### Axes of the moving frame

We have previously noted that events occurring on the $t'$-axis, must satisfy $x' = 0$ in the moving frame and $x - vt = 0$ in the stationary frame, and we worked out in the last section that events on the $x'$-axis are simultaneous, where $t' = 0$ in the moving frame and $t - \frac{vx}{c^2}$ in the stationary frame.

We chose to draw our diagrams with the scaling such that $$c \equiv 1$$ but now, we shall use this convention in our expressions as well. To recover expressions where $c \neq 1$, we can use the substitutions $$\begin{align*}t &\rightarrow ct\\v &\rightarrow \frac{v}{c}\end{align*}$$

The expressions for the $t'$-axis don't change $$x - vt= 0 \text{ or } x' = 0$$ but the $x'$-axis becomes $$t - vx = 0 \text{ or } t' = 0$$ and the symmetry of the expressions becomes clear.

We would like to find a general relationship between the two sets of coordinates, analagous to the Galilean transformations.

One thing we do know is that the origins of the two frames coincide, that is, $$t = x = 0 \;\Leftrightarrow \; t' = x' = 0$$

### Derivation

A first guess might be $$\begin{align*}t' &= t - vx\\x' &= x - vt\end{align*}$$ since at least these satisfy the above requirement. We can generalise this relationship to cases where $t \neq 0$, by trying $$\begin{align*}t' &= f(v)\left(t - vx\right)\\x' &= g(v)\left(x - vt\right)\end{align*}$$ where $f(v), g(v)$ are numbers that could depend on the relative speed between the two frames, since it is still true that $$t = x = 0 \;\Leftrightarrow \; t' = x' = 0$$

Nb. Note that we wouldn't expect these numbers to depend on the directional component of the velocity, since that would imply that there are special directions.

We can simplify these expressions somewhat by considering the equations of the path of a light-ray passing through the common origins at time zero, in the two different coordinates, $$\begin{align*}x &= t\\x' &= t'\end{align*}$$ which have exactly the same form since the speed of light is the same in both ($c \equiv 1$). Substituting the first of these into our expressions gives $$\begin{align*}t' &= f(v)\left(1 - v\right)t\\x' &= g(v)\left(1 - v\right)t\end{align*}$$ which we now equate using the second expression. Hence $$f(v)\left(1 - v\right)t = t' = x' = g(v)\left(1 - v\right)t$$ or simply $$f(v) = g(v)$$

Thus, we can write the transformations as $$\begin{align*}t' &= f(v)\left(t - vx\right)\\x' &= f(v)\left(x - vt\right)\end{align*}$$

A further consideration is that we can reverse the roles of the two frames by noting that if we were at rest in the second frame, the first frame would be moving with velocity $-v$. Apart from this difference, the form of the inverse equations should be exactly the same, $$\begin{align*}t &= f(v)\left(t' + vx'\right)\\x &= f(v)\left(x' + vt'\right)\end{align*}$$

We can substitute these into either of the first two expressions above to get a value for $f(v)$, $$\begin{align*}t' &= f(v)\left(t - vx\right)\\&= f(v)\left[f(v)\left(t' + vx'\right) - v f(v)\left(x' + vt'\right)\right]\\&= f(v)^2\left(t' + vx' - vx' - v^2t'\right)\\&= f(v)^2\left(1 - v^2\right)t'\end{align*}$$ hence we get $$f(v)^2\left(1 - v^2\right) = 1$$ or $$f(v) = \frac{1}{\sqrt{1 - v^2}}$$

### Summary

In summary, suppose we have two frames of reference, the second of which moves at velocity $v$ relative to the first, and whose origins $O, O'$ coincide. Then, the two sets of coordinates associated with the frames are related by the **Lorentz transformations**, $$\begin{align*}t' &= \frac{t - vx}{\sqrt{1 - v^2}}\\x' &= \frac{x - vt}{\sqrt{1 - v^2}}\end{align*}$$

The **inverse Lorentz transformations** are then, $$\begin{align*}t &= \frac{t' + vx'}{\sqrt{1 - v^2}}\\x &= \frac{x' + vt'}{\sqrt{1 - v^2}}\end{align*}$$

Further, if we set $c \neq 1$, then we can use the substitutions $$\begin{align*}t &\rightarrow ct\\v &\rightarrow \frac{v}{c}\end{align*}$$ so that $$\begin{align*}t' &= \frac{t - \frac{vx}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}}\\x' &= \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}\end{align*}$$

But, to recover the symmetry we should actually define our coordinates in the form $(ct, x)$ and then $$\begin{align*}ct' &= \frac{ct - \frac{v}{c}x}{\sqrt{1 - \frac{v^2}{c^2}}}\\x' &= \frac{x - \frac{v}{c}ct}{\sqrt{1 - \frac{v^2}{c^2}}}\end{align*}$$

### Justifications for the Lorentz transformations

The first justification is that the speed of light, $c$, has the same value in all reference frames. To see this, consider a light-ray passing through the origin(s) at $t = t' = 0$. The path in the stationary frame is given by the equation $$x = ct$$

Substituting this into the Lorentz transformations, we get $$\begin{align*}ct' &= c\frac{t - \frac{vx}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}}\\&= c\frac{t - \frac{vct}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}}\\&= \frac{c - v}{\sqrt{1 - \frac{v^2}{c^2}}}t\end{align*}$$ and $$\begin{align*}x' &= \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}\\&= \frac{ct - vt}{\sqrt{1 - \frac{v^2}{c^2}}}\\&= \frac{c - v}{\sqrt{1 - \frac{v^2}{c^2}}}t\end{align*}$$

Hence, the path of the light ray in the *primed* coordinates is $$x' = ct'$$ meaning that the speed $c$ is the same for both frames.

The second main justification is that the Lorentz transformations become the Galilean transformations in the limit where $$\frac{v}{c} \rightarrow 0$$ since then $$\begin{align*}t' = \frac{t - \frac{v}{c}\frac{x}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} &\rightarrow \frac{t - 0}{\sqrt{1 - 0}} = t\\x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}} &\rightarrow \frac{x - vt}{\sqrt{1 - 0}} = x - vt\end{align*}$$