Special Relativity (i'm confused)

In summary: Otherwise, this summary should be sufficient.In summary, you will see the light source sooner than the person who is stationary from where you left off if you are traveling towards the light source at a significant fraction of the speed of light. Keep in mind though, that this has nothing to do with the theory of relativity.
  • #1
newbiephysics
3
1
Does traveling towards a light source at a significant fraction of c mean that you will infact see the light source sooner than say someone who is stationary at the point you began traveling at the fraction of c from? Say you are traveling towards a light source a light year away, will you see this light source sooner than the person who is stationary from where you left off? I'm really confused lol.
 
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  • #2
yes, but this has nothing to do with relativity, the same applies in classical mechanics
 
  • #3
newbiephysics said:
Does traveling towards a light source at a significant fraction of c mean that you will infact see the light source sooner than say someone who is stationary at the point you began traveling at the fraction of c from? Say you are traveling towards a light source a light year away, will you see this light source sooner than the person who is stationary from where you left off? I'm really confused lol.
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).
 
  • #4
relativityfan said:
yes, but this has nothing to do with relativity, the same applies in classical mechanics
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.
 
  • #5
Ah, if you are between the light source and somebody else, you will always see the light source before the other person, even if you are traveling away from the light source, as long as the light hits you before you pass the other person. Your speed, relative to the light source or to the other person has no bearing on the answer to your question.

You are confused about something, do you want to explain why you thought the answer could be that the light hit the other person before it hit you? I suspect you really meant to ask a different question.
 
  • #6
ghwellsjr said:
You are confused about something, do you want to explain why you thought the answer could be that the light hit the other person before it hit you? I suspect you really meant to ask a different question.
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.
 
  • #7
Even if you forget about length contraction and relativity of simultaneity (or don't know about them), and even if you assume that the light from the source is traveling towards you at c in your new reference frame, the fact that you are closer to the light source than you were when you left your starting point (and the other observer) means the light has to travel past you before it reaches your starting point (and the other observer). I don't see how this could be confusing to anyone unless they are misunderstanding special relativity and/or physics in general.

I hope newbiephysics will respond and tell use why he was confused.
 
  • #8
ghwellsjr said:
Even if you forget about length contraction and relativity of simultaneity (or don't know about them), and even if you assume that the light from the source is traveling towards you at c in your new reference frame, the fact that you are closer to the light source than you were when you left your starting point (and the other observer) means the light has to travel past you before it reaches your starting point (and the other observer).
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.
 
  • #9
I've been mulling over this one trying to figure out why newbiephysics could be confused on this point and I think it must be that he is thinking that since he is traveling at a significant fraction of the speed of light, time is progressing more slowly for him and therefore it will take longer for the light to reach him. That's probably what you realized he was thinking, too, Jesse, without specifically saying it, so it passed right by me. In that case, I can see all your points and I agree with you.

Now, newbiephysics, can we hear from you?
 
  • #10
ghwellsjr said:
That's probably what you realized he was thinking, too, Jesse, without specifically saying it, so it passed right by me. In that case, I can see all your points and I agree with you.
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?
 
  • #11
First of all, newbiephysics stated that both observers were stationary at the time the light was emitted from a source one light year away. Isn't that what "at the point you began traveling at the fraction of c from" means? So both observers and the light source are all in the same frame of reference to begin with. Then the observer called "you" accelerates toward the light source until "you" achieve a high rate of speed. That means "you" are closer to the light source than the "someone" "you" left behind and therefore "you" will see the light before the other person does. This is an analysis that takes place in one reference and is the easiest to understand. I don't know why it has to go any further than this. Are you suggesting that there is some other analysis that would indicate that the light reaches the stationary observer before it reaches "you"?

But if our goal is to try and second-guess why this obvious answer was so confusing to newbiephysics that it prompted him to ask his question, then until he tells us why he was confused, your guess is as good as mine. I already stated why I think he was confused and if I'm right, or even if I'm not, I would tell him that he can analyze the entire problem in anyone reference frame but if he tries to analyze part of it in the frame of "you" (thinking that time is progressing more slowly and therefore taking longer for the light to get to him) and the other part in the frame of the stationary person, then he might not get the correct answer and that might be the source of his confusion.

Now if he wants to analyze the entire problem in the frame of the moving "you", it will get very complicated and even more confusing because "you" doesn't start out in that frame. So he can change the problem and say that "you" was traveling all the time so now he has to bring up the options about when the light was emitted and the synchronization issue and the length contraction issue and the constancy of the speed of light in each frame issue but not the time dilation issue?

Jesse, you asked me a question. It's a double negative. It's a very confusing question. I'll try to answer it. I realize I have chopped off the false assumptions before it but here it is:
JesseM said:
...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?
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.

Now for the second part: Do I believe that the light could take different amounts of time in each observer's frame to travel from the source to them? Of course it could. If, perchance, it took the same amount of time, all I have to do is speed up the moving observer and repeat and it will take different amounts of time. (Am I missing something?)

Now I have to work out that double-negative. I was told that you can change both negatives and you end up with the same thing: Do I agree that it's possible that both the above-analyzed statements could be true. Well, if you meant the round-trip measurement of the speed of light then it looks like the answer is yes, otherwise, no. I hope I got that right.

Honestly, Jesse, I'm not trying to be difficult, I'm doing the best I can.

Please don't forget to answer my question at the end of the first paragraph.

Newbiephysics, where are you?
 
  • #12
ghwellsjr said:
First of all, newbiephysics stated that both observers were stationary at the time the light was emitted from a source one light year away. Isn't that what "at the point you began traveling at the fraction of c from" means? So both observers and the light source are all in the same frame of reference to begin with. Then the observer called "you" accelerates toward the light source until "you" achieve a high rate of speed.
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:
That means "you" are closer to the light source than the "someone" "you" left behind and therefore "you" will see the light before the other person does. This is an analysis that takes place in one reference and is the easiest to understand. I don't know why it has to go any further than this. Are you suggesting that there is some other analysis that would indicate that the light reaches the stationary observer before it reaches "you"?
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:
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.
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:
Now for the second part: Do I believe that the light could take different amounts of time in each observer's frame to travel from the source to them? Of course it could. If, perchance, it took the same amount of time, all I have to do is speed up the moving observer and repeat and it will take different amounts of time. (Am I missing something?)
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:
Now I have to work out that double-negative. I was told that you can change both negatives and you end up with the same thing: Do I agree that it's possible that both the above-analyzed statements could be true. Well, if you meant the round-trip measurement of the speed of light then it looks like the answer is yes, otherwise, no. I hope I got that right.
I didn't really mean it that way, perhaps it will be clearer if I alter it to read:
do you disagree with my claim 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?
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).
 
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  • #13
I think there is a sense in which ghwellsjr right that 'this should seem obvious' without taking account of simultaneity, or length contraction. I don't know how this relates the OP's way of looking at it. This 'obvious' way of looking at it wrong in all the details, but a newbie might see nothing wrong with it.

What am I talking about? Assume stationary observer (SO), supernova (SN) a light year away at rest relative to SO, and moving observer (MO) who passes SO simultaneously with the SN event (as perceived by SO). Suppose MO moving at 2/3 c. MO moves .4 light years in .6 years , seeing the SN at that point. They see the light earlier; they see the light coming to them as having traveled .6 light years in .6 years, thus speed c. Later (.4 year later), SO sees the SN, and computes that speed was also c. At first glance, no contradiction or issue.

In fact, this is even 'correct' all the way through if MO chose to do all measurements and analysis from SO frame. You can even make it seem to work in MO frame. SN is moving toward MO at 2/3 c. If SN event occurred 1 light year from MO passing SO, and light is same speed in all frames, then MO now thinks light should take a year to reach them, having traveled a light year. Still ok, time different in MO frame, but c comes out same. SN signal will take 3 years to reach SO, having traveled 3 light years. So with this simplistic analysis, both observers see light moving at c, both think MO sees SN first, the only difference is time: SO thinks the two times are .6 years and 1 year, while MO thinks they are 1 year and 3 years.

So I can see a newbie wondering arriving at the idea you need time dilation (in the wrong amount), but what's this about simultaneity and length contraction?

Anyway, I think it could be worthwhile explaining in detail what is wrong with this picture. I don't have time to do it right now.
 
  • #14
JesseM said:
ghwellsr said:
Are you suggesting that there is some other analysis that would indicate that the light reaches the stationary observer before it reaches "you"?
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.
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.

So we are back to second-guessing why newbiephysics was confused and so, as I said before, your guess is as good as mine. We will have to wait for him to tell us why he was confused.
 
  • #15
ghwellsjr said:
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.
"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".

Aside from the question of newbiephysics' confusion, one of your previous comments suggested you may also be confused about an aspect of SR, could you respond to this part?
JesseM said:
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.
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.
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.
 
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  • #16
JesseM 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?
How do you propose to demonstrate that what you say is true?
 
  • #17
Passionflower said:
How do you propose to demonstrate that what you say is true?
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.
 
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  • #18
V approaching zero means that it would take an approaching infinite time to get to a given location a distance x away. When v is equal to zero it is not possible to transport to any distance. So you can extrapolate all you want but you would never get anywhere with a zero velocity.

As far as I know Einstein himself never claimed that the one way speed of light is c in an inertial frame, I am not sure why others are so eager to 'prove' that.
 
  • #19
Passionflower said:
V approaching zero means that it would take an approaching infinite time to get to a given location a distance x away. When v is equal to zero it is not possible to transport to any distance. So you can extrapolate all you want but you would never get anywhere with a zero velocity.
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:
As far as I know Einstein himself never claimed that the one way speed of light is c in an inertial frame
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.
 
  • #20
JesseM said:
You're familiar with the idea of limits from calculus, right?
We take an example so it is all math and so we have a no spin zone:

We use c=1 for velocity.

We transport the clocks a distance of 1.

To calculate the elapsed time of the transported clock arriving at the agreed on distance we us:

[tex]
{\frac {d\sqrt {1-{v}^{2}}}{v}}
[/tex]

We use 1 for d, but feel free to use, 5, 10 any number you like.

Now let's take the limit v->0.

I know the answer but because you need to question my ability to understand limits I let you answer that question.

By the way I simplified it, to be exact we should include two accelerations in opposite directions but it would not make difference for the answer.
 
  • #21
JesseM said:
...
Aside from the question of newbiephysics' confusion, one of your previous comments suggested you may also be confused about an aspect of SR, could you respond to this part?

[If you want to see the embedded quotes from the original, look at post #15. I don't know how you get those nested quotes to show up in your posts unless you are doing an awful lot of cutting, pasting, and editing.]

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?
...

In A's frame, SR defines the time intervals for both one-way light trips starting and stopping at A to be equal but the time intervals for both one-way light trips starting and stopping at B will be unequal. Likewise, in B's frame, SR defines the time intervals for both one-way light trips starting and stopping at B to be equal but the time intervals for both one-way light trips starting and stopping at A will be unequal. Furthermore, in the mirror's frame, both time intervals for both A and B will be unequal.

And, as a side note, as I have mentioned before, if A and B measure the round-trip speed of the light beam, they will both get the same answer, c. However, they cannot measure the the round trip speed of light using the mirror you propose, they have to each have their own additional mirror that is stationary with respect to themselves. We are assuming, of course, that the light beam is wide enough that it can reflect off of three different mirrors and the moving mirrors are positioned so they don't collide with each other.

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.

If you look on page 131 of the above linked book, you will read this quote from Einstein:

"That light requires the same time to traverse the path A->B as for the path B->A is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity." (Emphasis in original.)

In his 1905 paper, he also made it very clear that the one-way speed of light cannot be measured or observed but must be defined. In addition, he affirmed that the round trip speed of light can be measured and will always be exactly c.
 
  • #22
JesseM 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?

Hi Jesse,

You might want to check this claim. It appears you might have slipped up here. In A's frame the time between the light being sent out and hitting the mirror is not the same as the time between hitting the mirror and returning to A.

JesseM 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.
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.
 
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  • #23
Passionflower said:
We take an example so it is all math and so we have a no spin zone:

We use c=1 for velocity.

We transport the clocks a distance of 1.

To calculate the elapsed time of the transported clock arriving at the agreed on distance we us:

[tex]
{\frac {d\sqrt {1-{v}^{2}}}{v}}
[/tex]

We use 1 for d, but feel free to use, 5, 10 any number you like.

Now let's take the limit v->0.

I know the answer but because you need to question my ability to understand limits I let you answer that question.

By the way I simplified it, to be exact we should include two accelerations in opposite directions but it would not make difference for the answer.

I think what Jesse is saying is basically this:

The error (e) due to slowly transporting a clock a distance of [itex]\Delta x[/itex] can be calculated as:

[tex]e = \frac{\Delta x/v(1 - \sqrt{1-v^2})} {\Delta x/v} = \frac{\Delta x(1 - \sqrt{1-v^2})} {\Delta x} [/itex]

and in the limit that v goes to zero the above error goes to zero. However I agree that if v is exactly zero it would take an infinite time to transport the clock so we have to settle for e>0 but we can make e as arbitrarily close to zero as we like, as long as we are not one of those fussy people that insist that the only acceptable error margin is exactly zero :smile:.
 
  • #24
Passionflower said:
We take an example so it is all math and so we have a no spin zone:

We use c=1 for velocity.

We transport the clocks a distance of 1.

To calculate the elapsed time of the transported clock arriving at the agreed on distance we us:

[tex]
{\frac {d\sqrt {1-{v}^{2}}}{v}}
[/tex]

We use 1 for d, but feel free to use, 5, 10 any number you like.

Now let's take the limit v->0.

I know the answer but because you need to question my ability to understand limits I let you answer that question.
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 [tex]\frac {D\sqrt{1 - v^2}}{v}[/tex], 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 [tex]\frac{D(1 - \sqrt{1 - v^2})}{v}[/tex]. 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 [tex]-D(1/2)(1 - v^2)^{-1/2}(-2v) = \frac{Dv}{\sqrt{1 - v^2}}[/tex] (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 [tex]\frac{Dv}{\sqrt{1 - v^2}}[/tex] 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.
 
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  • #25
yuiop said:
Hi Jesse,

You might want to check this claim. It appears you might have slipped up here. In A's frame the time between the light being sent out and hitting the mirror is not the same as the time between hitting the mirror and returning to A.
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:
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.
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:
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.
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".
 
  • #26
ghwellsjr said:
In A's frame, SR defines the time intervals for both one-way light trips starting and stopping at A to be equal but the time intervals for both one-way light trips starting and stopping at B will be unequal.
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:
And, as a side note, as I have mentioned before, if A and B measure the round-trip speed of the light beam, they will both get the same answer, c. However, they cannot measure the the round trip speed of light using the mirror you propose, they have to each have their own additional mirror that is stationary with respect to themselves.
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?
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.
ghwellsjr said:
If you look on page 131 of the above linked book, you will read this quote from Einstein:

"That light requires the same time to traverse the path A->B as for the path B->A is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity." (Emphasis in original.)

In his 1905 paper, he also made it very clear that the one-way speed of light cannot be measured or observed but must be defined. In addition, he affirmed that the round trip speed of light can be measured and will always be exactly c.
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.

Also, although it's true that Einstein's definition of synchronization is a convention, it is a genuine physical prediction that the laws of physics will be invariant in a family of inertial coordinate systems that define simultaneity using this convention, whereas they would not be invariant in a family of inertial coordinate systems that used a different convention. And in fact the Lorentz-invariance of the laws of physics gives us the reason why the slow transport method must yield the same definition of synchronization as Einstein's method.
 
  • #27
JesseM 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.
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.
 
  • #28
JesseM 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.
Well apparently you don't want to talk about the elapsed time for a single clock.

You can go as slow as you want you cannot make the difference zero as there would be no movement. You either fail to see that or simply want to ignore that.

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.
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.

You simply refuse to address my question.

We take an example where the moving clock is going to go a distance 1 away from the inertial clock.
So what does the elapsed time of the clock read when it arrives at the agreed upon distance so we can subtract this time from the elapsed time of the home clock?
 
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  • #29
Passionflower said:
Well apparently you don't want to talk about the elapsed time for a single clock.
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:
You can go as slow as you want you cannot make the difference zero as there would be no movement. You either fail to see that or simply want to ignore that.
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:
No, you were making the claim that you can measure the one way speed of light.
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:
You also claimed that Einstein implied that was the case.
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:
You simply refuse to address my question.

We take an example where the moving clock is going to go a distance 1 away from the inertial clock.
So what does the elapsed time of the clock read when it arrives at the agreed upon distance so we can subtract this time from the elapsed time of the home clock?
I haven't "refused to address your question", I thought I had already addressed this satisfactorily:
JesseM said:
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 [tex]\frac {D\sqrt{1 - v^2}}{v}[/tex]
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.
 
  • #30
This is from post #26:
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.

And here is the bolded statement you want me agree or disagree with:

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.

Before I can answer this, I want to make it clear that when I talk about an aether theory (or an absolute theory) as an alternative to SR, I'm only talking about the one that is indistinguishable from SR except that it affirms one priviledged reference frame. This is what I'm saying some scientists believed prior to Einstein. This was articulated best by Henri Poincare and you can read about it in wikipedia or in this reference from a book I have:

http://books.google.com/books?id=tJ...d the future of mathematical physics"&f=false

That being said, I don't want to put my time and energies into figuring out whether I agree or disagree with something based on something else that I don't agree with in the first place. I hope this will lead to a better understanding of my position.
 
  • #32
newbiephysics said:
Does traveling towards a light source at a significant fraction of c mean that you will infact see the light source sooner than say someone who is stationary at the point you began traveling at the fraction of c from? Say you are traveling towards a light source a light year away, will you see this light source sooner than the person who is stationary from where you left off? I'm really confused lol.

Perhaps it might be easier to look at this in the frame of the light.

E---------------------------------------------------R

Where the light is emitted at E and takes 1 light year to reach the Recipient at R.
We have exactly the same circumstance in either case. As the light is traveling at c in either case.

However, if we add in the light source, which remains stationary at E for the stationary observer, we find that in the light's frame of reference the light source (but not the point of emission E) is moving toward the traveling observer, and we have the following depiction:

E--------------------------S------------------------R

Note that as in the light's frame of reference all distances and times are proper distances and times, the light will reach each recipient simultaneously.

It is only when we see it from the reference frame of one or other of the observers that JesseM's Relativity of Simultaneity, time dilation and length contraction come into play.
 
  • #33
You are confused by what Jesse was saying just like I was. I thought he was explaining a legitimate analysis that would conclude that both observers see the light at the same time but he wasn't. He was attempting to explain how a confused person could come to that conclusion. But any correct analysis will show that the observer moving toward the light will see the light before the stationary observer.

But, beyond that, you are also confused about other things. Light doesn't have a frame of reference. Also, the frame of reference in which the light source was emitted and the stationary observer are the same frame. You also make contradictory statements: "the light source, which remains stationary at E" and "the light source ... is moving ". And I don't at all follow your logic.
 
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  • #34
Grimble, can you help me understand your explanation, please?
Grimble said:
Perhaps it might be easier to look at this in the frame of the light.

E---------------------------------------------------R

Where the light is emitted at E and takes 1 light year to reach the Recipient at R.
We have exactly the same circumstance in either case. As the light is traveling at c in either case.
When you say "either case", what are the two cases that you are referring to?
Grimble said:
However, if we add in the light source, which remains stationary at E for the stationary observer, we find that in the light's frame of reference the light source (but not the point of emission E) is moving toward the traveling observer,
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:
and we have the following depiction:

E--------------------------S------------------------R
You haven't said what "S" is. Is that the "light source"? Where is the moving observer in your diagram?
Grimble said:
Note that as in the light's frame of reference all distances and times are proper distances and times, the light will reach each recipient simultaneously.
Are you saying this because you think that time stops for light and so everything is simultaneous?
Grimble said:
It is only when we see it from the reference frame of one or other of the observers that JesseM's Relativity of Simultaneity, time dilation and length contraction come into play.
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.
 
  • #35
ghwellsjr said:
Grimble, can you help me understand your explanation, please?

When you say "either case", what are the two cases that you are referring to?

Those of the stationary observer and the moving observer.

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)"?

I was making the distinction between the point where the light was emitted and the light source's subsequent position nearer the moving observer.

You haven't said what "S" is. Is that the "light source"? Where is the moving observer in your diagram?

Yes, and the moving observer is R the recipient. He has to be always the same distance from the point of emission of the light or else the speed of the light would not be 'c'.

Are you saying this because you think that time stops for light and so everything is simultaneous?

No, time does not stop in the light's frame of reference, only when observed from any other frame. It is the limit of time dilation, isn't it?

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.

Yes, the stationary observer's frame of reference is identical to the light source's frame of reference but NOT to the LIGHT's frame of reference, which is moving at the speed of light relative to the observer's frame of reference.

I was agreeing with Jesse that they are only relevant when using different frames of reference, whereas I am only using the light's frame of reference.

And thank you for your response.:smile:
 

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