# Is relativity of simultaneity measurable?

Discussion in 'Physics & Math' started by Pete, May 8, 2013.

1. ### LaymanTotally Internally ReflectedValued Senior Member

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It would be the time they would use in order to measure the speed of light as according to the equation c = d / t. The speed of light is constant so it is a given, so then they could divide that by the distance the flash is away from them to determine how long it took the flash to reach them. The other frame would use c = d' / t'.

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3. ### LaymanTotally Internally ReflectedValued Senior Member

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Okay, say one frame wants to measure the distance the beam has traveled. He would use the equation d = c t to figure that. Another frame would use the equation d' = c t'. The speed of light is the same value in both frames. The distance and times are different values. But, even though the distance and times are different values the photon itself is at exactly the same location ( disregarding other quantum mechanical effects). They have to be assigned different values so that the speed of light is the same even though they have both traveled a different apparent distance from each frame, but the real distance it traveled in both frames is the same it is just the apparent distance from one frame being in motion that makes it just seem different. So when you plug in the equation of the proper time you should always be finding the value of the other frames time when the photon is assumed to be at the same real location.

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5. ### Neddy BateValued Senior Member

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Yes, knowing the distance between the lighting strikes means you also know how long it takes for the light from the lighting strikes to reach the observers. How does this help you figure out whether the lighting strikes were simultaneous in both frames or not?

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7. ### LaymanTotally Internally ReflectedValued Senior Member

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You could simply plug both times into the proper time equation, and then if both sides equal each other the events should be happening at the same time. If both sides do not equal each other then they wouldn't be events that they would be experiencing at the same time.

8. ### Neddy BateValued Senior Member

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OK, so if the observer at M' were to see one lighting strike before the other, (even though both lighting strikes occurred a distance of d' away from the observer at M'), then M' would say the strikes did not occur simultaneously. Right?

9. ### LaymanTotally Internally ReflectedValued Senior Member

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But, they don't see one lightning strike before another due to their relative velocity. That is not what has been found from experiments. Instead, they found that it is Gauge Invariant.

10. ### Neddy BateValued Senior Member

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But IF (SOMEHOW) the observer at M' were to see the front lighting strike before the rear one, (even though both lighting strikes occurred a distance of d' away from the observer at M'), then M' would say the strikes did not occur simultaneously. Right?

11. ### LaymanTotally Internally ReflectedValued Senior Member

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If there was an actual experiment that found that two beams that are equal distance arrived at different times due to an objects relative velocity then I would concede my whole argument. I would have to rethink the whole thing, because frankly that is what all this is based on. But, I don't think that is ever going to happen.

12. ### Neddy BateValued Senior Member

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I'm just asking you to leave that possibility open, rather than assuming that the lighting strikes must be simultaneous. Like you said, if RoS were true, then it should be measurable in this way, right?

.....

Anyway, let's get back to the part of Einstein's experiment where you and I agree. I think you and I agree that M and M' are in the same place when the strikes occur, and that the light from the strikes reaches M simultaneously. Am I correct in that?

13. ### LaymanTotally Internally ReflectedValued Senior Member

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Well, opening that possibility completely undoes the entire basis of my argument. I might as well just say, yes you right, even though that I know it to be wrong. I think they may have made a mistake with the discovery of Gauge Invariance, by not pointing out that it does not fit with Einsteins theory of RoS. Perhaps it is a good thing they didn't or it might not have ever been discovered...

I think that the flashes of light will reach the observer in the middle at the same time from both frames, similar to what happens when a MME is set in motion. Einstein didn't base his theory on the MME, I am starting to think he really should have.

14. ### Neddy BateValued Senior Member

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Alright, that means you and I do agree that the light reaches M simultaneously. At that instant, if you asked M where M' is located, what would he say? (Assume the train moves left to right).

15. ### LaymanTotally Internally ReflectedValued Senior Member

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He would say it is at M'. Hah, I knew that was a trick question.

16. ### Neddy BateValued Senior Member

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Ha! Good one.

But seriously, I think you must agree that M would say that M' is off to the right of M at that instant. So, basic geometry tells us M' would have already seen the light from the right strike at that instant. Basic geometry also tells us M' would not yet have seen the light from the left strike at that instant.

The MME experiment does not have an observer like M'. There is no observer moving relative to the MME apparatus, and also intercepting its light signals.

17. ### LaymanTotally Internally ReflectedValued Senior Member

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That could be problem. When I solve for the proper time I don't even consider that M' is some M + x. Only that M and M' have different values that are not equal to each other, but then each of these values are only related to each other somehow through Pythagorean's Theorem. This is why sometimes I have claimed that space-time actually warps in order to maintain the speed of light from both frames. It is more like the two frames are connected in some way in order to maintain an absolute frame that is at the speed of light.

I know there have been variations of the MME that have been set in motion. I just don't remember who or where they were done. I think a little research into the discover of Gauge Invariance could shine some light on this part.

18. ### Neddy BateValued Senior Member

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How can you calculate any time (or times) for when the light arrives at M' if you don't consider the location of M'? Sure the observer at M' can carry a clock with him, and it must always display some time, whatever that may be. But we are trying to determine whether he will measure two different times, or one single time. We need to solve this geometrically.

The original MME was in motion with the earth. Observer M is also in motion with the earth. That doesn't mean M would measure RoS by himself. He needs M' to measure it.

19. ### LaymanTotally Internally ReflectedValued Senior Member

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How could you be certain that simply M + x = M'? When all values of d in M would not equal all values of d' in M'? How then could you justify just moving M' over from M? Any d + x =/= d', only by solving through the Pythagorean Theorem or the proper time and proper length could you know the true difference between d and d'.

I think read a long time ago that wasn't good enough for some people so they actually put it in motion and then determined the results as the surface of the Earth being at rest. Like I said, I don't remember exactly who are where these experiments were done.

20. ### Neddy BateValued Senior Member

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We can do all the calculations from the embankment frame.

At embankment time $t=0$ let M and M' be instantaneously co-located at embankment location $x=0$

At that same instant, let the left lighting strike occur at embankment location $-x_{strike}$ and let the right lighting strike occur at embankment location $+x_{strike}$

The velocity of the train is known to be $v$ therefore, as embankment time $t$ progrsses forward, the location of M' will be:
$x_{M'} = vt$
as measured by the embankment frame.

Light from the strikes travels toward M at a speed of $c$ so the location of the left wavefront will be:
$x_{Lwave} = -x_{strike} + ct$
and the location of the right wavefront will be:
$x_{Rwave} = +x_{strike} - ct$
as measured by the embankment frame.

After an embankment time of $\frac{x_{strike}}{c}$ the left wavefront will be located at $x_{Lwave} = 0$ and the right wavefront will be located at $x_{Rwave} = 0$ both of which are where M is located.

But by that time, M' will be located at:
$x_{M'} = \frac{vx_{strike}}{c}$
which is off to the right of M.

Thus, at embankment time $\frac{x_{strike}}{c}$ the observer at M' must have already seen the right wave front, and could not yet have seen the left wavefront.

21. ### LaymanTotally Internally ReflectedValued Senior Member

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I thought it was well known that geometric representations of relativity do not actually give accurate answers or solutions. I just don't think the mathematics used here is complex enough in order to describe the situation perfectly accurately. If you followed along this reasoning you would eventually find that since in frame M' it observes two flashes to arrive at different times that are starting out at the same distance away from an observer in M', then you could only conclude that one flash traveled faster than the other flash. That is the only way they could have arrived at different times. Or by saying that the events didn't happen at the same time and there is no way to determine when that time actually was, then it is just allowing a fudge factor to take the blame that is completely unknown, hence the one way speed of light problem.

22. ### eramSciengineerValued Senior Member

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A whole bunch of nonsense unfortunately.

23. ### arfa branecall me arfValued Senior Member

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I think the contributors to this thread have overlooked something, which is what "measurement" means.
It might not be possible to measure RoS directly (i.e. "simultaneously"), but I can't help thinking, what is stopping two observers from using, say, a high-speed camera each to record their observations? One records two simultaneous lightning strikes, the other records two non-simultaneous strikes. They compare recordings at a later time, et voila, RoS!

Or is there a problem with this scenario? If so, what is it?