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Here is a quote from "The Elegant Universe" by Brian Greene.

Imagine two countries that have been at war are sitting down to sign a treaty ending hostilities while traveling aboard a train that is moving at a constant velocity. The catch is that neither country's delegate wants to sign the treaty before the other delegate and thus, a simple system is devised to ensure that both delegates sign the peace treaty simultaneously. The solution involves setting a light bulb at the center of a table in such a way that the light bulb is exactly between the delegate from Forwardland (who is facing the direction the train is traveling) and the delegate from Backwardland (who has her back to the direction the train is traveling). When the light bulbs lights up, that is the signal for both delegates to sign the treaty.

This setup is agreeable to all parties on the train and to both security councils in the countries' respective capitals. Once the bulb lights up and the delegates have simultaneously signed the peace treaty, everyone on the train celebrates the cessation of hostilities, but they are perplexed to discover that fighting has broken out anew between the two countries. The reason given is that the delegate from Forwardland was tricked into signing the treaty before the delegate from Backwardland

Why do the delegates see the light at the same time? the forward land delegate IS approaching the light and the backwardland delegate IS moving away from it. so therefor the delegate from forward land SHOUlD see the light first. (because he traveled less distance to see the light.)

SO how come the observers on the platform outside the train see this but the ones ON the train don't??? Since the speed of light IS constant it makes sense that this would be how everybody saw it, on and off the train. I have searched for an answer to my question for about a month now and haven't a found a satisfying one. Please don't use too complex of vocabulary.

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Simply put, relativistic speeds cause for events previously thought of as simultaneous to no longer be simultaneous if the velocity of the reference frame of the event changes relative to the defined observer. The best way to wrap your head around this is to pictorially trace what is happening in space time.

A good visual representation of shifting of simultaneous events

The case you describe is $v>0$.

Think of $v$ in this case as the velocity of the train relative to you, and you, the observer of the train, are the white line, seeing what occurs in sliver of space $x$ as time marches forward, indicated by observing events at all points in $x$ at a time t. If the train appears to be getting further from you, then delegate Forwardland is point C and delegate Backwardsland is A.

Interestingly enough if the train was approaching you, you would have seen the opposite occur, where the delegate of Backwardsland signed first (like $v=-0.5c$). If $v=0$ you are traveling the same speed as the train, and you might as well be one of the people on the train observing the (now) simultaneous event.

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  • $\begingroup$ What a nice image! Did you make it? If not, where did you find it? $\endgroup$
    – rob
    Commented May 14, 2015 at 4:51
  • $\begingroup$ I should have added that this picture is available on wikipedia $\endgroup$
    – Skyler
    Commented May 14, 2015 at 16:05
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the forward land delegate IS approaching the light and the backwardland delegate IS moving away from it

Are they? Or are the delegates sitting perfectly still, and the Earth spinning quickly beneath them? Of course we have a convention that the Earth is stationary and the train moves across it, but we also know that the Earth is not stationary - it spins and orbits, and the sun it orbits is also circling the center of the galaxy, and the Milky Way galaxy is dancing with its galactic cluster, etc. We can choose any body+direction to designate an arbitrary "stationary" coordinate system, but because it's arbitrary, we can also easily choose a second coordinate system in which the first system is not stationary.

For example, let's take our delegates in the train: as they are preparing to sign the treaty, one of the aides asks the Forwardland delegate why he is moving so much, and reasonably the delegate replies "I am not moving, I'm sitting quite still at this table." Indeed, from the perspective of those in the train car, they are sitting still around the table with a stationary light halfway between them. If the curtains are drawn so that they can't see out the window, they have no way of knowing they're moving. The laws of physics work the same way for them sitting still in the train as they do sitting still on the ground; the speed of light is $c$, objects fall straight down rather than off to one side, etc. That's why the delegates see the light at the same time even though they are "moving toward" or "moving away" from it; because those assessments are only accurate as seen from the observer standing on the ground.

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  • $\begingroup$ that still doesn't make sense to me. The light does have to travel less distance to reach the forward land delegate than it does for the backward land delegate, Right? (if not, then I'm beyond confused) so why do they see the light at the same time? $\endgroup$ Commented May 15, 2015 at 0:58
  • $\begingroup$ @AlexTaylor They're sitting the same distance from the light bulb, e.g. if one delegate pulls out a yardstick and begins measuring, he'll find that both of them are sitting, say, two feet from the bulb. It doesn't matter whether he measures from one end of the table or the other because the light bulb is in the middle. So why would the light travel less distance one way than the other? $\endgroup$
    – Asher
    Commented May 15, 2015 at 7:39
  • $\begingroup$ @AlexTaylor And before you answer that, also consider this: is the train traveling east, or west? In your view, this would also make a difference, since at normal train speeds the spin of the Earth is much, much faster than the speed of the train; but on top of that, the orbit of the Earth is much faster than the spin, so the time of day also effects the travel speed; but if the train is traveling due east or west, the light should go off to one side because of the tilt of the solar system in its orbit around the galaxy... et cetera. $\endgroup$
    – Asher
    Commented May 15, 2015 at 7:46
  • $\begingroup$ @AlexTaylor What it boils down to is that the researcher standing in his laboratory in Berlin measures the same speed of light as the researcher traveling on the train from Berlin to Bern, because the speed of light is the same in all inertial reference frames (and to a high degree of precision, the spin and orbit of the Earth are small enough to consider the surface "inertial"). There's no "absolute" frame from which all other frames can be measured. $\endgroup$
    – Asher
    Commented May 15, 2015 at 7:52
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    $\begingroup$ @AlexTaylor That's exactly the root. Here's what it means: If you are measuring the speed of light in a laboratory or experiment that is not accelerating, you will always measure it at exactly c, whether you're measuring it in a stationary lab on Earth's surface or whether you're measuring it on a rocket speeding past Earth at a constant speed. This naturally leads to different observers citing different simultaneity results - just the problem you've quoted with the delegates signing - but the constancy of the speed of light has been experimentally verified enough to call fact, regardless. $\endgroup$
    – Asher
    Commented May 16, 2015 at 1:50
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I Like AlexTalyor am concerned with the lack of 'Post processing' done by Brian Greene on the observation/perception in the train. In the Train inertial frame of reference we are told to accept that according to Einstein that the observers on the Train are in effect stationary as they have no means determining if they are moving. However in this case the observers 'know' they are on a 'Moving Train' with velocity 'v'. I am not interested in what the observers on the platform or (other frame of reference) view or calculate only on what happens on the Train after the light goes on.

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  • $\begingroup$ What happens on the train is that light expands from the bulb at the same speed toward the front and rear of the train. Whether the people on the train know the train is moving, or don't know, makes no difference. So if two people on the train sit equidistant from the bulb, they will see it switched on at the same time. $\endgroup$ Commented Jan 3, 2022 at 22:57
  • $\begingroup$ "Whether the people on the train know the train is moving, or don't know, makes no difference." Its whether the 'Moving' Frame makes a diference to the light wave front $\endgroup$
    – JohnI
    Commented Jan 3, 2022 at 23:04
  • $\begingroup$ un out of time last comment "Whether the people on the train know the train is moving, or don't know, makes no difference." Its whether the 'Moving' Frame makes a difference to when the light wave front is met is concerning me. I accept the maths work and have been proved. However from the instant the photon leaves the light is the frame of reference 'stationary' for all velocties of that frame. Consider an isolated frame with no references to other frames with an ability to change speed. Say a space ship that can achieve changes in constant speed. $\endgroup$
    – JohnI
    Commented Jan 3, 2022 at 23:18
  • $\begingroup$ The rules of SR apply most straightforwardly in inertial frames of reference. If you now introduce the idea of accelerating frames of reference, the maths become much harder. You ask 'is the frame of reference stationary for all velocities of that frame'. That question is self contradictory if you are talking about inertial frames, as a change of velocity implies acceleration. $\endgroup$ Commented Jan 4, 2022 at 7:02
  • $\begingroup$ I was thinking that after the period of acceleration a steady state speed would exist when the experiment could be repeated. The observers within the spaceship might not know how fast they were moving but would conclude that having been under acceleration and now were not that they must be moving at some unknown speed. Ie no acceleration when undertaking the experiment. $\endgroup$
    – JohnI
    Commented Jan 4, 2022 at 12:06
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The answer is simply that two separated events that are simultaneous in one frame won't, generally, be simultaneous in another frame moving relative to the first.

On the train, the two key events are the light reaching each of the two delegates, and they happen simultaneously on the train, but on the platform they happen at different times.

It might help you to think about what is happening if you bear in mind that the people on the train and the people on the platform also disagree about where the centre of the expanding light is. For the sake of illustration, let's suppose that the bulb was a flashbulb sending out an instantaneous flash of light, and that at the exact same instant that the flash went off, a paint gun directly under the bulb on the bottom of the train sprayed a blob of paint on the tracks. In the frame of the carriage, the light emitted will move out as a spherical wave always centred on the flash bulb, whereas in the frame of the tracks the light will be a spherical wave always centred on the blob of paint. Given that the bulb and the blob are moving relative to each other, that means that the people on the train and the people on the platform will disagree about the position of the central point from which the light is spreading.

You should not imagine from that description that there really are two separate spheres of light. On the contrary, there is only one expanding wavefront, but the people on the train and the people on the platform see it on planes of constant time that are tilted relative to each other, so it appears to the platform group, whose plane of constant time is tilted one way, as a sphere centred on the blob of paint, while on the train the very same wavefront is seen as a sphere with a different centre.

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This thought experiment falsifies Einstein's 2nd postulate which states that the SOL is independent of the motion of its emitter because if that's true, then it'll travel at c but the president of backwardland will be moving away from it; hence it'll take longer to reach him than if he wasn't moving. Similarly, the president of forwardland will be moving towards the photon, hence he'll receive it before the president of backwardland. And this is using Einstein's 2nd postulate. For both moving presidents to receive the photons at the same time, the photon must acquire the speed of the train. That way the photon going towards backwardland will have speed c+v; but backwardland is moving away from it at speed v; hence relative velocity is c. Similarly, photon moving towards forwardland is moving with speed c-v; but forwardland is moving towards it at speed v; hence, relative speed is c i.e. Photons will reach both presidents at the same time. The same will be witnessed from outside. Both instances prove that Einstein's 2nd postulate is false.

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    $\begingroup$ Einstein's postulate of the constancy of speed of light on inertial reference frames is not falsified by this experiment. As explained in the other answers, the issue is that the concept of simultaneity is not invariant under change of reference frames. $\endgroup$ Commented Dec 5, 2021 at 1:17
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    $\begingroup$ This answer is incorrect and should be deleted. $\endgroup$ Commented Jan 3, 2022 at 22:22

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