WIKI and Time Dilation: The Possible Error in Relative Velocity

[JesseM]

(time interval in frame of observer who sees clock in motion) = t'

So, WIKI says t' = γt and you say the unprimed frame is at rest

I didn't say anything about either frame being "at rest", this is your own ill-defined terminology which everyone and their mother is telling you to stop using because it obviously causes you endless confusion. I just said the unprimed frame was the frame of the clock (the clock is at rest relative to the unprimed frame), and the primed frame was the frame of the observer (the observer is at rest relative to the primed frame). In this case, if we pick two events on the worldline of the clock, the times between these events in each frame will be related by t' = γt. This will be true regardless of which frame you choose to label "at rest" and which you choose to label "moving", as long as the conditions I mention above are met. Do you disagree with this or not? Yes or no?

Also, you seem to be rather evasive about answering my simple, oft-repeated question about whether you agree or disagree that the time dilation equation always works like this:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

If you disagree, please give me a counterexample.

 

[JesseM]

Maybe you are correct.

Let's follow through with light aberration and see if you are.

primed frame (in which the light clock appears to be moving) will be t' = t γ

You are wrong here. It should be t' = t/γ or t' γ = t.

What does this have to do with light aberration? And why do you preface that equation with "primed frame"? The equation t' = t γ isn't written from the point of view of any specific frame, it relates the times in two different frames. If the clock is moving at 0.6c in the primed frame, and we look at the pair of events on the clock's worldline where the clock reads a time of 0 and where it reads a time of 20, do you disagree that t (the time between these events in the unprimed frame) is 20 while t' (the time between these events in the primed frame) is 25, satisfying t' = t*1.25?

 

[chinglu1998]

I didn't say anything about either frame being "at rest", this is your own ill-defined terminology which everyone and their mother is telling you to stop using because it obviously causes you endless confusion. I just said the unprimed frame was the frame of the clock (the clock is at rest relative to the unprimed frame), and the primed frame was the frame of the observer (the observer is at rest relative to the primed frame). In this case, if we pick two events on the worldline of the clock, the times between these events in each frame will be related by t' = γt. This will be true regardless of which frame you choose to label "at rest" and which you choose to label "moving", as long as the conditions I mention above are met. Do you disagree with this or not? Yes or no?

Can you explain how time dilation is valid if a frame is not taken stationary? I mean, you seem to think this idea is dirty. I am sorry, but you must take one frame as stationary to determine time dilation.

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

If you disagree, please give me a counterexample.

Yes, this is how time dilation works. But, the reason I see not need to answer this question is your evasiveness to indicates a frame is stationary.

Are you ready to step up to the plate and claim the unprimed frame in WIKI is stationary and t' = γt. I want to see you declare this.

 

[chinglu1998]

What does this have to do with light aberration? And why do you preface that equation with "primed frame"? The equation t' = t γ isn't written from the point of view of any specific frame, it relates the times in two different frames. If the clock is moving at 0.6c in the primed frame, and we look at the pair of events on the clock's worldline where the clock reads a time of 0 and where it reads a time of 20, do you disagree that t (the time between these events in the unprimed frame) is 20 while t' (the time between these events in the primed frame) is 25, satisfying t' = t*1.25?

What does this have to do with light aberration?

Did you even read the article?

satisfying t' = t*1.25

Actually, you got me. When the unprimed frame is stationary, it elapses less time than the primed frame because of light aberration. Hence, the moving frame (primed frame) beats faster instead of slower as SR would demand for time dilation. You just refuted SR.

 

[JesseM]

Can you explain how time dilation is valid if a frame is not taken stationary? I mean, you seem to think this idea is dirty. I am sorry, but you must take one frame as stationary to determine time dilation.

I am sorry, but this a nonsensical notion of yours that no physicist would agree with, physicists do sometimes choose to label a frame as "stationary" but this is understood to be an arbitrary label for the sake of convenience with no relevance to any equation or physical result.

Let's say the clock is in the unprimed frame and the observer is in the primed frame. Pick two events on the clock's worldline, and say the time between them in the unprimed frame is t while the time between these same two events in the primed frame is t'. Then, if we label unprimed as "stationary" and primed as "moving" the correct time dilation equation is:

t' = t γ

But if we label the primed frame as "stationary" and unprimed as "moving", while keeping everything else the same, the correct time dilation equation is:

t' = t γ

Do you disagree? Yes or no, please. If you don't disagree, then you can see that the choice of which frame to label as "stationary" is utterly irrelevant to the equation, which is exactly the same either way. If you do disagree, please specify in which case you think the equation is wrong, and then I can go through a numerical example to show that the equation is still correct in that case.

Are you ready to step up to the plate and claim the unprimed frame in WIKI is stationary and t' = γt. I want to see you declare this.

There is no convention in physics that you must label one frame as "stationary", and in fact the wiki doesn't do so. But for the sake of argument, let's pretend the wiki did label the unprimed as "stationary". If they did, the correct equation given their description of the problem would still be t' = t γ, since as I said above it is utterly irrelevant to the equation which frame you choose to pin the arbitrary label "stationary" on.

 

[JesseM]

What does this have to do with light aberration?

Did you even read the article?

Yes, I did. I am not asking about the fact that light aberration is seen in the frame moving relative to the clock, I am asking about what light aberration has to do with your argument in the post I was responding to:

Let's follow through with light aberration and see if you are.

primed frame (in which the light clock appears to be moving) will be t' = t γ

You are wrong here. It should be t' = t/γ or t' γ = t.

You said "let's follow through with light aberration", but none of the subsequent sentences seemed to have anything to do with light aberration.

Actually, you got me. When the unprimed frame is stationary, it elapses less time than the primed frame because of light aberration. Hence, the moving frame (primed frame) beats faster instead of slower as SR would demand for time dilation. You just refuted SR.

You continually use vague ill-defined terminology, I don't know what it even means to say a frame "elapses less time", the time dilation equation deals with the time between a pair of events on the worldline of a physical clock. If we pick two events on the worldline of a clock, and t is the time between them in the clock's frame while t' is the time between those same events in the frame of an observer who is moving relative to the clock (though we are free to label the observer as "stationary" and the clock as "moving" rather than vice versa), then the equation will always be t' = t γ. As I said in my previous post, as long as those conditions are met this equation is correct, it's completely irrelevant which frame you label as "stationary" and which you label as "moving".

 

[chinglu1998]

I am sorry, but this a nonsensical notion of yours that no physicist would agree with, physicists do sometimes choose to label a frame as "stationary" but this is understood to be an arbitrary label for the sake of convenience with no relevance to any equation or physical result.

Einstein used the label "stationary" 62 times in his paper.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Are you calling Einstein wrong. Let's get it out now!

Let's say the clock is in the unprimed frame and the observer is in the primed frame. Pick two events on the clock's worldline, and say the time between them in the unprimed frame is t while the time between these same two events in the primed frame is t'. Then, if we label unprimed as "stationary" and primed as "moving" the correct time dilation equation is:

t' = t γ

But if we label the primed frame as "stationary" and unprimed as "moving", while keeping everything else the same, the correct time dilation equation is:

t' = t γ

Do you disagree? Yes or no, please. If you don't disagree, then you can see that the choice of which frame to label as "stationary" is utterly irrelevant to the equation, which is exactly the same either way. If you do disagree, I can go through a numerical example to show you're wrong.

I agree with all above. Where is your reciprocal time dilation as required by SR?

 

[JesseM]

Einstein used the label "stationary" 62 times in his paper.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Are you calling Einstein wrong. Let's get it out now!

Do you have zero reading comprehension? I said physicists do sometimes choose to label a frame as "stationary" but this is understood to be an arbitrary label for the sake of convenience with no relevance to any equation or physical result. Einstein of course calls one frame "stationary", but this is "understood to be an arbitrary label for the sake of convenience", that's why he says In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.'' Calling the system "stationary" is purely to distinguish it verbally, it has no physical relevance and doesn't affect any aspect of his mathematical derivation.

I agree with all above.

So you agree it makes no difference to the equation which frame we call "stationary"? And you agree the wiki article was correct to write t' = t γ since their scenario matched the one I described?

Where is your reciprocal time dilation as required by SR?

The reciprocity lies in the fact that if we picked a different clock which was in the primed frame rather than the unprimed frame, and picked two events on the worldline of that clock, then the time between these events in each frame would be given by t = t' γ. Again, it's all about which frame is the clock's frame, which is why I said it's best to think of the time dilation equation as (time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

 

[Dale]

How do you square your statement with Einstein's?

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''

http://www.fourmilab.ch/etexts/einstein/specrel/www/

In what way is my statement in conflict with Einsteins?

And no, you do not apply the metric to the stationary system. That is the light postulate defined as c²t² = x² + y² + z².

Yes, you do. The first postulate requires that the metric which applies to any inertial frame must apply to all of them. If it applies to one frame then it applies to all. The light cone is the set of all points:
ds² = -c²dt² + dx² + dy² + dz²
such that ds²=0
The same in one inertial frame as in all others. You cannot pick one metric for one inertial frame and another metric for another inertial frame without violating the first postualte.

 

[yuiop]

I agree with all above. Where is your reciprocal time dilation as required by SR?

Read post #90 for an example of reciprocal time dilation.

 

[darkhorror]

Can you explain how time dilation is valid if a frame is not taken stationary? I mean, you seem to think this idea is dirty. I am sorry, but you must take one frame as stationary to determine time dilation.




Yes, this is how time dilation works. But, the reason I see not need to answer this question is your evasiveness to indicates a frame is stationary.

Are you ready to step up to the plate and claim the unprimed frame in WIKI is stationary and t' = γt. I want to see you declare this.

I think I see your problem in reasoning and understanding of time dilation. I'll ask a few questions see if you agree or disagree and hopefully we can find exactly what the problem your having is. I made a post earlier that had most of the questions but i'll try to simplify.

You have the Earth and a ship. The ship is moving at a constant .6c away from the earth.

1. Is the ship "stationary" in the ship's frame of reference?

2. If in the Earth's frame of reference the ship's clock is ticking slower does this mean that in the ships frame of reference that the Earth's clock is ticking faster?

3. In which frame of reference is the Earth's clock ticking slower than the ship's clock?

4. Does the clock on the ship always tick slower than the clock on the Earth no matter which frame of reference you chose?

 

[darkhorror]

I think I might have figured out what is being said. How does this sound?

In the wiki article the observer is in the primed frame. So it doesn't matter what you label the frames as if it be "moving" or "stationary". To the observer which ever frame he is in he will always be at rest. He doesn't care if you call him "moving" or "stationary".

 

[grav-universe]

Okay, I think I finally figured out a way to clarify that part of the Wiki article with the light clock that fits in well with it and the diagrams, by adding a simple paragraph to the end. How does this sound?

This is the time dilation found only by the frame of reference that is considered stationary as it observes the light pulse which is bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to our frame of reference, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the stationary frame of reference observes of the moving frame's clock, as found earlier, is the ratio of the time that the stationary frame observes passing upon the moving frame's clock as compared to the time that passes upon the stationary frame's own clock.

 

[espen180]

OP, maybe the article

http://en.wikipedia.org/wiki/Introduction_to_special_relativity

is more suited to you. I suggest reading it closely before continuing the discussion, as you are demonstrating a misunderstanding of very fundamental topics in SR, and progression to more advanced topics will be impossible unless this is cleared up.

Okay, I think I finally figured out a way to clarify that part of the Wiki article with the light clock that fits in well with it and the diagrams, by adding a simple paragraph to the end. How does this sound?

It may just be me, but I don't think that paragraph, or any such clarification is needed.

 

[yuiop]

This is the time dilation found only by the frame of reference that is considered stationary as it observes the light pulse which is bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to our frame of reference, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the stationary frame of reference observes of the moving frame's clock, as found earlier, is the ratio of the time that the stationary frame observes passing upon the moving frame's clock as compared to the time that passes upon the stationary frame's own clock.

You really should stop talking about the "stationary frame" as it just confuses the heck out everyone here and you too.

The observer at rest in frame S considers himself to be at rest in his own reference frame and likewise the observer in frame S' considers himself to be at rest in his own reference frame. Frames S and S' are moving with a velocity magnitude of v relative to each other. There is no clear definition of whether S or S' is the stationary frame so as soon as you talk about the "stationary frame" no one knows what you are talking about and it just leads to confusion as it has in your case.

It is better to say something like the frame (S) in which the light clock is at rest, or the frame (S') in which the light clock has relative motion -v.

It *might* be OK to use phrases like "the stationary frame" as shorthand for "the reference frame in which the light clock is at rest" as long you clearly define the expression (as Einstein did) but using an identifier such as "frame S" is just a good a shorthand. Your expression "the frame of reference that is considered stationary as it observes the light pulse which is bouncing back and forth in the moving frame" gives the reader no clue as to whether "the stationary frame" is the frame in which the light clock is at rest or not.

 

[grav-universe]

You really should stop talking about the "stationary frame" as it just confuses the heck out everyone here and you too.

The observer at rest in frame S considers himself to be at rest in his own reference frame and likewise the observer in frame S' considers himself to be at rest in his own reference frame. Frames S and S' are moving with a velocity magnitude of v relative to each other. There is no clear definition of whether S or S' is the stationary frame so as soon as you talk about the "stationary frame" no one knows what you are talking about and it just leads to confusion as it has in your case.

It is better to say something like the frame (S) in which the light clock is at rest, or the frame (S') in which the light clock has relative motion v.

Right, okay. Maybe just the observing frame rather than the stationary frame, then? The paragraph I added has already been cut in the article, but I still want to clarify it. I also considered adding a second paragraph which shows the same time dilation can be found from either frame, but I think it still requires some rewording. Perhaps you can help with that. So far it reads

"If we were to now reverse the roles of the observing and moving frames, making the original frame of the light clock the observing frame while considering the other frame to be moving, while also placing another light clock in the other frame with a second light pulse bouncing back and forth between two mirrors in the same way as in the original frame, then the first frame will measure the same time dilation of the second frame's clock as well, found in the same way as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

 

[espen180]

Right, okay. Maybe just the observing frame rather than the stationary frame, then? The paragraph I added has already been cut in the article, but I still want to clarify it. I also considered adding a second paragraph which shows the same time dilation can be found from either frame, but I think it still requires some rewording. Perhaps you can help with that. So far it reads

"If we were to now reverse the roles of the observing and moving frames, making the original frame of the light clock the observing frame while considering the other frame to be moving, while also placing another light clock in the other frame with a second light pulse bouncing back and forth between two mirrors in the same way as in the original frame, then the first frame will measure the same time dilation of the second frame's clock as well, found in the same way as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

This is unneeded, because the following paragraph, titled "Time dilation due to relative velocity symmetric between observers" covers this.

 

[yuiop]

I also considered adding a second paragraph which shows the same time dilation can be found from either frame, but I think it still requires some rewording. Perhaps you can help with that.

Thanks, but I am probably one of the worst people to do this task as I use a very informal language myself. There are plenty of others skilled in the formal language here such as Jesse or Dalespam who would be much better suited to the task, but from what I can tell the Wikipedia article on time dilation is pretty accurate and well explained and does not need much, if anything adding to it.

 

[grav-universe]

This is unneeded, because the following paragraph, titled "Time dilation due to relative velocity symmetric between observers" covers this.

That isn't clear either. Okay, maybe just a quick mention on that, adding it to the overall paragraph. How about this?

"This is the time dilation found only by the frame of reference that observes the light pulse bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to the observing frame, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the observing frame observes of the moving frame's clock, as found earlier, is the ratio of the time that the observing frame views passing upon the moving frame's clock as compared to the time that passes upon the observing frame's own clock. In addition, if we were to now place another light clock in the second frame with a second light pulse bouncing back and forth between two mirrors in that frame in the same way while reversing the roles of the observing and moving frames, then the first frame will now measure the same time dilation of the second frame's clock as found as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

 

[yuiop]

That isn't clear either. Okay, maybe just a quick mention on that, adding it to the overall paragraph. How about this?

"This is the time dilation found only by the frame of reference that observes the light pulse bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to the observing frame, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the observing frame observes of the moving frame's clock, as found earlier, is the ratio of the time that the observing frame views passing upon the moving frame's clock as compared to the time that passes upon the observing frame's own clock. In addition, if we were to now place another light clock in the second frame with a second light pulse bouncing back and forth between two mirrors in that frame in the same way while reversing the roles of the observing and moving frames, then the first frame will now measure the same time dilation of the second frame's clock as found as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

"The moving frame" is as poorly defined as the "the stationary frame". An observer in frame S considers frame S' to be moving relative to his frame and the observer in frame S' considers frame S to be moving relative to his own frame. Which is the "the moving frame"? Frame S or frame S'? If someone reads your paragraph very carefully they can deduce that by "the moving frame" you mean the frame in which the light clock is at rest, but it is hard work because it is not made clear early on.

 

[grav-universe]

"The moving frame" is as poorly defined as the "the stationary frame". An observer in frame S considers frame S' to be moving relative to his frame and the observer in frame S' considers frame S to be moving relative to his own frame. Which is the "the moving frame"? Frame S or frame S'? If someone reads your paragraph very carefully they can deduce that by "the moving frame" you mean the frame in which the light clock is at rest, but it is hard work because it is not made clear early on.

Right. Okay then, how about this?

"This is the time dilation found only by the frame of reference that observes the light pulse bouncing back and forth in the frame of the light clock. To clarify this further, let us now place a mechanical clock at mirror A. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings of the mechanical clock to be Δt as that frame measures itself. According to the observing frame, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the observing frame observes of the other frame's clock, as found earlier, is the ratio of the time that the observing frame views passing upon the other frame's clock as compared to the time that passes upon the observing frame's own clock. In addition, if we were to now place another light clock in the second frame while reversing the roles of the frames, then the first frame will also measure the same time dilation of the second frame's clock, demonstrating that each frame measures the same time dilation of the other."

It probably doesn't matter much anyway at this point because I posted my suggestion in the discussion section and it was remarked by the person that cut it that I would require a reliable source. I'm not sure how I would find a reliable source that states it in a way that precisely fits the way it has already been presented as well as the diagrams, but I guess I'll look around.

 

[JesseM]

Right. Okay then, how about this?

"This is the time dilation found only by the frame of reference that observes the light pulse bouncing back and forth in the frame of the light clock. To clarify this further, let us now place a mechanical clock at mirror A. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings of the mechanical clock to be Δt as that frame measures itself. According to the observing frame, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the observing frame observes of the other frame's clock, as found earlier, is the ratio of the time that the observing frame views passing upon the other frame's clock as compared to the time that passes upon the observing frame's own clock. In addition, if we were to now place another light clock in the second frame while reversing the roles of the frames, then the first frame will measure the same time dilation of the second frame's clock as found as before from the second frame, demonstrating that each frame measures the same time dilation of the other."

It probably doesn't matter much anyway at this point because I posted my suggestion in the discussion section and it was remarked by the person that cut it that I would require a reliable source. I'm not sure how I would find a reliable source that states it in a way that precisely fits the way it has already been presented as well as the diagrams, but I guess I'll look around.

If you want a reference for the idea that another type of clock at rest next to the light clock must keep time with it, and therefore must also appear slowed down in the observer's frame, you could use p. 40 of Brian Greene's The Elegant Universe. My only editing suggestion here would be that "the frame of reference that observes the light pulse bouncing back and forth in the frame of the light clock" doesn't really make sense (how can the external observer observe something 'in the frame of the light clock'? The bouncing of the pulse isn't 'in' any frame, it's just a series of physical events observed in all frames). How about just "the frame of the observer who sees the clock in motion" or something like that?

 

[grav-universe]

If you want a reference for the idea that another type of clock at rest next to the light clock must keep time with it, and therefore must also appear slowed down in the observer's frame, you could use p. 40 of Brian Greene's The Elegant Universe.

Ah, beautiful. Thanks, JesseM. I also found a free image from Wikimedia that can be used. I've mentioned both of these in the talk page.

My only editing suggestion here would be that "the frame of reference that observes the light pulse bouncing back and forth in the frame of the light clock" doesn't really make sense (how can the external observer observe something 'in the frame of the light clock'? The bouncing of the pulse isn't 'in' any frame, it's just a series of physical events observed in all frames). How about just "the frame of the observer who sees the clock in motion" or something like that?

Yes, I was looking at that too. Your suggestion sounds perfect. So now we have

"This is the time dilation found only from the frame of the observer who sees the clock in motion. To clarify further, let us now place a mechanical clock at mirror A. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings of the mechanical clock to be Δt as that frame measures itself. According to the observing frame, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the observing frame views of the other frame's clock, as found earlier, is the ratio of the time that the observing frame views passing upon the other frame's clock to that of the time that passes upon the observing frame's own clock. In addition, if we were to now place another light clock in the second frame while reversing the roles of the frames, then the first frame will also measure the same time dilation of the second frame's clock, demonstrating that each frame measures the same time dilation of the other."

Any further suggestions from anybody? To really close things up, I'm thinking we might also need something that shows the time the observing frame views of the other frame is real time, nothing to do with the flight of light with observations.

 

[Vanadium 50]

I am sorry, but you must take one frame as stationary to determine time dilation.

You keep saying this. It is not correct - that is not how relativity works.

 

[IsometricPion]

Now, in the primed frame, it will see light aberration, hence the light path is longer in the view of the primed frame.

Apply the Pythagorean Theorem and you calculate t' = t γ just as WIKI said, but this is not time dilation since the unprimed frame is at rest.

By light aberration I assume you mean that while the path of the light beam is perpendicular to the mirrors in the rest frame of the clock (the unprimed frame), it is not perpendicular in the primed frame. Light aberration is due to motion in a direction perpendicular to the path of the light beam (so there is no way to have a frame that observes aberration and is at rest with respect to source). Objects in motion are observed to exhibit time dilation. So, if there is light aberration there will also be time dilation. The frame in which light aberration is observed measures the light clock to be in motion.

Now assume the primed frame is at rest. Same thing, light aberration will force that frame to conclude t' = t γ.

If you are using the light aberration via the Pythagorean theorem to calculate time dilation, you are implicitly assuming that the light clock is in motion, so you will always obtain the same answer (i.e., that for the time dilation of the unprimed frame due to its motion wrt the primed frame). It is clear that considering an equivalent light clock at rest in the primed frame will simply reverse the labels in the resulting equation \Delta{}t'=\gamma{}\Delta{}t\rightarrow{}\Delta{}t=\gamma{}\Delta{}t'. This is the formula arrived at for the time dilation of the primed frame wrt the unprimed frame (it could be done via the Pythagorean theorem as well, since the light clock at rest in the primed frame would exhibit light aberration in the unprimed frame).

 

[chinglu1998]

In what way is my statement in conflict with Einsteins?

Yes, you do. The first postulate requires that the metric which applies to any inertial frame must apply to all of them. If it applies to one frame then it applies to all. The light cone is the set of all points:
ds² = -c²dt² + dx² + dy² + dz²
such that ds²=0
The same in one inertial frame as in all others. You cannot pick one metric for one inertial frame and another metric for another inertial frame without violating the first postualte.

I am going to have to agree you are correct with the metric.

 

[chinglu1998]

Do you have zero reading comprehension? I said physicists do sometimes choose to label a frame as "stationary" but this is understood to be an arbitrary label for the sake of convenience with no relevance to any equation or physical result. Einstein of course calls one frame "stationary", but this is "understood to be an arbitrary label for the sake of convenience", that's why he says In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.'' Calling the system "stationary" is purely to distinguish it verbally, it has no physical relevance and doesn't affect any aspect of his mathematical derivation.

O can agree with you except, when the frame is taken as statonary, objects in that frame have Euclidian measurements. Those in the moving frame have Minkowski. Is that not a difference?

Though, whatever the frame, by the relativity postulate, each frame obeys this rule.

So you agree it makes no difference to the equation which frame we call "stationary"? And you agree the wiki article was correct to write t' = t γ since their scenario matched the one I described?

I am saying, it needs to be clear which frame is stationary.
If the unprimed frame is stationary as the article claims, then t' = t/γ. Is this false?

The reciprocity lies in the fact that if we picked a different clock which was in the primed frame rather than the unprimed frame, and picked two events on the worldline of that clock, then the time between these events in each frame would be given by t = t' γ. Again, it's all about which frame is the clock's frame, which is why I said it's best to think of the time dilation equation as (time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

I am not sure how to take you. But, your statement of a different clock and reciprocity would be the way I look at it and calculate.

Now, let me ask you a question. Assume one clock and only one.

Say it is a primed frame clock at a negative coordinate. Please calculate using LT the time result between the two frames.

 

[chinglu1998]

Read post #90 for an example of reciprocal time dilation.

Agreed, your post #90 is a perfect example of reciprocal time dilation.

Except, instead of two clocks for A and B, let's have one at say -k' in the primed system.

It moves to the unprimed origin. Does this create reciprocal time dilation?

 

[JesseM]

O can agree with you except, when the frame is taken as statonary, objects in that frame have Euclidian measurements. Those in the moving frame have Minkowski. Is that not a difference?

As I said I don't understand your terminology, all equations (Euclidean or Minkowski) that apply in one frame also apply in every other frame. You never answered my questions in post #84 about your ill-defined notion that this has something to do with "topological spaces" (I think maybe you need to look up the definition of 'topology').

I am saying, it needs to be clear which frame is stationary.
If the unprimed frame is stationary as the article claims, then t' = t/γ. Is this false?

Yes, it's false. Did you even read what I wrote in post #105 before saying "I agree with all of the above" in your response in post #107? I explicitly said that if the unprimed frame is stationary the equation is still t' = t γ. Read the first part again:

Let's say the clock is in the unprimed frame and the observer is in the primed frame. Pick two events on the clock's worldline, and say the time between them in the unprimed frame is t while the time between these same two events in the primed frame is t'. Then, if we label unprimed as "stationary" and primed as "moving" the correct time dilation equation is:

t' = t γ

Do you want to take back your agreement with this statement?

Now, let me ask you a question. Assume one clock and only one.

Say it is a primed frame clock at a negative coordinate. Please calculate using LT the time result between the two frames.

If the clock is in the primed frame (unlike in the wiki where it was in the unprimed frame) the relation will t = t' γ, regardless of whether we label the primed frame as "stationary" or "moving". Do you disagree? If you do I can certainly show that this is true using the LT (and with the clock at a negative position coordinate in the primed frame) if you like, but if you agree then there doesn't seem to be any point in going through the math.

 

[chinglu1998]

I think I might have figured out what is being said. How does this sound?

In the wiki article the observer is in the primed frame. So it doesn't matter what you label the frames as if it be "moving" or "stationary". To the observer which ever frame he is in he will always be at rest. He doesn't care if you call him "moving" or "stationary".

I agree if the observer is in the primed frame.

I do not know why I am the only one that can see the error in the article.

It says,

That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly.

How exactly are you going to be an observer moving relative to a clock and perform the calculation? If you are an observer in the primed frame, the clock is moving relative to you.

Do you see the logic error in considering the primed frame as the observer frame? An observer frame does not move relative to anything.

 

[chinglu1998]

OP, maybe the article

http://en.wikipedia.org/wiki/Introduction_to_special_relativity

is more suited to you. I suggest reading it closely before continuing the discussion, as you are demonstrating a misunderstanding of very fundamental topics in SR, and progression to more advanced topics will be impossible unless this is cleared up.



It may just be me, but I don't think that paragraph, or any such clarification is needed.

Then you do not understand SR and you need to start more simply.
The article says,

That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly.

which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself.

As an observer in SR, how do you know you are moving?

 

[chinglu1998]

You really should stop talking about the "stationary frame" as it just confuses the heck out everyone here and you too.

The observer at rest in frame S considers himself to be at rest in his own reference frame and likewise the observer in frame S' considers himself to be at rest in his own reference frame. Frames S and S' are moving with a velocity magnitude of v relative to each other. There is no clear definition of whether S or S' is the stationary frame so as soon as you talk about the "stationary frame" no one knows what you are talking about and it just leads to confusion as it has in your case.

It is better to say something like the frame (S) in which the light clock is at rest, or the frame (S') in which the light clock has relative motion -v.

It *might* be OK to use phrases like "the stationary frame" as shorthand for "the reference frame in which the light clock is at rest" as long you clearly define the expression (as Einstein did) but using an identifier such as "frame S" is just a good a shorthand. Your expression "the frame of reference that is considered stationary as it observes the light pulse which is bouncing back and forth in the moving frame" gives the reader no clue as to whether "the stationary frame" is the frame in which the light clock is at rest or not.

It is better to say something like the frame (S) in which the light clock is at rest, or the frame (S') in which the light clock has relative motion -v.

Yes, could not agree more.

Here is the article.

which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself.

This is saying the frame with the moving observer is viewing a clock at rest. Exactly under the language of SR is a statement like this true than an observer is moving?

 

[JesseM]

How exactly are you going to be an observer moving relative to a clock and perform the calculation? If you are an observer in the primed frame, the clock is moving relative to you.

"observer is moving relative to the clock" just means the observer is in motion in the clock's frame, and "clock is moving relative to the observer" just means the clock is in motion in the observer's frame. In relativity one necessarily implies the other, so the two statements are completely interchangeable, saying "the observer is moving relative to the clock" doesn't contradict the idea that when we perform calculations in the observer's frame, we treat the observer as being at rest and the clock as having a nonzero velocity.

 

[chinglu1998]

Originally Posted by chinglu1998
I am sorry, but you must take one frame as stationary to determine time dilation.

You keep saying this. It is not correct - that is not how relativity works.

Einstein:
Further, we imagine one of the clocks which are qualified to mark the time t when at rest relatively to the stationary system, and the time when at rest relatively to the moving system, to be located at the origin of the co-ordinates of k, and so adjusted that it marks the time . What is the rate of this clock, when viewed from the stationary system?
http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[JesseM]

Einstein:
Further, we imagine one of the clocks which are qualified to mark the time t when at rest relatively to the stationary system, and the time when at rest relatively to the moving system, to be located at the origin of the co-ordinates of k, and so adjusted that it marks the time . What is the rate of this clock, when viewed from the stationary system?
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Gee, where in that quote does Einstein say "you must take one frame as stationary to determine time dilation"? Nowhere. He says at the beginning that he chooses to label one frame as "stationary" purely for verbal purposes (the label is completely irrelevant to all his actual calculations, including calculations which 'determine time dilation'):

In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.''

 

[chinglu1998]

By light aberration I assume you mean that while the path of the light beam is perpendicular to the mirrors in the rest frame of the clock (the unprimed frame), it is not perpendicular in the primed frame. Light aberration is due to motion in a direction perpendicular to the path of the light beam (so there is no way to have a frame that observes aberration and is at rest with respect to source). Objects in motion are observed to exhibit time dilation. So, if there is light aberration there will also be time dilation. The frame in which light aberration is observed measures the light clock to be in motion.If you are using the light aberration via the Pythagorean theorem to calculate time dilation, you are implicitly assuming that the light clock is in motion, so you will always obtain the same answer (i.e., that for the time dilation of the unprimed frame due to its motion wrt the primed frame). It is clear that considering an equivalent light clock at rest in the primed frame will simply reverse the labels in the resulting equation \Delta{}t'=\gamma{}\Delta{}t\rightarrow{}\Delta{}t=\gamma{}\Delta{}t'. This is the formula arrived at for the time dilation of the primed frame wrt the unprimed frame (it could be done via the Pythagorean theorem as well, since the light clock at rest in the primed frame would exhibit light aberration in the unprimed frame).

In reality, it does not matter.

Assume you are the observer at rest with the light source. No matter what, the light moves in a straight line up the y-axis in your view.

But, if there was a moving frame, what do you believe it saw?

You think it moved and when the light beam reach a certain y coordinate, it applies the Pythagorean theorem based on its origin.

Now, assume you are an observer in the rest frame and the light source is moving.

You conclude light aberration and the moving light source is in straight lines.

It does not matter which frame you chose, the calculation is absolute. The frame at rest with the light source concludes less time.

 

[chinglu1998]

As I said I don't understand your terminology, all equations (Euclidean or Minkowski) that apply in one frame also apply in every other frame. You never answered my questions in post #84 about your ill-defined notion that this has something to do with "topological spaces" (I think maybe you need to look up the definition of 'topology').

I agree when you are the observer frame, you calculateexactly the same way any observer in any other frame would calculate.

Now, if you are the observer, do you calculate a moving observer will calculate events as you do? I know you will say no.

Yes, it's false. Did you even read what I wrote in post #105 before saying "I agree with all of the above" in your response in post #107? I explicitly said that if the unprimed frame is stationary the equation is still t' = t γ. Read the first part again:

Do you want to take back your agreement with this statement

?

No, I do not want to take back my agreement.
if the unprimed frame is stationary the equation is still t' = t γ

This is correct. In other places, my threads would be terminated for this. I agree 100%.

If the clock is in the primed frame (unlike in the wiki where it was in the unprimed frame) the relation will t = t' γ, regardless of whether we label the primed frame as "stationary" or "moving". Do you disagree? If you do I can certainly show that this is true using the LT (and with the clock at a negative position coordinate in the primed frame) if you like, but if you agree then there doesn't seem to be any point in going through the math.

No need, I agree. But, I want to make sure, by LT invertibility, in this case both frames calculate t = t' γ, is that your position?

 

[chinglu1998]

Gee, where in that quote does Einstein say "you must take one frame as stationary to determine time dilation"? Nowhere. He says at the beginning that he chooses to label one frame as "stationary" purely for verbal purposes (the label is completely irrelevant to all his actual calculations, including calculations which 'determine time dilation'):

Oh, I was not saying you must take one frame in that sense. I was saying you must take a frame as stationary to calculate time dilation.

Is this false?

 

[Vanadium 50]

I do not know why I am the only one that can see the error in the article.

Because you are the one who is wrong. (See the previous 100+ messages for details)

Then you do not understand SR and you need to start more simply.

No, it's you who does not understand it. Just because SR can handle a problem with one observer at rest, it doesn't mean that it provides a way to tell who is 'really' at rest. In fact, it doesn't even require that anyone be at rest: one can solve problems where everyone is in motion.

Understanding a theory does not mean being able to come up with a quote from someone else (even Einstein) which might support your position. It means being able to do the calculations. The words are window dressing - the mathematics is the theory.

 

[chinglu1998]

Because you are the one who is wrong. (See the previous 100+ messages for details)

It seems you are very good at math. Can you use math to support this?

Also, can you explain how an observer is moving under SR. How does an observer know it is moving? The WIKI article depends on this.

No, it's you who does not understand it. Just because SR can handle a problem with one observer at rest, it doesn't mean that it provides a way to tell who is 'really' at rest. In fact, it doesn't even require that anyone be at rest: one can solve problems where everyone is in motion.

Understanding a theory does not mean being able to come up with a quote from someone else (even Einstein) which might support your position. It means being able to do the calculations. The words are window dressing - the mathematics is the theory.

So, if I post a quote directly from Einstein that is the same as my postiion, then I am wrong? Can you explain this? You see, your position disagrees with the stated position of Einstein not mine.

 

[JesseM]

Now, if you are the observer, do you calculate a moving observer will calculate events as you do? I know you will say no.

No, but I fail to see the relevance to your statement "when the frame is taken as statonary, objects in that frame have Euclidian measurements. Those in the moving frame have Minkowski."

?

No, I do not want to take back my agreement.
if the unprimed frame is stationary the equation is still t' = t γ

This is correct. In other places, my threads would be terminated for this. I agree 100%.

Then why did you say "If the unprimed frame is stationary as the article claims, then t' = t/γ" in post #127? Was it a typo?

No need, I agree. But, I want to make sure, by LT invertibility, in this case both frames calculate t = t' γ, is that your position?

I don't get why you say "both frames calculate" because it's not like there's one derivation from the perspective of the primed frame and a different derivation from the perspective of the unprimed frame, you must use both frames to derive the equation. But yes, as long as the clock is in the primed frame and we are looking at the time between two events on the clock's worldline, then the equation will always be t = t' γ

 

[JesseM]

Oh, I was not saying you must take one frame in that sense. I was saying you must take a frame as stationary to calculate time dilation.

Is this false?

If "must" means you think it has any relevance to the actual mathematical calculation, then your statement is false. The only reason to identify one frame as "stationary" is for the purposes of verbal discussion as Einstein states, it has nothing to do with the mathematical calculations of time dilation. You can do the calculations just fine without ever adopting such a verbal convention.

 

[chinglu1998]

No, but I fail to see the relevance to your statement "when the frame is taken as statonary, objects in that frame have Euclidian measurements. Those in the moving frame have Minkowski."

Let's assume light pulse emitted when origins same.
I cannot know moving frame calculations. All I know is mine as observer at rest with this frame. I use my information and apply LT. Then I know what is in moving frame.

If you think you know what goes on in moving frame without having your light sphere information, then you do not understand SR.

The Minkowski space is so complex, it is not obvious what is going on from the rest frame.

So, this distinction of rest frame and Minkowski frame is crucial.

Then why did you say "If the unprimed frame is stationary as the article claims, then t' = t/γ" in post #127? Was it a typo?

No, not typo. That is what SR says. If unprimed frame is rest frame and clock is in ther other frame which is moving, it beats time dilated or slower. So, t' = t/γ or t'γ = t.

Is this false?

I don't get why you say "both frames calculate" because it's not like there's one derivation from the perspective of the primed frame and a different derivation from the perspective of the unprimed frame, you must use both frames to derive the equation. But yes, as long as the clock is in the primed frame and we are looking at the time between two events on the clock's worldline, then the equation will always be t = t' γ

Again, you are wrong. You supply rest frame data to LT to determine the moving frame conditions.

So, each frame calculates the other based on their own data.

So, "both frames calculate" is correct and consistent with LT.

 

[chinglu1998]

If "must" means you think it has any relevance to the actual mathematical calculation, then your statement is false. The only reason to identify one frame as "stationary" is for the purposes of verbal discussion as Einstein states, it has nothing to do with the mathematical calculations of time dilation. You can do the calculations just fine without ever adopting such a verbal convention.

OK, apply LT without taking a frame as stationary.

Oh, what data will you be using for input? Will you be getting your data from the other frame or yours?

Finally, if there is one clock are you claiming each frame will calculate time dilation for that one clock?

 

[JesseM]

Let's assume light pulse emitted when origins same.
I cannot know moving frame calculations. All I know is mine as observer at rest with this frame. I use my information and apply LT. Then I know what is in moving frame.

If you mean you can only measure things from the perspective of your frame, assuming you are using rulers and clocks at rest in your frame, then I suppose that's true. But you can certainly calculate what the coordinates would be in another frame, just apply the Lorentz transformation to the coordinates in your frame.

The Minkowski space is so complex, it is not obvious what is going on from the rest frame.

So, this distinction of rest frame and Minkowski frame is crucial.

Still don't know what you mean by "Minkowski space", again I have only seen physicists talk about Minkowski spacetime. A 3D spacelike cross-section of 4D Minkowski spacetime consisting of events that are simultaneous in some inertial frame (i.e. a surface of simultaneity) is just a 3D Euclidean space, regardless of which frame's definition of simultaneity is used.

No, not typo. That is what SR says. If unprimed frame is rest frame and clock is in ther other frame which is moving, it beats time dilated or slower. So, t' = t/γ or t'γ = t.

Is this false?

No, but now you say "clock is in ther other frame which is moving", i.e. the primed frame. But before you didn't say anything about the clock being in the primed frame, you just said "If the unprimed frame is stationary as the article claims, then t' = t/γ"--and if you are talking about what "the article claims", the article was talking about a clock in the unprimed frame, not the primed frame. Do you claim there is an error in the article or not? Let's call the unprimed frame "stationary", but also say the clock is in the unprimed frame, as the article does--in that case could you please tell me, yes or no, if you think there is any error in the equation that the article gives for this scenario, namely t' = t γ.

Again, you are wrong. You supply rest frame data to LT to determine the moving frame conditions.

So, each frame calculates the other based on their own data.

So, "both frames calculate" is correct and consistent with LT.

So is your idea of the significance of the "stationary" frame just that it's the data we start out with (based on measurements of clocks and rulers at rest in that frame, presumably), and then have to apply the LT to figure out what the corresponding data would be in other frames? This would be an OK convention to adopt, but you are mistaken if you think this is an established convention among physicists, rather than a novel convention that you invented.

 

[yuiop]

I do not know why I am the only one that can see the error in the article.

It says,

That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly.

How exactly are you going to be an observer moving relative to a clock and perform the calculation?

Let us say "the clock" in this case is the light clock, as the article does not explicitly mention any other clocks. Let us call the rest frame of the light clock the unprimed frame S and an observer A is at rest with the mirrors of the light clock. Another observer (B) moving relative to S is at rest in frame S'. B measures the time between bounces of the light clock using his own clocks. B's clocks can clockwork or digital or any kind of clock he likes. B is moving relative to A, so he is only alongside the A's light clock briefly. The only way to make reliable time measurements is to be right alongside the event when the measurement is made to eliminate complications due to light travel times. A therefore cannot be there at the time the light pulse is emitted and at the time the light returns after the reflection, so B needs an assistant with a synchronised digital clock and they compare notes afterwards. In the primed frame the time measurement is made using two clocks while in the unprimed frame the measurement is made using a single clock. Do you understand that?

If you are an observer in the primed frame, the clock is moving relative to you.

Do you see the logic error in considering the primed frame as the observer frame? An observer frame does not move relative to anything.

This is wrong. If anything is moving relative to the primed frame (e.g the light clock) then the primed frame is equally moving relative to the other object. This is what relativity is all about! There is no sense that one observer (e.g. A) is stationary and the other (e.g. B) is moving or vice versa. A is moving relative to B is the same as B is moving relative to A and neither has an absolute claim to being stationary or moving. That is absolute motion and there is no such thing in relativity, only motion relative to other objects. Right now, sitting at your computer reading this you have multiple relative motions. Your motion relative to your computer is zero, but your motion relative to the Sun or the galaxy or some passing atmospheric molecule is something else and you absolutely no way of knowing whether you are truly at rest or not in absolute way. that is why it is meaningless or at very least confusing to talk of the stationary or moving frame without further clarification.

Then you do not understand SR and you need to start more simply.
The article says,

That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly.

which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself.

As an observer in SR, how do you know you are moving?

The article said "That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly." so it made clear that the frame is moving relative to the clock. There is no ambiguity about relative motion because that is clearly defined.

You on the other hand said "which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself." which does not make it clear what "the moving observer" is moving relative to. You have now introduced an ambiguity and then confuse yourself and ask "how do you know you are moving?". The fact is in relativity you have no way of knowing if you are moving or not, only your motion relative to other objects, that themselves may or may not be moving in any absolute sense.

As Dalespam said, if you say something is moving in relativity, you must always say moving relative to (add something here) otherwise the statement is meaningless. You can also say something is moving from the point of view (add some identifiable observer here). Whether or not an object is moving depends on who is looking at it. That is fundamental to relativity, but you seem to missing that point.

 

[chinglu1998]

If you mean you can only measure things from the perspective of your frame, assuming you are using rulers and clocks at rest in your frame, then I suppose that's true. But you can certainly calculate what the coordinates would be in another frame, just apply the Lorentz transformation to the coordinates in your frame.

Exactly. You can use LT based on your data. So, is is important which frame you operate from, stationary.

Still don't know what you mean by "Minkowski space", again I have only seen physicists talk about Minkowski spacetime. A 3D spacelike cross-section of 4D Minkowski spacetime consisting of events that are simultaneous in some inertial frame (i.e. a surface of simultaneity) is just a 3D Euclidean space, regardless of which frame's definition of simultaneity is used.

It is the space calculated by using your stationary data and input into LT. The result is the Minkowski.

No, but now you say "clock is in ther other frame which is moving", i.e. the primed frame. But before you didn't say anything about the clock being in the primed frame, you just said "If the unprimed frame is stationary as the article claims, then t' = t/γ"--and if you are talking about what "the article claims", the article was talking about a clock in the unprimed frame, not the primed frame. Do you claim there is an error in the article or not? Let's call the unprimed frame "stationary", but also say the clock is in the unprimed frame, as the article does--in that case could you please tell me, yes or no, if you think there is any error in the equation that the article gives for this scenario, namely t' = t γ.

We have mixed features. On the pure WIKI article, it claims the moving frame beats time dilated but then claims the observer is moving. SR does not admit a moving observer unless there is another I guess watching it.

Well, the WIKI article did not have a 2nd observer. Hence, it is claiming an observer moves when it is the only observer, a contradiction.

So is your idea of the significance of the "stationary" frame just that it's the data we start out with (based on measurements of clocks and rulers at rest in that frame, presumably), and then have to apply the LT to figure out what the corresponding data would be in other frames? This would be an OK convention to adopt, but you are mistaken if you think this is an established convention among physicists, rather than a novel convention that you invented.

Yes, this is mine. Please explain additional views. Thanks. If you choose to give a link, I will read.

But, no matter, WIKI is wrong to have 1 observer moving. I hope your link is not like this.

 

[chinglu1998]

Let us say "the clock" in this case is the light clock, as the article does not explicitly mention any other clocks. Let us call the rest frame of the light clock the unprimed frame S and an observer A is at rest with the mirrors of the light clock. Another observer (B) moving relative to S is at rest in frame S'.

I am going to stop you right here. WIKI does not admit 2 observers. There is only 1 observer and that observer is moving.

Can you explain how that 1 observer is moving under SR?

 

[JesseM]

It is the space calculated by using your stationary data and input into LT. The result is the Minkowski.

But your data concerns the coordinates of events throughout spacetime, so if you plug it into LT you don't get a "space", you get another description of spacetime, where for example light from an event forms 4D cones with sides sloped at c just like in your frame.

We have mixed features. On the pure WIKI article, it claims the moving frame beats time dilated but then claims the observer is moving. SR does not admit a moving observer unless there is another I guess watching it.

The wiki article only says the observer is moving relative to the clock, it doesn't say the observer is "moving" in your sense that we have to start from the observations of a "stationary" frame and only later calculate how things would look in a "moving" frame. Again, this rule that we must start with the data from the "stationary" or "observer" frame is your own invention, not one that's a standard notion among physicists.

So is your disagreement with the wiki article not about their equations, but just about the way the wiki started out with data in a frame different from the observer's frame? i.e. you just don't like the order in which the data and equations were presented, but have no objection to the data and equations on their own?

So is your idea of the significance of the "stationary" frame just that it's the data we start out with (based on measurements of clocks and rulers at rest in that frame, presumably), and then have to apply the LT to figure out what the corresponding data would be in other frames? This would be an OK convention to adopt, but you are mistaken if you think this is an established convention among physicists, rather than a novel convention that you invented.

Yes, this is mine. Please explain additional views. Thanks. If you choose to give a link, I will read.

The usual approach is just to talk about how things look in different frames and use the LT to go between them, without any rule that we must start with data in a "stationary" frame or the frame of the "observer". When presenting a problem a textbook can start with how things look in any frame the author wants, even if it isn't the frame of the observer or is the one the author verbally refers to as "moving".

 

[yuiop]

We have mixed features. On the pure WIKI article, it claims the moving frame beats time dilated but then claims the observer is moving. SR does not admit a moving observer unless there is another I guess watching it.

Well, the WIKI article did not have a 2nd observer. Hence, it is claiming an observer moves when it is the only observer, a contradiction.

This is amusing. An observer is only moving if another observer is watching?