A train passes an observer on a platform at half the speed of light. Who measures the proper time?

Question

The proper time between two events is defined as the time measured in the reference frame in which both events occur at the same location. For example, a rocket that travels from Earth to Mars measures the proper time of the journey because landing and taking off both occur at the same location relative to the rocket (this is obviously a massive simplification which ignores acceleration, amongst other things).

I have also read that the proper time between two events is also the shortest time between those two events that can be measured by any inertial reference frame.

Now, consider the example of a train moving at half the speed of light past an observer on a platform. We want to find the time it takes for the train to pass the observer (the two events would be the front and back of the train being in line with the observer). According to the definition of proper time above, the observer should measure the proper time because both events occur at the same location relative to him/her. This should mean that the observer records a shorter time between the two events than someone on the train would.

However, this is false and the scenario above is quite a common example in special relativity classes and lectures to demonstrate that the passenger on the train would actually measure proper time and the observer on the platform wouldn't.

Where am I going wrong here?

Answer 1

To be sure, the proper time $\tau$ associated with two events is the elapsed time measured by an inertial clock with world line through both events. And yes, the elapsed coordinate time (in any inertial coordinate system) between these events is equal to or greater than the proper time

$$\Delta t_{AB} = \gamma\tau_{AB} \ge \tau_{AB}$$

This should mean that the observer records a shorter time between the two events than someone on the train would.

However, this is false

As you write, if event A (B) is the front (back) of the train in line with the clock at rest on the platform, then the proper time $\tau$ is measured by the clock on the platform.

Now, stipulate that at event A, both the clock on the platform and the clock at the front of the train read $t_A = t'_A = 0$. At event B, the clock on the platform reads

$$t_B = \frac{L/\gamma}{c/2} = \tau_{AB}$$

where $L$ is the length of the train in its rest frame (the train is length contracted to length $L/\gamma$ in the platform frame of reference). What time $t'_B$ will the clock at the back the train read at event B? The time dilation formula tells us that

$$t'_B = \Delta t'_{AB} = \gamma\tau_{AB} = \frac{L}{c/2}$$

which is just what is expected but is also greater than the proper time, i.e., the statement "However, this is false" is itself false.

Answer 2

You are mistaken that it is false. In fact it can be shown to be true quite easily from the Lorentz transform.

Without loss of generality, let the spacetime origin be the event that the front of the train passes the platform observer (front event) and choose units where c=1 and the proper time is 1 between the front and back events.

Then the front event has coordinates $(0,0)$ and the back event has coordinates $(t,x)=(1,0)$. Transforming to the train’s (primed) frame the front event is trivial, but the back event transforms to $(t’,x’)=(\gamma t, -\gamma v t)$. Since $\gamma\gt 1$ the platform observer does indeed record a shorter time than the train.