# Special Relativity - A Mathematical Approach

## Mathematically understanding how and why the effects of special relativity arise.

Summary

Rotations in the plane preserve a circle, and rotations in space preserve a sphere. It turns out that the postulates of relativity tell us that the coordinate transformation between two inertial reference frames preserves a hyperboloid. These transformations are sometimes refered as hyperbolic rotations; the first illustration below demonstrates the two dimensional version. The are a variety of algebraic structures that can be used to describe these transformations, one of them being quotients of linear functions over the complex numbers. The geometric objects that arise from these functions look quite interesting, and can be seen in the last illustration.

Hyperbolae

Let's start out by talking about a curve in the plane given by $x^2 - y^2 = 1$. Let's try and parametrize it with respect to a parameter $t$. The most common way to parametrize this curve is as follows: $$x(t)=\cosh(t)=\frac{e^t+e^{-t}}{2}$$ $$y(t)=\sinh(t)=\frac{e^t-e^{-t}}{2}$$ There are a few things that come in handy with this parametrization. The first is that it's the unique solution to $$x'(t)=y(t)$$ $$y'(t)=x(t)$$ Such that $x(0)=1$ and $y(0)=0$. Additionally, $$y(a+b)=x(a)y(b)+x(b)y(a)$$ $$x(a+b)=x(a)x(b)+y(a)y(b)$$ These last two properties are of interest because it means that for any real $\zeta$ $$\left( \begin{array}{ccc} \cosh(\zeta) & \sinh(\zeta) \\ \sinh(\zeta) & \cosh(\zeta) \end{array} \right) \left( \begin{array}{ccc} \cosh(t) \\ \sinh(t) \end{array} \right) = \left( \begin{array}{ccc} \cosh(t)\cosh(\zeta)+\sinh(t)\sinh(\zeta) \\ \sinh(t)\cosh(\zeta)+\sinh(\zeta)\cosh(t) \end{array} \right) = \left( \begin{array}{ccc} \cosh(t+\zeta) \\ \sinh(t+\zeta) \end{array} \right)$$ This last property is key when describing the geometry of 1+1 dimensional spacetime (note that m+n dimensions means m spatial dimensions and n temporal dimensions). Let's see how we can generalize it for a curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The parametrization for this curve is $(a\cosh (t),\, b\sinh (t))$. Thus, $$\left( \begin{array}{ccc} \cosh(\zeta) & \sinh(\zeta) \\ \sinh(\zeta) & \cosh(\zeta) \end{array} \right) \left( \begin{array}{ccc} a\cosh(t) \\ b\sinh(t) \end{array} \right) = \left( \begin{array}{ccc} a(\cosh(t)\cosh(\zeta)+\sinh(t)\sinh(\zeta)) \\ b(\sinh(t)\cosh(\zeta)+\sinh(\zeta)\cosh(t)) \end{array} \right) = \left( \begin{array}{ccc} a\cosh(t+\zeta) \\ b\sinh(t+\zeta) \end{array} \right)$$ The same can be shown for $\frac{y^2}{a^2}-\frac{x^2}{b^2}$. This means that the matrix above is a linear transformation that preserves hyperbolae the same way rotations preserve circles. This is why this type of transformation is called a "hyperbolic rotation." In a minute you'll find out why a transformation like this is useful for describing spacetime. For now, here's a way to visualize the transformations:

ΞΆ=0

On the left is the image of a square grid under a hyperbolic rotation, and on the right is the image of a square grid under an ordinary rotation. As you can see, any vertex of the grid that starts on the circle stays on the circle for the ordinary rotation. This is what is meant by "rotations preserve a circle." The same thing can be seen for the hyperbola on the grid that is being hyperbolically rotated.

The Postulates of Relativity

The first postulate of relativity, called the principle of relativity, states that all inertial reference frames are equivalent. Specifically, a reference frame is just a coordinate system and an inertial reference frame is one that is stationary or moving at a constant velocity. What is meant by equivalence in this case is that the laws of physics don't change when you switch between these reference frames. To be more explicit, let's define a reference frame with coordinates $(x,\, y,\, z,\, t)$. Let's say that this is an inertial reference frame and let's define another inertial reference frame with coordinates $(x',\, y',\, z',\, t')$. If we wrap up the coordinates into 4 dimensional vectors, $u$ and $u'$ respectively, the relations between these vectors must be of the following form: $$u'=Au+b$$ $A$ is a matrix which encodes the stretching, reflection/rotation and motion that one must undergo to get from $u$ to $u'$. Notice that by the word motion I don't mean translation. Think about the vector $$\left( \begin{array}{ccc} A_{00} \\ A_{10} \\ A_{20} \\ A_{30} \end{array} \right)$$ $A_{00}$ will be multiplied by $t$ in the expression for $t'$, we can think of it as a scaling factor between the two ways of measuring time. Specifically, for every 1 tick of a clock in the $u$ reference frame, there will be $A_{00}$ ticks of a clock in the $u'$ reference frame. The other components of the vector will also be multiplied by $t$ but to give spatial quantities, thus we can think of these elements as representing the velocity of the origin of $u'$ in the coordinates of $u$. The other elements of this matrix encode the stretching and rotation/reflection one must undergo to switch between frames. The vector $b$ represents a translation between the coordinate systems, i.e. the displacement between them at $t=0$.

So how do we determine what this transformation actually is? That's where the second postulate comes in, called the invariance of c. What it says is basically in the title: the speed of light is the same in all inertial reference frames. How can we interpret this mathematically? Say we have a ray of light that travels a displacement $\Delta r$ in the amount of time $\Delta t$. Then we have $$c\Delta t = \Delta r$$ $$\rightarrow c^2 \Delta t^2 = \Delta r^2$$ $$\rightarrow c^2 \Delta t^2 - \Delta r^2=0$$ Where $c$ is the speed of light. The same holds for any inertial reference frame, i.e. $c^2 \Delta t'^2 - \Delta r'^2=0$. The reason for the squaring of each displacement is that in $n$ dimensions $$\Delta r = \sqrt{\Delta x_1^2 + \Delta x_2^2 +...+\Delta x_n^2}$$ Dealing with the square root is annoying so instead we square it and deal with $\Delta x_1^2 + \Delta x_2^2 +...+ \Delta x_n^2$.

Returning to 1+1 dimensions, if we take these equations to the limit as the interval goes to zero, we get $$c^2 d t^2 - d r^2=0$$ $$c^2 d t'^2 - d r'^2=0$$ The left hand side of the first equation is called $ds^2$, the line element, and that of the second equation is $ds'^2$. Since both expressions are of the same degree as polynomials, we can deduce that $ds^2=kds'^2$ for some function $k$. Since spacetime is homogenous (the same at all positions) and isotropic (the same in all directions) the only thing $k$ can depend on is the relative speed of the reference frames. If we introduce a third reference frame with line element $ds''^2$ and $v_0$ is the velocity of $ds'$ moving away from $ds$, $\, v_1$ is the velocity between $ds$ and $ds''$, and $v_2$ is the velocity between $ds'$ and $ds''$ (here I've used the line elements of the reference frames, $ds$, $ds'$, and $ds''$, to denote the reference frames themselves) then $$ds^2=k(v_0)ds'^2 \quad ds^2=k(v_1)ds''^2 \quad ds'^2=k(v_2)ds''^2$$ $$\rightarrow \quad k(v_0)=\frac{k(v_1)}{k(v_2)}$$ Note that $v_0$ can be deduced given the magnitudes of $v_1$ and $v_2$ and the angle between them. However, $k$ does not depend on direction (meaning it doesn't depend on angle) so the expression $\frac{k(v_1)}{k(v_2)}$ also doesn't depend on angle. It follows that $k$ is constant and, from the equation above, is equal to 1. This means that $$ds^2=ds'^2$$ $$\rightarrow c^2 dt^2 - dr^2=c^2dt'^2 - dr'^2$$

Lorentz Transformations

Since $c^2 dt^2 - dr^2=c^2dt'^2 - dr'^2$ is true for any two reference frames, what we're looking for is an affine transformation that preserves $c^2 dt^2 - dr^2$.$\,$ All translations trivially preserve this expression, since we can rewrite it as $\lim_{t_1\rightarrow t_2,r_1\rightarrow r_2} c^2(t_1 - t_2)^2 - (r_1 - r_2)^2\,$ (in 1+1 dimensional spacetime). One can verify that adding any vector $\langle t_0,\, r_0 \rangle$ to both $\langle t_1,\, r_1\rangle$ and $\langle t_2,\, r_2\rangle$ will result in the same expression. So the question becomes, what kind of linear transformations preserve $ds^2$? And we already know the answer to this! Any matrix of the form $$\left( \begin{array}{ccc} \cosh(\zeta) & \sinh(\zeta) \\ \sinh(\zeta) & \cosh(\zeta) \end{array} \right)$$ Will preserve $ds^2$ when it acts on $\langle dt,\, dr \rangle$. We call this group of matrices the group of Lorentz transformations or Lorentz group for 1+1 dimensional spacetime.

So how is $\zeta$ detrmined? Long story short, $\zeta=\tanh^{-1} \frac{v}{c}$ where $v$ is the speed of one reference frame relative to the other, $c$ is the speed of light and $\tanh^{-1}$ is the inverse function of $\tanh$, which is defined as $\frac{\sinh(t)}{\cosh(t)}$. This can be verified by performing the matrix multiplication on $\langle dt,\, dr \rangle$, setting it equal to $ds^2$, regrouping terms, estabilishing what must equal zero and solving. I'll leave this as an exercise for the reader. I also recommend solving for the actual elements of the matrix in terms of relative velocity and the speed of light, and trying to figure out how the effects of length contraction and time dilation arise.

The only thing that I've done so far that was in more than 1+1 dimensions was justify why we squared $\Delta t$ and $\Delta r$ when deriving the invariant interval. Let's look at what the invariant interval would be for our 3+1 dimensional spacetime: $$ds^2=c^2dt^2-dx^2-dy^2-dz^2$$ What kind of transformations preserve $-dx^2-dy^2-dz^2$? Well, setting that expression equal to a (negative) constant yields the equation of a sphere (neglecting the fact that they're infinitesmals). So the transformations that preserve that part of the expression will be 3D rotations or reflections. Since two reflections is a rotation and an arbitrary reflection is a rotation composed with a set reflection, we can think of this subgroup of Lorentz transformations as having two connected components: one that is the group of all 3D rotations, and one that is any 3D rotation after some given reflection. 3D rotations have 3 degrees of freedom so we can say that both of these components are 3-dimensional. However, let's not forget the $dt^2$ term. Transformations that mess with time are hyperbolic rotations in 4 dimensions (which preserve hyperboloids instead of hyperbolae). The hyperbolic rotation is determined by the 3-velocity of the two reference frames; meaning they have 3 degrees of freedom. This means that the group of Lorentz transformations of 3+1 dimensional spacetime is 6 dimensional and has two connected components.

You might be thinking "great, but how do I do calculations?" That's fine, but I think the calculations involved are extremely cumbersome. It's nice to take a step back and think of everything topologically, even though the interpretation I've presented here is highly non-rigorous.

Relationship with Mobius Transformations (and more Fun with Mobius Transformtations)

This part of the article doesn't have anything to do with the physical phenomena involved in special relativity, as far as I know. Nonetheless, the connections we can draw with seemingly disparate topics in math are interesting.

The space in which the effects of special relativity take place, where hyperbolic rotations are isometries, is called Minkowski space. Minkowski space is $\mathbb{R}^4$ with a quadratic form $Q(v)$. That is, $Q$ is a function that acts on points in the space and allows us to have a notion of square distance, also known as quadrance. In 3D Euclidean space $Q(v)=v_1^2+v_2^2+v_3^2$ due to the Pythagorean theorem. However, in Minkowski space $$Q(v)=v_0^2-v_1^2-v_2^2-v_3^2$$ This way, if we want to calculate $\Delta s^2$ between two events $u$ and $v$ it is simply $Q(u-v)$ This allows us to write a vector in Minkowski space in the following way: $$v= \left( \begin{array}{ccc} v_0+v_1 & v_2 + iv_3 \\ v_2 - iv_3 & v_0-v_1 \end{array} \right)$$ It can be easily verified that the determinant of this matrix is the quadratic form on Minkowski space. If $P$ is a matrix in the special linear group SL(2, C), i.e. a complex 2x2 matrix with determinant 1, and $P^\star$ is the conjugate transpose of $P$ then $$\det(PvP^\star)=\det(v)$$ This action of $P$ on $v$ gives rise to a homomorphism from SL(2, C) to the subgroup of Lorentz transformations that don't include reflections. A homomorphism, by the way, just means that if we have another matrix, $Q\in SL(2,\, \mathbb{C})$ then the action $PQvQ^\star P^\star$ is equivalent to the action of $P$ as a Lorentz transformation after the action of $Q$ as a Lorentz transformation. This homomorphism maps any transformation composed with a reflection to just the bare transformation, and all other transformations are mapped to themselves. This means that the kernel, the elements that get mapped to the identity, are the identity itself and the reflection. Linear algebra and group theory tell us that quotienting a group by a kernel will yield a group that is isomorphic to the codomain of the homomorphism. In this case, quotienting by the kernel is saying we don't care about the sign of the determinant (recall that negative determinants mean reflections are involved) so what we get is called the projective special linear group, $PSL(2, \mathbb{C})$, which is the group of all non-zero determinant complex matrices where we don't care about determinant. In other words, if $P$ is an invertible complex 2x2 matrix and we're considering it as an element of $PSL(2, \mathbb{C})$ then $\forall a\neq 0 \quad aP=P$.

So what is this all leading up to? Think about functions of a complex variable of the form $$f(z)=\frac{az+b}{cz+d}$$ These are called Mobius transformations, they can be represented by a matrix like this: $$\left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)$$ Notice that multiplying this matrix by any non-zero scalar will yield the same Mobius transformation. Also, the Mobius transformation is invertible if and only if $ad-bc\neq 0$. Thus, there is a natural isomorphism between the group of Mobius transformations and PSL(2, C). Therefore, the Mobius group and the Lorentz group without reflections are isomorphic!

What's so exciting about that? Aside from the fact that the two types of transformations seem entirely unrelated initially, certain finite subgroups of the Mobius group have very interesting behavior. Specifically, if we let the subgroups act on the complex plane and look at which points have non-chaotic orbits, we get a certain type of fractal:

called an Apollonian Gasket. These can also be constructed starting with three circles each touching the other two and drawing circles that are tangent to at least three others!

The lesson: never think of any ideas in math as being unrelated.