Does a clock at the rear end of a train is ticking faster than a clock at the front end of a train?

Question

When a train zooms past us, we observe a clock at the rear end of the train ahead of a clock at the front of the train. I wonder if it implies that the clock at the rear end is ticking faster than the clock at the front. Or perhaps both clocks are tickling at the same rate. My intuition tells me that the former is true but I am not sure.

Answer 1

Although you can work through the math in detail (see this answer), there's an quick way to see that the clock ticks at the same rate. The formula for time dilation is:

$$ \Delta t' = \gamma \Delta t$$

where $\gamma$ is the Lorentz factor:

$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

Note this formula depends only on $v$ (the relative velocity between the two observers). Therefore, the question boils down to "are the clocks at the rear end of the train and the front end of the train travelling at the same speed?". It should be obvious that they are (since they're in the same train), and hence they show the same amount of time dilation.

Answer 2

Assume distance and time are measured in light seconds and seconds respectively so that $c=1$. Suppose the proper length of the train is $L=10$. Place yourself in an inertial frame where the train travels at $v=\frac35 c$ to the right. Coordinates of your inertial frame will be unprimed while coordinates of train's frame will be primed. Inside the train, at $t'=0$, a train conductor reports that clock readings of the train front (located at $x'=L$) and back (located at $x'=0$) are $t'_f=0$ and $t'_b=0$ respectively.

Now perform inverse Lorentz transforms:

$t_b=\gamma(t'_b+vx'_b)=\frac54\left(0+\frac35(0)\right)=0$

$x_b=\gamma(x'_b+vt'_b)=\frac54\left(0+\frac35(0)\right)=0$

$t_f=\gamma(t'_f+vx'_f)=\frac54\left(0+\frac3510\right)=\frac{15}{2}$

$x_f=\gamma(x'_f+vt'_f)=\frac54\left(10+\frac35(0)\right)=\frac{25}{2}$

Now at $t=0$ in your inertial frame, we have

$t_b=0$

$x_b=0$

$t_f=\frac{15}{2}-\frac{15}{2}=0$

$x_f=\frac{25}{2}-v\frac{15}{2}=\frac{25}{2}-\frac35\frac{15}{2}=8$

Then do Lorentz transform

$t_f'=\gamma(t_f-vx_f)=\frac54\left(0-\frac358\right)=-6$

So at $t=0$, we observe the clock reading of the back of the train is $t'_b=0$ and the clock reading of the front of the train $t_f'=-6$. The time difference is $\Delta t_0=t_f'-t_b'=-6-0=-6$.

Now at $t=\frac{15}{2}$, the coordinates of the back of the train are

$t_b=0+\frac{15}{2}=\frac{15}{2}$

$x_b=0+v\frac{15}{2}=\frac35\frac{15}{2}=\frac92$

Let's do Lorentz transforms once again

$t_b'=\gamma(t_b-vx_b)=\frac54\left(\frac{15}{2}-\frac35\frac92\right)=6$

Ok so at $t=\frac{15}{2}$, we observe the clock reading of the back of the train is $t'_b=6$ and the clock reading of the front of the train $t_f'=0$. The time difference is $\Delta t_1=t_f'-t_b'=0-6=-6$.

Conclusion: Your initial intuition is wrong. Since $\frac{\Delta t_1}{\Delta t_0}=1$, both clocks tick at the same rate.

Answer 3

Don't conflate the Doppler effect with the Lorentz transformation.

If the train is half-way past you at the moment when you measure, then the clock at the front will be receding from you, while the clock at the rear is approaching you. The Doppler effect will cause your raw observation of the approaching clock's tick rate to be greater than your raw observation of the receding clock. But if you correct your observations to account for the Doppler effect, then you should conclude that both tick rates actually are the same, and both clocks will appear to tick more slowly than your own clock because of the Lorentz transformation.