1. St. George’s College 2015 - Mathematical
Sciences Exploration Studies
Daniel Xavier Ogburn 1
School of Physics,
Field Theory and Quantum Gravity,
University of Western Australia
June 2, 2015
1Electronic address: daniel.ogburn@research.uwa.edu.au
5. Chapter 1
Introduction
1.1 Tutor List
For the year 2015, here is a list of tutors for mathematics, physics and related
areas:
• Myself (Daniel Ogburn)
• Ben Luo
• Theresa Feddersen .
1.2 Exploration Studies and Extension Problems
As a young (or old) individual, one should strive for ‘professionalism’ in their
pursuits. This means doing things ‘properly’ and diligently, without cutting cor-
ners. One aspect of professionalism in the mathematical sciences, is to develop a
forte for ‘problem solving’. Developing your capacity for problem solving is not
something that can really be philosophised or ‘rote-learned’. Certain principles
may help you, but at the end of the day the best way to develop this faculty is to
attempt many different problems. Your mission, should you choose to accept, is
to attempt the problems in these extension studies. Because they are unique and
‘different’ to the style of problems you will usually get at a university, they will
help you to develop in new ways.
Further to improving your ‘professionalism’ and problem solving capacity, these
studies should provide a fun and interesting side journey. We will try and explore
areas of mathematics and physics which are somehow glossed over. Hopefully
you will find that these areas are in fact rich and interesting, with much to be
explored. To a large extent, you will develop tools and insights to give you an
edge in your university work.
Lastly, note that the idea here is to have fun! Engage your peers and your tutors
as you work your way through these explorations. As the prince of mathematics
once said:
“It is not knowledge, but the act of learning, not possession but the act of getting
there, which grants the greatest enjoyment.”– Carl Friedrich Gauss.
5
6. 6 CHAPTER 1. INTRODUCTION
1.2.1 People you should know about
• Euclid
• Archimedes
• Muhammad Al-Kwarizmi
• Carl Friedrich Gauss: Gauss referred to mathematics as ‘The Queen of all
Sciences’ and is almost universally considered to be the ‘Prince of Math-
ematics’. Apart from being a prodigy, Gauss is responsible for the expan-
sion and developments in 18th century mathematics and physics. There-
fore, you will see his name behind many fundamental theorems from all
areas of mathematics.
• Bernard Riemann: Gauss’ best student. Along with Gauss, Riemann is
largely responsible for non-Euclidean geometry – the basis for all modern
theoretical physics. He is also responsible for the ‘Riemann Hypothesis’,
which has been long considered to be the greatest unsolved conjecture in
mathematics.
• Leonard Euler
• Joseph Lagrange
• Joseph Fourier
• Emmy Noether
• Henri Poincare
• David Hilbert
• Elie Cartan
• John von Neumann
• Srinivasa Ramanujan
• Alexander Grothendieck
• Grigori Perelman
• Terrence Tao
• Isaac Newton and Gottfried Wilhelm:
• Gallileo Galilei
• Michael Faraday
• James Maxwell
• Sir William Rowan Hamilton
• Ludwig Boltzmann
• Max Planck
• Paul Dirac
• Albert Einstein
• Richard Feynmann
• Lev Landau
7. 1.2. EXPLORATION STUDIES AND EXTENSION PROBLEMS 7
• Edward Witten
• Nima Arkani-Hamed
• Jin-Mann Wong (Jenny Wong).
1.2.2 Theorems and Theories you should know about
Here are a few of the ‘major’ results that you should know about by the end
of your mathematics / physics degree. Some of them you should understand in
detail – i.e. derivations, proofs and conceptual working knowledge. Other items,
you should at least have heard or come across and understand the essence of the
result if not the specifics. Please note that the list is far from exhaustive and there
are probably many important results that I have ommitted at this time.
• Euler’s Formula
• The Fundamental Theorem of Algebra
• Weirstrass Factorization Theorem
• The Generalized Stokes’ Theorem
• The Cauchy Residue Theorem
• The Gaussian Distribution and Hypothesis Testing
• Principle of Least Squares Regression and the L2
, l2
Hilbert Spaces
• The Multinomial Theorem
• Uniform, Binomial, Poisson and Chi-squared distributions
• Non-parametric statistics
• Fourier’s Theorem and its generalizations
• The Spectral Theorem
• The Shannon-Nyquist Theorem
• Parseval’s Theorem
• Shannon Entropy
• The Riemann Hypothesis and Prime Number Theorem
• The Navier-Stokes Equation and associated Millenium Problem
• Thurston’s Geometrization Theorem (Poincare Conjecture)
• Information Complexity and P = NP Conjecture
• Fermat’s Last Theorem
• Maxwell’s Equations in 3-dimensional and 4-dimensional form
• The Black-Scholes Equation
• Markov Chains and Monte Carlo Methods
• Game Theory and the Nash Equillibrium
• Measure Theory and the Lesbegue Integral
• Dedekind’s Construction of the Real Numbers
8. 8 CHAPTER 1. INTRODUCTION
• Proof of the Irrationality of
?
2.
• Cantor’s diagonalization argument and classification of infinities
• Godel’s Incompleteness Theorem
• The Axiom of Choice
• Lagrange’ Thereom (Group Theory)
• The First and Second Isomorphism Theorems
• Splitting Lemma
• The Jordan-Holder Theorem
• Hamilton’s Quaternion Formula
• Brouwer Fixed Point Theorems
• The Hairy Ball Theorem
• Banach-Tarksi Paradox
• The Lyapunov Exponent and Lorenz Attractor
• Noether’s Theorem and Killing’s Equation
• De-Rham Cohomology, Closed Forms and Conservation Laws
• Topological Invariants and the Atiyah-Singer Index Theorem
• Newton’s Second Law
• Newton’s Law of Gravity and Derivation of the Kepler Orbits
• The Lorentz Symmetry Group and Special Relativity
• Planck’s Radiation Law
• Boltzmann’s Equipartition Theorem and Proof
• The Ising Model and its solution
• The Bose-Einstein Distribution and Fermi Distribution
• The EPR Paradox
• Boltzmann’s Entropy Formula
• LASERS and Stimulated Emission
• Spherical Harmonics and Quantum Model of the Hydrogen Atom
• General Relativity, Einstein Field Equations and Einstein-Hilbert Action
• The Standard Model of Particle Physics
• Lie Groups and Lie Algebras
• Hilbert Spaces, Metric Spaces and Topological
• The Heisenberg Uncertainty Principle
• The Schrodinger and Dirac Equations for Quantum Mechanics
• The Klein-Kordon Equation and Feynmann Path Integral
• The Hawking-Bekenstein and Unruh Black Hole Thermodynamic Formu-
las
9. 1.3. LAYOUT AND CONVENTIONS 9
• The Hawking-Penrose Singularity theorems and Penrose black hole In-
equalities
• The Magnetohydrodynamic Equations
• Hamiltonian and Lagrangian Varitational Principles
• The Hamilton-Jacobi Equation and Louiville’s Theorem
• Sturm-Liouville Theory and Harmonic Analysis
• Gamma Function, Hypergeometric Function and Orthogonal Polynomial
Families
• Eigenvalue/Eigenvector Decomposition and Jordan Normal Form
• The Matrix Exponential
• The Standard Model of Cosmology
• Random Matrices and Random Walks
• The Cosmological Constant Problem and Dark Energy
• Approaches to Quantum Gravity
• Ogburn-Waters-Sciffer method for generating ellipsoidal harmonic func-
tions.
1.3 Layout and Conventions
For these exploration sessions, I will typically include excerises and problems in
increasing order of difficulty. Furthermore, I make the (somewhat grey) distinc-
tion between ‘exercises’ and ‘problems’ as follows.
Definition 1 (Exercise) ‘Exercises’ are essentially numerical or algorithmic cal-
culations to check whether you understand how to manipulate the mechanics of
the mathematics involved.
‘Problems’ are generally more difficult as they require more conceptual under-
standing – i.e. you have to be able to interpret the problem and formulate in such
a way that it is reduced to an ‘exercise’.
Definition 2 (Problem) A ‘problem’ is something which can be reduced to an
exercise, with the appropriate creativity and conceptual faculty.
One essential skill for any true scientist, is the ability to reduce real-world sce-
narios, models and problems into mathematical ones. In this manner, the most
powerful sciences (quantitative ones) are ones in which scientists turn problems
into exercises.
11. Chapter 2
Dimensional Analysis and
Fundamental Laws
2.1 Dimensional Analysis
2.1.1 Preamble: March 9, 2015
Dimensional analysis is a deceptively simple, but fundamentally powerful tool in
the mathematical sciences – one that is often overlooked! Furthermore, dimen-
sional is not just a study tool – it is a research-level tool that can allow one to
probe the unknown and construct or discover something new and tangible.
For student purposes, dimensional analysis serves as fast error-checking algo-
rithm for your calculations. It is also useful for extracting ‘physically meaning-
ful’ information out of your system. In this sense, being ‘dimensionally-aware’
throughout all of your calculations can help one to develop a deeper understand-
ing of the quantities and objects being manipulated.
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
. This has vast practical appli-
cations to mathematical modelling, fluid mechanics, thermodynamics, electro-
dynamics, cosmology and much more. For now, we begin with a few examples
then work through some questions 2
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 mathemat-
ics 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!
What was said for ‘dimensional consistency’ is also true of ‘structural consis-
tency’ – a concept which we shall briefly review below.
1
For those of you who have done (or will do) linear algebra, this is just a practical conse-
quence of the ‘rank-nullity’ theorem.
2
Thanks to Scott Meyer and Matthew Fernandez for feedback.
11
12. 12CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
Theorem 1 (Dimensional Consistency) Given any equation:
LHS = RHS (2.1)
In any set of variables, the equation is wrong if the dimensions of ‘LHS’ and
‘RHS’ are not equivalent.
An easy way to prove this theorem is to note that given any equation, we have an
associated ‘auxiliary dimensional equation’. That is, an equation consisting only
of the dimensions of LHS and RHS, which we shall denote with the ‘square-
bracket notation’ [LHS] and [RHS], respectively:
[LHS] = [RHS]. (2.2)
Hence for an equation to be correct, it must be both numerically correct and
dimensionally correct.For more complicated equations, such as tensor or spinor
equations, we must also consider ‘structural consistency’.
Notation: Throughout this tutorial we will use ‘logarithmic notation’ for dimen-
sional analysis. This is often used by cosmologists and particle physicists, but
can be easily converted to the more common ‘exponential notation’.
2.1.2 Examples
Example 1 (Structural Consistency) One simple example of structural consis-
tency is matrix mutiplication – we can only multiply two matrices A and B if the
number of columns of A is equal to the number of rows of B.
Any equation that is not structurally consistent is fundamentally nonsensical and
therefore wrong. Therefore, it is wise to be always be mindful of the structures
involved in an equation.
Now recall lengths, areas and volumes. The fundamental unit that characterizes
these quantities is length: L.
Example 2 (Spatial Dimensions) 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 (in ‘exponential
notation’) L3
. Of course, we already knew this!
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. (2.3)
13. 2.1. DIMENSIONAL ANALYSIS 13
Problem 1 (Hyperbox) Sometimes playing in a 3-dimensional world is boring
– which is why 10-dimensional superstring was invented 3
. Consider now a
N-dimensional hyperbox, which we shall refer to simply as ‘N-box’. So for ex-
ample, a cube can be considered to be a 3 − box. In determining the amount
of material required to mass-produce N-boxes, Microsoft comes up with the fol-
lowing equationw for the ‘hyper-volume’ and of an N − box with sides of equal
length ‘a’:
V ol(N − box) = aN−1
. (2.4)
Explain mathematically why this equation is incorrect, or (possibly) correct on
dimensional grounds.
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 whey protein in milk, it has units ML−3
of mass over
volume – hence dimensionally: [C] = M − 3L.
Now that we have done some simple exercises, lets see how dimensional analysis
can be used for error checking. Lets say someone tells us that the volume VS of
a sphere of radius R is given by
VS =
4
3
πR2
. (2.5)
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
3
π] = 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
3
πR2
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 3 Newton’s Second Law of Motion:
Force = Mass × Acceleration, (2.6)
or more generally,
F = m
dP
dt
, (2.7)
where P is the momentum of a particle of mass m. Newton’s Second Law of
motion is the fundamental postulate governing classical physics between the late
3
Disclaimer: String Theory may have been invented for other reasons.
14. 14CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
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
displacement 4
, has units of length over time:
[v] =[
dx
dt
]
=[dx] − [dt]
=[x] − [t]
=L − T , (2.8)
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, (2.9)
which means ‘a has units of length over time-squared: L
T2 . Finally, mass m
trivially 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,
(2.10)
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 dimension-
ally consistent
A =
1
2
b2
h. (2.11)
2. A circle has a radius r with dimensions of length L. Its area is given by
A =
1
2
πr2
. (2.12)
Is this dimensionally consistent? A stronger question to ask is whether
this formula is correct – if not, why not?
4
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].
15. 2.1. DIMENSIONAL ANALYSIS 15
A Euclidean ellipse, produced by slicing a cone, can be defined as: “the set of
points uch that the sum of the distances to two fixed points (the foci) is constant.”
In a parallel universe, Rene Descartes decides that he write the equation for an
ellipse with its major axis coincident with the x − axis of the Cartesian plane,
as:
x2
a2
+
y
b2
= 1, (2.13)
where the right-hand side is dimensionless.
Here a is the semi-major axis length and b is the semi-minor axis length. Noting
that one parametrisation in polar coordinates is x = a cos(θ), y = b sin(θ), we
can see that x and y have units of length. Therefore, prove that this equation is
wrong on dimensional grounds. Now suggest the correct equation.
There is one more rule of dimensional analysis which involves analysing equa-
tions 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], (2.14)
but rather, we have the consistency requirement that:
[A] = [B] = [C] = [D]. (2.15)
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 un-
known 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
2
bh + a2
, where b is the base length of the triangle, h is its verti-
cal length and a is the length of the sides of the square. Therefore, [AHouse] =
[ATriangle] = [ASquare] = 2L, hence [1
2
bh] = [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 prop-
erty – indeed, the dimensionless constants of a system describe a universality
class 5
. A university class is essentially a set of theories which have the same
‘critical behaviour’ – i.e. near a critical point (e.g. phase transition), each the-
ory in the same universality class will possess quantities which obey the same
scaling laws.
2.1.3 Problems
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!
Problem 2 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
. (2.16)
5
A more precise meaning of this statement can be found in the theory of ‘Renormalization
Groups’.
16. 16CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
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 ‘hyper-faces’.
Exercise 3 (Make Math, Not War) The U.S. Navy invests a significant amount
of money into acoustic scattering studies for submarine detection (SONAR). As
part of this research, the Dahlgren Naval Academy uses ‘prolate spheroidal har-
monics’ (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
generated 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
(2.17)
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))? (2.18)
Note, sin−1
is the ‘inverse sine’ or ‘arcsine’ function. It necessarily preserves
dimensionality, hence [sin−1
(e)] = [e]. The variable e is the ‘eccentricity’ 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
. (2.19)
A perfect sphere corresponds to e = 0, where as an infinitely stretched sphere
corresponds to e → 1.
Problem 3 (Twiggy) In a parallel-universe, Andrew Forrest has a dungeon with
BF flawless black opals inside it. From a financial point of view, these have di-
mensions 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. Sens-
ing the loss of his monopoly on the black opal market, Andrew Forrest employs
a competing physicist at Curtin University to create a quantum vacuum stabi-
lizer. This reduces the number of black opals that Ian can produce per minute by
RC black opals per minute, where |RC|≤ RUWA. Working on a broad concept
problem, a team of first year students at St. George’s college 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
)
(2.20)
where the constant τ (having dimensions of time T) denotes 6
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 depends on the num-
ber of avocados served at the college since the establishment of St. George’s
Avocadoes Anonymous up to the given time t.
6
This is the Greek letter tau – not the Roman letter t.
17. 2.1. DIMENSIONAL ANALYSIS 17
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
(2.21)
On dimensional grounds, list two reasons why this model incorrect.
Problem 4 (Spacetime Surfing) Don’t worry about the physics, just keep track
of dimensions and rules!
On March 17, 2014 the Harvard-Smithsonian Center for Astrophysics released
a press-conference tomorrow indicating the discovery of gravitational 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, dimension-
ally we have: [time] = [distance] and [c] = [distance/time] = L − T = 0. A
geometry which models gravitational 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 (2.22)
where η is a flat-space metric (describing an empty universe):
η := −dt + dx ⊗ dx + dy ⊗ dy + dz ⊗ dz (2.23)
and h is a symmetric-tensor, given in de-Donder gauge by
h := cos(k · r)A +
1
2
× trace(h) × η. (2.24)
Here is a small (<< 1) dimensionless parameter: [ ] = 0 and A is a symmetric
tensor field with dimensions of length-squared: [A] = 2L. Note that the trace
operation turns tensors into scalars, so it removes the dimensionality 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 multiplication 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 di-
mensionless). Therefore, pretending [A] is unknown, prove that [A] = 2L given
all the other information.
18. 18CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
After completing the last few 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!
Therefore, 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. 7
Problem 5 (Super-Fluffy Super-Symmetric Tensors) Despite successful ‘solar-
system tests’ of Einstein’s theory of gravity, it has severe shortcomings. One
fundamental issue with Einstein’s theory is that it is not consistent with quantum
theory (which we know is extremely accurate on short-distance scales) – leading
to problems such as the ‘cosmological constant problem’. Another problem is
that it predicts singularities where the laws of physics breakdown. To rectify
Einstein’s theory, many physicists have attempted to unify gravity with quantum
mechanics over the last century. As it turns out, creating a theory of quantum
gravity presents immense mathematical and experimental obstacles.
One approach to understanding quantum gravity, is to consider supersymmetric
theories containing particles of ‘higher spin’. Such theories are conjectured to
reduce back to superstring theory when some symmetry is broken. If so, such
theories (when constructed) will lead to a greater understanding of the global
structure of string theory – for example, dualities. To construct a supersymmetric
theory with massive higher spin particles, one must find a geometric object called
the ‘Super-cotton tensor’ – this describes the conformal8
(‘shapes and angles’)
structure of spacetime. To help find this tensor, Wα(2s), we know that it has
following dimensional form:
[W] = [(D ¯D)2s+1
]H, (2.25)
where Hα(2s) is the gravitational superfield (when s = 1) and Dα and ¯Dβ are
‘spinor-derivative’ operators.
Roughly speaking, along the lines of Roger Penrose, one can think of a (mass-
less) spinor as the ‘square root’ of some vector field. Therefore, the square of a
spinor must have the same dimensionality as a partial derivative:
[D2
] = [ ¯D2
] = [
∂
∂x
]. (2.26)
Furthermore, analysis of non-minimal and type (1,1) supergravity actions leads
us to conclude that:
[H] = −
1
2
M, (2.27)
where M is the unit mass. Note that in natural units with the speed of light
,c = 1, the unit of mass and unit of length are inverses of each other:
M = −L, (2.28)
or M = L−1
in exponential notation. Finally, note that a differential 1-form dx
can be thought of as a differential length element, hence has units of length:
[dx] = L. (2.29)
7
Dimensional analysis would have saved the present author about 100 hours of supergravity
calculations – time which was largely lost due to two dimensionally-inconsistent equations in a
published journal article.
8
Conformal symmetries are ones that preserve relative angles, but not lengths. For example,
a scaling transformation is an example of a conformal transformation.
19. 2.1. DIMENSIONAL ANALYSIS 19
Since ∂
∂x
can be thought of as a rate of change of length, it must have the inverse
units of dx.
I: With this information, deduce both the mass dimensions and length dimen-
sions of the Super-Cotton tensor W.
II: Using the flat superspace anti-commutation relation
{Dα, ¯Dβ} := Dα
¯Dβ − ¯DβDα = ∂αβ, (2.30)
where ∂αβ ∝
3
a=1
σa
αβ∂a is the spinor form of the vector derivative ∂a := ∂
∂xa ,
derive the relation:
[D] =
1
2
[
∂
∂xa
] (2.31)
which was assumed in part I.
Note that you may assume [D] = [ ¯D] and that the Clifford algebra generator
σa
is dimensionless. Also, note that superscript a in xa
is simply a coordinate
label. For a 3-dimensional manifold, a = 1, 2, 3 so we have local coordinates
x1
, x2
, x3
to keep track of points in space.
2.1.4 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 when-
ever 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 ...
20. 20CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
2.2 Dimensionless Constants and Fundamental Laws
2.2.1 Physical Systems, Fundamental Laws
One of the key concepts in dimensional analysis is that of dimensionless param-
eters. Dimensionless parameters are important, because they allow you to char-
acterise both physical and theoretical mathematical systems in a scale-invariant
way. Note that mastering the following concepts and exercises requires a good
understanding of the material in Session 1. For the more mathematically in-
clined, one of the examples and exercises illustrates how to mathematically prove
the π theorem by using the rank-nullity theorem from linear algebra – this is a
good exercise for understanding matrix equations and the correspondence be-
tween matrices and simultaneous equations.
In terms of applications, we will use dimensional analysis to reach a deeper
understanding of simple harmonic motion, viscous fluids, electromagnetism and
Einstein’s theory of gravity.
Definition 3 (Physical Systems) A physical system in the mathematical sciences
typically consists of:
1. A set of physical parameters.
2. A set of governing equations (fundamental laws) which describe the be-
haviour and evolution of the system.
3. A set of fundamental ‘units’ which describe the dimensionality of the sys-
tem.
Definition 4 (Fundamental Law) A fundamental law is a principle, mathe-
matical statement or an axiom used to describe a system, which cannot be de-
rived from any other principles, equations or axioms.
In this sense, we can view the ‘fundamental laws’ of the natural sciences as more
heauristic notion of ‘mathematical axioms’ (formal assumptions agreed upon by
utility and sensibility).
Arguably, the notion of a ‘fundamental law’ is relative to the context of one’s
analysis. For example, many classical laws, previously considered to be fun-
damental, are a macroscopic consequence of quantum dynamics or statistical
mechanics. However, if we are only looking for classical effects in our analysis,
we may often ignore the quantum mechanical details and treat our classical laws
as ‘fundamental’. The goal of the natural sciences – including mathematics, is to
reduce the number of ‘fundamental laws’ of nature to a minimum. In this sense,
one is able to capture nature in the ‘simplest way possible’9
. In this manner, per-
haps the largest and longest standing goal of theoretical physics, is to construct
and experimentally test a full theory of quantum gravity.
Fundamental laws often go hand-in-hand with one or more special ‘dimension-
less constants’ that capture information and deep insights about the mathematics
and physics of systems governed by that law. We shall now investigate such
examples.
9
Confer with ‘Occam’s Razor’.
21. 2.2. DIMENSIONLESS CONSTANTS AND FUNDAMENTAL LAWS 21
2.2.2 Examples and Problems
Example 4 Lets take a simple, but profound 10
example – the simple harmonic
oscillator. One example of a simple harmonic oscillator, is a mass placed on
a frictionless 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 posi-
tion x0 and initial velocity v0 of the mass m.
2. An equation of motion called ‘Hooke’s Law’ 11
, 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 (2.32)
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
d2
x
dt2
= −κx, (2.33)
where a = d2x
dt2 is the acceleration of the spring.
3. A set of 3 physical units: mass M, time T, length L (usually kilograms,
seconds, metres).
Ignoring microscopic and non-linear effects, we may naively view ‘Hooke’s law’
as a fundamental law (or definition) concerning systems in simple harmonic mo-
tion. Therefore, it must have some special ‘dimensionless constant’ attached to
it.
From the 4 parameters and 3 physical units in the simple harmonic oscillator
system, the Pi theorem claims that we can form one dimensionless constant. To
do this, one needs to know the dimensions of the parameters 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
d2
x
dt2
] = [−κx] =⇒
[m] + [
d2
x
dt2
] =[κ] + [x]
M + L − 2T =[κ] + L =⇒
[κ] =M − 2T. (2.34)
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
10
Despite its simplicity, the (quantum) harmonic oscillator is the cornerstone for modern quan-
tum 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.
11
After the famous pirate, Captain Robert Hooke.
22. 22CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
[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. (2.35)
Since the constant G has no formal name, we will claim it and call it the ‘Geor-
gian 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 (corresponding to the ‘dimensions’) and
at least one governing equation. Second, we showed how we can calculate the
units of an otherwise unknown constant by using dimensional analysis – this is
how we found the dimensions of the spring constant κ.
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 a dimen-
sionless quantity, is 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 2 (Buckingham Pi Theorem) Given a system specified by n inde-
pendent parameters and k different physical units, there are exactly n − k in-
dependent 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 de-
scribed 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 5 In fluid mechanics, the notion of the ‘thickness’ of a fluid is formal-
ized 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 12
and honey have a very high shear viscosity, whereas
superfluid Helium has zero viscosity 13
.
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
(2.36)
12
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.
13
The transition to the ‘superfluid’ phase occurs below 1 Kelvin – i.e. close to absolute zero
temperature.
23. 2.2. DIMENSIONLESS CONSTANTS AND FUNDAMENTAL LAWS 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 [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 (2.37)
whence η has units of M
LT
. Now, the kinematic viscosity ν 14
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
. (2.38)
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. (2.39)
In some set of scenarios, we can think of this fluid as parameterized by four
parameters: 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
(2.40)
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.
The Reynold’s number also controls the amount of ‘turbulence’ present in a fluid
system – with high Reynolds numbers corresponding to turbulent flow. There-
fore, by evolving the dimensionless Reynolds constant from low values to high
values, we will see a laminar flow turn into one with instabilities, vortices and
chaos.
14
This is the Greek letter ‘nu - not the Roman letter ‘v’.
24. 24CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
2.2.3 Buckingham Pi-Theorem
Formally, the rank-nullity theorem states that given a m × n matrix (m rows, n
columns) A, which maps n-dimensional vectors in Rn
to m-dimensional vectors
in Rm
, then the rank and nullity of the matrix A satisfy:
rank(A) + nullity(A) = n (2.41)
where the rank of A is defined as the number of linearly independent row vectors
(or column vectors) 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).
In the context of dimensional analysis and the π Theorem, we can think of math-
ematical or physical system with n parameters and k different types of funda-
mental units (dimensions) as a system of n linear equations (one for each pa-
rameter) in k variables (the units). In particular, we make use of the additive or
‘logarithmic’ notation which we have been using for dimensional analysis.
Problem 6 (π Day, π theorem) On 04/03/2015, Tibra Ali decides to have a Pi
battle at the Perimeter Institute for Theoretical Physics. On the same day, An-
gela Burvill and Joshua Bailey decide to have a Pi eating contest – where each
student has to eat one frozen meat pie for each correct digit of Pi that other re-
cites. However, realizing that becoming a mathematician requires thousands of
hours of diligence, William decides to prove the ‘Buckingham Pi theorem’ using
the ‘rank-nullity’ theorem from linear algebra – which he remembers from last
semester! To up the stakes, Ben Luo decides to dangle Rowan Seton from the
top of the college tower till William proves his theorem.
Assuming Ben has finite strength, save Rowan by proving the Pi theorem with
William.
Hint for proof: For the above problem, note that we can view a system with
paramters χ1, ..., χn, fundamental units u1, ..., uk and the following dimensions
for the parameters:
χ1 =λ11u1+ . . . + λ1kuk
χ2 =λ21u1+ . . . + λ2kuk
...
...
...
...
χn =λn1u1+ . . . + λnkuk , (2.42)
as a system of n linear equations in m variables. One can now apply the rank-
nullity theorem to this system that the number of dimensionless constants which
one can form from the corresponding physical system, should be equal to the
nullity of the n × k ‘dimensional matrix’ formed by the coefficients λij where
i = 1, ..., n and j = 1, ..k.
Example 6 As a simple example of the algebra required, say we have a system
with three parameters x, y, z and two fundamental physical units U1, U2. 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. Therefore, we let the first column correspond to the parame-
ter x, the second column to y and the third column to z.
The he 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 dimensions of x
25. 2.2. DIMENSIONLESS CONSTANTS AND FUNDAMENTAL LAWS 25
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.
Forming the ‘dimensional matrix’ D for this physical system, we have:
D =
a c e
b d f
(2.43)
To see that this makes sense, we can simply act15
the transpose of the dimen-
sional matrix DT
on the vector U =
U1
U2
containing the physical units to re-
cover all three of our dimensional equations [x] = aU1 + bU2, [y] = cU1 + dU2
etc. To find dimensionless constants, we have to solve the ‘nullspace equation’:
a c e
b d f
!
α
β
γ
(
) =
0
0
for all possible vectors !
α
β
γ
(
). In particular, dimension-
less constants will be a product of powers of the different physical parameters:
xα
yβ
zγ
, 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
matrix equation, coincides with the ‘nullity’ of the dimensional matrix D – it
is precisely equal to the number of dimensionless constants we can form. In
particular, since we have n = 3 independent physical parameters x, y, z corre-
sponding to three columns of our dimensional matrix D and k = 2 fundamental
units U1, U2 corresponding to the two (linearly-independent 16
) rows of D, the
rank-nullity theorem tells us that the nullity of D is given by
nullity(D) = n − k = 3 − 2 = 1. (2.44)
Since the nullity of D is precisely equal to the number of dimensionless con-
stants we can form for this physical system, this shows that the π Theorem for
dimensional analysis, is just a special instance of the rank-nullity theorem for
linear algebra.
2.2.4 Gravity, The Hierarchy Problem and Extra-Dimensional
Braneworlds
The following is an extended set of exercises which test all the skills the tuto-
rials have elucidated 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 relativ-
ity 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 investigate a solution to this peculiarity using
brane-world models of the universe.
15
By matrix multiplication.
16
These rows are necessarily linearly independent, since we assume our fundamental physical
units to be independent – by definition.
26. 26CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
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 experi-
mental quantum description of three of these forces – that is, we have constructed
quantum field theories to describe the ‘quanta’ (particles) which mediate these
forces. Gravity, despite our everyday experience of it, remains somewhat myste-
rious 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. If
there really any hope of long-distance interstellar space travel and other extreme
‘sci-fi’ technologies, a theory of quantum gravity will be the cornerstone.
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
description (Quantum Electrodynamics), this force is carried by massless
particles 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 inter-
actions 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.
• 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 tra-
jectories 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
example, the electromagnetic and weak nuclear forces unify to make the elec-
troweak force. Attempts to unify the electroweak and strong nuclear forces
have been partially 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 unsuccessful, with the only real promising candidate being String
Theory.
One of the biggest mysteries about the gravitational force, is why it is so weak
compared 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 charac-
terise the gravitational and electromagnetic forces with some special dimension-
less constants – then compare their strengths to prove this claim. Finally, we will
end on some very recent 17
advancements in theoretical physics which propose
an explanation of why gravity is the weakest of the four forces.
Exercise 5 (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 gravitational force law to explain the astronomical bservations of
17
The last 5-10 years.
27. 2.2. DIMENSIONLESS CONSTANTS AND FUNDAMENTAL LAWS 27
Johannes Kepler and Tycho Brae:
F = −GN
m1m2
r2
ˆr (2.45)
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.
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 funda-
mental units, hence [m1] = [m2] = M. Furthermore, by definition the unit
vector 18 ˆr = r2−r1
|r2−r1|
is dimensionless: [ˆr] = 0. Note that in general, the di-
mensions 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 – fundamentally 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 necessary 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 19
• 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
spacetime varies in different regions. It also measures how gravity distorts
parallel-transport. It is given roughly 20
as the anti-symmetrized second
tensor ‘gradient’ of the metric: Riem ∼ ⊗ ⊗ g, where are a type
of derivative operator and ⊗ is a type of multiplication for tensors.
18
Here r1 and r2 are the position vectors describing the location of the masses m1 and m2
with respect to some origin.
19
Note that most physicists do not understand differential geometry, hence when they speak
of tensors 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 com-
pared to their components and contravariant tensor have two factors less than their components
– which basically means adding ±2L to the dimensions.
20
Don’t ever show this to a differential geometer. If you want the real definition, see me.
28. 28CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
• 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 grav-
ity 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 21
. It is given by the taking the trace of Riemann
tensor twice: R = Trace(Trace(Riem)) = Trace(Ric).
QII:Using the above information, derive the dimensions of Newton’s gravita-
tional constant GN again, this time using Einstein’s law of gravity:
Ric −
1
2
Rg =
8πGN
c4
T. (2.46)
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 con-
stant – 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.
In particular, we have:
αG =
GN m2
e
¯hc
≈ 1.7518 × 10−45
(2.47)
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 22
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. (2.48)
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.
21
In retrospect, David Hilbert deserves almost the same level of credit as Einstein for the
theory of general relativity.
22
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.
29. 2.2. DIMENSIONLESS CONSTANTS AND FUNDAMENTAL LAWS 29
Now, for the last part of this problem, we introduce one more fundamental phys-
ical unit: the unit of electric charge, Q 23
. 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) pre-
dicted 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
(2.49)
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
electric permittivity of the vacuum. In these units, the fine-structure constant is
given by
αEM =
e2
4π
≈ 7.297 × 10−3
. (2.50)
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 con-
sistent 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
. (2.51)
This says that the electromagnetic force is about 42 orders of magnitude24
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 uni-
verse 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 ‘graviton’
particles) 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 25
, the introduction of large extra-dimensions also
solves the ‘Dark Energy’ or ‘Cosmological Constant’ problem – where Dark En-
ergy naturally 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.
23
The SI unit for charge is Coulombs.
24
Note, 42 is also the meaning of life.
25
Those investigated in the present author’s masters thesis.
30. 30CHAPTER 2. DIMENSIONAL ANALYSIS AND FUNDAMENTAL LAWS
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 super-symmetry
added – this enables bosons and fermions to transform into each other 26
. In
these models, the extra-dimensions take the form of some compact hypersur-
face. Newton’s gravitational constant GN is then derived from the relation 27
:
GN =
3κ2
16πS
(2.52)
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.
26
Supersymmetry removes the problem of Tachyons in String Theory and also stabilizes the
mass of the Higgs boson.
27
First derived in this generality by the present author in 2013.
31. Chapter 3
Geometry of Antiquity and The
Universe
3.1 Introduction: Conic Sections
Our scientific perception of the world today, is due largely to the great geometers
of antiquity. Pythagoras’ theorem for example, essentially defines the ‘straight-
line’ (Euclidean) distance between two points in space – giving us Euclidean
preconceptions of the world. In this manner, one of the most influential devel-
opements that the Greeks left us with, is the theory of conic sections. Developed
to a large extent by Appolonius and Archimedes, conic sections have provided a
core staple of the framework for the scientific renaissance instigated by Galileo
and Kepler – leading ultimately to Newton’s theory of gravity, the planetary or-
bits and a heliocentric view of the universe.
Definition 5 A traditional conic section is the curve of intersection, obtained by
slicing a cone with a plane. Geometricallys, a general conic C is a set of points
S whose distances to a fixed point (focus) F and a fixed line (directrix) l are in
a constant ratio (the eccentricity) . Algebraically:
p ∈ C ⇐⇒
d(p, F)
d(p, l)
= , (3.1)
where d is any metric (measure of distance).
Note that in the special case of a (Euclidean) circle, the focus is at the center
of the circle and the directrix is at infinity – hence the eccentricity = 0 for a
circle.
The following problem should be re-attempted at the end of each (conic) section
of this chapter.
Problem 7 (GoPro or Go Home) Frustrated by his attempts to retake Constantino-
ple from the neo-Ottoman empire, the cyborg Emperor Constantine decides to go
home. Getting into his skytaxi, which travels on fixed skylanes which permit only
perpendicular turns, Constantine realises that he travelling in an l1 metric space
– the ‘taxicab geometry’. Here the distance between two points P1 = (x1, y1)
and P2 = (x2, y2) in R2
is defined by the so-called taxi-cab metric:
d1(P1, P2) := |x1 − x2|+|y1 − y2|. (3.2)
1. Using the geometric definition of a parabola, sketch the graph of a few
parabolas with different focal lengths in the taxi-cab metric.
31
32. 32 CHAPTER 3. GEOMETRY OF ANTIQUITY AND THE UNIVERSE
2. Using the geometric definition of a circle, sketch the graph of a unit circle
in the taxi-cab metric.
3. Using the geometric definition of an ellipse, sketch the graph of a few
ellipses – with varying eccentricity, in the taxi-cab metric.
4. Using the geometric definition of a hyperbola, sketch the graph of a few
hyperbolae in the taxi-cab metric.
5. Compare the above ‘taxi-cab’ conic sections to graphs of the correspond-
ing Euclidean conic sections.
3.2 Parabolas and Geometric Optics
3.2.1 Overview
The first, most significant application of parabolas, was in Galilleo’s revolution-
ary projectile motion experiments. Sesequently, they served as ‘Victoria’s secret
models’ for Isaac Newton’s ‘Principia Mathematica – in particular, in his analy-
sis of conic sections and Kepler’s laws of planetary motion.
We shall begin first by giving a general geometric definition of the parabola,
then deriving the canonical (natural) equation for a Euclidean parabola in carte-
sian coordinates. Once this is established, we will investigate and derive a few
remarkable properties of parabolas – in particular, motivated by the ‘science of
light’ (optic). Finally, we will study some fun, practical applications of parabo-
las in regards to the natural world – geometric optics, projectile motion and
parabolic orbits.
3.2.2 The Parabola
Definition 6 (Geometric Definition) A parabola is the set of points which is
equistant from a focus (fixed point) and directrix (fixed line).
It follows from the definition, that a parabola is a conic section with eccentricity
= 1. In particular, it can be obtained by slicing a cone parallel to a plane
tangent to the cone.
See Whiteboard Diagram
By now, most of you will be familiar with algebraic forms of the parabola. For
example, as a rational normal curve with exponent 2 in algebraic geometry, or as
cartesian equation: y = x2
from Euclidean geometry. You will now derive the
canonical Euclidean equation y = x2
from the geometric definition.
Exercise 6 (Pachelbel’s Parabola and The Canon Equation) Having being am-
bushed by the ineffable weeping angels, Dr. Who is forced back to the Renais-
sance Era. In a severe misunderstanding, he accidentally replaces Pachelbel’s
Canon for a derivation of Canoncial Euclidean parabola equation – which he
needs to return to his own timeline. Trying to make sense of the parabola,
Pachelbel decides to invent the Cartesian coordinate system so he can graph
this technology of the ‘future’.
Help Palchelbel by deriving the canonical parabola equation, while sketching
every step clearly.
33. 3.2. PARABOLAS AND GEOMETRIC OPTICS 33
1. Draw horizontal (x) and vertical (y) coordinate axes. On the vertical axis
– the symmetry axis of the parabola, mark the origin, (0, 0) – this is the
vertex of the parabola. Upwards from the origin, on the symmetry axis,
mark the point F = (0, f) – this is the ‘focal point’ (focus) of the parabola
and f is the ‘focal length’. Below the x-axis, draw the line y = −f – this
is the ‘directrix’ of the parabola.
Note that the value, y = −f for the directrix, can be derived from the
definition of the parabola having chosen the origin 0 = (0, 0) to lie on the
parabola. In particuarly, the length OF is equal to the distance from O to
a point perpendicularly below O on the directrix.
2. Pick any point P in the plane – preferably one in the positive (x, y) quad-
rant. Now draw a line FP between this point and the focus F. Draw
another line PD from P to a point D perpendicularly below on the direc-
trix. By the definition of the parabola, the lines FP and PD should have
equal length. Therefore, using Pythagora’s theorem to compute the length
of FP, derive the following relation:
(y + f)2
= x2
+ (y − f)2
. (3.3)
3. Using the above equation, show that:
y =
x2
4f
. (3.4)
This is the cartesian equation for a Euclidean parabola with focal length
f, axis of symmetry along the y-axis and vertex (0, 0). Setting f = 1
4
, we
get the ‘canonical parabola equation’:
y = x2
. (3.5)
Problem 8 (The Doctor’s Cannon) Having finished his derivation, Pachelbel
returns to Dr. Who to verify his mathematical construction. At this point, Dr.
Who has added skrillex to Pachelbel’s cannon. Furious, Pachelbel demands that
Dr. Who remove all new additions to the cannon. Reluctantly, Dr. Who decides
that he shall acquiesce provided that Pachelbel removes all redundant steps from
his mathematical derivation and justify its generality.
Help save history from skrillex by helping Pachelbel in his derivation. In par-
ticular, some steps in the above derivation provided superfluous, ‘a-priori’ infor-
mation. Can you identify which ones?
Furthermore, we choose the origin to be the vertex and the y − axis to be the
symmetry axis – this made calculations easier. What ‘obvious’ properties of Eu-
clidean space allow us to do this, without losing any generality in our derivation?
Problem 9 (Constantine’s Plasma Cannon) On his way home, cyborg Emperor
Constantine’s taxicab is ambushed by the weeping angels who are hunting Dr.
Who throughout spacetime. As a result, the emperor is teleported back to Pachel-
bel’s study in the Renaissance era. Seeing this opportunity, Dr. Who and Pachel-
bel beg for the emperor’s help – in particular, his plasma cannon should give the
angels something real to weep about. To this end, Constantine decides he will
help Dr. Who and save the universe from skrillex music ... iff Dr. Who helps him
to sketch and derive the parabola equation in the taxi-cab metric.
Help Constantine help Palchelbel help Dr. Who, by writing down the cartesian
equation for a parabola of focal length f with distances defined by the taxi-cab
metric d1 instead of the euclidean metric.
34. 34 CHAPTER 3. GEOMETRY OF ANTIQUITY AND THE UNIVERSE
Now sketch this parabola.
3.2.3 Scale Invariance and Transcendality
Recalling from earlier exploration sessions, we studied several physical systems
and laws of the universe which exhibited very special constants – in particu-
lar, ‘dimensionless constants’ which characterised such laws or systems on all
length scales. The Reynold’s number for fluids and fine-structure constant for
quantum electrodynamics were two such constants. Now we present a mathe-
matical constant which characterises the ‘shape’ of all parabolas in a universal,
scale-invariant way. Since this constant is dimensionless, it is invariant under
conformal transformations. 1
First, we must define the ‘Latus rectum’ of the parabola. In particular, the latus
rectum of a parabola is the chord perpendicular to the symmetry axis (i.e. parallel
to the directrix) which passes through the focus F and intersects the parabola on
each side of the symmetry axis.
See Whiteboard Diagram
Exercise 7 (Parabolic Proctology) Using the geometric definition of a parabola,
prove that the latus rectum has a length of 4f, where f is the focal length of the
parabola.
Hint: For a parabola of the form y = x2
4f
, note the y-coordinates of the point at
which the latus rectum intersects the parabola.
Hint: Since you’re using the geometric definition of a parabola, you will have to
make use of the directrix – which is conveniently located at y = −f if you chose
the above parabola.
Definition 7 (Universal Parabolic Constant) The universal parabolic constant
P, is defined as the ratio (for any parabola), of the arc length S of the parabolic
segment formed by the latus rectum to the focal parameter 2f (half the latus-
rectum length) :
P =
S
2f
. (3.6)
See Whiteboard Diagram
Exercise 8 (Who would like to write a Fugue?) Whilst waiting for the cyborg
emperor to take care of the angels, Dr. Who picks up an renaissance guitar
ancestor and plays ‘While my guitar gently weeps’. Unsatisfied, he decides to
write a Fugue. Fugue’s, interpretted in the right sense, possess (almost) confor-
mal symmetry. One particular conformal symmetry is the ‘dilation/contraction’
operation – which shrinks or expands vectors (and hence objects).
If we let f have units of length, use dimensional analysis to prove that the uni-
versal parabolic constant is dimensionless.
Now, for a more serious derivation, we shall calculate the exact value of P and
prove a remarkable number-theoretic property – that it is transcendental.
Problem 10 (Transcendence (Hard)) 1. Simplifying the problem: Because
of translational and rotational symmetry, it suffices to consider a parabola
of the following form: y = x2
4f
with the y axis as its symmetry axis and
origin (0, 0) as the vertex.
1
Roughly, transformations that preserve relative angles but not lengths.
35. 3.2. PARABOLAS AND GEOMETRIC OPTICS 35
2. Calculating parabolic arc-length To calculate the arc-length of parabola
cut-off by the latus rectum, we express the parabola as a parametric curve
γ with curve parameter x:
γ(x) = (x,
x2
4f
), (3.7)
hence γ maps the parameter x to the corresponding point (x, y) = (x, x2
4f
)
on the parabola.
Since the tangent vector to this curve represents infinitesimal rates of change
along the curve (with respect to parameter x), it is given by the velocity
vector:
d
dx
γ(x) =
d
dx
(x,
x2
4f
) = (1,
x
2f
). (3.8)
In particular, an infinitesimal length element along the curve, is repre-
sented by the vector (differential 1-form):
dγ = (1,
x
2f
)dx, (3.9)
which has magnitude:
ds = (1,
x
2f
) dx =
d
1 +
x2
4f2
dx. (3.10)
Hence, if we integrate this length element from x = −2f to x = +2f (the
end points of the latus rectum), we get the parabolic arc length we desire:
S =
2f
−2f
d
1 +
x2
4f2
dx. (3.11)
Now, the universal parabolic constant was defined to be P = S
2f
, hence:
P =
1
2f
2f
−2f
d
1 +
x2
4f2
dx. (3.12)
3. Integration Step I Use a change of variables to prove that we can simplify
the arc-length integral to the following canonical form:
P =
1
−1
?
1 + t2dx. (3.13)
This form is ‘canonical’ in the sense that focal length f doesn’t appear
anywhere in the integral.
4. Integration Step II Use trigonometric substitution (or otherwise) to show
that:
P = arcsin(1) +
?
2. (3.14)
Hint: Recall the hyperbolic trigonometric identities:
cosh2
(θ) − sinh2
(θ) = 1 =⇒ 1 + sinh2
(θ) = cosh(θ), (3.15)
cosh(2θ) = cosh2
(θ) + sinh2
(θ) = 2 cosh2
(θ) − 1. (3.16)
36. 36 CHAPTER 3. GEOMETRY OF ANTIQUITY AND THE UNIVERSE
5. Algebraic Simplification Using the definition of hyperbolic sine:
sinh(θ) =
eθ
− e−θ
2
, (3.17)
along with the quadratic formula:
az2
+ bz + c = 0 ⇐⇒ z =
−b ±
?
b2 − 4ac
2a
, (3.18)
prove that
arcsin(1) = ln(1 +
?
2). (3.19)
Hint: let z = eθ
, then solve sinh(θ) = 1 for theta using the Euler expan-
sion for sinh given above.
6. Transcendality Recall that a real number α is transcendental if it is not the
root of any polynomial equation with rational coefficients. Real numbers
which are roots of polynomials with rational coefficients are ‘algebraic’
numbers. Hence if a number is transcendental it cannot be algebraic and
vice-versa. It follows that the sum of a transcendental number and an
algebraic number is necessarily transcendental.
To see that the universal parabolic constant P = ln(1 +
?
2) +
?
2 is
transcendental, it suffices to prove that ln(1 +
?
2) is transcendental. This
is because
?
2 is irrational, but not transcendental: in particular, we can
form a quadratic equation with rational coefficients: x2
− 2 = 0, of which?
2 is a root.
To see that ln(1 +
?
2) is transcendental, we do a proof by contradiction.
In particular, the Lindemann–Weierstrass theorem implies that if λ is alge-
braic (not transcendental), then eλ
is necesarily transcendental. Hence, if
ln(1 +
?
2) were algebraic, eln(1+
?2)
= 1 +
?
2 would be transcendental –
however, it is clearly not since this is a root of a quadratic with rational co-
efficients. Therefore, ln(1+
?
2) is transcendental and hence the universal
parabolic constant:
P = ln(1 +
?
2) +
?
2 2.295587, (3.20)
is a transcendental number.
Problem 11 (Tying loose ends) Prove the assertion that 1+
?
2 is an algebraic
number. In particular, find a polynomial with rational coefficients such that one
of its roots is equal to 1 +
?
2.
Hint: Recalling elementary polynomial theory, roots of the form: α +
?β –
where α, β are integers, come in pairs: λ = α ±
?β. Therefore, you should be
looking for a quadratic.
Problem 12 (Pi Day) If you didn’t celebrate Pi day, use the Lindelmann-Weirstrass
theorem to prove that π is transcendental. In particular, recall Euler’s formula:
eiπ
+ 1 = 0. (3.21)
Hint: Try assuming that iπ is algebraic.
Look at what we achived so far – we have proved that parabolas are charac-
terised by a transcendental dimnesional constant. Transcendental numbers are
extremely rare – e and π being the most famous examples.
37. 3.2. PARABOLAS AND GEOMETRIC OPTICS 37
3.2.4 Symmetries and Canonical Form
The natural symmetries of Euclidean space are symmetries which preserve the
Euclidean metric – that is, transformations of Rn
which leave lengths and rela-
tive angles (i.e. angles between vectors) unchanged. In elementary terms, these
are symmetries which leave the ‘dot-product’ unchanged. Because of this, as
parabolas are invariant under rotations and translations – their governing equa-
tions in a given coordinate system might change, but the parabola itself will be
unaffected. For example, translations simply correspond to a shift in the focus
F of the parabola, whilst rotations correspond to a rotation of the directrix D of
a parabola. Therefore, we can define a parabola more abstractly in the following
way.
Definition 8 (Ogburn’s Definition) Given a metric space (M, d) with set M
and metric d, a parabola is the ordered pair (F, D) where F ∈ M and D is a
straight line in (M, d), satisfying the following properties:
1. Focal Parameter: The minimum distance between F and D is 2f.
2. Parabolic Property: When (F, D) acts on any subset S of M, the result is
the collection of points U ⊂ S which is equidistant from F and D:
d(U, D) = d(U, F). (3.22)
In this manner, it becomes clear that if d is an inner product – such as the
Euclidean metric (dot-product), then a parabola (F, D) will be preserved by
isometries (rotations and translations for Euclidean space) since they preserve
the ‘parabolic property’ and ‘focal parameter’.
In Euclidean space, we can take any parabola and apply a sequence of transfor-
mations to it so that it becomes a canonical parabola y = x2
4f
. In particular, we
will need at most 2 translations to move the focus to F = (0, f), followed by at
most 1 rotation to rotate the directrix to coincide with the line y = −f . Proving
this for parabolas which have only been translated and/or rotated by multiples of
90 degrees, is relatively simple – which we shall do now.
Problem 13 (Transformations and Canonical Form) 1. Translations Given
a parabola of the form:
ay2
+ bx2
+ cy + dx + e = 0, (3.23)
where a, b, c, d, e are real constants and either a or b is zero, complete the
square to get a parabola of the form:
(y − y0) =
(x − x0)2
4f
, or (x − x0) =
(y − y0)2
4f
. (3.24)
In particular, find the vertices (x0, y0) and focal lengths f for these parabo-
las in terms of a, b, c, d and e.
2. Vertices Using the previous equations:
(y − y0) =
(x − x0)2
4f
, or (x − x0) =
(y − y0)2
4f
, (3.25)
prove that (x0, y0) is indeed the vertex of each of these parabolas.
Hint: It suffices to show that (x0, y0) is a minimum or maximum critical
point (turning point) of each the curves. Use calculus.
38. 38 CHAPTER 3. GEOMETRY OF ANTIQUITY AND THE UNIVERSE
Problem 14 (DIY) For parabolas which have been rotated through some arbi-
trary angle θ, we note that parabolas can be put into 1-1 correspondence with
quadratic forms. Using the quadratic form corresponding to a given parabola,
we can then apply change of basis transformations (rotation matrices) to rotate
the parabola back into the standard orientation with the y axis the symmetry
axis. Investigate this when you get the chance!
3.2.5 Optical Properties and Spherical Aberration
Due to their reflective properties, parabolas act as the ideal shape for many mir-
rors and lenses. In reality, parabolic lenses are difficult to construct, so ‘spherical
lenses’ are used instead. To this extent, one takes the radius of curvature of such
a lense be large relative to the length of the lense – then one can approximate the
portion of circle traced out by the lense as a parabola. Such an approximation is
the basis for a large amount of classical optics – for example, lens making.
Perhaps the most ‘physically’ important mathematical property of the parabola,
is its ‘parabolic reflection property’. To this extent, in the following, we shall
treat parabolas as ‘reflective surfaces’ and take it for granted that light travels in
straight lines (geodesics to be precise). Furthermore, we shall assume the law of
reflection: that is, that the angle between the normal to a surface and the incident
light ray is equal to the angle between the reflected light ray and the normal to
the surface. Mathematically:
θincidence = θreflection. (3.26)
For the purpose of reflection, we look at the tangent plane to a surface at a point
– the point where the incident light ray strikes the surface. This allows us to
apply the law of reflection to arbitrary differentiable surfaces.
Theorem 3 (Parabolic Reflection) Light rays incident on a reflective parabola,
parallel to the axis of symmetry are reflected back through the focus. Conversely,
light rays incident on the parabola which travel through the focus, are reflected
from the parabola along a line parallel to the symmetry axis.
Problem 15 (Reflective Moments) To prove this theorem we must do the fol-
lowing:
1. Simplify Since parabolas are characterised by a universal constant, it suf-
fices to prove the reflection property for a simple parabola of the form
y = x2
– i.e. f = 1
4
.
2. Diagrams Draw the focus F, vertex O and point P = (x0, y0) on the
parabola which the light ray hits. Now draw a line PD from P to the
point D perpendicularly below P, lying on the directrix. Draw the line
FP – this has the same length as PD, via the geometric definition of a
parabola.
3. Bisector = Tangent Draw a point M as the mid-point of the line con-
necting F and D. Then, using the law of reflection and some congruent
triangles, you should be able to show that MF bisects the angle FPD –
in particular, MF is perpendicular to FD. Now locate the x coordinate
of the point M – you should be able to prove (again using the geometric
definition of the parabola), that x = 1
2
x0 – i.e. the midpoint of the line
OD.
39. 3.2. PARABOLAS AND GEOMETRIC OPTICS 39
Now use calculus to calculate the slope of the tangent to the parabola at
the point of light intersection, P. Prove that the slope of the bisector MP
is equal to the slope of the tangent at P – hence identifying the bisector as
the tangent to parabola at P.
4. Fin At this point, the theorem has been proved. Do the necessary trigonom-
etry to and ray diagrams to see why this is so (unless it’s already obvious
to you). If you’re still stuck, as your tutor to draw the diagrams for you!
So far, we have demonstrated that (reflective) parabolas have the unique prop-
erty of reflecting light rays which are parallel to their symmetry axis, through the
focus of the parabola and vice-versa. Therefore, for many practical applications
– where a single focal point is required, parabolic lenses are the ideal lens. In
reality however, it is hard to make perfectly parabolic lenses so spherical or ‘cir-
cular’ lenses are used instead. The idealised performance of such lenses depends
on the ratio of the tangential length L of the lens, to the radius of curvature R of
lens. In particular, for the lens to ‘behave like a parabola’, its length L must be
much smaller than the radius of curvature and hence the focal length f (noting
that R = 2f). The deviation or ‘error’ arising from this parabolic approxima-
tion is the essence of ‘spherical aberration’ – that is, the blurring and loss of
resolution of images formed by the lens.
Problem 16 (Parabolic Approximation and Spherical Abberation in Lenses)
To quantify the previous statements, we shall now investigate spherical aberra-
tion mathematically. Let yp define a segment of a parabola – i.e. an ideal lens,
with focal length f and tangential length L. Now let ys define the lower segment
of a semi-circle whose center lies a distance R = 2f directly above the vertex of
the parabola – this represents a circular lens. Therefore, we have
yp =
x2
4f
ys =R − R
™
1 − (
x
R
)2, −L ≤ x ≤ L
∆y :=ys − yp, (3.27)
where ∆y is the difference between the y coordinate of the lower semi-circle,
ys, and the parabola, yp. Approximating a circular lens – a lower semi-circle,
by a parabola whose vertex (0, 0) coincides with the edge of the semi-circle and
whose focal length f = 1
2
is half the radius of curvature R of the circle, we get an
error ∆y which grows the further away we are from the vertex of the parabola.
1. Taylor Expanding the Semi-Circle Using a Taylor expansion about zero,
in the variable z := x
R
, show that we can write semi-circle equation as:
ys =
∞
k=1
(−1)k
¢1
2
k
(
x
R
)2k
=
1
4
(
x
R
)4
+
1
8
(
x
R
)4
+
1
16
(
X
R
)6
+ ... (3.28)
Hint: You can use binomial theorem instead. In particular, this says that
for any real constant α and variable z with |z| 1:
(1 + z)α
=
∞
n=0
¢
α
n
zn
, (3.29)
where the binomial coefficients are defined by:
¢
α
n
=
α!
n! (α − n)!
. (3.30)
40. 40 CHAPTER 3. GEOMETRY OF ANTIQUITY AND THE UNIVERSE
When α is non-integer, the binomial coefficients are generalized by the
‘Gamma funtion’ Γ – or equivalently, for real-valued α, the ‘Pochammer’
symbol’ (α)(n) := α(α − 1)...(α − n + 1). In particular, we have:
¢
α
n
:=
Γ(α + 1)
Γ(n + 1)Γ(α − n)
=
α(α − 1)...(α − n + 1)
n!
. (3.31)
Note that for integer n, Γ(n + 1) = n!.
2. A Parabola: To be or not to be Using the series expansion of the cir-
cle equation, ys, show that the error in the parabolic approximation for a
spherical(circular) lens, is given by:
∆y =R − R
™
1 − (
x
R
)2 −
x2
2R
=
∞
k=2
(−1)k
¢1
2
k
(
x
R
)2k
=
1
8
(
x
R
)4
+
1
16
(
x
R
)6
+ ... (3.32)
This shows that the ‘spherical abberration’ that occurs in the parabolic
approximation of a circular lens, is of the order O(( x
R
)4
) – where x is the
distance from the vertex of the lens in the direction parallel to the directrix
– i.e. perpendicular to the symmetry axis of the parabola. In particular,
the maximum error we have is:
Max[∆y] =
∞
k=2
(−1)k
¢1
2
k
(
L
R
)2k
= O((
L
R
)4
), (3.33)
where xmax = L is the length of the lens measured by a line tangent to
the vertex of the lens. In our case, this is the length of the line tangent to
the parabola at (0, 0) when we are trying to approximate the parabola near
its vertex by a circular arc. Thus, one way to keep the spherical aberration
small is make the radius of curvature R of the lens large with respect to
the length L of the lens.
We shall now apply the last result to obtain a differential error estimate which
quantifies how the spherical-aberration of a lens (‘non-parabolicness’) generates
a ‘fuzziness’ or spread in the in focus. In particular, instead of the focus being a
single (ideal) point, it now becomes a small line segment – physically leading to
a blurriness of images formed by the lens.
Problem 17 (Losing Focus!) Thanks to Emperor Constantine’s ‘plasma inter-
vention cannon’, the weeping angels are now a thing of the past. However, the
‘past’ is relative! This means that the weeping angels still lurk in one of many
universes. Not to fear, ‘The Doctor’ 2
decides it is time to return to the future
– leaving Pachelbel’s (musical) cannon unspoiled by The Doctor’s attempted
Skrillex additions.
To travel to the future, The Doctor needs to fire up his ‘Alcubierre’ warp drive
– this will allow him to generate a faster-than-light warp-bubble which he can
travel through spacetime with. However, during the angel attack, one of his
synchronising lasers was damaged. In order to fix the laser, he must ground a
2
Thanks ‘The’ Dr. Ashleigh Punch for noting that ‘Dr. Who’ should always be referred to
as ‘The Doctor’. Good luck to her when she finally meets, marries him and has time-travelling
babies.
41. 3.2. PARABOLAS AND GEOMETRIC OPTICS 41
new optical lens – such that its focal point, F = (0, f), shifts by a maximum of
1 micrometer: ∆f = 1µm = 10−6
m under the effect of spherical aberration.
Assuming he needs a lens of length L = 1mm = 10−3
m, we can help The
Doctor, as follows.
1. Mathematical Constructions By re-writing the standard parabola equa-
tion, we can express the focal length f as a function of the (x, y) coordi-
nates:
f =
x2
4y
. (3.34)
Now, we note that a ‘linear approximation’ to the error in the focal length,
is given by the ‘total differential’, df. In particular, by viewing f = f(x, y)
as a function of two-variables x and y, show that its total differential (ex-
terior derivative) is given by:
df =
x
2y
dx −
x2
4y
dy. (3.35)
Hint: Recall that the total differential of a function f(x, y) of two variables
is given by:
df(x, y) :=
∂f
∂x
dx +
∂f
∂y
dy. (3.36)
2. Physical Estimates Now, we replacing the differential df, dx and dy by
their finite counterparts: ∆f, ∆x, ∆y – i.e. the ‘error’ in f,x and y, we
get:
∆f =
x
2y
∆x −
x2
4y
∆y. (3.37)
To get the maximum error in the focal length however, we need to consider
the magnitude of error contributions from ∆x and ∆y, hence we re-define
∆f as:
∆f :=|
x
2y
||∆x|+|
x2
4y
||∆y|. (3.38)
Therefore, the error ∆f is maximized when ∆x and ∆y are maximized
(for a fixed coordinate (x, y)).
Physically, we set ∆x = 0 since there is no ‘error’ in the x-coordinate of
our lens – the tangent to circle aligns with the tangent to parabola vertex.
Now, we recall from the last problem that the maximum error in our y
coordinate is given when x takes its maximum value x = L – i.e. the
‘spherical aberration’ is maximized at the edges of the lens (away from
the vertex):
Max[∆y] =
∞
k=2
(−1)k
¢1
2
k
(
L
R
)2k
= O((
L
R
)4
), (3.39)
where R = 2f is the ‘radius of curvature’ of the lens (the radius of the
circle).
Using x = L, y = L2
4f
, ∆x = 0 and the maximum value for ∆y, show that:
Max[∆f] = fMax[|∆y|] = f|
∞
k=2
(−1)k
¢1
2
k
(
L
R
)2k
|= O((
L
R
)4
).
(3.40)
42. 42 CHAPTER 3. GEOMETRY OF ANTIQUITY AND THE UNIVERSE
3. Experimental Solution Ignoring higher-order contributions to the error in
f, we have:
∆f ≈
1
8
(
L
R
)4
=
1
16
L4
R3
. (3.41)
Show this by taking the first term in the binomial expansion above.
With this leading-order estimate for ∆f, we want ∆f ≤ 10−6
m to achieve
the accuracy desired for The Doctor’s laser. For the given lens length
L = 0.001m, calculate the minimum radius of curvature Rmin for the lens
required to achieve the accuracy: ∆f ≤ 10−6
m.
4. Checking Validity By taking the next term in the binomial expansion, we
can compute the next order contribution to the error in f:
f
1
16
(
L
R
)6
=
1
32
L6
R5
. (3.42)
Rather than adding this to the error ∆f, we can instead use the value of
Rmin we calculated (which gave ∆f = 10−6
m) to estimate the relative
magnitude of the leading error term: 1
16
L4
R3 and the next correct term , 1
32
L6
R5 .
Compute the ratio of these error terms and argue whether or not it was
justified to ignore the next correction term when calculating an approxi-
mation for Rmin. For example, if the ratio is less than 0.01 (or 1%), we
can justify ignoring the correction term.
