# Train example of special relativity

My friend Eric is at the center of the train. The train is moving forward. The front end and back end of the train flash a light at the same time. From Eric's perspective, both light arrives at him at the same time. From my perspective, because Eric is moving forward, light from front arrives at Eric first, then light from back arrives at Eric later.

I get the above example that time of event is relative from different perspectives. What if Eric and I agree when he sees light, he will raise hand, and will only raise hand once every time he sees light, regardless from how many light sources. Then, in the above example, from Eric's perspective, he will raise hand once, because he sees one time light flashing, from both ends of the train. But from my perspective, will I see him raise hand once or twice? If once, is it at the time light from front or at the time light from back hit him? If twice, then I see 2 events while Eric sees only 1 event. So not only timing of events are different from each perspective, but number of events also depends on perspective?

• Your question is one of extremely many duplicates. You are simply assuming wrongly that Eric's simultaneous lightning flashes are also simultaenous lightning flashes at two positions to you. That is wrong. You will think that one lightning flash happened earlier and one happened later. You will deduce that Eric received both flashes at the same time. There is thus only one hand raise. Commented Apr 8 at 4:08
• What does "perspective" mean, and how could the number of hand raises possibly depend on it? Commented Apr 8 at 4:11

If the lights on the train flash simultaneously in the frame of the train, then the light from each will arrive at Eric at the same time and he will raise his hand once. In that scenario in your reference frame, the light at the rear of the train flashes before the light at the front, which explains why Eric sees them at the same time.

Conversely, if the lights flash simultaneously in your frame, then in Eric's frame the light at the front of the train flashes before the light at the rear. He will put up his hand twice, first to denote seeing the light from the front of the train and later to denote seeing the light from the rear.

The key point is that if the lights flash simultaneously in one frame, they will flash at different times in the other.

This is what you get for making a disordered SR problem in which the moving thing (the train) is stationary, and the stationary thing (the station) is moving.

I always suggest we follow a protocol:

Introduce the stationary (in our mind of course, we view buildings as stationary) first in a frame, $$S$$.

Now talk events in that frame: some lights flash--we'll get back to it

Intro the "moving" frame, $$S'$$. It's primed, it moves, it the train, and idk the direction: it moves forward. Avoid spurious coordinates.

It's pretty clear that if the front and back end lights flash simultaneously in $$S'$$, then the REAR light flashes 1st in $$S$$.

Since you set the problem up backwards, it's hard to do. I refuse to do an inverse Lorentz transform.

Rather I will introduce a third frame, $$S''$$, that is a plane flying at $$-v'$$ in $$S'$$ (oh, and always state the velocity...just out of courtesy --not the value, just the variable. Or else I have to do it).

So now forget about $$S$$ since $$S=S''$$, and drop a prime, so the train is stationary in $$S$$ and the plane ($$S'$$) is moving at $$-v$$ there.

I'm setting $$c=L/2=1$$. Now define events $$(t, x)$$ in $$S$$:

Front/Rear light flashes:

$$E_{\pm} = (0, \pm L)$$

Now transform to the balloon frame (since the plane is stationary in the station frame, I changed it to a balloon...remember: you set the problem up backwards, not me): $$E'^{\pm} = \Big(\gamma(t-(-v)(\pm L)), \gamma(\pm L-vt)\Big)$$

allow care about is the time component $$E_t'^{\pm} = \pm\gamma L$$

showing that the rear light flashes first, so you're whole premiss was backwards, as was the set-up.

Now:

finish the LT, and define new events: (when, where) the light are are received in $$S$$. (The new $$S$$, which was the old $$S'$$).

Don't use "Eric". The 1st frame should be a 'A' name: Alice, then the moving frame's second name 'B' (Bob)--then (at least up to 17 hours ago), you could use pronouns to refer to them, so instead of "the observer riding in the center of the train", you just say "he", after Bob's initial intro.

So the Receive ($$R^{\pm}_X$$) events: figure them out and you'll find they're an it: since the whole problem stipulates that in $$S$$:

$$R^+_X = R^-_X$$

you should be able to convince yourself that's true in all frames.

Relativity is confusing, and if you don't clarify your problem you will not make paradoxes, you'll make straight contradictions.