# Photon Spheres and Rotating Black Holes

This is another post on my current theme of black holes. Recently, I wrote a blog post about light orbiting a Schwarzchild black hole. These so called Photon spheres are sufficiently interesting to wonder whether they exist for more complicated black holes. They do, and we’re going to find out about them. This post is based on a fantastic paper [1] I read the other week. If you find this stuff interesting, and you’ve got an introductory level of GR you should definitely check out the paper here.

One of the coolest thing about black holes is you can describe any black hole with just three numbers: Mass $(M)$, Charge $(Q)$, and Angular Momentum $(J)$. Of course, if we’re being completely general we also need to describe where the black hole is and how fast it’s moving through space – but we can always perform a Lorentz transformation so that we’re in the rest-frame of the black hole and so we really only care about $M$, $Q$ and $J$. The fact that you can describe black holes with only three numbers is one of the main reasons I find them so interesting. Consider for example, two black holes, A and B. Black hole A is made entirely from matter – protons, neutrons, electrons and the like. On the other hand, black hole B is made entirely from anti-matter – anti-protons, anti-neutrons, anti-electrons and the like. Further, suppose A and B have the same charge, angular momentum and mass. How can we tell A from B? That is, what experiment can we do to tell the difference between the matter black hole or the anti-matter black hole? The answer is, we can’t tell the difference. For all intents and purposes, there is no way of telling whether a black hole is made of matter or anti-matter. Indeed, a third black hole C, with the same mass, charge and angular momentum, but made entirely from light, would be similarly indistinguishable from the first two.

As promised in the title, we’re going to be looking at rotating black holes. A rotating black hole is, like the Schwarzchild black hole, uncharged ($Q=0$). Unlike the Schwarzchild solution however, the rotating black hole has a non-zero angular momentum $J$. Rotating black holes are also called Kerr black holes, after Roy Kerr who was the first to write down an exact solution for one. We will consider an uncharged black hole which is rotating with constant angular momentum. For convenience, we define the angular momentum per unit mass $a = \frac{J}{M}$. Since the black hole is rotating, we lose the spherical symmetry we have with the Schwarzchild metric, so we expect the solution to be a little more complicated. We do, however, still have axial-symmetry, i.e. rotational symmetry around the axis of rotation.

# Avian Flow: Continuum Boid Dynamics

I’m excited for this, if only because of the enormous amount of humorous nomenclature that will come of it. Ok, here we go.

Have you ever heard of Boids? Boids (‘Birds’ with a Brooklyn accent) refers to a certain type of algorithm for simulating the flocking of large numbers of animals – for example, the eponymous birds. The word refers both to the algorithm itself and to the individual elements being simulated, which are rendered onscreen and whose motion through simulated space evolves according to a set of three rules. The particular details of the algorithm may vary from implementation to implementation, but these three main features of the Boids’ are usually the same.

Boids, as entities, possess a position and a velocity, and obey the following physical principles:

1. Alignment. Boids wish to conform to the flight paths of their neighbours, and so will rotate to align their own velocity with the average velocity of neighbouring boids which are sufficiently close.
2. Cohesion. Boids like to fly in groups, and they like to be as close to the centre of the group as possible. For this reason they also steer towards the mean position of those sufficiently close neighbours.
3. Separation. Boids don’t, however, like to collide, so they will actively avoid any of their neighbours that come too close.

If you want to have a look at an actual implementation and some wonderful, in-depth visualisations of all of these things, then you can do no better than this post over at Harry Brundage’s blog. Seriously, go over there and have a play. I’ll wait here.

Neat, huh? But why should we care? Why am I talking about boids? Well, the first thing that came into my head when I learned of boids was:

Why not make this a continuous system?

# Sum-Product Cycles of Positive Integers Pt. 3

To recap, we are attempting to find general solutions to the system of $n$ coupled Diophantine equations with the form:

$x_1 + y_1 = x_2 y_2$

$x_2 + y_2 = x_3 y_3$

$\vdots$

$x_n + y_n = x_1 y_1$.

We have already found that a solution is either compound (the concatenation of two shorter solutions), or it is simple (no consecutive subset is a solution to any shorter cycle). We have a canonical simple cycle, which for any pair $(i, j)$ is the $(i - 1)(j - 1)$-cycle given by

$[(i, j)]_{Can} = [(i, j), (1, i+j), (1, i+j+1), \ldots, (1, ij-1)]$

As we shall see in this third (and final) post, the canonical cycle is very significant in the classification of simple solutions, and hence of all solutions.

# Sum-Product Cycles of Positive Integers Pt. 2

Recall the sum-product $n$-cycle that we examined in this post. We seek simultaneous solutions to the coupled system of $n$ Diophantine equations

$x_1 + y_1 = x_2 y_2$

$x_2 + y_2 = x_3 y_3$

$\vdots$

$x_n + y_n = x_1 y_1$.

We found some particular instances of complete descriptions of $B_n$, the set of equivalence classes of solutions (equivalence under commutation of the operations or of cyclic permutation of the indices $i = 1, 2, \ldots, n$). Now we seek a more powerful machinery with which to find and classify solutions.

# Sum-Product Cycles of Positive Integers

## Introduction

So in this post I’m going to talk a bit about a nice little number-theoretic problem that has been rattling around in my head a bit over the last few weeks. I’m essentially looking at the solutions (for each integer $n$) to a system of $n$ coupled second-order Diophantine equations. The equations involved have a nice internal symmetry and I have a couple of theorems giving a complete classification of the solutions for each $n$ – at least for positive integer solutions.

Let’s look at the prototypical equation we want to generalise:

$a + b = ab \qquad a, b \in \mathbb{Z}^+$.

Clearly the only viable solution to this is setting $(a, b) = (2, 2)$ – it’s a very strong condition for two natural numbers to satisfy. However, it is a very aesthetically pleasing equation, so we would like to consider some extensions, and indeed a couple present themselves immediately.

One possible extension, which I am not going to consider here, is to find $n$-tuples of positive integers $(x_1, x_2, \ldots, x_n)$ which satisfy

$\sum_{i = 1}^n x_i = \prod_{i = 1}^n x_i$.

((We observe immediately that the $n$-tuple $(1, 1, \ldots, 1, 2, n)$ always works – but can you find how to generalise this to other $n$-tuples? What about those guaranteed to contain a certain subset of positive integers $\{x_i, x_j, \ldots, x_k\}$? What about finding the number of unique solutions for each $n$? You can read more about this in [1].))

The extension I will be looking at is the following: consider $n$ many Diophantine equations of the form

$x_1 + y_1 = x_2 y_2$

$x_2 + y_2 = x_3 y_3$

$\vdots$

$x_n + y_n = x_1 y_1$.

Does such a system of equations always have a solution? How many solutions are there for a given $n$? What structure exists on the set of solutions? Clearly the solutions themselves are invariant under pair commutation ($(x_i, y_i) \mapsto (y_i, x_i)$) and under cyclic permutation of the indices ($i \mapsto i + k (\mathrm{mod})_n$). This kind of structure – where we have both an overall permutation of blocks and an underlying, independent permutation within each block – is realisable as the action of a wreath product on our space of objects. Let’s formalise this.

Definition 1

Let the $n$ many coupled Diophantine equations of the above form be called the sum-product $n$-cycle, or just the $n$-cycle. Let $A_n$ denote the set of $n$-tuples of pairs of positive integers which satisfy the $n$-cycle. That is,

$A_n \subseteq ((\mathbb{Z}^+)^2)^n$.

Let $\Omega = S_2 \wr \mathbb{Z}_n$ denote the regular wreath product of $S_n$, the permutation group of two elements, and $\mathbb{Z}_n$, the group of integers modulo $n$. Furthermore let elements of $\Omega$ act upon $A_n$ with the primitive wreath action. We take the quotient space by this action,

$B_n = A_n / \Omega$,

to be our space of $n$-cycle solutions.

And let’s immediately disregard this formalism for the time being. We need to focus first on coming to grips with what an $n$-cycle is, and how it’s solved. Let’s look at some small-$n$ solutions. Continue reading

# Photon Spheres

Ok, now that I have finished the preamble over here, I can finally start talking about what I originally intended to blog about: Photon spheres.

The defining feature of a black hole is that the gravitational attraction beneath the event horizon is so strong that even light can’t escape. That is, the curvature of spacetime in the vicinity of the black hole is so intense that there are no geodesics which are able to leave. This is what makes black holes so interesting, since anything (including light) which is dropped into a black hole is lost forever (we’re talking classical black holes at the moment, so no Hawking radiation). On the other hand, light (or matter) travelling near a black hole can escape, provided it stays outside of the event horizon.

Light from the blue object is deflected towards the grey massive object

General relativity predicts that light passing near a heavy object will be deflected by the gravitational field of that object. Equivalently, light will follow a geodesic in the curved spacetime around the heavy object.

This situation strongly resembles the situation we have in orbital mechanics, where a small object like a satellite or asteroid is moving near a large object, like a planet. Depending on the velocity of the object, it either escapes the planet, falls into the planet, or starts orbiting the planet. It seems like a natural question, then, to ask:

Can light orbit a black hole? Continue reading

# Preamble to Black Holes

I’ve been reading a lot about General Relativity this past week and so I thought I would do a couple of posts on Black Holes, since they are well and truly one of the most interesting things about General Relativity. This first post is really just a (very) brief introduction to General Relativity. My main goal is to write about some of the cool things I’ve come across lately, so think of this as setting the theme for the next few posts.

General Relativity (GR) is a geometric theory of gravitation proposed by Einstein. In GR, the flat background spacetime of special relativity is replaced by a curved spacetime. The curvature of spacetime is greater around objects with a higher mass, and particles (including light) travel along paths called geodesics (essentially the ‘shortest’ path between two points) in this curved spacetime.

The standard analogy here is bowling ball on a rubber sheet: If you take a big rubber sheet stretched flat and put a bowling ball in the middle, the sheet stretches and curves around the bowling ball, but as you get further away from the ball the sheet becomes less curved.

The rubber sheet analogy – heavy objects stretch the spacetime around them

Now imagine a tiny little marble moving on the rubber sheet. As the marble gets closer to the bowling ball, the curvature of the sheet causes the marble to accelerate towards the bowling ball. The marble feels this acceleration as the gravitational force of the bowling ball – gravity is just the curvature of spacetime. The path that the marble traces out on the rubber sheet is called a geodesic. In flat space, a geodesic is simply a straight line, but on a curved surface geodesics can be more complicated. Archibald Wheeler described GR rather poetically, saying “Spacetime tells matter how to move, matter tells spacetime how to curve”

Light following a geodesic through curved spacetime

The rubber sheet analogy is all well and good, but let’s get into some honest mathematics!

General Relativity is described, rather succinctly, with the Einstein Field Equations (EFE).

$R_{\mu \nu} - \frac{1}{2}Rg_{\mu \nu} + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4}T_{\mu \nu}$

The EFE relates the local curvature of spacetime with the local energy and momentum in that spacetime. Solutions to the EFE are metrics $g_{\mu \nu}$ which describe the geometry of spacetime. It should be noted that the apparent simplicity of this equation is incredibly misleading. The Einstein Field Equations are actually a set of 10 coupled, non-linear, hyperbolic-elliptic, partial differential equations – solving them is a highly non-trivial task and exact solutions are only known for the simplest cases.

The EFE determine how a given distribution of matter/energy influences the geometry of spacetime. To describe the motion of freely falling matter through curved spacetime, we also need the geodesic equation.

$\frac{\text{d}^2 x^{\mu}}{\text{d}\lambda^2} + \Gamma^{\mu}_{\alpha \beta} \frac{\text{d} x^{\alpha}}{\text{d} \lambda} \frac{\text{d} x^{\beta}}{\text{d} \lambda} = 0$

The simplest solution to the Einstein Field Equations is the Minkowski metric – that is, flat space. Traditionally denoted by $\eta_{\mu \nu}$ rather than $g_{\mu \nu}$, the Minkowski metric describes spacetime with no matter, no energy, no curvature – nothing. The metric has, in cartesian coordinates, the following form

The metric is related to the line element (spacetime interval) in the following way

$ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu}$

where the repeated indices are implicitly summed over. For the Minkowski metric in cartesian coordinates $(ct, x, y, z)$, this becomes

$ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$

A metric can either be specified by its components in a particular coordinate system $g_{\mu \nu}$, or the line element $ds^2$. Note that I’m using physicists terms like ‘line element’ and ‘spacetime interval’ in this article. There are two reasons for this. Firstly, I don’t want to go into the mathematics of differential geometry in this post – my goal is to describe some cool physics, and the clearest way to do that is to use physics notation. Secondly, this is the notation used in the literature.

I’m going to finish off this post by describing another solution to the Einstein Field Equations – the Schwarzchild metric. This metric describes spacetime around a spherically symmetric, uncharged, non-rotating massive object. In polar coordinates $(ct, r, \theta, \phi)$, the metric takes the form

$ds^2 = -(1-\frac{R_s}{r})c^2 dt^2 + (1-\frac{R_s}{r})^{-1}dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2$

where $R_s$ is the Schwarzchild radius, a constant related to the mass of the object.

If any spherically symmetric, uncharged, non-rotating massive object has a radius less than its Schwarzchild radius, the object forms a black hole and is called a Schwarzchild black hole. The physics of black holes is an incredibly cool area of General Relativity, and there are a number of surprising results that you can get. In the next blog post I will describe one of the lesser known results – the photon sphere.

# The Pi Proof of Penzance

I was watching the Pirates of Penzance the other day and was reminded of a fantastic example of a constrained proof – a proof that $\pi$ is irrational set to the tune of the Major General’s Song that was written by Kevin Wald. Having nothing better to do I decided to memorise it. I also really like it as a proof, it’s nifty, but also incredibly easy, you could easily make it an assignment for a first-year calculus course.

Here are the lyrics, written by Kevin Wald and available here  (I also recommend checking out some of the other things on his site – I particularly liked Cole Porter does Indo-European).

That pi must be irrational, I claim, is demonstratable:

Assume that with a quotient of whole numbers it’s equatable

Say, m o’er n. Define $a_k$, by fiat dictatorial,

“And it is, it is, a glorious thing to be a mathematician”

For every natural k to be one over k factorial

Times integral from naught to pi of (n times (t)(pi – t))

To power k, times sine (or for you Latin scholars, sinus) t,

dt. These a’s are *positive*, with *finite sum* (indeed, it e-

Quals integral exp(n times (t)(pi – t)) sin t dt).

Chorus: It’s integral exp(n times (t)(pi – t)) sin t dt!

It’s integral exp(n times (t)(pi – t)) sin t dt!

It’s integral exp(n times (t)(pi – t)) sin t dt, dt!

But integrate by parts — each a’s the sum of the preceding two

Times integers, $a_naught$ is 2, $a_1$’s 4n, thus leading to

(since *all* must then be integers) a contradiction statable,

And thus that pi’s irrational, you see, is demonstratable!

Chorus: Since *all the a’s are integers*, a contradiction’s statable,

And thus that pi’s irrational, we see, is demonstratable! Continue reading

# Homotopy Groups? We don’t need no stinking Homotopy Groups! (at least if we want to prove Brouwer’s Fixed point theorem)

One of the things that annoyed me while I was doing honours in economics, Brouwer’s Fixed Point Theorem was used in several proofs, but never proven (except in one-dimension, where it’s almost trivial). It was used in the proof of theorems such as Nash’s Equilibrium Theorem, when in reality the existence of a fixed point was the most non-intuitive component of the proof (actually I’ve heard folktales that Nash’s Equilibrium Theorem is actually equivalent to Brouwer’s Fixed Point Theorem, which, if true, would make me more annoyed). However, this is understandable, as the standard proof of this theorem uses homotopy groups, the introduction of which fails almost any cost benefit analysis for a lecturer trying to teach economics (although, they do have some truly awesome applications to Game Theory – the study of which is fairly high on my “to learn” list). However I recently found a very easy proof, that uses Sperner’s Lemma. It should be accessible to anyone who has some basic analysis, and at least an intuitive understanding of graph theory. In fact I’ve also written up the proof as a problem set, designed to let people prove the result “quasi-independently”. I’ve posted it on my website at https://sites.google.com/site/matthewphillipcalvin/writing.

I first saw this idea on a blog post of Vadim Kulikov’s blog post http://blogs.helsinki.fi/kulikov/2010/02/03/brouwers-fixed-point-theorem-many-in-one-post/. I also used some notes written by Jonathan Huang at http://www.stanford.edu/~jhuang11/research/old/sperner.pdf.

Warning: There is a lot of discussion here before I actually get to any maths. If you want to skip ahead to the Handshake Lemma you won’t miss anything relevant to the proof (this was intended to be a short introduction, but it rather ballooned out of proportion).

Some years ago I read “The Strange Case of Mrs. Hudson’s Cat” by Colin Bruce. The premise of the book is that Sherlock Holmes, still in Victorian (or Edwardian) Britain is presented with several cases that involve, in some way or another, principles based on modern physics, most of which make interesting cases in and of themselves, and all of which make principles of modern physics seem intuitive (when thought about the right way). One case involves a man being shot twice in quick succession. At the time he happens to be embracing a woman. At the first shot the  woman notices the impact of the shot and then hears the shot, seconds late the woman hears the second shot then feels the impact. The question is how is this possible (the same gun was used both times). This simple puzzle sets the scene for a discussion of the wave-particle duality in light and an introduction to relativity. It’s a fantastic book, and one I think about a surprising amount. (In case this plug hasn’t been clear enough, I’d recommend reading the book see links here or here, or, you know, get it out the library).

Anyway the reason I brought up this book, is that there’s a particular example that’s stayed with me. In one of the cases Sherlock and Watson are presented with two (minor) problems to solve (these aren’t quite the same problems as were in the book but they’re close enough):

CASE ONE: There are four stones half buried in the ground with only one face visible. These stones have a letter and a numeral engraved on opposite sides. Suppose we can see that the stones read A2W9. We want to determine whether any stone with a vowel on one side has an even number on the other, but can only dig up two stones. Can we determine this, and which stones do we check.

CASE TWO: Watson is chaperoning a party. The party serves two drinks lemonade and beer. In an effort to counteract youth drinking people have to buy a wristband, yellow if they want only lemonade, orange if they want the choice of beer as well. Only people over 18 may have beer. Holmes is presented in quick succession by a sixteen year old, a man who wants a yellow band, a twenty-three year old, and a woman who wants an orange band. Who does Holmes have to check IDs/ask what colour band they want?

These puzzles probably aren’t difficult for maths students, we deal with the idea of how to check whether P implies Q everyday. But for most people the second question is much easier than the first, although really the two questions  are isomorphic (the rule is P implies Q, where P is either a vowel or under-eighteen, and Q is an even number or gets a yellow wristband. To check this we only need to check that the cases where we observe P or not Q obey the rule. That is A and 9, or the sixteen year old and the person who wants an orange band).

So what do we take from the fact that most people find the second case significantly easier. Humans are very well suited to thinking about other people, and less well suited to thinking about problems in the abstract (not that we’re bad at that). Ceteris paribus I enjoy abstract problems that can be stated in human terms, which I why I particularly enjoy this proof’s explanation of the Handshaking Lemma. This is a graph theoretic result that can be presented as a “human” problem, and is very tidy to solve.

## Handshaking Lemma

Handshaking Lemma Show that the number of those, who have shaken their hands with others an odd number of times, is even. (Note to pure mathematicians – assume the number of people in the world is finite.). Continue reading

# The Joy of Hex and Brouwer’s Fixed Point Theorem

About a month ago I ran a workshop on Game Theory for Mathematicians. A group of people in the Maths Society were discussing how the best way to learn mathematics is to have someone sit down with you (or stand by a whiteboard) and explain it. I think a large part of that is that you can let them know when you don’t follow something they say (and if you start to look too confused, they pick up on that and slow down). Anyway, we decided that it would be a good idea to run a workshop, aimed primarily at second and third year students, where we would try to teach them some basic advanced ideas (e.g. groups, topology, graph theory) as Socratically as possible. I ended up volunteering to run a workshop on Game Theory, which worked well because it’s a great mathematical area, that isn’t really covered much by undergraduate courses. We had three hours planned and I broke these up into three (roughly equal) segments, Hackenbush, Strategic Game Theory, and Hex. The Hex segment was by far my favourite and I suspect the students’ favourite as well. We didn’t end up proving the main result I wanted to get to, but we had some great discussions about game theoretic ideas. I enjoyed the proof though, and I thought I’d write it down here so I don’t forget it. It’s an interesting game in its own right, and gives a very cool, and relatively easy to understand, proof of Brouwer’s Fixed Point Theorem. (This also lets us prove the Jordan Curve Theorem, which I’ll try to get to in another post).

## Brouwer’s Fixed Point Theorem

Brouwer’s Fixed Point Theorem is a handy little thing that pops up all over economics and mathematics. In fact two Nobel Prizes have essentially been awarded to economists for just applying a generalisation of the theorem (Kakutani’s Fixed Point Theorem) to economic problems (Arrow in 1972 and Nash in 1994). (My tongue’s reasonably far in my cheek here, the prize was really awarded for the economic ideas that lay behind this work, the fixed point theorem was just the method by which they proved it). I’ll talk in another post (probably) about why I’m frustrated with how economics treats the theorem (they use it but don’t prove it), and give a very cute (and quick) little proof of it using Sperner’s Lemma, but today I want to show the most intuitive proof I know.

The proof I’m showing here was first shown by David Gale in the paper:

Gale, David. “The game of Hex and the Brouwer fixed-point theorem.” The American Mathematical Monthly 86, no. 10 (1979): 818-827.

I was first introduced to it at the end of a course in higher dimensional real analysis at Auckland Uni. It was the last lecture of term and we were told it would be unassessed (the unassessed stuff always seems more fun). We spent the first quarter of an hour or so just playing Hex (as I remember I was consistently beaten), and then looked at the proof of the Brouwer Fixed Point Theorem. The next year I looked a little bit into topological games (following a very neat seminar by the same lecturer who had run the higher dimensional analysis course) and refound the proof. So when I decided to run a course for mathematicians looking at game theory, it was an obvious choice and a good excuse to force myself to understand the proof.

It all begins by looking at the game of Hex, which is played on a board, much like the one below:

###### Source http://commons.wikimedia.org/wiki/File:Hex11.PNG, licensed under the Creative Commons Attribution-Share Alike 3.0 Unported

It can be played on various sized boards – John Nash, the inventor, supposedly recommended a 14 by 14 board as the optimal game size. (I’ve wondered whether Nash, who would use Brouwer’s fixed point theorem in his Nobel prize winning work, had noticed the connexion between the two). Continue reading