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Feynman diagrams without any physics

Joseph Wilson — Sun, 19 Apr 2026 00:00:00 +0000

A tale from elementary probability theory

Feynman diagrams are associated with complicated physics, but they can also be found hiding in relatively simple questions about Gaussian random variables.

For example, suppose I asked you to calculate the expectation value

where each is a (not necessarily independent) normal random variable with zero mean. Say I want the answer in terms of the covariances between the variables.

If you are especially visually-minded (and considerably brilliant) you may find yourself eventually drawing the following diagrams…

…on your way to producing the answer, which is in this case is

Here’s some Julia code approximately showing that this formula is correct.

Layout algorithms for flexigrids

Joseph Wilson — Mon, 08 Sep 2025 00:00:00 +0000

This post is a development journal entry for thefletcher package forTypst. Fletcher is a diagramming package that lets users placenodes (content) on aflexigrid (a special coordinate system).

Flexigrids are a cross between a table layout and a coordinate grid. Like tables, the rows and columns grow automatically to account for the size of nodes, but unlike tables, flexigrids also allow fractional coordinates (so a node can be at column $2.5$, for example).

This leads to some interesting puzzles in terms of the best layout algorithm for computing row and column sizes given a set of nodes and their coordinates.

Method 0: Normal coordinates

The simplest way to lay out nodes on a page is to place them at Cartesian coordinates on a fixed grid.

Figure 1: Nodes laid out with Cartesian coordinates.

Method 1: Standard table layouts
If we want a table-like layout in which cells grow to accommodate nodes, the simplest approach is to require nodes to have integer coordinates. The row/column sizes are set to the maximum height/width of the nodes present in that row/column.
Cell gutter may then applied by adding a uniform gap between rows and columns.
In the presence of cell gutter, the centre positions $x_0$ and $x_1$ of flexigrid cells of widths $w_0$ and $w_1$ are constrained by
$$ x_1 - x_0 = g + \frac{w_0 + w_1}2 $$
where $g \ge 0$ is the gutter between adjacent cells.
As a layout algorithm, for each node placed we update the cell widths as
$$ w_u \leftarrow \max(w_u, W) $$
where $W$ is the width of the node at position $u \in \mathbb{Z}$ and $w_u$ is the original cell width at column $u$.
Method 2: Linear interpolation of cell sizes
To allow fractional positioning of nodes, we can define anelastic coordinate system which linearly interpolates between row/column centre points. If $u \in \mathbb{R}$ is an elastic coordinate, then the corresponding physical coordinate is
$$ x_u = (1 - t)x_{\lfloor n\rfloor} + t x_{\lfloor u\rfloor + 1} $$
where $t = u - \lfloor u\rfloor \in [0, 1)$ and $x_i$ are the row/column centres for $i \in \mathbb{Z}$.
If we place a node of width $W$ at $x_u$, how do we update the cell sizes?
A simple method is to linearly interpolate the allocation of the width into each cell, according to
$$ \begin{align} w'_0 &= (1 - t)W \\ w'_1 &= t W \\ \end{align} $$
which agrees with Method 1 when $u \in \mathbb{Z}$. The actual cell size is then the maximum of the new size $w'_i$ and the original size $w_i$ without the node:
$$ w_i \leftarrow \max(w_i, w'_i) $$
This works well when there is no gutter between cells:
However, placing nodes with cell gutter leads to cells being unnecessarily large:
Figure 2: Limitations of the linear layout algorithm.
Notice that the first column is too wide for $u = 0.2$ and the last column is too wide for $u = 0.8$. Both columns are too wide for $u = 0.5$.
Another way to visualise the problem above is shown in Figure 3. The cell centres $x_0$ and $x_1$ and widths $w_0$ and $w_1$ on the horizontal axis are shown as a function of the gutter $g$ on the vertical axis.
Figure 3: The effect of a node of width $W$ being placed at $x_t$ between two cells of width $w_0$ and $w_1$, as a function of cell gutter $g$.
Notice that the cells in Figure 3 (blue and red) reach outside the span of the node, $W$. As the cell gutter $g$ increases, the cells contain less of the node yet do not shrink.
Figure 4 shows what happens when we vary the node’s position, $x_0$. Notice how the first and second cells are sometimes wider than necessary to enclose the white node.
Figure 4: Cell widths computed with the linear layout algorithm, Method 2, for different coordinates $u$ of the white node.
Method 3: Thetight layout algorithm
To fix the aforementioned problem, we must define the desired layout behaviour taking into account cell gutter.
Figure 5 demonstrates the desired layout algorithm for various values of $u$, to be compared to Figure 4. Notice that the centre of the node at $u$ is exactly linearly interpolated between cell centres, and that the cells are never too large (the cells contain nodes tightly).
Figure 5: Cell widths computed with the tight layout algorithm, Method 3.
What is the layout rule that gives rise to this? Consider placing a node of width $W$ at elastic coordinate $0 \le u \le 1$. If a cell is fully contained in the width of the node, we want the cell to expand by as much as it can while still being fully contained. If a cell is not fully contained in the node’s span, it does not change size when the node is placed. This results in the scenario in Figure 6, where the cells separate to respect increasing gutter $g$ but also shrink to ensure they stay within the node $W$.
Figure 6: Node at $u = 1/3$ of width $W = 6$ placed with the tight layout algorithm.
If the cells have an existing positive width, then our picture looks like Figure 7.
Figure 7: Node at $u = 1/3$ of width $W = 6$ where the blue and red columns have existing widths $w_0 = 2$ and $w_1 = 1$.
To calculate the new cell widths $w'_0$ and $w'_1$ for a given $u$, $W$ and $g$ along with the original widths $w_0$ and $w_1$, we first write down the coordinates of the inner edges of the cells.
For the left cell (blue), there are two cases to consider.
Fully contained in the node: In this case, the left end of the cell is at the left end of the node, at coordinate $-W/2$. The cell width is $2x_0 + W$, so the inner end (right) of the cell is at
$$ x_0^\text{inner} = 2x_0 + W/2. $$
Not fully contained: In this case, the cell is unaffected and keeps its initial width $w_0$. The inner (right) end of the cell is then
$$ x_0^\text{inner} = x_0 + w_0/2. $$
The correct end point is the rightmost or maximum of these two. Similar reasoning applies to the right node (red).
$$ \begin{align} x_0^\text{inner} &= \max{2x_0 + W/2, x_0 + w_0/2} \\ x_1^\text{inner} &= \min{2x_1 - W/2, x_1 - w_1/2} \\ \end{align} $$
The gap between the cells $g = x_1^\text{inner} - x_0^\text{inner}$ is therefore
$$ \begin{align} g = \min\{ & x_1 - x_0 - (w_1 + w_0)/2, && x_1 - 2x_0 - (w_1 + W)/2, \\ & 2x_1 - x_0 - (W + w_0)/2, && 2x_1 - 2x_0 - (W + W)/2\}. \end{align} $$
Remembering that we set $x_t = 0$, we may use $(1 - t)x_0 + t x_1 = 0$ to eliminate $x_1 = (t - 1)/t x_0$.
$$ \begin{align} g = \min\bigg\{ & \left(\frac{t - 1}t - 1\right) x_0 - \frac{w_1 + w_0}2, && \left(\frac{t - 1}t - 2\right) x_0 - \frac{w_1 + W}2, \\ & \left(2\frac{t - 1}t - 1\right) x_0 - \frac{W + w_0}2, && \left(2\frac{t - 1}t - 2\right) x_0 - \frac{W + W}2\bigg\} \\ \end{align} $$
Simplifying this gives
$$ \begin{align} -t g = \max\{ & x_0 + t(w_1 + w_0)/2, && (1 + t) x_0 + t(w_1 + W)/2, \\ & (2 - t) x_0 + t(W + w_0)/2, && 2 x_0 + t W \} \end{align} $$
where the $\min$ turns into a $\max$ because we multiplied by $-t \le 0$. Notice that the right-hand side is a linear function of $x_0$ with positive slope. We can therefore solve for $x_0$ by solving each case independently and taking the maximum solution (prove this to yourself by drawing some lines).
$$ \begin{align} x_0 = -\frac{t}{2} \max\bigg\{ & 2g + w_1 + w_0, && \frac{2g + w_1 + W}{1 + t}, \\ & \frac{2g + W + w_0}{2 - t}, && \frac{2g + 2W}{2} \bigg\} \end{align} $$
The new width of the cells are then:
$$ \begin{align} w'_0 &= 2x_0 + W \\ w'_1 &= 2\left(\frac{t - 1}{t}\right) x_0 - W \\ \end{align} $$
Summary of layout algorithms
Finding row sizes is identical to finding column sizes (except node width is replaced by height, etc).
To compute column widths, for each node, take its position $u$ and width $W$ along with the widths of the adjacent columns $w_0 \coloneqq w_{\lfloor u_i\rfloor}$ and $w_1 \coloneqq w_{\lfloor u_i\rfloor + 1}$ . Update the column widths according to
$$ w_i = \max(w_i, w'_i) $$
where the new sizes $w'_0$ and $w'_1$ are given by the layout algorithm,
$$ (w'_0, w'_1) = \text{Sizer}(t, W, w_0, w_1, g) , $$
which takes the fractional part $t = u - \lfloor u \rfloor$ of the node’s position, the node’s width $W$ and the original column widths $w_i$ as well as the column gutter $g$.
Linearly interpolated sizes:
$$ \text{Sizer}_\text{Linear}(t, W, w_0, w_1, g) = ((1 - t)W, tW) $$
Tight layout:
$$ \text{Sizer}_\text{Tight}(t, W, w_0, w_1, g) = \begin{cases} (W, 0) & \text{ if $t = 0$} \\ (0, W) & \text{ if $t = 1$} \\ \left(2x_0 + W, W - 2\left(\frac{t - 1}{t}\right)x_0\right) \end{cases} $$
where
$$ \begin{align} x_0 = -\frac{t}{2} \max\bigg\{ & 2g + w_1 + w_0, && \frac{2g + w_1 + W}{1 + t}, \\ & \frac{2g + W + w_0}{2 - t}, && \frac{2g + 2W}{2} \bigg\} \end{align} $$
It’s interesting how ugly the second method is on paper. But hopefully I’m not alone in believing that it produces nicer results!
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Expression Synth 🎶

Joseph Wilson — Mon, 17 Mar 2025 00:00:00 +0000

Fullscreen version

French verb conjugations quizzer

Joseph Wilson — Sun, 15 Sep 2024 00:00:00 +0000

Practice conjugating French verbs in specific tenses and moods. By changing the menus and toggles, you can practice conjugating verbs, distinguishing tenses by how they sound or are written, or even your pronunciation.

Fullscreen version, source code

Basic general relativity with Julia

Joseph Wilson — Sat, 19 Aug 2023 00:00:00 +0000

figure img { max-height: 350px; }

Four years after I took general relativity, my professor asked me to send him an old animation I had made during his course showingparallel transport of vectors around a line of latitude on the sphere. Actually, I wasn’t the one who had made it (though I was flattered that he had thought so!)

The tool I reached for was Julia. To my delight, the whole thing was rather simple to make. This was the humble result:

While taking that course, I remember thinking that numerically computing geodesics or simulating parallel transport were frightfully involved and required a mess of technical computing. Not so. If you have similar impression, read on.

This post describes how you might roll your own code for doing basic numerical general relativity (or rather, differential geometry) while only depending on external packages for dual numbers, differential equation solvers, and plotting.

Step 0: Maps between manifolds
Step 1: Automatic differentiation
Step 2: Metrics and Christoffel symbols
Step 3: Geodesics
Step 4: Parallel transport

Step 0: Maps between manifolds

The very first task is to decide how to represent the relevant mathematical objects: manifolds, homomorphisms between them and tensor fields. These things can seem complicated, but given a choice of coordinates, a point in a manifold is just aVector of coordinates, a homomorphism is a function taking aVector argument and returning aVector, and tensor fields are the same thing, but returning a multidimensionalArray.

For example, here’s an embedding $f: 𝕊^2 → ℝ^3$ of a sphere into 3D space using the usual spherical coordinates $(θ, φ)$, as a Julia function:

f((θ,φ))=[sin(θ)cos(φ)sin(θ)sin(φ)cos(θ)]

Notice this has one vector argument which is de-structured into two scalars.

julia>f([π/2,0])# meridian point on the equator3-elementVector{Float64}:1.00.06.123233995736766e-17

Here’s how you might use this embedding to plot a surface withMakie.jl:

usingGLMakiepoints=[f([θ,φ])forθ∈range(0,π,32),φ∈range(0,2π,64)]x,y,z=eachslice(stack(points),dims=1)wireframe(x,y,z,transparency=true,alpha=0.1,color=:blue)surface!(x,y,z,transparency=true,alpha=0.2)

You could imagine doing something more involved like defining an object-oriented zoo of types for manifolds, coordinate charts, transition functions, homomorphisms, tensor fields, and so on. Don’t bother. Dead simple is good.

Step 1: Automatic differentiation

Differential geometry has lots of “$∂$”s. We want to have derivatives of homomorphisms such asf be computedautomatically. There are many approaches to doing this with Julia, ranging from symbolic differentiation withSymbolics.jl to usingZygote.jl to differentiate code.

In our case, we’re not doing anything fancy — just building vectors with mathematical functions linesin — so all we’ll need is an implementation of dual numbers.

Dual numbers

If you extend the reals by adding an element $ε ≠ 0$ satisfying $ε^2 = 0$, then you can get derivative information for free when you evaluate a function with $+ε$, since

$$ f(x + ε) = f(x) + ε f'(x) $$

by Taylor’s theorem. The function value is the real part, and its derivative is the coefficient of $ε$. If we denote this coefficient by $⟨\quad⟩_ε$, we have our dual number derivative formula:

$$ f'(x) = ⟨f(x + ε)⟩_ε $$

To do this in Julia, theTaylorSeries.jl package is sufficient. Here’s computing the derivative of $\sin(x)$ at $x = \frac{π}{3}$:

julia>usingTaylorSeriesjulia>sin(Taylor1([π/3,1]))0.8660254037844386+0.5000000000000001t+𝒪(t²)julia>ans.coeffs[2]# get the ε term0.5000000000000001julia>ans==cos(π/3)true

To make accessing the $ε$ coefficient clearer, we may define

epsilonpart(a::Taylor1{T})whereT=get(a.coeffs,2,zero(T))

which works even if, for some reason,a is truncated to $𝒪(1)$ so thata.coeffs has no second component.

We can pack this into a function to take the derivative of functions on $ℝ$.

∂(f,x::Number)=epsilonpart.(f(Taylor1([x,1])))

which would be used like so:

julia>∂(sin,1)≈cos(1)true

Computing Jacobians

This trick works for functions of vectors, too. We just add $+ε$ to the component we want to differentiate with respect to:

$$ ∂_μ f(\vec x) = \bigg⟨ f\left(\begin{bmatrix}x^1 \\ \vdots \\ x^μ + ε \\ \vdots \\ x^n \end{bmatrix}\right) \bigg⟩_ε $$

For a vector-valued function $f : ℝ^m → ℝ^n$, we have a matrix of derivatives (a Jacobian).

$$ ∂f(\vec x) ≔ \mqty[ ∂_μ f^ν(\vec x) ]_{μν} = \mqty[ ∂_1 f(\vec x) \\ \vdots \\ ∂_m f(\vec x) ] = \mqty[ ∂_1 f^1(\vec x) & \cdots & ∂_1 f^n(\vec x) \\ \vdots & \ddots & \vdots \\ ∂_m f^1(\vec x) & \cdots & ∂_n f^n(\vec x) ] $$

Note. The Jacobian is conventionally the transpose of this, but we would like to have the first dimension (along columns) to run along the first index that appears, $μ$, and the second dimension (along rows) along the the second index, $ν$. This means indices in code will appear in the same order as in the maths, making the generalisation to higher order tensors easier. {: .aside}

Spelling this out with our dual number derivative formula:

$$ ∂f(\vec x) = \mqty[ \bigg⟨f^1\qty(\mqty[x^1 + ε\\\vdots\\x^n])\bigg⟩_ε & \cdots & \bigg⟨f^m\qty(\mqty[x^1 + ε\\\vdots\\x^n])\bigg⟩_ε \\ \vdots & \ddots & \vdots \\ \bigg⟨f^1\qty(\mqty[x^1\\\vdots\\x^n + ε])\bigg⟩_ε & \cdots & \bigg⟨f^m\qty(\mqty[x^1\\\vdots\\x^n + ε])\bigg⟩_ε \\ ] $$

Here’s a literal Julia implementation:

function∂(f,x::AbstractArray)I=eachindex(x)derivs=[epsilonpart.(f([Taylor1([x[i],i==j])fori∈I]))forj∈I]stack(derivs,dims=1)# stack vectors as rows in a matrixend

Let’s test it on our sphere embedding mapf.

julia>∂(f,[π/2,0])2×3Matrix{Float64}:6.12323e-170.0-1.00.01.0-0.0

The quantity $∂_μ f^ν (x)$ is then given by∂(f, x)[μ,∨]. We can check it is correct by comparing it to the analytic Jacobian:

julia>∂f((θ,φ))=[+cos(θ)cos(φ)+cos(θ)sin(φ)-sin(θ)-sin(θ)sin(φ)+sin(θ)cos(φ)0];julia>all(∂(f,x)≈∂f(x)forxineachcol(rand(2,100)))true

So far so good.

Higher order derivatives

But what about second order derivatives?

Luckily, since this is Julia,TaylorSeries.jl is generic and lets you construct Taylor series whose coefficients are themselves Taylor series.

This means our implementation for∂ works even on functions which call∂, letting us compute higher order derivatives in the obvious way. For example, $∂_μ ∂_ν f^λ(\vec x)$ may be evaluated as:

∂∂f(x)=∂(x′->∂(f,x′),x)

Notice that this produces a three dimensional array, since $∂_μ ∂_ν f^λ(\vec x)$ has three indices.

julia>∂∂f([1,2])2×2×3Array{Float64,3}:[:,:,1]=0.350175-0.491295-0.4912950.350175[:,:,2]=-0.765147-0.224845-0.224845-0.765147[:,:,3]=-0.540302-0.0-0.0-0.0

This shows how∂ diverges slightly from other Jacobian implementations: ours doesn’t flatten extra dimensions in order to produce a matrix.

Step 2: Metrics and Christoffel symbols

To do things like compute geodesics and simulate parallel transport, we need the metric $g_{μν}$ and Christoffel symbols $Γ^λ{}_{μν}$ for our manifold.

Induced metrics for embedded manifolds

If our manifold happens to be a parametric surface embedded in $ℝ^n$, such as our sphere embedding $f : 𝕊^2 → ℝ^3$, then we should be able to compute the metric on the surface induced by the ambient Euclidean metric, $η = \operatorname{diag}(1, \dots, 1)$. Specifically, the induced metric $g$ is thepullback of $η$ by $f$,

$$ g ≔ f^*η .$$

The pullback $f^*$ is defined so that the above is equivalent to

$$ g(\vec u, \vec v) ≔ η(f_*\vec u, f_*\vec v) $$

for any input vectors, where $f_*$ is thepushforward — nothing but the Jacobian matrix from before.

$$ f_* \big|_{\vec x} ≡ ∂f(\vec x) $$

With coordinates, the action on a vector may be written as

$$ (f_* \vec u )^ν\big|_{\vec x} = ∂_μ f^ν(\vec x) \, u^μ $$

so that the component form of $g(\vec u, \vec v)$ is:

$$ \begin{align*} g_{μν} u^μ v^ν &= η_{ρσ} \, ∂_μ f^ρ u^μ \, ∂_ν f^σ u^ρ \\ g_{μν} &= ∂_ρ f^μ \, η_{μν} \, ∂_σ f^ν \\ \end{align*} $$

Really, this is just matrix-matrix multiplication.

$$ g = f_* η (f_*)^\intercal = f_* (f_*)^\intercal $$

The Julia translation is just as pretty:

g(x)=let∂f=∂(f,x)∂f*∂f'end

This should give us the metric on the sphere. Let’s sanity check it again:

julia>g_sphere((θ,ϕ))=[100sin(θ)^2];julia>all(g(x)≈g_sphere(x)forxineachcol(rand(2,100)))true

Christoffel symbols

Now comes the interesting maths. The Christoffel symbols are defined in terms of the metric and its first derivatives as

$$ Γ^λ{}_{μν} = \frac12g^{λσ}\qty(\partial_μ g_{σν} + \partial_ν g_{σμ} - \partial_σ g_{μν}) .$$

Implementing this in Julia is just a matter of summing over all the right components. We could do this withsum(... for μ ∈ 1:n), but a nice way of doing this is to use a package likeTensorOperations.jl which provides a macro enabling Einstein summation notation.

usingTensorOperationsfunctionΓ(g,x)∂g=∂(g,x)# 3-index array@tensorΓ[λ,μ,ν]:=2\inv(g(x))[λ,σ]*(∂g[μ,σ,ν]+∂g[ν,σ,μ]-∂g[σ,μ,ν])end

Within the@tensor expression, any Julia identifier inside[ ] is treated as an index, whose range is automatically computed at runtime, with repeated indices summed. The:= means a new array is created, whereas= would write into an existing array.

To test this out and check that the basic symmetry $Γ^λ{}_{μν} = Γ^λ{}_{νμ}$ holds:

julia>Γ(g,[1,2])2×2×2Array{Float64,3}:[:,:,1]=0.00.00.00.642093[:,:,2]=0.0-0.4546490.6420930.0julia>ans==permutedims(ans,(1,3,2))true

Step 3: Geodesics

Now we get to the cool stuff — geodesics.

A geodesic is a curve $\vec x(t)$ which satisfies thegeodesic equation,

$$ \dot x^μ(t) ∇_μ \dot x^λ(t) = 0 .$$

In words, the velocity vector $\dot x$ is covariantly constant along the direction it points. Unpacking the covariant derivative, this becomes:

$$ \dot x^μ(t) ∂_μ \dot x^λ(t) + \dot x^μ(t) Γ^λ{}_{μν} \big|_{\vec x(t)} \dot x^ν(t) = 0 $$

Simplifying using the chain rule $\frac{\dd x^μ}{\dd t}\frac{∂\dot{x}^λ}{∂x^μ} = \ddot x^λ$ and leaving the position and $t$-dependence implicit, we have a system of second order ordinary differential equations

$$ \ddot x^λ = -Γ^λ{}_{μν} \dot x^μ \dot x^ν .$$

Julia has good support for numerically solving ODEs. We will use theDifferentialEquations.jl package. It requires the ODE to be put into standard form,

$$ \ddot{x} = f_\text{geodesic}(\dot{x}, x, g, t) .$$

The third argument $g$ can contain any extra data needed to evaluate $\ddot{x}$, which is just the metric in our case.

Additionally, the first argument of the Julia function is the arrayẍ to write the derivatives into (optional, but leads to better performance).

functiongeodesic!(ẍ,ẋ,x,g,t)@tensorẍ[λ]=-Γ(g,x)[λ,μ,ν]*ẋ[μ]*ẋ[ν]end

_Solving ODEs in Julia._

Here’s an example of numerically solving the geodesic equation with some initial data. Take our sphere, and imagine launching a particle eastward along the equator, $θ = π/2$.

julia>usingDifferentialEquationsjulia>ẋ0=[0.,1.];x0=[π/2,0.];tspan=(0,10);julia>prob=SecondOrderODEProblem(geodesic!,ẋ0,x0,tspan,g)ODEProblemwithuTypeArrayPartition{Float64,Tuple{Vector{Float64},Vector{Float64}}}andtTypeInt64.In-place:truetimespan:(0,10)u0:([0.0,1.0],[1.5707963267948966,0.0])julia>sol=solve(prob);

The solution objectsol is matrix-like, with four rows and $n$ columns

$$ \mqty[ \dot{\vec x}(t_1) & \cdots & \dot{\vec x}(t_n) \\ \vec x(t_1) & \cdots & \vec x(t_n) \\ ] ,$$

so we can extract just $\vec x(t)$ by taking the bottom two rows.

julia>sol[3:4,:]2×6Matrix{Float64}:1.57081.57081.57081.57081.57081.57080.00.001413060.01554360.1568491.5699110.0

We can see that our geodesic curve stays at the equator $θ = π/2$ and orbits around with $φ$ increasing at one radian per unit time.

Boring, but correct!

Interesting example: Doughnut geodesics

Let’s combine what we have so far to compute some geodesic curves on a torus.

The code is presented as a script for readability, but if you wanted to use it, you should pack everything into functions to avoid global variables and use something likeRevise.jl.

usingTaylorSeriesusingTensorOperationsusingDifferentialEquationsusingGLMakieepsilonpart(a::Taylor1{T})whereT=get(a.coeffs,2,zero(T))function∂(f,x)I=eachindex(x)derivs=[epsilonpart.(f([Taylor1([x[i],i==j])fori∈I]))forj∈I]stack(derivs,dims=1)# stack vectors as rows in a matrixendfunctionΓ(g,x)∂G=∂(g,x)@tensorΓ[λ,μ,ν]:=2\inv(g(x))[λ,σ]*(∂G[μ,σ,ν]+∂G[ν,σ,μ]-∂G[σ,μ,ν])endfunctiongeodesic!(ẍ,ẋ,x,g,t)@tensorẍ[λ]=-Γ(g,x)[λ,μ,ν]*ẋ[μ]*ẋ[ν]endR=2r=1.5# parametric torusf((θ,φ))=[(R+r*cos(φ))*cos(θ),(R+r*cos(φ))*sin(θ),r*sin(φ),]# embedded torus metricg(x)=letdf=∂(f,x)df*df'end# plot toruscircle=range(0,2π,length=64+1)x,y,z=eachslice(stack([f([θ,φ])forθ∈circle,φ∈circle]),dims=1)wireframe(x,y,z,transparency=true,alpha=0.1,color=:blue)surface!(x,y,z,transparency=true,alpha=0.2)# compute geodesicprob=SecondOrderODEProblem(geodesic!,[0.06,1],[0.,0.],(0,20),g)sol=solve(prob)# plot geodesicpath=mapslices(f,sol(0:0.1:30)[3:4,:],dims=1)lines!(path,linewidth=3,color=:black)

Step 4: Parallel transport

The other cool thing you can do once you have a metric is slide things along a path $γ(t)$ irrotationally.

If $\vec U(t)$ is a vector defined everywhere on the path $γ(t)$, then it is the parallel transport of some initial vector $U(0)$ if it satisfies theparallel transport equation,

$$ \dot γ^μ(t) ∇_μ U^λ(t) = 0 .$$

Like for the geodesic equation, unpacking the covariant derivative leads to the system of differential equations

$$ \dot γ^μ(t) ∂_μ U^λ(t) + \dot γ^μ(t) Γ^λ{}_{μν}\big|_{γ(t)} U^ν(t) = 0 $$

which, more succinctly, is

$$ \dot U^λ = -Γ^λ{}_{μν} \dot γ^μ U^ν .$$

This is only a first order differential equation, but now we have two pieces of data to provide as parameters: the metric $g$ and the path, $γ$.

In Julia, the differential equation function looks like

functionparalleltrans!(u̇,u,(g,γ),t)γ̇=∂(γ,t)@tensoru̇[λ]=-Γ(g,γ(t))[λ,μ,ν]*γ̇[μ]*u[ν]end

and solving it, for the sphere example, looks like:

julia>f((θ,φ))=[sin(θ)cos(φ),sin(θ)sin(φ),cos(θ)];julia>g(x)=let∂f=∂(f,x)∂f*∂f'end;julia>γ(t)=[π/4,t];# circle of latitudejulia>U0=[0.,1.];# initial vector pointing eastjulia>prob=ODEProblem(paralleltrans!,U0,10,(g,γ))ODEProblemwithuTypeVector{Float64}andtTypeInt64.In-place:truetimespan:(0,10)u0:2-elementVector{Float64}:0.01.0julia>sol=solve(prob);

The resulting solution objectsol is matrix-like, as before. But we can also call it likesol(t) to get the interpolated approximation for a given time.

This means we can animate it, with a little extra setup code:

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Defining Entropy with a Composition Law

Joseph Wilson — Mon, 24 Apr 2023 00:00:00 +0000

The mathematical definition ofentropy is often hard to motivate, coming across as rather mysterious. But it can be uniquely characterised with a very natural and pretty ‘composition law’, which anyone could come up with.

Theorem: The formula forinformation entropy

$$ H(𝒑) = -K \sum_{i} p_i \log_2 p_i $$

is theunique real-valued function of a finite probability distribution $𝒑$ which:

is continuous with respect to $𝒑$,
obeys theComposition Law,
increases for uniform distributions $𝒑 = (\frac1n, ..., \frac1n)$ as the width $n$ increases,
has some fixed value $K$ for a fair coin toss $𝒑 = (\frac12, \frac12)$.

The last assumption is merely a choice of units — without it, the function is defined up to an overall scaling factor.

Motivating the Composition Law

If we wanted to define the ‘uncertainty’ associated with a distribution, what properties should our measure have?

You can picture a finite probability distribution as a decision tree with weighted branches:

The same distribution can also be expressed by dividing the decision tree up into a composition of sub-trees:

Notice that each final outcome still has the same overall probability — all we’ve done is add an extra ‘step’ in the random process.

Since these two pictures represent the same scenario,we want our uncertainty measure to be the same for both. But the second picture has many parts. How do you measure the total uncertainty? By taking aweighted sum of the uncertainties of each sub-tree.

Why a weighted sum? Consider what should happen when one of the branches’ probabilities goes to zero: the total uncertainty should not include the uncertainty of the sub-tree under that branch, since it is certain that its outcomes won’t happen.

For example, the probability of $\{5, 6\}$ is $.1$ in the picture, so the total uncertainty contains only a small contribution from the green sub-tree.

Expressing this mathematically, the uncertainty measure $H$ should satisfy

where the left hand side is the uncertainty of the distribution in the first picture, and the right hand side is the total uncertainty in the second picture.

Note that the probabilities on each branch are still important, but are just too small to draw.

Formally

Now we can codify thiscomposition law more formally. Let

$$ \begin{aligned} 𝒑: Ω &→ [0, 1] \\ i &↦ p_i \end{aligned} $$

be a finite probability distribution which is normalised so that $\sum_{i ∈ Ω} p_i = 1$. Let $\sim$ be some equivalence relation on the set of outcomes, $Ω$. Define thequotient distribution

$$ \begin{aligned} 𝒑/{\sim}: Ω/{\sim} &→ [0, 1] \\ [i] &↦ \textstyle\sum_{j \sim i} p_j \end{aligned} $$

where $[i]$ is the equivalence class of $i$. This corresponds to taking a distribution and binning the outcomes. Also define therestricted distributions

$$ \begin{aligned} 𝒑|_{[i]} : [i] &→ [0, 1] \\ j &↦ \frac{p_j}{\sum_{k \sim j} p_k} \end{aligned} $$

which correspond to keeping only a subset of outcomes and renormalising the probabilities.

These fit into the example above where $Ω = \{1, ..., 6\}$ like so:

Expressed in this language, the composition law is

$$ H(𝒑) = H(𝒑/{\sim}) + \sum_{e ∈ Ω/{\sim}} (𝒑/{\sim})(e) \, H(𝒑|_e) $$

where $(𝒑/{\sim})(e) = \sum_{i ∈ e} p_i$.

Proof of Uniqueness

We will prove that any ‘uncertainty’ function $H(𝒑) ≡ H(p_1, ..., p_n)$ satisfying

continuity
the composition law
the property that $H(\frac1n, …, \frac1n)$ increases with $n$

must be equal to the Shannon entropy

$$ H(𝒑) = -K \sum_{i} p_i \log_2 p_i $$

where $K = H(\frac12, \frac12)$. The proof is in three steps, one for each property above:

Show that $H(𝒑)$ is uniquely defined for all distributions $𝒑$ if it is known for allrational distributions $𝒒 ∈ ℚ^n$.
Show that $H(𝒒)$ is uniquely defined for all rational distributions $𝒒$ if the entropy of the uniform distribution $U(n) ≔ H(\underbrace{\textstyle\frac1n, ..., \frac1n}_n)$ is known for all $n$.
Show that $U(n)$ is uniquely defined for all $n$ if we fix $U(2) = K$.

Step 1.

This follows by the assumption of continuity. If $𝒑 ∈ ℝ^n$ is the limit a sequence of rational distributions $𝒒_i ∈ ℚ^n$, then by continuity $H(𝒑) = \lim_i H(𝒒_i)$.

Step 2.

Let $𝒒 ∈ ℚ^n$ be a rational distribution, and let $D$ be the lowest common denominator of all the probabilities $q_i$, so that

$$ 𝒒 = (d_1/D, d_2/D, ..., d_n/D) $$

where $d_i$ are non-negative integers. Now consider the set $Ω = \{1, 2, ..., D\}$, and let $𝒓(i) ≔ 1/D$ be the uniform distribution on $Ω$. Define an equivalence relation $\sim$ which partitions $Ω$ into $n$ different sets $\{e_1, e_2, ..., e_n\}$, where the $i$th set contains $d_i$ elements. The size of the $i$th equivalence group, as a fraction of the whole, is given by $|e_i|/|Ω| = d_i/D = q_i$, and we have $𝒓/{\sim} = 𝒒$ by construction.

From the composition law, we have

$$ \begin{aligned} H(𝒓) &= H(𝒓/{\sim}) + \sum_{i = 1}^n \frac{d_i}D \, H(𝒓|_{e_i}) \\ U(D) &= H(𝒒) + \sum_{i=1}^n \frac{d_i}D \, U(d_i) \end{aligned} $$

since $H(𝒓) = U(D)$ and $H(𝒓|_{e_i}) = U(d_i)$. This shows that $H(𝒒)$ is uniquely defined if $U(n)$ is known for all $n$.

Step 3.

We will now show that $U(n)$ is uniquely defined by $U(2)$ by showing that the only possible functions are

$$ U(n) = K\log_2 n $$

where $K$ is a free parameter.

Consider a uniform distribution $r$ on $\{1, 2, ..., nm\}$ partitioned by $\sim$ into $n$ groups of $m$, so that the $i$th equivalence class is $[ni] = \{ni, ni + 1, ni + m - 1\}$. Writing down the composition law for this partition yields

$$ \begin{aligned} H(𝒓) &= H(𝒓/{\sim}) + \sum_{i = 1}^n \frac1n H(𝒓|_{[ni]}) \\ U(nm) &= U(n) + \sum_{i = 1}^n \frac1n U(m) \end{aligned} $$

and hence $U(nm) = U(n) + U(m)$. The only increasing functions $U(n)$ satisfying this property are multiples of $\log n$. By choosing $U(2) ≔ K$, we fix $U(n) = K\log_2 n$.

This completes the proof!

In summary, under the three assumptions, knowing $U(2) ≡ H(\frac12, \frac12)$ is enough to uniquely define $U(n)$ for all $n$, which is in turn enough to uniquely define $H(𝒒)$ for all rational distributions $𝒒$, which is enough to define $H$ completely.

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Daylight Hours: A spherical geometry problem

Joseph Wilson — Mon, 09 Jan 2023 00:00:00 +0000

How many hours are there between sunrise and sunset? The Earth’s tilt makes this a fun geometry problem…

Hours of daylight at the solstice

Let $ε$ be the Earth’s tilt, orobliquity. This is the angle between Earth’s axis of rotation $N$ and the normal to Earth’s orbital plane around the sun, $O$. At first, let’s assume that the Earth tilts directly toward the sun (or directly away, for $ε< 0$).= This= direct= alignment= occurs= twice= a= year,= at= the= summer= and= winder= solstices.= We= can= consider= seasons= later;= for= now,= assume= we’re= at= summer= solstice.

In this scenario, the boundary between day and night is a great circle whose angle from the north pole is $ε$. In the figure below, the view of Earth is aligned to the north pole, so that changing the tilt $ε$ changes the relative position of the sun.

Now, let $φ$ be the latitude of a reference point on Earth, so that $φ = 0$ is on the equator, and $φ > 0$ is in the northern hemisphere. Looking down on the north pole, we see that the number of daylight hours is determined by the arc length of the circle of latitude which lies outside the shadow of night (indicated in orange).

To find this arc length in terms of $ε$ and $φ$, we must find the points of intercept of the circle of latitude with the ellipse defined by the top-down view of the day/night boundary.

Dawn and dusk intercept points

Assuming that the Earth is the unit circle, the circle of latitude has radius $\cos φ$, and the day/night ellipse has horizontal radius $\sin ε$ and unit vertical radius. Parametrically, the ellipse is represented as

$$ (x, y) = (\sin ε \cos λ, \sin λ), λ ∈ [0, 2π) .$$

This intercepts the circle when the length of the vector $(x, y)$ is the circle’s radius, $\|(x, y)\| = \cos φ$. Squaring this, we get the intercept condition

$$ \sin^2 ε \cos^2 λ + \sin^2 λ = \cos^2 φ $$

which, using $\cos^2 λ = \frac12(1 + \cos 2λ)$ and $\sin^2 λ = \frac12(1 - \cos 2λ)$, is the same as

$$ 1 + \sin^2ε + (\sin^2 ε - 1)\cos 2λ = \cos^2φ .$$

Solving for the parameter value $λ$, we get

$$ \cos 2λ = \frac{2\cos^2φ - \sin^2 ε - 1}{\sin^2 ε - 1} = 2 \left( \frac{\sin^2 φ}{\cos^2 ε} \right) - 1 $$

where we use $\cos^2 θ + \sin^2 θ = 1$ twice to simplify. Instead of solving directly for $λ$ here, note that the right-hand side matches the identity $\cos 2λ = 2\cos^2 λ - 1$, so we must have

$$ \cos λ = \frac{\sin φ}{\cos ε} .$$

Inverting this in generality gives us

$$ λ = ±\arccos\left(\frac{\sin φ}{\cos ε}\right) + 2nπ $$

for some $n ∈ ℤ$. If we take just the positive value for $n = 0$ then we have $λ ∈ [0, π]$, since this is the conventional range of $\arccos$. This means our intercept point $(x, y) = (\sin ε \cos λ, \sin λ)$, shown by the highlighted dot, always falls in the upper half-plane:

This is one of four possible intercepts, but it’s the one we want: when $εφ > 0$, the intercept is in the first quadrant, and when $εφ< 0$= it= is= in= the= second,= which= is= consistent= with= out= three-dimensional= picture= of= Earth,= where= the= intercept= we= want= depends= on= which= hemisphere= we’re= in.

We’re almost done, but the parameter value $λ$ isn’t the actual polar angle of the intercept point — we need that to find the arc length which lies in daylight. The polar coordinate $θ$ of the intercept point $(x, y)$ is

$$ θ = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{\sinλ}{\sin ε \cos λ}\right) = \arctan\left(\frac{1}{\sin ε} \tan λ\right) .$$

The arc length is thus twice this angle, $2θ$, while the total circumference is $2π$. Therefore, written in full, the fraction of the day in daylight is

$$ L = \frac1π\arctan\left(\frac{1}{\sin ε} \tan \left[\arccos\left(\frac{\sin φ}{\cos ε}\right)\right]\right) .$$

Incorporating seasons

In reality, the Earth’s tilt is not exactly toward or away from the sun, because it orbits the sun while its titled axis of rotation remains fixed.

At the northern summer solstice, the north pole is tilted toward the sun. Let $ω$ be the angle of the Earth’s course around the sun from this point.

Looking down the axis the orbital plane, $O$, with the sun always to the right, we see the north pole trace out a circle as $ω$ varies over the course of a year.

The Earth’s obliquity $ε_0$ remains fixed (the angle of the dashed green arc). However, the true angle the north pole makes from the day–night boundary (the solid green arc which we called $ε$ before) varies with $ω$. This is the seasonal dependence of day length.

Our task now is to find $ε$ in terms of $ε_0$ and $ω$.

The points $O$, $N$ and $P$ form a spherical triangle in which each side is an arc of a great circle. To find $ε$ (the angle of the arc $\overline{PN}$) we may use the spherical law of sines:

$$ \frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C} $$

Specifically, reading off the figure above, this gives us

$$ \frac{\sin\overline{PN}}{\sin(\frac{π}{2} - ω)} = \frac{\sin\overline{ON}}{\sin\frac{π}{2}} $$

which reduces to

$$ \frac{\sin ε}{\cos ω} = \sin ε_0 .$$

For completeness, the full expression for the fraction of the day in daylight is

$$ L = \frac1π\arctan\left(\frac{1}{\sin ε_0\cosω} \tan \left[\arccos\left(\frac{\sin φ}{\sqrt{1 - (\sin ε_0\cosω)^2}}\right)\right]\right) $$

where $ε_0 = 24.5^\circ$ is the obliquity of the Earth, $φ$ is the latitude of the reference point and $ω$ is the season expressed as the arc angle along Earth’s orbit since the solstice.

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Billiard geometry: Visually solving elastic collisions

Joseph Wilson — Fri, 11 Nov 2022 00:00:00 +0000

Conservation of momentum

Momentum is always conserved, so that

$$ 𝒑_1 + 𝒑_2 = 𝒑_\text{total} = 𝒑'_1 + 𝒑'_2 .$$

We can more easily visualise this constraint by drawing the momentum vectors tip-to-tail. The initial and final vector pairs must both extend to the same $𝒑_\text{total}$.

The collisions we can describe look more realistic now that momentum is conserved, but we can stillcreate ordestroy kinetic energy…

Conservation of energy

Energy conservation is a bit more interesting. In an elastic collision, the total kinetic energy is conserved:

$$ \frac{(𝒑_1)^2}{2m_1} + \frac{(𝒑_2)^2}{2m_2} = T_\text{total} = \frac{(𝒑_1')^2}{2m_1} + \frac{(𝒑_2')^2}{2m_2} $$

If the masses are equal, this implies

$$ \sqrt{(𝒑_1)^2 + (𝒑_2)^2} = \ell = \sqrt{(𝒑'_1)^2 + (𝒑'_2)^2} $$

, with the geometric interpretation that the total of the “lengths” of the momentum vectors is the same before and after the collision. This is the defining property of an ellipse with the origin and $p_\text{total}$ as foci:

Now, all the scenarios we can describe are plausible elastic collisions.

Angle of contact

To complete the picture, the tangent line at which the bodies kiss is found geometrically by connecting the ellipse points.

Now, everything looks right, no matter what you do to the points! (Except perhaps scenarios withintersecting balls…)

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Geometric Algebra for Special Relativity and Manifold Geometry

Joseph Wilson — Sat, 12 Mar 2022 00:00:00 +0000

My master’s thesis in mathematical physics completed in March 2022 at Victoria University of Wellington.

The thesis centres around geometric algebra and its applications to special relativity.

Table of Contents

I. Geometric Algebra and Special Relativity

Introduction
Preliminary Theory
- Associative Algebras
- The Wedge Product: Multivectors
- The Metric: Length and Angle
The Geometric Algebra
- Construction and Overview
- Relations to Other Algebras
- Rotors and the Associated Lie Groups
- Higher Notions of Orthogonality
- More Graded Products
The Algebra of Spacetime
- The Space/Time Split
- The Invariant Bivector Decomposition
- Lorentz Conjugacy Classes
Composition of Rotors in terms of their Generators
- A Geometric BCHD Formula
- BCHD Composition in Spacetime
Calculus in Flat Geometries
- Differentiation of Fields
- Case Study: Maxwell’s Equations

II. Geometric Algebra and Special Relativity

World Flags Quizzer

Joseph Wilson — Sun, 06 Mar 2022 00:00:00 +0000

Learn world flags by getting quizzed!

Flags are chosen randomly weighted by the country’s population. Press Check or type enter to submit your answer, and type ? to show the flag’s country name. Flags you get right in one try are “memorized” and won’t be shown again.

Fullscreen version

A Simple Example of a Connection With Torsion

Joseph Wilson — Tue, 19 Oct 2021 00:00:00 +0000

While learning general relativity as an undergrad, I was uneasy with the idea oftorsion of an affine connection. Familiar phrases like “torsion measures how a frametwists as it undergoes parallel transport” seemed too lofty to serve as a helpful mental model.

The following example is one which gave me anaha! moment.

A “rolling” connection on $ℝ^2$.

Consider a connection on the plane defined in such a way that a vector undergoing parallel transport rotates in the $xy$-plane in proportion to its motion along the $x$-direction. That is, define a connection so that the vector field defined by

$$ \vec u(x, y) = \mqty( \cos kx \\ \sin kx) ,$$

which spins counter-clockwise when moving in the $+x$-direction, is covariantly constant, $∇_μ \vec u = 0$.

We can write the covariant derivative

$$ ∇_μu^a = ∂_μu^a + Γ^a{}_{μb}u^b $$

more easily by reading the Latin indices as the two components of a matrix. Then, for differentiation in the $x$-direction, we have

$$ ∇_x \vec u = ∂_x \vec u + Γ_x \vec u = \mqty(-k\sin kx \\ +k\cos kx) + Γ_x \mqty( \cos kx \\ \sin kx) $$

where

$$ Γ_x \vec u$$

is matrix multiplication. Enforcing $∇_x \vec u = 0$ implies that the matrix $Γ_x$ must be

$$ Γ_x = \mqty(Γ^x{}_{xx} & Γ^x{}_{xy} \\ Γ^y{}_{xx} & Γ^y{}_{xy}) = \mqty(0 & -k \\ +k & 0) .$$

The $y$-direction is trivial, since $\vec u$ does not depend on $y$. And so,

$$ ∇_y \vec u = ∂_y \vec u + Γ_y\vec u = Γ_y\vec u $$

is zero only when

$$ Γ_y = \mqty(Γ^x{}_{yx} & Γ^x{}_{yy} \\ Γ^y{}_{yx} & Γ^y{}_{yy}) = \mqty(0 & 0 \\ 0 & 0) .$$

This connection has zero curvature…

If we calculated the Riemann curvature

$$ R = \dd Γ + Γ ∧ Γ ,$$

we would find that this connection is flat, since $\dd Γ = 0$ (all Christoffel symbols are constant) and

$$ Γ ∧ Γ % = (Γ_x \dd x + Γ_y \dd y) ∧ (Γ_x \dd x + Γ_y \dd y) = (Γ_x Γ_y - Γ_y Γ_x) \, \dd x ∧ \dd y = 0 $$

also vanishes since $Γ_y = 0$.

…but non-zero torsion

However, the torsion

$$ T^λ{}_{μν} = Γ^λ{}_{μν} - Γ^λ{}_{νμ} $$

does not vanish:

$$ T_x = \mqty(T^x{}_{xx} & T^x{}_{xy} \\ T^y{}_{xx} & T^y{}_{xy}) = \mqty(0 & -k \\ 0 & 0) ,\qquad T_y = \mqty(T^x{}_{yx} & T^x{}_{yy} \\ T^y{}_{yx} & T^y{}_{yy}) = \mqty(+k & 0 \\ 0 & 0) .$$

We can see that the torsion measures the wavenumber $k$ controlling the rate at which vectors spin when parallel transported along the $x$ axis.

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Timeline of Scientists

Joseph Wilson — Mon, 16 Aug 2021 00:00:00 +0000

This is an interactive timeline of famous mathematicians and physicists throughout history. The data is pulled from Wikidata.

Showing scientists, in decreasing order of number of Wikidata entities named after them. Hold `alt` and scroll to zoom.

Explicit Baker–Campbell–Hausdorff–Dynkin formula for Spacetime via Geometric Algebra

Joseph Wilson and Matt Visser — Mon, 12 Jul 2021 00:00:00 +0000

In brief. In $≤4$ dimensions, there’s a simple formula for the bivector $σ_3$ in terms of bivectors $σ_1$ and $σ_2$ such that $e^{σ_1}e^{σ_2} = ±e^{σ_3}$.

Abstract. We present a compact Baker–Campbell–Hausdorff–Dynkin formula for the composition of Lorentz transformations $e^{σ_i}$ in the spin representation (a.k.a. Lorentz rotors) in terms of their generators $σ_i$:

$$ \ln(e^{σ_1}e^{σ_2}) = \tanh^{-1}\qty(\frac{ \tanh σ_1 + \tanh σ_2 + \frac12\qty[\tanh σ_1, \tanh σ_2] }{ 1 + \frac12\qty{\tanh σ_1, \tanh σ_2} }) $$

This formula is general to geometric algebras (a.k.a. real Clifford algebras) of dimension $≤ 4$, naturally generalising Rodrigues’ formula for rotations in $ℝ^3$. In particular, it applies to Lorentz rotors within the framework of Hestenes’ spacetime algebra, and provides an efficient method for composing Lorentz generators. Computer implementations are possible with a complex $2×2$ matrix representation realised by the Pauli spin matrices. The formula is applied to the composition of relativistic $3$-velocities yielding simple expressions for the resulting boost and the concomitant Wigner angle.

Full paper. Atdoi.org/10.1142/S0219887821502261 and onArχiv.

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Mathematical One-liners

Joseph Wilson — Fri, 09 Jul 2021 00:00:00 +0000

function toggleSpoilers() { document.body.classList.toggle("spoiled") } function toggleNames() { document.body.classList.toggle("shownames") }

This is a collection of succinct but wonderfully satisfying theorems. Leave a comment if you have a good one!

The punchlines are hidden; try to guess the blanks!
Toggle theorem names.
Toggle spoilers.

Liouville’s theorem
Any bounded complex differentiable function isconstant.
Jordan normal form
Any matrix is similar to anupper triangular matrix.
Cauchy’s integral theorem
The integral of a holomorphic function around a loopvanishes identically.
Riemann Mapping Theorem
Strict subsets of $ℂ$ which arenonempty, open, and simply connected are related by a conformal mapping.
A complete ordered field isisomorphic to the real line.
Every field is a subfield ofthe surreal numbers.
Cayley’s theorem
All groups arepermutation groups.
Lagrange’s theorem
The order of a finite group isdivisible by the orders of its subgroups.
Fundamental theorem of finite Abelian groups
A finite Abelian group is a direct sum ofprime-order cyclic groups.
The $ℤ$-modules are exactly theAbelian groups.
Cayley–Hamilton theorem
A square matrixsatisfies its own characteristic polynomial.
Green–Tao theorem
There arearbitrarily long arithmetic progressions of primes.
Dirichlet’s Theorem
Every proper arithmetic sequence containsinfinitely many primes.
Two random infinite graphs areisomorphic with probability one.
Hurwitz’s Theorem
Every normed division algebra isisomorphic to $ℝ$, $ℂ$, $ℍ$ or $𝕆$.

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Light installation for Gatherings Restaurant

Joseph Wilson — Sun, 27 Dec 2020 00:00:00 +0000

A projected light show made for a multi-sensory dining experience atGatherings restaurant.

Interactive fullscreen version.

Act 0: Arrival

Pet nat // sparkling white wine, honey, fielding, clover, shoreline, waipara champagne, popping.
Canapes, kina w potato, ricotta and seaweed tart // refined, interesting, saline, creamy green.

Act 1: Genesis

Saltlick chardonnay, north Otago, 2015 // acid, hazy orange, green gaze, plums, transparent, open.
Crayfish and tomatoes // capsicum, smoky sweetness.

Act 2: Richness and Shimmer

Limestone hills, Pinot noir // ash, wood, aromatic, Umeboshi, black olives, floral aroma.
Lado and raw tuna // red iron, flowers that grow on coastlines, native species, chewy fish, soft fat.

Act 3: Catholic churches and Earth

Arden pinot noir 2018 // clay in soil, volume, roundness, saline, freshness.
Pumpkin, black garlic, yoghurt, onion // hemp seeds, nuttiness and fat, depth and earthiness.

Act 4: Crescendo

Limestone Syrah - truffles, elderberries, black currants // wholesome, big notes, rich, chocolate.
Venison dish // autumnal forests, barbecue, elderberries, mushrooms, deep in the bush, raining.

Act 5: Dessert

Dessert wine, 2011, still in barrel, convert Viognier // cloudy sweet.
Rhubarb cooked in black boy peach compot, kafir infused with kawa kawa, citrus granita.

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An Overview of the Strong CP Problem and Axion Cosmology

Joseph Wilson — Sat, 07 Nov 2020 00:00:00 +0000

Literature review supervised by Dr. Jenni Adams at the University of Canterbury.

I wanted to learn more about particle physics after my Bachelor’s, so a year of part-time study culminated in this literature review. I learned basic classical (and a little quantum) field theory, and read about the “strong $CP$ problem” of quantum chromodynamics and its solution via “axions”. The report is aimed at the mathematically-inclined graduate who is uninitiated in particle physics.

FullPDF.

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The Many Faces of Stokes’ Theorem

Joseph A. Wilson — Tue, 17 Mar 2020 00:00:00 +0000

{% assign project_url = site.github.url | append: “/projects/stokes-theorem” %}

Thegeneralised Stokes theorem has the remarkably compact form

$$ \boxed{ \int_Ω \dd ω = \int_{∂Ω} ω } $$

where $Ω$ is a $k$-dimensionalmanifold with boundary $∂Ω$, and $ω$ is a $(k - 1)$-form on $Ω$.

While it may seem alien at first when expressed in full generality, you may recognise some of the many special cases of Stokes’ theorem, especially from vector calculus.

Special cases of Stokes’ theorem:

Adjust the parameters below or select a preset to see the associated Stokes theorem.

Ambient Space:				Metric signature:
Differential Form:				Hodge dual:

The metric signature is relevant to theHodge dual operation (which requires the notion of a metric). Divergence-type theorems arise naturally when $ω$ is the Hodge dual of a $1$-form.