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NPTEL – Physics – Mathematical Physics - 1
Lecture 2
Physical examples of gradient
1. An electric dipole moment 𝑝⃗ located at origin, creates a potential at π‘Ÿβƒ— given
by,
𝑝.π‘Ÿβƒ—
V(π‘Ÿβƒ—) = K , where K is a constant.
π‘Ÿ3
Find the electric field 𝐸⃗⃗ = - βˆ‡βƒ—βƒ— V (π‘Ÿβƒ—). Where βˆ‡βƒ—βƒ— is an operator given by βƒ—βˆ‡βƒ—= ( iΛ† 𝑑
+ Λ†j 𝑑
+ kΛ† 𝑑
)
𝑑π‘₯ 𝑑𝑦 𝑑𝑧
Solution
𝐸⃗⃗ = βˆ’π‘ (π‘₯Μ‚
𝑑
+ 𝑦̂
𝑑
+
𝑑
)
𝑑π‘₯ 𝑑𝑦 𝑑𝑧
(π‘₯ + 𝑦 + 𝑧)
(π‘₯2 + 𝑦2 + 𝑧2) ⁄2
3
The π‘₯Μ‚ component yields,
= βˆ’π‘ [
1
(π‘₯2 + 𝑦2 + 𝑧2) ⁄2 (π‘₯2 + 𝑦2 + 𝑧2) ⁄2
3 5
βˆ’
3π‘₯2
] xΜ‚
Putting the contribution from yΜ‚ and zΜ‚
𝐸⃗⃗ =
3𝑝
(π‘₯2 xΛ† + 𝑦2 yΛ† + 𝑧2 zΛ† )
βˆ’
𝑝 rΛ†
π‘Ÿ5 π‘Ÿ3
2. For a position vector π‘Ÿβƒ—, calculate βˆ‡βƒ—βƒ—π‘Ÿπ‘› and use this result to calculate
βƒ—βˆ‡βƒ—( ).
1
π‘Ÿ
Solution
βˆ‡βƒ—βƒ—π‘Ÿπ‘› = ( π‘₯Μ‚ 𝑑
+
yΛ† 𝑑π‘₯
𝑑
𝑑𝑦 𝑑𝑧
+ zΜ‚
𝑑
)π‘Ÿπ‘›
= π‘₯Μ‚ π‘›π‘Ÿπ‘›βˆ’1 π‘‘π‘Ÿ
+ yΛ† π‘›π‘Ÿπ‘›βˆ’1 π‘‘π‘Ÿ
+ 𝑧̂ π‘›π‘Ÿπ‘›βˆ’1 π‘‘π‘Ÿ
𝑑π‘₯ 𝑑𝑦 𝑑𝑧
= π‘›π‘Ÿπ‘›βˆ’1[ xΛ† π‘‘π‘Ÿ
+
yΛ† 𝑑π‘₯
π‘‘π‘Ÿ
+ zΜ‚ π‘‘π‘Ÿ
]
𝑑𝑦 𝑑𝑧
Using π‘Ÿ2 = π‘₯2 + 𝑦2 + 𝑧2
π‘‘π‘Ÿ π‘₯ π‘‘π‘Ÿ
𝑑π‘₯ π‘Ÿ 𝑑𝑦 π‘Ÿ 𝑑𝑧 π‘Ÿ
= , = , =
𝑦 π‘‘π‘Ÿ 𝑧
βˆ‡βƒ—βƒ—π‘Ÿπ‘› = π‘›π‘Ÿπ‘›βˆ’2 (π‘₯ xΛ† + 𝑦 yΛ† + 𝑧
zˆ )
= π‘›π‘Ÿπ‘›βˆ’2π‘Ÿβƒ—
Joint initiative of IITs and IISc – Funded by MHRD Page 12 of 32
NPTEL – Physics – Mathematical Physics - 1
To calculate βˆ‡ ( ), put n =-1
βƒ—
βƒ—
1
π‘Ÿ
βˆ‡βƒ—βƒ—
(
1 π‘Ÿ
) =
π‘Ÿβƒ—
π‘Ÿ3
The potential due to a point change goes as , so electric field goes as
1
π‘Ÿ
π‘Ÿβƒ—
π‘Ÿ3
Exercise
Two identical point charges of magnitude Q at (1, 0, 0) and (-1, 0, 0). The potential in the xy
plane is given by,
V(x, y, z =0) = [ +
𝑄 1 1
4πœ‹πœ–0 √(π‘₯βˆ’1)2+𝑦2 √(π‘₯+1)2+𝑦2 ]
Find the electric field using, 𝐸⃗⃗ = - 𝛻⃗⃗𝑉.
Formal definition and properties of a Gradient
Thus gradient represents a direction that is perpendicular to the surface 𝑔(π‘Ÿβƒ—). βƒ—βˆ‡βƒ—π‘”(π‘Ÿβƒ—) = 0
denotes a minimum or a maximum or a saddle point (saddle point is defined as the one that is minimum in
one direction, but is maximum in the other direction.
Let us consider a scalar function πœ‘(π‘₯, 𝑦, 𝑧). A change in πœ‘ is given by,
π‘‘πœ‘ = (πœ•πœ‘
) 𝑑π‘₯ + (πœ•πœ‘
) 𝑑𝑦 + (πœ•πœ‘
) 𝑑𝑧 βˆ—
πœ•π‘¦ πœ•y πœ•z
Example
Find βˆ‡βƒ—βƒ— πœ‘ and |βˆ‡βƒ—βƒ— πœ‘| for πœ‘ = 2π‘₯𝑧4- π‘₯2y at the point (2,-2,-1). Solution
πœ‘ = 2π‘₯𝑧4 - π‘₯2𝑦
βˆ‡βƒ—βƒ—πœ‘ = xΛ† [2𝑧4 βˆ’ 2π‘₯𝑦] + yΛ† [βˆ’π‘₯2] + zΛ† [8π‘₯𝑧3]
|βƒ—βƒ—βˆ‡βƒ—βƒ— πœ‘|(2,βˆ’2,βˆ’1) = 10 xΛ† βˆ’4𝑦 βˆ’ 16𝑧3
Then |βˆ‡βƒ—βƒ—πœ‘| = 2√93
Joint initiative of IITs and IISc – Funded by MHRD Page 13 of 32
NPTEL – Physics – Mathematical Physics - 1
Exercise problem
The above equation can be written as,
π‘‘πœ‘ = βˆ‡βƒ— πœ‘. 𝑑𝑙 with βˆ‡βƒ— πœ‘ = xΛ† πœ•πœ‘
+ yΛ† πœ•πœ‘
+ zΛ† πœ•πœ‘
and 𝑑𝑙 is a differential length
element
πœ•π‘₯ πœ•π‘¦ πœ•π‘§
given by,
𝑑𝑙 = π‘₯Μ‚ 𝑑π‘₯ + 𝑦̂ 𝑑𝑦 + 𝑧̂ 𝑑𝑧
The βˆ‡βƒ— πœ‘ is a vector where magnitude is the maximum value of π‘‘πœ‘
and whose direction is
the 𝑑𝑙
direction in which is maximum.
Joint initiative of IITs and IISc – Funded by MHRD Page 14 of 32
π‘‘πœ‘
𝑑𝑙
Exercise
Find a unit vector perpendicular to the surface π‘₯2 + 𝑦2 + 𝑧2 = 1 at the point (4,2,3).
Before we wind up our discussion on gradient, we wish to remind the reader that gradient of a function πœ‘,
that is βˆ‡βƒ—βƒ—πœ‘ is perpendicular to the surface πœ‘ = constant. Thus, considering a physical situation that four
sound boxes are located at four corners of the room, each emitting a music of different frequency and
intensity, if we ask a question that what is the distribution of sound intensity in the room. Gradient of this
intensity function will tell you the direction along which the change is most rapid.

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lec2.ppt

  • 1. NPTEL – Physics – Mathematical Physics - 1 Lecture 2 Physical examples of gradient 1. An electric dipole moment 𝑝⃗ located at origin, creates a potential at π‘Ÿβƒ— given by, 𝑝.π‘Ÿβƒ— V(π‘Ÿβƒ—) = K , where K is a constant. π‘Ÿ3 Find the electric field 𝐸⃗⃗ = - βˆ‡βƒ—βƒ— V (π‘Ÿβƒ—). Where βˆ‡βƒ—βƒ— is an operator given by βƒ—βˆ‡βƒ—= ( iΛ† 𝑑 + Λ†j 𝑑 + kΛ† 𝑑 ) 𝑑π‘₯ 𝑑𝑦 𝑑𝑧 Solution 𝐸⃗⃗ = βˆ’π‘ (π‘₯Μ‚ 𝑑 + 𝑦̂ 𝑑 + 𝑑 ) 𝑑π‘₯ 𝑑𝑦 𝑑𝑧 (π‘₯ + 𝑦 + 𝑧) (π‘₯2 + 𝑦2 + 𝑧2) ⁄2 3 The π‘₯Μ‚ component yields, = βˆ’π‘ [ 1 (π‘₯2 + 𝑦2 + 𝑧2) ⁄2 (π‘₯2 + 𝑦2 + 𝑧2) ⁄2 3 5 βˆ’ 3π‘₯2 ] xΜ‚ Putting the contribution from yΜ‚ and zΜ‚ 𝐸⃗⃗ = 3𝑝 (π‘₯2 xΛ† + 𝑦2 yΛ† + 𝑧2 zΛ† ) βˆ’ 𝑝 rΛ† π‘Ÿ5 π‘Ÿ3 2. For a position vector π‘Ÿβƒ—, calculate βˆ‡βƒ—βƒ—π‘Ÿπ‘› and use this result to calculate βƒ—βˆ‡βƒ—( ). 1 π‘Ÿ Solution βˆ‡βƒ—βƒ—π‘Ÿπ‘› = ( π‘₯Μ‚ 𝑑 + yΛ† 𝑑π‘₯ 𝑑 𝑑𝑦 𝑑𝑧 + zΜ‚ 𝑑 )π‘Ÿπ‘› = π‘₯Μ‚ π‘›π‘Ÿπ‘›βˆ’1 π‘‘π‘Ÿ + yΛ† π‘›π‘Ÿπ‘›βˆ’1 π‘‘π‘Ÿ + 𝑧̂ π‘›π‘Ÿπ‘›βˆ’1 π‘‘π‘Ÿ 𝑑π‘₯ 𝑑𝑦 𝑑𝑧 = π‘›π‘Ÿπ‘›βˆ’1[ xΛ† π‘‘π‘Ÿ + yΛ† 𝑑π‘₯ π‘‘π‘Ÿ + zΜ‚ π‘‘π‘Ÿ ] 𝑑𝑦 𝑑𝑧 Using π‘Ÿ2 = π‘₯2 + 𝑦2 + 𝑧2 π‘‘π‘Ÿ π‘₯ π‘‘π‘Ÿ 𝑑π‘₯ π‘Ÿ 𝑑𝑦 π‘Ÿ 𝑑𝑧 π‘Ÿ = , = , = 𝑦 π‘‘π‘Ÿ 𝑧 βˆ‡βƒ—βƒ—π‘Ÿπ‘› = π‘›π‘Ÿπ‘›βˆ’2 (π‘₯ xΛ† + 𝑦 yΛ† + 𝑧 zΛ† ) = π‘›π‘Ÿπ‘›βˆ’2π‘Ÿβƒ— Joint initiative of IITs and IISc – Funded by MHRD Page 12 of 32
  • 2. NPTEL – Physics – Mathematical Physics - 1 To calculate βˆ‡ ( ), put n =-1 βƒ— βƒ— 1 π‘Ÿ βˆ‡βƒ—βƒ— ( 1 π‘Ÿ ) = π‘Ÿβƒ— π‘Ÿ3 The potential due to a point change goes as , so electric field goes as 1 π‘Ÿ π‘Ÿβƒ— π‘Ÿ3 Exercise Two identical point charges of magnitude Q at (1, 0, 0) and (-1, 0, 0). The potential in the xy plane is given by, V(x, y, z =0) = [ + 𝑄 1 1 4πœ‹πœ–0 √(π‘₯βˆ’1)2+𝑦2 √(π‘₯+1)2+𝑦2 ] Find the electric field using, 𝐸⃗⃗ = - 𝛻⃗⃗𝑉. Formal definition and properties of a Gradient Thus gradient represents a direction that is perpendicular to the surface 𝑔(π‘Ÿβƒ—). βƒ—βˆ‡βƒ—π‘”(π‘Ÿβƒ—) = 0 denotes a minimum or a maximum or a saddle point (saddle point is defined as the one that is minimum in one direction, but is maximum in the other direction. Let us consider a scalar function πœ‘(π‘₯, 𝑦, 𝑧). A change in πœ‘ is given by, π‘‘πœ‘ = (πœ•πœ‘ ) 𝑑π‘₯ + (πœ•πœ‘ ) 𝑑𝑦 + (πœ•πœ‘ ) 𝑑𝑧 βˆ— πœ•π‘¦ πœ•y πœ•z Example Find βˆ‡βƒ—βƒ— πœ‘ and |βˆ‡βƒ—βƒ— πœ‘| for πœ‘ = 2π‘₯𝑧4- π‘₯2y at the point (2,-2,-1). Solution πœ‘ = 2π‘₯𝑧4 - π‘₯2𝑦 βˆ‡βƒ—βƒ—πœ‘ = xΛ† [2𝑧4 βˆ’ 2π‘₯𝑦] + yΛ† [βˆ’π‘₯2] + zΛ† [8π‘₯𝑧3] |βƒ—βƒ—βˆ‡βƒ—βƒ— πœ‘|(2,βˆ’2,βˆ’1) = 10 xΛ† βˆ’4𝑦 βˆ’ 16𝑧3 Then |βˆ‡βƒ—βƒ—πœ‘| = 2√93 Joint initiative of IITs and IISc – Funded by MHRD Page 13 of 32
  • 3. NPTEL – Physics – Mathematical Physics - 1 Exercise problem The above equation can be written as, π‘‘πœ‘ = βˆ‡βƒ— πœ‘. 𝑑𝑙 with βˆ‡βƒ— πœ‘ = xΛ† πœ•πœ‘ + yΛ† πœ•πœ‘ + zΛ† πœ•πœ‘ and 𝑑𝑙 is a differential length element πœ•π‘₯ πœ•π‘¦ πœ•π‘§ given by, 𝑑𝑙 = π‘₯Μ‚ 𝑑π‘₯ + 𝑦̂ 𝑑𝑦 + 𝑧̂ 𝑑𝑧 The βˆ‡βƒ— πœ‘ is a vector where magnitude is the maximum value of π‘‘πœ‘ and whose direction is the 𝑑𝑙 direction in which is maximum. Joint initiative of IITs and IISc – Funded by MHRD Page 14 of 32 π‘‘πœ‘ 𝑑𝑙 Exercise Find a unit vector perpendicular to the surface π‘₯2 + 𝑦2 + 𝑧2 = 1 at the point (4,2,3). Before we wind up our discussion on gradient, we wish to remind the reader that gradient of a function πœ‘, that is βˆ‡βƒ—βƒ—πœ‘ is perpendicular to the surface πœ‘ = constant. Thus, considering a physical situation that four sound boxes are located at four corners of the room, each emitting a music of different frequency and intensity, if we ask a question that what is the distribution of sound intensity in the room. Gradient of this intensity function will tell you the direction along which the change is most rapid.