1. St. George’s College 2014 - Mathematical Sciences
Tutorials (Broad Concept Problems)
Daniel Xavier Ogburn ∗
School of Physics,
Field Theory and Quantum Gravity,
University of Western Australia
December 22, 2014
∗
Electronic address: daniel.ogburn@research.uwa.edu.au
1
6. 1 Introduction
At this present moment, the suggested layout for tutorials will be: 20 minutes of
‘broad concept problems’ and 40 minutes of subject-specific help with questions
from your coursework. If the there is a large turnout of students and more time is
necessary, the tutorials will be extend to 30 minutes of ‘broad concept problems’
and 60 minutes of subject-specific help. In addition, students may approach tutors
outside of tutorial times for help with specific coursework problems or concepts –
but they will have to arrange this themselves with the individual tutors.
2 Tutor List
For the year 2014, here is a list of tutors and the respective subjects they are dedi-
cated to. Note that each tutor will probably be able to help you with other mathe-
matics or physics related enquiries.
• Murdock Grewar – PHYS1021
• Tessa McGrath – MATH1711
• Ben Luo – PHYS1001
• Jake Miller – MATH1001
In addition, students with any mathematics, physics or statistics enquiries are wel-
come to seek me for assistance.
3 Broad Concept Problems
These problems are designed to help expose you to important material and concepts
outside the scope of a standard curriculum. They are also designed to help you
think about applications of your mathematical powers to the world at large. In this
manner, the hope is that students will develop a higher level of critical thinking,
logical reasoning and mathematical intuition for investigating different scenarios
and solving new problems.
Note that I will generally aim to cover problems that you wouldn’t usually see in
your lectures – or focus on topics which are (by student and professional experi-
ence) useful and important, but otherwise overlooked or just briefly glossed-over
6
7. in typical university courses. Since the tutorials are targeted at people from both
pure and applied mathematics, or physical and non-physical sciences, the assumed
physical science knowledge will be kept to a minimum. In cases where physics or
engineering examples are used, the prerequisite concepts will be introduced – but
only to emphasize the take-home message.
To get the most out of these tutorials, you should attempt all the broad-concept
problems. Some weeks, we will continue a certain theme from the previous week.
If you can’t finish the problems in the tutorial, or decide to finish them after the
tutorial in your own-time – feel free to ask questions during the week. The tutors
appreciate that students are busy with their coursework and assessed homework
problems, so the broad concept problems should be fairly quick to solve. As an
incentive, doing these ’extra-curricular’ questions should give you an edge over
your rival Tommy Moore, Uni Hall, St. Cats and Trinity students.
4 Tutorial 1 - Dimensional Analysis and the Buckingham
Pi Theorem (part I)
4.1 Prologue: March 15, 2014
Dimensional analysis is a deceptively simple, but fundamentally powerful tool in
the mathematical sciences – one that is often overlooked! There will be a day
when the importance of dimensional analysis is forgotten and lost in the education
system, but today is not that day.
Ultimately, dimensional analysis serves as fast error-checking algorithm for your
calculations. It is also useful for extracting ‘physically meaningful’ information
out of your system. In particular, given a large set of parameters describing a
system, one can often form a smaller number of dimensionless parameters which
completely characterize that system – hence removing any redundant information.
The precise statement of the last idea is known as the ‘Buckingham Pi Theorem’1,
which we shall investigate next week – don’t worry about the formality of the
name, it has vast (but simple) practical applications to fluid mechanics, thermody-
namics, electrodynamics, cosmology and much more. For now, we begin with a
few examples then work through some questions 2.
1
For those of you who have done (or will do) linear algebra, this is just a practical consequence
of the ‘rank-nullity’ theorem.
2
Thanks to Scott Meyer and Matthew Fernandez for feedback
7
8. The main idea of the following examples and problems is two-fold: first inspect
an equation and work out the dimensions (or units) of each variable and constant,
given some starting information. We then check whether or not the equation is
dimensionally consistent. Any equation from any area of science and mathematics
must be dimensionally consistent – if it isn’t, then it’s wrong. In this sense, you
don’t need to understand the science or theory behind an equation to deduce when
it is incorrect on dimensional grounds.
4.2 Examples and Problems
Recall lengths, areas and volumes. The fundamental unit that characterizes these
quantities is length: L. Given a rectangular box, with sides of length a, b, c the
volume is VB = a × b × c. Since each of the sides has the dimensions of length:
[a] = [b] = [c] = L, the volume has dimensions
[VB] =[a × b × c]
=[a] + [b] + [c]
=L + L + L
=3L ,
which we interpret as length-cubed: L3. The notation [ ] is used to denote the
dimensions of whatever quantity is inside the brackets. Notice also, that when we
were looking for the dimensions of a product of variables [a × b × c], we added the
dimensions of each variable: [a×b×c] = [a]+[b]+[c] = L+L+L = 3L.Finally,
we ended up with [VB] = 3L, which means that the volume V has 3 factors of the
unit length L – hence volume V has dimensions of length-cubed L3. Of course,
we already knew this!
Similarly to the multiplication rule, if we are inverting quantities we invert their
units – hence: [1
a] = −[a], [ 1
a2 ] = −[a2] = −2[a], etc. Combining this with the
multiplication rule, we get the division rule: [a
b ] = [a] − [b]. For example, if C
is the concentration of protein in milk, it has units ML−3 of mass over volume –
hence dimensionally: [C] = M − 3L.
Exercise 1 Use the rectangular box example to calculate the dimensions of the
area of a rectangle of sides with length ‘a and ‘b , given the area formula
AR = ab. (1)
Now that we have done some simple problems, lets see how dimensional analysis
can be used for error checking. Lets say someone tells us that the volume VS of a
8
9. sphere of radius R is given by
VS =
4
3
πR2
. (2)
Obviously, this is wrong – but if you’ve forgotten the correct formula, there’s an
easy way to see why it is wrong using dimensional analysis. First of all [R] = L,
since radius has dimensions of length. Furthermore, [4
3π] = 0 since this is just a
pure number (so it is dimensionless). Therefore,
[VS] =[
4
3
πR2
]
=[
4
3
π] + [R × R]
=[
4
3
π] + [R] + [R]
=0 + L + L
=2L.
But wait a minute, volume has units of length cubed, hence [VS] = 3L. We
then conclude by dimensional arguments that the formula VS = 4
3πR2 is incor-
rect!
Although the last example was easy, the same principles can be applied to much
more complicated formulas in the mathematical sciences – indeed, it is used in
research and in practice when doing estimates, checking articles or performing
large derivations and calculations. Lets do one more example.
Example 1 Newton’s Second Law of Motion: Force = Mass × Acceleration, or
F = ma, is the fundamental postulate governing classical physics between the
late 17th and early 19th centuries. It is vastly important today as the law defines
what the force is, for an object of mass ‘m moving with an acceleration ‘a . The
three fundamental units here are mass M, time T and length L. Displacement ‘x
has dimensions of length L, hence velocity ‘v – which is the rate of change of
9
10. displacement 3, has units of length over time:
[v] =[
dx
dt
]
=[dx] − [dt]
=[x] − [t]
=L − T , (3)
hence v has units L
T . Similarly, acceleration a is the rate of change of velocity,
hence
[a] =[
dv
dt
]
=[dv] − [dt]
=(L − T) − T
=L − 2T, (4)
which means ‘a has units of length over time-squared: L
T2 . Finally, mass m triv-
ially has units of mass: [m] = M (note that here we use the capital M to denote
the fundamental unit of mass, where as the lower-case m is mass variable that we
insert into Newton’s 2nd Law). Therefore, force F has the following dimensions
[F] =[m][a]
=[m] + [a]
=M + L − 2T,
(5)
whence F has units of (mass × length)/ (time-squared): ML
T2 .
Exercise 2 Use dimensional analysis to conclude which formulas are incorrect on
dimensional grounds – i.e. which of the following formulas are dimensionally
inconsistent. Show your working.
1. A triangle has a base b and a vertical height h, each with dimensions of
length L. Check whether the following formula for its area is dimensionally
consistent
A =
1
2
b2
h. (6)
3
For those of you unfamiliar with the definition of velocity and acceleration in terms of calculus,
you can think of dx
dt
as the change in displacement x over an ‘infinitesimally small amount’ of time
dt. Then dx carries dimensions of length and dt has dimensions of time: [dx] = L , [dt] = T. Note
that in general, for an arbirtrary quantity y, the ‘infinitesimal quantity’ dy carries the dimensions:
[dy] = [y].
10
11. 2. A circle has a radius r with dimensions of length L. Its area is given by
A =
1
2
πr2
. (7)
Is this dimensionally consistent? A stronger question to ask is whether this
formula is correct – if not, why not?
There is one more rule of dimensional analysis which involves analysing equations
which include a sum of terms. In particular, given a quantity A = B + C + D, to
compute the dimensions [A] of A, we don’t just add the dimensions of B, C and
D:
[A] = [B] + [C] + [D], (8)
but rather, we have the consistency requirement that:
[A] = [B] = [C] = [D]. (9)
This is because B, C and D should all separately have the same units. As such,
this observation is very useful for determining the dimension of multiple unknown
quantities in an equation that involves a sum of different terms. For example, the
area of a toddler house drawing is given by: AHouse = ATriangle + ASquare =
1
2bh + a2, where b is the base length of the triangle, h is its vertical length and
a is the length of the sides of the square. Therefore, [AHouse] = [ATriangle] =
[ASquare] = 2L, hence [1
2bh] = [a2] which implies [b] + [h] = 2[a] = 2L.
One last concept: A dimensionless constant, C, is defined to be a quantity which
has no dimensions – hence [C] = 0. These are fundamentally important in the
description of a physical system since they do not depend on the units you choose.
Thus, in some manner they are represent a ‘universal’ quantity or property – indeed,
the dimensionless constants of a system describe a universality class 4.
To answer the following questions, try not to worry too much about terminology
or new and abstract concepts. We are only interested in dimensions – so if you
stay focused and don’t get distracted by the extra information, you can finish them
quickly with no prerequisite knowledge!
Exercise 3 1. A hypercube living in d dimensions has d sides, each with length
a and dimensions of length L. Its hyper-volume has units of Ld and is given
by the formula
V = aD
. (10)
4
A more precise meaning of this statement can be found in the theory of ‘Renormalization
Groups’.
11
12. Verify that this is dimensionally consistent – i.e. show that [V ] = L + ... +
L = d × L. What dimensions would its surface area have? Hint: this would
be same the as dimensions of the area of one of its ‘faces’.
2. The U.S. Navy invests a significant amount of money into acoustic scatter-
ing studies for submarine detection (SONAR). As part of this research, the
Dahlgren Naval Academy uses ‘prolate spheroidal harmonics’ (vibrational
modes of a ‘stretched sphere’) to do fast, accurate scattering calculations. In
this process, a submarine can be approximated to be the shape of a ‘prolate
spheroid’ or ‘rugby ball’. A prolate spheroid is essentially the surface gen-
erated by rotating an ellipse about its major axis. Given a prolate spheroid
with a semi-major axis length a and semi-minor axis of length b, its volume
is
V =
4π
3
ab2
(11)
Is this formula dimensionally-consistent? What about the following formula
for the surface area (it should have units of length-squared):
S = 2πb2
(1 +
a
be
sin−1
(e))? (12)
Note, sin−1
is the ‘inverse sine’ or ‘arcsine’ function. It necessarily pre-
serves dimensionality, hence [sin−1
(e)] = [e]. The variable e is the ‘ec-
centricity’ of the spheroid. It is a dimensionless quantity: [e] = 0, which
measures how ‘stretched’ the spheroid is – i.e. how much it deviates from a
sphere. It is given by the (dimensionally-consistent!) formula:
e2
= 1 −
b2
a2
. (13)
A perfect sphere corresponds to e = 0, where as an infinitely stretched
sphere corresponds to e → 1.
3. In a parallel-universe, Andrew Forrest has a dungeon with BF flawless black
opals inside it. From a financial point of view, these have dimensions of
money $ – i.e. [BF ] = $. A machine recently designed by Ian McArthur,
head of physics at UWA, uses quantum fluctuations of the spacetime vacuum
to produce black opals at a rate of RUWA black opals per minute. Sensing
the loss of his monopoly on the black opal market, Andrew Forrest employs
a competing physicist at Curtin University to create a quantum vacuum sta-
bilizer. This reduces the number of black opals that Ian can produce per
minute by RC black opals per minute, where |RC|≤ RUWA. Working on
12
13. a broad concept problem, a team of first year students at St. George’s col-
lege come up with the following model to predict the value V of shares in
Forrest BlackOps inc. on the stockmarket as a function of time t (time has
dimensions T):
V = β
D
BF
− λ(RUWA + RC)τDe−λ(1− t
τ
)
(14)
where the constant τ (having dimensions of time T) denotes 5 the time at
which European Union is predicted to collapse. Furthermore, D is a function
that measures the market demand for black opals (with no dimensions) and
β is an economic constant predicted by game theory with units of money-
squared: $2. Finally, λ is a dimensionless parameter (so [λ] = 0) that de-
pends on the number of avocados served at the college since the establish-
ment of St. George’s Avocadoes Anonymous up to the given time t.
Is this model dimensionally consistent – i.e. does [V ] = $?
What about the following formula, proposed by students from St. Catherines
College (who didn’t practice dimensional analysis)?
V =
D
BF
− D2
e−t
(15)
On dimensional grounds, list two reasons why this model incorrect.
4. Bonus Question (Don’t worry about the physics, just keep track of dimen-
sions and rules)
The Harvard-Smithsonian Center for Astrophysics is about to release a press-
conference tomorrow (March 17, 2014), indicating the discovery of gravi-
tational waves. Gravitational waves are ripples through spacetime created
by large gravitational disturbances in the cosmos – for example, exploding
stars and coalescing black-holes. These are predicted by Einstein’s theory
of General Relativity – a theory in which gravity is a simple consequence
of the geometry (shape) of spacetime. In this theory, choosing natural units
for the speed of light: c = 1, time and spatial length become dimensionally
equivalent: T = L. Therefore, dimensionally we have: [time] = [distance]
and [c] = [distance/time] = L − T = 0. A geometry which models gravi-
tational waves is described by the following metric (an abstract object which
tells you how gravity and measures of time and length vary at each point in
spacetime):
g = η + h (16)
5
This is the Greek letter tau – not the Roman letter t.
13
14. where η is a flat-space metric (describing an empty universe):
η := −dt + dx ⊗ dx + dy ⊗ dy + dz ⊗ dz (17)
and h is a symmetric-tensor, given in de-Donder gauge by
h := cos(k · r)A +
1
2
× trace(h) × η. (18)
Here is a small (<< 1) dimensionless parameter: [ ] = 0 and A is a sym-
metric tensor field with dimensions of length-squared: [A] = 2L. Note that
the trace operation turns tensors into scalars, so it removes the dimensional-
ity of a tensor: [trace(h)] = 0. Furthermore, consider · as another form of
multiplication. Since the wave vector k and position vector r have inverse
units, we have [k] = −L, [r] = +L – hence [k · r] = 0. For the purposes
of dimensional-analysis, we can treat the tensor product ⊗ as ordinary mul-
tiplication also. The differential quantities have the following dimensions:
[dt] = [dx] = [dy] = [dz] = L, hence [dx ⊗ dx] = 2[dt] = 2L for
example. Since x, y, z, t represent coordinates in spacetime, we also have
[x] = [y] = [z] = [t] = L.
Show that the metric g demonstrates a dimensionally-inconsistent solution
to the Einstein field equations. Where is the error? Suggest what could be
done to this metric to ‘fix’ it and give a dimensionally-consistent solution.
Remark: If you were certain that the equation for h was correct, it would be
unnecessary to tell you the dimensions of A – you could work it out since you
already know [cos(k · r)] = 0 (the function cos(something) is necessarily
dimensionless). Therefore, pretending [A] is unknown, prove that [A] = 2L
given all the other information.
After completing the last two problems, one should realize that much time can be
saved by ignoring most of the information and concentrating only the dimensions
of the variables and constants in the given formulas. This is true in general! There-
fore, to do dimensional analysis, one need not necessarily understand the science
or mathematics behind an equation – but simply the dimensions of the quantities
involved. Therefore, it is an easy way to show when something is wrong without
knowing what you are talking about. 6
6
Dimensional analysis would have saved the present author about 100 hours of supergravity cal-
culations – time which was largely lost due to two dimensionally-inconsistent equations in a pub-
lished journal article.
14
15. 4.2.1 Moral of the story
Dimensional analysis can tell you when an equation is wrong, but it doesn’t nec-
essarily imply that an equation is correct – even though its dimensions might be
consistent. As a student, you should make use of dimensional analysis whenever
you can – try it on all formulas you get which have dimensionful quantities. This
will help you to gain a strong intuition of whether or not statements and equations
are sensible and consistent. This helps you to be a fast calculator and it will also
help you to pick up errors in your lecture notes ...
5 Tutorials 2 - Dimensional Analysis and the Buckingham
Pi Theorem (part II)
5.1 Background
One of the key concepts in dimensional analysis is that of dimensionless parame-
ters. Dimensionless parameters are important, because they allow you to charac-
terise both physical and theoretical mathematical systems in a scale-invariant way.
Note that mastering the following concepts and exercises requires a good under-
standing of the material in Tutorial 1. For the more mathematically inclined, one
of the examples and exercises illustrates how to mathematically prove the π theo-
rem by using the rank-nullity theorem from linear algebra – this is a good exercise
for understanding matrix equations and the correspondence between matrices and
simultaneous equations! For the applied minds, we use dimensional analysis to in-
vestigate and form dimensionless constants to characterise the harmonic oscillator,
viscous fluids, electromagnetism and Einstein’s theory of gravity.
BIG DISCLAIMER: Notation
Note that for the most part, we have used ‘additive notation’ to denote the dimen-
sions of some quantity – e.g. [Force] = M + L − 2T. However, in engineering
and sometimes in physics 7, you will often see multiplicative notation being used –
meaning F has dimensions ML
T2 . For these tutorials, we have referred to the later as
the ‘units’ of F, rather than its dimensions. Technically speaking, both are correct
– although units typically refer to some standard of measure, such as kilograms
or kg for the standard SI unit of mass. Here we’ve just taken M, L, T to refer to
both dimensions and their respective standard units. After some practice, it should
7
In particle physics and quantum field theory, additive notation is common for computations as it
is the smarter way to do things.
15
16. be easy interchange between conventions – the reason we use additive notation
is that it’s faster to calculate dimensions this way and it is less prone to mistakes
(since you are adding and subtracting instead of multiplying and dividing). Fur-
thermore, additive notation makes it easier to prove things like the π theorem for
dimensional
A physical system in the mathematical sciences typically consists of:
1. A set of physical parameters.
2. A set of governing equations which describe the behaviour or evolution of
the system.
3. A set of fundamental ‘units’ which describe the dimensionality of the sys-
tem.
5.2 Examples and Problems
Example 2 Lets take a simple, but profound 8 example – the simple harmonic
oscillator. One example of a simple harmonic oscillator, is a mass placed on a fric-
tionless tabletop attached to a spring. This string is either stretched or compressed,
then released so that the mass proceeds to undergo simple harmonic motion. This
physical system is therefore described by
1. A set of 4 physical parameters: the spring constant κ and the initial position
x0 and initial velocity v0 of the mass m.
2. An equation of motion called ‘Hooke’s Law’ 9, which says that when you
stretch or compress the spring, the force acting to restore the spring to its
natural length is given by:
F = −κx (19)
where x is the displacement of the mass attached to the spring. Combining
this with Newton’s 2nd Law, F = ma, we get the equation of motion for the
spring:
m
d2x
dt2
= −κx, (20)
where a = d2x
dt2 is the acceleration of the spring.
8
Despite its simplicity, the (quantum) harmonic oscillator is the cornerstone for modern quantum
field theory and particle physics. In this picture, a quantum field is an infinite continuum of simple
harmonic oscillators, whose motion is captured by Fourier theory, Lie algebras and Special Relativity.
9
After the famous pirate, Captain Hooke.
16
17. 3. A set of 3 physical units: mass M, time T, length L (usually kilograms,
seconds, metres).
Now, from these 4 parameters and 3 physical units, I claim that we can form one
dimensionless constant. To do this, one needs to know the dimensions of the pa-
rameters involved. Clearly initial displacement has dimensions of length and initial
velocity has dimensions of length /time: [x0] = L, [v0] = L − T. To work out the
dimensions of the spring constant κ, we inspect the equation of motion.
Since acceleration has dimensions of length over time-squared, we have [d2x
dt2 ] =
L − 2T. Therefore, we have
[m
d2x
dt2
] = [−κx] =⇒
[m] + [
d2x
dt2
] =[κ] + [x]
M + L − 2T =[κ] + L =⇒
[κ] =M − 2T. (21)
Note that the mathematical symbol ‘ =⇒ ’ means ‘implies’. Now that we have the
dimensions of all parameters in this system, we can form a dimensionless product.
In particular, we need one inverse mass factor and two factors of time to cancel the
dimensions in [κ] = M −2T. We can get an inverse unit of mass from [ 1
m ] = −M
and two inverse time units by combining [x0] = L and [v0] = L − T. In particular,
[(x0
v0
)2] = 2[x0] − 2[v0] = 2L − 2(L − T) = 2T. Hence, we get the dimensionless
constant:
G :=
k
m
(
x0
v0
)2
=⇒
[G] =[
k
m
(
x0
v0
)2
]
=[k] − [m] + 2([x0] − 2[v0])
=M − 2T − M + 2T = 0. (22)
Since the constant G has no formal name, we will claim it and call it the ‘Georgian
Constant’ after St. George – the patron saint of dimensional analysis.
The last example illustrated a few important concepts. First of all, we showed
that mathematically all the information about a physical system is giving by a set
of parameters, a set of physical units or dimensions and at least one governing
equation. Second, we showed how we can work the units of an otherwise unknown
constant by using dimensional analysis – this is how we found the dimensions of
the spring constant κ.
17
18. Finally, we showed in this particular case, having 4 parameters and 3 physical
units, we were able to form one dimensionless constant: G . Although we could
have taken any multiple or power of this constant and still arrived at dimensionless
quantity, there essentially only one independent product that we can form out of
the parameters in the simple harmonic oscillator. This is because G, 1
G , G2 or 2G
for example, all contain the same ‘information’.
The last observation is one example of the ‘fundamental theorem of dimensional
analysis’, also known as the ‘π theorem’.
Theorem 1 (Buckingham Pi Theorem) Given a system specified by n indepen-
dent parameters and k different physical units, there are exactly n−k independent
dimensionless constants which can be formed by taking products of the parameters.
Thus in the last example, we saw that the simple harmonic oscillator was described
4 parameters and 3 physical units – hence as claimed, there was indeed only 4−3 =
1 independent dimensionless constant that we could have formed. Hence, any other
dimensionless constant in this system must be some multiple or some power of G.
Before doing the exercises, here is one more example from fluid mechanics.
Example 3 In fluid mechanics, the notion of the ‘thickness’ of a fluid is formalized
by defining its ‘viscosity’. In particular, the dynamic or shear viscosity of a fluid
measures its ability to resist ‘shearing’– an effect where successive layers of the
fluid move in the same direction but with different speeds. For example, relative
to water, glass 10 and honey have a very high shear viscosity, whereas superfluid
Helium has zero viscosity 11.
Given a fluid trapped between two parallel plates–the bottom plate being station-
ary and the top plate moving with velocity v parallel to the stationary plate, the
magnitude of the force required to keep the top plate moving at constant velocity
is given by:
F = ηA
v
y
(23)
Here v is the speed (magnitude of the velocity) of the top plate, A is its surface
area and y is the separation distance between the plate. The parameter η is defined
to be the shear viscosity of the fluid. We can calculate its units using dimensional
analysis. First, from Newton’s 2nd law we know that the force has the dimensions:
[F] = M + L − 2T. Furthermore, the area A has dimensions of length-squared
10
The myth about old church windows sagging is not due to the fact that glass can be modelled as
a viscous liquid, but rather due to the glass-making techniques of past centuries.
11
The transition to the ‘superfluid’ phase occurs below 1 Kelvin – i.e. close to absolute zero
temperature.
18
19. [A] = 2L, the speed v has dimensions [v] = L − T and the separation y has
dimensions [y] = L. Hence
[F] =[η] + [A] + [v] − [y] =⇒
[η] =[F] − [A] − [v] + [y]
=(M + L − 2T) − 2L − (L − T) + L
=M − L − T (24)
whence η has units of M
LT . Now, the kinematic viscosity ν 12 of the fluid is defined
as the ratio of the dynamic viscosity η and the density ρ (mass per volume) of the
fluid:
ν =
η
ρ
. (25)
Since density has units of mass per length-cubed, we have [ρ] = M − 3L and thus
[ν] = [
η
ρ
] = [η] − [ρ] = M − L − T − (M − 3L) = 2L − T. (26)
In some set of scenarios, we can think of this fluid as parameterized by four pa-
rameters: density ρ, shear viscosity η , kinematic viscosity ν and the fluid speed v
(assuming the fluid only travels in the horizontal direction). Since we have three
different physical units – mass, length and time, the Pi theorem tells us we can
form one independent dimensionless constant. This special, widely-used constant
is called the ‘Reynolds number’ of the fluid and is defined by:
R =
ρvl
η
=
lv
ν
(27)
where l is the ‘characteristic length scale’ for the fluid system (e.g. for a fluid
flowing in a pipe, this length scale would be the diameter of the pipe).
In essence, the Reynolds number expresses the ratio of inertial forces to the viscous
forces. In this manner, it describes relative importance of these two types of forces
in different scenarios. Since it is dimensionless, the Reynolds number is scale
invariant – meaning it characterises the way a fluid will flow on all length scales
(within the valid regime of your theory).
Exercise 4 We defined the Reynolds number R in two ways – one in terms of its
dynamic viscosity η and the other in terms of its kinematic viscosity ν. Show that
the Reynolds number is dimensionless using both of its definitions.
12
This is the Greek letter ‘nu - not the Roman letter ‘v’.
19
20. Example 4 (Mathematical Challenge: Proving the π Theorem) Here is a walk-
through of a proof of the Pi Theorem, using the ‘rank-nullity’ theorem from linear
algebra. For those of you who haven’t encountered matrices before, you can still
make sense of the following in terms of systems of linear equations – but that will
be trickier ... so either save it for later, or talk to your tutor.
Formally, the rank-nullity theorem states that given a m × n matrix (m rows, n
columns) A, which maps n-dimensional vectors to m-dimensional vectors, then
the rank and nullity of the matrix A satisfy:
rank(A) + nullity(A) = n (28)
where the rank of A is defined as the number of linearly independent rows of A
and the nullity of A is defined as the dimension of the kernel of A – i.e. the number
of linearly independent n-dimensional vectors which get mapped to 0 by A. Note
that m ≤ n necessarily (or the system is over-determined).
Now, in the context of dimensional analysis and the π Theorem, we can think a
mathematical or physical system with n parameters and k different types of fun-
damental units (dimensions) as a system of k linear equations in n unknowns, as
follows. Say for example, we have three parameters x, y, z and two fundamental
physical units U1, U2. Then we can represent the dimensions of our parameters
as a matrix by letting each column correspond to different parameters and letting
each row correspond to different fundamental units. So in this example, we let the
first column correspond to the parameter x, the second column to y and the third
column to z. Then the first row corresponds to the unit U1 second row to the unit
U2. Then the entry in the first row and column corresponds to the number of di-
mensions of x has in the unit U1. So if for example, x has the units Ua
1 Ub
2 then it
has dimensions: [x] = [Ua
1 ] + [Ub
2] = aU1 + bU2. Similarly, let y have units Uc
1Ud
2
and z have units Ue
1 Uf
2 : hence [y] = cU1 + dU2 and [z] = eU1 + fU2. We can
form the ‘dimensional matrix’ D for this physical system, which is represented as:
D =
a c e
b d f
(29)
To see that this makes sense, we can simply act13 the transpose of the dimensional
matrix DT on the vector U =
U1
U2
containing the physical units to recover all
three of our dimensional equations [x] = aU1 + bU2, [y] = cU1 + dU2 etc. To find
dimensionless constants, we have to solve the ‘nullspace equation’:
13
By matrix multiplication.
20
21. a c e
b d f
!
α
β
γ
(
) =
0
0
for all possible vectors !
α
β
γ
(
). In particular, dimen-
sionless constants will be a product of powers of the different physical parame-
ters: xαyβzγ, where the exponents α, β, γ are components of a vector !
α
β
γ
(
) which
solves the nullspace equation.
The number of linearly independent vectors !
α
β
γ
(
) which solves the null-space ma-
trix equation, coincides with the ‘nullity’ of the dimensional matrix D – it is pre-
cisely equal to the number of dimensionless constants we can form. In particular,
since we have n = 3 independent physical parameters x, y, z corresponding to
three columns of our dimensional matrix D and k = 2 fundamental units U1, U2
corresponding to the two (linearly-independent 14) rows of D, the rank-nullity the-
orem tells us that the nullity of D is given by
nullity(D) = n − k = 3 − 2 = 1. (30)
Since the nullity of D is precisely equal to the number of dimensionless constants
we can form for this physical system, this shows that the π Theorem for dimen-
sional analysis, is just a special instance of the rank-nullity theorem for linear al-
gebra.
Exercise 5 (Challenge: Finish proving the π Theorem) In the previous example,
we set-up the proof of the π theorem for the general case ... but really only proved
it for the case of 3 parameters and 2 fundamental units. By extending the argu-
ment to n parameters x1, ...., xn and k units U1, ..., Uk, prove the π theorem for the
general case of arbitrary n and k.
Hint: Sketching this proof simply amounts to keeping tracking of your indices and
labels. As a suggestion, try denoting the units of x1 by Ua11
1 ...Ua1k
k and the units
of x2 by Ua21
1 ...Ua2k
k etc.
If you have completed and understood these exercises, you are well on your way
to becoming an expert in dimensional analysis. Soon you’ll be better than your
lecturers (possibly).
14
These rows are necessarily linearly independent, since we assume our fundamental physical
units to be independent – by definition.
21
22. 6 Tutorial 3 - Return of Dimensional Analysis: Gravity,
The Hierarchy Problem and extra-dimensional Braneworlds
6.1 Introduction
The following is an extended exercise which test all the skills the tutorials have elu-
cidated so far in dimensional analysis. It will also you introduce to some concepts
which may be new and bizarre, whilst linking them back to everyday reality. The
overall goal will be to derive a dimensionless constant that characterises classical
gravity on all length scales (no knowledge of relativity is required)! By comparing
this constant to another dimensionless constant from electromagnetism, we will see
why gravity is so much weaker than the other three forces in nature – then investi-
gate a solution to this peculiarity using brane-world models of the universe.
6.2 Background
As far as we understand, all interactions in nature take place through four funda-
mental forces. At present, we have a rather ‘successful’ theoretical and experimen-
tal quantum description of three of these forces – that is, we have constructed quan-
tum field theories to describe the ‘quanta’ (particles) which mediate these forces.
Gravity, despite our everyday experience of it, remains somewhat mysterious and
theoretically elusive in several ways – in particular, because it is highly resistant to
all attempts to turn it into a quantum theory like the other forces. As a reminder,
the four forces dictating our universe are the
• Electromagnetic Force: Which governs electromagnetic radiation (such as
light) as well as interactions between charged particles. In the quantum de-
scription (Quantum Electrodynamics), this force is carried by massless par-
ticles known as ‘photons’.
• Weak Nuclear Force: In the quantum description, this force is mediated by
massive particles known as the Z and W± bosons. It is involved in quark
transformations as well as some interactions between charged particles.
• Strong Nuclear Force: In the quantum description (Quantum Chromodyan-
mics), this force is mediated by ‘gluons’ and is responsible for the interac-
tions between quarks, which are the particles making up hadrons such as the
proton and neutron. In this manner, it is responsible for processes such as
fusion, which is the source of energy for our sun.
22
23. • Gravitational Force: In the attempted quantum descriptions, this force is
mediated by a massless particle known as the ‘graviton’. It is responsible for
the interactions of all particles with mass, but also determines the trajectories
of massless particles (e.g. gravitational bending of light) since it warps the
spacetime continuum.
At higher energies, these four forces start to unify into one single force – for ex-
ample, the electromagnetic and weak nuclear forces unify to make the electroweak
force. Attempts to unify the electroweak and strong nuclear forces have been par-
tially successful and fall under ‘The Standard Model’ of particle physics. On the
other hand, attempts to unify gravity with the other forces have been largely un-
successful, with the only real promising candidate being String Theory.
One of the biggest mysteries about the gravitational force, is why it is so weak com-
pared to the other forces in nature. In some sense this is ‘unnatural’, hence suggests
that on some deeper level, gravity is fundamentally different form the other forces.
As the goal of this tute, we will use dimensional analysis to characterise the grav-
itational and electromagnetic forces with some special dimensionless constants –
then compare their strengths to prove this claim. Finally, we will end on some very
recent 15 advancements in theoretical physics which propose an explanation of why
gravity is the weakest of the four forces.
6.3 Extended Problem
Exercise 6 (Newton, Einstein and Braneworlds: The Gravitational Coupling Constant)
Of the many things that Isaac Newton is famous for, one of them is coming up with
multiple mathematical proofs of the fact that the planets orbit the sun in elliptical
paths – and that this elliptical motion is a direct consequence of an inverse square
law. Thus, by planar geometry and calculus he came up with the following gravi-
tational force law to explain the astronomical observations of Johannes Kepler and
Tycho Brae:
F = −GN
m1m2
r2
ˆr (31)
where GN is Newton’s gravitational constant, m1 and m2 are the masses of two
objects separated by a distance r and ˆr is a ‘unit vector’ (vector with magnitude 1)
pointing from one object to the other. This tells us the gravitational force that one
massive object exerts on another massive object.
15
The last 5-10 years.
23
24. QI:Using Newton’s 2nd Law, F = ma, deduce the dimensions or units of GN .
Note that you are working with mass, length and time (M,L,T) as your fundamental
units, hence [m1] = [m2] = M. Furthermore, by definition the unit vector 16
ˆr = r2−r1
|r2−r1| is dimensionless: [ˆr] = 0. Note that in general, the dimensions or
units of a vector quantity are always the same as the units of the magnitude (and
components) of that vector – hence [r] = [r] for example.
Now that we have the dimensions of GN , we are ready to consider Einstein’s theory
of gravitation. Einstein’s theory differs from Newton’s theory in many ways – fun-
damentally it explains gravity as a consequence of spacetime curving around any
object with mass, where the ‘amount’ of curvature being greater for greater masses
(e.g. the Sun). On an astrophysical level, it is important as it helps to explain the
big bang, solar fusion and the existence of the black holes – objects which are nec-
essary for the stability of some galaxies such as the Milk Way. In terms of everyday
living, general relativity is essential for the operation of GPS satellites – without
the gravitational corrections to the timing (gravitational time-dilation) offered by
Einstein’s theory, the GPS system would not be accurate enough to work.
In Einstein’s theory, spacetime is modelled by the following objects 17
• A energy-momentum tensor T which contains information about ‘sources’
of curvature – matter and energy. It’s components have dimensions of an
energy-density: [Tab] = [ Energy
V olume ] = M − L − 2T. Since the tensor itself is
a second-rank covariant tensor, we have: [T] = [Tabdxa ⊗ dxb] = [Tab] +
[dxa ⊗ dxb] = M − L − 2T + 2L = M + L − 2T.
Note that the dimensionality of energy can be deduced from the relation:
Work = Force × Distance and hence [Energy] = [Work] = [Force] +
[Distance] = M + L − 2T + L = M + 2L − 2T.
• A metric tensor g describing how gravity distorts measures of length and
time. This has units of length-squared: [g] = 2L.
• The Riemann Curvature tensor, Riem, describes how the curvature of space-
time varies in different regions. It also measures how gravity distorts parallel-
16
Here r1 and r2 are the position vectors describing the location of the masses m1 and m2 with
respect to some origin.
17
Note that most physicists do not understand differential geometry, hence when they speak of ten-
sors they usually are talking about components of tensors. This won’t matter here, but for reference,
if you ever want to compare: covariant tensors have two extra factors of length compared to their
components and contravariant tensor have two factors less than their components – which basically
means adding ±2L to the dimensions.
24
25. transport. It is given roughly 18 as the anti-symmetrized second tensor ‘gra-
dient’ of the metric: Riem ∼ ⊗ ⊗ g, where are a type of derivative
operator and ⊗ is a type of multiplication for tensors.
• The Ricci tensor, Ric, is given by taking the trace of the Riemann tensor:
Ric = Trace(Riem). It describes how gravity distorts volumes and is also
related to how different geometries evolve under the heat equation.
• The Ricci Scalar R – this quantity is a function which measures how gravity
locally distorts volumes. Einstein’s theory can be derived by saying that
nature minimizes this quantity – an approach due to a mathematician named
David Hilbert 19. It is given by the taking the trace of Riemann tensor twice:
R = Trace(Trace(Riem)) = Trace(Ric).
• The speed of light, c. This universal speed limit quantifies how fast mass-
less particles can move and also how fast gravitational disturbances (gravity
waves) can propagate. It has dimensions of speed: [c] = L − T.
QII:Using the above information, derive the dimensions of Newton’s gravitational
constant GN again, this time using Einstein’s law of gravity:
Ric −
1
2
Rg =
8πGN
c4
T. (32)
You will need the following facts: the derivative operator reduces the length
dimension of a tensor by one factor, whereas the tensor product ⊗ raises it by one
factor (in this case). Hence [Riem] = 2[ ] + 2[⊗] + [g] = −2L + 2L + 2L = 2L.
Furthermore, the trace of a (covariant) tensor reduces its length dimension by two
factors, hence for example: Trace[Riem] = [Riem] − 2L.
Tip: To ease calculations, you may use so-called ‘natural units’ where the speed
of light c = 1. In these units length and time have the same dimensionality, hence
[c] = [Distance] − [Time] = 0 and T = L. You will then get the dimensions of
GN in natural units which you can compare to your value of GN using Newton’s
Law, after you set T = L.
Finally, we are in a position to understand a very special dimensionless constant –
the ‘gravitational coupling constant’, αG. Since it is dimensionless, this constant
characterises the strength of the gravitational force on all length scales (within the
regime of validity of Einstein’s theory). It can be defined in terms of any pair of
stable elementary particles – in practice, we use the electron.
18
Don’t ever show this to a differential geometer. If you want the real definition, see me.
19
In retrospect, David Hilbert deserves almost the same level of credit as Einstein for the theory
of general relativity.
25
26. In particular, we have:
αG =
GN m2
e
¯hc
≈ 1.7518 × 10−45
(33)
where c is the speed of light, GN is Newton’s gravitational constant and me is the
mass on an electron. The quantity ¯h = h
2π is the reduced Planck constant which
characterises the scale at which matter exhibits quantum behaviour such as wave-
particle duality 20
QIII:Show that the gravitational coupling constant αG is indeed dimensionless.
Note that [me] = M. To work out the dimensions of ¯h = h
2π , you will need the
Planck-Einstein relation which relates the energy of a photon (particle of light) its
frequency:
E = hf. (34)
Then [h] = [E] − [f]. Since the frequency of light is the number of oscillations
of the electromagnetic wave per unit time, we have [f] = −T. You can get the
dimensions , [E] of energy E from the calculation shown above for the energy-
momentum tensor.
Now, for the last part of this problem, we introduce one more fundamental phys-
ical unit: the unit of electric charge, Q 21. Similar to the gravitational coupling
constant, there is a dimensionless constant which characterises the strength of the
electromagnetic interaction (which is responsible for almost all of chemistry) – the
‘fine structure constant’ αEM . The value of this constant is (accurately) predicted
and measured using the theory of Quantum Electrodynamics, which is a type of
quantum field theory largely due to Richard Feynmann and Freeman Dyson. It is
given by
αEM =
1
4π 0
e2
¯hc
(35)
where 0 is electric permittivity of the vacuum. It has units [ 0] = [Farads/Meter] =
[Seconds4 Amps2 Meters−2 kg−1] = 4T + 2Q − 2T − 2L − M. Hence
[ 0] = 2T + 2Q − 2L − M. The parameter e is the charge of an electron, with
dimensions [e] = Q.
Using ‘natural units’ – a popular convention in particle physics, we set all of our
previous parameters to equal 1. Thus, 4πGN = c = ¯h = 0 = 1, where 0 is
20
If ¯h was really large – say ¯h ≈ 1 for example, then we would observe wave-particle duality on
a macroscopic scale and the universe would be a scary, crazy place. Bullets would diffract through
doorways and Leanora’s fists could quantum tunnel through walls.
21
The SI unit for charge is Coulombs.
26
27. electric permittivity of the vacuum. In these units, the fine-structure constant is
given by
αEM =
e2
4π
≈ 7.297 × 10−3
. (36)
QIV:Choosing natural units: 4πGN = c = ¯h = 0 = 1, is the same as forcing
these parameters to be dimensionless. Show that this is equivalent to setting all the
fundamental units to be the same T = L = M = Q. Hint: you should get four
equations for the dimensions of these parameters.
Note that you can calculate the values of the fine-structure and gravitational cou-
pling constants yourself by Googling their values in SI units (or any other consis-
tent set of units you choose). Taking their ratio, we see that (in natural units):
αEM
αG
= (
e
me
)2
≈
7.297 × 10−3
1.752 × 10−45
≈ 4.16 × 1042
. (37)
This says that the electromagnetic force is about 42 orders of magnitude22 stronger
than the gravitational force. In a similar fashion, the weak-nuclear force is about 32
orders of magnitude (1032) times stronger than gravity. The challenge to explain
why gravity is so weak compared to the other forces is known as ‘the heirarchy
problem’.
One class of attempts to solve the heirarchy problem, involves the visible universe
being confined to a 4-dimensional ‘brane’, which is basically a 4-dimensional slice
living in a larger spacetime. Such models are called ‘braneworld models’. In this
view, the electromagnetic, weak and strong nuclear forces take place on the 4-
dimensional brane – but gravitational interactions (mediated by particles known as
‘gravitons’) take place in 4-dimensions and in the ‘large extra dimensions’. This
then gives a natural explanation to the gravitational coupling constant being so
small. In some variations 23, the introduction of large extra-dimensions also solves
the ‘Dark Energy’ or ‘Cosmological Constant’ problem – where Dark Energy nat-
urally arises as the ‘surface tension’ of the 4-dimensional brane. Using braneworld
models, we can derive (!) Newton’s gravitational constant directly from the size
(‘hyper-volume’) of the extra dimensions in our universe.
A very special class of braneworld models , known as known as theories with
‘Supersymmetric Large Extra Dimensions’ envisions spacetime as 6-dimensional
(4-dimensional brane + 2 large extra dimensions) with some extra symmetry added
(super-symmetry) that enables bosons and fermions to transform into each other
22
Note, 42 is also the meaning of life.
23
Those investigated in the present author’s masters thesis.
27
28. 24. In these models, the extra-dimensions take the form of some compact hypersur-
face. Newton’s gravitational constant GN is then theoretically explained using the
formula 25:
GN =
3κ2
16πS
(38)
where S is the surface-area of the extra dimensions and κ is Einstein’s constant,
with dimensions [κ] = [GN ].
QV:The above formula for GN is correct, even though it may look dimensionally
incorrect. What units would S need to have for dimensional consistency? In that
case, what quantity does the surface-area S actually represent? Hint: Recall the
‘unit vector’ in Newton’s law of gravity.
The last problem illustrates a common theme in engineering, physics and math-
ematics – normalization. Normalized quantities are typically dimensionless! As
such, they are very useful and friendly to work with.
24
Supersymmetry removes the problem of Tachyons in String Theory and also stabilizes the mass
of the Higgs boson.
25
First derived in this generality by the present author in 2013.
28
29. 7 Tutorial 4: 50 Shades of Error, Shade I – Multivariable
calculus and The Total Differential
In this tutorial, we revise some elementary concepts from multivariable calculus
– partial differentiation and the ‘total differential’ or ‘exterior derivative’. If you
haven’t formally studied these topics then don’t worry – as long as you are comfort-
able differentiating functions of a single variable, the rest will follow easily.
After revising these mathematical tools, we will see how they are used in error
analysis. In particular, the total differential provides an elegant way to compute
the absolute error for any derived quantity, in terms of your experimental preci-
sion error. This is extremely useful for the applied sciences and engineering. For
those of you who are only interested in pure mathematics, then note that the tech-
niques used here are precisely the same techniques that are used when you study
linear approximations 26 and Taylor series expansions for functions of more than
variable.
Note that this tutorial is the first of a sequence of tutorials that will be dedicated to
error analysis, least-squares regression (e.g. line of best fit) and other techniques
that you will use frequently in statistics and the applied sciences to determine the
value of derived quantity, along with an estimate of its corresponding error. As
such they will successively build on each other.
7.1 Russian Playpen: Functions of more than one variable
Given a function f of one variable, which maps real numbers 27 R to real numbers
R, we can formally28 express it as:
f :R → R
:x → f(x),
which says that f sends the number x another number f(x). For example, if we
have the function f(x) = x2 whose graph is a parabola, then we write:
f :R → R
:x → x2
.
26
Linear or ‘tangent plane’ approximations are just a special case of a Taylor series expansion.
27
The ‘blackboard font’ r, denoted as R, symbolizes the set of ‘real numbers’. This is includes all
integers, rational numbers, irrational and transcendental numbers (such as π) etc.
28
Technically, you restrict the set f is mapping from to its domain and the set it is mapping into, to
its range.
29
30. So for example, in this case we have f(1) = 12 = 1 and f(7) = 72 = 49
etc.
We now generalize this as follows. A function f of more than one variable, maps
several copies of the set of real numbers to several copies of the set real numbers.
For example, a function f of two variables, x and y, can be formally expressed
as
f :R × R → R × R
:(x, y) → f(x, y).
Here the notation R × R means the set of all ordered pairs of real numbers (x, y).
So for example, if we have the circular function given by: f(x, y) = x2 + y2, then
we have f(1, −1) = 12+(−1)2 = 1+1 = 2 and f(π,
?3) = π2+(
?3)2 = π2+3
e.t.c.
Note, there is nothing strange about functions of several variables. You see them
everyday. For example, we can view the volume of a rectangular box with sides of
length x, y and z as a function of three variables:
V (x, y, z) = xyz. (39)
Or, as another example, the concentration C of a substance dissolved in water will
depend on the amount (‘mass’ or any other measure) m of the substance dissolved
and the amount (volume) of water (or any other liquid) v it is being dissolved into.
Thus we can consider the blood alcohol concentration C of a student at PROSH as
a function of two variables: C = C(m, v), where m is the amount of alcohol and
v is the amount of blood in that person.
7.2 Russian Daycare: Partial Differentiation
To some, partial differentiation may sound hard. However, it is actually extremely
simple – hence why it is taught to children at daycare in Russia. All you need to do,
is differentiate your function with respect to some chosen variable, while treating
all the other variables as constants.
The notation for partial derivatives is given by ‘del’ symbol, ∂. So for example,
if we are taking the usual total derivative with respect to x, we have the Leibniz29
29
Leibniz was Austrian version of Newton, or Newton was the English version of Leibniz. Leibniz
developed calculus at the same time as Newton as well as several other fields of mathematics – such
as binary numbers.
30
31. notation d
dx. If we are taking a partial derivative with respect to x, we use the
notation ∂
∂x instead. The best way to illustrate is with a few examples.
Example 5 (Return of the Box) Our rectangular box has now followed us into
Tutorial 4. Having being stalked by this sentient box, we decide to partially differ-
entiate its volume. Denoting the length of each of its sides by the variables x, y
and z respectively, its volume is given by the following function of three variables:
V (x, y, z) = xyz. Partially differentiating it with respect to x, we find:
∂
∂x
V (x, y, z) =
∂
∂x
(xyz) = (
∂x
∂x
)yz = yz. (40)
What we did here was to treat y and z as constants, while differentiating with
respect to x. Since the derivative of x with respect to x is just 1, we arrived at the
above result. We show similarly that:
∂
∂y
V (x, y, z) =xz
∂
∂z
V (x, y, z) =xy. (41)
Now, if we differentiate twice with respect to x, or twice with respect to y, we get:
∂2
∂x2
V (x, y, z) =:
∂
∂x
∂
∂x
V (x, y, z) =
∂
∂x
(yz) = 0
∂2
∂y2
(xyz) =:
∂
∂y
∂
∂y
(xyz) =
∂
∂y
(xz) = 0, (42)
where the notation ∂2
∂x2 denotes the ‘second partial derivative’ with respect to x.
Note that differentiating the volume with respect to x the second time, gives zero
since the first derivative of V (x, y, z) with respect to x no longer depends on x –
i.e. the product (yz) is a constant with respect to x, hence its partial derivative with
respect to x vanishes.
We can also take mixed derivatives. For example, differentiating V (x, y, z) first
with respect to x and then with respect to y, gives:
∂
∂y
∂
∂x
V (x, y, z) =
∂
∂y
(yz) = z. (43)
Now, if take the derivatives in reverse order – y first, then x, we get
∂
∂x
∂
∂y
V (x, y, z) =
∂
∂x
(xz) = z, (44)
31
32. which is exactly the same as taking the derivatives in original order. This illustrates
an importantly and extremely consequential property of functions of more than
one variable – in general, for ‘nice’30 functions (most functions you will ever deal
with), the order in which you take two partial derivatives doesn’t matter. That is,
for nice functions f, we have
∂
∂y
∂
∂x
f(x, y, ...) =
∂
∂x
∂
∂y
f(x, y, ...). (45)
This observation is formalized as ‘Clairaut’s Theorem’ (or ‘Schawrz’s Theorem’)31.
Exercise 7 (Sir Steven – The Suspicious Spheroid) A solid oblate spheroid (‘squashed
sphere’)32, by the name of Sir Steven, suspiciously follows our friend – the rect-
angular box, into Tutorial 4. Sir Steven was produced by rotating a filled-ellipse
about its minor (shorter) axis. At his present age, Sir Steven has a minor axis
length of 2b and a major axis length of 2a. Since being knighted, Steven has taken
to a gluttonous lifestyle (hence a b). During Lent, Sir Steven decides to read
Allen Mandelbaum’s translation of Dante Alighieri’s Divine Comedy – and in the
midst of an epiphany, he decides to calculate his own volume, which is given as
a function V (a, b) of two variables a and b (the semi-major and semi-minor axes
lengths):
VSteven =
4π
3
a2
b. (46)
30
This means functions with continuous second-partial derivatives. More general, the ability to
commute the order of partial derivatives holds at any given point provided that the function has
continuous second partial derivatives in some open neighbourhood about that point.
31
After the French and German mathematicians, Alexis Clairaut and Hermann Schwarz, respec-
tively.
32
This problem is dedicated to Nicholas Jones, University of Bristol and his love of spheroids.
32
33. Q: Compute the following partial derivatives of V (a, b):
∂
∂a
V (a, b) =
∂
∂b
V (a, b) =
∂2
∂a2
V (a, b) =
∂2
∂b2
V (a, b) =
∂
∂b
∂
∂a
V (a, b) =
∂
∂a
∂
∂b
V (a, b) =
∂3
∂a3
V (a, b) =. (47)
Challenge Q (Russian Grade 1): With the help of his intelligent friend, Pappus 33
the Prolate Spheroid, Sir Steven manages to compute his surface area as a function
S(a, b) of two variables a and b:
SSteven = 2πa2
{1 +
1 − e2
e
tanh−1
(e)} (48)
where the eccentricity e of the generating-ellipse is defined via e2 = 1− b2
a2 . Using
this surface area formula, compute the following partial derivatives:
∂
∂a
S(a, b) =
∂
∂b
S(a, b) =
(49)
Hint: You will need to use the product (Leibniz) rule for differentiation along
with chain rule and the following identity for the derivative of arc-hyperbolic tan34
(hyperbolic tan inverse):
d
dx
tanh−1
(x) =
1
1 − x2
. (50)
33
Pappus claimed that Hippasus – a student of the Ancient Greek Pythagorean school of geometry,
was drowned for proving (or sharing) the ‘secret’ irrationality of
?2.
34
Sometimes the notation artanh(x) is used instead of tanh−1
(x) to the inverse hyperbolic tan-
gent function.
33
34. This derivative is well-defined for all real-values of x such that |x| 1. Thus,
you should replace x with e then use the chain rule to get partial derivatives of
tanh−1
(e) with respect to a or b, since e is a function of a and b.
Extra-Challenging Q (Russian Grade 1.1): If you think you have what it takes
to pass Grade 1 in Soviet Russia, compute the following partial derivatives then
check them using Wolfram alpha or Mathematica / Computer Algebra Software of
Choice:
∂2
∂a2
S(a, b) =
∂
∂b
∂
∂a
S(a, b) =
∂2014
∂a2014
S(a, b) = . (51)
Hint: If you can derive an expression for ∂n
∂an S(a, b) where ‘n’ is an arbitrary
positive integer (n = 1, 2, 3....) the last equation should be easy.35
7.3 Russian Kindergarten: The Exterior Derivative (Total Differen-
tial)
The ‘total differential’ or ‘exterior derivative’ of a function f, is denoted by df
– the resulting object is known as an ‘exact differential 1-form’ or ‘co-vector’. We
will see why it has the latter name shortly. To illustrate how to compute df, we
give a few examples then state the general case.
Given a function f = f(x) of a single variable x, its total differential is given
by:
df =
df
dx
dx. (52)
The quantity df
dx is a function (the derivative of f with respect to x), however the
quantity dx can be thought of in several ways. Formally, dx is a ‘differential 1-
form’ or basis ‘co-vector’ analogous to the standard basis vectors you may have
seen 36 e1, ˆx or ex. Informally, it can be thought of as infinitesimal quantity or
length in the x-coordinate. You will also recall that when you integrate a function
f = f(x) with respect to the variable x, you write it as:
f(x)dx. (53)
35
Disclaimer: It’s probably not easy, relatively speaking.
36
These are some of the more common notations.
34
35. If we replace f with its derivative df(x)
dx , then we have
df(x)
dx
dx = f(x) + c (54)
where c is some constant of integration; this is just a consequence of the funda-
mental theorem of calculus. Note that however, we said that df = df(x)
dx dx, so we
can actually view this statement as:
df(x)
dx
dx = df = f + c. (55)
In this manner, we can think of as a formal inverse37 of the ‘exterior derivative’
or ‘total differential’ operator d.
For a function f = f(x, y) of two variables x and y, computing its total differential
requires partial derivatives. In particular, we have
df =
∂f
∂x
dx +
∂f
∂y
dy. (56)
The object df is still a differential 1-form, but now it has two components: ∂f
∂x is
the component in the dx direction and ∂f
∂y is the component in the dy direction.
Alternatively we say ∂f
∂x is the coefficient of dx and ∂f
∂y is the coefficient of dy.
Hence we see that the total differential df of the function f, behaves similarly to a
2-dimensional vector (when f is a function of two-variables) – which motivates the
name ‘co-vector’ to describe df.
We generalise this now, in the most natural way. For a function f of n variables
x1, x2, ..., xn, its total differential is given by:
df =
∂f
∂x1
dx1 +
∂f
∂x2
dx2 + ... +
∂f
∂xn
dxn. (57)
This says that we partially differentiate f with respect to each of its variables,
then multiply that derivative by basis 1-form corresponding to the coordinate you
are differentiating with respect to. Adding all of these together gives the total
differential, shown in the equation (57). This may seem a little abstract, so its
best illustrated with a few examples – which we will return to next week when we
proceed with error analysis!
Note that the exterior derivative operator d obeys the following general properties
when acting on functions:
37
This is a very simple case of the so-called “generalized Stokes’ Theorem from differential
geometry.
35
36. 1. Linearity: d(c1f + c2g) = c1df + c2dg for any two constants c1, c2 and any
two (differentiable) functions f, g.
2. Product (Leibniz) Rule: d(fg) = g(df) + fdg, for any two (differentiable)
functions f, g.
Example 6 (Rocky the Rectangular Box) Unable to stay down, Rocky the rect-
angular box has returned to help with exterior derivatives. Rocky’s volume V is
given as a function of three variables: V = V (x, y, z), where x, y, z are the lengths
of its sides. Since V (x, y, z) = xyz, the total differential of the volume is given
by:
dV (x, y, z) :=
∂V
∂x
dx +
∂V
∂y
dy +
∂V
∂z
dz = yzdx + xzdy + xydz. (58)
Observation: Notice that coefficient of dx is equal to yz, which is the surface area
of the face of the box in the plane perpendicular to the x-direction. Similarly, the
coefficient xz of dy is the area of the face of the box in the plane perpendicular
to y-direction etc. Depending on the symmetry of an object, its surface area and
volume are usually related in some manner by the operations of differentiation and
integration.
For example, Snorlax the Sleepy Sphere, has a volume V = V (r) which is a
function of its radius. In particular, the exact differential of its volume is given
by:
dV (r) = d(
4π
3
r3
) =
4π
3
d
dr
r3
= 4πr2
dr. (59)
The coefficient of dr is the surface area of the sphere, perpendicular to the dr
direction (recall that the surface of a sphere is perpendicular to its radius). In
particular, the quantity
dV
dr
= 4πr2
(60)
is the surface area.
7.3.1 Exercises
The following exercises are split into some purely mathematical exercises – geom-
etry, along with some applied exercises (thermodynamics) for physicists, engineers
and chemists. Bonus neural connections for those who complete both sets!
Exercise 8 (Geometry of Solids)
36
37. 1. Given a circular cylinder of radius r and height h, we can view its volume V
and surface area S as functions of two variables:
V (r, h) =πr2
h
S(r, h) =2π(r2
+ rh). (61)
Compute the exterior derivatives dV and dS.
2. An elliptical cylinder is a cylinder with elliptical cross-sections – you can
think of its as ellipses stacked on top of each other ... 38 Given an elliptical
cylinder with height h, cross-sectional ellipses with semi-major axes lengths
a and semi-minor axes lengths b, its volume V and surface area S can be
viewed as functions of three variables
V (a, b, h) =πabh
S(a, b, h) =2πab + 2πph. (62)
where p is the perimeter of the elliptical cross-sections. To express p exactly,
one requires an infinite series:
p = 2πa(1 −
∞
n=1
(2n)!2
(2nn! )4
e2n
2n − 1
) (63)
where e =
?a2−b2
a is the eccentricity of the ellipse. Using the Ramanujan39
approximation: p ≈ π[3(a+b)−
—
(3a + b)(a + 3b)], compute the exterior
derivatives dV and dS.
3. For those of you who have studied infinite series, compute dS using the exact
expression for the perimeter of an ellipse stated above.
Given these examples, in addition to the previous exercises, complete the following
problems.
Exercise 9 (Exact Differentials: Thermodynamics/Thermochemistry) Thermodynamics
is a broad theory, originally explaining the phenomenon that we know as ‘heat’.
More generally, it governs a vast range of macroscopic phenomena in nature –
from reaction rates in thermochemical processes to the surface area of blackholes.
The most famous abstraction of thermodynamics, due to Steven Hawking, Bill Un-
ruh and Jacob Bekenstein, is that the surface area of a black hole is proportional to
38
Puns – bringing English lit and mathematics together since 1600.
39
A famous Indian child prodigy and mathematical genius who made great rediscoveries and con-
tributions to number theory, estimations and analysis in isolation.
37
38. its entropy and its temperature is inversely-proportional to its mass 40. One of the
fundamental concepts in thermodynamics, is the minimization of different types of
so-called ‘state functions’ or ‘thermodynamic potentials’ – representing different
types of energies.
• Internal Energy: U := U(S, V, Ni) = dU, where
dU = TdS − pdV +
i
µidNi. (64)
• Hemholtz Free Energy: F(T, V, Ni) = U − TS.
• Enthalphy: H(S, p, Ni) = U + pV .
• Gibbs Free Energy: G(T, p, Ni) = U + pV − TS.
Here we stated the natural variables for each function, U, F, H and G in the brack-
ets (..). These variables are entropy S, temperature T, volume V , pressure P and
number (amount) Ni of the i th reactant species (i.e. substance, chemical etc). The
chemical potentials µi are all fixed constants. Note, for those of you haven’t seen
the sigma41 notation for summation,
i
simply means the sum over all species
labelled by the index i.
By keeping track of which variables each function is strictly dependent on and
noting the expression for dU, prove that we get the following exact differentials:
dH(S, p, Ni) =TdS + V dP +
i
µidNi
dF(T, V, Ni) = − sdT − pdV +
i
µidNi
dG(T, p, Ni) = − SdT + V dp +
i
µidNi. (65)
Exercise 10 (Mathematical Proof: Cyclic Reciprocity Rule and Thermodynamics)
The goal of this exercise is to understand the following proof, memorize the main
steps (tricks) and then reproduce it from memory 42.
Say we are looking at the level sets of a function of three variables – for instance,
one of the thermodynamical potentials from the last exercise. In particular, suppose
40
Physicists that the present author has had the privilege of talking to in person :P.
41
Σ is the symbol for the Greek capital letter, ‘sigma’.
42
This problem is dedicated to Aston Williams, Engineer of Chemicals.
38
39. we have a function f = f(x, y, z) of the three variables x, y, z. If we have the
additional constraint that:
f(x, y, z) = 0 (66)
(e.g. zero Hemholtz free energy), then the implicit function theorem from multi-
variable calculus tells us that we can write any one of the variables x, y, z in terms
of the two other variables. WLOG43 lets take the variable z to be a function of the
two variables x and y: z = z(x, y).
The exterior derivative (total differential) of z is then given by
dz =
¢
∂z
∂x
y
dx +
¢
∂z
∂y
x
dy, (67)
where as usual, the partial derivative ∂z
∂x is taken while y is held constant and the
partial derivative ∂z
∂y is taken while x is held constant. This is made explicit by
the notation ( ∂z
∂x) )y, where the brackets and subscript denote the variables we are
keeping constant while differentiating 44
Taking dz = 0 (holding z constant), we can use the implicit function theorem again
to view y as function of x (when dz = 0): y = y(x), hence we have
dy =
¢
∂y
∂x
z
dx. (68)
Substituting this relation into the equation dz = 0, we see that:
0 = dz =
¢
∂z
∂x
y
dx +
¢
∂z
∂y
x
¢
∂y
∂x
z
dx. (69)
Since this equality is actually a co-vector (differential 1-form) equality, we use the
fact that a co-vector is identically zero if and only if its components are zero – i.e.
the coefficients of dx in this case. Hence
0 =
¢
∂z
∂x
y
+
¢
∂z
∂y
x
¢
∂y
∂x
z
=⇒ (70)
43
This is a common mathematical acronym for ‘Without Loss of Generality’.
44
The reason for this pedantry now, is that usually we differentiate with respect to variables which
are independent of each other. However, in the following step of the proof, y may also be related to z
except in the special circumstance that dz = 0 – thus we must explicitly denote that z is being held
fixed, hence this notation.
39
41. 8 Tutorial 5: Absolute Error and Game of Thrones
In this tutorial we will investigate the task of computing the ‘absolute error’ in a
given quantity, as a function of the precision in your measuring devices and mea-
suring ability. The problems and examples will have a Game of Thrones theme, to
celebrate (*spoiler alert*) the death of King Joffrey.
8.1 Absolute Error
Most quantities that we measure in science and engineering are ‘derived quanti-
ties’. This means that we measure them indirectly – in particular, we measure
some set of basic properties of a system or environment, then use some mathe-
matical model or formula to relate these properties to the quantity we are trying to
measure.
For example, if we want to measure the surface are of a basket ball, one would
probably measure the circumference46 with a tape measure or string. Then, using
the relation:
C = 2πR = πD (72)
where C is the circumference, R is the radius and D = 2R is the diameter, one
can then compute the radius of the ball. Once the radius is known, the surface area
S can be calculated:
S = 4πR2
. (73)
In this manner we have only performed a length measurement, yet we have ob-
tained a measurement of a ‘derived property’ of the ball – it’s surface area. If
you think carefully about the tools we use to measure things, one quickly comes
to the conclusion that almost all measurements we perform are those of derived
quantities. The question then arises – how do we obtain an estimate of the error in
our final measurement? To do so, one would have to relate the error in a derived
quantity to the error in the basic quantities which we directly measure.
One general procedure for obtaining a ‘total error’ or ‘absolute error’ estimate,
involves three ingredients:
• A knowledge of the precision of your measuring ability (inherently restricted
by the precision of your instruments).
46
That is, the circumference of a great circle – a circle which passes through the centre and divides
the ball into two equal hemispheres.
41
42. • A mathematical function relating your derived quantity to the quantities you
directly measure.
• The ‘total differential / exterior-derivative / exact derivative’ formula (Tuto-
rial 4).
Mathematically, we proceed as follows.
Definition 1 (Absolute Error) Let x1, ..., xn be a set of n quantities which are
to be measured (with their respective units). Now, let f(x1, ..., xn) be a func-
tion of n variables, representing some derived quantity which is to be measured.
If ∆x1, ..., ∆xn are the errors associated to the measurements of x1, ..., xn (e.g.
instrument precision) then the corresponding ‘absolute error’ in f(x1, ..., xn) is
given by the linear estimate:
∆f(x1, ..., xn) = |
∂f
∂x1
∆x1|+|
∂f
∂x2
∆x2|+... + |
∂f
∂xn
∆xn|. (74)
which is evaluated at the measured values x1, ..., xn.
Note that the formula for ∆f is similar to the total differential, df, where the dif-
ference is that we have replaced the covectors (1-forms) dx1, ..., dxn with the mea-
surement errors ∆x1, ..., ∆xn. The absolute value of each term is also taken – this
is because when looking to estimate the ‘Maximum Probable Error’, each error
should add up. When quoting the value of f as (derived) measurement, we say that
the quantity f has the value:
Measured Value of f = f(x1, ..., xn) ± ∆f. (75)
Therefore, (with some probability) we say that the true value of f lies in the interval
[f − ∆f, f + ∆f].
Note that in the case of perfect measurement technique, one would attribute the
errors ∆x1, ..., ∆xn to the instrumental precision. So for example, if you are mea-
suring the height h of Tyrion Lannister with a tape measure, the error ∆h would be
equal to half the width of the gradings in the tape measure. Finally, one should note
that this ‘absolute error’ formula only takes deterministic errors into account (i.e.
precision e.t.c) – it does not factor in wrong measurement technique or external
errors which one has not accounted for.
Before attempting the problems and examples, consider the following philosoph-
ical note. Because of Quantum Mechanics – in particular, the Heisenberg Un-
certainty Principle and the inherent non-deterministic nature of the universe, it is
inherently impossible to measure anything with 100% accuracy or certainty. This is
42
43. not due to imperfect craftsmanship (imperfect measuring devices) or human imper-
fection – it is because the process of observation and measurement itself, requires
interacting with the entity which we are trying to measure. This interaction alters
the state of the entity we are trying to measure and is necessarily constrained by
the Heisenberg uncertainty principle.
8.2 Examples and Problems
Example 7 (Thinking Ahead with Ned) In a sadistic rage, the false King Joffrey
decides to measure the surface area of Ned Stark’s head after decapitation. Being
a boy of elementary means, he approximates the Lord of Winterfell’s head as a
sphere. Using a string and ruler, he measures the circumference of Ned’s head by
to be C = 24 inches. He does this by marking the string, then measuring the string
with the ruler. The gradings on the ruler are spaced 1
4 of an inch apart – hence
precision of the ruler is 1
8 in. Assuming his technique is correct, this means that
the error associated to the circumference measurement is ∆C = 1
8 in. Therefore,
Joffrey deduces the surface area of Ned’s head to be:
S = 4πR2
= 4π(
C
2π
)2
=
1
π
C2
=
1
π
(24in)2
=
574
π
in2
. (76)
Viewing S = S(C) as a function of the measurement C, the absolute error in S is
given by:
∆S =|
∂S
∂C
∆C|= |
2
π
C∆C|
=|
2
π
24 ×
1
8
|in2
=
6
π
in2
. (77)
Hence, with the equivalent sphere approximation, the surface area of Ned Stark’s
head is:
S ± ∆S = (
574
π
±
6
π
)in2
≈ (183 ± 1.91)in2
. (78)
Exercise 11 (Thinking Ahead: Part II) Using his previous measurement of the
circumference of Ned’s head, compute the volume V of Ned’s head along with the
absolute error ∆V . Recall that the volume of a ball of radius R is given by:
V =
4
3
πR3
. (79)
Hint: First write V in terms the circumference C, using the relation C = 2πR.
43
44. Dry Humour: To account for dehydration-related shrinkage of Nedard’s head, add
5% ± 1% of the measured volume of Ned’s head. Note, you add the ±1% of V to
the previously calculated error ∆V – that is: ∆VNew = ∆VOld + 0.01V .
Exercise 12 (Thinking Ahead: Part III, Return of The King) After receiving tu-
ition help from St. George’s College tutors, King Joffrey decides to further his
skills by measuring the volume of Ned Stark’s head – this time, using more sophis-
ticated estimates. In particular, he approximates the Lord of Winterfell’s head to
be that of a prolate spheroid47, with its major axis aligned with the symmetry axis
(vertical axis) of Ned’s head. Again, using a string and ruler (this time in metric
units), Joffrey proceeds to measure the circular circumference of Ned’s head (pro-
late spheroids have circular cross sections along their minor axis) as well as the
elliptical circumference of Ned’s head (elliptical cross-sections along the major
axis).
Joffrey makes the following measurements:
CCircular =55cm = 2πR
CElliptical =62cm = 4aE(e) (80)
where a is the semi-major axis length of the ellipse, b is t the semi-minor axis
length, e =
˜
1 − (b
a)2 is the eccentricity of the ellipse and E(e) is a complete
elliptical integral of the second kind (computed numerically or as an infinite series
expansion in e). Note that the semi-minor axis length b of a prolate spheroid, is
equal to the radius of the circular cross-section along the minor axis of the spheroid
:
b = R, (81)
since the spheroid is generated by revolving the ellipse about the axis perpendicular
to the minor axis.
Given that Joffrey’s newfound metric ruler has 1mm = 0.001m spacings, the
precision error in his measurements in now given by: ∆CC = ∆CE = 0.5mm =
0.5 × 10−4m. Using the volume formula for a prolate spheroid:
V (a, b) =
4
3
πa2
b, (82)
compute the volume of Ned’s head, along with the associated absolute error ∆V .
This requires Russian Grade 1 skills.
47
A prolate spheroid was chosen over an oblate spheroid after using Microsoft Paint to compare
the width and height of Sean Bean’s (the actor playing Ned Stark) head.
44
45. Hint: To proceed, you should write the volume V in terms of the circular and
elliptical circumferences: V = V (CC, CE). This requires writing the semi-minor
and semi-major axis lengths in terms of the Circumferences. We already know that
b = R, hence b = CC
2π . To get the semi-major length a in terms of CE, one needs
an approximation for the elliptical integral E(e). Recalling from Russian Grade 1
in Tutorial 4, we have the Ramanujan approximation:
CE ≈ π[3(a + b) −
—
10ab + 3(a2 + b2)]. (83)
By bringing the 3(a + b) term to the left-hand side and squaring both sides, we can
obtain a quadratic equation for a in terms of b and CE. The positive root of this
equation is given by:
a =
3CE − 4bπ +
˜
3C2
E + 12bCEπ − 20b2π2
6π
. (84)
Substituting these expressions for a and b into V , one can then compute V and its
partial derivatives, required for computing ∆V .
Exercise 13 (La forma de la espada – “The Shape of the Sword”) The goal of
this problem, is to be able to reproduce all the steps and arguments to derive the
volume estimate – then compute the volume measurement and absolute error at the
end. Disclaimer: there may be errors in this error analysis!
To add further insult to the Stark family, Tywin Lannister – Hand of the King
and head of the Lannister family, decides to melt down Edard Stark’s greatsword,
“Ice. Being a pragmatic man, Tywin decides to calculate the volume of this sword
in order to work out how much Valyrian steel he will have to forge two new swords,
for his sons.
Not being as clever as Archimedes, Tywin doesn’t think to use water displacement
to measure this volume. Instead he proceeds as follows. We can approximate the
blade (Valyrian steel part) of the sword to be that of a shallow rhomboidal prism,
with maximum width at the hilt of the sword, decreasing in thickness down to the
pointed tip. A rhomboidal (diamond-shaped) prism, means that the width-wise
cross-sections of the sword of are shaped like rhombuses with very narrow (acute)
angles α in the plane parallel to the cutting edges and very large (obtuse) angles
β length in the plane perpendicular to the cutting edge. Despite the decreasing
thickness, the angles in the rhombus cross-sections will remain the same48.
Say that the rhomboidal cross-sections are measured to linearly decrease in area,
down from the hilt to the tip – reached zero area at the pointed end of the blade. By
48
So we could in-fact view the blade as a continuous conformal map of a rhombus.
45
46. knowing the length of the blade and the cross-sectional area at the hilt and at the tip
(zero), we can construct a linear function, A(x) (where x is the distance down the
blade, measured from the hilt), from which we can interpolate the cross-sectional
area of the blade anywhere between the hilt and the tip. The volume will then be
‘sum’ of these cross-sectional areas stacked on top each of each other – i.e. the
integral:
V =
xtip
xhilt
A(x)dx. (85)
Because of his war with the Stark family, Tywin has run out of protractors and is
thus left only with Joffrey’s string and ruler to carry out his measurements – to
which he proclaims, “FML! Tywin now summons the help of his educated son,
Tyrion Lannister. In a stroke of cleverness, to work out the angles of the rhombus
cross-section at the blade hilt, Tyrion measures the circumference of the blade.
Because of sword symmetry (the rhombus consisting of two mirrored isosceles
triangles), this circumference C is equal to four times the length of each side of the
cross-sectional rhombus at the hilt:
Crhombus(xhilt) = 4Lhilt. (86)
If the blade was completely flat, it would have a width of 2Lflat at the hilt, instead
of the string-measured value of 2Lhilt. Thus, by holding the string tangential to the
corner of the rhombus (which runs down the center of the blade), Tyrion measures
the ‘flat width’ of the blade: 2Lflat. He then computes the ‘flat circumference’ at
the hilt:
Cflat = 4Lflat (87)
and concludes that the deviation:
Crhombus − Cflat = 4(Lhilt − Lflat) (88)
must be due to the entirely to rhomboidal geometry49 of the cross-sections. Using
planar geometry that he learned while in his mother’s womb, Tyrion realises that
2Lflat is equal to the central diameter of the rhombus. Forming a right-angle
triangle in the rhombus, with hypotenuse Lhilt, acute angle α
2 , adjacent side Lflat.
49
The key concept here is that of a ‘defect angle’. The rhomboidal geometry introduces a non-zero
angular defect away from zero-angle describing flat cross-sections (straight lines).
46
47. Figure 1: Cross-sectional rhombus of idealized broadsword.
Therefore, simple trigonometry gives:
cos(
α
2
) =
Lflat
Lhilt
sin(
β
2
) =
Lflat
Lhilt
, (89)
which allows Tyrion to deduce the interior angles α and β of the cross-sectional
rhombus. By symmetry, the area of the cross-sectional rhombus at the hilt is simply
four times the area of this triangle (using Pythagoras’ theorem since we want all
quantities in terms of the measured quantities Lh, Lf )
Arhombus(xhilt) =4 ×
1
2
× base × height = 4 ×
1
2
Lflat
˜
L2
hilt − L2
flat
=2Lflat
˜
L2
hilt − L2
flat . (90)
To work out A(x) for any x ∈ [xhilt, xtip], Tyrion lays the sword flat. Overhead, the
sword looks like an isosceles triangle, with base 2Lflat and height Lblade. Splitting
these into two right-angled triangles, we get the following diagram:
Figure 2: Top view of broadsword laid flat.
In particular, Tyrion finds that tan(γ) =
Lflat
Lblade
. Setting up a coordinate system
with xhilt := 0 at the hilt and x = xtip = Lblade at the end of the blade, the height
y of the triangle at any point along the blade, can then be computed as a function
47
48. of the position x along the blade. Trigonometry shows that:
y = (Lblade − x) tan(θ) = (Lblade − x)
Lflat
Lblade
= Lflat −
1
Lblade
x. (91)
To work out the area A(x) of the cross-sectional rhombus at any point x along
the blade, one uses the previous formula: 2Lflat
˜
L2
hilt − L2
flat, but makes the
following replacement Lflat → y and Lhilt → y
tan(α
2
) = Lhilt
Lflat
y, since tan(α
2 ) =
Lflat
Lhilt
(recall the first diagram). Hence we have:
Arhombus(x) =2y
d
(
Lhilt
Lflat
y)2 − y2 = 2y
d
y2((
Lhilt
Lflat
)2 − 1) = 2y2
d
(
Lhilt
Lflat
)2 − 1
=2Lblade(Lflat −
1
Lblade
x)2
d
(
Lhilt
Lflat
)2 − 1. (92)
Having learned calculus from the ‘Principia Mathematica’, Tyrion concludes that
the volume is therefore given by the following function of three measured variables
50 — Lflat, Lhilt and Lblade:
V (Lflat, Lhilt, Lblade) =
x=Lblade
x=0
A(x)dx = 2
d
(
Lhilt
Lflat
)2 − 1
x=Lblade
x=0
(Lflat −
1
Lblade
x)2
dx
=2
d
(
Lhilt
Lflat
)2 − 1
1
3
Lblade
1 − 3Lflat + 3L2
flat
¨
=2Lblade
d
(
Lhilt
Lflat
)2 − 1
¢
L2
flat − Lflat +
1
3
. (93)
Given Tywin’s measurements of the broadsword along with the corresponding pre-
cision error
Lblade = 42in±
1
8
in, Cflat = 4Lflat = 4in±
1
8
in, Crhombus = 4Lhilt = (3+
7
8
)in±
1
8
in,
(94)
50
As a consistency check, one should note that we expect the volume to be a function of exactly
three variables. This is because the cross-sectional area of the sword is parametrised by two-variables
(being a non-square rhombus), whilst the length of the sword is parametrised by another independent
variable. If the cross-sectional rhombus was turned into a square, the sword would be reduced to
a spike and the volume would be a function of two measured variables – the blade length and the
length of one of the sides of the cross-sectional square.
48
49. one can deduce the following measurements and (reduced) errors for the L vari-
ables 51
Lblade = [42 ±
1
8
] in, Lflat = [1 ±
1
2
] in, Lhilt = [
1
4
(3 +
7
8
) ±
1
2
] in. (95)
Problem I From these measurements, compute the volume (in units of inches
cubed), V , of Ned’s broadswoard along with the corresponding absolute error, ∆V .
Convert these measurements into metric units, using the conversion: 1 inch =
2.54 cm.
In a thoughtful moment, Tyrion decides to calculate the financial worth of the
sword in terms of pure Valyrian steel. Given that Valyrian steel is worth 100
times its weight in gold, calculate the total worth W of the broadsword in terms
of kilograms of gold. To do this, use the fact that density of Valyrian steel 52 is
ρ = 7.85 g/cm3 = 0.284 lb/in3. Remember to use consistent units – either stick
with imperial units or convert everything to metric units.
Problem II Given that mass M = V olume × Density = V ρ, compute the
error in the amount of gold Tyrion will make by selling the steel smelted from
the broadsword blade. Assume that the density ρ given is accurate to the num-
ber of decimal places quoted – i.e. the precision error in density is given by:
∆ρ = 0.005g/cm3. Hence deduce the minimum and maximum amount of gold
(W = 100M) Tyrion will make, based on Tywin’s measurements – i.e. compute
W − ∆W and W + ∆W.
• The dimensions in the last question were computed using slightly larger-
than-average dimensions for Claymores and Two-handed swords from the
medieval ages.
• The last exercise illustrates an important technique in making measurements:
by measuring the circumference of the sword rather than just the edge of
the cross-sectional rhombus, the precision error in determining Lflat was
reduced by a factor of 4. In general when making measurements, it is better
to make measurements of quantities which are much larger than the precision
limitation set by your instrument – from these measurements, you can then
deduce measurements for quantities you need with lower precision error. So
for example, in determining the area of a circle with string, it is better to
measure its circumference rather than its radius (since the former is larger) –
this way, one may reduce the precision error in determining the radius by a
factor of 2π.
51
See the remark after this exercise.
52
The density of Carbon 1060 Steel used to make “Ice” replicas for crazy Game of Thrones fans.
49
50. Exercise 14 (Littlefinger’s bane – Lord Tyrion, Master of Coin) Having been given
the responsibility of managing the Kingdom’s finances, Tyrion Lannister finds that
he has inherited some ‘financial discrepancies’ from Littlefinger – that is, he has
found some mathematical ‘short-comings’ 53 in Littlefinger’s bookwork. In partic-
ular, apart from pocketing coin from time to time, Tyrion finds that his predecessor
Littlefinger has been using the wrong interest rate formula to calculate the king-
dom’s debt. Furthermore, Tyrion finds that Littlefinger has been ‘inflating’ the
recorded expenses, so as to inflate his own pockets. Being clever, Littlefinger ran-
domized the expenses which he had inflated and also kept all ‘inflations’ to within
2% of the true expense.
Littlefinger used discrete compound interest, compounded quarterly, to compute
the interest S(t) − S0 that the kingdom owes to a certain bank t years after taking
an initial loan S0. Given an annual interest rate of 7% – i.e. r = 0.07, the amount
owed to the bank is given by
S(t) = S0(1 +
r
m
)mt
(96)
where m = 4 is the number of times the interest was assumed to be compounded
per year. However, driven by avarice, the bank in fact changed the terms of the loan
so that interest was compounded continuously – i.e. m → ∞. With this correction,
Tyrion finds the actual amount owed after t years:
S(t) = lim
m→∞
S0(1 +
r
m
)mt
= S0ert
(97)
where e is the exponential function. First, Tyrion must correct the size of the debt
blackhole that Littlefinger’s endless borrowing has brought the kingdom into. To
do this, Tyrion must estimate the true total expenses E of the kingdom along with
an ‘absolute error estimate’ ∆E to account for the amount of money Littlefinger
has stolen. Once this is done, Tyrion must calculate the amount of interest that
the Kingdom will owe in the next financial year, along with an error estimate to
account how much of this may be due to Littlefinger.
Problem I: First make sure that you understand why discrete compound interest is
given by a geometric sequence and why continuous compounded interest is given
by the exponential function. Now,given an initial loan of S0 = 4, 000, 000gc (gc
= ‘gold coins’) taken t = 12.5 years ago along with a second (separate) loan of
˜S0 = 6, 000, 000gc taken t = 2 years ago, compute the total amount of money,
S + ˜S, that the Kingdom currently owes to the bank.
53
Pun intended.
50
