Griffiths discusses an interesting gedanken experiment to explain the that two events which are simultaneous in one inertial frame are not, in general, simultaneous in another.

Let a freight car is moving along a smooth, straight track with a velocity $v$ as observed by an observer on the ground. Let the length of the car be $2L$ and from the center of the car hangs a light bulb. Let us analyse the motion from the perspective of an observer standing on the ground. At $t=0$, when the bulb is switched on, the light travels from the center to the front-end and the back-end of the car with the same velocity $c$ (postulate of relativity).

What are the times (as measured by the ground observer) at which the light reaches the front end and the back end? Let the light reaches the front end at time $t_1$ and the back end at time $t_2$. Then, $t_1=(L+vt_1)/c$ and $t_2=(L-vt_2)/c$. Is this correct or does one have to take length contraction intro account and change from $L$ to $L/\gamma$? Can we derive properly derive this using the Lorentz transformation formula?


2 Answers 2


Let's first analyze it in the frame of the car. In this frame the event that light reaches the end of the car (event 1) and the even that the light reaches the front of the car (event 2) are given by the 4-vectors

$$ x_1 ^\mu = \begin{pmatrix}c t_1 \\ x_1 \end{pmatrix}=\begin{pmatrix}L \\ -L \end{pmatrix} \qquad x_2 ^\mu = \begin{pmatrix}c t_2 \\ x_2 \end{pmatrix}=\begin{pmatrix}L \\ L \end{pmatrix} $$

Now we Lorentz transform into a frame that moves with velocity $v$. Then in this (primed) frame:

$$ c t_1' = \frac{c t_1 -x_1 v/c}{\sqrt{1-v^2/c^2}}=\frac{L -L v/c}{\sqrt{1-v^2/c^2}}= \frac{L -vt_1}{\sqrt{1-v^2/c^2}}$$

where I used in the last step $L/c=t_1$. Similarly,

$$ c t_2' = \frac{c t_2 -x_1 v/c}{\sqrt{1-v^2/c^2}}=\frac{L +L v/c}{\sqrt{1-v^2/c^2}}= \frac{L +vt_2}{\sqrt{1-v^2/c^2}}$$

If the velocities are small such that $v/c \ll 1$ then $\sqrt{1-v^2/c^2} \approx 1$ and you get the result in Griffiths. Note that already the first order correction in $v/c$ is included in this calculation, which should tell you that this effect is small.


If we begin in a frame that observes two simultaneous events at spatial location points $\vec{r}_1$ and $\vec{r}_2$, then we can always choose our spatial coordinate axis such that $\vec{r}_1$ and $\vec{r}_2$ are collinear. Let $r=|\vec{r}_1-\vec{r}_2|$, then the corresponding spacetime interval $x^{\mu}$ is given by

$$x^{\mu} = (0,r,0,0),$$

which has a vanishing component in the time direction. Using the lorentz transform $\Lambda_{\mu\nu}$ we can transform into another frame according to

$$\tilde{x}^{\mu} = \Lambda_{\mu\nu}x^{\nu}.$$

From the matrix representation of $\Lambda$, we see that, in the generic case, Lorentz transformations mix space and time. If we choose to boost along the 1-direction with speed $v$, then

$$\tilde{x}^0 = \gamma \frac{vr}{c^2},$$

which shows that the tilde frame does not observe the events simultaneously.


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