How to understand simultaneity and the lack of simultaneity?

Question

Albert is at rest with respect to the ground. Hermann is in a carriage that is moving with speed v relative to Albert in the direction shown. Two flashes of light are emitted from the back and the front of the carriage. According to Hermann’s clock they arrive at Hermann’s position simultaneously.

Explain with reference to the concept of proper time, why the arrival of the light pulses at Hermann will also be simultaneous to Albert.

I always thought this experiment is to demonstrate the lack of simultaneity? From Albert's frame, doesn't the light emitted from the back of the carriage is observed first, since the distance this light has to cover is shorter as the result by the carriage moving to the right?

UPDATE::

This is the answer provided? Is this correct?

Answer 1

The detection of 2 light pulses by Herman occur at the same time, and at the same place. Let's call it:

$$ E_3 = (0, 0)_{\rm Herm} $$

This is one event. An event is a single point in spacetime.

I can transform it into any frame (a primed frame, $S'$), and it will be:

$$ E_3' = (t', x')_{S'} $$

That is a single event, period. It must be simultaneous and co-located. It is by definition.

If we go to Al's frame, it can be:

$$ E_3 = (0, 0)_{\rm Al} $$

where, as with Herman, I used it to define the origin of the coordinates.

For there to be a discrepancy with simultaneity, we need a spatial separation of two distinct events.

That would fall on the emission of the light. In Hermann's frame, those events occur at:

$$ E_1 = (-L/2, -L/2)_{\rm Herm} $$ $$ E_2 = (-L/2, +L/2)_{\rm Herm} $$

where $L$ is the length of car, and $c=1$. We see that $t_1=t_2= -L/2$; the emission is simultaneous.

They cannot be simultaneous in Al's frame. A Lorentz Transformation by $-v$ shows:

$$ E_1 = \big(\gamma(-L/2-(-vL/2)), \gamma(-L/2-v(-L/2)\big)_{\rm Al}$$ $$ E_1 = \big(-\frac 1 2 \gamma L(1+v), -\frac 1 2 \gamma L(1+v)\big)_{\rm Al} $$

$$ E_2 = \big(\gamma(-L/2-(+vL/2)), \gamma(+L/2-v(-L/2)\big)_{\rm Al}$$ $$ E_2 = \big(-\frac 1 2 \gamma L(1-v), \frac 1 2 \gamma L(1-v)\big)_{\rm Al} $$

So we see that for Albert, $E_1$ occurs before $E_2$, that is, the trailing photon is emitted 1st and has to catch up with Hermann, and the leading photon is emitted second, with Hermann running into it. Hence the $\gamma(1\pm v)$ difference.

Answer 2

Since the question explicitly asks you to use the concept of proper time, let's do that.

Suppose that Hermann is carrying a stopwatch, and he will start it when the first pulse arrives and stop it when the second pulse arrives. This means that the stopwatch measures the proper time between these two events. (It is the proper time because in Hermann's frame of reference the two events happen at the same position.)

Now, in this particular case, and in Hermann's frame of reference, we know that the time between the two events was zero - the two pulses arrived simultaneously, so he started the stopwatch and then immediately stopped it, so that the stopwatch will read zero.

What will the stopwatch read in Albert's frame of reference? Well, after the experiment is complete, the reading on the stopwatch is independent of your frame of reference. If it reads zero to Hermann then it must also read zero to Albert.

Now, in Albert's frame, the stopwatch is moving, so Albert might want to correct for time dilation by multiplying the measured time by $\gamma$ - but zero times $\gamma$ is still zero, i.e., the two events are indeed simultaneous in Albert's frame of reference.

It is important to keep in mind that the two events in question are "the light flash from the left of the train arrives at Hermann's position" and "the light flash from the right of the train arrives at Hermann's position", and even though we're asking about these two events from Albert's frame of reference, it is still Hermann's position - not Albert's - that matters. We haven't even been told where Albert is standing relative to Hermann, because it doesn't matter; we're only interested in what Albert sees Hermann seeing.

In particular, the question does not assert:

• That Albert will see the two light flashes at the same time, i.e., that they will arrive at Albert's position simultaneously.

• That the light flashes are sent simultaneously in Albert's frame of reference.

• That the light flashes are sent simultaneously in Hermann's frame of reference. (We don't know whether Hermann is standing in the exact middle of the train carriage, after all!)

Answer 3

Assume Hermann is in the middle of the carriage. He sees the light flashes emitted simultaneously.

But due to relativistic effects, Albert sees the front flash emitted first in a reddened shade, with the rear flash emitted later in a bluer shade; they are not simultaneous in his reference frame. He also sees Hermann appear to be standing in the rear half of the carriage, at the point they are opposite each other. The initial red flash takes longer to travel the longer distance, by which time the later blue flash has just travelled the longer distance.

Thus, it is simultaneity or otherwise of the spatially-separated (emission) events which vary according to the observer, it does not not vary for events which occur at a single point in spacetime.

Albert calculates that the number of wavelengths travelled by the short-wave blue pulse is exactly the same as the number of wavelengths travelled by the long-wave red light. Hermann agrees with that number for his two equal green pulses.