# General transformations

### Matrix notation

It is sometimes useful to express the Lorentz transformation in terms of 4x4-matrices.

For the usual situation - the primed frame moving at speed $v$ relative to the unprimed frame along their common $x$-axis - we have $$\begin{bmatrix}t'\\x'\\y'\\z'\end{bmatrix} = \begin{bmatrix}\gamma & -\gamma v & 0 & 0\\-\gamma v & \gamma & 0 & 0\\0&0&1&0\\0&0&0&1\end{bmatrix} \begin{bmatrix}t\\x\\y\\z\end{bmatrix}$$ where $\gamma = \frac{1}{\sqrt{1 - v^2}}$ as usual.

We could also write this as $$\vec{x}' = \mathbf{L} \vec{x}$$ where $L$ the Lorentz transformation matrix and $\vec{x}$ is a vector with four components.

### Rotations in space

We would like to describe transformations under which the laws of physics don't change. In other words, transformations which are symmetric with respect to the laws of physics.

The condition for the symmetry was that the square of the separation between two events was invariant under the given transformation.

For the vector $\vec{x} \rightarrow (t, x, y, z)$ this means that $$S^2 = t^2 - x^2 - y^2 - z^2$$ must be invariant under the given transformation.

Along with Lorentz transformations (also known as boosts), the full set of space-time symmetric transformations include rotations in space, where we assume that the time component is unchanged and doesn't get mixed up with the spatial components.

Figure 5.1 - Rotation about the z-axis

The simplest rotation we can consider is about the $z$-axis.

All rotations are valid, but this is has especially simple matrix, $$\mathbf{R} = \begin{bmatrix}1&0&0&0\\0& \cos \theta & \sin \theta & 0\\0& -\sin \theta & \cos \theta & 0\\0&0&0&1\end{bmatrix}$$ which leads to \begin{align*}t' &= t\\x' &= x \cos \theta + y \sin \theta\\y' &= -x \sin \theta + y \cos \theta\\z' &= z\end{align*}

Now, \begin{align*}S'^2 &= t'^2 - x'^2 - y'^2 - z'^2\\&= t^2 - \left(x \cos \theta + y \sin \theta\right)^2 - \left(-x \sin \theta + y \cos \theta\right)^2 - z^2\\&= t^2 - \left(x^2 \cos^2 \theta + y^2 \sin^2 \theta + 2xy\cos\theta \sin\theta\right) - \left(x^2 \sin^2 \theta + y^2 \cos^2 \theta - 2xy\cos\theta \sin\theta\right) - z^2\\&= t^2 - \left(x^2 \left(\cos^2 \theta + \sin^2 \theta\right) + y^2 \left(\cos^2 \theta + \sin^2 \theta\right) + 2xy\cos\theta \sin\theta - 2xy\cos\theta \sin\theta\right) - z^2\\&= t^2 - \left(x^2 + y^2 \right) - z^2\\&= t^2 - x^2 - y^2 - z^2\\&= S^2\end{align*}

So, rotations satisfy the symmetry condition. Furthermore, it is easy to show that compositions of transformations are also valid.

For example, we might need to do a Lorentz transformation in the $x'$-direction of figure 5.1, rather than in the $x$-direction, as usual. So we rotate the frame $(t,x,y,z)$ through the angle $\theta$ to the frame $(t',x',y',z')$ and then boost this frame to the third frame $(t'', x'', y'', z'')$.

We can write \begin{align*}\begin{bmatrix}t''\\x''\\y''\\z''\end{bmatrix} &= \mathbf{L}\begin{bmatrix}t'\\x'\\y'\\z'\end{bmatrix} = \mathbf{L}\mathbf{R}\begin{bmatrix}t\\x\\y\\z\end{bmatrix}\\&= \begin{bmatrix}\gamma & -\gamma v & 0 & 0\\-\gamma v & \gamma & 0 & 0\\0&0&1&0\\0&0&0&1\end{bmatrix} \begin{bmatrix}1&0&0&0\\0& \cos \theta & \sin \theta & 0\\0& -\sin \theta & \cos \theta & 0\\0&0&0&1\end{bmatrix}\begin{bmatrix}t\\x\\y\\z\end{bmatrix}\\&= \begin{bmatrix}\gamma & -\gamma v \cos \theta & -\gamma v \sin \theta & 0\\-\gamma v & \gamma \cos \theta & \gamma \sin \theta & 0\\0& -\sin \theta & \cos \theta & 0\\0&0&0&1\end{bmatrix}\begin{bmatrix}t\\x\\y\\z\end{bmatrix}\end{align*} which leads to \begin{align*}t'' &= \gamma \left(t - v \left(x \cos \theta + y\sin \theta\right)\right)\\x'' &= \gamma \left(\left(x \cos \theta + y \sin \theta\right) - vt\right)\\y'' &= -x \sin \theta + y \cos \theta\\z'' &= z\end{align*}