Relativity of simultaneity
To illustrate the concept of relativity of simultaneity in special relativity, Einstein proposed a thought experiment where lightning strikes at the head and tail of a running train (points H and T, in short). There are two observers, Alice on the ground and Bob on the train exactly at the middle between H and T. The two events that lightning strikes H and T occur simultaneously to Alice, but not to Bob.
In the analysis of this thought experiment, it is important to distinguish the time that the observers receive the lights (received time) and the time that the light sent from both ends of the train (sent time). Both Newton and Einstein agree that Bob should receive the light from H earlier than the light from T since Bob is moving toward H and faraway from T. The disagreement is the sent time. Suppose in Alice's frame (ground),
- the train's length is $2L$;
- the train's speed is $v$ and the speed of light is $c$;
- Bob receives the light sent from H (T) at the time $t_H$ ($t_T$) assuming the lightning strikes at both H and simultaneously at $t=0$.
Newton's view
In Newton's world, there is absolute time and absolute space. So in Bob's frame, Bob still receives the light from H at the time $t'_H = t_H$ and receives the light from T at the time $t'_T = t_T$. Furthermore, in Bob's frame, both light travels the same distance from H and $T$ to Bob, which $L'=L$.
However, the light's speed is different with respect to Bob: the speed of the light from H is $c+v$ while the speed of the light from $T$ is $c-v$. Therefore, it takes time durations $\frac{L'}{c+v}$ and $\frac{L'}{c-v}$ for lights travel from H and T to Bob.
As a result, in Bob's frame, the sent time from H and T can be deduced as $t'_H-\frac{L'}{c+v}$ and $t'_T-\frac{L'}{c-v}$, which are all equal to zero (simultaneity).
Einstein's view
In Einstein's view, the speed of light is the same $c$ in both Alice and Bob's frames. Therefore, in Bob's frame, the time durations from the lights to travel from H and T to Bob are all equal to $\frac{L'}{c}$. So the sent time is deduced as $t'_H-\frac{L'}{c}$ and $t'_T-\frac{L'}{c}$, which are not equal anymore!
For more quantitative calculations, we need to apply two relations from special relativity: one is time dilation and the other is length contraction (See the appendix of this post for the classical proof that can be understood by friends who are not in physics major): \begin{equation} t_H = \frac{t'_H}{\sqrt{1-\frac{v^2}{c^2}}}\,,\quad L=L'\sqrt{1-\frac{v^2}{c^2}}\,.\end{equation} The sent time from $H$ is actually \begin{equation} t'_H-\frac{L'}{c} = t_H \sqrt{1-\frac{v^2}{c^2}} - \frac{L}{c\sqrt{1-\frac{v^2}{c^2}}} = \frac{L}{c+v} \sqrt{1-\frac{v^2}{c^2}} - \frac{L}{c\sqrt{1-\frac{v^2}{c^2}}} = \frac{-\frac{v}{c^2}L}{\sqrt{1-\frac{v^2}{c^2}}}\,. \tag{1}\end{equation} Similarly, the sent time from $T$ can be calculated as \begin{equation} t'_T-\frac{L'}{c} = \frac{+\frac{v}{c^2}L}{\sqrt{1-\frac{v^2}{c^2}}}\,. \tag{2}\end{equation}
Lorentz Transformation
One can calculate the sent time from H in one line if knowing the Lorentz transformation \begin{equation} t'=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}\,.\end{equation} The lightning strikes at H in Alice frame occurs at $t=0$ and $x=L$. Submitting $t=0$ and $x=L$ into the above equation immediately gives the same result as in Eq. (1).
Appendix
The following proofs does not relies on the Lorentz transformation. We only need to accept the fact that the speed of light is the same in all inertial frames, and be clear on how to measure the time and length in physics.
Time dilation
The time dilation $t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$ can be derived by considering a moving photon clock. Note that any periodic motion can used as a clock. The photon clock here is a light pulse bouncing vertically and periodically between two mirrors.
From the diagram, we have equations \begin{equation} t_0=\frac{2L_0}{c}\,,\quad t=\frac{2D}{c}\,, \quad D^2=L_0^2 + \left(v\frac{t}{2}\right)^2\,,\end{equation}
from which we can get the correct equation between $t$ and $t_0$.
Length contraction
Length contraction $L=L_0\sqrt{1-\frac{v^2}{c^2}}$ is usually proved by considering a clock travel from one place to another place on the earth. The key point is that the length between two places with respect to the earth is the proper length.
Here, we provide an alternative proof that replies on the same photon clock used in proving the time dilation. This time we rotate the photo clock 90 degree so that the light pulse bounces horizontally along the moving direction. Suppose the light pulse leaves the mirror A at time $0$, reaches the mirror $B$ at time $t_1$ and comes back to the mirror A at the time $t$. From time dilation, we know that \begin{equation} t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\,,\quad t_0=\frac{2L_0}{c}\,.\end{equation} Since the light pulse travels along the same direction of clock motion, we have \begin{equation} c t_1 = L + vt_1\,,\quad c(t-t_1)+v(t-t_1)=L\,.\end{equation} With some algebra, we can get the correct relation between $L$ and $L_0$.
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