Rotations in spacetime

...or, How to Guess Special Relativity from Geometry.

Non-relativistic physics takes place on a background of 3-dimensional Euclidean space. Time is incorporated into this picture as a global parameter used to describe motion in this space. We can write things in terms of a 4-dimensional "space-time", but this is something of a triviality as space and time have very little to do with each other. If space (and its contents) at any given instant could be described as a flat slice, then our "space-time" would be like a stack of slices with time increasing vertically, but this picture is not really any different to looking at each slice individually (in the correct order). This artificiality is reflected in the Galilean transformations, which describe how inertial (non-accelerating) observers interpret space and time coordinates - in 3 spatial dimensions, these are given by $(t',\mathbf{x}')=(t,\mathbf{x}-\mathbf{v}t)$, with $t',t\in\mathbb{R}$, $\mathbf{x}',\mathbf{x},\mathbf{v}\in\mathbb{R}^3$ for observers with a relative velocity of $\mathbf{v}$. Note that the time coordinate is unaffected by the motion, leading to a sort of asymmetry between space and time.

The Galilean transformations are important from a physical perspective because they relate to how observers (for example, experimental physicists) perceive the universe. However, there are important "pre-physical" geometric transformations of interest as well. The ones of primary interest are the translation in time, the translations in space and the rotations. Translations are given by $(t',\mathbf{x}')=(t+s,\mathbf{x}+\mathbf{y})$, where $s\in\mathbb{R}$ and $\mathbf{y}\in\mathbb{R}^3$, with $s$ and $\mathbf{y}$ carrying the interpretations as shifts in time and space respectively. We won't focus on these as they're basically the same in the relativistic case (and are arguably harder to deal with). Rotations however take the apparently even more simple form of $(t',\mathbf{x}')=(t,\mathbf{Rx})$, for $\mathrm{SO}(3)$, but this is really something of a sleight of hand of notation. To see why, let's investigate rotations a little more deeply.

Rotations can be thought of as (linear) transformations which preserve lengths and angles, so in order to talk about rotations it's necessary to impose some way to measure those things. The way we will do that is using a metric, which is a bilinear (takes two inputs and is linear in both) form (the objects it takes is vectors and it spits out a scalar). As one can tell from this description, when acting on pairs of vectors the metric looks much like an inner product, and in fact for the case of $\mathbb{R}^n$ the Euclidean metric (which we will henceforth use) defines the standard inner product in this way. Calling our metric $\delta$, for vectors $\mathbf{u}$ and $\mathbf{v}$ we have $\delta(\mathbf{u},\mathbf{v})=\mathbf{u}\cdot\mathbf{v}\equiv\langle\mathbf{u},\mathbf{v}\rangle\equiv\mathbf{u}^\mathrm{T}\mathbf{v}\in\mathbb{R}$, to give an overview of common notations.

First, we note that the norm induced by the inner product/metric is given in the standard way by $\lVert\mathbf{u}\rVert=\delta(\mathbf{u},\mathbf{u})$. If a transformation (/operator/matrix), denoted by $\mathbf{M}$, preserves the length of $\mathbf{u}$ then $\lVert\mathbf{Mu}\rVert=\lVert\mathbf{u}\rVert$. If $\mathbf{M}$ is length-preserving, then the requirement for it to be angle-preserving is expressed as $\delta(\mathbf{Mu},\mathbf{Mv})=\delta(\mathbf{u},\mathbf{v})$ (note that in this case, length-preservation is a special case of angle-preservation). Using transpose/matrix notation, this can be written as $(\mathbf{Mu})^\mathrm{T}(\mathbf{Mv})=\mathbf{u}^\mathrm{T}\mathbf{M}^\mathrm{T}\mathbf{Mv}=\mathbf{u}^\mathrm{T}\mathbf{v}$, which clearly implies that $\mathbf{M}^\mathrm{T}\mathbf{M}=\mathbf{I}$, or in other words, $\mathbf{M}^\mathrm{T}=\mathbf{M}^{-1}$. Finally, we will also impose that the determinant of $\mathbf{M}$ is equal to $+1$ - the length-preservation requirement demands $\mathrm{det}(\mathbf{M})=\pm1$, but the $-1$ case corresponds to reflections, which we are not interested in here. As a side-note, as a matrix, the identity $\mathbf{I}$ is exactly equivalent to the metric in Cartesian coordinates $\mathbf{\delta}$, so we could equally write this condition in the somewhat unusual form $\mathbf{M}^\mathrm{T}\mathbf{\delta M}=\mathbf{\delta}$ (although one could argue the identity matrix was secretly there all along in the definition of the inner product $\delta(\mathbf{u},\mathbf{v})=\mathbf{u}^\mathrm{T}\delta\mathbf{v}$). In any case, all of the above is what it means for $\mathbf{M}$ to be a rotation, also known as a "special orthogonal transformation", hence the name of the space $\mathrm{SO}$. For a rotation in 3 spatial dimensions, we write this as $\mathbf{M}\in\mathrm{SO}(3)$.

Now, what happens when instead of 3-dimensional Euclidean space, we instead examine 4-dimensional Minkowski space(time)? A priori there is little reason to do this in particular, outside of a certain limited sense of curiosity, but let us continue regardless. In this case, rather than position vectors $\mathbf{x}$, we have position 4-vectors $x$, which we can coordinatise by $x=(x^0,\mathbf{x})$ where $x^0$ is a time coordinate baked into the space, and the spatial coordinates are as in the 3-dimensional Euclidean case. This is like, rather than a collection of space-slices stacked time-wise, a single block of spacetime we can choose to slice horizontally, or diagonally, or at whatever angle we choose (up to 45°), picking our space and time directions as appropriate. Of course, this picture isn't yet completely justified, but it can be with the formalism we will have developed by the end of this post. In any case, we must note that what distinguishes any viable choice of $x^0$ from $x^i$, $i=1,2,3$ is that, in matrix form, our metric takes the form $\eta=\mathrm{diag}(-1,1,1,1)$ - this is critical, as the fact that the first entry (the "00" component) is negative is what makes the time coordinate different to the spatial ("$ii$") ones (recall for comparison that $\mathbf{I}_3=\delta=\mathrm{diag}(1,1,1)$).

The fact that our metric is different immediately suggests that we are free (or perhaps required) to modify our idea of a rotation, which we defined using the Euclidean 3-metric. What is the Minkowski equivalent? Let's consider first the angle-preservation requirement. Now we have $\eta(Mu,Mv)=\eta(u,v)$, but this time the matrix form of the inner product is given by $\eta(u,v)\equiv u^\mathrm{T}\eta v$, so we have $u^\mathrm{T}M^\mathrm{T}\eta Mv=u^\mathrm{T}\eta v\implies M^\mathrm{T}\eta M=\eta$. This is encouraging, because it's exactly the same as the requirement for angle-preservation in the 3-dimensional case but with the Euclidean metric swapped for the Minkowski metric (and the $3\times3$ rotation matrices swapped out for these mysterious $4\times4$ "rotation" matrices). As before, we will demand that $\mathrm{det}(M)=+1$, and having done so we can write $M\in\mathrm{SO}(1,3)$, the set of "indefinite special orthogonal matrices (of signature (1,3))". A bit more of a mouthful this time around!

Having names and labels is convenient, but it doesn't give us much of a feel for what we've found. To do that, let's simplify, and consider rather than the case of 1 time dimension and 3 space dimensions (a (1,3) signature), the case of 1 time dimension and 1 space dimension (a (1,1) signature). This might seem odd at first, given you can't exactly rotate in 1 spatial dimension, but it will pay off! Let's consider an arbitrary $2\times2$ matrix $A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. The angle-preservation requirement then means
\begin{equation*}
A^\mathrm{T}\eta A=\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c^2-a^2&cd-ab\\cd-ab&d^2-b^2\end{pmatrix}=\eta=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.
\end{equation*}
This gives us 3 equations:

1. $c^2-a^2=-1$ or equivalently $a^2-c^2=1$,
2. $d^2-b^2=1$,
3. $cd-ab=0$ or $ad=bc$.

Three equations for four variables means we must express the solution in terms of a parameter, which we will call $\phi$. In that case, we have 2 options (up to signs) which solve the above equations equally well: $a=d=\cosh(\phi)$ and $b=c=\sinh(\phi)$, or $a=-d=\cosh(\phi)$ and $b=-c=\sinh(\phi)$. We can distinguish between them by demanding that $\det(A)=+1$, which is satisfied by the former option. So, now we've found the explicit form for matrices in $\mathrm{SO}(1,1)$... what can we make of them?

Figure 1: On the left, two circular sectors, each of equal Euclidean area subtending an equal circular angle $\theta_0$. Each can be considered as the other one but (circularly) rotated (red arrows).
On the right, three hyperbolic sectors, each of equal Euclidean area subtending an equal hyperbolic angle $\phi_0$. Each can be considered the same as the others but (hyperbolically) rotated (red arrows). Note that the hyperbolic sector subtends varying circular angles as it's hyperbolically rotated through, but the hyperbolic angle subtended by the sector remains constant. Be careful that this does not lead to confusion.

As it happens, the matrices $\left(\begin{smallmatrix}\cosh(\phi)&\sinh(\phi)\\ \sinh(\phi)&\cosh(\phi)\end{smallmatrix}\right)$ describe what are called hyperbolic rotations. Just like a circle can be parametrised by the (circular) angle $\theta\in[0,2\pi)$, and (circular) rotations leave circles invariant, rectangular hyperbolas can be parametrised by the hyperbolic angle $\phi\in(-\infty,+\infty)$ and rectangular hyperbolas are left invariant by hyperbolic rotations (Fig. 1). These hyperbolic rotations are sometimes called squeeze mappings, because they preserve (Euclidean) areas without rotating or shearing by squeezing (or stretching) them in some sense (Fig. 2). In fact, analogous to the circular case, the exact parametrisation of a hyperbola by $\phi$ can be related to the angle subtended by a hyperbolic sector under hyperbolic rotation, although the details are not important for our purposes.

Figure 2: The reason hyperbolic rotations are sometimes called "squeeze mappings" is that a shape which is hyperbolically rotated becomes 'squeezed' taller and thinner or wider and flatter, depending on the sign of the rotation. One way to think of this is by taking a shape by its bounding box, holding one corner in place and moving its opposite corner along a hyperbola. In any case, note that the Euclidean area is constant throughout the hyperbolic rotation!

What we are more interested in are the physical interpretations of this hyperbolic rotation. In order to investigate, let's consider the action of $A$ on a position vector $(x^0,x^1)$ in Minkowski space, noting that our Cartesian coordinates are given by $x^0=ct,x^1=x$ in dimensionful units ($c$ is a constant with dimensions of [length/time] (the same as speed) such that $ct$ has units of length to match $x$):
\begin{equation*}
\begin{pmatrix}ct'\\x'\end{pmatrix}=\begin{pmatrix}\cosh(\phi)&\sinh(\phi)\\ \sinh(\phi)&\cosh(\phi)\end{pmatrix}\begin{pmatrix}ct\\x\end{pmatrix}=\begin{pmatrix}\cosh(\phi)ct+\sinh(\phi)x\\\sinh(\phi)ct+\cosh(\phi)x\end{pmatrix}.
\end{equation*}
The meaning of this is not immediately obvious, so let's consider a special case of interest: the world-line of an observer at rest within their own frame, i.e., the line given by $x=0$. This line, when transformed, clearly takes the form $(ct',x')=(\cosh(\phi)ct,\sinh(\phi)ct)$, or written differently, the pair of equations $t'=\cosh(\phi)t$, $x'=\sinh(\phi)ct$. We can divide one equation by the other to give $x'/t'=\tanh(\phi)c$ (noting that, comparable to the ordinary trigonometric case, $\sinh(\phi)/\cosh(\phi)=\tanh(\phi)$). This is significant because it strongly suggests that $\tanh(\phi)c$ is a speed, in particular the speed of the worldline $(ct',x')$ with respect to the $(ct,x)$ frame of reference (Fig. 3). This is born out by the standard definition of speed in one-dimension, $v\equiv x/t$, which in this case would mean that $\tanh(\phi)=v/c$, which agrees with the fact that $\tanh(\phi)$ ranges from $-1$ to $+1$ as $\phi$ ranges from $-\infty$ to $+\infty$. One should note that if this is correct, then it implies that the speed of an inertial (straight) worldline with respect to a stationary observer ($v$) can never exceed $c$ (because $\lvert\tanh(\phi)\rvert<1$ for all $\phi$)! Furthermore, we note that $\tanh(\alpha+\beta)=\frac{\tanh(\alpha)+\tanh(\beta)}{1+\tanh(\alpha)\tanh(\beta)}$, which immediately yields the otherwise peculiar-looking relativistic velocity addition formula $u=\frac{v+w}{1+vw/c^2}$ (constrast with the non-relativistic $u=v+w$).

Figure 3: Three reference frames, $(ct,x)$ (dark colours), $(ct',x')$ (medium colours) and $(ct'',x'')$ (light colours) on a 1+1-dimensional spacetime diagram. The frames are related by Lorentz transformations/hyperbolic rotations, and correspond to three observers with differing velocities who all pass each other at a point which is the origin of this diagram. Note that for each frame, the time and space coordinate directions are orthogonal with respect to the Minkowski metric (though clearly not with respect to the Euclidean metric - circular angles are not hyperbolic angles!).
 Their worldlines correspond to the paths $x=0$, $x'=0$ and $x''=0$ - each considers themselves to be stationary, though clearly their (red) worldlines depict differing velocities. In fact, on this diagram, the dark observer would consider the other two frames to be Lorentz transformed by hyperbolic angles/rapidities of $\phi'\simeq0.28$ and $\phi''\simeq0.66$. These correspond to velocities of approximately $0.27c$ and $0.54c$ respectively. Note that twice the velocity does not give twice the rapidity, as per the velocity addition formula! It should also not be overlooked that in the dark colour frame, the other time coordinates are quite literally hyperbolically rotated into the spatial direction and the other space coordinates are equally hyperbolically rotated into the temporal direction!
Some additional information about this spacetime diagram can be found in the footnotes.

Before going further, it is worthwhile to remark upon just how remarkable this is. From purely geometrical concerns and the coordinate identification $(x^0,x^1)=(ct,x)$ with dimensional arguments we have managed to link rotation (albeit the strange hyperbolic version of it) with something as concrete as velocity. What's weirder, the (hyperbolic) rotation we're talking about is the rotation of a spatial coordinate into the time direction, and vice versa! Not just that, we seem to have stumbled on a universal speed limit, entirely by accident. That's pretty cool.

In this context, the hyperbolic angle $\phi=\tanh^{-1}(v/c)$ is referred to as the rapidity. It's natural to ask next what our hyperbolic rotation matrix looks like in terms of velocity rather than rapidity. Substituting in, we find $\cosh(\phi)=\frac{1}{\sqrt{1-(v/c)^2}}$ and $\sinh(\phi)=\frac{v}{c}\frac{1}{\sqrt{1-(v/c)^2}}$. If we denote by the "Lorentz gamma" $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$ then we can write our hyperbolic rotation matrix as
\begin{equation*}
\Lambda=\begin{pmatrix}\gamma&\gamma v/c\\ \gamma v/c&\gamma\end{pmatrix}.
\end{equation*}
This transformation is known as a Lorentz transformation (hence its renaming $\Lambda$). The Lorentz transformations are the main transformations under which the equations of electromagnetism are invariant, and using this fact one can determine that the speed $c$ in fact corresponds to the speed of light in a vacuum, although that exercise is beyond the scope of this post. In terms coordinates, the Lorentz transformations can be written as $t'=\gamma(t-vx/c^2)$ and $x'=\gamma(x-vt)$. It is worthwhile to consider these equations in 'natural units' where $c=1$ and compare their forms to the Galiliean transformations we shoehorned in at the start of the post. In exactly the way the Galilean transformations were awkwardly asymmetric between space and time, the Lorentz transformations are beautifully symmetric, reflecting the fact that they emerged not a posteriori, inserted by hand, but as a natural property of the geometry of the Minkowski spacetime. And, in fact, this 1+1-dimensional case extends naturally to the 1+3-dimensional one; rather than $\Lambda\in\mathrm{SO}(1,1)$ we instead consider $\Lambda\in\mathrm{SO}(1,3)$, with all the requisite changes in machinery to go along with the change in spatial dimension. Note that $\mathrm{SO}(1,3)$ (thankfully) contains as a subgroup $\mathrm{SO}(3)$, the group of ordinary rotations in 3-dimensions, just as we expect it should if Minkowski spacetime is to have any meaningful correspondence to our experienced reality! This should give us even more confidence that $\mathrm{SO}(1,3)$ correctly generalises the characterising space of rotations $\mathrm{SO}(3)$ on non-relativistic space to its equivalent on relativistic spacetime.

Now, all of the above is a long, long way from a complete characterisation of the spaces $\mathrm{SO}(p,q)$ and $\mathrm{SO}(n)$ (even in the dimensions we looked at) and their relations to Minkowski and Euclidean space(time)s. It's also a long way from describing the full richness of the theory of relativity and its many important consequences and subtleties. In fact, all of the information I've touched on above is really just the first scratch of the surface a student might get in both of these areas; it just happens that they turn out to be connected in a way which most students won't find out about until much later, and I hope I've managed to spark some interest by bringing that connection to light perhaps a little earlier than might otherwise happen.

Footnotes

The diagonal dotted line corresponds to the speed of light and is at $\phi=+\infty$ according to all observers, and appears exactly diagonal in all observers' spacetime diagrams. This reflects the invariance of the speed of light with respect to all reference frames.

The points where the hyperbolas intersect the coordinate lines denote equal invariant distances from the origin in the time and space directions in each frame. As a side note, the diagram gives an intuitive explanation for length contraction - the distances from the origin to $x_0$, and the origin to $x''_0$ have the same invariant length (they have the same length in their respective rest frames) but $x''_0$ is measured to be shorter than $x_0$ by the dark colour observer and $x_0$ is measured to be shorter than $x_0''$ by the light colour observer. A similar construction can be made for time dilation.