If both you and the other person have your clocks set to the same time at the moment you depart from them, then your clock will read an earlier time when the light reaches you than theirs reads when the light reaches them. Keep in mind though, because of length contraction, in your frame the distance between the position where you depart from the other person and the position of the source of the light will be shorter than it is in the frame of the other person (assuming they're at rest relative to the light source). Likewise because of the relativity of simultaneity (discussed more here and here), if we assume the source sent out the light "at the same time" you were departing from the other person according to the definition of simultaneity in their frame, in your own frame the source actually sent out the light somewhat before you departed from the other person (assuming 'your frame' refers to an inertial frame where you are at rest after departing, so that this frame was in motion relative to the first person even before you departed--it's simplest to imagine you have always been moving inertially and just passed by the first person, rather than originally having been at rest relative to the first person and then accelerating to depart).newbiephysics said:
The difference is that in classical physics, if the light is moving at c in the frame of the other person, it won't be moving at c in your rest frame. In relativity it must move at the same speed of c in both frames, I think that's probably why newbiephysics was confused about how this scenario would work.relativityfan said:
Again the reason for the confusion seems pretty straightforward, I presume it's about the fact that both observers are supposed to measure the speed of the light beam relative to themselves as c. So if you forget about length contraction and the relativity of simultaneity, you might think that if both observers were right next to each other when the source sent out the light, then since they were both the same distance D from the source they should both see the light after a time of D/c.ghwellsjr said:
But the idea that you see the light earlier would contradict the idea that the light still moves at c in your frame, if you (falsely) assume that the source emitted the light at the moment you were at the same position as the other observer, and that the distance to the source at the time it emitted the light was D for you just like the other observer. By definition speed is just distance/time, so if both of you agreed the distance to the source was D at the time the light was emitted but it took different times to reach you, then that would mean the 'speed' cannot have been the same in both your frames. This apparent contradiction (resolved by realizing about length contraction and the relativity of simultaneity) is what I assume the poster was confused about.ghwellsjr said:
No, my point didn't have to do with time dilation. I was just saying that if you forget about length contraction and the relativity of simultaneity (so you think both observers agree the light was emitted at the same moment they were at the same position, at the same distance from the source), then any difference in when they see the light would contradict the idea that the light moves at c in each observer's rest frame. Given these incorrect assumptions about distance and simultaneity, do you disagree that it's impossible that it could both be true that the light moves at c in each observer's rest frame and that the light could take different amounts of time in each observer's frame to travel from the source to them?ghwellsjr said:
Do I believe that the light moves at c in each observer's rest frame? I believe that if each observer measures the speed of the light coming from the source (by putting a mirror a known distance behind them, starting a stopwatch when the light reaches them, stopping the stopwatch when the reflection reaches them, and dividing twice the distance by the time on the stopwatch), they will both get exactly the same answer. If that is what you are asking about, the answer is yes. But if you are asking about the one way speed of light as it passes each observer (in other words, when they measured the roundtrip speed of light, did the light take the same time to go from the observer to the mirror as it took for the reflection to go from the mirror back to the observer), then the answer is no, it's possible possible for these two times to be equal for one observer but not for both.JesseM said:
Well, in problems like this we usually treat the acceleration as quasi-instantaneous, so you can look at the situation at the moment after this quasi-instantaneous acceleration, when the two observers are still at approximately the same position but have a significant velocity relative to one another, and assume that simultaneous with this (again failing to take into account relativity of simultaneity) the source emits a light flash.ghwellsjr said:
No, but there's an analysis that would indicate the light reaches both observers at the same time, since each observer conceives of themselves as stationary in their own frame. So if you fail to take into account relativity of simultaneity and length contraction, then in each observer's rest frame the source emitted a flash at a moment when the source was at a distance of D, so if it's true that light travels at c in both frames, each observer would have to see the light when their own clock read a time of D/c.ghwellsjr said:
I'm not sure I understand you, are you saying that even if both observers are at rest in inertial frames, and in each frame we assign time coordinates to the light leaving the observer, the light bouncing off a mirror, and the light returning to the observer, that the difference in time coordinates between the first two events will be different from the difference in time coordinates between the second two? If so that's a misunderstanding of SR, the second postulate of SR demands that the one-way speed of light is always c in every inertial frame.ghwellsjr said:
Yes, you're missing that regardless of the relative speeds, we can pick either observer's rest frame to calculate the time on the observer's clock when the light reaches them. Again I'm assuming newbiephysics' confusion related to not taking into account length contraction and relativity of simultaneity, so if you forget about those things and think both observers agreed the event of the source emitting the flash was simultaneous with both their clocks reading 0, and both agreed the distance to the source at that moment was D, then since the one-way speed of light is c in both frames you'd get the conclusion that both observers' clocks should show a time of D/c when the light reaches them.ghwellsjr said:
I didn't really mean it that way, perhaps it will be clearer if I alter it to read:ghwellsjr said:
i.e. if you think it's possible both statements could be true simultaneously, then you are disagreeing with me, because I'm saying the two statements contradict each other (again under the incorrect assumption that the two observers both agree the light flash was emitted simultaneously with them departing from one another, and that both agree the source was at a distance D at that moment).
When I asked this question, I meant, is there a correct, legitimate analysis that would conclude that the light reaches the stationary observer before the moving one and so when you said not before but at the same time, I thought you believed that this could happen but instead you were trying to explain to me how a confused person could incorrectly analyze the situation and come to that conclusion. You could have saved me a lot of time if you had answered "Yes", with a different incorrect analysis, a confused person could come to the incorrect conclusion that the stationary observer sees the light before the moving observer.JesseM said:
"Saved you a lot of time?" How was I supposed to know you were asking if there was a correct analysis? Previously we were talking about the "reason for the confusion" (posts 5, 6, 7), and I specified in every later post that I was talking about what happens if you forget about the relativity of simultaneity and length contraction, so it should have been obvious I thought we were still talking about this issue. If you wanted to totally change the subject you could have said something like "leaving aside the issue of what incorrect assumptions newbiephysics may or may not have been making, do you think there is any correct analysis that yields a paradox where both observers measure the same time for the light to reach them" and my answer would be "of course not".ghwellsjr said:
To clarify, suppose that observers A and B are moving inertially relative to one another, and at the time that their positions coincide, a light beam is sent out from that same position towards a distant mirror (say A is approaching the mirror while B is moving away from it). In A's frame, the time between the light being sent out and hitting the mirror is the same as the time between hitting the mirror and returning to A; likewise in B's frame the time between the light being sent out and hitting the mirror is the same as the time between hitting the mirror and returning to B. Do you agree? It occurs to me that maybe you're just saying that in B's frame, the time between light being emitted and hitting the mirror is different than the time between the light hitting the mirror and hitting A, and vice versa in A's frame.JesseM said:
How do you propose to demonstrate that what you say is true?JesseM said:
The experiment couldn't really be done at present because of the problem of getting clocks to move at relativistic speeds, but I was just talking about what should be true according to the theory of SR. For a theoretical description of how such an experiment would work, since my claim involved one-way speeds you would actually need two pairs of clocks A1-A2 and B1-B2, with A1 and A2 at rest relative to one another and synchronized using the slow transport method (using the light signal definition of 'synchronization' would make the experiment pointless), and likewise for B1 and B2. Then you could arrange things so that A1 and B1 were at the same position when the light signal was sent out from that position, and A2 and B2 were at the same position as the mirror at the time the light bounced off the mirror. This would give each pair a local way to assign times to the events of the light departing, the light hitting the mirror, and the light returning to the clock it departed from. Hopefully you agree that SR would predict in this case that (time on A2 when light hits mirror) - (time on A1 when light departs) = (time on A1 when light returns) - (time on A2 when light hits mirror), and likewise that (time on B2 when light hits mirror) - (time on B1 when light departs) = (time on B1 when light returns) - (time on B2 when light hits mirror). Likewise, hopefully you agree that this is a nontrivial physical prediction, since alternatives to SR like an aether theory without all moving clocks being subject to time dilation, or a theory where the speed of a light beam depends on the speed of the emitter/reflector, would predict something different.Passionflower said:
You're familiar with the idea of limits from calculus, right? The idea is that in the limit as v approaches zero in the frame where the clocks will be at rest once separated, the clocks approach being synchronized in exactly the same way as if Einstein's synchronization procedure was used. Another way of putting this is that you can pick any small offset from Einstein synchronization--say, 0.001 nanoseconds--and there will always be a small (but nonzero) velocity v such that transporting them at v (or at velocities of v or less, it's not actually necessary for slow transport that both be moved at precisely the same speed) results in their times differing from Einstein synchronization by less than this amount (this is basically just the epsilon-delta definition of a limit).Passionflower said:
He may not have used the precise words "one way speed" but it's clear he meant it to be, for example his famous train thought experiment demonstrating the relativity of simultaneity (see sections 8 and 9 of his book Relativity: The Special and General Theory) doesn't make sense unless you assume both beams from the lightning flashes have the same one-way speed in each frame. The derivation of the Lorentz transform itself doesn't work if you don't assume the one-way speed is supposed to be c in each inertial frame, if you only assume the two-way speed is the same in each frame there would be other possible coordinate transformations that would satisfy that. But the physical prediction of SR is that the laws of physics are invariant under the Lorentz transformation, not some other transformation where only the two-way speed is c in every frame.Passionflower said:
We take an example so it is all math and so we have a no spin zone:JesseM said:
JesseM said:
JesseM said:
JesseM said:
If the "one way speed" of light is measured using the Einstein light signal clock synchronisation method then the one-way speed of light is a built in assumption of the method. For example, if we assumed the speed of light going East is twice as fast as the speed of light going West and synchronised our clocks using this assumption, then when we measured the speed of light using these synchronised clocks we would find that the speed of light going East is indeed twice as fast as the speed of light going West! Whatever we assume about the one-way speed of light when we synchronise our clocks using light signals, will be what we measure for the one-way speed of light using those same clocks. The two way speed of light on the other hand can be unambiguously determined using a single clock. Having said all that, I would agree that the synchronisation of clocks using the slow clock transport method would in principle give us a sensible method of measuring the one-way speed of light to a reasonable degree of accuracy. There are of course complications on how we define "slow" as this requires us to measure the speed of the transported clock which requires us to have synchronised clocks in the first place to measure the velocity, but these problems should not be insurmountable.JesseM said:
Passionflower said:
But we don't want the "elapsed time" for a single clock, we want the difference in elapsed time for two different clocks, to show that this difference approaches 0 in the limit as the velocities of the clocks approach 0. Suppose we synchronize the two clocks at a common location, then we leave one clock at rest, and move the other away at v for some distance D, which takes a coordinate time of D/v in this frame. Then as you say, the time elapsed on the clock that is moved will be \frac {D\sqrt{1 - v^2}}{v}, but the time elapsed on the clock that's left at rest will just be D/v, so the difference in elapsed time for the two clocks is \frac{D(1 - \sqrt{1 - v^2})}{v}. You can see that both top and bottom approach 0 as v approaches 0, so we need to use rule[/url] and take the derivative of both top and bottom, with the derivative of the top being -D(1/2)(1 - v^2)^{-1/2}(-2v) = \frac{Dv}{\sqrt{1 - v^2}} (using the chain rule), and the derivative of the bottom is just 1, so L'Hôpital's rule says we should take the limit of \frac{Dv}{\sqrt{1 - v^2}} as v approaches 0, which is 0. Thus in the limit as the velocity with which one clock is moved approaches 0, the difference between their readings at the end of the moving process also approaches 0.Passionflower said:
How do you figure? If in A's rest frame the mirror is at a distance D at the moment the light is reflected off it, then since the one-way speed is c in both directions and A remains at rest at a single location, the time to go from A to the the mirror must be D/c, and the time to go from the mirror back to A must also be D/c.yuiop said:
Well, in a universe with different laws of physics where light had a two-way speed of c in some aether frame but rulers and clocks moving relative to the aether frame were not subject to length contraction or time dilation, then Einstein's clock synchronization method would guarantee that all frames measured the one-way speed to be the same in all directions, but the actual value of the speed would be different in different frames. But yes, assuming a universe with Lorentz-symmetric laws, then using Einstein's method does make claims about the one-way speed true by definition. That's why I said in post #17 "with A1 and A2 at rest relative to one another and synchronized using the slow transport method (using the light signal definition of 'synchronization' would make the experiment pointless)".yuiop said:
Yes, for example we could just define "slow" by saying that the clock should be moved by giving it a constant proper acceleration for a brief amount of proper time on that clock, then let it coast inertially to its destination for the rest of its trip, and if we keep the proper acceleration constant (which could be measured by an accelerometer attached to the clock) while considering the limit as the proper time in which it experiences the proper acceleration goes to zero, this would give a non-simultaneity-dependent definition of "slow transport".yuiop said:
OK, we agree which time intervals will be equal and which will be unequal in A's frame. And my other point is that it's a physical prediction of SR that using the slow transport method will approach perfect agreement with the standard synchronization method, so if we use the slow transport method to synchronize clocks which are at the location of the mirror when the light bounces off of it, then it becomes an actual physical prediction that the time intervals will be equal.ghwellsjr said:
The motion of the mirror is irrelevant since the light only reflects off it instantaneously, there is no need for the mirror to be at rest in either A's frame or B's frame. And certainly the motion of the mirror in A's frame won't affect the timing in A's frame, if A sent out a wide beam of light and half of it reflected off a mirror at rest in A's frame while the other half reflected off a mirror moving at 0.99c in A's frame, but with both mirrors happening to be at the same location in spacetime at the moment the beam hit them, then both halves of the beam would return to A at the same moment. Do you disagree?ghwellsjr said:
JesseM said:
Saying that the one-way speed is defined to be the same in both directions under Einstein's definition of synchronization is not a rebuttal to my statement above, since in my statement I am talking about using a different synchronization method, the slow transport method, and predicting that what is true of the times under Einstein's method is also true of the times under the slow transport method. That's what I called a "nontrivial physical prediction", so if you want to address this point, please tell me whether you agree or disagree with the bolded statement above.ghwellsjr said:
Upon reflection, it looks like it was me that slipped up. Oops! I was thinking of the rest frame of the mirror where the one way times are different, but yes, in the A's rest frame, the one way times are the same. Sorry,it looks like it was me was that was confused.JesseM said:
Well apparently you don't want to talk about the elapsed time for a single clock.JesseM said:
No, you were making the claim that you can measure the one way speed of light. You also claimed that Einstein implied that was the case.JesseM said:
I don't want to talk about that in isolation (though I did address the issue of elapsed time for each clock, as I mention at the bottom of this post) because the issue under discussion was synchronization, which by definition involves two clocks.Passionflower said:
I never said the difference should be precisely zero--I made clear that I was talking about limits, didn't I? Do you disagree that in the limit as the velocity approaches zero, the difference between slow transport and Einstein's method approaches zero as well?Passionflower said:
Yes, you can measure the one way speed of light relative to clocks synchronized with the slow transport method. Do you think the physicists who did the experiments listed here were not performing a valid physics experiment?Passionflower said:
Actually I said nothing about Einstein talking about "measuring" the one-way speed of light, I just responded to your comment "Einstein himself never claimed that the one way speed of light is c in an inertial frame" by pointing out that yes, Einstein did claim that. Do you disagree that by Einstein's definition of an "inertial frame", light must have a one-way speed of c in an inertial frame? Do you disagree that it is a real physical prediction about the laws of physics that they will be invariant under the specific type of frames defined by Einstein, not some other possible coordinate transformation with a different definition of simultaneity?Passionflower said:
I haven't "refused to address your question", I thought I had already addressed this satisfactorily:Passionflower said:
If for some reason you don't think this answers your question, please state your question a little more clearly. I always do my best to answer questions put to me, and I hope you will do the same with mine above.JesseM said:
JesseM said:
newbiephysics said:
When you say "either case", what are the two cases that you are referring to?Grimble said:
The term "light source", used by the original poster, is your point "E". What do you then mean by "the light source (but not the point of emission E)"?Grimble said:
You haven't said what "S" is. Is that the "light source"? Where is the moving observer in your diagram?Grimble said:
Are you saying this because you think that time stops for light and so everything is simultaneous?Grimble said:
Since the stationary observer's frame of reference is identical to the light source's frame of reference, do you think we need to consider Relativity of Simultaneity, time dilation and length contraction in that frame to understand the situation and get the right answer? Jesse was saying that you would need to take all three into account if you used the moving observer's frame of reference and he was speculating that the original poster could have gotten confused if he forgot about Relativity of Simultaneity. But if you do it correctly, you will not conclude that both observers see the light simulaneously but rather, the moving observer will see it before the stationary observer.Grimble said:
ghwellsjr said:
I'm not sure what the first question was but here are the facts:SinghRP said:
Grimble said:
I can see lots of things wrong with this picture. Were you attempting to illustrate another way that newbiephysics could have gotten confused on this situation? Are you ever going to have time to explain what is wrong? I can't tell what you were trying to do in this post.PAllen said:
ghwellsjr said:
Grimble said:
Grimble said:
Grimble said:
ghwellsjr said:
PAllen said:
Yes, this is correct (and important).Grimble said:
Passionflower said:
You are right, the formula should be:yuiop said:
This graph is comparing the distances from the perspective of different observers, it is not a traditional spacetime diagram.yuiop said:
PAllen said:
