From Maxwell's Equations to Special Relativity

Postulates

Maxwell's equations suggest light is caused by fluctuation of electric and magnetic field and its speed is determine by $$c = \frac{1}{\sqrt{\epsilon\mu}}$$ where $\epsilon$ and $\mu$ depend on the medium light propagates through.
In classical mechanics, the speed of an object to a moving observer is $$v' = v + u$$ based on the assumption of absolute time and space. When an observer is moving relative to the medium of a wave, the speed of the wave observed by the moving observer should be vector sum of the observer's velocity relative to the medium and the wave velocity. $$v'_{\text{wave}} = v_{\text{wave speed}} + u_{\text{medium speed relative to observer}}$$ When light propagates in vacuum, what should be the medium relative to the moving observer to calculate the relative light speed? Einstein proposed

• all laws of physics have the same form in all inertial frames
• light speed is constant for all inertial frames

Time Dilation

The distances of light traveled between two points in different inertial frames are different. If all inertial frames share the same speed of light, it follows logically they would observe different time intervals for the same event. Let's see how can we quantify the difference between time elapsed in different inertial frames.

When a light signal is sent from the point $O$ to the point $P$, in the frame {$S$}, the time duration is $t$

Let $t'$ be the time duration of this event in another inertial frame {$S'$}, which is moving to right at a speed $v$ relative to {$S$} and $O'$ overlapped with $O'$ at the time when the signal was sent at $O$. $P$ in {$S$} is observed to have moved to the left when light signal has reached $P$

As the speed of light is the same for both frames, $$\begin{align} (ct')^2 &= (ct)^2 + (vt')^2 \\ t' &= \frac{t}{\sqrt{1-\frac{v^2}{c^2}}} \\ \end{align}$$ For the same event, the observed time duration is different for different inertial frames. $$\bbox[5px,border:2px solid #666] { \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c2}}} }$$ For one unit of time elapsed in the observer's clock, he observes a less than one unit of time elapsed in the object travelingat speed $v$ relative to him. So, in the observer's frame, the observer observe slower time flow in the moving object. This is called time dilation.

Length Contraction

When an object moving at speed $v$, its head passes a mark in the frame {$S$} at time $t_1$ and its tail passes the mark at time $t_2$. the observed length in {$S$} is $$L' = v\Delta t = v(t_2 - t_1)$$

In the object's frame {$S'$}, the mark passes its tail at time $t_1'$ and its head at $t_2'$. The length of the object in {$S'$} is $$L = v\Delta t' = v(t_2'- t_1')$$

As we know from above that the duration of this event in the two frames is related by $$\begin{align} \Delta t' &= \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c2}}} \\ L &= v\frac{\Delta t}{\sqrt{1- \frac{v^2}{c^2}}} \\ &= \frac{L'}{\sqrt{1- \frac{v^2}{c^2}}} \end{align}$$ The length observed in inertial frames other than the inertial frame of the length is always shorter than the original length in its own inertial frame $$\bbox[5px,border:2px solid #666] { L' = \sqrt{1- \frac{v^2}{c^2}}L }$$ This is called length contraction.

Lorentz Transformation

After knowing time dilation and length contraction, we want to know how coordinates should be changed when switching between inertial frames.

By length contraction and time dilation mentioned above, $$\begin{align} x' &= \frac{x}{\sqrt{1-\frac{v^2}{c^2}}} - vt' \\ &= \frac{x}{\sqrt{1-\frac{v^2}{c^2}}} - \frac{vt}{\sqrt{1-\frac{v^2}{c^2}}} \end{align}$$ Define $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ We have $$x' = \gamma(x - vt)$$ On the other hand, {$S$} is moving at $-v$ in {$S'$}, so we have $$x = \gamma(x' + vt')$$ Combining the two equations, $$\begin{align} x &= \gamma(\gamma(x - vt) + vt') \\ &= \gamma^2 x - \gamma^2 vt + \gamma vt' \\ \gamma vt' &= (1-\gamma^2)x + \gamma^2 vt \\ &= -\frac{v^2/c^2}{1-\frac{v^2}{c^2}} x + \gamma^2vt \\ &= -\gamma^2\frac{v^2x}{c^2} + \gamma^2 vt \\ t' &= \gamma(t - \frac{vx}{c^2}) \end{align}$$ Thus, for a set of coordinates, when observed in another inertial frame, the coordinates would be changed to $$\begin{cases} x' = \gamma(x - vt) \\[2ex] y' = y \\[2ex] z' = z\\[2ex] t' = \gamma(t - \frac{vx}{c^2})\\[2ex] \end{cases}$$ or in matrix form $${ \begin{bmatrix}ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix} }$$ So simple and beautiful.