Repositioning clocks: continuing from the switch paradox

[JesseM]

...which is exactly what I have been talking about all along.

But it's not what I have been talking about all along, and I was the one who made the original statement in post #10 that "Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame, then they are out-of-sync by vL/c^2 in the frame that sees them moving at v", which you objected to in post #13 by saying "They are out of sync by $$\gamma(v)\frac{vL}{c^2}$$". Given what I meant by the phrase "out-of-sync by" (which matches what physicists would normally mean by that type of phrase in this context), my statement was perfectly correct.

This is why mathematics is much better than natural languages. I have said all along $$dt=0$$. I did not say $$dt'=0$$.

And I never said your math was wrong, I just said that your objection to the statement "they are out-of-sync by vL/c^2" was misguided, since the statement doesn't concern events that have a time difference of dt=0 but rather events that have a time difference of dt'=0. If you acknowledge natural language is ambiguous, maybe you shouldn't be so quick to raise objections to natural language statements that you don't understand, and should instead ask questions to try to clarify what the accepted terminology is.

 

[starthaus]

two clocks are a distance L apart and synchronized in their mutual rest frame,

...meaning $$dt=0$$

 

[starthaus]

And I never said your math was wrong, I just said that your objection to the statement "they are out-of-sync by vL/c^2" was misguided, since the statement doesn't concern events that have a time difference of dt=0 but rather events that have a time difference of dt'=0. If you acknowledge natural language is ambiguous, maybe you shouldn't be so quick to raise objections to natural language statements that you don't understand, and should instead ask questions to try to clarify what the accepted terminology is.

If this is what you wanted to say, then you needed to phrase it precisely.
Two events separated by $$dt,dx$$ in frame F are perceived simultaneously in frame F'. What is their time separation in frame F?

$$0=dt'=\gamma(dt-vdx/c^2)$$

so:

$$dt=vdx/c^2$$

The observer in frame F' does not "measure" a de-synchrnization of $$vdx/c^2$$. The observer in F' only infers a de-synchronization from the fact that the events appeared simultaneous in F'.

 

[JesseM]

two clocks are a distance L apart and synchronized in their mutual rest frame,

...meaning $$dt=0$$

No, it doesn't mean that, not if we aren't considering a pair of events that are simultaneous in the clocks' rest frame, like the pair "front clock reads T=0" and "back clock reads T=8". If the clocks are synchronized in frame F, then for these two events dt=8. It's only meaningful to give a value of dt when you have first specified what two events you want to calculate the time-interval between.

If this is what you wanted to say, then you needed to phrase it precisely.

I phrased it in the standard way that other physicists phrase similar claims. Again, you might have asked what I meant before jumping into accuse me of saying something incorrect. Anyway, I explained almost immediately what I meant in more precise language, like this from post #16:

But that's not telling us the amount they're out-of-sync in the observer's frame! To find that, we need to find two events on each worldline that are simultaneous in the observer's frame, and then find the time interval between them in the rest frame of the clocks (since the observer judges how out-of-sync the clocks are by looking at what each clock reads at a single moment in his own frame).

In the observer's frame, the clocks are only separated by a distance of L/gamma(v), so suppose in the observer's frame the two simultaneous readings occur at these coordinates:

x=0, t=0
x=L/gamma, t=0

In that case the time of the first event in the clock frame is t'=0, while the time of the second event is:

t' = gamma*(0 - vL/gamma*c^2) = -vL/c^2

I don't think this is at all ambiguous, if you had been paying attention rather than rushing to object you might have seen that I was just using "out-of-sync in the observer's frame" in a way that didn't match how you interpreted this phrase, so that your objection was a purely semantic one.

Two events separated by $$dt,dx$$ in frame F are perceived simultaneously in frame F'. What is their time separation in frame F?

$$0=dt'=\gamma(dt-vdx/c^2)$$

so:

$$dt=vdx/c^2$$

The observer in frame F' does not "measure" a de-synchrnization of $$vdx/c^2$$. The observer in F' only infers a de-synchronization from the fact that the events appeared simultaneous in F'.

Again you seem to be assuming a certain use of language which doesn't match how physicists normally talk. It is standard in relativity to talk about anything that is true in one frame to be what is "measured" by an observer at rest in that frame (for example, the bottom of p. 1 of http://web.mit.edu/sahughes/www/8.022/lec11.pdf earlier saying "Suppose there is a clock at the planet P synchronized in S with Homer's clock at Earth. In S', these clocks are unsynchronized by the amount L0V/c^2.") In any case I didn't actually use the word "measure" to describe how out-of-sync the clocks are in frame F', for example my original post #10 which you objected to just said "Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame, then they are out-of-sync by vL/c^2 in the frame that sees them moving at v."

 

[starthaus]

No, it doesn't mean that, not if we aren't considering a pair of events that are simultaneous in the clocks' rest frame, like the pair "front clock reads T=0" and "back clock reads T=8". If the clocks are synchronized in frame F, then for these two events dt=8.

Synchronized means showing the same exact time, i.e. $$dt=0$$. The above does not qualify. What you mean is "running at the same rate" (which is trivial, since the clocks are at rest wrt each other). This is why natural language is not an appropriate language for physics.

In any case I didn't actually use the word "measure" to describe how out-of-sync the clocks are in frame F', for example my original post #10 which you objected to just said "Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame,

...which, once again, means $$dt=0$$ (see above)

then they are out-of-sync by vL/c^2 in the frame that sees them moving at v."

I think we've beaten this one to death.

 

[JesseM]

Synchronized means showing the same exact time, i.e. $$dt=0$$.

They do always show the same time at any given time in F. At t=0, the trailing clock reads T=0, and the leading clock reads T=0. At t=8, the trailing clock reads T=8, and the leading clock reads T=8. That's what any physicist would understand the phrase "synchronized in frame F" to mean--that if you compare their readings at a single time in frame F, both clocks show the same reading at this time. Do you interpret "synchronized in frame F" to mean something different?

If not, just note that in the equation dt' = gamma*(dt - v*dx/c^2), dx and dt can stand for the distance and time interval between any two events I choose, so if I choose event #1 to be "the leading clock reads T=0" and I choose event #2 to be "the trailing clock reads T=8", then since event #1 occurs at t=0 in frame F and event #2 occurs at t=8 in frame F, the dt for this specific pair of events is dt=8, despite the fact that the events both occur on the worldlines of two clocks which are "synchronized in frame F" according to the definition above.

 

[starthaus]

They do always show the same time at any given time in F. At t=0, the trailing clock reads T=0, and the leading clock reads T=0. At t=8, the trailing clock reads T=8, and the leading clock reads T=8. That's what any physicist would understand the phrase "synchronized in frame F" to mean--that if you compare their readings at a single time in frame F, both clocks show the same reading at this time. Do you interpret "synchronized in frame F" to mean something different?

No, what gives you the idea?

If not, just note that in the equation dt' = gamma*(dt - v*dx/c^2), dx and dt can stand for the distance and time interval between any two events I choose, so if I choose event #1 to be "the leading clock reads T=0" and I choose event #2 to be "the trailing clock reads T=8", t

I chose "trailing clock AND leading clock at the same time T", hence $$dt=0$$

 

[JesseM]

No, what gives you the idea?

OK, so do you agree that for the synchronized clocks I described, the event "leading clock reads T=0" occurs at time t=0 in frame F, and the event "trailing clock reads T=8" occurs at time t=8 in frame F, so between this specific pair of events we have dt=8, and this doesn't contradict the fact that the two clocks are "synchronized" in frame F?

I chose "trailing clock AND leading clock at the same time T", hence $$dt=0$$

I'm not talking about the events you chose, I'm talking about the events "leading clock reads T=0" and "trailing clock reads T=8". Do you disagree that in frame F dt=8 for these events, and in frame F' dt'=0 for these events? If you don't disagree with that, do you disagree that according to the definition of "unsynchronized by" I gave above, it is correct to say that in frame F' the two clocks are "unsynchronized by 8 seconds", since in frame F' the event of the leading clock reading T=0 occurs simultaneously with the event of the trailing clock reading T=8?

You may not like this definition of "unsynchronized by", but that's just a semantic quibble, and I already showed that physicists tend to use "unsynchronized" in the same way. So please tell me if you disagree that according to this definition, it is correct to say that the clocks are unsynchronized by 8 seconds in frame F'.

 

[starthaus]

OK, so do you agree that for the synchronized clocks I described, the event "leading clock reads T=0" occurs at time t=0 in frame F, and the event "trailing clock reads T=8" occurs at time t=8 in frame F, so between this specific pair of events we have dt=8, and this doesn't contradict the fact that the two clocks are "synchronized" in frame F?

I'm not talking about the events you chose, I'm talking about the events "leading clock reads T=0" and "trailing clock reads T=8". Do you disagree that in frame F dt=8 for these events, and in frame F' dt'=0 for these events? If you don't disagree with that, do you disagree that according to the definition of "unsynchronized by" I gave above, it is correct to say that in frame F' the two clocks are "unsynchronized by 8 seconds", since in frame F' the event of the leading clock reading T=0 occurs simultaneously with the event of the trailing clock reading T=8?

If you insist on this choice then:

$$dt'=\gamma(dt-vdx/c^2)$$

 

[JesseM]

If you insist on this choice then:

$$dt'=\gamma(dt-vdx/c^2)$$

Yes, that's the equation I already mentioned I was using, in post #41 for example:

If not, just note that in the equation dt' = gamma*(dt - v*dx/c^2), dx and dt can stand for the distance and time interval between any two events I choose, so if I choose event #1 to be "the leading clock reads T=0" and I choose event #2 to be "the trailing clock reads T=8", then since event #1 occurs at t=0 in frame F and event #2 occurs at t=8 in frame F, the dt for this specific pair of events is dt=8, despite the fact that the events both occur on the worldlines of two clocks which are "synchronized in frame F" according to the definition above.

And also earlier, in post #24:

Your equation dt' = gamma*v*dx/c^2 is incorrect for those two events, because dt was nonzero for the events I picked. The full Lorentz transformation equation is

dt' = gamma*(dt - v*dx/c^2)

If dt is not zero, you can't reduce this to dt' = gamma*v*dx/c^2. For the two events I chose (the event of the leading clock reading T=0, and the event of the trailing clock reading T=8), dt was not zero, here dt=8.

 

[starthaus]

Yes, that's the equation I already mentioned I was using, in post #41 for example:

Good, because this is exactly the equation I showed early in post 17 in this thread.
Now, you can get a physically measurable result from it by accepting that "synchronized" means $$dt=0$$ or you can get a non-measurable entity by continuing to insist that the "measurement" is done by imposing $$dt'=0$$. Of course, in the latter case, there is no physical measurement that can be done, just a mathematical inference since you have already imposed $$dt'=0$$.

 

[JesseM]

Now, you can get a physically measurable result from it by accepting that "synchronized" means $$dt=0$$

The way physicists talk, "synchronized" is a property of the clocks themselves, not a property of the pair of events on each clocks' worldline we choose to analyze. If at every time coordinate t, both clocks read T=t in frame F (like my example where both clocks read T=0 at time coordinate t=0, and both read T=8 at time coordinate t=8), the clocks themselves are said to be "synchronized" in frame F, and that remains true of the clocks even if we choose to look at a pair of events on each clock's worldlines which occur at different times in F. If you define "synchronized" relative to the pair of events we choose as opposed to the clocks themselves, you are using language in a bizarre way that you shouldn't expect any physicist to understand without a detailed explanation.

Besides, you seem to contradict yourself, since I earlier asked:

They do always show the same time at any given time in F. At t=0, the trailing clock reads T=0, and the leading clock reads T=0. At t=8, the trailing clock reads T=8, and the leading clock reads T=8. That's what any physicist would understand the phrase "synchronized in frame F" to mean--that if you compare their readings at a single time in frame F, both clocks show the same reading at this time. Do you interpret "synchronized in frame F" to mean something different?

And you replied:

No, what gives you the idea?

But you can see that according to the definition of "synchronized in F" I give above, all that matters is that the clocks always show the same reading at any single time in F, that remains true even if we choose to analyze two events on their worldlines which don't occur at the same time in F.

or you can get a non-measurable entity by continuing to insist that the "measurement" is done by imposing $$dt'=0$$. Of course, in the latter case, there is no measurement that can be done.

You don't think it's possible for an observer in F' to measure what both clocks read at a single time t'? If not, you must be using some weird definition of "measure" which again doesn't match the terminology used by physicists. As I said earlier in post #39:

It is standard in relativity to talk about anything that is true in one frame to be what is "measured" by an observer at rest in that frame (for example, the bottom of p. 1 of http://web.mit.edu/sahughes/www/8.022/lec11.pdf says "inertial observers in different frames of reference measure different intervals of time between events")

 

[starthaus]

The way physicists talk, "synchronized" is a property of the clocks themselves, not a property of the pair of events on each clocks' worldline we choose to analyze. If at every time coordinate t, both clocks read T=t in frame F (like my example where both clocks read T=0 at time coordinate t=0, and both read T=8 at time coordinate t=8), the clocks themselves are said to be "synchronized" in frame F, and that remains true of the clocks even if we choose to look at a pair of events on each clock's worldlines which occur at different times in F.

The point is that you don't even have to "look at them" from frame F' "(if you ever accept that "synchronized" means $$dt=0$$).

For chuckles and grins, how do you look at two clocks separated by say a proper distance of 100m whizzing past you at 0.8c? Please explain.I am an experimental physicist and I must confess that I have no clue how one can achieve what you are suggesting.

 

[JesseM]

The point is that you don't even have to "look at them" from frame F' "(if you ever accept that "synchronized" means $$dt=0$$).

It doesn't mean that according to how physicists use the term "synchronized", no, it only means that at the clocks always show the same reading simultaneously in their rest frame. I can provide quotes showing them using it my way rather than your way if you like.

For chuckles and grins, how do you look at two clocks separated by say 100m whizzing past you at 0.8c? Please explain.

The standard textbook procedure for physically defining the meaning of an inertial frame's coordinates involves a grid of measuring-rods and clocks at rest in that frame, with the clocks synchronized according to the Einstein synchronization convention, and with the coordinates of any event being defined in terms of local readings on this system of measuring-rods and clocks. So, suppose F' has a ruler at rest relative to him which defines his x'-axis, and at each marking on the ruler there's a clock which is also at rest in F'. Then if he's observing a pair of clocks moving relative to himself in the -x' direction, he might observe through his telescope that at the moment the leading clock passed the x'=0 light-second marker on his ruler, the leading clock read T=0, while his own clock at rest at that marker read t'=0 seconds. Then he might also observe through his telescope that at the moment the trailing clock passed the x'=6 light-second marker on his ruler, the trailing clock read T=8, while his own clock at rest at that marker read t'=0 seconds. Regardless of how long it takes the light from these events to reach his telescope (wherever he is sitting), because of these local readings on his measuring-rod/clock system, he will assign the event of the leading clock reading T=0 coordinates of x'=0, t'=0 and he will assign the event of the trailing clock reading T=8 coordinates of x'=6, t'=0. That's the standard meaning of how inertial observers "measure" the coordinates of different events in SR, using local readings on this sort of system of measuring rods and clocks.

 

[starthaus]

he might observe through his telescope that at the moment the leading clock passed the x'=0 light-second marker on his ruler, the leading clock read T=0, while his own clock at rest at that marker read t'=0 seconds. Then he might also observe through his telescope that at the moment the trailing clock passed the x'=6 light-second marker on his ruler, the trailing clock read T=8,

You mean the observer manages to swivel the telescope instantaneously such that he can observe both clocks simultaneously (as mandated by your condition $$dt'=0$$) while they whizz by him at 0.8c? And with no motion blur such that he can read the clock faces?

 

[Aaron_Shaw]

You mean the observer manages to swivel the telescope instantaneously such that he can observe both clocks simultaneously (as mandated by your condition $$dt'=0$$) while they whizz by him at 0.8c?

I'm sure one could manufacture a device like a telescope which is able to observe in two different directions at once.

 

[starthaus]

I'm sure one could manufacture a device like a telescope which is able to observe in two different directions at once.

Indeed. What about the motion blur? You think that they can do something about the 0.8c?
You must also solve the problem that the two light paths from the two clocks aren't equal, so you will need to adjust the clock readings (that is, after you manage to de-blur the images you are getting at 0.8c).

 

[JesseM]

You mean the observer manages to swivel the telescope instantaneously such that he can observe both clocks simultaneously (as mandated by your condition $$dt'=0$$) while they whizz by him at 0.8c?

Well, for one thing the observer might be at different distances from the x'=0 mark and the x'=6 mark, in which case the light from the two events wouldn't reach him at the same time. But you can also just imagine there is a camera next to each mark that takes continuous footage of events in the vicinity of that mark, and that each camera is continuously sending data to the observer's computer via cables, or that the observer periodically flies out to the different cameras to see what they've recorded. This is just meant to be a theoretical procedure for defining each frame's coordinates anyway, as long as it'd be possible in principle you don't have to worry about how practical it would be (since in practice you can use other methods like parallex to gauge the distance of different events, and divide distance by speed of light to figure out the time interval between when the event occurred in your frame and when the light from the event actually reached you).

 

[starthaus]

Well, for one thing the observer might be at different distances from the x'=0 mark and the x'=6 mark, in which case the light from the two events wouldn't reach him at the same time. But you can also just imagine there is a camera next to each mark that takes continuous footage of events in the vicinity of that mark, and that each camera is continuously sending data to the observer's computer via cables, or that the observer periodically flies out to the different cameras to see what they've recorded.

...and how does the observer ensure that the snapshots are simultaneous? How does he physically realize the condition $$dt'=0$$?

This is just meant to be a theoretical procedure for defining each frame's coordinates anyway, as long as it'd be possible in principle you don't have to worry about how practical it would be

In my profession, I do.

 

[JesseM]

Indeed. What about the motion blur? You think that they can do something about the 0.8c?

You have to distinguish between theoretical statements about what would be true in idealized experiments which don't violate any fundamental laws of physics, and practical statements about what we might actually measure. In practice we've never been able to accelerate a pair of clocks to relativistic speed to observe how if they are synchronized in their own frame according to Einstein's procedure they will be out-of-sync in our frame, but we know this is a theoretical prediction of SR.

You must also solve the problem that the two light paths from the two clocks aren't equal, so you will need to adjust the clock readings (that is, after you manage to de-blur the images you are getting at 0.8c).

You don't have to worry about light delays if you are using local readings on your own set of synchronized clocks. If I am sitting at position x'=100, and at t'=94 I see the image of an event happening right next to the x'=6 marker on my ruler, and my clock at that marker reads t'=0 in the image, then I will assign that event a time coordinate of t'=0 even though I didn't actually learn about it until t'=94.

 

[JesseM]

...and how does the observer ensure that the snapshots are simultaneous? How does he physically realize the condition $$dt'=0$$?

Again, by looking at local clocks at rest in his frame which are in the immediate vicinity of where the snapshot was taken, clocks which have been previously "synchronized" in the observer's frame according to the Einstein synchronization convention.

In my profession, I do.

Well, things work different in theoretical physics, where physicists often explore the predictions of the laws of physics for situations that would be too difficult to realize as practical experiments for now.

 

[starthaus]

Well, things work different in theoretical physics, where physicists often explore the predictions of the laws of physics for situations that would be too difficult to realize as practical experiments for now.

In other words, the experiment is not physically realizable with the convention $$dt'=0$$ for labeling the clocks as synchronized.

 

[Austin0]

Desynchronization is not dependant on specific events but can be calculated purely as a function of velocity and spatial separation between clocks in their own rest frame. Yes??

If you think it would apply to L as per JM i.e. proper length, then either you are simply wrong or I have to relearn relative simultaneity.

it is the latter

I think I may hold off on any adjustments to my understanding of relative simultaniety.

Given an inertial frame F1 with two clocks separated by 10 units of 1 ls ,,as measured in that frame and an observation frame F2 ...v= 0.8 c
WOuld you question that as observed and measured in F2 that the degree of desynchronization between the clocks in F1 would be 0.8* 10 = (+) or( - ) 8.0 sec. ?

You are maintaining that the desynch would be 1.666667 * 0.8*10 =13.3333 ?

I am not "maintaining". 13.33ns is the correct result.

With your understaning of the Lorentz math it should be blindingly obvious that simply given a relative velocity of 0.8 c it can be stated with certainty that the relative degree of desynchronization between the two frames must be (+) or (-) 0.8 sec per light second of proper spatial separation of clocks . Without further information or recourse.
This applies reciprocally from either frame.

Your 13.33333 sec is only correct for specific locations x',0 and x', 16.6667=dx'= 16.6667 . I t has no general truth regarding the desynchronization between the frames and is only correct at all because you have idiosyncratically turned the conditions as stated upside down and insisted on working from the other frame.
I can accept correction when warranted but I fail to discern either point or purpose in your exercise as it educates none , apparently not even yourself IMHO

 

[JesseM]

In other words, the experiment is not physically realizable with the convention $$dt'=0$$ for labeling the clocks as synchronized.

It's "physically realizable" in the sense that it could be realized without violating any laws of physics, but it's not practically realizable with present technology, no. But the point is that no experiment comparing two events on clock worldlines from the perspectives of two frames moving at relativistic velocities relative to one another is practically realizable, because we can't really achieve relativistic velocities relative to the Earth. So, it's also not practically realizable to look at two events with dt=0 in the clock rest frame and then show experimentally that the time between these events in a frame moving at relativistic velocity relative to the clocks is dt'=gamma*L*v/c^2.

 

[starthaus]

It's "physically realizable" in the sense that it could be realized without violating any laws of physics, but it's not practically realizable with present technology, no.

Which means that such a definition of synchronization has no practical use.

But the point is that no experiment comparing two events on clock worldlines from the perspectives of two frames moving at relativistic velocities relative to one another is practically realizable, because we can't really achieve relativistic velocities relative to the Earth. So, it's also not practically realizable to look at two events with dt=0 in the clock rest frame and then show experimentally that the time between these events in a frame moving at relativistic velocity relative to the clocks is dt'=gamma*L*v/c^2.

This is incorrect, the synchronization definition employing $$dt=0$$ does not imply any observation of the clocks. All you need to know is that they had been accelerated slowly enough such that they did not get out of synch. For that extent, the clocks may be totally enclosed in arocket, unobservable, you will still be able to determine their desynchroonization in frame F' based only on two things : proper length L and coordinate speed v.

 

[JesseM]

Which means that such a definition of synchronization has no practical use.

So do you also think the Lorentz transformation equation dt' = gamma*(dt - v*dx/c^2) has no practical use? Is there any practical way to verify that for two events that have a dx of 10 light-seconds in our frame and a dt of 8 seconds in our frame, the events would have a dt' of 0 in a frame moving at 0.8c relative to us?

This is incorrect, the synchronization definition employing $$dt=0$$ does not imply any observation of the clocks.

I don't know what "synchronization definition" you are using that's different from mine. My definition had nothing to do with a specific pair of events so I am not considering any particular value of dt, my definition is just that if two clocks show the same reading at every time-coordinate in a given frame, they are "synchronized" relative to that frame. Are you using a different definition?

All you need to know is that they had been accelerated slowly enough such that they did not get out of synch.

"Did not get out of sync" in your frame, or in the rest frame of the clocks? If you accelerate both clocks at the same coordinate acceleration in your rest frame, then both clocks will remain synchronized in your frame, which means according to my definition of synchronization above, they will become out-of-sync in their new rest frame.

For that extent, the clocks may be totally enclosed in arocket, unobservable, you will still be able to determine their desynchroonization in frame F' based only on two things : proper length L and coordinate speed v.

OK, suppose that while the rocket is accelerating up to relativistic speed, both clocks are right next to each other in the middle of the rocket. This means we can know theoretically that they remain "synchronized" in all frames (both our frame and other frames) according to my definition above, as long as they are together. Then once the rocket is coasting, there's a device which automatically pushes one clock slowly along a track towards the front of the rocket, and which automatically pushes the other slowly along a track towards the back (the device is designed to push both clocks at the same slow speed relative to the rocket). In that case, we can determine theoretically that the after reaching the front and back the two clocks should still be "synchronized" in the current rest frame of the rocket (according to my definition above), and that this means that two readings on the clocks which have a separation of vL/c^2 should occur simultaneously in our frame.

 

[starthaus]

So do you also think the Lorentz transformation equation dt' = gamma*(dt - v*dx/c^2) has no practical use?

I never said such a thing, please stop putting words in my mouth and try to stick to the subject being discussed.

Is there any practical way to verify that for two events that have a dx of 10 light-seconds in our frame and a dt of 8 seconds in our frame, the events would have a dt' of 0 in a frame moving at 0.8c relative to us?

No, there isn't. You, yourself have admitted earlier on that there isn't any practical way.

I don't know what "synchronization definition" you are using that's different from mine.

You are using $$dt'=0$$ which , according to your own admission is impossible to implement practically.
I am using $$dt=0$$ which is very easy to implement (and is implemented in prcatice).

OK, suppose that while the rocket is accelerating up to relativistic speed, both clocks are right next to each other in the middle of the rocket.

This is not the original definition of the problem and has nothing to do with the original experiment. In the original experiment the clocks start separated by the distance L. Please stick with the experiment under discussion.

 

[JesseM]

I never said such a thing, please stop putting words in my mouth and try to stick to the subject being discussed.

I didn't put words into your mouth, I asked about what your position on this is.

No, there isn't. You, yourself have admitted earlier on that there isn't any practical way.

So, do you think we can ever verify that the equation dt' = gamma*(dt - v*dx/c^2) is correct practically in a case where we pick two events with dt not equal to zero? If not, does this equation have any "practical use" if we already know dt' = gamma*v*dx/c^2 applies in cases where dt=0?

You are using $$dt'=0$$ which , according to your own admission is impossible to implement practically.
I am using $$dt=0$$ which is very easy to implement (and is implemented in prcatice).

I don't know what you mean by "implement". It is not possible in practice to ever actually measure any value for dt' if the primed frame is moving at a relativistic speed relative to us, since in practice we can't actually get a system of rulers and clocks moving at relativistic speed relative to us. But if you allow theoretical conclusions based on knowledge of the time between events dt in our frame, like the conclusion that two events with dt=0 in our frame must occur with a separation of dt'=gamma*v*dx/c^2, then there's no justification for not allowing it for events with nonzero dt. For example, you don't even need to accelerate clocks at all, you can just consider two clocks at rest in F which are 10 light-seconds apart and synchronized in F. In this case if you define your events to be "left clock reading T=0" and "right clock reading T=8", then here dx=10 and dt=8--since these clocks are at rest relative to us it isn't hard to measure their positions and times. And then as a theoretical conclusion, we know it must be true that dt' = gamma*(dt - v*dx/c^2), so if the primed frame has v=0.8 this implies dt'=0 for this particular pair of events.

This is not the original definition of the problem and has nothing to do with the original experiment.

In what post was an "original experiment" specified?

In the original experiment the clocks start separated by the distance L.

I don't remember any post that said they "start" separated by L, it was just said that the clocks are at a distance of L in their rest frame F and are moving at v relative to the observer's frame F'. In my example this will be true as soon as the mechanism moves the clocks to the front and back of the rocket. We can define the "start" as some time when the rocket is already coasting and the mechanism has finished moving the clocks.

If you somehow think we must define the "start" as before the rocket leaves Earth, in this case your example doesn't meet the specified conditions at the "start" either, since at that point the clocks weren't yet at rest in F' which was one of the conditions discussed.

 

[starthaus]

But if you allow theoretical conclusions based on knowledge of the time between events dt in our frame, like the conclusion that two events with dt=0 in our frame must occur with a separation of dt'=gamma*v*dx/c^2, then there's no justification for not allowing it for events with nonzero dt.

Yes, there is. Not a theoretical disproof but a practical one, the measurement in the latter case requires the realization of the condition $$dt'=0$$ in the observer frame F'. By your own admission, this is not realizable.

 

[JesseM]

Yes, there is. Not a theoretical disproof but a practical one, the measurement in the latter case requires the realization of the condition $$dt'=0$$ in the observer frame F'. By your own admission, this is not realizable.

My own claim was about the practicality of measuring dt' in frame F', which is equally impossible in practice for dt'=0 and dt'=gamma*v*L/c^2, because we can't practically get rulers and clocks accelerated to the relativistic speed of F'. Your own argument about dt'=gamma*v*L/c^2 was based on the wholly impractical notion of a rocket being accelerated to a relativistic speed, and then on a purely theoretical calculation of what would be true in F' based on what we know is true in F. If this practically impossible scenario counts as a "realization" of dt'=gamma*v*L/c^2 in your increasingly bizarre terminology, then simply having two synchronized clocks at rest in F and picking two events on their worldlines with coordinates x=0,t=0 and x=10,t=8 in F (like the event of the first clock reading T=0 and the event of the second clock reading T=8) should also count as a "realization" of dt'=0, since a purely theoretical calculation of what would be true in F' also shows that dt'=0 for this pair of events.

Likewise, if you want a "realization" of dt'=0 not just for any pair of events but for a pair of events on the worldlines of clocks actually at rest in F' (as opposed to clocks at rest in F as above), then if "realization" can include the wholly impractical notion of a rocket that can accelerate to relativistic speeds, then my thought-experiment with the rocket that has a mechanism that slowly moves the clocks to either side should also count as just as much of a "realization" as your "realization" of dt'=gamma*v*L/c^2.

 

[starthaus]

My own claim was about the practicality of measuring dt' in frame F', which is equally impossible in practice for dt'=0 and dt'=gamma*v*L/c^2,

This is not true if you assume $$dt=0$$ instead of $$dt'=0$$. The impractical aspect goes away once you replace your condition $$dt'=0$$ with my condition $$dt=0$$.

 

[JesseM]

This is not true if you assume $$dt=0$$ instead of $$dt'=0$$.

Oh really? If F' is moving at 0.8c relative to our frame F, do you have a practical experimental method to verify that dt'=gamma*v*L/c^2 for two events which have dt=0 in F, using clocks that are actually at rest in F' to experimentally measure dt'? Or are you resorting to a purely theoretical calculation of the value of dt' in F'?

 

[starthaus]

Oh really? If F' is moving at 0.8c relative to our frame F, do you have a practical experimental method to verify that dt'=gamma*v*L/c^2 for two events which have dt=0 in F, using clocks that are actually at rest in F' to experimentally measure dt'? Or are you resorting to a purely theoretical calculation of the value of dt' in F'?

Sure I do, I am an experimentalist, remember? All I need to do is to determine v. The beauty of the method is that it works even with variable v. As long as you accept that SR is correct, the determination holds.

 

[JesseM]

Sure I do, I am an experimentalist, remember? All I need to do is to determine v. The beauty of the method is that it works even with variable v. As long as you accept that SR is correct, the determination holds.

So you are just doing a theoretical calculation in F', not actually experimentally measuring dt' with clocks at rest in F'? If so, am I also allowed to just measure dx and dt for two events in F, then do a theoretical calculation to find dt' in F?

 

[starthaus]

So you are just doing a theoretical calculation in F', not actually experimentally measuring dt' with clocks at rest in F'? If so, am I also allowed to just measure dx and dt for two events in F, then do a theoretical calculation to find dt' in F?

No, you are not since your experiment depends directly on $$dt'=0$$.