# 31 Geometry of Spacetime

We begin by defining spacetime to be the set \(\RR^4=\{(x,t)\mid x\in\RR^3,t\in\RR\}\) of all possible points in space, at all possible moments of time. We call a point \((x,t)\) an *event*, and we call a point \(x\in\RR^3\) a *location*, and a point \(t\in\RR\) a *moment*. Our goal is to try to understand the mathematical structures that determine the behavior of spacetime - a lofty goal with very little to go off of initially!

## 31.1 Axiomatization

Like many things in mathematics, one way to study spacetime is to study its group of symmetries. This is analogous to how we study Euclidean space by discovering its group of isometries, and then use our newfound knowledge of translations rotations and reflections to simplify all further calculations.

But how do we go about finding the symmetries to start with? In Euclidean geometry, perhaps we start by realizing that the Euclidean plane has a particular mathematical structure on it - it has a *distance function* (induced by a dot product on all tangent spaces), and we can rigorously define the isometry group as the set of all transformations that preserve this dot product. But for spacetime, the situation seems much more difficult - we don’t know what sort of math structure best describes spacetime, that’s one of the things we are looking to discover! So we can’t just explicitly define spacetime symmetries to be the things that preserve this (unknown!) structure.

However, if we dig back deep enough into history, we can find a close analogy between our current predicament and geometry. The greeks after all did not know about infinitesimal tangent spaces and all that: instead, they described geometry *axiomatically*, by specifying properties that they observed to be true of space, and then using these as the foundations of their mathematical theory. Us moderns then could take the axioms and try to rigorously find which mathematical spaces satisfy them: we saw in Geometry class that if you take Axioms 1-4 there are two possible ways the world could be (Euclidean or Hyperbolic), but once you have all the Axioms 1-5, there is a unique structure (Euclidean dot product on every tangent space) which instantiates them.

Could we attempt an axiomatic description of spacetime? That is, could we list some rules we claim to be true (based on our observations of the world around us) and use these as *constraints* on our mathematical theory? Perhaps, if we choose a good set of axioms, we will be able to find a small number of possible solutions (like Euclidean / Hyperbolic space, for Euclid’s Axioms 1-4). Then, if we were left without a unique solution, we could try to find *more axioms* to add (based on other observations of the world around us) which further narrow down to a unique mathematical structure, which would allow us to begin a rigorous study of spacetime.

Here’s a proposal for three first axioms: in the first we take the ideas of Euclid, in the second the ideas of Galileo, and while the third observation doesn’t have a name, it certainly predates the others, going back to the first humans to imagine themselves as living *in* space while time *passes*.

**Definition 31.1 (Axioms of Spacetime Symmetries)**

**Symmetries of space and time are symmetries of spaceitme:**If \(A\colon\mathbb{R}^3\to\mathbb{R}^3\) is a Euclidean isometry, then \(\mathcal{A}(p,t)=(A(p),t)\) is a symmetry of spacetime. Similarly, if \(B\) is any isometry of \(\mathbb{R}\) (a translation, reflection or combination) then \(\mathcal{B}(p,t):=(p,B(t))\) is a symmetry of spacetime.**Galileo’s Principle:**All symmetries of spacetime must preserve the class of constant speed trajectories. And, if \(\ell_1\) and \(\ell_2\) are the worldlines of any two constant speed observers, there is a symmetry of spacetime taking \(\ell_1\) to \(\ell_2\).**Space is different than time:**There is no symmetry of spacetime which takes the vertical axis \((0,t)\) to a line in space \((\ell,0)\).

The symmetries of spacetime form some group \(G\) of transformations \(\RR^4\to\RR^4\). But which group? These axioms put strong constraints on this group of transformations, and so our immediate goal is to try and discover which groups \(G\) satisfy these axioms. Perhaps surprisingly, the list is small! There are only two different groups, up to isomorphism, whose transformations satisfy Axioms (1)-(3). Said colloquially, these axioms alone imply there is only two possible ways that spacetime could logically behave.

### 31.1.1 Step 1: Interesting Symmetries Change Velocity

By Axiom (1) we know that spacetime is homogeneous, as Euclidean space is homogeneous, and the real line is homogeneous. We can use this to reduce the class of isometries we are interested in. Let \(\phi\colon\RR^4\to\RR^4\) be an arbitrary symmetry of spacetime, and let \(\phi(0,0)=(x_0,t_0)\). Then we can find a Euclidean isometry \(A\colon\RR^3\to\RR^3\) with \(A(x)=0\), and the time isometry \(B(t)=t-t_0\) for which \(B(t_0)=0\), and build (via axiom (1)) a spacetime symmetry \(\psi(x,t)=(A(x),B(t))\). Then notice that \[\psi\phi(0,0)=\psi(x,t)=(0,0)\] So \(\eta=\psi\phi\) is a symmetry of spacetime (both \(\phi\) and \(\psi\) were in \(G\), so \(\eta =\psi\phi\in G\)) where \(\eta(0,0)=(0,0)\). But now, taking an inverse, we see

\[\phi = \psi^{-1}\eta\]

So an arbitrary symmetry of spacetime is the composition of a symmetry which fixes the origin, and a symmetry which is just a translation in space and time. Since we understand the symmetries of space and time separately (from Euclidean geometry!) this implies that if we can understand the spacetime symmetries that fix the origin, we can understand the entire group \(G\).

So, what are the symmetries fixing the origin? Some of these we also already understand: if we take a Euclidean rotation \(R\) that fixes \(0\in\RR^3\) for instance, the symmetry \(\mathcal{R}(x,t)=(R(x),t)\) fixes the origin. Thus, what we are *really* interested in are spacetime symmetries that fix the origin of *spacetime* but do not fix the origin of space for all time (as this is a Euclidean isometry, just applied at each time!).

This means we are looking for isometries that fix the origin but do not fix the \(t\) axis! These are the only sort that we do not already understand. But by Galileo’s principle (axiom 2) the \(t\) axis is a constant speed trajectory (moving at speed zero) and so any symmetry of spacetime must send it to another constant speed trajectory, which is another affine line. But, since without loss of generality our isometry fixes the origin, this is just some other line through \((0,0)\)! All such lines are of the form \((vt, t)\) for \(v\in\RR^3\), or of the form \((v t,0)\) if they lie directly in the Euclidean space. However Axiom (3) rules out this second class (this would be a symmetry that took the time direction to a space direction) so, we know that \((0,t)\) must be sent to \((vt,t)\). Such a trajectory represents something that moves \(v\) units of space for every \(t\) units of time, and thus represents someone moving at velocity \(v\).

Thus, the only isometries we are interested in are the ones that take \((0,t)\) to \((vt,t)\): the rest we already understand! Let’s call such a symmetry \(S(v)\).

One thing to ask ourselves here; \(S(v)\) unique (if we are trying to give it a name, after all)? What if we had two symmetries \(A\) and \(B\) which both took \((0,t)\) to \((vt,t)\)? Then \(B^{-1}\) would take \((vt,t)\) to \((0,t)\), and so the combination \(B^{-1}A\) fixes the \(t\) axis pointwise - it restricts to a Euclidean isometry only in space! That is, we can write \(B^{-1}A=\mathcal{R}\) where \(\mathcal{R}(x,t)=(R(x),t)\) is just a Euclidean transformation. Composing with \(B\) yields \(A=B\mathcal{R}\), so \(A\) and \(B\) differ by a Euclidean isometry. BUT we already understand Euclidean isometries! So \(A\) and \(B\) are ‘essentially’ the same. (Precise note: really what we’ve seen here is that if \(A\) and \(B\) both take \((0,t)\) to \((vt,t)\) then they lie in the *SAME COSET* of the Euclidean group inside the group of spacetime symmetries!)

%Note to future self: prove that things fixing the t axis pointwise must be the SAME euclidean isometry on each slice

### 31.1.2 Step 2: The Group is Linear

The first step is to show that the only groups that satisfy these axioms are groups of linear transformations, that is, \(G<\mathrm{GL}(4,\RR)\). We did not do a full and rigorous job of this in the independent study, (let me know if we should return, and dot all our i’s and cross our t’s): instead we pointed to two ways one can prove this with more work

- Using Galileo’s axiom, we see that if \(\phi\colon\RR^4\to\RR^4\) is a symmetry it must take constant speed trajectories to constant speed trajectories. Since constant speed trajectories are affine lines, this means it must preserve (at least some subset of) affine lines. \(\phi\) also fixes the origin, so it sends affine lines through the origin to affine lines through the origin: like a linear map! (It’s more work, but in fact this is the only option, it is a linear map)
- We could instead proceed and just look for the
*subgroup*of \(G\) that is linear - and accept the fact that there could also be nonlinear symmetries out there! We will end up finding a bunch of linear symmetries, and in the end, can go back and argue that we have actually found*all the symmetries*: there were no nonlinear ones out there to be worried about!

### 31.1.3 Step 3: We can Work with 1 Space Dimension

Let \(S(v)\) be a symmetry of spacetime taking \((0,t)\) to \((\vec{v}t,t)\), and let \(v=\|\vec{v}\|\) be the speed. Then, just in \(\RR^3\) there is a Euclidean isometry rotating about \(0\) which sends \(\vec{v}\) to \((0,0,v)\). For a proof recall that Euclidean space is isotropic, so we can take any direction on the unit sphere to any other direction on the unit sphere. Thus there’s an isometry taking \(\vec{v}/\|\vec{v}\|\) to \((0,0,1)\). But, isometries preserve length, so this same isometry must send the vector \(\vec{v}\) to a vector of the same length in the direction of \((0,0,1)\): that is, by definition \((0,0,v)\). If we call this Euclidean rotation \(R\), we can build an associated spacetime symmetry \(\mathcal{R}(x,t)=(R(x),t)\).

Using this, consider the \(X=\mathcal{R}^{-1}S(v)\): this takes the \((0,t)\) axis to \(((0,0,s)t,t)\). Then \(\mathcal{R}X= S(v)\), so we can understand our arbitrary isometry \(S(v)\) as a composition of \(X\), which takes \((0,t)\) to a velocity along the \(z\) axis, and a Euclidean rotation. Since we already understand Euclidean isometries, this lets us *further simplify* what we are interested in: it is enough if we understand the isometries which take a stationary observer to one moving along the \(z\) axis!

Using the fact that we *also* know that such resulting transformations are linear, we can write this as a matrix:

\[S=\begin{pmatrix} \star &\star &\star &\star\\ \star &\star &\star &\star\\ \star &\star &\star &\star\\ \star &\star &\star &\star\\ \end{pmatrix}\]

Our goal is to fill in the missing entries! (right now, all of them!). One first thing to notice, is that with our motion in the \(z\) direction, the \(x\) and \(y\) directions don’t get mixed up with \(z\) and \(t\) after the transformation (an argument from axiom 2 goes as follows: if distances in the \(x\) and \(y\) directions were affected by me moving in the \(z\) direction, I could tell whether or not I was moving - violating Galileo’s principle! - by seeing how the length of something I was carrying with me changed! I would like a better argument here though)

This implies the matrix is block diagonal: the \(xy\) do not mix with the \(zt\), so we have

\[S=\begin{pmatrix} \star &\star & 0&0\\ \star &\star &0&0\\ 0&0&\star &\star\\ 0&0&\star &\star \end{pmatrix}\]

Now, we have a transformation which translates along the \(z\)-axis, and does something in the \(x,y\) direction. Say that it does the Euclidean rotation \(R\) in the \(xy\) plane, which acts on spacetime as \(R(x,y,z,t)=(R(x,y),z,t)\). Because we are free to use Euclidean isometries to simplify our situation, we can compose our map above with \(R^{-1}\) without changing the property that we care about (that it translates along the \(z\) axis) so we can without loss of generality assume that the \(2\times 2\) euclidean block is the identity! Thus, our matrix is

\[S=\begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&\star&\star\\ 0&0&\star&\star \end{pmatrix}\]

This leaves a very manageable sized problem - everything about the symmetries of spacetime can be totally understood so long as we know how they work with one space and one time dimension!

### 31.1.4 Step 4: Some Matrix Calculations

This is where all the real work is!!

Now we have gotten ourselves into a sufficiently restricted situation that we can do some actual calculations. We are only interested in symmetries fixing the origin, and taking \((0,t)\) to \((vt,t)\) for some \(0\neq v\in\RR^3\). We know these symmetries are linear, and after a Euclidean isometry can actually assume without loss of generality that \(v\) is parallel to the \(z\) axis, so the matrix representing our symmetry is block diagonal with only one \(2\times 2\) undetermined block:

\[S(v)=\begin{pmatrix} a(v) & b(v)\\ c(v) & d(v) \end{pmatrix}\]

We then did a bunch of matrix calculations that I do not feel like typing tonight (sorry!) but I emailed out a handwritten copy of. Together, this implies that there are two possibilities for the group of symmetries of \(\RR^4\) satisfying axioms (1) to (3).

**Possibility 1**: The group of symmetries of spacetime consists of the following matrices \[G=\left\{\begin{pmatrix}
1& -v \\
0& 1
\end{pmatrix}\Big | v\in\RR \right\}
\]

**Possibility 2**: The group of symmetries of spacetime consists of the following matrices, for \(c\) some positive constant. \[L=\left\{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\begin{pmatrix}
1& -v \\
-v/c^2& 1
\end{pmatrix}\Bigg| v\in(-c,c)\right\}
\]

## 31.2 Implications

Our next goal, as mathematicians is to try and study these two possible worlds, and derive some properties they have. We will refer to Possibility I as the Galilean world, and Possibility II as the Lorentzian world.

**Proposition 31.1 (All Lorentzian Worlds are Isomorphic)** At first, it appears that there are really *uncountably many* different possibilities for spacetime: one possible Galilean world, but a continuum of Lorentzian worlds, one for each value of \(c\in(0,\infty)\). But, it turns out that this entire continuum of Lorentzian worlds are qualitatively the same: we can make this formal by saying that the group of symmetries for any two lorentzian worlds are *isomorphic*

**Exercise 31.1 (Prove This:)** When \(c=1\) we have the Lorentz \(\mathcal{L}\) group with matrices \[\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1&-v\\-v&1\end{pmatrix}\] For the Lorentz group \(\mathcal{L}_c\) with an arbitrary \(c\), show that the map \[\mathcal{L}_c\to\mathcal{L_1}\] defined by sending \(v\) to \(v/c\) is

- Injective
- Surjective
- A group homomorphism

### 31.2.1 Velocity Addition

These two worlds have very different rules for velocity addition: in the Galilean world, velocities add:

**Proposition 31.2** For any \(v,w\in\RR\), \[G(v)G(w)=G(v+w)\]

But in the Lorentzian world, velocities satisfy a rather different formula, which makes sure that the overall velocity always remains within \((-c,c)\) as the formula seems to require.

**Proposition 31.3** For any \(v,w\in(-c,c)\), \[L(v)L(w)=L\left(\frac{v+w}{1+\frac{vw}{c^2}}\right)\]

**Exercise 31.2** Do these calculations.

**Exercise 31.3** Then, using the velocity addition for \(\mathcal{L}\), show that no matter what speed you start out moving at, and no matter how much you speed up by, you will never go faster than \(c\). That is, our mystery constant is actually a *universal speed limit*.

### 31.2.2 Existence of a Constant Speed

Show that in the Galilean world, if there is an object moving at any speed \(v\) you can catch up to it: there’s a symmetry of spacetime that boosts your velocity to \(v\), and now the object is standing still. In the Lorentzian world, we already know that it is impossible to boost an *initially stationary* observer to any speed beyond \((-c,c)\). But what if we had an object moving at the universal speed \(c\) to begin with? This would follow a trajectory \((x,t)=(ct,t)\) through spacetime.

**Exercise 31.4** Say you observe an object to be moving at speed \(c\).

What would happen if you speed up? Would you see its speed change at all? Show that no matter what speed you go (so, no matter which Lorentz transformation \(L(v)\) you apply to \((ct,t)\)) you will always see this object move at the same speed.

This is an incredibly weird prediction: first, we saw that it is impossible to start out stationary and move at any speed outside of \((-c,c)\). So, you may have thought that if there were an object moving at speed \(c\), even though you could never catch up to it,

### 31.2.3 Metric

The type of mathematical structure spacetime has in these two possibilities is also rather different. Perhaps surprisingly, it turns out that the Lorentzian spacetime (with the more complicated looking matrices!) actually has a nicer mathematical structure.

What should we be looking for here? Well, if Geometry taught us anything, the geometric properties of a space are usually stored in the form of a dot product on infinitesimal tangent vectors (we saw this even for strange geometries, like Minkowski space). So, now that we have two potential symmetry groups of spacetime, we should ask if these correspond to any sort of geometric structure.

**Exercise 31.5 (The Galilean World)** Show that if \((x,y)\star (u,v)=\alpha xu + \beta yv\) is any inner product on the space \(\RR^2=(z,t)\) of \(1+1\) dimensional Galilean spacetime which is preserved by all transformations \(G(v)\) then either \(\alpha = 0\) of \(\beta=0\). That is, its not really an inner product on spacetime at all but rather only notices information about space, or information about time. This tells us that in the Galilean world, there is no geometry of spacetime, but rather space and time are just separate entities.

**Exercise 31.6 (The Lorentzian World)** Show that the inner product \((x,y)\star (u,v)= xu - c^2yv\) is preserved by all lorentzian transformations \(L(v)\). This is a Minkowski dot product (just scaled by the quantity \(c^2\)): and says that this spacetime cannot be thought of as space and time separately any more than the \(x\) and \(y\) axes in the plane can be thought of separately - they are just parts of one larger geometric object (this time with the geometry of Minkwoski space!)

This was Hermann Minkowski’s big realization upon reading Einstein’s work: in 1908 he said in a lecture, announcing this that

“Gentlemen! The views of space and time which I wish to lay before you … They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.”

*Remark 31.1* (Reminder: Showing a dot product is preserved). If \(L\) is a transformation and \(v,w\) are tangent vectors based at \(p\), to show that \(L\) preserves the dot product \(\star\) one must show that \(DL_p(v)\star DL_p(w)=v\star w\). For us, since \(L\) is linear we know \(DL_p=L\) and so we just need to compute \(L\) applied to vectors \(v\) and \(w\) and then take the dot product. Alternatively, remember we can just show infinitesimal length is preserved, so \(\|Lw\|^2=Lw\star Lw = w\star w\) for a single arbitrary vector \(w\).

This proves our main theorem: we’ve discoverd the mathematical model for spacetime in this case, and identified it with something we already understand!

**Theorem 31.1** The geometry of spacetime equipped with the Lorentz symmetries is isomorphic to Minkowski space.