|r|⋅|sinϕ|=|ω × r|. The vector velocity v is directed perpendicular to the plane of ω and r, so that ω, r, and v for a right-handed system. Hence, the cross product ω × r gives not only the magnitude of v but the direction as well. ω r ϕ 2014 MRT A physical interpratation of the curl of a v O z P If we now take the point O as the origin of coordinates, we can write r=xi + yj +zk and ω = ω1i +ω2j+ ω3k. Hence, the equation v =ω×r can be written at length in the form v=(ω2z −ω3y)i −(ω1z− ω3x)j+ (ω1y− ω2x)k. If we take the curl (i.e., ∇× …) of v, we therefore get: |r|⋅|sinϕ| ˆ ˆ ˆ ˆ ˆ ˆ Expanding this, remembering that ω is a constant vector, we find∇× v= 2ω1 i + 2ω2 j+2ω3 k= 2ω which implies that ω = ½∇× v. The angular velocity of a uniformly rotating body is thus equal to one-half the curl of the linear velocity of any point of the body. The name curl in this context is now ˆ ˆ ˆ ˆ ˆ ˆ ω xyxzyz zyx 213132 )( ˆˆˆ ωωωωωω −−−− ∂∂∂∂∂∂= kji v×∇
• As a general example (and refresher for those new to this vector stuff) of how to use vectors in calculations, consider two vectors, A=i+2j+3k and B=3i+2j+k and find:ˆ ˆ ˆ ˆ ˆ ˆ 74.31412374.314321 222322 ==++===++= BA and Magnitude of A and B: kjikjikjikjiBA ˆ4ˆ4ˆ4ˆ)13(ˆ)22(ˆ)31()ˆˆ2ˆ3()ˆ3ˆ2ˆ( ++++++++++ =+++== Addition of A and B (or of B and A): kikjikjikjiBA ˆ2ˆ2ˆ)13(ˆ)22(ˆ)31()ˆˆ2ˆ3()ˆ3ˆ2ˆ( +++++−++− −=−−−== Substraction of B from A (but not A from B – in that case B−A=−(A−B)): 10343)13()22()31()ˆˆ2ˆ3()ˆ3ˆ2ˆ( =++=⋅+⋅+⋅=•=• kjikjiBA ++++ Scalar product of A with B (or of B with A): kjikji kji BA ˆ4ˆ8ˆ4ˆ)]32()21[(ˆ)]33()11[(ˆ)]23()12[( 123 321 ˆˆˆ −++−× −=⋅−⋅⋅−⋅⋅−⋅== Vector product of A with B (but not B with A – in that case B× A=−A× B): 2014 MRT kjikji kji A A A ˆ80.0ˆ53.0ˆ27.0ˆ 14 3ˆ 14 2ˆ 14 1 14 ˆ3ˆ2ˆ ˆ ++++ ++ ==== A unit vector A in the direction of A:ˆ
• mN⋅=+=−⋅−⋅= − == iikji kji Fr ˆ5ˆ)41(ˆ0ˆ0ˆ)]22()11[( 120 210 ˆˆˆ +−× The angle θ between A and B: )4.44(714.0 14 10 cos10cos14cos1414cos  or==⇒====• θθθθBABA A unit vector n perpendicular to both A and B : since r =x i+yj+zk and length |∇r2 |=2r and direction ê(∇r2 )=êr (i.e., the direction of r). 2014 MRT ˆ kji kjikjikji BA BA n ˆ 6 1ˆ 6 2ˆ 6 1 616 ˆ4ˆ8ˆ4 96 ˆ4ˆ8ˆ4 )4(8)4( ˆ4ˆ8ˆ4 ˆ 222 −+ −+−+−+ × × −= ⋅ − = − = −++− − == Another example, this time physical, would be to calculate the moment (i.e., torque τ ) of the force F=(−2j+k)N about the origin if the force acts at a point [0,1,2]m. Here we have r=(j+2k)m so the moment in this case is: y x zjˆ iˆ kˆ τ F ˆ ˆ ˆ ˆ r The gradient of a scalar function φ(r)=φ(r)= r2 =x2 +y2 +z2 (called a 3-sphere): ˆ ˆ ˆ θ )90()0(cos0cos5cos55cos 022)1)(2()2)(1( 1  or− =⇒====• =+−=+−=• θθθθFrFr Fr Since the scalar product of the two vectors r and F is 0 this means the two vectors are perpendicular (i.e. 90°) to each other! rkjikji 2ˆ2ˆ2ˆ2ˆ)(ˆ)(ˆ)( )( 2 222 =++=⇒ ∂ ∂ + ∂ ∂ + ∂ ∂ = zyxr z r y r x r r ∇∇φ
• As an example,letusevaluatethecurlofafield φ orientedinthex-direction: ∇ × iφ (x,y,z). In this case, we have (e.g., use the formula for ∇ × F and substitute F for iφ): φ φφφφ φ φ       ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ =      ∂ ∂ −      ∂ ∂ −−= ∂ ∂ ∂ ∂ ∂ ∂ = kjkjkji kji i ˆˆˆˆˆ0ˆ0ˆ)00( 00 ˆˆˆ ˆ zzzzzzzyx −−+−×∇ 2014 MRT To plot this surface (or any other curve) visit www.wolframalpha.com and type x^2+y^2-z=1. You should obtain this ‘infinite paraboloïd’. kji kji kji ˆˆ2ˆ2 ˆ)(ˆ)(ˆ)( ˆˆˆ 222222 −+ ++ ++∇ yx z zyx y zyx x zyx zyx = ∂ −+∂ ∂ −+∂ ∂ −+∂ = ∂ ∂ ∂ ∂ ∂ ∂ = φφφ φ Assume we have a field φ(x,y,z) =1= x2 + y2 – z; therefore: As another example, let us find a unit vector that is normal to the surface x2 + y2 – z=1 at the coordinate P(−1,−1,0). Now, all that remains is to evaluate ∇φ at P(−1,−1,0): kjikji ˆˆ2ˆ2ˆˆ)1(2ˆ)1(2),,( )0,1,1( −−+∇ −=−−−=−−P zyxφ then find the unit normal vector n to φ(−1,−1,0):ˆ kji kjikji n ˆ 3 1ˆ 3 2ˆ 3 2 3 ˆˆ2ˆ2 9 ˆˆ2ˆ2 ˆ −− −−−− ∇ ∇ −= − = − == φ φ P(−1,−1,0) nˆ x y z ˆ ˆ
• zd z J J z z ∂ ∂ +→→zJ As yet another example, physical this time, let us establish a physical meaning for the divergence of a vector J (i.e., ∇ • J) by use of an illustration from hydrodynamics. 2014 MRT Consider the flow indicated in the Figure (a flow occuring from the left face to the right face and reduced to Fz) and let J(r)=ρ(r)v(r), that is, J represents the mass of fluid (i.e., the mass flux density) flowing through a unit area normal to side ABCD per unit time. The density of the fluid is denoted by ρ, and v is its velocity, both functions of position, r. The z-component of J through the area ABCD (i.e., using the product of the differentials dx and dy to indicate a surface element) indicated in the Figure per unit time is given by: ydxdJz The flow through the area EFGH per unit time may be represented by the following (Taylor series) expansion: ydxdzd z J zJydxddzzJ z zz       + ∂ ∂ +=+ )()( where we neglect higher-order terms in this expansion (i.e., the ‘+…’ terms). Vd qd =ρ y x z dx dy A B C D E F G H An infinitisimal quantity of charge dq A unit volume dV=dxdydz dV J dz vq Fluid density z-component flux NOTE: The ABCDEFGH box should actually be within the limits of the Jz ‘stream’. For clarity, it is magnified.
• The net increase in the mass of the fluid inside the volume element dV=dxdydz per unit time due to the flow through the two opposite faces (of the cube in the Figure) is: 2014 MRT Vd z J zdydxd z J ydxdJydxdJydxdzd z J JydxdJ zz zz z zz ∂ ∂ −= ∂ ∂ −−=      ∂ ∂ +− Similarly, the net increase in the mass of fluid per unit time due to the flow through BFDH and AECG is: Vd y J zdydxd y J zdxdJzdxdJzdxdyd y J JzdxdJ yy yy y yy ∂ ∂ −= ∂ ∂ −−=        ∂ ∂ +− and that through CDGH and ABEF is: Vd x J zdydxd x J zdydJzdydJzdydxd x J JzdydJ xx xx x xx ∂ ∂ −= ∂ ∂ −−=      ∂ ∂ +− The total increase in the mass of the fluid (as calculated above) per unit volume (i.e., dV) per unit time due to the excess of inward flow over the outward flow is: Jkjikji •−=•      ∂ ∂ ∂ ∂ ∂ ∂ −=        ∂ ∂ + ∂ ∂ + ∂ ∂ −=         ∂ ∂ − ∂ ∂ − ∂ ∂ − ∇++++ )ˆˆˆ(ˆˆˆ zyx zyx zyx JJJ zyxz J y J x J Vd Vd z J y J x J which is just the rate of increase of the density (i.e., dρ/dt) of the fluid inside of the volume element dV.
• In other words, this last expression is equivalent to the physical principle: 2014 MRT t∂ ∂ =•− )( )( r rJ ρ ∇ The above equation is called the continuity equation, it describes the transport of a conserved quantity – in this case the density ρ (i.e., mass per unit volume). The continuity equation is also a stronger, more local form of a conservation law (i.e., nothing is created and/or nothing is lost in the flow of matter in a ‘closed’ physical system). t∂ ∂ −=•+•=• ρ ρρρ vvv ∇∇∇ )( where we used the identity ∇ •(φA)=A•∇φ +φ∇ •A. Substituting J=ρv, we get: )(00 solenoidor =•=•= ∂ ∂ vJ ∇∇ t ρ and so it is also required that ∇ρ =0. In this case, the excess or outward flow over inward flow is zero. In other words, there is no noticeable compression or expansion of the fluid. For an incompressible fluid (i.e., a fluid in which there is no change in density of the fluid particle as its motion is followed) we have:
• dV V S dSd nS ˆ= ∇T y x z r O Gauss’ Divergence Theorem for heat conduction (transfer): The closed manifold dV is the boundary of V oriented by outward- pointing normals, and n is the outward pointing unit normal field of the boundary dV. We now consider the problem of the transfer of heat Q through a body by means of conduction. We assume (with material conductivity k [W·m−1 ·K−1 ]) 1. The temperature, T, within the body is a finite and continuous point function T(x,y,z,t); 2. Through any point, a surface on which the temperature is everywhere the same (isothermal surface) may be drawn; 3. The temperature gradient, ∇T, and the direction of heat flow are normal to the isothermal surface at the point in question; 4. ∇T is a finite and continuous function and is in the direction of increasing T, and; 5. The rate of heat flow per unit area across the isothermal surface, F, is given by: (Fourier’s Law - 1822) TT c k Tk ct T 21 )( 1 ∇=•=•= ∂ ∂ σρρ ∇∇∇∇ If S is an arbitrary closed surface drawn entirely within a certain region of the substance under investigation, then the total amount of heat flowing out of S in time ∆t is given by: [ ]∫∫∫∫∫ •∆−≡•−∆=⇒ •= ∆ ∆ VS VdTktSdTktQ d t Q )(ˆ)( ∇∇∇ n SF 2014 MRTnˆ Tk∇−=F The following quantity also represents the amount of heat flowing out of S in time ∆t, provided c is the specific heat of the material, ρ is it’s density, and ∂T/∂t is the rate of increase in temperature within S: ∫∫∫       ∂ ∂ ∆−= V dV t T ctQ ρ σ is the diffusivity which equals cρ/k (for thermal conductivity k, specific heat c, and density ρ.)
• It is assumed that the following transformations exist:xi = fi(u1 ,u2 ,u3 )(for i, j=1,2,3), and: s rrrr r dud u ud u ud u ud u d i i i ≡ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ = ∑= 3 1 3 3 2 2 1 1 ++ The square of the length of this displacement is given by the interval: ∑∑= = ∂ ∂ • ∂ ∂ =•= 3 1 3 1 2 i j ji ji udud uu ddsd rr rr 2014 MRT where gi j = xi • xj. Since xi • xj = xj • xi, gi j = gj i (i.e., gi j is symmetric in the interchange of i and j indices). The quantities gi j are called metric coefficients and characterize the relative nature of the space. When an object at P is described using coordinates other than cartesian, say using coordinates ui =Fi(xj ) we can generalize using curvilinear coordinates. x2 P Note that the ∂r/∂ui are tangent to the ui respectively, since ∂r/∂u1 means u2 and u3 are fixed and r=r(u1 ) is constrained to move along the u1 surface. Letting the three coordinate axes be represented by xi, where xi =∂r/∂ui (see Figure). Hence the equation above for ds2 =ds•ds becomes: ê3 x3 x1 ∑∑∑ =•= ji ji ji i j ji ji ududgududsd xx2 ê1 ê2 u1 u2 u3 3 3 3 2 2 2 1 1 1 ˆˆˆ x x e x x e x x e === and; The position vector of a point P is a function of the ui , that is, r=r(u1 ,u2 ,u3 ). An element of displacement is given by (using the Chain Rule): ),,(),,(),,( 321 3 3321 2 2321 1 1 xxxFuxxxFuxxxFu === and,
• In an orthogonal curvilinear coordinate system (i.e., the three coordinate surfaces are everywhere mutually perpendicular) we have xi • xj =0 (for i≠j). Hence the square of the element of length becomes: 23 33 22 22 21 11 2 )()()( udgudgudgsd ++= Note that du2 =du3 =0 when the element of length ds is along u1 . We may therefore write: 3 3 3 333 2 2 2 222 1 1 1 111 udhudgsdudhudgsdudhudgsd ====== and; where hi =√(gii) are called scale factors. 2014 MRT We now develop a method for determining gii (i.e., hi) when the xi are known. In Cartesian coordinates, g11=g22=g33=1 (e.g., we get ds2 =(du1 )2 +(du2 )2 +(du3 )2 ) and: ∑∑ ∑∑ ∑∑∑ = = ===         ∂ ∂ ∂ ∂ =                 ∂ ∂         ∂ ∂ == 3 1 3 1 3 1 3 1 3 1 2 i j ji k j k i k k j j j k i i i k k kk udud u x u x ud u x ud u x xdxdsd Comparing this equation with ds2 =Σij gij dui du j , we see that: ∑= ∂ ∂ ∂ ∂ = 3 1k j k i k ji u x u x g ∑=         ∂ ∂ = 3 1 2 k i k ii u x g or when i=j:
• The area and the volume elements are given by (no sum implied): )( ˆ )( ˆ )( ˆ ˆˆˆ 321 21 3 231 31 2 132 32 1 332211 Fhh hh Fhh hh Fhh hh FFF eee eeeF ++++ == With this, the divergence of the vector F is given by:         ∂ ∂ + ∂ ∂ + ∂ ∂ = •+•+•=• )()()( 1 )( ˆ )( ˆ )( ˆ 321323121321 321 321 21 3 231 31 2 132 32 1 Fhh u Fhh u Fhh uhhh Fhh hh Fhh hh Fhh hh ∇∇∇∇ eee F where we have used the vector identity ∇ •(φX)=X•∇φ +φ ∇ •X and the fact that ∇ •(ê1/h2h3)=∇ •(ê2/h1h3)=∇ •(ê3/h1h2)=0 (Exercise - Hint: First show that êi = hi∇ui ). 2014 MRT An arbitrary vector F, in terms of general curvilinear components, can be written as: 321 321 udududhhhdududhhd ji jiji == τσ and Since ∂f /∂s is the component of ∇ f in the ds direction (remember that we said earlier that ‘the gradient of f is a vector whose component in any direction, dr, is the derivative of f with respect to r’), the gradient of an arbitrary function f, ∇ f, is given by: 3 3 3 2 2 2 1 1 1 3 3 2 2 1 1 3 1 ˆˆˆ ˆˆˆˆ u f hu f hu f hs f s f s f s f f ii i ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ = ∑= eee eeee ++++∇ where êi are units vectors along dsi and dsi =hi dui =√(gii) dui .
• When the Laplacian operator is applied to a scalar function f , it is given by:                 ∂ ∂ ∂ ∂ +         ∂ ∂ ∂ ∂ +         ∂ ∂ ∂ ∂ =         ∂ ∂ + ∂ ∂ + ∂ ∂ •=•=∇ 3 3 21 32 2 31 21 1 32 1 321 3 3 3 2 2 2 1 1 12 1 ˆˆˆ u f h hh uu f h hh uu f h hh uhhh u f hu f hu f h ff eee ∇∇∇ 2014 MRT In a similar way (Exercise), we find the curl of the vector F: 332211 321 332211 321 ˆˆˆ 1 FhFhFh uuu hhh hhh ∂ ∂ ∂ ∂ ∂ ∂ = eee F×∇ since F=(ê1/h1)(h1F1)+(ê2/h2)(h2F2)+(ê3/h3)(h3F3) and ∇ × (ê1/h1)=∇ × (ê2/h2)=∇ × (ê3/h3)=0 (Exercise). In other words, the Laplacian is ‘the divergence of a gradient’ – in this case of a scalar function f – given in orthogonal curvilinear coordinates (Exercise - Verify in Cartesian). In case you were wondering, here are a few other gradient related vector identities: ∇×∇φ =0,∇•∇×A=0and∇×∇×A=∇∇•A− ∇2 Aandpositionones:∇•r=3and∇×r=0.
• 11cossin cos)sin(cossin cossinsincossin )cos()sinsin()cossin( 111 22 11 2222 22222 222 2222 1 3 2 1 2 2 1 13 1 2 111 ===+= ++= ++=       ∂ ∂ +      ∂ ∂ +      ∂ ∂ =       ∂ ∂ +      ∂ ∂ +      ∂ ∂ =         ∂ ∂ +         ∂ ∂ +         ∂ ∂ =         ∂ ∂ = ∑= ghg r r r r r r r z r y r x u x u x u x u x g k k andθθ θϕϕθ θϕθϕθ θϕθϕθ We will now invest some time expanding the details of this method in an example that will come back frequently – spherical coordinates (r,θ andϕ) and their associated ope- rators and/or functions – and proceed to calculate the resulting Laplacian operator. Use: iii k i k ii gh u x g =         ∂ ∂ = ∑= and 3 1 2 θϕθϕθ cossinsincossin rzryrx === and; which are the transformation relations between x, y & z and r,θ &ϕ. and defining the coordinates as x1 =x, x2 =y & x3 =z and u1 =r, u2 =θ & u3 =ϕ we get: 2014 MRT First, recall that:
• rghrrg r rrr rrr zyx u x u x u x u x g k k ===+= ++= ++=       ∂ ∂ +      ∂ ∂ +      ∂ ∂ =       ∂ ∂ +      ∂ ∂ +      ∂ ∂ =         ∂ ∂ +         ∂ ∂ +         ∂ ∂ =         ∂ ∂ = ∑= 222 2222 22 22222 22222222 222 2222 2 3 2 2 2 2 2 13 1 2 222 )sin(cos ]sin)sin(cos[cos sinsincoscoscos )cos()sinsin()cossin( andθθ θϕϕθ θϕθϕθ θ θ ϕθ θ ϕθ θ θθθ θθ ϕϕθ ϕθϕθ θ ϕ ϕθ ϕ ϕθ ϕ ϕϕϕ sinsin )]cos(sin[sin 0cossinsinsin )cos()sinsin()cossin( 333 22 33 2222 222222 222 2222 3 3 2 3 2 2 3 13 1 2 333 rghrg r rr rrr zyx u x u x u x u x g k k === += ++=       ∂ ∂ +      ∂ ∂ +      ∂ ∂ =       ∂ ∂ +      ∂ ∂ +      ∂ ∂ =         ∂ ∂ +         ∂ ∂ +         ∂ ∂ =         ∂ ∂ = ∑= and 2014 MRT then: and:
• 2014 MRT Secondly, recall that the Laplacian in general curvilinear coordinates was given by:                 ∂ ∂ ∂ ∂ +         ∂ ∂ ∂ ∂ +         ∂ ∂ ∂ ∂ =         ∂ ∂ + ∂ ∂ + ∂ ∂ •=•=∇ 3 3 21 32 2 31 21 1 32 1 321 3 3 3 2 2 2 1 1 12 1ˆˆˆ uh hh uuh hh uuh hh uhhhuhuhuh eee ∇∇∇ or under the guise of a more compact form: where: 2 2 22 1 ϕθϕθ ∇+∇=∇ r rr         ∂ ∂ +      ∂ ∂ ∂ ∂ =∇      ∂ ∂ ∂ ∂ =∇ 2 2 2 22 2 2 sin sin 11 ϕθ θ θθ ϕθ rr r rr r and 2 2 222 2 2 2 2 2 2 sin 1 sin sin 11 sin 1 sinsin sin 1 sin 1sin1 1 sin sin1 1 ϕθθ θ θθ ϕθϕθ θ θ θ θ ϕθϕθ θ θ θ θ ∂ ∂ +      ∂ ∂ ∂ ∂ +      ∂ ∂ ∂ ∂ =∇               ∂ ∂ ∂ ∂ +      ∂ ∂ ∂ ∂ +      ∂ ∂ ∂ ∂ =               ∂ ∂⋅ ∂ ∂ +      ∂ ∂⋅ ∂ ∂ +      ∂ ∂⋅ ∂ ∂ ⋅⋅ =∇ rrr r rr r r rr r r r r r rr rrr The rest is trivial to calculate. The Laplacian in spherical coordinates is now given by:
• k i ki i izyx xx v z F y F x F evF ˆ ∂ ∂ =• ∂ ∂ =• ∂ ∂ + ∂ ∂ + ∂ ∂ =• σ σ∇∇∇ and, Point P at the tip of the distance vector r is given in Rectangular (Cartesian) Coordinates by the intersection of constant x, constant y and constant z planes. kji ˆˆˆ z f y f x f f ∂ ∂ ∂ ∂ ∂ ∂ = ++∇ kj i kji F ˆˆ ˆ ˆˆˆ         ∂ ∂ − ∂ ∂       ∂ ∂ − ∂ ∂         ∂ ∂ − ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ = y F x F z F x F z F y F FFF zyx xyxz yz zyx ++ −×∇ 2 2 2 2 2 2 z f y f x f fff ∂ ∂ + ∂ ∂ + ∂ ∂ =•=∇≡ 2 ∇∇ kjiFF ˆˆˆ 2222 zyx FFF ∇∇∇=∇≡ ++ The Laplacian of a vector function F= F(x,y,z): The Laplacian of a scalar function f = f(x,y,z): As a vector product of a vector function F=F(x,y,z): As a scalar product of a vector function F= F(x,y,z) and vector field v and a second-order tensor σ (e.g., stress): The gradient of a scalar function f = f (x,y,z): 2014 MRT y x z r O P Trajectory p vˆ y x r P ˆ k j i ˆ ˆi kˆ Constant x plane Constant y plane Constant z plane y x z O d d y d x d z zdydxddV = z 2222ˆˆˆ zdydxddzdydxdd ++==  andkji ++ d is an infinitesimal differential increment of length: dV is an infinitesimal differential increment of volume: r jˆ ˆ êz σx y êy σx z êx σy z dV ∫∫∫∫∫ •=• VS Vdd σ∇σ S The Divergence Theorem is: ∫∫∫∫ •= SS kki dSd Se σˆσ The normal surface integral of σ over a closed boundary: ∫∫∫∫∫ ∫∫∫∫∫ •+∇= ∂ ∂ •=• VS VS VddS n VddS ) )(ˆ 2 φψψ φ ψ φψφψ ∇∇( ∇∇∇ n Green’s Theorem is: and since ∇φ •n = ∂φ/∂n is the directional derivative: ˆ S ),,(),,( ),,( ),,( ˆˆˆˆ ),,( ˆˆˆˆ ),,( 222 222 zyxzyx zyx zyx vvv vvvzyx zyx zyxzyx k jzyx zyxj i i rv aa r vvkji vvv rrkji rrr   == = = == ++=⇒= == ++=⇒= ++ ++ φ  
• x x yP ζˆ r ρˆ ϕˆ ϕ θ Constant ζ plane Constant ϕ planeConstant ρ cylinder ζϕρρ ρ ρ ζϕρ ∂ ∂ + ∂ ∂ + ∂ ∂ =• FFF 1)(1 F∇ ζρ ˆˆˆ ζϕρρ dddd +ϕ+= ϕρζρ ddddV = 2222 )( ζϕρρ dddd ++= Point P at the tip of the distance vector r is determined in Cylindrical Coordinates by the intersection of a constant ζ (i.e., a plane), constant ϕ (i.e., a half-plane) and constant ρ (i.e., a cylinder) surfaces. y z z r ζρrr ˆˆ),,( ζρζϕρ +== d is an infinitesimal differential increment of length. ζρ ˆˆ 1 ˆ ζϕρρ ∂ ∂ ∂ ∂ ∂ ∂ = fff f +ϕ+∇ ζ ρF ˆ)(1 ˆˆ 1         ∂ ∂ − ∂ ∂         ∂ ∂ − ∂ ∂         ∂ ∂ − ∂ ∂ = ϕρ ρ ρ ζρζϕρ ρϕ ρζϕζ FF FFFF + +ϕ−×∇ )2 FFFF ×(∇×∇−∇∇ )•(=∇= dV is an infinitesimal differential increment of volume. 2014 MRT 2 2 2 2 2 2 11 ζϕρρ ρ ρρ ∂ ∂ + ∂ ∂ +      ∂ ∂ ∂ ∂ =∇ fff f Laplacian of a vector function F = F (ρ,ϕ ,ζ ) (identity) The Laplacian of a scalar function f = f (ρ,ϕ ,ζ ) The vector product of a vector function F= F(ρ,ϕ ,ζ ) The scalar product of a vector function F= F(ρ,ϕ ,ζ ) Produces another vector perpendicular to the plane formed by ∇ and F with unit vectors given. O • ζ ρ x z y O ϕ ζ ρdϕ dρ d dϕ d ζ ρ • • We can make the following geometric objects into physical realities if we substitute the scalar current I for f = f (ρ,ϕ ,ζ ) and the current density J for the vector F=F (ρ,ϕ ,ζ ). The gradient of a scalar function f = f (ρ,ϕ ,ζ ) •P 
• x z yP θˆ r rˆ ϕˆ ϕ θ Constant r sphere Constant ϕ plane Constant θ cone x z y O ϕ θ • r rdθ dr dθ d rsinθdϕ dϕ ϕθθ θ θ ϕθ ∂ ∂ + ∂ ∂ + ∂ ∂ =• F r F rr Fr r r sin 1)sin( sin 1)(1 2 2 F∇ ϕ++ ˆsinˆˆ ϕθθ drrddrd θr= ϕθθ dddrrdV sin2 = θθ ϕθθ ϕθθ 222222 222222 2222 sin sin )sin()( +=ΩΩ+= ++= ++= dddrdr drdrdr drdrdrd where  • Point P at the tip of the distance vector r is determined in Spherical Coordinates by the intersection of a constant θ (i.e., a cone), constant r (i.e., a sphere) and constant ϕ (i.e., a half-plane) surfaces. x x y r rrr ˆ),,( rr == ϕθ P d is an infinitesimal differential increment of length. ϕ++∇ ˆ sin 1ˆ1 ˆ ϕθθ ∂ ∂ ∂ ∂ ∂ ∂ = f r f rr f f θr ϕ+ +− −×∇ ˆ )(1 ˆ sin 1)(1 ˆ )sin( sin 1       ∂ ∂ − ∂ ∂         ∂ ∂ − ∂ ∂         ∂ ∂ − ∂ ∂ = θ ϕθ ϕθ θ θ θ ϕ θϕ r r F r rF r F r rF r FF r θ rF )2 FFFF ×(∇×∇−∇∇ )•(=∇= • ˆ j i O kˆ • dV is an infinitesimal differential increment of volume. 2014 MRT 2 2 22 2 2 2 2 sin 1 sin sin 11 ϕθ θ θ θθ ∂ ∂ + +      ∂ ∂ ∂ ∂ +      ∂ ∂ ∂ ∂ =∇ f r f rr f r rr f Laplacian of a vector function F=F (r,θ ,ϕ ) (identity) The Laplacian of a scalar function f = f (r,θ ,ϕ ) The vector product of a vector function F=F(r,θ ,ϕ ) The scalar product of a vector function F= F(r,θ ,ϕ ) The gradient of a scalar function f = f (r,θ ,ϕ ) Produces another vector perpendicular to the plane formed by ∇ and F with unit vectors given. We can make the following geometric objects into physical realities if we substitute the scalar charge density ρ for f = f (r,θ,ϕ ) and the current density J for the vector F = F (r,θ,ϕ ). Manifolds ˆ 
• The line integral, IΓ , of a vector F along a curve Γ from A to B is defined as the definite integral of the scalar component of F in the direction of the tangent to the curve at P(r) where dr is an element of displacement at P(r): We can show that the work done on an object by a resultant force F during a displace- ment from A to B is equal to the change in the kinetic energy (T=½mv2 ) of the object. 2014 MRT since d(v•v)/dt =(dv/dt)•v +v•dv/dt = 2v•dv/dt. If the force F is the gradient of a single-valued scalar function V(r) and the path of integration is closed, we obtain Work = 0. When an object at P is subjected to a conservative (E=T+V=Constant) force F(r), it is said to possess potential energy V(r) relative to some fixed point, B. ∫ •= B A drrF )(Work y x z O Tvvmd m dmtd td d m AB t t t t t t B A B A B A ∆=−=•=•=•= ∫∫∫ )()( 2 22 2 1vvvvv v Work r A P dr F(r) = F1 + F2Γ Zero work means that the final kinetic energy of the object is equal to the initial kinetic energy. Hence energy is conserved; such a force, F(r) = −∇V , is said to be a conservative force. The conservation of total energy (E) is expressed by: E = T +V. The work is given by ∫ F(r)•dr where F= Σi Fi = ma= mdv/dt (Newton’s second law above) and dr= (dr/dt)dt = vdt. The expression for the work done above can be put in the form: Note also that the line integral from A to B is the negative of that from B to A. For instance, the work done by a variable force F(r) in moving an object from A to B is defined as (see Figure): rF∫Γ Γ •= dI ∫ •= B P dV rrFr )()( ∫ •=∆ B A dK rrF )( VTE += The Total Energy is: B F2 F1 The Line Integral
• The normal surface integral of a vector function F(r) over a closed boundary is defined as the surface (double) integral of the scalar component of F in the direction of a normal to the surface S. This is written: where dS is the area of an element of the boundary surface and n is a unit outward normal to this element of area. The symbol means double (in general, over any two coordinates) integration around a closed surface (indicated by the loop around ∫∫S ). The sign convention for the unit normal n is a follows: 1)For a closed surface (i.e., a surface which incloses a volume), the outward normal is called positive; 2)For an open surface, the right-hand screw rule is used; the direction of rotation is the same as that in which the periphery is traversed. The divergence theorem due to Gauss (1777-1855) is as follows: The normal surface integral of a vector function F over the boundary of a closed surface dS of arbitrary shape is equal to the volume (triple) integral of the divergence of F (i.e., ∇ • F) taken throughout the enclosed volume dV. In equation form, we write: 2014 MRT ∫∫∫∫ •=•≡ SS S dSdI SFnF ˆ ˆ ˆ ∫∫∫∫∫ •=• VS dVd FSF ∇ Vector Theorems
• To prove this theorem, we first expand the right-hand side of the divergence theorem and obtain: Although the theorem is valid for an arbitrary shaped closed surface, we choose the volume in the Figure for convenience (weird – but it is easier than a 3-sphere!) In the Figure, we have: ∫∫∫∫∫∫       ∂ ∂ + ∂ ∂ + ∂ ∂ =• V zyx V zdydxd z F y F x F dVF∇ 4 ˆ Si dzdyd =−y x z dx dz dy −iˆ −kˆ jˆ iˆ kˆ x″ y″ (x′,y′,z′) z″ (x″,y″,z″) 1 ˆ Sk dydxd =− 2014 MRT zdxdd zdydd ydxdd jS iS kS ˆ)( ˆ)( ˆ)( 3 2 1 −= += −= back top left zdxdd ydxdd zdydd jS kS iS ˆ)( ˆ)( ˆ)( 6 5 4 += += −= front right bottom (1) (See Figure)
• (2) (4) Integrating the last term on the right-hand side of Eq. (1) with respect to z from z′ to z″, we obtain: ∫∫∫∫∫∫∫∫∫ ′−′′=′−′′= ∂ ∂ ′′ ′ 15 ),,(),,()],,(),,([ S z S z S zz z z z ydxdzyxFydxdzyxFydxdzyxFzyxFzdydxd z F 2014 MRT ∫∫∫∫∫∫∫∫∫ •=•′−•′′=      ∂ ∂ ′′ ′ S z S z S z z z z SdFdzyxFdzyxFydxdzd z F nkSknSkn ˆˆˆˆ),,(ˆˆ),,( 15 1155 Note that: )(ˆˆ )(ˆˆ 55 11 right left ydxdd ydxdd +=• −=• Skn Skn Using Eq. (3) in Eq. (2), we write: (3) where: 0)( =∫∫sides  since k is perpendicular to dS2, dS3, dS4, and dS6.ˆ
• (6) (5) Similarly, we can show that: 2014 MRT ∫∫∫∫∫∫∫∫∫ •=•′−•′′=      ∂ ∂ ′′ ′ S x S x S x x x x SdFdzyxFdzyxFzdydxd x F niSinSin ˆˆˆˆ),,(ˆˆ),,( 42 4422 and: ∫∫∫∫∫∫∫∫∫ •=•′−•′′=        ∂ ∂ ′′ ′ S y S y S y y y y SdFdzyxFdzyxFzdxdyd y F njSjnSjn ˆˆˆˆ),,(ˆˆ),,( 36 3366 Comparing Eqs. (5), (6), and (4), respectively (i.e., with the right-hand side of Eq. (1)), we obtain: QED ˆ)ˆˆˆ( ˆˆˆˆˆˆ ∫∫ ∫∫ ∫∫∫∫∫∫∫∫∫∫∫∫ •= •= •+•+•=      ∂ ∂ + ∂ ∂ + ∂ ∂ =• ′′ ′ ′′ ′ ′′ ′ S S zyx S z S y S x z z y y x x zyx V d SdFFF SdFSdFSdFzdydxd z F y F x F dV SF nkji nknjniF ++ ∇ Hence the theorem is proved. (7)
• Green’s theorem (1793-1841) is an important corollary of the divergence theorem, and it has numerous applications in various branches of physics. 2014 MRT Let ψ and φ be two scalar functions of position with continuous derivatives within a certain region bounded by a closed surface S. On applying the divergence theorem to the vector ψ ∇φ in this region, we obtain (i.e., substitute F=ψ ∇φ and dS=ndS in Eq. (7)): ˆ ∫∫∫∫∫∫∫∫∫∫∫ •+=•+•=•=• VVVS dVdVdVSd )()()(ˆ 2 φψφψφψφψφψφψ ∇∇∇∇∇∇∇∇∇∇ n or: ∫∫∫∫∫ •+= ∂ ∂ VS dVSd n )( 2 φψφψ φ ψ ∇∇∇ where: φψφψφψ ∇∇∇∇∇∇ •+•=• )( and, given the definition of the directional derivative: n∂ ∂ =• φ φ nˆ∇ Eq. (8) is known as Green’s theorem of the first form. (8) (9)
• A second form of Green’s theorem is obtained if we consider the following two equations: 2014 MRT φψφψφψ ∇∇∇∇∇∇ •+•=• )( On substrating Eq. (11) from Eq. (10) and integrating over an arbitrary volume, we obtain: ∫∫∫∫∫∫ −=−• VV dVdV )()( 22 ψφφψψφφψ ∇∇∇∇∇ Converting the left-hand side of Eq. (12) to a surface integral by use of the divergence theorem, we obtain: Eq. (13) is the Green’s theorem of the second form. (10) and: ψφψφψφ ∇∇∇∇∇∇ •+•=• )( (11) (12) ∫∫∫∫∫ −=      ∂ ∂ − ∂ ∂ VS dVSd nn )( 22 ψφφψ ψ φ φ ψ ∇∇ (13)
• A third and equally important vector integration theorem is due to Stokes (1819- 1903). 2014 MRT ∫∫∫ •=• S dd SFΓF ×∇ Γ The curl theorem is as follows: If F and its first derivatives are continuous, the line integral of F around a closed curve Γ is equal to the normal surface integral of curlF (i.e., ∇ × F) over an open surface bounded by Γ. In other words, the surface integral of curlF taken over any open surface S is equal to the line integral of F around the periphery Γ of the surface. In equation form, we write: So, we see that the surface integral of the curl of a vector field over an open surface is equal to the closed line integral of the vector along the contour bonding the surface. In a nutshell, Stoke’s theorem converts a surface integral of the curl of a vector to a line integral of the vector, and vice versa. There are two special cases to consider. 0∇×∇∇ =≡ )( ffcurl 1) The curl of a gradient of any scalar field f is identically zero: 2) The divergence of the curl of any vector field F is identically zero: 0)(div =•≡ FFcurl ×∇∇
• To prove this theorem, we first expand the right-hand side of the curl theorem above (i.e., ∫ΓF•dΓ =∫∫S∇ × F•dS); it becomes: 2014 MRT (14) ∫∫∫∫ •=• S zyx S SdFFFSd )ˆˆˆ(ˆˆ kjinnF ×∇+×∇+×∇×∇ The last integral on the right-hand side of Eq. (14) reduces to: ∫∫∫∫∫∫       ∂ ∂ •− ∂ ∂ •= ∂ ∂ ∂ ∂ ∂ ∂ •=• S zz S z S z x F Sd y F SdSd F zyx SdF )ˆˆ()ˆˆ( 00 ˆˆˆ ˆ)ˆ(ˆ jnin kji nkn ×∇ Note that the projection of dS onto the yz-plane (see Figure) leads to: zdydSd +=• in ˆˆ (15) (16) Now we have to find the n•jdS term.ˆ ˆ y x z Back dz dy iˆjˆ iˆ kˆ A B Front ˆn dS Γ dx jˆ
• We now let the line segment AB be the intersection of the surface S with a plane that is parallel to the xy-plane at a distance z from the origin (see Figure - Left). Along the strip AB, we find that (notice the order of terms in the negative version, c.f., Eq. (19)): 2014 MRT xd x F yd y F Fdyd y F xd x F Fd zz z zz z ∂ ∂ − ∂ ∂ =−⇒ ∂ ∂ − ∂ ∂ = and: jir ˆˆ ydxdd −= (17) The vector dr is tangent to AB at P (see Figure - Right) and perpendicular to n. We may therefore write: ˆ ydxdd jninnr ˆˆˆˆ0ˆ •−•==• or: (18)       =•=• Sd zdyd yd xd yd xd injn ˆˆˆˆ or better yet, our sought out answer – the projection of dS onto the xz-plane: zdxdSd +=• jn ˆˆ A(x1, y1, z) B(x2, y2, z) A jir ˆˆ ydxdd −= Plane of constant z A B ˆn dS P +z −z
• On substituting Eqs. (16) and (18), respectively, into Eq. (15), then Eq. (17) we obtain: 2014 MRT ∫∫ ∫∫∫ ∫∫∫∫∫∫ −−=−=−=       ∂ ∂ − ∂ ∂ =      ∂ ∂ − ∂ ∂ =• zdzyxFzyxFFdzdzdFd zdxd x F yd y F x F zdxd y F zdydSdF zz zyxB zyxA z S z S zz S zz S z )],,(),,([)( )()()ˆ(ˆ 1122 ),,( ),,( 22 11 kn ×∇ (19) The sense of the periphery at A is positive (i.e., dz=dΓz) and it is negative at B (i.e., dz=−dΓz) (see Figure - Right). Hence Eq. (19) becomes: ∫∫∫∫∫ =+=• Γ 1122 ΓΓ),,(Γ),,()ˆ(ˆ zz A zz B zz S z dFdzyxFdzyxFSdF      )(partfront)(partback kn ×∇ (20) Similarly, we find: ∫∫∫∫∫∫ =•=• ΓΓ Γ)ˆ(ˆΓ)ˆ(ˆ yy S yxx S x dFSdFdFSdF jnin ×∇×∇ and Combining the two Eqs. (21) and Eq. (20) with Eq. (14), we obtain: (21) QED )ΓΓΓ()ˆˆˆ( Γ Γ ∫ ∫∫∫∫∫ •= ++=•=• ΓF SkjiSF d dFdFdFdFFFd zzyyxx S zyx S ×∇+×∇+×∇×∇ (22) and the theorem is proved.
• y x z S dΓ R ˆn dS dA ∇ × F Γ As an example in the use of the curl theorem, let us evaluate ∫Γ (zdx+xdy+ydz), where Γ is the projection as the cylinder y2 +z2 =1 in the plane x+z=2. According to the right- hand rule, you should orient Γ counter-clockwize as viewed from above (see Figure). kjiF ˆˆˆ ++×∇ = We recall Stoke’s theorem: ∫∫∫ •=• S Sdd nFΓF ˆ)( Γ ×∇ and identify F(x,y,z) =zi + xj +yk (as well dΓ = dxi + dy j+ dzk) so this allows us to evaluate ∇ × F readily: The given orientation of Γ corresponds to an upward orientation of the surface S. Thus, if f (x,y,z)= x+z –2= 0 defines the plane, then with this the gradient of f (x,y,z) is ∇ f =i + k and the upper normal n is: kiki ki n ˆ71.0ˆ71.0ˆ 2 1ˆ 2 1 2 ˆˆ ˆ ++ + ∇ ∇ ==== f f Hence, from Stoke’s theorem: ∫∫∫∫∫∫∫∫ ==               •=• SSSS dSdSdSSd 2 2 2ˆ 2 1ˆ 2 1 )ˆˆˆ(ˆ)( kikjinF +++×∇ which will give us the extra √2 factor (e.g., (∂g/∂y)2 = 0 while (∂g/∂z)2 =(–1)2 =1) with π/4 the result of the integral over R (i.e., for y2 +z2 = π21842222 1 0 2 1 0 1 0Γ 2 =−=         ⋅===• ∫∫ ∫∫∫∫∫∫ − zdzzdyddASdd z RS ΓF Now, if z =g(x,y) is the equation of a surface S, then the differential of the surface area is dS= √[1 +(∂g/∂x)2 + (∂g/∂y)2 ]dA. Thus, considering the y and z plane here (i.e., g(y,z) =x=2 – z) we use the relation ∫∫S f (x,y,z)dS =∫∫R f (g(y,z),y,z)√[1 + (∂g/∂y)2 + (∂g/∂z)2 ]dA:2014 MRT ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ     )( :Since kiΓkjF ji ji F F F kjiF ˆˆ)0,1,2(ˆˆ)0,1,2(ˆ ˆˆ 2 ˆˆ ),0,( ),0,( ),0,(ˆ ˆˆˆ),,( 45.0 5 5 89.0 5 52 08.1 4 23 36.0 4 22 3 2 1 2 1 2 3 2 1 2 3 2 1 2 3 zdxdd yxzzyx ++ + + ++ −== === = 122 =+ zy 2=+ yx ),0,(ˆ 2 1 2 3 F jF ˆ)0,0,2(ˆ = )0,1,2(ˆF iF ˆ)1,0,0(ˆ =   jiF ˆˆ)1,0,1(ˆ 71.0 2 2 71.0 2 2 += ∫ ∫∫∫ − = 1 0 1 0 2 4 1 zdyddA z R
• We now present the proofs of two recurring integral relations. ∫∫∫∫∫∫∫ == SSV dSdVd Sn φφφ ˆ∇ To prove this, let F=φC, where φ is a constant scalar and C is a constant vector. On substituting this for F into the divergence theorem (c.f., Eq. (7)), we obtain: ∫∫∫∫∫ •=• SV SdVd nCC ˆ)( φφ∇ Note that ∇ •(φC)=C•∇φ, since ∇ •C=0. By using ∇ •(φC)=C•∇φ, we may write Eq. (23) in the form: (23) ∫∫∫∫∫ •=• SV SdVd )ˆ( nCC φφ∇ or: 0ˆ =         −• ∫∫∫∫∫ SV SdVd nC φφ∇ or: since C is an arbitrary constant vector. 2014 MRT QEDˆ ∫∫∫∫∫ = SV SdVd nφφ∇ The first relation is: (25) (24)
• The second relation is: ∫∫∫∫∫∫∫ −== SSV dSdVd SAAnA ×××∇ ˆ To prove this, this time let F=A× B, where A and B is are arbitrary constant vectors, and on substituting for F into the divergence theorem (c.f., Eq. (7)), we obtain: ∫∫∫∫∫ •=• SV SdVd nBABA ˆ)()( ××∇ The integrand on the left-hand side of Eq. (26) may be written as ∇ •(A× B)=B•∇ × A− A•∇ × B=B•∇ × A, since B is a constant vector. Also note that A× B•n=B•(n× A). If we substitute these former and latter relations into Eq. (26), it becomes: (26) ∫∫∫∫∫ •=• SV SdVd )ˆ( AnBAB ××∇ since B is an arbitrary constant vector. 2014 MRT ˆ ˆ (27) 0ˆ =         −• ∫∫∫∫∫ SV SdVd AnAB ××∇ QEDˆ ∫∫∫∫∫ = SV SdVd AnA ××∇ (28) or: or:
• Let us consider two examples. 2014 MRT In the first example, if S is a closed surface which encloses a volume V, prove that: 0n =∫∫S Sdˆ In Eq. (25), let φ=1. We therefore obtain: QEDˆ ∫∫∫∫∫ == SV SdVd n0φ∇ and the result is proved. In the second example, if S is a closed surface which encloses a volume V, prove that: 0Sr =∫∫S d× In Eq. (28), let A=r and ndS=dS . We therefore obtain: since ∇ × r=0 and the result is proved. QED0rSr ==− ∫∫∫∫∫ VS Vdd ×∇× ˆ (29) (30)
• Complex Functions One’s first exposure to complex numbers probably came in connection with a study of the general solution to a quadratic equation of the form ax2 +bx+c=0: When 4ac >b2 , we write: a acbb x 2 42 −±− = since i=√(−1) ⇒ i2 =−1 and (i/i)⋅i=−1/i. The latter development is made necessary by the requirement that every quadratic equation must have two roots (i.e., real solutions). a bacib a bacib a bacb a bacb x 2 4 2 )4( 2 )4( 2 4 22222 −±− = +±− = −−±− = +−±− = A complex variable z=x+iy, (and its complex conjugate z* =x−iy) is an ordered pair of real variables x and y which satisfy certain laws of operation. In general, x+iy≠y+ix; hence the term ‘ordered pair’ is important. The real and imaginary parts of z are Rez=x and Imz=y, respectively. Note that z=0 implies that x=y=0, and that z1 =z2 means that x1 =x2 and y1 =y2. 2014 MRT The algebraic operations for two complex variables are as follows: )()()()( 2121221121 yyixxyixyixzz ±+±=+±+=± )()()()( 12212121221121 yxyxiyyxxyixyixzz ++−=+⋅+=         + − + + + = − − + + = + + = 2 2 2 2 2112 2 2 2 2 2121 12 12 22 11 22 11 2 1 )( )( )( )( )( )( yx yxyx i yx yyxx yix yix yix yix yix yix z z 1) Addition (Substraction): 2) Multiplication: 3) Division(z2 ≠0):
• A convenient graphical representation for complex numbers was derived by Argand (1768-1822); it is illustrated in the Figure. The principle argument of z is denoted by θp. A convenient graphical representation for complex variable was devised by Argand. It is a vector representation for z where the iy-axis is an imaginary axis. 2014 MRT Also note that: z = x + iy | z | θ x iy π20 ≤≤ pθ The respective notations are r= |z|= mod z and θ =arg z. We note that the argument of z is not unique (determined up to a multiple of 2π – this often explains the prevalence of factors of 2π in the arguments in many relations involving i ). Hence we may write: kp π2+= θθ with k= 0, ±1, ±2, … and where: 22222 )(Im)(Re zzzzyxr +=≡+= and An Argand diagram is simply a vector representation for z where the y-axis is an imaginary axis – that is to say, z is just the vector sum of x and iy. In plane polar coordinates (x,y)→ (r,θ ), we have: θθ sincos ryrx == and In terms of r and θ , the complex variable z above becomes: )sin(cos θθ iriyxz +=+= where r= |z|, is called the modulus (i.e., the absolute value) of z and θ =tan−1 (y/ x) denotes the argument (i.e., the phase) of z.
• The complex conjugate of z, z∗ , is z∗ =x–iy and its graphical representation is given in the Figure. 2014 MRTz∗ = x – iy |z ∗ | θ x iy To find the complex conjugate of a complex quantity, change the sign of the imaginary part (or imaginary terms). In connection with the complex conjugate, we have: * 2 * 121 * 2 * 121 2 *)( *)( ))((* zzzz zzzz ziyxiyxzz = +=+ =−+= and * 2 * 1 * 2 1 z z z z =      Comparing zz∗ =|z|2 above with |z|2 =(Rez)2 +(Imz)2 we observe that zz∗ is always real.
• An Example is provided given z=(1+i)/(2–3i). (a) Put z in the standard form, x+iy. Find (b) Rez, (c) Imz, (d) modz and (e) argz. (f) Give the graphical representation of z. (a) 13 5 13 1 94 )23(32 94 3232 32 32 32 1 i iii i i i i z +−= + ++− = +−+ −++ = + + ⋅ + + = ii 66 (f) See Figure. Plot of z=(1+i)/(2–3i) on an Argand diagram with r=√(26)/13 and θ =–tan–1 (5). 2014 MRT x iy (d) so (b)       − 13 5 , 13 1 385.0 13 5 077.0 13 1 1 ==−= and,i 0.2−0.2 −0.2i +0.1i 392.026 13 1 13 5 13 1 )(Im)(Remod 22 22 ==      +      −= =+= rzzz 13 1 Re −=z (c) 13 5 Im =z (e)  5.11)5(tan)5(tantanarg 111 −=−=−=== −−− x y z θ 38.0 13 5 sin08.0 13 1 cos ===−=−== θθ ryrx and −0.1 0.1 +0.2i −0.1i 0.3 0.4 +0.3i +0.4i r
• We now develop an extremely useful relation due to Euler (1707-1783). For real θ, we may write: 2014 MRT  +−+−=+−+−= !6!4!2 1cos !7!5!3 sin 642753 θθθ θ θθθ θθ and For complex variables, we assume that the power-series expansion is similar to that for real variables: N N N N N x N x N x         +== ∞→ ∞ = ∑ 1lim ! e 0 where e(x=1) ≡exp(x=1)=limN→0(1+1N /N)N is Euler’s number e=2.71828. Hence we write: ∑ ∞ = = 0 ! e N N z N z where with z=iθ in the above expansion,it becomes (i.e.,whereeiπ⁄2 =√−1≡i comes from): 101π2sinπ2cose101πsinπcosesincos !5!3!4!2 1 !5!4!3!2 1 ! )( e π2π 53425432 0 =⋅+=+=−=⋅+−=+=⇒+=         ++−+         ++−=+++−−+== ∑ ∞ = iiiii i ii i N i ii N N i &θθ θθ θ θθθθθθ θ θθ  This is Euler’s formula. In polar coordinates, i.e., using r and θ, it is given by: θ θθ i rirz e)sin(cos =+=
• Two outcomes are needed now: N-th products and N-th roots or powers of a complex number. First, the product of two complex variables, z1 and z2, may be written as: 2014 MRT )]( 21 21212122211121 21 e )]sin()[cos()sin(cos)sin(cos θθ θθθθθθθθ + = +++=++= i rr irririrzz In the case of N terms, the product is z1z2zN=rN[cos(θ1 +θ2 ++ θN)+i sin(θ1 +θ2 ++θN). When z1=z2==zN, the product becomes: θ θθ NiNNN rNiNrz e)sin(cos =+= which is known as de Moivre’s (1667-1754) Theorem.                 + +        + =      += N k i N k r N i N rz ppNNN π2 sin π2 cossincos 111 θθθθ Secondly, the de Moivre’s theorem may be used to develop a general relation for extracting the N-th root of a complex number or raising of a complex quantity to the N-th power. The N-th root of z may be written as: where the principal argument is constrained by 0≤θp ≤2π and k=0, 1, 2,…, (N −1) since we require only N distinct roots of z.
• The graphical representation of this root is shown in the Figure. Plot of the square root of i, √i, on an Argand diagram. 2014 MRT For k=0, we obtain: x iy For k=1, we obtain (we use the Ptolemy identities, the sum and difference formulas for sine and cosine, i.e., cos(α+β)=cosαcosβ − sinαsinβ and sin(α+ β)= sinαcosβ +cosαsinβ): )1( 2 1 2 1 2 1 0 2 1 1 2 1 0 2 1 1 2 1 πsin 4 π cosπcos 4 π sinπsin 4 π sinπcos 4 π cosπ 4 π sinπ 4 π cos 1 21 iii iii k +−=−−=      ⋅+−⋅+      ⋅−−⋅=       ++      −=      ++      += = since cos(π/4) =sin(π/4)= 1/√2, cosπ =−1 and sinπ=0. The result is: )1( 2 1 2 1 2 1 4 π sin 4 π cos 0 21 i iii k += +=+= = As an Exercise, let us find the square root of i (i.e., √i ≡√√−1=[(−1)½ ]½ =(−1)¼ ). From the previous N-th root of the z equation,the specific equation for this problem is (i.e., N=½ and z=0+i⋅1 & θ =tan−1 (y/x) =tan−1 (1)=π/4=θp/2 for k=0): since the principal argument is θp =π/2 and z=i (x=0, y=1) means that r=√(x2 +y2 )=1.       ++      +=      + +      + =                 + +        + = kik k i kk i k ri pp π 4 π sinπ 4 π cos 2 π22π sin 2 π22π cos 2 π2 sin 2 π2 cos2121 θθ )1( 2 1 i+ )1( 2 1 i+− 707.0 2 1 414.121 ==−= and,i 1−1 −i +i )1( 2 121 ii +±=
• The word ‘matrix’ was introduced in 1850 by Sylvester and represents a rectangular array of quantities:             = mnmm n n aaa aaa aaa M     21 22221 11211 where aij are called elements (of the i-th row and j-th column): they may be real (or complex) numbers or functions. The matrix M has m rows and n columns and is called a matrix or order m×n (m by n). If m=n, the matrix is called a square matrix and in this case the main diagonal of the square matrix consists of the elements a11, a22, …, ann. ( )1 10 01 01 10 0 0 01 10 321 −=      − + =         −+ −− =      + − =      + + = i i i REEL/DIAG PARITY TICOMPLEX/ANREEL/SYM and, σσσ These are the Pauli (1900-1958) ‘spin’ matrices and they have the following properties: 3122121 2 3 2 2 2 1332211 2],[1 σσσσσσσσσσσσσσσσ i=−≡=++=++ and e.g., the direct products σ1⊗σ3 and σ1⊗σ2 and their addition (σ1⊗σ3)⊕(σ1⊗σ2) isgivenby:             − − − − =             − − ⊕             − − =      ⋅⋅ ⋅⋅ ⊕      ⋅⋅ ⋅⋅ =⊗⊕⊗ 001 001 100 100 000 000 000 000 0010 0001 1000 0100 01 10 01 10 )()( 22 22 33 33 2131 i i i i i i i i σσ σσ σσ σσ σσσσ 2014 MRT Matrix Operations As a general example in what follows, consider the following 2×2 complex matrices:
• Transpose (or Inverse) Matrix – MT (or M−1 ): The transpose (or inverse) of an arbitrary matrix M is written MT (or more written conveniently as M−1 ) and is obtained by inter- changing corresponding rows and columns of M, that is, M=aij and MT ≡M−1 =aji. e.g.       − =≡      − = − 0 0 0 0 1 222 i i i i σσσ T and If M* =M, the matrix is real. Complex Conjugate Matrix – M* : The complex conjugate of an arbitrary matrix M is formed by taking the complex conjugate of each element. e.g.       − =      − = 0 0 0 0 * 22 i i i i σσ and Hermitian Conjugate Matrix – M † : The Hermitian conjugate of an arbitrary matrix M is obtained by taking the complex conjugate of the matrix and then the transpose of the complex conjugate matrix. e.g.       + − =⊕=⊕      − + =⊕      + − =⊕ 01 10 )(])[( 01 10 )( 01 10 † 21 * 21 * 2121 i i i i i i σσσσσσσσ T finallyandthen If M† =M the matrix is said to be a Hermitian matrix. 2014 MRT
• Unit Matrix: The unit (identity) matrix I (or plainly 1 as used in Group Theory) is given by I=δij where IM=MI=I. Recall that the Kronecker delta δij has the following property:    ≠ = = ji ji ji when when 0 1 δ 2014 MRT Diagonal Matrix: Here we write: Singular Matrix: If the determinant is null, i.e., when detA≡|A|=0, then A is said to be a singular matrix. e.g. 0det 00 01 =≡      = AAA then jijiDD δ= e.g.           −= 400 010 002 D Since, by definition, the determinant is given by (Exercise): 12 12 11 11 2 1 1 1 2221 1211 AaAaAa aa aa A j j j +=== ∑= where: 1221 2112 2222 1111 )1()1( aaAaaA −=−==−= ++ and On substituting A11 and A12 into the expression for |A|, we obtain: 21122211 aaaaA −=
• Adjoint of a Matrix: The adjoint of a matrix is written as adjA; it is defined as the cofactor transpose, that is:       − − =      − − =      = 13 31 13 21 12 31 cc T hencethenIf AAA e.g. (Exercise) 2014 MRT Tc adj AA = Cofactor Matrix: The cofactor matrix is written as Ac and is defined by: ji AA =c e.g.           =           = 332331 322221 312111 c 333231 232221 131211 AAA AAA AAA A aaa aaa aaa A thenIf where: .,, ,,, ,,, 2221 12113333 2321 13112332 2322 13121331 3231 12113223 3331 13112222 3332 13121221 3231 22213113 3331 23212112 3332 23221111 )1()1()1( )1()1()1( )1()1()1( aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A aa aa A +++ +++ +++ −=−=−= −=−=−= −=−=−=
• Self-Adjoint Matrix: If adjA = A, A is said to be self-adjoint. e.g. 2014 MRT Symmetric and Antisymmetric Matrices: If ST =S, S is said to be a symmetric matrix whereas if AT =−A, A is said to be an antisymmetric (skew) matrix. e.g. 222111 0 0 0 0 01 10 01 10 σσσσσσ −=      − =      − =+=      =      = i i i i TT then:tricAntisymme;then:Symmetric Hermitian Matrix: If H† =H, H is said to be a Hermitian matrix. e.g. 2 † 2 * 2 * 22 0 0 )( 0 0 0 0 σσσσσ =      − ==      − =      − = i i i i i i T henceandIf In quantum mechanics, all physical observables (e.g., the Hamiltonian or total energy operator) are represented by Hermitian operators (matrices). Unitary Matrix: If UU† =I (or U† =U−1 ) U is said to be a unitary matrix. e.g. I=      =            ==      =      =      = 10 01 01 10 01 10 01 10 )( 01 10 01 10 † 11 † 1 * 1 * 11 σσσσσσ andhencethenIf T Unitary matrices are important in quantum theory (e.g., the scattering matrix) and even more important in quantum field theory in that they ensure the conservation of probability and further ensures that particle ‘ghosts’ do not crop into the equations describing the scattering processes. AAAAA =      − − ==      − − =      − − = 10 01 adj 10 01 10 01 cc T hencethenIf
• Orthogonal Matrix: If OOT =I, O is said to be an Orthogonal matrix. e.g. I=      =            =+=      =      = 10 01 01 10 01 10 01 10 01 10 11111 TT thenand σσσσσ Trace of a Matrix: The trace of a matrix A is given by the sum of its diagonal components: ∑= k kkaATr Inverse Matrix: For the inverse matrix, A−1 , we require that: IAA =−1 We will have to work out an explicit expression for A−1 . The value of a determinant is: 2014 MRT e.g. 972Tr 73 42 2211 2 12221 1211 =+=+==      =      = ∑= aaaA aa aa A k kkand ∑∑ == == n j jk jiki n j ji ji AaAAaA 11 δor Let bjk =Ak j , that is, B=AcT , so that this expression for the determinant becomes: T or c 1 AAABAIbaA n j jkjiki === ∑= δ where δik =I.
• On dividing I|A|=AAcT by |A|, we obtain: The quantity in brackets must be A−1 because of AA−1 =I. For A, we have |A|=1−6=−5 for the determinant, and:       − − =         = 13 21 2221 1211 c AA AA A and: 2014 MRT       = 12 31 A Hence the inverse of A is:         = A A AI Tc Let us try out an example and find the inverse of:       − − = 12 31cT A       − − −==− 12 31 5 1c 1 A A A T Check: QED 10 01 50 05 5 1 12 31 12 31 5 1c 1 I A A AAA =      =      − − −=      − −       −=         =− T
• x v y u y v x u ∂ ∂ −= ∂ ∂ ∂ ∂ = ∂ ∂ and Derivative of f(z) Let us return to complex functions. A single-valued function f (z) is said to be analytic in a certain region of the complex plane if it has a unique derivative at every point in the region. The derivative of f (z), like the function of a real variable, is defined by: where       ∆+∆ ∆+∆ =      ∆ −∆+ =      ∆ ∆ ==′ →∆→∆→∆ yix viu z zfzzf z f zd zfd zf zzz 000 lim )()( limlim )( )( yixzyixzviufyxviyxuzf ∆+∆=∆+=∆+∆=∆+= &and&),(),()( There are an infinite number of ways of obtaining ∆z →0 in the z-plane. 2014 MRT Case I: ∆ x= 0 and i ∆ y→0: x iy which are known as the Cauchy (1789-1857)-Riemann (1826- 1866) conditions; they constitute the necessary conditions for a unique derivative of f (z) in Γ. y v y u i y v i i y u iyi viu zf yiyiyi ∂ ∂ + ∂ ∂ −=      ∆ ∆ +      ∆ ∆ =      ∆ ∆+∆ =′ →∆→∆→∆ 000 I limlim 1 lim)( 0 0 →∆ =∆ yi x Γ 0 0 =∆ →∆ yi x Case II: ∆ x→0 and i ∆ y= 0: x v i x u x v i x u x viu zf xxx ∂ ∂ + ∂ ∂ =      ∆ ∆ +      ∆ ∆ =      ∆ ∆+∆ =′ →∆→∆→∆ 000 II limlimlim)( If f (z) is analytic in the region Γ (see Figure), we require that fI′(z) = fII′(z) (i.e., ∂v/∂y −i ∂u/∂y=∂u/∂x+ i ∂v/∂x). As a result of this requirement, we obtain the following relations (in the complex plane): Case I Case II
• As an aside, let us see what happens when we differentiate the Cauchy-Riemann conditions ∂u/∂x=∂v/∂y and ∂u/∂y=−∂v/∂x with respect to x and y, respectively. We obtain: where uxx =∂2 u/∂x2 and uyy =∂2 u/∂y2 , &c. If u and v possess continuous partial derivatives up to second order, this leads to: yxyyxyxx uuuu xy v y u yx v x u −== ∂∂ ∂ −= ∂ ∂ ∂∂ ∂ = ∂ ∂ and and 2 2 22 2 2 2014 MRT 00 00 2 2 2 2 2 2 2 2 =+=+ = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ yyxxyyxx vvuu y v x v y u x u and and Therefore u and v are solutions of Laplace’s equation ∂2 φ/∂x2 + ∂2 φ/∂y2 =0 in two dimensions (e.g., recall the three dimensional one as ∇2 φ =0 where ∇2 is the Laplacian operator ∇2 ≡∇ •∇ =∂2 /∂x2 +∂2 /∂y2 +∂2 /∂z2 ); they are called harmonic (or conjugate) functions. Incidently, in polar coordinates the Cauchy-Riemann conditions become: r vu r v rr u ∂ ∂ −= ∂ ∂ ∂ ∂ = ∂ ∂ θθ 11 and
• For example, (a) Show that v(x,y)=3x2 y–y3 is harmonic; (b) Find the conjugate function, u(x,y); and (c) Find the analytic function f (z)=u(x,y)+iv(x,y). (a) If ∂2 v/∂x2 + ∂2 v/∂y2 =0 then v is said to be harmonic. So: By adding the right-hand side of ∂2 v/∂x2 and ∂2 v/∂y 2 , we see that v is harmonic (i.e., 0). y y v yx y v y x v xy x v 63366 2 2 22 2 2 −= ∂ ∂ ⇒−= ∂ ∂ = ∂ ∂ ⇒= ∂ ∂ and (b) Comparing the Cauchy-Riemann conditions ∂u/∂x=∂v/∂y and ∂u/∂y=−∂v/∂x we get: Integrating the two above equations with respect to x and y, respectively, we obtain: y u xy x v x u yx y v ∂ ∂ −== ∂ ∂ ∂ ∂ =−= ∂ ∂ 633 22 and )(3)6(),()(3)33(),( 22322 xgxyydxyyxuyfxyxxdyxyxu +−==+−=−= ∫∫ and )(3)(3 223 xgxyyfxyx +−=+− or In the above equation, we must have f (y)=0 (e.g., since by inspection we see that −3y2 x + f (y)=−3xy2 ) and also g(x)=x3 . The required conjugate function is therefore given by: xyxyxu 23 3),( −= (c) The corresponding analytic function is given by (Exercise - Hint: z3 =(x+iy)3 ): 2014 MRT3323233223 )(33)3(3)( ziyxxyyiyxixyyxixyxviuzf =+=−−+=−+−=+=
• Contour Integrals The integral (in the Riemannian sense) of a function f (z) is defined, as in the case of real variables, by: where the path C has been divided into N segments and ξj is some point on the curve between zj−1 and zj (see Figure). ...))((lim))((lim))((lim)()( 122 0 011 0 1 1 0 z 1201 1 o +−+−=−≡= →−→− = − →− ∞→ ′ ∑∫∫ − zzfzzfzzfdzzfdzzf zzzz N j jjj zz NzC jj ξξξ Simply connected with a closed curve drawn within Co and multiply connected regions where all points within C3 do not belong to the region between C and C . 2014 MRT iy ξj zo z′ z1 z2 zjzj−1 In complex variable theory, ∫C f (z)dz is called a contour integral of f (z) along the contour C from zo and z′. x THEOREM: If f (z) is analytic and its partial derivatives are continuous throughout some simply connected region, then for every closed path C within this region: Co C1C2 C3 Here is a definition of a simply connected region: A region in the complex plane is called simply connected if all closed paths within the region contain only points that belong to the region (i.e., all closed paths within the region can be shrunk to a point) Simply Connected Multiply Connected 0)( =∫C zdzf The symbol ∫ means the integral is around a closed path, and the symbol ∫ indicates that the path is traversed in a positive (counterclockwise) manner. This convention will be to traverse the path in such a way that the region of interest lies to the left. This equation is called Cauchy’s integral theorem.
• Cauchy’s Integral Formula Another important and extremely useful relation concerning the integral of a function of a complex variable is Cauchy’s integral formula. It can be written as: where zo is within C. The function f (z) is assumed to be analytic within C; however, f (z)/(z–zo) is clearly not analytic at z=zo. For r→0 (see Figure) the left part of the Figure (a) is equivalent to the right part (b). 2014 MRT iy Around the path C′, setting z– zo =r eiθ and dz = ir eiθdθ we obtain: x From (b) it can be seen that f (z)/(z– zo) is analytic in the region between C and C′. Hence we may apply Cauchy’s integral theorem and obtain: )(π2 )( o o zfizd zz zf C = −∫ for r→0. The formula for the Cauchy integral is thus established. In summary, we may write: zo (a) (b) r zo z C ′ CC ≡ ∫∫∫∫ ′′ − = − = − + − CCCC zd zz zf zd zz zf zd zz zf zd zz zf oooo )()( 0 )()( or )(π2)e()e( e )e()( o π2 0 o π2 0 o o zfidrzfidri r rzf zd zz zf ii i i C =+= + = − ∫∫∫ θθ θθ θ θ    = −∫ Cz Czzf zd zz zf i C outsidefor insidefor o oo o 0 )()( π2 1 and for the N-th derivative of f (z) with respect to z, we have: ∫ += − =≡ C Nzz NN d z f i N zfzf ξ ξ ξ 1 o )( o )( )( )( π2 ! )()( o when f (ξ) is analytic within C (Exercise - Hint: f ′(z)= lim∆z→0(…)).
• We keep using Taylor’s Series expansion so might as well develop it… If f (z) is analytic in some region Γ and C is a circle within Γ with center at zo, then f (z) can be expanded in a Taylor’s (1685-1731) series. Here z is any point interior to C (see Figure), that is: This series for f (z) converges absolutely and uniformily for |z–zo|<R where R is the radius of convergence. 2014 MRT iy where |z–zo|<R and R=|ξ–zo|. To prove that the expansion holds, a series expansion for 1/ξ–z will be developed. For convenience, we set KT =(z–zo)/(ξ–zo). On substracting from unity on both sides of KT, we obtain 1− KT =1 −[(z–zo)/(ξ–zo)]=(ξ–z)/(ξ–zo). By inverting both sides, we have 1/(1 −KT )=(ξ–zo)/(ξ–z )=ΣNKT N (for |KT |<1). The required expansion for 1/(ξ–z) is obtained by dividing the 1/(1 −KT )=ΣNKT N by ξ–zo; the result is 1/ξ–z = 1/(ξ–zo)ΣN[(z–zo)/(ξ–zo)]N =ΣN[(z–zo)N /(ξ–zo)N+1 ]. Substituting this into f (z) above, we obtain the required exansion for f (z): x From the Cauchy integral formula, we note that: ∑ ∞ = − = − ++−′+= 0 o )(ooo )( ooo )( ! )( ! ))(( ))(()()( N N NNN zf N zz N zzzf zzzfzfzf  zo C ∫ − = C d z f i zf ξ ξ ξ)( π2 1 )( Γ z z − zo z −ξ ξ − zo ξ ∑ ∑ ∫∑ ∫∫ − =                 − −=         − − = − ++ N N N N C N N N C N N C zf N zz d z f i zzd z zzf i d z f i )( ! )( )( )( π2 1 )( )( ))(( π2 1)( π2 1 o )(o 1 o o1 o o o ξ ξ ξ ξ ξ ξ ξ ξ ξ
• As an example, let us expand 1/1–z in a Taylor’s series about zo = i and find, as a bonus to highlight the dimensions, the radius of convergence R=|1–zo|. We have the complex function f (z)=1/(1–z)≡(1– z)–1 and its first derivative with respect to z is f ′(z)≡ d[f (z)]/dz=d[(1– z)–1 ]/dz= –(1– z)–2 d(– z)/dz = (1– z)–2 while the other two are f "(z)=2⋅1(1– z)–3 and f ′′′(z)=3⋅2⋅1(1 – z)–4 . The general term is thus obvious by inspection: ∑ ∑∑∑ + +− = − − = − − = − = − = N N N N N N N N Niz N N N i iz iN N iz if N iz zf N zz zf 1 )1()( o )(o )1( )( )1(! ! )( )( ! )( )( ! )( )( o where N⋅…⋅3⋅2⋅1=N! (i.e., N factorial). The required expansion around zo = i is therefore: )1()( )1(!)( +− −= NN zNzf The radius of convergence of the above series is: 2014 MRT 21lim 1 =−== +∞→ i a a R N N N by use of zz∗ =(x+iy)(x–iy)=|z|2 where: 1 )1( 1 + − = NN i a
• As we just saw with Euler’s formula, the theory of the representation of a function of a real variable by means of series of sines and cosines is an indespensible technique in mathematical physics. Fourier analysis is such a theory and in 1807, Fourier (1768- 1830) stated without proof (and used it in developing a solution to the heat equation) the following THEOREM: Any single-valued function f (x) defined on the interval [−π,+π] may be represented over this interval by the (trigonometric) ‘Fourier’ series*: 2014 MRT ∑ ∞ = ++= 1 o )sincos( 2 )( n nn nxbnxa a xf provided the expansion coefficients, an and bn, are determined by use of Euler’s formulas: ...),3,2,1(sin)( π 1 ...),2,1,0(cos)( π 1 π π π π ==== ∫∫ −− ndxnxxfbndxnxxfa nn and The interval [−π,+π] is limitative to say the least and to solve many physical problems it is necessary to develop a Fourier series that will be valid over a wider interval. To obtain an expansion valid in the interval, say [−l,+l], we let: and determine φ such that f (x)= f (x+2l). * I will forgo the proof outlined by Dirichlet (1805-1859) in 1829 or the (5 or so) Dirichlet conditions imposed on the theorem. ∑ ∞ = ++= 1 o )sincos( 2 )( n nn xbxa a xf φφ Fourier Series and Transforms
• In this case, φ=nπ/l; hence, we obtain: 2014 MRT ∑ ∞ =       ++= 1 o π sin π cos 2 )( n nn l xn b l xn a a xf The complex form of the Fourier series is obtained by expressing both cos(nπx/l) and sin(nπx/l) in exponential form, that is: or: ∑ ∞ −∞= = n lxni ncxf π e)( for −l ≤ x ≤+l (such that f (x) satisfies Dirichlet’s conditions in the interval). The corresponding Euler coefficients are: ...),3,2,1( π sin)( 1 ...),2,1,0( π cos)( 1 =      ==      = ∫∫ −− ndx l xn xf l bndx l xn xf l a l l n l l n and ∑∑∑ ∞ = −∞ = −∞ =         − +         + +=            +      += 1 ππ 1 ππ o 1 o 2 ee 2 ee 2 π sin π cos 2 )( n lxnilxni n n lxnilxni n n nn i ba a l xn b l xn a a xf which is the complex form of Fourier series which is valid over [−l,+l] and where: )( 2 1 )( 2 1 2 1 oo nnnnnn biacbiacac +=−== −and;
• On multiplying both sides of the complex form of Fourier series by exp(−inπx/l) and integrating with respect to x, we obtain: 2014 MRT m n mnn n mnn n l l lmnxi n l l lxni lc cl lc xdxfcxdxf 2 2 )2( e)(e)( )(ππ = = = = ∑ ∑ ∑ ∫∫ ∞ −∞= ∞ −∞= ∞ −∞= − − − − δ δ Therefore the cn coefficient in of the complex form of Fourier series is given by: ∫− − = l l lxni n xdxf l c π e)( 2 1
• Now we investigate a particular aspect of the theory of Fourier transforms. When the method of separation of variables (i.e., assuming that a solution – or field that has a symmetrical x and y dependence)φ(x,y) – can be written as φ(x,y)=X(x)Y(y)) is applied to certain partial differential equations of mathematical physics, integrals of the form: 2014 MRT ∫− = b a xdxKxfF ),()()( αα often occur. The function f (α ) is said to be the integral transform of f (x) by the kernel F(α ,x). The kernel associated with the Fourier transform is: ∫− − = l l xi xdxfF α α e)( π2 1 )( while other kernels associated with Laplace, Fourier-Bessel (Henkel) and Mellin also do exist but will not be used. To make the transition l→∞ in the complex form of the Fourier series, we introduce l n k π = where (∆k/π)l=1 since ∆n=1. Hence we write the complex form of Fourier series and the cn coefficient in the following forms: where Cl (k)=lcn /π. ∫∑ − − ∞ −∞== − =∆= l l xki l lkn xki l xdxfkCkkCxf e)( π2 1 )(e)()( π and
• If we let l→∞, we obtain: 2014 MRT ∫ ∞ ∞− − = xdxfkC xki e)( π2 1 )( ∫ ∞ ∞− = kdkCxf xki e)()( and: We let F(k)=√(2π)C(−k) and obtain: ∫ ∞ ∞− = xdxfkF xki e)( π2 1 )( ∫ ∞ ∞− − = kdkFxf xki e)( π2 1 )( and: These two relations form the Fourier transform pair.
• As an example, let us find the Fourier transform, F(k), of the Gaussian distribution function, f(x)= Nexp(−αx2 ), where N and α are constants. 2014 MRT (we used exp(iξ)=cosξ+isinξ) where the second integral is zero since it is the integral of an odd function from −∞ to +∞ and: Note that both f (x) and F(k) are Gaussian distribution functions with peaks at the origin. The standard deviation, width, is defined as the range of the variable x (or k) for which the function f (x) (or F(k)) drops by a factor of exp(−½)=0.606 of its minimum value. So, for f(x)= Nexp(−αx2 ), the standard deviation is given by: By using our formula (i.e., F(k)=[1/√(2π)]∫ f (x)exp(ikx)), we find that: αα α 422 e π cose)( kx xdxkkF − ∞ ∞− − == ∫ α σ 2 1 2 = ∆ = x x α αα αα 42 22 22 e π2 sinecose π2 e π2 e)e( π2 1 )( k xx xxkixkix N xdxkixdxk N xd N xdNkF − ∞ ∞− − ∞ ∞− − ∞ ∞− − ∞ ∞− − =       += == ∫∫ ∫∫
• For: 2014 MRT Note that ∆x∆k=(2/√α)(2√α)=4. If α→0 (small), then ∆x→∞ and ∆k→0. For α→∞ (large), ∆x→0 and ∆k→∞. the standard deviation is given by: ασ 2 2 = ∆ = k k α α 42 e 2 )( kN kF − = As another example, let us find the Fourier transform for a box function f (x) where:    > <≤− = ax axa xf for for 0 1 )( By use of F(k)=[1/√(2π)]∫ f (x)exp(ikx), we obtain: k ak ki xdxdxdxdkF a a xkia a xki a xki a a xki a xki sin π 2 e π2 1 e π2 1 e)0(e)1(e)0( π2 1 )( =         ==      ++= − − ∞ − − ∞− ∫∫∫∫ So, the Fourier transform of a ‘coordinate’ box function, f (x)=1 (if −a≤ x<a) or 0 (if |x|>a) is a somewhat complicated function (of say momentum k): F(k)=[√(2/π)][sin(ka)]/k!
• ∫∫ ∞ ∞− −− ∞ ∞− − == xdxxdtxtk txkixki x xx )ω( e)( π2 1 e),( π2 1 ),( ψΨΦ In physics, k (or kx) is the magnitude (or x-component) of the wave vector k (where in three dimension, k• r is used) its dimensions of inverse length, and x (or r) is a distance (or position). The combination of ωt, where ω is the angular (circular) frequency with dimensions of inverse time and t is the time, is also used procuring a total phase (the argument of the exponential function) of k• r−ωt is widely in quantum mechanics. 2014 MRT where px =hkx /2π (p =hk/2π) is the momentum (along the x-direction) and E=hω/2π is the total energy. ∫∫ ∞ ∞− −− ∞ ∞− − == xdx h xdtx h tp htExpihxpi x xx )(π2π2 e)( 1 e),( 1 ),( ψΨΦ or and ∫∫ ∞ ∞− − ∞ ∞− == x txki xx xki x kdkkdtktx xx )ω( e)( π2 1 e),( π2 1 ),( ϕΦΨ ∫∫ ∞ ∞− − ∞ ∞− == x htExpi xx hxpi x pdp h pdtp h tx xx )(π2π2 e)( 1 e),( 1 ),( ϕΦΨ or Here (and in Quantum Mechanics) the corresponding transform pairs are:
• ∑ ∞ −∞= = n nnn xcxF )()( ψ Now since we’re investigating properties of the Fourier transforms that can be easily (and piecemeal) be exported to Quantum Mechanics, let us investigate what is called the Dirac (1902-1984) delta function. If the set of functions {ψn(x)} is the orthonormal basis for a vector space En, and F(x) is an arbitrary function in En, then: 2014 MRT Substituting cn from the above into F(x)=Σ±∞ cnψn(x), we find that: nnnn xdxx δψψ =∫ ∞ ∞− )()(* where the cn are called expansion coefficients and (with ‘*’ for ‘complex conjugate’): The orthonormality property expressed in the above integral will now be used to develop an expression for cn in the above series. On multiplying the series by ψ*n(x) and integrating over the range of x, we obtain: n n nnn n nnnnn ccxdxxcxdxFx === ∑∑ ∫∫ ∞ −∞= ∞ −∞= ∞ ∞− ∞ ∞− δψψψ )()(*)()(* xdxxFxxF n n n )()()(*)( ψψ∑ ∫ ∞ −∞= ∞ ∞− = The expression in this last equation is called a generalized Fourier series.
• Interchanging the sum with the integral in the generalized Fourier series, we may write: 2014 MRT where ∫∫ ∑ ∞ ∞− ∞ ∞− ∞ −∞= −=         = xdxxxFxdxxxFxF n nn )()()()(*)()( δψψ ∑ ∞ −∞= =− n nn xxxx )()(*)( ψψδ The δ(x−x) quantity is the one-dimensional Dirac delta function, and its most important property is expressed by F(x)=∫±∞ ψ*n(x)δ(x−x)dx. The following are the various notations for the Dirac δ -function used in the literature: δ(x−x), δ(x), and δ(x,x). The Dirac δ -function is defined such that: 1)( =−∫ ∞ ∞− xdxxδ and    =−∞ ≠− =− 0 00 )( xx xx xx for for δ
• lxni n l x π e 2 1 )( =ψ We will develop some other useful relations involving the Dirac δ-function since its concept is extremely important in analyzing physical quantities of systems involving impulsive-type forces. Consider the case of harmonic functions: 2014 MRT or On substituting these into δ(x−x)=Σ±∞ ψ*n(x)ψn(x), we obtain: where k=nπ/l and 1/2l=∆k/2π since ∆k=0. If l→∞ (i.e., ∆k→0) in the last term of the above equation, the sum changes into an integral, that is: ∑∑ ∞ =−∞= − ∞ −∞= − ∆==− π )()(π e 2 1 e 2 1 )( lkn xxki n lxxni k ll xxδ ∫ ∞ ∞− − =− kdxx xxki )( e π2 1 )(δ )(π )](sin[ lim π2 1 elim π2 1 )( )( xx xxl kdxx l l l xxki l − − ==− ∞→− − ∞→ ∫δ These last two equations are two widely used representations for the Dirac δ-function. For example, in three Cartesian dimensions, we write: 1)()0(0)(e π)2( 1 )()()()( 3 =≠==≡ ∫∫ ∞ ∞− ∞ ∞− • rrrrkr rk ddzyx i δδδδδδ andwhere
• One more thing… Perseval’s relation! The integral of the product of two functions f (x) and g∗(x), the complex conjugate of g(x), may be written as: 2014 MRT where the δ (kx′–kx) quantity is, in this case, the one-dimensional Dirac delta function, applied to the standard momentum kx′, and its most important property is expressed by: ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫∫∫ ∞ ∞− ∞ ∞− ∞ ∞− ∞ ∞− ∞ ∞− ∞ ∞− −′ ∞ ∞− ∞ ∞− ′ ∞ ∞− − ∞ ∞− ′= −′′′=       ′′=               ′′×      = x xxxx kkxi xx x xki x xki kdkGkF kkkGkdkFkd xdkGkFkdkd xdkdkGkdkFxdxgxf xx xx )(*)( )()(*)( e π2 1 )(*)( e)(* π2 1 e)( π2 1 )(*)( )( δ ∫ ∞ ∞− ′′−′= xdxxxFxF )()()( δ The following are the various notations for the Dirac δ-function used: δ (x–x′), δ (x), and δ (x,x′). The Dirac δ-function is defined such that:    =′−∞ ≠′− =′−=′′−∫ ∞ ∞− 0 00 )(1)( xx xx xxxdxx for for and δδ
• As an example, let us use Perseval’s formula to show that: 2014 MRT Given that Perseval’s formula: 1)()(* =∫ ∞ ∞− xxx pdpp ϕϕ since, as indicated above: if (the normalization condition): 1)()(* =∫ ∞ ∞− xdxx ψψ Here we write (taking the last transform pairs for t=0): ∫ ∞ ∞− = x hxpi x pdp h x xπ2 e)( 1 )( ϕψ ∫∫ ∞ ∞− ∞ ∞− ′= xkdkGkFxdxgxf )(*)()(*)( when applied to ψ and ϕ, leads to: 1)()(*)()(* == ∫∫ ∞ ∞− ∞ ∞− xxx pdppxdxx ϕϕψψ 1)()(* =∫ ∞ ∞− xdxx ψψ
• The cosine and sine transform pairs are defined, respectively, by: 2014 MRT ∫ ∫ ∞ ∞ = = 0 cosine 0 cosine cos)( π 2 )( cos)( π 2 )( kdxkkFxf xdxkxfkF and Note that the Fourier transform of f (x) (i.e., F(x)=[1/√(2π)]∫±∞ f (x)eikx dx) may be written as: If f(x) is an even function of x (i.e., f(x) = f(−x)), then we see from the relation above (i.e., F(x)=[1/√(2π)]∫±∞ f (x)(coskx+isinkx)dx) that the cosine transform is equal to the Fourier transform. If, however, f(x) is an odd function of x (i.e., f(x) = −f(−x)), then we see from the same relation above that the sine transform is equal to the Fourier transform provided −F(k) is replaced by Fsine(k). ∫ ∫ ∞ ∞ = = 0 sine 0 sine sin)( π 2 )( sin)( π 2 )( kdxkkFxf xdxkxfkF ∫∫ ∞ ∞− ∞ ∞− +== xdxkixkxfxdxfkF xki )sin(cos)( π2 1 e)( π2 1 )(
• Now we turn to the transforms of derivatives. Fourier transforms, cosine transforms, and sine transforms can often be used to transform a differential equation (ordinary or partial) which describe a complicated hysical problem into a simpler equation (algebraic or ordinary differential) that can be easily solved. The required solution of the orginal differential equation is then obtained by finding the inverse transform of the solution of the simpler equation (in transform or phase space). In order to use the transform method to solve first- and second-order differential equations, the transforms of first- and second-order derivatives are needed. 2014 MRT We now develop the transforms of first- and second-order derivatives. The Fourier transforms of first- and second-order derivatives will be represented by F(1) (k) and F(2) (k), respectively. That is to say: ∫ ∞ ∞− = xd xd xfd kF xki e )( π2 1 )()1( On integrating this equation by parts, we obtain: where f(x)→0 as x→±∞. Similarly we find that: )(e )( π2 e )( e)( π2 1 )()1( kFkixd xd xfdki xd xd xfd kixfkF xkixkixki −=−=      −= ∫∫ ∞ ∞− ∞ ∞− ∞+ ∞− )(e )( π2 1 )( 2 2 2 )2( kFikxd xd xfd kF xki −== ∫ ∞ ∞− where f(x)→0 and f ′(x)→0 as x→±∞.
• A note now on one-dimensional Fourier transforms of a function of two and three independent variables. They are given below: 2014 MRT ∫∫ ∞ ∞− − ∞ ∞− == kdykKyxfxdyxfykF xkixki e),( π2 1 ),(e),( π2 1 ),( and The transforms of derivatives of two independent variable are correspondingly given by: ),(e ),( π2 1 ),()1( ykFkixd x yxf ykF xki −= ∂ ∂ = ∫ ∞ ∞− and ∫∫ ∞ ∞− − ∞ ∞− == kdzykKzyxfxdzyxfzykF xkixki e),,( π2 1 ),,(e),,( π2 1 ),,( and for f(x)→0 as x→±∞, and ),(e ),( π2 1 ),( 2 2 2 )2( ykFkxd x yxf ykF xki −= ∂ ∂ = ∫ ∞ ∞− where f(x,y)→0 and ∂f (x,y)/∂x →0 as x→±∞. Things apply similarly for cosine and sine transforms (e.g., Fc(x,y)=√(2/π)∫0 to ∞ f (x,y)× coskxdx) except for the derivatives (Exercise. Solution: Fc (1) (k,y)=kFs(k,y)−√(2/π) f(0,y), Fc (2) (k,y)=−kFc(k,y)−√(2/π) f ′(0,y) for the cosine derivative and Fs (1) (k,y)=−kFc(k,y), Fs (2) (k, 2
• In linear response theory, the general equation for the one-dimensional transform (i.e., F(α )=∫a to b f(x)K(α ,x)dx), takes the form: 2014 MRT ∫∫ ∞ ∞− −== xdxfxKxdxKxfF b a )()(),()()( ααα which is consistent with the properties of the Dirac delta function. In this latter case, K(α −x) is called the impulse response of the system. When the equation F(α )= ∫−∞ to ∞ K(α −x) f(x)dx above is written in the form: where K(α −x) is called the response of the linear system, f(x) is the input (signal) to the linear system, and F(α ) is the output (signal). If K(α −x)=δ(α −x), then: it is called the one-dimensional convolution integral of two integrable functions f(x) and g(x). ∫ ∞ ∞− −= xdxfxF )()()( αα δ ∫ ∞ ∞− −== ξξξ dgxfgfxF )()( π2 1 *)(
• Now, let F(k) and G(k) be the Fourier transforms of f(x) and g(x), respectively. For these functions, the convolution integral becomes: 2014 MRT ∫ ∫ ∫ ∫∫ ∞ ∞− ∞ ∞− − ∞ ∞− ∞ ∞− −− ∞ ∞− =       =− ξξ ξξξξξξ ξ ξ dgkdkF ddkFgdgxf kikxi xki e)(e)( π2 1 e)( π2 1 )( π2 1 )()( π2 1 )( This is the convolution theorem for Fourier transforms. It means that the Fourier transform (inverse transform) of the product F(k)G(k), the right-hand side of the above result, is the convolution (i.e., folding) of the original functions. f *g. ∫∫ ∞ ∞− − ∞ ∞− =− kdkGkFdgxf kxi e)()( π2 1 )()( π2 1 ξξξ hence
• As an example, let us consider the Heat Equation. The temperature distribution T(x,0) in a very long heat-conducting rod is described by the one-dimensional heat equation: 2014 MRT Taking the Fourier transform of the heat equation we obtain: x txT x txT ∂ ∂ = ∂ ∂ ),(1),( 2 2 σ By use of F(2) (x,y)=−k2 F(k,y), the above equation reduces to: where the initial condition is given by T(x,0)= f (x). σ is called the diffusivity which equals 1/σ =k/cρ (i.e., for thermal conductivity k, specific heat c, and density ρ.) Here it is assumed that T(x,t)→0 and ∂T/∂x→0 as x→±∞, and the Fourier transform of f(x) exists. From our result obtained earlier (i.e., F(x,y)=[1/√(2π)]∫±∞ f (x,y)eikx dx) we generalize it to include time t and let: ∫ ∞ ∞− = xdtxTtkT xki e),( π2 1 ),( ∫∫ ∞ ∞− ∞ ∞− ∂ ∂ = ∂ ∂ xdtxT t xd x T xkixki e),( π2 11 e π2 1 2 2 σ t tkT tkTk ∂ ∂ =− ),(1 ),(2 σ where T(x,t)→0 and ∂T/∂x→0 as x→±∞ (the temperature at zero at the ends of the very long rod.)
• Now let us digress a bit and look at how this heat equation is solved traditionally. Writing this equation for say a 1 m long bar as (using the shorthand notation ux =∂u/∂x) uxx =(1/σ)ut (for ux(0,t)=ux (1,t)=0 and ut (x,0)= f(x)) and using the academic ‘method of separation of variable’ where we assume u(x,t)=X(x)T(t), we obtain: 2014 MRT λ σ == ′′ T T X X 1 or 0=−′′ XX λ where X(x)=C1exp[−√(λ)x]+C2 exp[√(λ)x] and 0=− TT λ where T(t)=C3exp(σ λt). The general solution is (with the Cs arbitrary constants): txx CCCtxu λσλλ e)ee(),( 321 += − Using the above condition ux(0,t)=0, where ux(x,t)={−kC1exp[−√(λ) x]+kC2 exp[√(λ)x]}× C3exp(σλx) we see that C1 =C2. Hence the solution becomes: txx CCtxu λσλλ e)ee(),( 31 − += From the second condition ux(1,t)=0, where ux(1,t)=C1√(λ){exp[√(λ)]−exp[−√(λ)]}× C3exp(σλx)=0 we find that λ<0 since C1 ≠0and λ≠0for a nontrivial solution. For λ<0, we have: t DxBxAtxu λσ λλ − += e)sincos(),(
• In the above solution, ux(0,t)=0=(Bk)Dexp(−σ λx) which implies that B=0 and ux(1,t)=0= [−A√(λ)]sin[√(λ)]×Dexp(−σλx). Since A≠0is a nontrivial solution, we take: 2014 MRT ( )...,2,1,0π0sin === nn forseigenvalueor λλ The solution now becomes: The condition on time, ut(x,0)= f(x), leads to the specific form for An′. Here we have ut(x,0)= f(x)= Σn=0to∞(−n2 π2 σ )(An′cosnπx) which we can write as: where ∑∑ ∞ = − ∞ = ′== 0 π 0 ]e)πcos[(),(),( 22 n tn n n n xnAtxutxu σ ∑ ∞ = = 0 π cos)( n n l xn axf n l l n Andx l xn xf l a ′−== ∫− σ22 π π cos)( 1 ∫− −=′ 1 122 πcos)( π 1 dxxnxf n An σ or We recognize the Fourier series instantly with these last steps. As such, the above development assumes that f(x) satisfies the Dirichlet condition.
• Ok. Back to our initial train of thought for solving the heat equation using Fourier transforms. The solution of the heat equation −k2 T(x,t)=(1/σ)∂T(k,t)/∂t is (i.e., take λ→k2 in the previous academic developpement): 2014 MRT Since the solution in k-space becomes: tk AtkT 2 e),( σ− = )(e)( π2 1 )0,( kFdxxfAkT xki === ∫ ∞ ∞− ∫∫ ∞ ∞− ∞ ∞− −− −== ξξξσ dxgfkdkFtxT xktk )()( π2 1 e]e)([ π2 1 ),( 2 The object is to develop an expression for T(x,t). Here we need the inverse transform of F(k)exp(−σ k2 t) which can be obtained by use of the convolution theorem, that is: We need the original form (in x-space) of the second function, g(x–ξ). It is just the Fourier transform of exp(−σ k2 t) which becomes: tk kFtkT 2 e)(),( σ− = ∫ ∞ ∞− −− = kdxg xkitk e)(e π2 1 )( 2 σ
• If u2 =2σ k2 t where dk=du/√(2σ t), then: 2014 MRT The exression for T(x,t) in terms of the convolution integral now becomes: The specific form of T(x,0)= f(x)→ f (ξ), the initial temperature distribution, must of course be given before the above integral can be evaluated. ∫ ∫ ∞ ∞− −− ∞ ∞− −− =         = ξξ σ ξξ σ σξ σξ df t df t txT tx tx 4)( 4)( 2 2 e)( π2 1 e)( 2 1 π2 1 ),( txxtuiu t ud t xg σσ σσ 4)2(2 22 e 2 1 e)(e π2 1 )( − ∞ ∞− −− == ∫
• Let us program such an initial distribution. A square wave! Such as you could imagine it my pool heater would turn on for 12 hours a day only. Technically Propane P (say to warm up a pool to an ideal Fahrenheit temperature of 84°F) is consumed in the heater to apply temperature T(P,ξ) to the pool’s water (e.g., in the heater there is a controlled fire), this is called putting on the heater, at a set maximum temperature, to obtain 84 degrees once per day with a day equal to one Earth’s revolution – 24 hours or 2π. Let us expand a Fourier series according to the time of day ξ so to that initial temperature distribution: The graphical representation of f (ξ) in [−π, π] and its periodic extension outside of [−π, π] are shown in the Figure.    ≤≤ <≤− = (Noon)(Midnight)forON (Midnight)(Noon)forOFF π0 0π0 )( ξ ξ ξ P f The expansion coefficients are: A square wave representing a pool heater 24 hour cycle devided along interval [−π,π]: day- time when the heater is OFF and nightime when the heater is ON (i.e., no monitoring ξ f (ξ ) Midnight P(TPool=85) P P dPdfda = ==+⋅= ∫∫∫− π 0 π 0 π 0 0 π o ππ 1 )( π 1 0 π 1 ξξξξ and 2014 MRT P P Noon 3π2π P(TPool=75)       ==       −==== +⋅=+⋅= ∫∫ ∫∫∫∫ − > − > for venfor and odd e n n P n n nP dnP P n P dnP P dnfdnbdnfdna nn π 2 0 0 cos π sin π ][sin π cos π cos)( π 1 sin0 π 1 cos)( π 1 cos0 π 1 π 0 π 0 π 0 π 0 π 0 0 π 0 π 0 0 π 0 ξ ξξξξξ ξξξξξξξξξξ −2π P T(P,ξ ) Fourier Theorem Region [−π,π] π−π NoonNoon
• The required Fourier series for the initial temperature distribution is: 2014 MRT ∑ ∑ ∞ = ∞ = += ++= odd odd 1 1 o sin π 2 2 ]sincos[ 2 )( n n nn n xnPP xnbxna a f ξ . e3sin π3π esin πππ4 e 3 3sin sin πππ4 e sin πππ4 e sin π 2 2π2 1 ),( 4)(4)( 4)( 4)( 1 4)( 1 22 2 22    = +++=       +++= +=           += ∫∫ ∫ ∫ ∑∫ ∑ ∞ ∞− −− ∞ ∞− −− ∞ ∞− −− ∞ ∞− −− ∞ = ∞ ∞− −− ∞ = ξ σ ξ σσ ξ σσ ξ σσ ξ σ ξ σξσξ σξ σξσξ dx t P dx t P t P d x x t P t P d n xn t P t P d n xnPP t PT txtx tx tx n tx n oddodd The temperature T is then: (Exercise - Hint: When in doupt, use Mathematica® ). The next step is to calculate σ and numerically find the temperature profile that will represent the pool’s warming over a 24 hour period.
• As another example, let us use the three-dimensional Fourier transform method to solve Poisson’s (1781-1840) equation for the electrostatic (scalar) potential function: 2014 MRT with d r≡d3 r=dxdydz (triple integral in Cartesian coordinates). The Fourier transform of Φ(r) is given by: )( ε 1 )(2 rr ρφ −=∇ with dk≡d3 k=dkx dky dkz (triple integral in Cartesian k-space coordinates). (when it is given by ∇2 φ (r)=0 it is called Laplace’s equation) where ρ is the charge density and ε is the permittivity (CGS units used – the kind the American Colleges use.*) In three-dimensions, we write: ∫ ∫∫∫ ∞ ∞− • ∞ ∞− ∞ ∞− ∞ ∞− = ⋅⋅=Φ rr k rk d zdzydyxdx i zkiykixki zyx e)( π)2( 1 e)( π2 1 e)( π2 1 e)( π2 1 )( 3/2 φ φφφ ∫ ∞ ∞− •− Φ= kkr rk di e)( π)2( 1 )( 3/2 φ * Formulas in SI units can be converted to their gaussian equivalents by the substitutions: εo→¼π , µo→4π/c2 and B→B/c .
• On taking the Fourier transform of both sides of Poisson’s equation, we obtain: 2014 MRT The required solution of the original differential equation,ϕ (r), is obtained by taking the inverse transform of Ρ(r)/k2 ε; we have: )( ε 1 )(2 kk Ρ−=Φ− k where or where φ(r)→0 and ∂φ/∂r→0 as r→±∞. The expression for Ρ(r) is: ∫ ∞ ∞− • =Ρ rrk rk di e)( π)2( 1 )( 3/2 ρ ∫∫ ∫∫ ∞ ∞− ∞ ∞− ∞ ∞− −•−∞ ∞− •− −== Ρ = rrrrrk r k k r rrk rk dGdd k d k i i )()( επ)2( 1 ε e)( π)2( 1 e ε )( π)2( 1 )( 32 )( 323/2 ρ ρ φ )( ε 1 )( 2 kk Ρ=Φ k ∫ ∞ ∞− −•− =− krr rrk d k G i 2 )( e )(
• Now we delve into the details of performing an important integral (funny but it comes back again and again even in quantum field theory when considering interactions of the formofaCoulombpotential:A0(x)=(e/)δ3 (r)=−(e/)δ3 (r)⇒A0=[1/(2π)3 ]∫(e/k2 )exp(ik•r)dk). If spherical polar coordinates are chosen for the variable in k-space, we have: 2014 MRT since sinθ =[exp(iθ)–exp(–iθ)]/2i (note also for future calculations that cosθ =[exp(iθ)+ exp(–iθ)]/2.) The final form for φ(r) is: The polar axis is the direction of r–r. The expression for G(r–r) becomes: kdddkddkdkkdd ϕθϕθθ )(cossin 223 −===k rrrr rr rr rr rrrr rrrr −− − − −− 2 00 1cos 1cos 0 2 2 cos1cos 1cos π2 0 0 2 2 cos π2sin π4 ee π2 )(cos e π2)(cos e )( == − − = −=−=− ∫∫ ∫∫ ∫ ∫ ∞∞ −−− −= += ∞= = −−= += = = ∞= = − kd k k kd ki kddk k kdddk k G kiki k k kik k ki θ θ θθ θ ϕ ϕ θ θϕθ ∫ ∫∫ ∞ ∞− ∞ ∞− ∞ ∞− = =         = r rr r r rr r r rr rr d dd − −− )( επ4 1 )( επ)2(2 π)2(π2 )( επ)2( 1 )( 3 22 3 ρ ρ ρφ The full meaning of the |r–r| vector will come about in the Electromagnetism section.
• O r The relation between the labels of the points of three-dimensional space for two observers whose coordinate systems are rotated with respect to one another about a common origin is given by: A rotation of the x-y plane about the z-axis by an angle ϕ. In the bottom part of this figure, β = ϕ since ϕ +γ = π/2 and γ +β = π/2 (180°/2 = 90°) because a and b are similar triangles. The relations between the ‘bared’ and ‘un-bared’ axes are: 2014 MRT ϕϕ ϕϕ ϕϕϕ sincos sin)(cos sinsincos 21 12 yxx x xx += ++= ++=   ϕϕ ϕϕ ϕ ϕ sincos coscos cos)( cos 2 2 1 xyy y y y −= −= −= =    since y= 1 +2 and since 1 = y−2 and 2cosϕ = xsinϕ and since we rotate the x-y plane around the z-axis, it remains unchanged: zz = In matrix form, these equations become:           =                     −=           ⋅+⋅+⋅ ⋅++− ⋅++ =           z y x R z y x zyx zyx zyx z y x )( 100 0cossin 0sincos 100 0cossin 0sincos 3 ϕϕϕ ϕϕ ϕϕ ϕϕ where R3(ϕ) denotes the 3×3 rotation matrix as a function of the angle ϕ . It is to be noted that infinitesimal rotations commute with one another whereas finite rotations do not. rrr r )(ϕR=→ whereRr(ϕ) isarotationbyanangleϕ andr=xi+yj+zk if Cartesian coordinates are used.ˆ ˆ ˆ y x z x y ϕ P(x,y) P(x,y) Typically, the positive +ϕ direction is that of the right- hand screw but to facilitate the math a bit we show here ϕ ≡−ϕ . For the true convention, replace ϕ by −ϕ . Which gives cos(−ϕ) = cosϕ and sin (−ϕ) = −sinϕ . Rotation Transformations             1            2 2sinϕ 1sinϕ β xcosϕ 1cosϕ γ y x x ϕ P(x,y) P(x,y) • b a M M α ϕ ϕϕ αϕαϕαϕ ϕϕ αϕαϕαϕ sinsin )sincoscos(sin)sin( sincos )sinsincos(cos)cos( yx OPOPy yx OPOPx += +=+= −= −=+= y
• So, the rotation matrix about the z-axis by and finite angle ϕ is: Similarly, the rotation matrix about the x-axis and y-axis by and finite angle ϕ are: In general, a rotation is defined as a linear transformation which leads the scalar product of two vectors invariant and whose determinant is +1. 2014 MRT           −= =+−=+= 100 0cossin 0sincos )( cossincoscos 3 ϕϕ ϕϕ ϕ ϕϕϕϕ R zzyxyyxx and,           − =           − = ϕϕ ϕϕ ϕ ϕϕ ϕϕϕ cos0sin 010 sin0cos )( cossin0 sincos0 001 )( 21 RR and Suppose an observer O is located in a coordinate system(x,y,z). Owoulddescribe, say, the temperature T which varies from point to point according to some law described by T = f(x,y,z)≡ f(r). There is nothing special about the coordinate system (x,y,z); the law expressing the variation of temperature from point to point could equally well be refered to by an observer O in the coordinate system O(x,y,z). However, the form of thelaw, i.e., the dependence on the coordinates, will generally be different in the two systems. If we let T=g(x,y,z)≡ g(r) in the (x,y,z) system, then the temperature at any point – say described by T= f(x,y,z) and T=g(x,y,z) – must be the same. The physics describing the temperature field will not change because of a rotation (assuming it is not to fast).
• We therefore require that (i.e., to preserve coordinate invariance): Thus D(R) acting on a function f(r) produces a new function g(r) such that g(r) is the transformed function under the coordinate transformation r=Rr r. The main point to keep in kind is that Rr transforms coordinates while D(R) transforms functions of coordinates. Now, a function f (r) is said to be an invariant function under a coordinate transforma- tion r=Rrr if: 2014 MRT where D(R) is a rotation operator – an engine that transforms functions of coordinates. )()()()()( 1 rrrr fRffRg === − D )()( rr gf = We now ask the following question: Given a scalar function f(r)= f(x,y,z) and a coordinate transformation r =Rr r, what is the formal method for finding g(r) such that f(r) =g(r)? For this purpose we define an operation D(R) by the relation: )()()()( 1 rrr fRffR == − D           == === − 100 010 001 1 EE zzyyxx and, Now, we state a few special rotation transformations (and do notice the symmetries): E – The identity transformation:
• 2014 MRT           − −== −=−== − 100 010 001 1 AA zzyyxx and, B – A rotation of coordinates about the O2-axis through 180°:                   − − == −=−=+−= − 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 BB zzyxyyxx and, A – A rotation of coordinates about the x-axis through 180°: C – A rotation of coordinates about the O3-axis through 180°:                   − − −− == −=+−=−−= − 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 CC zzyxyyxx and, O y x z 2 1 3 z y O y x z 2 1 3 z O y x z 2 1 3 z y x y 30° x 60° x
• D – A positive (counterclockwise) rotation of coordinates about the z-axis through 120°: 2014 MRT FDD zzyxyyxx =                   − − −− =                   − −− − = =+−=+−= − 100 0 2 1 2 3 0 2 3 2 1 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 and and, F – A negative (clockwise) rotation of coordinates about the z-axis through 120°: E A B C D F E E A B C D F A A E D F B C B B F E D C A C C D F E A B D D C A B F E F F B C A E D DFF zzyxyyxx =                   −− − =                   − −− = =−=−−= − 100 0 2 1 2 3 0 2 3 2 1 100 0 2 1 2 3 0 2 3 2 1 2 1 2 3 2 3 2 1 1 and and, Below is a Multiplication Table fortheelementsof D3 (e.g.,AC=F but CA=D and F−1 F=E). O y x z 2 1 3 z y x 120° O y x z 2 1 3 x y z 120° Applied Second AppliedFirst This Table establishes that the rotations E, A, B, C, D and F are the elements of a group – the D3 group.
• As an example,suppose we have a hydrogen atom surrounded by three unit charges q located at positions 1, 2, and 3 which form an equilateral triangle of side a; the proton is at the origin O and the electron at r (see Figure). The Hamiltonian H of the electron is: in which |r1|, |r2|, and |r3| are the distances O1, O2, and O3, respectively, and         ++−         +∇−=++=+= 2 3 2 2 2 1oo 2 2 2 o 111 επ4επ42 )( rrrrrr −−− qe r Ze m WVTWHH e  )( 232 )( 232 )( 3 3 2 22 2 32 2 22 2 21 22 2 2 1 rrrrrrrrr fz a y a xfz a y a xfzy a x ≡+      ++      −=≡+      −+      −=≡++      += −−− and, Hydrogen atom surrounded by three charges located at the corners of an equilateral triangle.         −−−=⇒= − zyxyxfzyxfDDffD , 2 1 2 3 , 2 3 2 1 ),,()()()()( 1 DD rr                   −− − = 100 0 2 1 2 3 0 2 3 2 1 D zzyxyyxx =−−=+−= &, 2 1 2 3 2 3 2 1 )()()( )()()( )( 22 1 2 3 322 3 2 1 )()( 21 13 3 2 22 2 rr rr rr ffD ffD fz a yx a yxfD = = =+         −−+         −−−= D D D and Ho is invariant under all three-dimensional rotations. so that the function f (r) is indeed invariant since: )()()()]()()()[( 321321 rrrrrr ffffffD ++=++D                   − −− =− 100 0 2 1 2 3 0 2 3 2 1 1 D 2014 MRT O y x z a r2 1 3 e Ze q q q O r 1 r1 |r – r1| O r 3 r3 |r – r3| O r 2 r2 | r – r2 | a a The interaction potential W remains to be investigated. Let us take, as a practical example, the coordinate transformation D which, as we just saw before, is a positive (counter clockwise) rotation of coordinates about the z-axis through 120°. We have: Therefore [If not installed, the English font will show up as D ]:
• We find, experimentally, that the ‘force’ that acts on an arbitrary ‘charge’ – however many there are or how many that are moving – depends only on the position of this charge, its velocity and the quantity of charge. We can write the (resultant) force F on this charge q moving with velocity v by the Lorentz force law: F=q(E+v× B)=qE⊕qv× B (Electric force ⊕ Magnetic force) Within electromagnetism, a ‘field’ is a physical quantity that takes different values at different points of space. We consider a static field then as a mathematical function of space (and time if the field is dynamic) and to visualize this field we draw many points in a space and at each should be given the intensity and direction of the vector field at that location. As we step back and look over the space, the field is then defined and mapped physically – it has become a manifold – and it will help provide a representation that will be done using some advanced mathematical concepts which we will turn to next. where we call E the electric field and B the magnetic field at the position (e.g., in simple Cartesian coordinates x, y & z) of the charge q(x,y,z). We can think of E(x,y,z) and B(x,y,z) as producing the forces that can be seen at a time t by charge q situated at [x,y,z] with the condition that: having placed the charge in its place it is not perturbed any further. We also associate with any coordinate [x,y,z] of space two physical orientated quan- tities (i.e., vectors), E and B, that can vary with time t. The electric and magnetic fields are then visualized as vector functions of x, y, z and t. Since a vector is specified by its components, each of the fields E(x,y,z,t) and B(x,y,z,t) represents three mathematical functions of x, y, z and t. We will now generally refer to the position vector r (e.g.,[x,y,z].) 2014 MRT Electromagnetism and Relativity
• • y x z r r rrR=+ − ( )       −+−+−==∝= ==∝ +−+−+− 222 22 32 )()()()( ˆ)(ˆ)( zzyyxx R Qq k Qq F Qq k Qq R R rF RRRrr rr R R rF E E − − Three-dimensional spatial representation of the interaction of two point charges q and Q, R away. 2014 MRT E Also, R= r–r is the separation distance and kE is a proportionality constant – a positive force implies it is repulsive, while a negative force implies it is attractive. The proportionality constant (i.e., the coupling intensity between charges) kE, called the Coulomb constant, is related to defined properties of space and can be calculated based on knowledge of empirical measurements of the speed of light: 22 /CoulomberNewton.met rHenry/mete 9 72o 2 o 10987.8 10 π4π4 1 ×= ×=== − c c ε k µ E q Q Rˆ R r r r rR= rrR =⇒ R = R Rˆ=R Rˆ Direction: Magnitude: By experience, we know that a force (that either attracts or repels) will be exerted on a point charge q when it is placed in the vicinity of another point charge Q. Mathematically: This force acts at a distance R away (in the direction of a unit vector parallel to R).Rˆ To aid in visualizing the E field, we introduce the ‘concept’ of field lines which gives the direction of the electrical force generated by a positive test charge. If another test charge was released, it would move in the direction of the field Electric Vector (i.e. force field per unit charge) Field Lines Manifold of coupling kE O R rr − =Rˆ
• Q q E FC +q1 −q2 v(r) = 0 Gaussian Surface Point charge Q = +q1 attracts q = −q2. )(3 1 2 C rr rr F E E − − q k q =−= Coulomb’s Law (Charles Coulomb – 1783): The magnitude of the ‘electrostatic’ force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them. (wikipedia.org) The scalar form of Coulomb's law is an expression for the magnitude and sign of the electrostatic force between two idealized point charges, small in size compared to their separation. This force (FC) acting simultaneously on point charges (Q=+q1) and (q=−q2), is given by: 2 21 2 12 CC 3C )( )()( rrrr rF rr rr rF EE E −− − − qq k qq kF Qq k −= − == = +• Illustration of the electric field surrounding a positive (red) and a negative (blue) charge. Direction: Magnitude: 2014 MRT r − r +q1 +q2 +q1 −q2 Q = +q1 attracts q = −q2 Q = +q1 repels q = +q2 2 2 2 2 1 C 1 C 3 C )( )( )( )( )( rrrr rF rE rr rr rF rE EE E −− − − q k q k q F q E Q k q −= − == + == == The direction of the electric field vector E is (always) from the positive to the negative charge whereas the magnitude of the electric field E=|E| at a point is the force per unit charge on a positive test charge at the point: kE(εo) Field
• Flux ΦE through a spherical (Gaussian) surface of radius r (ASphere = 4πr2 ) from an electric field with at it center a point positive Gauss's law also has a differential form: ∫∫∫= V dddrrQ ϕθθρ sin)( 2 r Gauss’s Law (Carl Gauss – 1834): The electric flux through any closed surface is proportional to (i.e., ‘∝’) the enclosed electric charge. (wikipedia.org) ϕθθ θ θε ρ ϕθ ∂ ∂ + ∂ ∂ + ∂ ∂ =•= E r E rr Er r r sin 1)sin( sin 1)(1 2 2 o E∇ where ∇ •E is the divergence of the electric field, and ρ is the charge density. The electric flux ΦE through an arbitrary spherical surface due to a point charge Q can be written as: oε Q d S =•=Φ ∫∫ SEE Gauss's Law states that the (net) electric flux through any closed surface is equal to the charge inside that surface divided by the dielectric constant εo. Gauss’s Law may be expressed in its integral form: Qd S ∝•∫∫ SE Here ΦE = ∫∫S E•dS is a surface integral denoting the electric flux through a closed surface S and Q denotes the total charge enclosed by volume V : (Spherical Coordinates) 2014 MRT rE ˆ π4 1 2 o r Q ε = Gaussian Surface dS dΦEQ r Constant electrical field lines r r r=ˆ rS ˆπ4 2 rd = Er r EdrrErdrErdEdQ r rr S 2 o 0 2 o 0 o 0 o 2 oo π4 2 π8π8)π8()ˆˆ()ˆπ4()ˆ( εεεεεε =⋅=⋅=•=•=•= ∫∫∫∫∫ rrrrSE
• Compasses reveal the direction of the local magnetic field. As seen here, the magnetic field points towards a magnet’s SS (south) pole and away from its NN (north) pole. B B B field lines never end! Magnetic field lines exit a magnet near its north pole and enter near its south pole, but inside the magnet B field lines continue through the mag- net from the south pole back to the north. If a B field line enters a magnet somewhere it has to leave somewhere else; it is not allowed to have an end point. Magnetic poles, therefore, always come in NN (North) and SS (South) pairs. Cutting a magnet in half results in two separate magnets each with both a north and a south pole. (wikipedia.org) Since all the magnetic field lines that enter any given region must also leave that region, subtracting the ‘number’ of field lines that enter the region from the number that exit gives identically zero. Conceptually thinking, this is equivalent to saying that the flux of B through any closed surface S is identically zero, and mathematically it is equivalent to the integral form: 0=•∫∫S dSB where the integral is a surface integral over the closed surface S (a closed surface is one that completely surrounds a region with no holes to let any field lines escape). Since dS points outward, the dot product in the integral is positive for B field pointing out and negative for B field pointing in. It also has a differential form: 2014 MRT 0 sin 1)sin( sin 1)(1 2 2 = ∂ ∂ + ∂ ∂ + ∂ ∂ =• ϕθθ θ θ ϕθ B r B rr Br r r B∇
• The relation between the current density J (current per unit area) and the current I in a cylindrical conductor (with conductivity σ ). A familiar example of a current density is that in a cylindrical conductor (say a wire of conductivity σ ) having radius r and carrying a constant current I , i.e. is the amount of charge transferred across the conductor's cross sectional surfaces per unit time: One could express the density current J(r) and a velocity distribution v(r) if the velocity of each element of charge at each point in space (not time as in space-time which would greatly complicate things!) were known; thus one would have: vS I vSt Q tSvtvSVQ vv vJ ∆ = ∆       ∆ ∆ ==∴ ∆∆=∆∆=∆=∆ 1 ][ ρ ρρρ )(ˆlim)()()( 0 0 rJrvrrJ tS Q t S ∆∆ ∆ == →∆ →∆ ρ )()()( rvrrJ ρ= Knowing J(r) is equivalent to knowing v(r) if ρ (r) is given and it is more convenient to work with the current density than with the velocity distribution. J=ρ v gives us a notion of the meaning of J(r): it is the amount of charge ∆Q passing across a surface ∆S perpendicular to J (and therefore also to v) at the point r during time interval ∆t – as ∆S and ∆Q tend to vanish (that ‘ → 0’ symbol): 2 2 π π)( r I J rJd td dQ I S = =•== ∫∫ SrJ 2014 MRT J Sr σ y x z v ∆S O Wire v∆t r tQ rJSJI rS ∆∆= == = 2 2 π π In a nutshell, the ‘electric’ current is the rate of flow of electric charge.
• An additional experimental fact, known as Ohm’s Law, states that the current density J(r) is proportional to the electrical field E(r). This rule is found to be an accurate description of the behavior of a vast number of conducting materials. Mathematically, it is written: Ohm’s Law (Georg Ohm – 1827): The current through a conductor between two points is directly proportional to the potential difference or voltage across the two points, and inversely proportional to the resistance between them. (wikipedia.org) )(σ)( rErJ = where σ is the conductivity of the conducting material (e.g., 18-gauge speaker wire.) Current flowing through a uniform cylindrical conductor (such as a round wire) with a uniform field applied. S is the cross-sectional area (for a round wire we have S = π r2 .) The potential difference V12 between two points P1 and P2 is defined as ∫ •−= 2 1 12 P P dV E d is the element of path along the integration of electric field vector E. If the applied E field is uniform and oriented along the length of the conductor as shown in the Figure, then defining the voltage V in the usual convention of being opposite in direction to the field (see Figure), and the above vector equation reduces to the scalar equation V = E . The electrical resistance of a uniform conductor is given in terms of resistivity ρ (ρ = 1/σ ) by: 2 π 1 )( rSS EV SE V SJ V I V R  ρρ σσ ====== 2014 MRT SI V R EIRV SE R V I   σ σ 1 == == == I I V12 J E  Wire P1 P2 22 2 πσπ π rErJSJI rS === = )( π4 1 3 1 o rr rr E − − = q ε σ
• Faraday noticed first that a closed loop of conducting material placed in static external electric or magnetic fields did not carry any current – static fields do not generate a net flow of current within closed conductors. This can be expressed by the equation: Law of Electromagnetic Induction (Michael Faraday – 1831): The induced electromotive force (EMF) in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit. (wikipedia.org) 0)( =•∫ d C rJ for any closed loop C. d is an element of length along the closed loop. Now if a changing magnetic field is introduced, this equation no longer is correct and Faraday found an experimental law that can be expressed mathematically as: t d t t d SC ∂ Φ∂ −=• ∂ ∂ −=• ∫∫∫ B S rB rJ σ ),( σ)(  where S is a surface bounded by the loop C formed by the conductor and σ is a constant that is characteristic of the conductor. When the magnetic flux ΦB changes, Faraday's Law of Induction says that the work E done (per unit charge) moving a test charge around the closed curve C – called the electromotive force (EMF) – is given by t∂ Φ∂ = B E and the flux is just: Flux = [average normal component] × [surface area]. 2014 MRT
• This Figure shows the velocity (dv) induced at a point P by an element of vortex filament (d) of strength Γ. Biot–Savart’s Law (J.-B. Biot & F. Savart – 1820): The magnetic vector field B depends on the magnitude, direction, length, and ‘proximity’ of the electric current I, and also on a fundamental constant called the magnetic constant µo. (wikipedia.org) With the definition of the current, I=J • S, the law is: where the integral sums over the wire length where vector d is the direction of the current, µo is the magnetic constant, r− r is the distance between the location of the infinitesimal length d and the location at which the magnetic field is being calculated. ∫= rr rr B − −× )( π4 μo   dI The Biot–Savart Law is used to compute the magnetic field generated by a steady current (i.e., a continual flow of charge – e.g., through 1 m long 18-gauge copper speaker wire − which is constant in time and in which charge is neither building up nor depleting at any point.) Enclosedoμ Id C =•∫   B where the line integral is over any arbitrary loop and IEnclosed is the current enclosed by that loop. d P dv r − r dFBS 2014 MRT Vortex filament of strength Γ A slightly more general way of relating the current I to the magnitic field B is through Ampère's Law:         = 3 o BS )( )( π4 )(μ )( rr rr rv r rF − −× ×   dI d ρ
• Force due to charge in motion (v ≠ 0): Force due to charge at rest (static): Consider an application where we are situated at an origin O of a reference frame S (i.e., aCartesianCoordinateReferenceFrame) that has volume V. Consider also an other reference frame S (i.e. a spherically symmetric Gaussian Surface or even equipotential flow or vortex lines) also with it’s own origin O which can be quite a distance r away from O. Coordinate system used to describe the density of force at point P. Both the densities of charge ρ(r) and current J(r) generate from O the E(r) and B(r) fields at point P from a distance r − r away – and measured a distance r from O. (Biot-Savard’s Law)         − − = ∫∫∫ r rr rrrJ rJrF 3 3 o BS )()(ˆ )( π4 )( d V × × µ (Coulomb’s Law) ∫∫∫ − − = V d r rr rrr rrF 3 3 o C ))(( )( π4 1 )( ρ ρ ε FC is thus the force density arising from a static distribution of charges whereas FBS is the force density attributable to moving charges or currents and the permittivity of free space (the dielectric constant) εo and permeability of free space (or magnetic constant) µo are respectively given by: εo = 8.854×10−12 Faraday/meter µ o = 4π ×10−7 Newton/Ampere 2014 MRT ˆ As a reference, the current I at P′ is given by: I = J • S We then do – from O – an observation of the force densities a distance r away at point P (produced within O by a density of charge ρ and current J) situated at a distance r − r away from O. A mathematical generalization of the experimental facts is given by the two following expressions for the force per unit volume V felt by an electrical charge density ρ(r) and an electrical current density J(r) (in the direction of the unit vector J/ |J|): • y x z )(rJ )(rJ)(rρ O P′ )(rv O r – r )(rρ r rr+ − Gaussian Surface Flow Line P ˆ
• )()()(C rErrF ρ= )()()(BS rBrJrF ×= or, evidently, after wrapping everything with an integral over the whole volume V in S : ∫         − − = V d r rr rr rrE 3 3 o ).( π4 1 )( ρ ε ∫         − − = V d r rr rr rJrB 3 3 o )(ˆ π4 )( × µ It is convenient to express the force densities as functions of the two vector fields: the electrical field E(r) and the magnetic field B(r). The relation between the densities of force and the vector fields is given by (notice how the density – i.e., the amount per unit volume – of charge ρ and of current J factor into the definition of FC(r) and FBS(r)):         − −= − − rrrr rr 1 3 ∇ 2014 MRT 53 3 1 rr rr rr − − −= − ∇& with the unit vector = J/|J| that is parallel and normal to a unit element of the surface (e.g. sphere) through which the vector representing the current density would pass and these relations will provide benefit as variation representations: The undisturbed charge q will be subject to E(r) and B(r) in proportion to both the densities of charge ρ (r) and current J(r) at point P: r away from O and r − r away from O. The velocity v(r) factors in for F(v) = qE ⊕ qv × B and J(v) = ρ v. Since ρ (r) is the charge per unit volume, the total charge in volume V (e.g., in spherical coordinates): where the (e.g. spherical) gradient ∇ is given by: ∫∫∫= V drddrrq Spherical 2 sin),,( ϕθθϕθρ ϕ∇ ˆ sin 1ˆ1 ˆ ϕθθ ∂ ∂ + ∂ ∂ + ∂ ∂ = rrr θry z )(rJ)rρ )(rv q q Charge ( − )(rJ )(rρ )(ˆ rrJ −× q )(rJ )(rρ ˆ • x )(rJ )(rρ r Q Jˆ J ) ) J + rr ˆ r P′ )(rρ )(rBˆ(r × (E r )(rFC )(rFBS P
• dV V S dSd nS ˆ= J(r) y x z r O Gauss’ Divergence Theorem: The closed manifold dV is the boundary of V oriented by outward-pointing normals, and n is the outward pointing unit normal field of the boundary dV. Another important piece of experimental information in which a vast amount of accumulated experimental data supports the hypothesis that electrical charges cannot be created or destroyed spontaneously within a closed physical system – this embodies the Law of conservation of electrical charge and can be expressed mathematically by the continuity equation: (The Continuity Equation) The outward flux of a vector field through a closed surface is equal to the volume integral of the divergence of the region inside the surface. Intuitively, it states that the sum of all sources minus the sum of all sinks gives the net flow out of a region. Law of conservation of electrical charge (Benjamin Franklin – 1747): Change in the amount of electric charge in any volume of space is exactly equal to the amount of charge flowing into the volume minus the amount of charge flowing out of the volume. (wikipedia.org) J•−= ∂ ∂ ∇ t ρ The continuity equation means that the current density emerging from a point in space (∇ •J) is equal to the rate of decrease of the charge density at that point (−∂ρ/∂t). Continuity equations offer more examples of laws with both differential and integral forms, related to each other by the divergence theorem: ∫∫∫∫∫∫∫ •≡•=• VSS dVdSd JnJSJ ∇ˆ 2014 MRT nˆ t∂ ∂ −=• )( )( r rJ ρ ∇
• (with here since ‘cartesian’ throughout.) In free space (i.e., for a system without any electrical charge) we would have ρ=0, and without current density either J=0. But with ρ≠0 and J≠0, we have: o div ε ρ = ∂ ∂ + ∂ ∂ + ∂ ∂ =•= z E y E x E zyx EE ∇ ty E x E z E x E z E y E EEE zyx xyxzyz zyx ∂ ∂ −=        ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ −        ∂ ∂ − ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ =≡ B kji kji EE ˆˆˆ ˆˆˆ rot ×∇ oε Q d S =•=Φ ∫∫ SEE oε ρ constantdielectricThe volumeclosedthewithinchargeNet surfaceclosedanythroughoffluxThe V S =E         • ∂ ∂ −=• ∫∫∫ SC d t d SBE  Maxwell’s Second Set of Equations (Differential and Integral forms, respectively): Maxwell’s First Set of Equations (Differential and Integral forms, respectively): ∫∫∫= V dzdydxQ )(rρ [ ]SC surfacethethroughofFluxof timetorespectwith ationdifferentiPartial ofNegativecurvearoundofncirculatioThe BE = 2014 MRT
• (with here since ‘cylindrical’ throughout.) o 1)(1 div ε ρ ζϕρρ ρ ρ ζϕρ = ∂ ∂ + ∂ ∂ + ∂ ∂ =•= EEE EE ∇ t EEEEEE EEE ∂ ∂ −=      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ == B ζρ ζρ EE ˆ)(1 ˆˆ 1 ˆˆˆ rot ϕρ ρ ρρζζϕρ ρ ζϕρ ρ ρϕζρϕζ ζϕρ ϕ ϕ ×∇ oε Q d S =•=Φ ∫∫ SEE oε ρ constantdielectricThe volumeclosedthewithinchargeNet surfaceclosedanythroughoffluxThe V S =E         • ∂ ∂ −=• ∫∫∫ SC d t d SBE  Maxwell’s Second Set of Equations (Differential and Integral forms, respectively): Maxwell’s First Set of Equations (Differential and Integral forms, respectively): ∫∫∫= V dddQ ζϕρρρ )(r [ ]SC surfacethethroughofFluxof timetorespectwith ationdifferentiPartial ofNegativecurvearoundofncirculatioThe BE = 2014 MRT In free space (i.e., for a system without any electrical charge) we would have ρ=0, and without current density either J=0. But with ρ≠0 and J≠0, we have:
• (with here since ‘spherical’ throughout.) In spherical coordinates, things get a bit dicier but the math greatly takes advantage of the inhenrent symmetries (notably spherical) throughout the study of electromagnetism. o 2 2 sin 1)sin( sin 1)(1 div ε ρ ϕθθ θ θ ϕθ = ∂ ∂ + ∂ ∂ + ∂ ∂ =•= E r E rr Er r r EE ∇ t E r rE rr rEE r EE r rr ∂ ∂ −=      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ == B θrEE ϕ×∇ ˆ )(1ˆ)( sin 11 ˆ )sin( sin 1 rot θϕθϕθ θ θ θϕθϕ oε Q d S =•=Φ ∫∫ SEE oε ρ constantdielectricThe volumeclosedthewithinchargeNet surfaceclosedanythroughoffluxThe V S =E         • ∂ ∂ −=• ∫∫∫ SC d t d SBE  [ ]SC surfacethethroughofFluxof timetorespectwith ationdifferentiPartial ofNegativecurvearoundofncirculatioThe BE = Maxwell’s Second Set of Equations (Differential and Integral forms, respectively): Maxwell’s First Set of Equations (Differential and Integral forms, respectively): ∫ ∫ ∫ = = = = = = = max 0 π 0 π2 0 2 sin)( rr r dddrrQ θ θ ϕ ϕ ϕθθρ r 2014 MRT
• 0div = ∂ ∂ + ∂ ∂ + ∂ ∂ =•= z B y B x B zyx BB ∇ ty B x B z B x B z B y B BBB zyx xyxzyz zyx ∂ ∂ =      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ −      ∂ ∂ − ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ == E kji kji BB oo ˆˆˆ ˆˆˆ rot εµ×∇ Maxwell’s Forth Set of Equations (Differential and Integral forms, respectively): Maxwell’s Third Set of Equations (Differential and Integral forms, respectively): 0=•=Φ ∫∫S dSBB ZerosurfaceclosedanythroughoffluxThe =SB         •+         • ∂ ∂ =         • ∫∫∫∫∫ SSC dd t dc SJSEB o 2 1 ε  [ ] [ ] o 2 ] ))( εconstantdielectricThe throughcurrentelectricalof[Flux plus throughofFluxof timetorespectwith ationdifferentiPartial aroundofncirculatioThetimeslightofSpeed S SC EB = 2014 MRT
• 0 1)(1 div = ∂ ∂ + ∂ ∂ + ∂ ∂ =•= ζϕρρ ρ ρ ζϕρ BBB BB ∇ t BBBBBB BBB ∂ ∂ =      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ == E ζρ ζρ BB oo ˆ)(1 ˆˆ 1 ˆˆˆ rot εµ ϕρ ρ ρρζζϕρ ρ ζϕρ ρ ρϕζρϕζ ζϕρ ϕ ϕ ×∇ Maxwell’s Forth Set of Equations (Differential and Integral forms, respectively): Maxwell’s Third Set of Equations (Differential and Integral forms, respectively): 0=•=Φ ∫∫S dSBB ZerosurfaceclosedanythroughoffluxThe =SB         •+         • ∂ ∂ =         • ∫∫∫∫∫ SSC dd t dc SJSEB o 2 1 ε  [ ] [ ] o 2 ] ))( εconstantdielectricThe throughcurrentelectricalof[Flux plus throughofFluxof timetorespectwith ationdifferentiPartial aroundofncirculatioThetimeslightofSpeed S SC EB = 2014 MRT
• 0 sin 1)sin( sin 1)(1 div 2 2 = ∂ ∂ + ∂ ∂ + ∂ ∂ =•= ϕθθ θ θ ϕθ B r B rr Br r r BB ∇ t B r rB rr rBB r BB r rr ∂ ∂ =      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ +      ∂ ∂ − ∂ ∂ == E θrBB ooˆ )(1ˆ)( sin 11 ˆ )sin( sin 1 rot εµ θϕθϕθ θ θ θϕθϕ ϕ×∇ Maxwell’s Forth Set of Equations (Differential and Integral forms, respectively): Maxwell’s Third Set of Equations (Differential and Integral forms, respectively): 0=•=Φ ∫∫S dSBB ZerosurfaceclosedanythroughoffluxThe =SB         •+         • ∂ ∂ =         • ∫∫∫∫∫ SSC dd t dc SJSEB o 2 1 ε  [ ] [ ] o 2 ] ))( εconstantdielectricThe throughcurrentelectricalof[Flux plus throughofFluxof timetorespectwith ationdifferentiPartial aroundofncirculatioThetimeslightofSpeed S SC EB = 2014 MRT
• t∂ ∂ −= B E×∇ t∂ ∂ = E B ooεµ×∇ We take the ‘curl’ (i.e., ∇ × …) of the first equation: We illustrate the great usefulness of the differential form of the Maxwell (1831-1879) equations by showing how they give rise (in free space) to a traveling wave. We write: A well-known vector identity states that: This is the differential equation of a plane wave. The velocity of propagation of the electromagnetic wave is thus the speed of light (in free space the permittivity of free space is εo =8.854×10−12 F/m and the permeability of free space is µo =12.566 ×10−7 N/A): and 2 2 oooo ) ) ttttt ∂ ∂ −=      ∂ ∂ ∂ ∂ −= ∂ (∂ −=      ∂ ∂ −=( EEBB E εµεµ ×∇ ×∇×∇×∇ EEE 2 )) ∇−•(=( ∇∇×∇×∇ But in the absence of a charge (i.e., when ρ = 0) – ∇ •E = 0 – the divergence of the electric field is zero. We thus arrive at: 0 1 2 2 22 2 oo = ∂ ∂ −∇⇒ ∂ ∂ =∇ 22 tct E E E E εµ 101.079103458792299 1 98 oo km/hm/sm/s ×=×≅== εµ c or if you like the physical image: light travels 1 ‘billion’ kilometers every single hour! 2014 MRT
• Then we show that surrounding the loop B(t), loops of E(t) are generated at each point along the loop B(t)… E ⇒=• 0E∇ ⇒=• 0B∇ B E B(t) E(t) • • E(t) • E(t) • E(t)⇒−= ∂ ∂ ⇔ ∂ ∂ −= )( )()( )( t t t t t t E BB E ×∇×∇ ⇒′+= ∂ ∂ ⇔ ∂ ∂ =′ )( 1)()( )( oo oo t t t t t t B EE B ×∇×∇ εµ εµ Applying this interpretation to the first three Maxwell’s Equations, we can say that the lines of the electric and magnetic field form closed loops in space… To physically interpret the equations above, we start by specifying that when the divergence of a vector field V is ‘zero’ (i.e., when ∇ •V = 0), this means that each line of the field will form a closed loop and consequently, the lines of force will not cross. The electric field in response, generates magnetic fields according to Maxwell’s ∇ × B(t) equation, which suggests that the time variation of the magnetic field will produce a rotation around it… B(t) B′(t) • E(t) • • • • B′(t) B′(t) B′(t) 2014 MRT
• An electromagnetic wave is visualized using the physical interpretation of Maxwell’s equations. The three dimensions of light (i.e., or any other electromagnetic radiation) are its intensity I∝|A2 | (where A is the amplitude of the electromagnetic wave), is tied to human perception of the apparent brightness of light, its frequency f (where λ is the wavelenght tied to this frequency by this relation: λ=c/f ), percieved by humans as the color of the light, and its polarization εµ (p), a function of the momentum of the electromagnetic wave and which humans do not percieve under normal conditions. Not to scale An Electromagnetict Wave – Travels at c2 = 1/εoµo ! … • B(t) E(t) B′(t) E′(t) E(εo) λ B(µo) c = 299,792,458 m/s • Polatization • • B •E E B • Elliptical Speed of light in vacuum: oo 1 µε λ == fc • B″(t) z x y 2π == fω 2πc/λ = ck ••• • 2014 MRT
•  DENSITY CHARGE OF DIVERGENCE ρ ε ε ⋅=• =•∫∫ o o 1  E E SE ∇ Q d S 0 0 =• =•∫∫  B B SB OF DIVERGENCE ∇ S d THE ELECTRIC FIELD E DIVERGES OUT FROM POSITIVE CHARGES AND IN TOWARD NEGATIVE CHARGES. THE TOTAL FLUX OF E (REPRESENTED BY ∫∫E•dS) THROUGH ANY CLOSED SURFACE S IS PROPORTIONAL TO THE CHARGE Q INSIDE. E CURLS AROUND CHANGING MAGNETIC B FIELDS (FARADAY’S LAW) IN A DIRECTION THAT WOULD MAKE A CURRENT I THAT WOULD PRODUCE A B FIELD TO OPPOSE (−) THE CHANGE IN B FLUX (LENZ’S LAW).  CHANGINGIS RATEOFCURL BE B E S B E td d d t d SC −= • ∂ ∂ −=• ∫∫∫   ×∇ B NEVER DIVERGES. IT JUST LOOPS AROUND ON ITSELF. B CURLS AROUND CURRENTS AND CHANGES IN E FIELDS.  CHANGINGIS RATE DENSITY CURRENTOFCURL EB E JB S E B td d d t Id SC ⋅⋅= • ∂ ∂ +=• ∫∫∫ ooo ooo εµµ εµµ +×∇   INTEGRAL: DIFFERENTIAL: N S B −+ E S B I E S N INCREASING B FLUX I B I B B INCREASING E FLUX + + + + + + + + + + + + − − − − − − − − − − − − ∫∫∫= V dVQ )(rρ 2014 MRT
• The principle of Galilean invariance then asserts that the ‘laws of nature’ are the same for two observers, i.e. that the form of the equations of motion is the same for both observers. The equations of motion must therefore be covariant with respect to the transformations r=r +vt and t =t. Unfortunately, this invariance principle applies only in situations where the velocity v is much lower than that of light. We need more… We shall now analyze the role played by invariance principles in the formulation of physical theories. Experiments have also yielded the fact that space is isotropic so that the orientation in space of an event is an irrelevant initial condition and this principle can be translated into the statement that: ‘the laws of motion are invariant under spatial rotations.’ Newton’s law of motion further indicated that the state of motion, as long as it is uniform with constant velocity, is likewise and irrelevant initial condition. This is the principle of Galilean invariance which assets that the laws of nature are independent of the velocity of the observer, and more precisely, that the laws of motion of classical mechanics are invariant with respect to Galilean transformations: τ+→+→ ttandarr tt t = −= vrr The laws of nature are independent of the position of the observer or, equivalently, that the laws of motion are covariant with respect to displacements in space and time, i.e., with respect to the transformations: 2014 MRT
• Evidently, the equation governing electromagnetic wave propagation has a different form in the ‘bared’ reference frame, with coordinate ri, from that which it has in the ‘unbared’ reference frame, with coordinate ri. Looking into the Galilean invariance as it applies to – especially Maxwell’s equations and the wave equation governing electromagnetic waves – the telegrapher’s equation: 0),( 1 2 2 2 2 =        ∂ ∂ −∇ tf tc r ),()()(2 1 ),( 1 2 2 2 2 2 2 2 2 2 tg ttc tf tc rvvr                 •+ ∂ ∂ •+ ∂ ∂ −∇=        ∂ ∂ −∇ ∇∇ where f (r,t) is a scalar function of r and t. Under Galilean transformations, ri →ri −vit =ri, and since f (r,t) is a scalar function, it must have the same numerical value in both coordinate systems. By letting f (r,t)=g(r,t)=g[r(r,t),t], the last member demonstrates that g depends of t implicitly through the dependence of r(r,t) on t. It can be showed that ∇2 f (r,t)=∇2 g(r,t) – which is good since they remain unchanged! Unfortunately however, it can also be shown that ∂f (r,t)/∂t=v•∇ g(r,t)+∂g(r,t)/∂t and: Based on this argument alone, Galilean invariance can not be considered further. which clearly highlights undesired couplings between the velocity v and the gradient ∇. Therefore, the telegrapher’s equation transforms according to the rule:    incouplingordersecondandFirst ∇ ]∇[∇[∇ • ••+∂∂•+∂∂=∂∂ v rvvrvrr ),(]),(2),(),( 2222 tgttgttgttf 2014 MRT
• It is a fundamental postulate of Physics that the laws of nature be expressed by equations that are valid for all coordinate systems (i.e., locally intertial reference frames). Alternatively, we say that the laws of nature are covariant, which means that they have the same forms in all coordinate systems. A systematic method of investi- gating the behavior of quantities that undergo a coordinate transformation is the subject matter of tensor analysis. In developing the mathematical subject of absolute differential geometry, Gauss, Riemann, and Christoffel (1829-1900) introduced the concept of a tensor. The subject of absolute tensor calculus (i.e., tensor analysis) was introduced and developed by Ricci (1853-1925) and his student Levi-Civita (1872-1941). Einstein (1879- 1955) made extensive use of tensors (i.e., technically called differential tensor calculus) in his formulation of the general theory of relativity. A tensor consists of a set of quantities called components whose properties are independent of the coordinate system used to describe them. The components of a tensor in two different coordinate systems are related by the characteristic transformation as discussed below. 2014 MRT Now a word on notation. A collection of indices (subscripts and/or superscripts) is used to make the mathematical development of tensor analysis compact. The superscripts, contravariant indices, are used to denote the contravariant components of a tensor, Tij... . The subscripts, covariant indices, are used to represent the covariant compon-ents of a tensor, Tij.... The components of a mixed tensor are specified by indicating both subscripts and superscripts (or a jumble of superscripts and subscripts), Ti k j m k n ... .... For ever more will we will use this notation (latin letters or greek symbols; lower or upper case; bold, italized or not, to denote the components of a tensor or the tensor itself.)
• defines a new coordinate system specified by the mutually independent variables: x1 , x2 , …, xn . The symbol φ i (e.g., a temperature distribution or field of some sort) are ass- umed to be single-valued real functions of the coordinates with continuous derivatives. The rank (order) of a tensor is the number (without counting an index which appears once as a subscript and once as a superscript) of indices in the letter or symbol representing a tensor (or the components of a tensor). Here a few examples of tensor (and their rank) which will make an appearance very soon: S is a tensor of rank zero (scalar – e.g., action); xi is a covariant vector of rank one (covariant vector – e.g., three- dimensional Cartesian coordinate); Pµ is a contravariant tensor of rank one (contra- variant vector – e.g., spacetime momentum); Tµ ν is a mixed tensor of rank two (e.g., energy-momentum-stress tensor) ; Eµν is a (contravariant–e.g., Einstein’s gravitational field tensor) tensor of rank two; Rµν ≡Σλ Rλ µλν (note the contraction on the index λ) is a tensor (e.g., Ricci curvature tensor) of rank two; Rµ νρσ is a mixed tensor (e.g., action Riemann curvature tensor) of rank four; and finally R≡ΣµΣν gµν Rµν (i.e., another contrac- tion on both µ andν) is a tensor of rank zero (scalar – e.g., curvature scalar). In an n- dimensional space, the number of components of a tensor of rank n is nr . ( )nixxxx nii ,,2,1),.,,( 21  == φ Consider a ordered set of n mutually independent real variables x1 ,x2 ,…,xn = [xi ], called the coordinate of a point, Pn(x1 ,x2 ,…,xn ). The collection of all such points corresponding to all he sets of values [xi ] forms an n-dimensional linear space (i.e., a manifold – French for variété) which we specify by Vn. The set of n equations: 2014 MRT
• On differentiating xi =φ i (x1 , x2 , …, xn ) with respect to xj , we obtain the following representation for an infinitesimal displacement in the original coordinate system, x j , in terms of the new coordinate system, xi : ∑∑ == ∂ ∂ = ∂ ∂ = ∂ ∂ ++ ∂ ∂ + ∂ ∂ = n j j j in j j j i n n iii i xd x x xd x xd x xd x xd x xd 11 2 2 1 1 φφφφ  A set of components, Aj , which transform according to the law above, which is given by the same process: 2014 MRT If det|∂xi /∂x j |≠0(i.e., the Jacobian – Jacobi (1804-1851) – is non zero) then the inverse transformation exists. ∑= ∂ ∂ = ∂ ∂ ++ ∂ ∂ + ∂ ∂ = n j j j i n n iii i A x x A x x A x x A x x A 1 2 2 1 1  forms the transformation law for the components of a contravariant tensor of rank one. As a quick example of its use, let us find the transformation for rotation in two dimensions. In this case, we have: x1 =x1 cosθ + x2 sinθ and x2 =x2 cosθ − x1 sinθ. The Jacobian of the transformation is given by: 1sincos)sin()(sincoscos cossin sincos det 22 2212 2111 =+=−⋅−⋅= − = ∂∂∂∂ ∂∂∂∂ = ∂ ∂ θθθθθθ θθ θθ xxxx xxxx x x j i
• In this last equation (i.e., Ai =Σj (∂xi /∂xj )Aj ) Aj is a function of the coordinates xj (i.e., Aj = Aj (xj )) and Ai is a function of the coordinates xi (i.e., Ai =Ai (xi )), where x j and x i refer to the old and new coordinate systems, respectively. 2014 MRT where the Kronecker delta,δ k j, is given by: or changing the dummy index k to j: i i j j A x x A ∂ ∂ =     ≠ = = kj kj k kk j 0 1 when )for,(when δ δ e.g. k k n k kk k n j jk j n n j j j kn i n j j j i i kn i n j j j i i k i i k A AA AA x x A x x A x x x x A x x = ⋅== = ∂ ∂ = ∂ ∂ = ∂ ∂ ∂ ∂ = ∂ ∂ ∑ ∑∑ ∑∑∑∑ ∑ = == == == = 1 1 11 11 11 1 δ δδ i i i x x ∂ ∂ The equation Ai =Σj (∂xi /∂xj )Aj may be solved for Aj if it is multiplied by ∂xk /∂xi and summed over i. In this case, we obtain:
• A set of quantities Bk is called the components of a covariant tensor of rank one if: 2014 MRT for an arbitrary contravariant tensor with components Ak . If Ak from Ak =Σi(∂xk /∂xi )Ai is substituted into Σk Ak Bk =Σk Ak Bk, the result is: ∑∑∑ ∑∑∑ = == === ∂ ∂ =         ∂ ∂ == n k n l l k k ln k k n i i i kn k k k n k k k BA x x BA x x BABA 1 11 111 or or ∑∑ == == n k k k n k k k BABA 11 scalar)(ainvariant 0 1 1 =         ∂ ∂ −∑ ∑= = n k n l lk l k k B x x BA ∑= ∂ ∂ = n l lk l k B x x B 1 since the Ak are arbitrary. Bk =Σl(∂xl /∂xk )Bl is the transformation law for the compo- nents of a covariant tensor of rank one. An easy to remember mnemonic for the placement of ∂xi is ‘CO BELOW’ for covariant components. Now for an application…
• Maxwell’s equations (the foundation of the theory of electromagnetism) are equa- tions that describe how electric field intensity (E) and magnetic field intensity (H) are generated and altered by each other and by charge density (ρ) and current density (J): 2014 MRT t t ∂ ∂ += =• ∂ ∂ −= =• D JH B B E D ×∇ ∇ ×∇ ∇ 0 ρ In MKS units, the quantity B is the magnetic induction: with εo (i.e., 8.854×10−12 C2 /N⋅m2 ) the permittivity of the vacuum. HB oµ= ED oε= with µo (i.e., 4π×10−7 kg⋅m/C2 ) being the permeability of the vacuum and quantity D is the electric displacement: These provide a set of partial differential equations that, together with the Lorentz force law (i.e., F=q(E+v× B)), form the foundation of classical electrodynamics, classical optics (with isotropic medium effects neglected), and electric circuits. Coulomb’s Law: Faraday’s Law: Absence of free magnetic poles: Ampère’s Law (∇ •J=0):
• If the Lorentz (1853-1928) condition: 2014 MRT where the d’Alembertian operator, , is defined by: is satisfied, the basic equations for the vector and scalar potentials are: The definitions of the vector A (i.e., the electromagnetic vector potential) and the scalar φ (i.e., the electromagnetic scalar potential), which are important parameters when studying Lorentz invariance concepts and non-local phenomena, respectively: tt ∂ ∂ −=•⇔= ∂ ∂ +• φφ ΑΑ ∇∇ 0 JA o 2 µ−= o 2 ε −= ρ φ and ∑= ∂ ∂ = ∂ ∂ −∇= 4 1 2 2 2 2 2 2 1 ν νxtc  t∂ ∂ −−== A EAB φ∇×∇ and
• The components of the four-vectors (four-dimensional orthogonal coordinate system – Minkowski (1869-1904) space – with the notation x4 =ict which demonstrates that a Lorentz transformation is a spacetime rotation) for position, current density, and potential are respectively taken to be: 2014 MRT The electromagnetic field tensor is defined by: ( ) [ ]       =≡=≡ −===≡ c iAAAAciJJJJ ictictizyxx zyxzyx φ ρ χα β ,,,,,, 1],[],,,[ AJ X and r ( )4,3,2,1, = ∂ ∂ − ∂ ∂ =≡ σρσ ρ ρ σ σρ x A x A FF where x xx xxx y xz z xy E c i t A xc i t A c i xc i t A cii i xc i t A cixc i ict A x c i x A x A F B z A x A x A x A FB y A x A x A x A F FFF x A x A F −=      ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ = ∂ ∂ ⋅ ⋅ − ∂ ∂ = ∂ ∂ ⋅− ∂ ∂ = ∂ ∂ − ∂       ∂ = ∂ ∂ − ∂ ∂ = −= ∂ ∂ − ∂ ∂ = ∂ ∂ − ∂ ∂ == ∂ ∂ − ∂ ∂ = ∂ ∂ − ∂ ∂ = ==== ∂ ∂ − ∂ ∂ = φφ φφ φ 111 )( 0 4 1 1 4 14 3 1 1 3 132 1 1 2 12 4433221 1 1 1 11 and ,
• 2014 MRT                     −− −− −− = 0 0 0 0 zyx zxy yxz xyz E c i E c i E c i E c i BB E c i BB E c i BB F σρ Consider the tensor equation generated by differentiating cyclically and adding terms: 0= ∂ ∂ + ∂ ∂ + ∂ ∂ σ ρµ ρ σµ µ σρ x F x F x F The remaining components of Fρσ may be obtained in a similar way; the result in matrix form is:
• Since Fµν is an antisymmetric tensor (look at the matrix above), it can be shown that only four independent equations result from the sixty-four possible equations that are indicated by the above tensor equation. Typical values for α, β, and µ are Set #2: 1,2,3 and then Set #3a: 4,2,3; Set #3b: 4,3,1; and Set #3c: 4,1,2. The resulting equations are: or 0 1,2,3,, 2 31 1 23 3 12 = ∂ ∂ + ∂ ∂ + ∂ ∂ = x F x F x F µβα:#2Set ( )equationsecondsMaxwell'0=•B∇ or 0 )( 4,2,3,, 2 34 4 23 3 42 = ∂       −∂ + ∂ ∂ + ∂       ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ = y E c i ict B z E c i x F x F x F z x y µβα:3aSet t B z E y E xyz ∂ ∂ −= ∂ ∂ − ∂ ∂ 2014 MRT or 0= ∂ ∂ + ∂ ∂ + ∂ ∂ y B x B z B yxz
• 2014 MRT or t B y E x E zxy ∂ ∂ −= ∂ ∂ − ∂ ∂ or t B x E z E yzx ∂ ∂ −= ∂ ∂ − ∂ ∂ 0 )( 4,1,2,, 1 24 4 12 2 41 = ∂       −∂ + ∂ ∂ + ∂       ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ = x E c i ict B y E c i x F x F x F y z x µβα:3cSet 0 )( 4,3,1,, 3 14 4 31 1 43 = ∂       −∂ + ∂ ∂ + ∂       ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ = z E c i ict B x E c i x F x F x F x y z µβα:3bSet or, finally, after combining all three subsets of Set #3, we get: ( )equationthirdsMaxwell' t∂ ∂ −= B E×∇
• The remaining two Maxwell equations may be obtained from the following tensor equation (i.e., 4 separate equations, one for each value of µ =1,2,3,4): 2014 MRT µ ν ν νµµµµµ J x F x F x F x F x F o 4 1 4 4 3 3 2 2 1 1 µ= ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ ∑= (Exercise: Show that the tensor equation just above, where Fµν and Jµ are the components of the electromagnetic field tensor and four-vector current density, respectively, reduces to Maxwell’s first Set #1 (i.e., ∇ •E=ρ/εo) and fourth Set #4 (i.e., ∇ × E=J+εo∂E/∂t) equations.) In a more compact form they can be written as: µ ν νµνλνµµλννµλ JFFFF o 4 1 0 µ=∂=∂+∂+∂ ∑= and where ∂µ ≡∂/∂xµ . When the presence of sources is non existant (i.e., being left with Homogeneous equations where ρ=0 and J=0), the Maxwell’s equation on the left above can be written even more compactly as: µ ν ν νµ λνµ JFF o,],[ 0 µ== ∑and If the comma-index notation (i.e., V,µ ≡∂V/∂xµ ) and the square brackets represent the antisymmetrization (i.e.,V[µν] =½(Vµν –Vνµ)).
• Let us try to figure out how Einstein saw stuff now that we have figured out how to formulate Maxwell’s equations in a so called ‘relativistic’ form. Three considerations will be studied now. 2014 MRT Consider two equal point charges q moving with the same velocity. In a frame moving with the charges, they are at rest (see Figure (a)) and experience only an electrostatic repulsion, FE=qE. In our ‘laboratory’ frame, in which the charges q are moving at speed v=|v| (see Figure (b)), each charge creates a magnetic field. The force between the charges is therefore reduced by magnetic attraction, FB=qv× B. The force between the charges depends on the frame of reference employed (i.e., the observer’s point of view.) + +q q FE = qE v + +q q FB = q v× B v (a) In a frame in which two equal charges are at rest, they experience only electric repulsion. (b) In a frame in which both charges have the same velocity, they also exerience a magnetic (a) (b) FE = qE FE = qE FE = qE FB = q v× B
• We already considered when a Galilean transformation is applied to Maxwell’s wave equation, its form changes completely. So if Galilean transformation equations are correct, Maxwell’s equations are valid in only one special frame – that of the ether. However, there is no evidence that Maxwell’s equations are restricted in this way. 2014 MRT Consider a short wire moving at constant velocity across the pole of a magnet. In the magnet’s frame (see Figure (a)), the magnet is at rest and the wire moves at velocity +v. An observer in the frame says that a charge q in the wire experiences a magnetic force. In the wire’s frame (see Figure (b)), the wire is a rest and the magnet has velocity −v. Since the charges are at rest in the wire’s frame, an observer in this frame would say that the charge q is subject to an electric force. We know experimentally that it is only relative motion of the wire and the source of the magnetic field that matters. Yet merely switching from one inertial frame to another requires a change from magnetic field to electric field. Even if both observers agree on the phenomenon, they use different laws to describe it. (a) The charges in a rod moving across the face of a magnet experiences a magnetic force. (b) If the rod is at rest and the magnet moves in the opposite direction, the charges in the rod experience only an electric force. × × × × × FE =qEFB =q v× B × × × × × (a) (b) × v v q q
• As a student, Einstein was aware of these and other problems. Indeed, as a boy of 16, he conceived of an intriguing question: What would one see if one travels with a beam of light? One should see stationary sinusoidal variations in space and the electrical and magnetic fields that constitute the wave. But this is not an acceptable solution of Maxwell’s wave equation – which requires a wave moving at the speed of light, c. Could the laws for the traveler be different from those for an observer at rest? Although by 1904 Einstein had found out about the Michelson-Morley experiment through the work of Lorentz, this experiment did not lay a significant role in the formulation of his theory. 2014 MRT Einstein had to make a choice. If the Galilean transformation and the laws of mechanics are correct, then Maxwell’s equations had to be reformulated. If Maxwell’s equations were correct, then the laws of mechanics were not exactly correct, even though no exception had yet been encountered. The sticking success of Maxwell’s theory made improbable that it was incorrect, so he decided that the Galilean transformations and the laws of mechanics had to be modified. Einstein believed that there must exist some powerful ‘universal principle’ that would guide him to the ‘true’ laws of physics. In June 1905, in a paper entitled On the Electrodynamics of Moving Bodies, Einstein introduced the special theory of relativity. Here is the opening passage: “It is known that Maxwell’s electrodynamics – as usually understood at the present time – when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. Take for example, the reciprocal electro- dynamic action of a magnet and a conductor. The observable phenomenon here depends only on the relative motion of the conductor and the magnet,
• whereas the customary view draws a sharp distinction between the two cases in which either the one, of the other of the bodies, is in motion. Examples of this sort, together with the unsuccessful attempts to discover any motion of the earth relative to the ‘light medium,’ [n.d.l.r., the ‘ether’] suggest that the phenomenon of electrodynamics as well as of mechanics possess no properties corresponding to the idea of absolute rest.” (A. Einstein, Ann. Physik, 17, 891 (1905)) 2014 MRT The fact that the speed of light, c, is an unattainable speed for a material particle resolves Einstein’s boyhood question regarding what he would see if he were to ride along with an electromagnetic wave. He would not see a stationary sinusoidal variation of electric and magnetic fields because he could never catch up (e.g.,with speedu) with a light wave (i.e., traveling at the speed of light,c). The issue raised in the quote from Einstein’s 1905 paper above. Einstein was uneasy about the use of an electric field or a magnetic field depending on one’s choice of reference frame (see Figure). The road and the car (i.e., Lamborghini Aventador LP700-4) serve as reference frames O and O, respectively. O y x O y x +u
• 2014 MRT (a) A charge moving relative to a wire. The positive and negative charges in the wire are equally spaced and have equal and oposite velocities. (b) In the frame of the charge q, the positive and negative charges in the wire have different speeds. The different factors for length contraction mean that the negative charge density is greater than the positive charge density. q (a) (b) + u FB = quB + − + − + − + − + − + − − + − − q + FE = qE Figure (a) shows a positive charge q moving at velocity u relative to a stationary wire that carries a current I. For simplicity we assume that the current in the wire arises from both positive and negative charges moving with opposite velocities, ±v . In the frame of the wire, the charge q experiences a magnetic force toward the wire, but no net electrical force. In the frame in which the charge q is at rest (see Figure (b)), it does not experience and magnetic force. In this frame, the positive charges in the wire move more slowly than v, where as the negative charges move faster than v. Electric charge is invariant in special relativity. Hence, because of length contraction, the negative charge density is greater than the positive charge density. The wire has a net negative charge in the rest frame of charge q. We see that an electrostatic field in the rest frame of charge q transforms into a magnetic field in another frame. v v I
• Now, back to tensors. The n2 quantities, Cij , transform according to: 2014 MRT ∑∑ ∂ ∂ ∂ ∂ = k l lk l j k i ji C x x x x C means that Ek l are the components of a second-rank mixed tensor. then Ckl are the components of a second-rank contravariant tensor. If the n2 quantities, Dij , transform according to: If Sk rs =Sk sr or Ak rs =−Ak sr then Sk rs (or Sk sr) or Ak rs (or −Ak sr) are said to be symmetric or antisymmetric (skew-symmetric), respectively, in the r and s indices. The symmetric (or antisymmetric) property is conserved (i.e., it does not change) under a transformation of coordinates and may be extended to higher-rank tensors for any contravariant or covariant indices. ∑∑ ∂ ∂ ∂ ∂ = k l lkj l i k ji D x x x x D then Dkl are the components of a second-rank covariant tensor. Similarly: ∑∑ ∂ ∂ ∂ ∂ = k l l k j l k i j i E x x x x E
• Now for the fun stuff: generating new tensors form the result of operations such as addition (and its inverse – subtraction), multiplication or outer product (and its inverse – division) and contraction. All of this is called tensor algebra. 2014 MRT The outer product of two tensors with components Ui k j m k n ... ... and Vλ ρ µ σ ν τ ... ... is defined by: As is clear from the general (or fundamental) transformation law: The operation of contraction is the process by which the number of covariant and contravariant indices of a mixed tensor is reduced by one. For example, consider the contraction of a mixed tensor with components Ti k j lm (e.g., Rj lm=Σi Ti i j lm). Here the components of Rj lm are obtained from the components of Ti k j lm by contracting the indices i and k (i.e., for equal i=k indices). Any index of the contravariant set and any index of the covariant set may be used to form the components of the new tensor. ∑∑∑ ∑∑∑ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = λ µ ν ρ σ τ νµλ τσρ τσρ νµλ      T x x x x x x x x x x x x T nml kji kji nml Two tensors of the same type (having the same number of covariant and contravariant indices) can be added (or subtracted) to produce a single tensor. For example:       kji nml kji nml kji nml TTU ∆±=       µνλ τσρ µνλ τσρ VUW kji nml kji nml ⊗=
• The process of combining outer multiplication and contraction to produce a new tensor is called inner multiplication, and the resulting tensor is called the inner product of the two tensors involved. For example, we may write Ti k j lm =Σn Y i n j Rn klm for the multiplication (i.e., the outer product) and contraction of two tensors with components Y i n j and Rn klm. 2014 MRT By use of the fundamental transformation law for the components of a tensor, we see that the quantities ΣN δ N M xN transform like the components of a tensor; hence xN are the components of a tensor and δ N M are the components of a second-rank mixed tensor. The transformation equation for δ N M is: since dxN =ΣM δ N M dxM . Therefore the components of the five-dimensional Kronecker delta, δ N M, have the same values in all coordinate systems within that worldsheet. For an arbitrary tensor with components xM (e.g., the set of five-dimensional worldsheet coordinates [xM]=[x1, x2, x3, x4, x5]), we may for the inner product: M N N N M xx =∑= 5 1 δ N MM N M N R R MR N R S R SM S R N N M x x x x x x x x x x x x δδδδ = ∂ ∂ =         ∂ ∂ =         ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ = ∑∑ ∑∑∑= = R R R S S S x x    ∂ ∂ 4 0 4 0 SAME THING
• The generalized form of the element of length (arc length or interval), ds, between two coordinate points xM and xM +dxM as defined by Riemann is given by the following quadratic differential form: 2014 MRT where it is assumed that: 1) the γMN are functions of the xM ; 2) γMN=γNM (symmetric); and 3) det|γMN|≠0. In this case, the space is called Riemannian space, and the quadratic form ΣMN γMN dxM dxN is called the metric (e.g., of a five-dimensional universe). ∑∑= M N NM MN xdxdsd γ2 and in turn a ‘flat’ space includes Euclidian geometry represented by the space metric gij (e.g., [xi]=[x1, x2, x3]=[x, y, z] – the basis being Coordinate Geometry): The tensor γMN is (physically) reducible to γMN =ηµν +γ55 with γ55 (i.e., the fifth diagonal component) and whereηµν is called the Minkowski metric (a.k.a. the spacetime metric with [xµ]=[x0, x1, x2, x3]=[ct, xi] or [ xi ,ict] if x4=ict, where c is the speed of light and t is the time travelled –a new ‘coordinate’ that forms the basis of this is Special Relativity):       = 550 0 γ η γ µν MN       = jig0 000η ηµν
• If rectangular coordinates are introduced in a Euclidean space we have gij =δij: 2014 MRT which is the familiar form of Pythagoras’ theorem in a three-dimensional world. For example, Einstein theory of General Relativity demands that ds be independent of the system of coordinates (e.g., free-falling from a ladder with Fair =0.) Hence we may write: Since the differentials dxρ dxσ in the above equation are arbitrary, we require that: 2222322212 3 1 3 1 2 )()()()( zdydxdxdxdxdxdxdxdxdxdgds i i i i i j ji ji ++=++==== ∑∑ ∑∑∑= = i j j jδ ∑∑∑ ∑∑∑∑ ∂ ∂ ∂ ∂ =         ∂ ∂         ∂ ∂ == νµ σρ σρ σ ν ρ µ νµ νµ σ σ σ ν ρ ρ ρ µ νµ νµ νµ νµ σρ σρ σρ xdxd x x x x gxd x x xd x x gxdxdgxdxdg ∑ ∂ ∂ ∂ ∂ = νµ νµσ ν ρ µ σρ g x x x x g 00 =         ∂ ∂ ∂ ∂ −⇒= ∂ ∂ ∂ ∂ − ∑ ∑∑∑∑ σρ σρ νµ νµσ ν ρ µ σρ νµ σρ σρ σ ν ρ µ νµ σρ σρ σρ xdxdg x x x x gxdxd x x x x gxdxdg Hence the components of gµν transform like the components of a second-rank covariant tensor, and gµν are called the components of the fundamental (or metric) tensor. (We say that gµν ≡g is the fundamental tensor.) In Cartesian coordinates, gij=0 for i≠j and a g =g =g =1 (i.e., for i=j ). Note that Σ g gµν =g g T =1+1+1+1=4 . or
• x 1 = 1 x 1 = 2 x 0 = 1 x 0 = 2 n 0 n 1 ∑= µν νµ µν dxdxg2 ds At each point in spacetime, we can define a set of unit vectors nµ – each of them is in a direction of one of the coordinate axes. The nµ are unit vectors correspond to incremental units of the coordinates. Then the four-vector u from initial point x µ to an adjacent point xµ +dxµ is given by (here for µ,ν = 1,2): We define the metric tensor gµν – or metric g – with its components by introducing the scalar product of the unit vectors The interval between points xµ and xµ + dxµ is: The components of the metric act as potentials at all points of spacetime: 10 potentials that weigh and scale the measurement of the spacetime intervals of proper time and proper lengths. It’s a totally different world in 4-D! A generalized set of coordinates on a ‘curved’ 2-D plane. Only the lines x1 = 1, x1 = 2, x2 = 1, x2 = 2 are drawn. Also shown are the unit vectors of the coordinate at point [1,1]. If a GPS were used, a basis could be Latitude and Longitude. For someone in the middle of Ottawa with only a compass, North/South and East/West could be used to orient oneself in Now we generalize Pytharoras’s Theorem by considering the square of the distance between two space-time coordinates and define our standard for measuring in space- time and appreciating the properties offered by such a 4-dimensional view of things… ∑=+= µ µ µ nnnu dxdxdx 2 2 1 1 νµµννµ nnnng •== g),( ∑∑∑ •=•=•= µν νµ νµ ν ν ν µ µ µ dxdxdxdx )()()( nnnnvuds2 [1,1] n2 x2 = 1 x2 = 2 x1 = 2 x1 = 1 n1 I’m in Ottawa! EAST NORTH 2014 MRT u dx2 dx1 Chateau Laurier xµ xµ + dxµ g ∑=⊕= µ µ µ nnnu dxdxdx 2 2 1 1
• The associate tensor to an arbitrary tensor Pν, pµ, is the result of the inner product of Pν and the fundamental tensor, gµν , that is: 2014 MRT We now show that pµ in pµ =Σν gµν Pν is just a new form of Pν . Note that: ∑∂ ∂ = ν νµ ν µ p x x p which is the transformation law for a covariant vector. Hence assuming the pµ in pµ =Σν gµν Pν is just a new form for Pµ is consistent with the required transformation law for the components of a tensor. or ∑= ν ν νµµ Pgp where pµ , which is a new form (covariant) of the old (contravariant) tensor Pν, is called: associate to Pν . The process in pµ =Σν gµν Pν is called lowering the subscript. ∑∑∑∑∑ ∑∑∑∑ ∑∑∑ ∂ ∂ = ∂ ∂ = ∂ ∂ = ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ =         ∂ ∂         ∂ ∂ ∂ ∂ == ρ ρµ ρ ρ ρµ ρ σρ σρ σ µ ρ τσρ τ σρτ σ µ ρ τσρ τ σρτ σ µ ρ ν τ τ τ ν σρ σρν σ µ ρ ν ν νµµ δ p x x Pg x x Pg x x Pg x x x x Pg x x x x P x x g x x x x Pgp σ σ σ τ τ τ ν ν ν )( x x ∂ ∂
• If pµ =Σν gµν Pν is solved for Pν, we obtain: 2014 MRT for g=det|gµν |≠0. Here [gµν]cT means cofactor transpose of the matrix gµν . On applying Pµ =Σν gµν pν we see that the quantities gµν form the components of a second rank contravariant tensor. The process highlighted by Pµ =Σν gµν pν is known as rising the subscript. The tensor Pµ is associate to pν, and gµν is called the reciprocal tensor to gµν . ν λ µ νµ µλ δ=∑ gg The process of lowering and raising indices may be performed on higher-rank tensors (e.g., an operation on the Riemann-Christoffel tensor Rτ νρσ gives Στ gµτ Rτ νρσ =Rµνρσ , the Riemann curvature tensor.) Multiplying Pµ =Σν gµν pν by gλµ and summing over µ, we get: ∑= ν ν νµµ pgP where g g g Tc ][ νµνµ = ∑∑∑ === νµ ν νµ µλ νµ ν νµ µλλ µ µ µλ pggpggpPg ][)( or
• n 0 n 1 e 0 e 1 (e 0)0 (e 0)1 At each point of spacetime, xµ , we introduce also a locally inertial coordinate system: it’s the ‘weightlessness’ system in free fall (i.e., the system where there is no gravitational force – imagine an astronaut a few hundred kilometers away from Earth.) Here are the three postulates of special relativity (Einstein 1905): 1. All the laws of nature have the same ‘form’ in all inertial systems; 2. The speed of light c is a universal constant – which is also the same in all inertial systems; 3. The speed of light c is independent of the speed of the source. We define this system by a set of vectors en (with n = 1,2,3,0) that will satisfy (let un assume the sped of light n =1): where ηαβ is the metric of the flat spacetime (with components on the diagonal only):ηαβ = (+1, +1, +1,−1). We show the choice of e1, e2 at point [1,1]. The components (e1)1 , (e1)2 are also indicated (see projection for e1). The hashed perimeter delimits what is termed “locally inertial” – the ‘locality’ condition – spacetime is flat in that area and symmetry is assured. A further basis is used: ‘Elgin’ and ‘Laurier’ act as part of a tetrad to αββα ηη ηη η =•=• ==•=• =• ⇒+= = += eeee eeee ee , 1 1 2222 21121221 1111 0, e2 e1 (e1)2 µ µ µ αα ne ∑= )(e 250 m huh! [1,1] EAST NORTH Laurier Elgin x1 = 1 x2 = 1 2014 MRT v Chateau Laurier η11 η22 (e1)1 n1 n2 η and introducing the vierbeins (eα)µ we have: ∑µν gµν (eα)µ (eβ)ν = ηαβ = η (eα ,eβ ) g (nµ ,nν ) = gµν = ∑αβ (eµ )α (eν )β ηαβ We call the set of four-vectors eα , a tetrad. We can express any member of a tetrad as a function of unit vectors of the generalized coordinate system by posing:
• ds2 =u • v αβ µν ν β µ αµν µν ν β µ ανµν ν ν βµ µ µ α η==•=•=•= ∑∑∑∑ )()()()()()()( eegeeee nnnnvuds2 2014 MRT u The interval ds is thus represented somewhat geometrically in the following Figure. We will generalize this concept to higher ‘real-life’ dimensions since we live in a world with 4 dimensions: 3 of Space (i.e., North/South, East/West and Up/Down) and an immutable one of Time (i.e.,wecan’tchangeitsrateorits direction to go in the past.)
• ( )22 00332211 2 2 00 33 22 11 ,1,1,1),1,1,1( 000 0100 0010 0001 000 000 000 000 lightofSpeed−=−=+=+=+=⇒−+++=→←             − + + + =             == cc c ηηηηηηη η η η η η µναβ µν  η ν = 1 ν = 2 ν = 3 ν = 0 µ = 0µ = 3µ = 2µ = 1 Space only (isotropic and homogeneous space – same in all directions!) Time only (the forth dimension – minus the square of the speed of light!) Stating obvious reasons based on symmetry (e.g., the componants of the diagonal are symmetrical, ηµν =ηνµ ) we are left with normalizing to include the speed of light… So, Special Relativity considers that the speed of light c is the same in all inertial frames. The diagonal metric η of flat spacetime (x1 , x2 , x3 and x0 ) is given by (indices µ,ν = 1,2,3,0): or Space coupling to time (identically zero – coupling does not happen!) and with the correct ηµν (i.e., µ for the matrix ‘column’ and ν for its ‘row’) in the right slot: Time coupling to space (identically zero – coupling does not happen!) Sphère de lumière 4th Dimension: x0 = ct 2014 MRT S S x1 x2 x3 1 3 2x ‡ ‡ v = cte t t x x 2222232221 )()()()( dtcdcdtdxdxdx =⇒=++ r P O S 1 x , 1 x 2 x , 2 x 3 x , 3 x 1 x 3 x 2 x S,S v t = t = 0 S O r r ( 1 x , 2 x , 3 x , t) ( 1 x , 2 x , 3 x ,t ) 3 x 2 x 1 x Sphere of light • •
• 02222 ==−=∑ sddtcddxdx r µν νµ µνη A light wave traveling at the speed of light satisfies |dr/dt|=c, or in other words: .,1,1,1 )( )( )( )( )( 2 00332211 03 30 02 20 01 10 00 3332 23 31 13 30 03 23 32 2221 12 20 02 13 31 12 21 1110 01 3 0 0 0 3 0 3 3 3 0 2 2 3 0 1 1 3 0 0 0 3 3 2 2 1 1 3 0 3 0 c dxdxηdxdxηdxdxdxdx dxdxdxdxdxdxdxdx dxdxdxdxdxdxdxdx dxdxdxdxdxdxdxdx dxdxdxdxdxdxdxdx dxdxdxdxdxdxdxdxdxdx −=+=+=+= ++++ +++++ +++++ ++++= +++= =+++=         ∑∑∑∑ ∑∑ ∑ ==== == = ηηηη η ηηη ηηη ηηη ηηηη ηηηηη µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ µ ν νµ µν :metrictheofvaluesdiagonalthesubstitutenow weAnd 00 22 11 η η η 33η Since the sum can be expanded in all generality – the same applicable everwhere repeated indices are seen: In flat spacetime (the perview of special relativity) the Spacetime Interval is reduced to: ∑=⇔⊗= µν νµνµ µνη dxdxsd µνη2 dxdx2 ds 2014 MRT The diagonal elements of the metric are highlighted in bold italic ηµν .
• η=⊗⇒= ∂ ∂ ∂ ∂ ⇒= ∑∑∑∑∑ νµ µνρσ νµ σ ν ρ µ νµ µ ν νµ µν µ ν νµ νµ ηηηηη dxdx x x x x dxdxdxdx Since the proper time dτ =0 and if xµ are the coordinates in one inertial frame then in any other inertial frame, the coordinate xµ must satisfy (sum rule over repeated indices): µµµµµµ ν νµ ν µ axxxxaxx +Λ+Λ+Λ+Λ=+Λ= ∑ )( 0 0 3 3 2 2 1 1 Any coordinate transformation xν →xµ that satisfies the equation above is linear: where aµ are arbitrary translation constants (typically just the translation vector a) and the 4×4 Λµ ν (β ) matrix is the set of one-to-one transformations from one frame xµ to the next xµ. As you can see – if you’ve seen it – things can get pretty tricky in 4-D! 2322212022 )()()()( dxdxdxdxcd +++−=r Remembering that in flat spacetime coordinates [x0 ,x1 ,x2 ,x3 ] and µ,ν =0,1,2,3 we have: 2014 MRT 202232221 03020100 33323130 23222120 13121110 3 0 3 0 )()()()( )000( )000( )000( )000( dxcdxdxdx dxdxdxdxdxdxdxdx dxdxdxdxdxdxdxdx dxdxdxdxdxdxdxdx dxdxdxdxdxdxdxdx dxdx −++= +++−+ +++++ +++++ ++++ =         ∑ ∑= = 2 1 1 1 cµ ν νµ µνη Continuing the expansion, we get:
• Length Contraction:    == vacuum)a(inlightofSpeed objecttheofSpeed where  c v β 212 2 0 0 )1( 1 1 )( − −= − =Λ= β β βγ ( )lengthpropertheiswhere LL L LL 21 1 β γ γ −=== − ( )timepropertheiswhere τ β τ τγ 2 1− ==t Time Dilatation: with rapidity β beingaparameter within special relativity.The Lorentz factor is also used: The Lorentz Transformation from the reference frame O(xν ) to the new frame O(xµ):                   −− − − − − =Λ= 22 22 1 1 1 00 11 1 00 0010 0001 )( ββ β β β β βµ ν c c Λ e.g., at v=0.42c (i.e., achieving 42% of the speed of light), γ =1.10, which means that the effects of Relativity become noticeable. Notice that for β =1 (assuming v=c), γ →∞! Space Space-Time Coupling Time only 2014 MRT In the direction of the axis Ox3.
• Geodetic Precession Frame-dragging Precession       −•⊕×= ωRω R vRΩ )( 3 )( 2 3 23232 RRc GJ Rc GM Ω v⊕ ∼ 1,674.4 km/h h ∼ 650.4 km 2014 MRT
• 2014 MRT Boost in an arbitrary direction. 2 )( v vvr r • =|| 2 )( v vvr rrrr • −=−= ||⊥ x y z Equations of transformation from reference frame O(x,y,z) to another frame O(x,y,z) in the most general case where the axis Ox = Oz is not in the direction of the velocity v. x z y tt 22 2 1 1 )(1 1 1 1 β γ − = − = • − = c c vvv       • −= •−= 2 ))(1( c tt t vR vvvRRr γ γγ −+ 2 c tt t || ⊥⊥|||| • = == rv rrvrr γγ γγ − − and tt = = Rr µ νΛ R ⊥||+= rrr v r|| r⊥ ˆ ˆ   r Ω
• 2014 MRT Boost in an arbitrary direction. x y z Boost in an arbitrary direction. n1 , n2 , and n3 are parameters and n2 ≡(n1 )2 +(n2 )2 +(n3 )2 =1. x z y tt  a        Λ+Λ= +         Λ+Λ= ∑ ∑ = = tt t i it i t tj t i ij i j xxx xxx x 3 1 3 1 a µ µµµ µ ν νµ ν µ axx axx t t i i i +         Λ+Λ= +Λ= ∑ ∑ = 3 1     = == tx x x i 0 Rxν µ νΛ iii t t i pp ˆ1ˆ −=−=Λ=Λ γγβ jijij i i j pp δγ +−=Λ=Λ ˆˆ)1( ])([ andbyreplacedwithbutassame ββ−Λ=Λ βµ ν ν µ M Mt t 22 2 1 1 + = − ==Λ p β γ pii pp ≡ˆ v  R Ω
•       +−+=−+=+=Λ ∑∑∑∑ ρσ ρσ ρσ ρ ρ ρ ρ ρ ρ ρσ ρσ ρσ )(ω 2 1 )(ε1)ε()ω( 2 1 1)εω,1(),( JP i P i J i UaU  ]ωinTerms[)ωω()ωδ)(ωδ( 2 O+++=++=ΛΛ= ρσσρσρ ν σ ν σ µ ρ µ ρµν ν σ µ ρµνρσ ηηηη µµµ ν µ ν µ ν εωδ =+=Λ a&             − − =                         − − =Λ==             )( )( 00 00 0010 0001 zct ctz y x t z y x c c xx t z y x βγ βγ γβγ βγγ νµ ν µ The Lorentz transformation transforms frame xν (say x,y,z,t) into frame xµ (say x,y,z,t) assuming the inertial frame is going in the x3 (+z) direction. If both ωµ ν and εµ are taken to be an infinitesimal Lorentz transformation and an infinitesimal translation, respectively: this allows us to study the spacetime transformation: in contrast to the linear unitary operator U=1+(i/)[Gt] can then be constructed in Quantum Mechanics: where the generators G of translation (Pµ ) and rotation (Jµν ) are given by P1 , P2 , and P3 (i.e., the components of the momentum operator), J23 , J31 , and J12 (the components of the angular momentum vector), and P0 is the energy operator (i.e., the Hamiltonian H.) 2014 MRT x2 = y x3 = z x1 = x x2 = y x3 = z x1 = x x0 = ct β = v/c
• Now the fun stuff begins... Tensor Calculus. Consider the invariant φ(x) = φ(x); differentiating both sides of this equation with respect to x µ , we obtain: 2014 MRT In the latter case, we see that the presence of the second term means that ∂Aσ/∂xτ do not form the components of a tensor (otherwise we would only have the first term on the right hand side, i.e., ∂Aκ /∂x ρ =Σστ (∂xκ/∂xσ )(∂xτ/∂x ρ)∂Aσ/∂xτ .) ∑∑∑ ∂ ∂ ∂ ∂ = ∂ ∂         ∂ ∂ = ∂ ∂ = ∂ ∂ = ρ ρµ ρ ρ µρ ρ ρ µ ρ ρµ φφφ δ φ xx x xx x xx 3 0 In this equation, we see that the ∂φ/∂xρ are the components of a first-rank covariant tensor. The above differentiation enabled us to generate a tensor of rank one from a tensor of rank zero. If we try to extend this process to obtain a tensor of rank two by differentiating a tensor of rank one (i.e., Aκ =Σρ (∂xκ /∂xρ )Aρ ), we find that: We now develop a scheme for a new kind of derivative, the covariant derivative, which enables us to obtain new tensors from the differentiation of other tensors by expressing the second term in the last equation in terms of the Christoffel symbols. The process of developing a scheme such that the derivative of a tensor always leads to a tensor is the chief aim of tensor calculus. ∑∑ ∂∂ ∂ + ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ σ σ σρ κ τσ τ σ ρ τ σ κ ρ κ A xx x x A x x x x x A 2
• The Christoffel symbols of the first and second kind are defined, respectively, by: From the above definitions, we see that two Christoffel symbols are symmetric with respect to µ and ν indices. Hence we may write Γµν,λ =Γνµ,λ and Γρ µν =Γρ νµ. Also note that:         ∂ ∂ − ∂ ∂ + ∂ ∂ =Γ λ νµ µ λν ν λµ λνµ x g x g x g 2 1 , and σνµ λ λνµ λ σ ρλ λνµ λρ σρ ρ ρ νµσρ δ ,,, Γ=Γ=Γ=Γ ∑∑∑ ggg ∑∑         ∂ ∂ − ∂ ∂ + ∂ ∂ =Γ=Γ λ λ νµ µ λν ν λµλρ λ λνµ λρρ νµ x g x g x g gg 2 1 , where µ, ν, and λ are regarded as subscripts and ρ is to be considered as a superscript when the summation is used. Γρ µν is not always a tensor as suggested by the symbol (and we will prove that soon enough.) Recall also that gµν = [gµν]cT /g with g=det|gµν |. Here is a brief word of caution… In the early years of Einstein using tensors and guys like Eddington (1882-1944) writing books about it, you would find Γµνλ or [µν, λ] and Γρ µν = {µν, ρ}={µ ρ ν}. (It’s the same mess as with the metric. Nowadays, particle physicists have pretty much firmed up the convention – into East coast ds2 =–dt2 +dx2 =ηµν dxµ dxν and West coast USA ds2 =dt2 –dx2 from what I hear. The latter has p2 =E2 −p2 =m2 if c=1). 2014 MRT
• Let us investigate the transformation laws for the Christoffel symbols. In terms of new coordinates x µ, we may write Γρσ,τ = ½(∂gρτ /∂x σ +∂gστ /∂x ρ −∂gρσ /∂x τ ). We had gρσ = Σµν (∂xµ /∂xρ )(∂xν /∂xσ )gµν. On differentiating gρσ with respect to xτ , we obtain: 2014 MRT Since gρσ =gσρ , the above equation becomes: ∑∑∑ ∑         ∂ ∂ ∂ ∂ ∂ ∂ +         ∂ ∂ ∂ ∂ ∂ ∂ + ∂ ∂         ∂ ∂ ∂ ∂ =         ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ νµ τ νµ σ ν ρ µ νµ νµσ ν τρ µ νµ νµσ ν ρ µ τ νµ νµσ ν ρ µ ττ σρ x g x x x x g x x xx x g x x x x x g x x x x xx g ∑∑ ∑∑∑ ∂ ∂ ∂ ∂ ∂ ∂ +         ∂∂ ∂ ∂ ∂ + ∂ ∂ ∂∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ + ∂∂ ∂ ∂ ∂ + ∂ ∂ ∂∂ ∂ = ∂ ∂ νµ τ νµ σ ν ρ µ νµ νµστ µ ρ ν σ ν ρτ µ νµ τ νµ σ ν ρ µ νµ µνστ µ ρ ν νµ νµσ ν ρτ µ τ σρ x g x x x x g xx x x x x x xx x x g x x x x g xx x x x g x x xx x x g 22 22 Cyclic permutation of ρ, σ, and τ (i.e., ρστ →τρσ) in the above equation leads to: ∑∑ ∂ ∂ ∂ ∂ ∂ ∂ +         ∂∂ ∂ ∂ ∂ + ∂ ∂ ∂∂ ∂ = ∂ ∂ νµ σ νµ ρ ν τ µ νµ νµρσ µ τ ν ρ ν τσ µ σ ρτ x g x x x x g xx x x x x x xx x x g 22
• For another cyclic permutationτρσ →στρ in the above equation, we obtain: 2014 MRT ∑∑ ∂∂ ∂ ∂ ∂ +Γ ∂ ∂ ∂ ∂ ∂ ∂ =Γ νµ νµσρ ν τ µ λνµ λνµτ λ σ ν ρ µ τσρ g xx x x x x x x x x x 2 ,, Substituting these results into Γρσ,τ = ½(∂gρτ /∂x σ +∂gστ /∂x ρ −∂gρσ /∂x τ ), we find that: This is the transformation law for the Christoffel symbol of the first kind and it shows that Γµν,λ transforms like a tensor only if the second term in the transformation equation vanishes. When this second term vanishes, the transformation is said to be affine. ∑∑ ∂ ∂ ∂ ∂ ∂ ∂ +         ∂∂ ∂ ∂ ∂ + ∂ ∂ ∂∂ ∂ = ∂ ∂ νµ ρ νµ τ ν σ µ νµ νµτρ µ σ ν τ ν σρ µ ρ τσ x g x x x x g xx x x x x x xx x x g 22 Now for the transformation law for the Christoffel symbol of the second kind. For the contravariant fundamental tensor, we had gρσ =Σµν (∂xµ /∂xρ )(∂xν /∂xσ )gµν . Inner multiplication of both sides of the transformation law for the Christoffel symbol of the first kind with the corresponding sides of the prior equation for gρσ leads to: ∑∑ ∂∂ ∂ ∂ ∂ +Γ ∂ ∂ ∂ ∂ ∂ ∂ =Γ ν σρ ν ν τ λνµ λ νµσ ν ρ µ λ τ τ σρ xx x x x x x x x x x 2 which is the transformation law for the Christoffel symbol of the second kind. It is not a tensor according to our definition – the second term ruins the invariance of Γλ µν .
• We need ∂2 xν/∂xρ ∂xσ ! Inner multiplication of the above equation with ∂xκ/∂xω yields: 2014 MRT ∑∑∑ Γ ∂ ∂ ∂ ∂ −Γ ∂ ∂ = ∂∂ ∂ τκνµ κ νµσ ν ρ µ κ κ τ τ τ τ σρτ κ τκ σρ κ κ κ τ τ δδδδ x x x x x x xx x2 Using the fact that ∂xτ /∂xω =δ τ ω we bring these like terms together, then throughout we change ω (a dummy index) to τ , then substituting ∂xκ /∂xν =δ κ ν and ∂xκ /∂xλ =δ κ λ: ∑∑ ∂∂ ∂ ∂ ∂ ∂ ∂ +Γ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ =Γ ∂ ∂ ν σρ ν ν τ ω κ λνµ λ νµσ ν ρ µ λ τ ω κ τ σρω κ xx x x x x x x x x x x x x x x x 2 ∑∑ Γ ∂ ∂ ∂ ∂ −Γ ∂ ∂ = ∂∂ ∂ νµ κ νµσ ν ρ µ τ τ σρτ κ σρ κ x x x x x x xx x2 The Delta functions and left over summations over κ and τ reduce everything to an expression for the second-degree differential:
• By using this last equation in the second term of an equation we obtained earlier (i.e., ∂Aκ /∂x ρ =Σστ (∂xκ/∂xσ )(∂xτ/∂x ρ)∂Aσ/∂xτ +Σσ (∂2 xκ/∂x ρ ∂xσ )Aσ ), and inserting in the following unit tem: Σω∂xω /∂xω =Σωδ ω ω , we get: 2014 MRT Substituting it into ∂Aκ /∂xρ =Σστ (∂xκ /∂xσ )(∂xτ /∂x ρ )∂Aσ /∂xτ +Σσ (∂xω /∂xρ )(∂2 xκ /∂xω ∂xσ )Aσ , we obtain: or by rearranging things (especially changing the index µ for ρ and summing): ∑ ∑∑∑         Γ ∂ ∂ ∂ ∂ −Γ ∂ ∂ ∂ ∂ =         ∂∂ ∂ ∂ ∂ ωσ σ νµ κ νµσ ν ω µ η η σωη κ ρ ω ωσ σ σω κ ρ ω A x x x x x x x x A xx x x x 2 ∑ ∑∑∑         Γ ∂ ∂ ∂ ∂ −Γ ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ ωσ σ νµ κ νµσ ν ω µ η η σωη κ ρ ω τσ τ σ ρ τ σ κ ρ κ A x x x x x x x x x A x x x x x A ∑∑∑ Γ ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ ∂ ∂ =         ∂ ∂ Γ+ ∂ ∂ ωση ση σωη κ ρ ω τσ τ σ ρ τ σ κ σν σ σ ν κ νρρ κ A x x x x x A x x x x A x x x A ∑ ∑∑         Γ+ ∂ ∂ ∂ ∂ ∂ ∂ =Γ+ ∂ ∂ τσ η ησ ηττ σ ρ τ σ κ ν νκ νρρ κ A x A x x x x A x A With Aν =Σσ (∂xν /∂xσ )Aσ , and making use of various Delta functions judiciously, the above equation reduces to a workable form and we finally get:
• Introducing the notation Aκ ;ρ which is defined as: 2014 MRT This last equation has the form as the basic transformation law for the components of a tensor. Hence Aκ ;ρ is a second-rank mixed tensor, and it is called the covariant derivative of the contravariant tensor Aκ with respect to xρ . By use of a similar procedure, we find the covariant derivative of the covariant tensor Bκ with respect to xρ : ∑∑ Γ−=         ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ ≡ κ κ κ τντν κλ κλ τν ν λτ τ λνλκ τ ν τν BBB x g x g x g g x B B ,; 2 1 where ∑∑ Γ+=         ∂ ∂ − ∂ ∂ + ∂ ∂ + ∂ ∂ ≡ ν νκ νρ κ ρ νλ ν λ νρ ρ λν ν λρλκ ρ κ κ ρ AAA x g x g x g g x A A ,; 2 1 where Aκ ,µ ≡ ∂Aκ /∂xµ so alternatively Aκ ;ρ =Aκ ,ρ +Σν Γκ ρ ν Aν . We see that: ∑ ∂ ∂ ∂ ∂ = τσ σ τρ τ σ κ κ ρ ;; A x x x x A ∑ ∂ ∂ ∂ ∂ = τν τνρ τ κ ν ρκ ;; B x x x x B Note the ± for Aκ ;ρ and Bν ;τ , respectively. The covariant derivative of a scalar is defined to be the same as the ordinary derivative, that is φ;µ≡φ,µ=∂φ/∂xµ.
• The above relations for the covariant derivatives of a tensor may be extended in a natural way to the case of a general mixed tensor. Consider, for example, the general mixed tensor Ti k j m k n ... ...; here we have: 2014 MRT                  ∑∑∑ ∑∑∑ Γ+Γ+Γ+ −Γ−Γ−Γ− ∂ ∂ = α α α α α α α α α α α α α α α α α α ji nml kki nml jkj nml i kji mln kji nlm kji nml kji nmlkji nml TTT TTT x T T ωωω ωωωωω;
• As an supplement showing reduction, let us tie things into something we know in three- dimensions as vectors but now is formulated in terms of their components. Imagine the scalar field φ and the vector field A. Here is the form for gradφ, divA, curlA, and ∇2 φ in terms of components (i.e., x1 ,x2 ,x3 =[xi ]): 2014 MRT ∑ ∑∑ ∑ ∑∑         ∂ ∂         ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂∂ ∂ =         ∂ ∂ Γ− ∂∂ ∂ ≡∇ ∂ ∂ − ∂ ∂ ≡         ∂ ∂ − ∂ ∂ + ∂ ∂ +=Γ+ ∂ ∂ ≡ ∂ ∂ ≡ ji lk kl ji i lj j lilk ji ji ji k k k jiji ji j i i j ij lki k l ki i lk k lilii i ki ki kii i j xx g x g x g g xx g xxx g x A x A A x g x g x g gAA x A x φφφφ φ φ φ 2 1 )( 2 1 div 22 2 , Α Α curl grad (Exercise: a) Show that divA is a scalar; b) Show that the above expressions yield the appropriate results for Cartesian [xi ] and spherical coordinates [xj ].)
• In three-dimensional Euclidean space, a straight line is the shortest distance between two points (e.g., in the Calculus of Variations section we found y=Ax+B.) We now generalize this fundamental concept to Riemannian space. 2014 MRT where Since the distance is independent of the system of coordinates used to describe it, a curve drawn from P1 and P2 is such that the integral over all increments of travel ds (i.e., ∫P1 to P2 ds) is stationary (i.e., a minimum) and is also independent of the system of coordinates used; this curve is called a geodesic. We now develop the differential equation of the geodesic in Riemannian space. Note that: Now, in spacetime, the curve xµ = xµ (λ) joins two fixed points P1(λ1) and P2(λ2), then the distance travelled along this curve between the two points P1 and P2 is given by: ∫∫ == 2 1 2 1 ),,( λ λ µµ λλ dxxFsds P P  λλλ µ µ νµ νµ νµ d xd x d xd d xd gF == ∑ and 0 2 1 2 1 === ∫∫ λ λ λδδδ dFsds P P
• By use of the result derived in the Calculus of Variations section we may write: 2014 MRT then Here we may take λ to equal the distance s along the curve in question and dF/ds=0 since s is an arbitrary parameter. In this case: λλ λλλλ µµµ µµµµµµµ d Fd x F Fx F x F d d x F Fx F d d Fx F Fd d x F Fx F Fd d x F d d x F   ∂ ∂ − ∂ ∂ −         ∂ ∂ = ∂ ∂ −        ∂ ∂ + ∂ ∂       = ∂ ∂ −        ∂ ∂ =         ∂ ∂ − ∂ ∂ = 2 1 2 11 2 1 2 1 2 1 0 2121212121 2121 ∑∑ === νµ νµ νµ νµ νµ νµ µ µ xxg sd xd sd xd gF sd xd x  and ∑∑∑ ∑∑ ∂ ∂ − ∂ ∂ += ∂ ∂ −         = ∂ ∂ −         ∂ ∂ = σρ σρ µ σρ νλ νλ λ νµ ν ν νµ σρ µ σρσρ ν ν νµµµ sd xd sd xd x g sd xd sd xd x g sd xd g x g xxxg sd d x F x F sd d 2 1 20 2 2   0 Our Euler-Lagrange equation, ∂F/∂xµ –d[∂F/(dxµ /ds)]/dx=0, becomes: ∑∑ ∂ ∂ = ∂ ∂ = ∂ ∂ σρ µ σρσρ µ ν ν νµµ x g xx x F xg x F   and2
• By using the Christoffel symbols of the first kind, we may write the previous equation in the form: 2014 MRT And since Σρ gρσ Γρ µν = Γµν,σ we use it with inner multiplication with respect to τ: This equation is the required equation of the geodesic in Riemannian space. ∑∑∑∑ Γ+=Γ+= σρµ σρ µσρ τµ νµ ν νµ τµ σρ σρ µσρ ν ν νµ sd xd sd xd g sd xd gg sd xd sd xd sd xd g ,2 2 ,2 2 0 0 2 1 2 2 2 2 =Γ+=         ∂ ∂ − ∂ ∂ + ∂ ∂ + ∑∑ σρ σρ τ σρ τ σρλ σρ λ σρ ρ λσ σ λρλτ τ sd xd sd xd sd xd sd xd sd xd x g x g x g g sd xd As an example, let us find the equation of the geodesic for a three-dimensional Cartesian coordinate system. For Cartesian coordinates in a three-dimensional Euclidean space, we have gij =1 when i =j and gij =0 when i ≠j (i=1,2,3 and Γk ij =0.) The equation of the geodesic in this special case becomes d2 xk /ds2 =0 (k=1,2,3.) If we consider an observer who is traveling with an object that is moving from P1 to P2 (the object is at rest with respect to the observer), ds=dτ where dτ is called the proper time. Since dvk /dτ =0, then vk =dxk /dτ is a constant. On integrating dxk /dτ =A, we obtain BAx k += τ where A and B are constants. This 3-D geodesic is the equation of a straight line!
• Riemann assumed that the quadratic form ds2 =Σµν gµν dxµdx ν defined the metrical properties of the world and should be regarded as a physical reality. However, it was Einstein, in his theory of general relativity, who attached physical significance to gµν by assuming that gµν formed the components of the gravitational potential. 2014 MRT First, let us consider the covariant derivative of an arbitrary covariant vector Aρ: Einstein assumed that the phenomena of gravitation are intimately connected with geometry and that the laws by which matter (which has it’s own ‘Energy- Momentum’ tensor, Tµν , associated to it’s material – mass density, cosmological or stellar pressure, stress related warp effects, &c – properties and spacetime distribution) affects measurements are the laws of gravitation. Here it must be noted that the gravitational potential φ (or the 16 ‘potentials’ gµν ) will then have an invariant quadratic differential form, while electromagnetic phenomena are governed by a potential (four- potential), Aµ, which has an invariant linear differential form given byΣµ Aµ dxµ . The existence of these two separate invariant forms is the source of the difficulty involved in developing a theory which can even unify gravitational and electromagnetic phenomena. ∑Γ− ∂ ∂ = ρ ρ ρ νµν µ νµ A x A A ; By using this again we see that: ∑∑ Γ−Γ− ∂ ∂ = ρ ρµ ρ νλ ρ νρ ρ µλλ νµ λνµ ;; ; ;; AA x A A
• σ σ σ λρ ρ µν ρ σ νρ ρ µλλ σ νµ ν σ λµ νλµλνµ A xx AA ∑ ∑         ΓΓ−ΓΓ+ ∂ Γ∂ − ∂ Γ∂ =− )(;;;; If we substitute Aµ;ν (with the appropriate change of indices) into Aµ;ν ;λ , we get: 2014 MRT Subtracting the latter from the former we obtain the result: ∑∑∑∑∑ ∑∑∑∑∑ ΓΓ+ ∂ ∂ Γ−ΓΓ+ ∂ ∂ Γ− ∂ Γ∂ + ∂∂ ∂ − ΓΓ+ ∂ ∂ Γ−ΓΓ+ ∂ ∂ Γ− ∂ Γ∂ − ∂∂ ∂ =− σρ σ σ ρµ ρ λν ρ ρ µρ λν σρ σ σ λρ ρ µν σ λ σσ µν σ σν σ λµ λν µ σρ σ σ ρµ ρ νλ ρ ρ µρ νλ σρ σ σ νρ ρ µλ σ ν σσ µλ σ σλ σ νµ νλ µ νλµλνµ A x A A x A A xxx A A x A A x A A xxx A AA 2 2 ;;;; Rearranging like terms and using the symmetry property of the Γs (i.e., Γρ µν =Γρ νµ) we obtain an important result: ∑ ∑∑ ∑∑         Γ− ∂ ∂ Γ−         Γ− ∂ ∂ Γ−         Γ− ∂ ∂ ∂ ∂ = ρ σ σ σ ρµρ µρ νλ ρ σ σ σ νρν ρρ µλ ρ ρ ρ νµν µ λλνµ A x A A x A A x A x A ;; ∑ ∑∑ ∑∑         Γ− ∂ ∂ Γ−         Γ− ∂ ∂ Γ−         Γ− ∂ ∂ ∂ ∂ = ρ σ σ σ ρµρ µρ λν ρ σ σ σ λρλ ρρ µν ρ ρ ρ λµλ µ ννλµ A x A A x A A x A x A ;; Similarly we obtain:
• Second, since Aµ;ν ;λ − Aµ;λ ;ν is a tensor and Aσ is an arbitrary tensor, the expression in brackets is a mixed tensor of contravariant rank one and covariant rank three (fourth rank mixed tensor) which can be conveniently expressed as: 2014 MRT Expanded to show the gµν s we get: )()( σ λρ ρ µν ρ σ νρ ρ µλλ σ νµ ν σ λµσ νµλ ΓΓ−ΓΓ+ ∂ Γ∂ − ∂ Γ∂ =Γ ∑xx R where the Riemann-Christoffel tensor, is given by a function of the Γρ µν: ∑=− σ σ σ νµλνλµλνµ ARAA ;;;; A student once said: ‘This Riemann curvature stuff is really hard!’ I agree. Imagine how hard it is to compute this tensor in four dimensions (e.g., xµ =[ct,r,θ,ϕ])! But in taking one’s time and using a few simplifications to come, the Riemann-Christoffel tensor is computed easily while taking a long time to compute all 64 components of Rσ λµν .             ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ −             ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ +     ∂∂ ∂ + ∂∂ ∂ − ∂∂ ∂ − ∂∂ ∂ − ∂∂ ∂ + ∂∂ ∂ + +     ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∑ ∑ τ λρ τ µν ρ τλ τ µν λ τρ τ µν τ λρ ν τµ ρ τλ ν τµ λ τρ ν τµ τ λρ µ τν ρ τλ µ τν λ τρ µ τντστρ τρ τ νρ τ µλ ρ τν τ µλ ν τρ τ µλ τ νρ λ τµ ρ τν λ τµ ν τρ λ τµ τ νρ µ τλ ρ τν µ τλ ν τρ µ τλτστρ τλ νµτσ µλ τντσ νλ τµτσ τν λµτρ µν τλτρ λν τµτρ τ τ νµ λ τσ µ τν λ τσ ν τµ λ τσ τ λµ ν τρ µ τλ ν τρ λ τµ ν τρ σ νµλ x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g gg x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g x g gg xx g g xx g g xx g g xx g g xx g g xx g g x g x g x g x g x g x g x g x g x g x g x g x g gR 4 1 2 1 )( 222222
• The Riemann-Christoffel tensor is formed exclusively from the fundamental tensor gµν . If a coordinate system is selected such that gµν are constants, then all components of the Riemann-Christoffel tensor vanish in this system and all other systems. The curvature tensor, Rκλµν , is obtained by use of the inner product of the fundamental tensor gκσ and the Riemann-Christoffel: 2014 MRT Contracting the Riemann-Christoffel tensor with respect to the σ and ν indices, we obtain the Ricci tensor: ∑= σ σ νµλσκνµλκ RgR ∑ ∑∑         ΓΓ−ΓΓ+ ∂ Γ∂ − ∂ Γ∂ == σ σ λρ ρ µσ ρ σ σρ ρ µλλ σ σµ σ σ λµ τσν σ νµλκ τν τσµλ )( xx RggR Explicitly, displaying the dependence on the gµν s, the curvature is given by: ∑ ΓΓ−ΓΓ+         ∂∂ ∂ + ∂∂ ∂ − ∂∂ ∂ − ∂∂ ∂ = σρ σ µλ ρ κν σ νλ ρ κµσρκµ νλ λµ νκ κν µλ λν µκ νµλκ )( 2 1 2222 g xx g xx g xx g xx g R
• We conclude this section on Tensor Calculus by writing down Einstein’s set of equations for the general relativistic theory of gravitation; they are: 2014 MRT where Gµν is called the Einstein tensor, Rµν is the Ricci tensor, gµν is the fundamental metric tensor, the curvature scalar R is given by R=Σµν gµν Rµν , Tµν is the Energy- Momentum Tensor (in matter-free space, Tµν =0) and κ is Einstein’s universal constant (κ =8π G/c4 .) νµνµνµνµ κ TRgRG =−= 2 1 νµνµνµνµ T c G gRgR 4 π8 2 1 =Λ+− Shortly after presenting the General Theory of Relativity to the world in 1916, Einstein was presented with data that could suggest the world was expanding. To meet with the requirements (at the time) that a universe must not grow (i.e., static), Einstein (1917) introduced the concept of a cosmological constant Λ that could be set to put on the brakes on an expanding universe. The resulting tensor equation ended up being: Einstein abandoned the concept as his ‘greatest blunder’ after Hubble’s (1889-1953) 1929 discovery that all galaxies outside our own Local Group are expanding away from each other, implying an overall expanding Universe. Now, however, the cosmological constant has the same effect as an intrinsic energy density of the vacuum, ρvac.
• Exercise: The four-dimensional spherical coordinates dsµ =[d(ct),dr,dθ,dϕ] form a curvilinear but orthogonal coordinate system with the orthonormal vectors êµ (µ=(ct,r,θ,ϕ) and note that in the following formulas repeated indices are not summed.) 2014 MRT 2222 22222 )sin(1 sin1 sin1ˆˆ ˆsinˆˆˆˆ −−−− ===== ===== ===•==•= ====⇒= ∂ ∂ = θ θ θ θ ϕϕθθ ϕϕθθ ϕθ ϕϕθθµµµµ rgrggchg rgrggchg rhrhhch rrch x rr t tt rrttt rrrttt rrtt ,,, and ,,, ;and,, ;and,, nn nnnn ds n ee eeeee a) From the line element ds=c2 dtêt +drêr +rdθêθ +rsinθêϕ , show that: ;and,, θ ϕϕθθ µ µµ sin ˆˆ ˆˆ ˆ rr c h r r t t ee ee e ====⇒= nnnnn 12 =•=•=•=• ϕ ϕ θ θ nnnnnnnn r r t t c and with ;22222222 sin ϕθθ drdrtdcsd ++=•= dsds and gives:
• b) For êi (i=r,θ,ϕ) show that the surface and volume elements are: 2014 MRT ϕθθϕθτ θ θσϕθσϕθθσ ϕθ ϕθθϕϕθ ddrdrddrdgd rgg g drdrdrddrdddrd ji r rrr sin sindet ;ˆˆsinˆsin 2 24 2 == == =• === ; , and, nnn × eee b) The affine connections can be calculated readily by using the Christoffel symbols of the first kind. However, you can practice your mathematical skill in curvilinear coordinates by verifying the following results for spherical coordinates directly from the basis vectors, ni, ni (i=r,θ,ϕ) expressed in terms of the unit vectors êi in spherical and Cartesian coordinates: ϕθϕϕθϕϕϕ θθθϕϕθθ θθ θθθ ee eeee ˆcosˆsin ˆ)ˆsinˆ(cossinˆ r rr rr rryxr =∂=∂=∂=∂ =∂=∂+−=∂−=∂ nnnn nnnn and ,, All other members of these two groups of symbols are zero. 4-D case is trivial now. So we finally get: θθθθ θ θ θ θ ϕϕϕϕ ϕ θϕ ϕ ϕθ ϕ ϕ ϕ ϕ θ θ θ θθθ 2sin 2 1 cossinsin cot sin cos1 2 −=−=Γ−=Γ ==Γ=Γ=Γ=Γ=Γ=Γ−=Γ and ,,, r r r r rrrr r
• 2 m/swhere 81.92 Earth apple 2 Earthapple ==== = ⊕ ⊕⊕ ⊕ R M GggmWF R Mm GF g g According to Einstein, an observer in an enclosed box in the middle of space would not be able to distinguish between the external effects of gravity applied to the box and motion applied to the box (since there are no windows). Newton cannot explain this… µνµν T c GG 4 π8 = Sir Isaac Newton’s view of gravity: Every particle in the universe attracts every other particle with a force F that is directly proportional to the product of their masses m and M and inversely proportional to the square of the distance R between them (incl. Earth’s). Albert Einstein’s view of gravity: Inertial motion occurs when objects are in free-fall instead of when they are at rest with respect to a massive object such as the Earth. Spacetime Geometry Gµν is curved by the presence of matter Tµν, and free-falling objects are following the spacetime geodesics. In physics, gravitation (or gravity) is the tendency of objects with mass to accelerate toward each other. Gravitation is 1040 times weaker than the electromagnetic force but it acts over great distances. Contrary to identical charges pushing each other (as evident when doing electrostatic experiments using a hair comb) it is always attractive. 2014 MRT
• TG κµνµνµνµν =⇔=−= T c G RgRG 4 π8 2 1 Gµν (or G ) is the Einstein Tensor which represents the geometry of the gravitational field in spacetime, and Tµν (or T ) is the Energy-Momentum Tensor (e.g., represents the distribution of energy and momentum (and stress) within the gravitational field). κ is Einstein’s universal constant. Generally, for a perfect fluid: where G is Newton’s Gravitational Constant (e.g. characterizing the strength of gravitation) and c is the speed of light in the vacuum. Uµ and Uν are the velocity quadri- vectors, ρ is the density of the perfect fluid and p is its pressure, and gµν represent the components of the space-time metric. The term −pgµν above is a new energy term which factors in the pressure exerted by a perfect fluid present within a gravitational field. Matter tells spacetime how to curve – Curved spacetime tells matter how to move. J.A. Wheeler Conclusion: an object moves along the shortest path between two points in spacetime. Geometry = [Einstein’s Universal Constant] × Energy In a more simpler way, Einstein’s equation mean (G for Geometry and T for Energy): µννµµν ρ pgUUpT −+= )( 2014 MRT This is Einstein’s General Relativity equation:
• Geometry = 0 (= λ gµν with cosmological constant λ ) When no matter is present – spacetime is flat (i.e., no curvature) and Einstein’s Equations reduce to their simplest form : κ κκκκκ κ κκκ x g g x g g x g g x g x g x g g ∂ ∂ − ∂ ∂ + ∂ ∂ =      ∂ ∂ − ∂ ∂ + ∂ ∂ =Γ νµλ ν µλ µ νλνµ ν µ µ νλλ µν 2 1 2 1 2 1 2 1 where These represent a complicated set of 10 non-linear partial differential equations of the second degre which are coupled in the components of the metric of spacetime gµν . g x g x − ∂ ∂ =      ∂ ∂ =Γ=Γ lnln 2 1 νλλν λ λν λ νλ λ λ x g g ∂ ∂ −=Γ µµ λ λ µµ 1 2 1 0=ΓΓ−ΓΓ+ ∂ Γ∂ − ∂ Γ∂ == λ λη η η η λ λλ λ µν λ νµλ µν ν µλ µλνµν xx RR Contraction on the Riemann-Christoffel Tensor Rσ µνκ above with the index σ and ν (i.e., replacing them both with λ then relabling κ for ν ) we obtain the Ricci Tensor : or 0=µνR 2014 MRT
• The circumferential radius of the event horizon is known as the Schwarzschild radius. If we put in the appropriate numerical values we find that the Schwarszchild radius of the Sun is RS = 3 Km. In other words, if the Sun were compressed into a radius smaller than this, light could no longer escape from its surface and it would become invisible (a black hole). If we play the same game with our own planet we find that the Earth’s Schwarzschild radius is R = 9 mm – yes, barely a centimeter (half an inch) radius! 2 2 c GM RS = Schwarzshild geometry defined by a radius r and a mass M. Schwarzshild’s Radius RS is defined as the radius for a given mass where, if that mass could be compressed to fit within that radius, no force could stop it from con- tinuing to collapse into a gravitational singu- larity and is a function of the mass M : 2014 MRT sin11 22222 1 222 ϕθθ ddrdr r R dtc r R ds SS +−      −−      −= − 2 r
• 222 ddds +=             +             = τ elementline theof variation malInfinitesi timeproper theof variation malInfinitesi       ++ − +−= 22222 2 2 222 sin 1 )()( ϕθθ drdr kr dr tRcdtds A 4-dimensional metric for space-time is a generalization of Pythagoras’ Theorem: 2 00 222 )()1,2,3)()()( tdgdidgd ii ===′ τ(and d′ The expansion of spacetime (The Big Bang) can be repre- sented by the geometry above. The differential lengths change from d to d′ as space-time changes by a proper time dτ . The coefficients of the Robertson Metric are: and Choosing k for the proper geometry (closed, open or hyperbolic) allows us to find the manner in which lengths and times change over eons. With µ,ν = [i,t]:                 − − = 2 222 22 2 2 000 0sin)(00 00)(0 000 1 )( c rtR rtR kr tR g θ νµ ν = 1 ν = 2 ν = 3 ν = 0 µ = 0µ = 3µ = 2µ = 1 gµν dr kr tR         − 2 1 )( R(t) σ (t)σ (to) ∆σ (t) ∆σ (to) R(t) = 2014 MRT
• )sin( 2 1 2 1 22222 1 2 22 2 2 ϕθθ ddrdr rc GM dtc rc GM ds +−      −−      −= − Another solution gives the gravitational field outside a spherically symmetric star: This solution to Einstein’s Equation hinted at the existence of black holes (term coined in the 1967). Notice how the r in the denominator suggests that as r→0, 2GM/c2 r →∞.       ++ − +−= 22222 2 2 2222 sin 1 )()( ϕθθ drdr kr dr tRdtctds A familiar solution is given by the Robertson-Walker solution to Einstein’s Equations: It represents a solution in the ‘ideal’ case where the whole of spacetime is (viewed say from Earth) is isotropic (i.e., looks the same in all directions) and homogeneous (i.e., no clusters or voids.) The scaling is now governed (a function of universal time) by R(t). To some extent, the scalar curvature represents the class of curvature under consideration. For example, consider the general arbitrary solution given by      −= = += = )1for(sinh )0for( )1for(sin )( k k k f χ χ χ χ )sin()( 222222 ϕθθχχσ ddfdd ++= This scaling function f (χ) depends on the sign of the curvature constant k (+1, 0 or −1): 2014 MRT
• Einstein’s theory of General Relativity provided the framework to understand space- time better and allow Physicists a path to understanding how the universe evolved that is supported by data and proof. This data supports the prediction made by the age of the universe to solutions provided by the Robertson-Walker model. ∫               Λ +Ω−+Ω= − 1 0 21 2 2 o oo o o 3 )1( 11 χ χ χ H d H t c o o ρ ρ =Ω Using the Robertson Metric, we can find the age of the universe: where the ratio between observed density and critical density is given by the density parameter : With a ‘zero’ cosmological constant (i.e., for λ=Λ/8πG=0), we get for Ωo =1 (i.e., ρo =ρc): 19 1 0 23 o 1 0 21 o 1 0 oo 21 o o )1018( 1 3 2 23 11 )1( 1 − × =         == Ω−+Ω = ∫∫ yr χ χχ χ χχ H d H d H t to = 12×109 yr which, theoretically, is comparable with to(WMAP 2009) = 13.7×109 yr. assuming Hubble’s constant to be Ho = [dR(t)/dt]/R(t) = 55 km/s/Mpc = 18×109 yr. 2014 MRT
• M is the mass, α is the angular momentum per unit mass (J/Mc) and Q, the charge. 22 S 2 QrRr ++−=∆ α θαρ 2222 cos+= r ϕ ρ θα ϕθθα ρ αθρ ρ ρ ddtc rR d rR rddrdtc rR ds SSS 2 2 2222 2 22222 2 22 2 2 sin2 sinsin1 +        ++−− ∆ −        −=                               −         ++− − ∆ − = 2 22 2 2 2 222 2 22 2 2 1 sin 00 sin sinsin00 000 000 c rR c rR c rRrR rg SS SS ρρ θα ρ θα θθα ρ α ρ ρ µν The coefficients of the Kerr Metric are: with the parameters: Given a spherically symetric grid around the black hole with time-space coordinates ct, r, θ, and ϕ : 2014 MRT
• Geodetic Effect + Frame-dragging       −•⊕= ⊕⊕ ⊕⊕ ωRω R vRΩ )( 3 )( 2 3 23232 RRc GJ Rc GM × 2014 MRT Evidence of frame-dragging and geodetic effects are confirmed using the Kerr Metric. Y Z X R ~ 650.4 km v⊕ = ~1675 km/h ω⊕ Ω
• Planck’s Lλ ∝Τ distribution ∴Luminescence [W/m²/sr ] is the is the heat flux density or irradiance [W/m2 ] ; per solid angle [Sr]. Energy [kJ] is work during a period of time [kW⋅s] over a given wavelength range [nm]. Of the three primary colors, green is most luminous, followed by red then blue. Luminescence L is the intensity of a source in a given direction divided by the apparent area in the same given direction (i.e., a unit of luminescence is the the watt per square meter per stradian – W/m²/sr) in MKS units. Planck’s Law defines the distribution of the luminosity of monochromatic electromagnetic (i.e., luminance) of thermal radiation of a black body as a function of the thermodynamic temperature T. where cλ = c/nλ is the velocity of the electromagnetic radiation in a medium supporting the propagation, the refraction index of the medium, nλ, the speed of light in free space (c = 299,792,458 m/s), Planck’s constant (h = 6.62617×10−34 J•s), Boltzmann’s constant (k = 1.38066×10 −23 J/K) and T is the temperature at the surface of the black body (in degrees kelvin [K = °C − 273.15]). Lλ is a flux of radiation energy (i.e., the power or luminosity ∴ [W]) per unit area ( A ∴ [m2 ] ), per unit of solid angle (dΩ ∴[sr] ) and per unit of wavelength (λ ∴[m] ); expressed in units of:         − = 1e 21 2 5 Tkhc hc L λλ λ λ λ [W/m2 /sr/m] Surface temperature of the Sun – 5780K: YELLOW ! 2014 MRT
• C. Harper, Introduction to Mathematical Physics, Prentice Hall, 1976. California State University, Haywood This is my favorite go-to reference for mathematical physics. Most of the complex variable and matrix definitions, &c. are succinctly spelled out and the wave packet discussions served as a primer followed by more in-depth discussion reference of Appendix A of Sakurai’s book below and also the solutions to the wave function (i.e. Radial, Azimutal and Angular) are very well presented and treated in this very readable 300 page volume. J.J. Sakurai, Modern Quantum Mechanics, Revised Edition, Addison-Wesley, 1994. University of California at Los Angeles Only half (Chapters 1-3 or about the first 200 pages) of this book makes for quite a thorough introduction to modern quantum mechanics. Most of the mathematical and angular momentum framework (i.e. orbital, spin and total as well as spherical harmonics discussed earlier) is based on Sakurai’s post-humanous presentation of the subject. S.S. Schweber, An Introduction to Relativistic Quantum Field Theory, Dover, 1989. Brandeis University Quite a voluminous 900 page tome written by a colleague of Hans Bethe. Most of the postulates and rotation invariance discussion, which was really the base or primer for the rest, is based on the first 50 pages of this book. The other 50 pages formed the basis of the relativistic material as well as the Klein-Gordon and Dirac equations. S. Weinberg, Lectures on Quantum Mechanics, Cambridge University Press, 2014. Josey Regental Chair in Science at the University of Texas at Austin Soon to become the seminal work on Quantum Mechanics, this ‘course’ from Weinberg fills-in all the gaps in learning and exposition required to fully understand the theory. Including the Historical Introduction, I would definitely suggest Chapter 2-4 and do the problems! Once you do, you will have truly mastered the theory to handle the next set of books from Weinberg – especially QFT I & II. S. Weinberg, The Quantum Theory of Fields, Volume I, Cambridge University Press, 1995. Josey Regental Chair in Science at the University of Texas at Austin Volume 1 (of 3) introduces quantum fields in which Chapter 1 serves as a historical introduction whereas the first 5 sections of Chapter 2 make for quite an elevated introduction to the relativistic quantum mechanics discussed here. Weinberg’s book was crucial in setting up the notation and the Lorentz and Poincaré transformations and bridging the gap for the relativistic one-particle and mass zero states using Wigner’s little group. Very high level reading! M. Weissbluth, Atoms and Molecules, Academic Press, 1978. Professor Emeritus of Applied Physics at Stanford Weissbluth’s book starts by knocking you off your chair with Angular Momentum, Rotations and Elements of Group Theory! The text is mathematical physics throughout but where more than one sentence occurs, the physics (or quantum chemistry) abounds. Chapter 15 starts Part III – One-electron Atoms with Dirac’s equation and in Chapter 16, it offers most of 2014 MRT References / Study Guide
• And now, let us enjoy some humor… As quoted by Sidney Coleman*: Class: Why not use Feynman’s lecture notes? Murray Gell-Mann: Because Feynman uses a different method than we do. Class: What is Feynman’s method? Murray Gell-Mann: You write down the problem. Then you look at it and you think. Then you write down the answer. *Murray Gell-Mann (1929-) is an American physicist who received the 1969 Nobel Prize in physics for his work on the theory of elementary particles. Sidney Richard Coleman (1937-2007) was an American theoretical physicist who studied under Murray Gell- Mann. 0.029−= +−=                   = +++++= ++++++=++++= +−=+= ∑ ∞ = 969.01002.0 969.0 03125.006250.012500.025000.050000.0 64 1 32 1 16 1 8 1 4 1 2 1 2 1 2 1 2 1 2 1 2 1 )1e(002.0 4321 1 π   n n i
• GETA HALF-LIFE! th yty )/2(ln e)0()( − =
• And God said ∫∫∫ ∫∫∫ ∫∫ ∫∫ • ∂ ∂ += Φ +=• • ∂ ∂ −= Φ −=• =• =• SC SC S S d t I td d Id d ttd d d d Q d S E B S B E SB SE E B oooooo o 0 εµµεµµ ε   and there was