- Maxwell's equations
- Einstein's formulation
- Synchronising clocks
- Space-time diagrams
- Breakdown of simultaneity
- Simultaneity in moving frames

### Maxwell's equations

Maxwell's equations describe electromagnetism; electric and magnetic forces, electric and magnetic fields, and the relationships that bind them together.

One of the predictions of Maxwell's equations is that the velocity of electromagnetic waves, or light, is always measured to have the same value, regardless of the frame in which it is measured.

If Maxwell's equations *are* assumed to be laws of physics - meaning that they should have the same form in every reference frame - then it follows that we should all agree on the speed at which light travels.

But this contradicts Galilean relativity, since that requires that velocity is not an invariant quantity. If a particle of light is moving at speed $c$ say, in the $x$-direction, then in the stationary frame, its spatial coordinate is $$x(t) = ct$$

In a frame moving at speed $v$, the spatial coordinate of the particle is $$\begin{align*}x'(t) &= x(t) - vt\\&= ct - vt\\&= \left(c - v\right)t\\&= c't\end{align*}$$

So, in Galilean relativity, we have $$c' = c - v \neq c$$ and the speed of light in the moving frame should be slower than in the stationary frame, directly contradicting Maxwell.

Scientists before Einstein thought that Galilean relativity was correct and so supposed that there had to exist a special, universal frame (called the aether) in which Maxwell's equations would be correct.

However, over time and many experiments (including Michelson-Morley) it was shown that that the speed of light did not depend on the velocity of the observer measuring it, so that $$c' = c$$ or $$c' \neq c - v$$

### Einstein's formulation

Einstein guessed that Maxwell was right, and the equations *were* laws of physics, which meant Galilean relativity had to be reviewed.

His special form of relativity had *two* principle assumptions:

- the laws of physics should be the same in every reference frame
- the speed of light is the same in all reference frames

The first consequence of this formulation is the breakdown of simultaneity. That is, Einstein showed events observed to be simultaneous in one frame will *not* necessarily be observed to be simultaneous in other frames.

### Synchronising clocks

Suppose we have two clocks, at locations $O, A$, on the $x$-axis, separated by a distance $2L$, and we want to synchronise them.

We use the fact that the speed of light is constant and that light pulses sent out from $O,A$ would have to travel the same distance to the midpoint $P$.

An observer at $P$ could get the pulses coming from $A$ to be synchronised to the pulses coming from $O$, by telling $A$ to nudge her clocks forward or back until both pulses arrived at $P$ simultaneously.

### Space-time diagrams

In order to draw practical space-time diagrams involving the speed of light, $c$, we choose units for length and time so that $$c \equiv 1$$

For example, we could use *seconds* for time and *light-seconds* for distance, or *years* for time and *light-years* for distance.

This means that light-rays travelling in the positive $x$-direction are at a $45^{\circ}$ angle *counter-clockwise* to the horizontal, and rays travelling in the negative $x$-direction are at a $45^{\circ}$ angle *clockwise* to the horizontal.

In figure 1.7, the synchronised wavefronts of light from flashes emanating from $O, A$ at $t = 0$ and an observer at rest at $P$ have spatial coordinates $$\begin{align*}x_O(t) &= ct\\x_A(t) &= 2L - ct\\x_P(t) &= L\end{align*}$$

All three coincide at the event $$E_P \rightarrow \left\{T_P, X_P\right\}$$ where we set $$x_O(T_P) = x_A(T_P)$$ to find $$\begin{align*}T_P &= \frac{L}{c}\\X_P &= L\end{align*}$$

So, an observer at $P$ will consider these flashes to be simultaneous.

### Breakdown of simultaneity

We can now ask if flashes sent out simultaneously in this stationary frame will be observed to be simultaneous in a different frame moving at speed $v$ say.

Suppose that an observer, with spatial coordinate $x_Q$, at rest in the moving frame, is exactly at $P$ at time $t = 0$ (as measured in the stationary frame) and exactly when the flashes of light are sent out from $O,A$.

Clearly, the moving observer encounters the flash from $A$ before the flash from $O$.

We can quantify this difference as measured by the stationary observer. In this frame, the moving observer has coordinate $$x_Q(t) = L + vt$$ which coincides with the wavefront from $A$ when $x_Q(T_A) = x_A(T_A)$ or $$T_A = \frac{L}{c + v}$$ and the wavefront from $O$ when $x_Q(T_O) = x_O(T_O)$ or $$T_O = \frac{L}{c - v}$$

So, in the stationary frame coordinates, $T_A \neq T_O$.

### Simultaneity in moving frames

We can ask which event would be measured to be simultaneous by the moving observer of the last section.

In figure 1.9, a frame is moving at speed $v$ relative to a stationary frame, such that their origins, $O, O'$ coincide at time $$t = t' = 0$$

Light sources have been placed in the moving frame, one at the origin and another at arbitrary distance. An observer, at rest in the moving frame, is at the midpoint of the two light sources.

Note that the light source at the origin has the same path as the $t'$-axis corresponding to the equation $$x - vt = 0$$

As measured in the stationary frame, at $t = 0$, one light source is at the origin, the other is at $x = 2L$ and the observer is at $x = L$ as before.

At $t = 0$, a flash emanates from the source at $O'$ and arrives at the observer at the event $E$.

To find the event $A$, which would be a flash from the second source which the observer would consider to be simultaneous with the flash from $O'$, we drop a perpendicular to the first light path, through $E$, and then find the intersection of this light path with the path of the second light source, at $A$.

Nb. The second path is perpendicular to the first path because it's travelling in the opposite direction.

The event $E$ has coordinates in the stationary frame that satisfy the pair of equations $$\begin{align*}x - ct &= 0 & \text{ light path}\\x - vt &= L & \text{ observer path}\end{align*}$$ which lead to \begin{align*}

t_E &= \frac{L}{c - v}

\\

x_E &= \frac{cL}{c - v}

\end{align*}

Light rays travelling in the negative $x$ direction have the form, $$x + ct = k$$ for some $k$, which we can determine since this line passes through $E$. Hence, we must have $$k = \frac{cL}{c - v} + c\left(\frac{L}{c - v}\right) = \frac{2cL}{c - v}$$ and the equation for the second light path is then $$x + ct = \frac{2cL}{c - v}$$

The event $A$ has coordinates (in the stationary frame) that satisfy the pair of equations $$\begin{align*}x + ct &= \frac{2cL}{c - v}\\x - vt &= 2L\end{align*}$$ which lead to \begin{align*}

t_A &= \frac{2vL}{c^2 - v^2}

\\

x_A &= \frac{2 c^2 L}{c^2 - v^2}

\end{align*}

This is the event that the observer considers to be simultaneous with the event from the origin.

Notice we can relate the two variables by writing $$\frac{t_A}{v} = \frac{2L}{c^2 - v^2} = \frac{x_A}{c^2}$$ so that $$t_A - \frac{vx_A}{c^2} = 0$$

Since the distance between the two light sources was chosen arbitrarily, changing it - and hence the value of $L$ - won't change the expressions for ${t_A, x_A}$, so we can drop the subscript and say that the line $$t - \frac{vx}{c^2} = 0$$ is a **line of simultaneity** with the origin $O'$.

This is the line the observer considers to be simultaneous with the origin. That is, events occurring on this line will occur at the same time $t' = 0$ as measured by the moving observer, which means we can label this line as the $x'$-axis.

So, we have found that lines of simultaneity in special relativity are not horizontal, as they were in Galilean relativity, hence we cannot infer that simultaneous events in one frame are simultaneous in another.