A thought experiment

  • #1
actionintegral
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I stand shoulder to shoulder with a group of people, we all face the same direction. We all have synchronized watches and nobody is moving.

We face a second group of people, facing us. They all have synchronized watches and nobody is moving.

Now suppose that my group feels like they are in a gravitational field that runs along the group from left to right. Pretend somehow the other group does not.

Given that the only difference between the two groups is the presence of a gravitational field for one group, how will their watches compare?
 

Answers and Replies

  • #2
lalbatros
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The group in the gravitational field cannot have synchronized watches, assuming each member is at a different potential.
 
  • #3
actionintegral
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The group in the gravitational field cannot have synchronized watches, assuming each member is at a different potential.

That's very interesting. Where can I learn what you just stated?
 
  • #4
cesiumfrog
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Look up tests of general relativity, particularly the ones with atomic clocks at different heights (eg. aboard planes and GPS satellites).
 
  • #5
pervect
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That's very interesting. Where can I learn what you just stated?

For another source, look at "gravitational time dilation", http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

As is predicted by GR, and confirmed by experiments (for instance the Harvard tower experiment section in the above URL, aka the Rebka-Pound experiment), clocks at different heights run at different rates.

Thefore if you synchoronize two clocks at different heights, they don't stay synchronized - because one clock, the higher clock, "ticks faster".

It doesn't matter whether the "different heights" occur because the clocks are in an actual gravitational field, or in the pseudo-gravitational field of an accelerating rocket, the same effecthappens.

To really describe what we mean by "tick faster" and "tick slower" demands that we describe how we compare the clocks. The answer in this case is fairly simple - light signals always have a constant round-trip propagation delay between the two points in question (the high clock and the low clock) as measured by either observer. (The observers won't agree on the propagation delay, but both observers will agree that the round trip propagation delay is constant).

If we assume that the propagation delay in each direction is also constant (given that we observe that the round trip propagation delay is constant) we can simply use light signals to compare the rates of the two clocks, taking into account some fixed propagation delay for the signal. Doing this, we find that the two clocks do not tick at the same rate.
 
  • #6
nakurusil
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That's very interesting. Where can I learn what you just stated?

http://www.citebase.org/fulltext?format=application/pdf&identifier=oai:arXiv.org:gr-qc/9811036 [Broken]
Chris, do you know of a more up to date version?
 
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  • #7
yogi
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Its easy to see why an observer located at the position of the higher clock will conclude that the lower clock ticks slower in the case of an accelerating rocket - successive pulses transmitted at one second intervals by the lower clock (in the rocket tail) will travel an additional distance to reach the front because the higher clock is moving away from the source so each light pulse must travel further - likewise for the lower clock - pulses transmitted by the higher clock will arrive sooner at the lower clock because each pulse travels a shorter distance (the lower clock is rising to meet the pulses) - but in a uniform G field the changing distance explanation isn't applicable - so what physical principle is at work? If the pulses emitted by the lower clock actually take longer to arrive at the upper clock in a uniform G field we have a violation of the tenants of SR. And if they don't what is the physics? The experiments support Einstein's equivalence - so is the effective distance dependent upon the way it is measured relative to the gravitational potentials?
 
  • #8
Hurkyl
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so what physical principle is at work?
Gravitational time dilation, of course.

The result of this experiment is almost literally the definition of what it means for (one particular component of) the curvature of space-time to be nonzero.
 
  • #9
Chris Hillman
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Eh?

http://www.citebase.org/fulltext?format=application/pdf&identifier=oai:arXiv.org:gr-qc/9811036 [Broken]
Chris, do you know of a more up to date version?

Are you asking (me?) if there is a more recent review than the 1998 arXiv eprint by Clifford M. Will? If so, there is a 2003 update at the LRR website
http://relativity.livingreviews.org/Articles/subject.html
 
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  • #10
yogi
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Gravitational time dilation, of course.

The result of this experiment is almost literally the definition of what it means for (one particular component of) the curvature of space-time to be nonzero.

Thanks for your response Hurkyl. What puzzles me is that the mechanism that leads to time dilation in the Schwarzschild solution is not the same as the mechanism that determines the frequency shift for the accelerating rocket.
In the case of the rocket, since acceleration per se does not alter the rate of an ideal clock, it would seem that the frequency shift is observational - so that a later comparison of the two clocks would show they have run at the same rate during the entire acceleration phase even though the lower clock has observed the upper to run at a faster rate and the upper clock has observed the lower to run at a slower rate. In the case of different gravitational potentials, a later comparison of the clocks would show that the lower clock has logged less time
 
  • #12
pervect
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Thanks for your response Hurkyl. What puzzles me is that the mechanism that leads to time dilation in the Schwarzschild solution is not the same as the mechanism that determines the frequency shift for the accelerating rocket.
In the case of the rocket, since acceleration per se does not alter the rate of an ideal clock, it would seem that the frequency shift is observational - so that a later comparison of the two clocks would show they have run at the same rate during the entire acceleration phase even though the lower clock has observed the upper to run at a faster rate and the upper clock has observed the lower to run at a slower rate. In the case of different gravitational potentials, a later comparison of the clocks would show that the lower clock has logged less time


I don't see why or how you could think this.

An example might help. Suppose we have a rigid (Born rigid) rod, and we acclerate the tail end of the rod at 1g for a proper time of one minute (i.e. a time of one minute as measured by a local clock at the tail end of the rod) and then we stop accelerating the rod.

The top end of the rod will, in order to satisfy the rigidity conditions, acclerate for a longer time at a lower acceleration (in terms of proper time and proper acceleration). For example, for some particular length of the rod, the top end of the rod will accelerate at a proper accleration of 1/2 g for a proper time of two minutes.

A rough space-time diagram of the situation can be seen at

http://en.wikipedia.org/wiki/Image:Bell_observers_experiment2.png

The lower magnitude of the acceleration of the second spaceship isn't really drawn correctly on this diagram though, I'm recylcling this diagram from a different application, namely Bell's spaceship paradox, which is closely related.


Putting it another way: if the spaceships acclerate at the same rate, a string connecting them will break (they won't maintain the same distance).

If you demand that the spaceships maintain the same separation (born rigid motion) the lead spaceship has to accelerate at a lower rate so that it doesn't break the string.


You might also try the article by H Nikolic aka Demystifier at http://arxiv.org/abs/physics/9810017
 
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  • #13
Hurkyl
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Thanks for your response Hurkyl. What puzzles me is that the mechanism that leads to time dilation in the Schwarzschild solution is not the same as the mechanism that determines the frequency shift for the accelerating rocket.
Actually, "uniformly accelerating rocket" and "tower in uniform gravitional field" are geometrically identical -- they are the same situation represented in different coordinate charts.


In the case of the rocket, since acceleration per se does not alter the rate of an ideal clock, it would seem that the frequency shift is observational - so that a later comparison of the two clocks would show they have run at the same rate during the entire acceleration phase even though the lower clock has observed the upper to run at a faster rate and the upper clock has observed the lower to run at a slower rate.
I assume you are doing an SR-style analysis in an inertial reference frame? You are right that acceleration doesn't affect the (coordinate) rate of a clock. But (coordinate) speed does: the lower clock has been travelling with a greater speed than the upper clock throughout the entire journey.

A quick heuristic way to see this is that, as measured by this inertial frame, the rocket is length-contracting as it speeds up. That can only happen if the front of the rocket moves more slowly than the tail of the rocket. (As pervect has pointed out)

Since the lower clock has been moving faster, it's been running more slowly than the upper clock, and happily we have agreement with the gravitational case:
In the case of different gravitational potentials, a later comparison of the clocks would show that the lower clock has logged less time
 
  • #14
yogi
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With due respect, I would at this point side with the Cern theoretic group in concluding that the string would not break. But, as Bell points out, those who first reach this conclusion frequently change their mind after further consideration. The analysis depends upon the reality of a physical length contraction in the comoving frame to explain the time difference between the clocks - I have a problem with that - but perhaps the light will be better in the morning.

Regards

Yogi
 
  • #15
Hurkyl
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With due respect, I would at this point side with the Cern theoretic group in concluding that the string would not break. But, as Bell points out, those who first reach this conclusion frequently change their mind after further consideration. The analysis depends upon the reality of a physical length contraction in the comoving frame to explain the time difference between the clocks - I have a problem with that - but perhaps the light will be better in the morning.

Regards

Yogi
Minor correction -- the analysis assumes that for every comoving inertial frame, at the instant1 when the frame is comoving with (some part of) the rocket, the length1 of the rocket is exactly its rest length.

1: as measured in said comoving frame
 
  • #16
nakurusil
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With due respect, I would at this point side with the Cern theoretic group in concluding that the string would not break. But, as Bell points out, those who first reach this conclusion frequently change their mind after further consideration. The analysis depends upon the reality of a physical length contraction in the comoving frame to explain the time difference between the clocks - I have a problem with that - but perhaps the light will be better in the morning.

Regards

Yogi

There is no length contraction in the comoving frame.
 
  • #17
RandallB
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the lower clock has been travelling with a greater speed than the upper clock throughout the entire journey.

A quick heuristic way to see this is that, as measured by this inertial frame, the rocket is length-contracting as it speeds up. That can only happen if the front of the rocket moves more slowly than the tail of the rocket. (As pervect has pointed out)
No that would mean someone traveling with the rocket could physically see the rocket change length. That can’t happen, the only place that sees the length of the rocket change is some other reference frame not comoving with the accelerating frame of the rocket both the front, back and those inside the rocket all moving in a common frame.
The key that even Bell himself missed here is IMO really quite simple in principle:
How do you figure out when the front of the rocket starts to move, so that is starts at the same time as the back of the rocket? And no, that does not happen as a result of there being attached to each other! The structure that separates the two points is not rigid enough to serve as a true “ridged” connection, any normal structure will compress at these forces, so each point must be accelerated independently and at an appropriate simultaneous coordination.
If done correctly the simple part is recognizing that any outside observer in a frame other then the rocket ship should be able to make two observations:
First, the distance of space between point A and B. This will always decrease for other frames and by the appropriate gamma function.
Second the length of the string attached to points A and B. Again this will decrease for other frames and by the appropriate gamma function.
Now by what logic could anyone rationally explain how the gamma function that predicts the change in distance and length would not be identical for both!
IMO Bell got this wrong, and if he were still available I’d ask him to provide the formula to detail the differences between the functions to describe length vs. distance. I think asking him that would have been be enough for him to be convinced the string would not break as well.

The hard part is seeing at what times the two ends should start, there is no reason to expect the start time to be coordinated by the rocket ship clock. The relativistic problem of simultaneity is still real and we should not expect the front and back clocks to remain synchronized once acceleration starts.
So how do we establish the correct start time for each end?
We will have to find a proper “preferred frame”!
This will NOT violate the SR rule against there being A Preferred Frame. Because once we establish such a “local preferred frame” to get our ship started by, we would not be able to coordinate that frame with a similarly defined “local preferred frame” for some rocket ship some 10 billion light years away. It would be nice if we could, but Hubble expansion demands that the two “local preferred frames” cannot be a common frame. Therefore, the idea of no universal preferred reference frame is still preserved as correct
 
  • #18
nakurusil
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No that would mean someone traveling with the rocket could physically see the rocket change length. That can’t happen, the only place that sees the length of the rocket change is some other reference frame not comoving with the accelerating frame of the rocket both the front, back and those inside the rocket all moving in a common frame.
The key that even Bell himself missed here is IMO really quite simple in principle:
How do you figure out when the front of the rocket starts to move, so that is starts at the same time as the back of the rocket? And no, that does not happen as a result of there being attached to each other! The structure that separates the two points is not rigid enough to serve as a true “ridged” connection, any normal structure will compress at these forces, so each point must be accelerated independently and at an appropriate simultaneous coordination.
If done correctly the simple part is recognizing that any outside observer in a frame other then the rocket ship should be able to make two observations:
First, the distance of space between point A and B. This will always decrease for other frames and by the appropriate gamma function.
Second the length of the string attached to points A and B. Again this will decrease for other frames and by the appropriate gamma function.
Now by what logic could anyone rationally explain how the gamma function that predicts the change in distance and length would not be identical for both!
IMO Bell got this wrong, and if he were still available I’d ask him to provide the formula to detail the differences between the functions to describe length vs. distance. I think asking him that would have been be enough for him to be convinced the string would not break as well.

The hard part is seeing at what times the two ends should start, there is no reason to expect the start time to be coordinated by the rocket ship clock. The relativistic problem of simultaneity is still real and we should not expect the front and back clocks to remain synchronized once acceleration starts.
So how do we establish the correct start time for each end?
We will have to find a proper “preferred frame”!
This will NOT violate the SR rule against there being A Preferred Frame. Because once we establish such a “local preferred frame” to get our ship started by, we would not be able to coordinate that frame with a similarly defined “local preferred frame” for some rocket ship some 10 billion light years away. It would be nice if we could, but Hubble expansion demands that the two “local preferred frames” cannot be a common frame. Therefore, the idea of no universal preferred reference frame is still preserved as correct



Here is a mathematical solution to Bell's paradox:

http://www.ph.utexas.edu/~gleeson/NotesChapter13.pdf
 
  • #20
pervect
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With due respect, I would at this point side with the Cern theoretic group in concluding that the string would not break. But, as Bell points out, those who first reach this conclusion frequently change their mind after further consideration. The analysis depends upon the reality of a physical length contraction in the comoving frame to explain the time difference between the clocks - I have a problem with that - but perhaps the light will be better in the morning.

Regards

Yogi

With all due respect, we have only hearsay evidence that the Cern group ever came to such a conclusion. We certainly have no published papers by the Cern group which claim the string does not break. We do, however, have numerous published papers which show that the string does break.

The moral of the story is (or should be) that people do put more care into their published papers than an informal discussion over lunch.

There really isn't any question that according to SR, the string breaks. It's easier to see this if one is capable of following the arguments - but the proof isn't hard, if you can solve a triangle, you can show that the string breaks. My own modest effort is online in the Wikipedia article on the topic (and the talk pages have an endless argument with Rod Ball over this very issue.)
 
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  • #21
yogi
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Pervect - thank you again for the comments -

With regard to the Cern group, there are many exceptions to the hearsay rule - Its a little known point of law that hearsay evidence is admissible if it occurs at lunch after several beers
 
  • #22
Hurkyl
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No that would mean someone traveling with the rocket could physically see the rocket change length. That can’t happen, the only place that sees the length of the rocket change is some other reference frame not comoving with the accelerating frame of the rocket both the front, back and those inside the rocket all moving in a common frame.
Such as the inertial reference frame in which I said the measurement is being made? :tongue:

How do you figure out when the front of the rocket starts to move, so that is starts at the same time as the back of the rocket?
It's not necessary for the statement I made in the post you're quoting... but one typically makes the simplifying assumption that
for every comoving inertial frame, at the instant1 when the frame is comoving with (some part of) the rocket, the length1 of the rocket is exactly its rest length.

1: as measured in said comoving frame
 
  • #23
RandallB
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Such as the inertial reference frame in which I said the measurement is being made? :tongue:

It's not necessary for the statement I made in the post you're quoting... but one typically makes the simplifying assumption that
You will have to stick your tongue out at yourself. Because your inertial reference frame assumptions in addition to having clocks on both ends of the string synchronized also assumes they can somehow track events separated in space as simultaneous! You cannot assume both ends start moving at the same proper time. If you understand SR it should be clear the times on those clocks can not be used to judge events as simultaneous. Just try to define based on those clock times what time they start to accelerate.
Simultaneity is critical to this problem and is not addressed in the treatment linked and referred to by nakurusil & pervect.
If you follow though that treatment rigorously starting from the top instead of the bottom acceleration, you will find the sting would compress or go slack. Again, as most that look at this issue, it fails to consider how to reconcile the simultaneity issue. And without doing that, it is simply incomplete.

All that is needed to be convincing that the string will break is a clear and direct conversion of how from the reference frame from one point on the string or from any other single reference frame the distance between the ends of the string change with respect to the length of the string. It just is not going to happen.
 
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  • #24
Hurkyl
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You will have to stick your tongue out at yourself.
Huh, None of what you wrote makes sense at all. First off, I was not discussing the string problem: I was discussing why, in a uniformly accelerating rocket, the rear clock runs more slowly than the front clock (as measured by any inertial reference frame).

Because your inertial reference frame assumptions in addition to having clocks on both ends of the string synchronized also assumes they can somehow track events separated in space as simultaneous!
The condition I stated in #15 says nothing about there being clocks on the rockets.

At any point along its worldline, a particle determines a comoving inertial reference frame, which is uniquely determined up to spatial orientation. That frame is defined so that that particular point lies at the origin, and the instantaneous coordinate velocity of the particle is zero at that point.

Any inertial reference frame determines a notion of simultaneity, which is invariant under a change of spatial orientation.

Thus, at any point along the worldline of the tail of the rocket, we have uniquely defined notion of simultaneity defined by any inertial reference frame that is comoving with the tail at that point. Using such a comoving frame, we can compute the length of the rocket at t=0. The condition stated in #15 asserts that this length is equal to the rest length of the rocket.

As for the string problem...
You cannot assume both ends start moving at the same proper time.
Sure I can: I can instruct both pilots to start their stopwatches the instant their rocket starts accelerating.

But, more importantly, the problem selects an inertial reference frame, and the problem states that both rockets are to being accelerating at the same time, as measured by that inertial reference frame.

So I can certainly assume both ends start moving at the same coordinate time. (And this is an easy enough condition to arrange in real life)


If you understand SR it should be clear the times on those clocks can not be used to judge events as simultaneous.
Which is why I did not.


Simultaneity is critical to this problem and is not addressed in the treatment linked and referred to by nakurusil & pervect.
Yes it is. Did you look at the picture prevect linked?

http://en.wikipedia.org/wiki/Image:Bell_observers_experiment2.png

You see that slanted dashed line? In any inertial reference frame where both rockets have decelerated to a stop, that dashed line is a line of simultaneity. Of course, the horizontal dashed line is a line of simultaneity of any reference frame where the rockets were initially at rest.

Did you look at the PDF nakurusil linked? The diagram on page 315 also depicts that line of simultaneity.


All that is needed to be convincing that the string will break is a clear and direct conversion of how from the reference frame from one point on the string or from any other single reference frame the distance between the ends of the string change with respect to the length of the string. It just is not going to happen.
I can't figure out if you think the string will break, or if you think it doesn't!
 
  • #25
nakurusil
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There really isn't any question that according to SR, the string breaks. It's easier to see this if one is capable of following the arguments - but the proof isn't hard, if you can solve a triangle, you can show that the string breaks. My own modest effort is online in the Wikipedia article on the topic (and the talk pages have an endless argument with Rod Ball over this very issue.)

Ahh, the whole can of worms. The articles in discussion have been published in Am.Jour.of Physics, not a stellar journal by any extent.
As to "yogi" ' s claim, it appears that the "Cern group" is only one person who makes an armwaving kind of argument (no math, just citations from previous papers): http://arxiv.org/find/physics/1/au:+Field_J/0/1/0/all/0/1
 
  • #26
George Jones
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The articles in discussion have been published in Am.Jour.of Physics, not a stellar journal by any extent.

The American Journal of Physics is a pedagogical journal, not a research journal, but this doesn't mean that publishing there is completely trivial.

In 2000, AJP was looking for a new editor, and, in the March issue, the outgoing editor wrote about what the job entailed, and what the new editor would have to be and do:

American Journal of Physics said:
For one reason or another, AJP turns down 75% of the manuscripts received, and this unfortunately means that you will have to say "No" with some frequency.
 
  • #27
quantum123
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So the AJP does not present new physics, but old stuff in a more educational form? Is it possible that a new breakthrough of physics ideas may accidentally emerge there?
 
  • #28
George Jones
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So the AJP does not present new physics, but old stuff in a more educational form? Is it possible that a new breakthrough of physics ideas may accidentally emerge there?

M. S. Morris and K. S. Thorne, “Wormholes in spacetime and their use for interstellar travel: A tool for teaching General Relativity”, Am. J. Phys. 56, 395 (1988).

This paper stimulated the creation of many papers that were (and still are) published in research journals.
 
  • #29
RandallB
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I was discussing why, in a uniformly accelerating rocket, the rear clock runs more slowly than the front clock (as measured by any inertial reference frame).
The condition I stated in says nothing about there being clocks on the rockets.

Any inertial reference frame determines a notion of simultaneity, which is invariant under a change of spatial orientation.

Thus, at any point along the worldline of the tail of the rocket, we have uniquely defined notion of simultaneity defined by any inertial reference frame that is comoving with the tail at that point. Using such a comoving frame, we can compute the length of the rocket at t=0. The condition stated in #15 asserts that this length is equal to the rest length of the rocket.

As for the string problem...

Sure I can: I can instruct both pilots to start their stopwatches the instant their rocket starts accelerating.

But, more importantly, the problem selects an inertial reference frame, and the problem states that both rockets are to being accelerating at the same time, as measured by that inertial reference frame.
........
I can't figure out if you think the string will break, or if you think it doesn't!
If the distance is the same as the length how could the string break?

String or Rocket how do you it not about telling the pilots to start a stopwatch with the engine - it how do you get them to start all of them the two stop watches and the two engines at the same time.
For your rocket with a clock at both ends starting a uniform acceleration at t = 0 for both can also have none moving clocks reading t = 0 at both ends. But once you establish a significant speed for the rocket will also see the two stationary clocks no longer agree with each other from the rocket’s view so how can you look to either point as a reference to make any judgment about the two moving clocks time simultaneously let alone the rate of each simultaneously. As the rocket looks back from it’s new reference frame seeing the two clocks that were next to the front and back of the rocket are no longer in sync, should question whether or not the front and back of the rocket actually did start at the same time based only on the settings of the two clocks that are now so badly out of sync.

Einstein’s whole point about simultaneity and SR is that although the time of distant points can by synchronized and t = 0 will appear to be simultaneous in that frame; there is no reason to assume that it is in reality simultaneous. Any fixed motion against a preferred frame that can correctly establish events as simultaneous will require that such an assumption is incorrect for the individual frame that has some motion.
None of these solutions address simultaneity as a reality, just an assumption for a local frame. Unless at least a Local Preferred frame and the motion of the rocket or string against it can be established a realistic answer cannot be given. I sure a global preferred frame cannot be established – and I have doubts about an appropriate “Local Preferred Frame” even using CMBR being definable, to solve these problems.

However, since if we did have a uniform vertical gravitational field a higher clock and lower clock would both run the same, I have no doubt the two clocks at either end of an accelerating rocket would run the same as well.
 
  • #30
Hurkyl
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If the distance is the same as the length how could the string break?
Because the length of the string (which is equal to the distance between the rockets) would be longer than the maximum length of the string.

(this length and distance measured in any inertial frame where the rockets are at rest after the experiment)



String or Rocket how do you it not about telling the pilots to start a stopwatch with the engine - it how do you get them to start all of them the two stop watches and the two engines at the same time.
Very easily. I have a particular inertial frame in which I want both rockets to start. So, I give each pilot access to a clock that is synchronized with the coordinate time of my chosen inertial frame, and I instruct them to start their engine at a designated time on the clock I provided.

Or, I could compute the midpoint between the rockets, as measured by my chosen inertial frame. Then, I could send a light signal to both rockets, and the pilots will start their engines the instant they receive the signal.

Or any number of different ways -- it's not a hard problem. And for the thought experiment, it's not even a necessary problem: it doesn't matter if we are able to guarantee a simultaneous launch (as measured by the chosen inertial frame) -- all that matters is that a simultaneous launch is possible.


But once you establish a significant speed for the rocket will also see the two stationary clocks no longer agree with each other from the rocket’s view so how can you look to either point as a reference to make any judgment about the two moving clocks time simultaneously let alone the rate of each simultaneously.
You are the only person in this thread who is talking about using the clocks on the rocket to judge anything. You're arguing against a strawman.


However, since if we did have a uniform vertical gravitational field a higher clock and lower clock would both run the same,
Er, you do realize that GR predicts the clocks run at different rates, and that experimental evidence confirms GR, right? (Such evidence has been provided in this very thread!)
 
  • #31
lalbatros
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Hi,

I have now the feeling that I was wrong in my answer at the beginning of the thread.
I said that synchronisation was not possible.
I have doubt about the meaning of that now.
I would like to know what is the usual meaning for synchronisation.

My doubt came from reading Landau.
By a change of variables, if I understood correctly, it is always possible to bring the component of the metric so that: gi0 = 0 and to also g00 = 1.
Looks like synchronisation in some sense, isn't it?

Michel
 
  • #32
RandallB
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Hurkyl said:
Er, you do realize that GR predicts the clocks run at different rates, and that experimental evidence confirms GR, right? Such evidence has been provided in this very thread!
What are you talking about? You’re the one that called for a uniform gravitational field! Which to be uniform in a vertical direction means it would provide a constant acceleration through the vertical fall. Gravity on earth doesn’t do that right. You do know you weight less at the top of a mountain than you do at the bottom of a valley right. To build a uniform vertical gravitational field you need to collect a large mass of some depth or thickness that extends out flat in all directions to about infinity. Then you will weigh the same no matter how high you are and accelerate at one uniform rate instead of an accelerating acceleration as you get when you pile up your mass in a ball like the earth. And the high clock runs the same as the lower. (No mountains or valleys we need to keep the surface flat.)

As to as to figuring out when things are truly simultaneous (other than co-located events) and being a “straw man”, I’ll leave it as my evidently taking Einstein’s points about simultaneity too seriously for current science and leave it at that.
 
  • #33
Hurkyl
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And the high clock runs the same as the lower. (No mountains or valleys we need to keep the surface flat.)
Er, no?

A coordinate chart where proper time is computed by

[tex]
c^2 d\tau^2 = A^2 z^2 c^2 dt^2 - dx^2 - dy^2 - dz^2,
[/tex]

is one with a constant, uniform gravitational field oriented along the z axis. (A is a constant depending on the strength of the field, and I assumed that z is strictly positive)

Let "up" denote the direction of the positive z-axis.

Clearly, in this chart, the higher clock a clock is, the faster it ticks1. Of course, we'd rather have an experiment than talk about coordinates.

Because the metric is symmetric under time translation, it's easy to see that if we performed an experiment

(1) Start with two stationary1 clocks located at the same place, and simultaneously reset them.
(2) When both clocks read 1 hour, lift clock A 10 meters1.
(3) When clock B reads 2 hours, lift clock B 10 meters1 in the same way that you lifted clock A.

then clock A will have a greater reading than clock B.

Surely this result is inconsistent with whatever you actually mean by the two clocks running at the same speed?


1: As measured in (t, x, y, z)-coordinates.
 
Last edited:
  • #34
pervect
Staff Emeritus
Science Advisor
Insights Author
10,165
1,326
Er, no?

A coordinate chart where proper time is computed by

[tex]
c^2 d\tau^2 = A^2 z^2 c^2 dt^2 - dx^2 - dy^2 - dz^2,
[/tex]

is one with a constant, uniform gravitational field oriented along the z axis. (A is a constant depending on the strength of the field, and I assumed that z is strictly positive)

This is the standard defintion of a uniform gravitational field in the literature. But I thought I should point out that this standard defintion can be confusing, because the accleration measured with an accelerometer as a function of z will not be constant. I.e. if one defines an orthonormal basis of one forms

(A z c) dt, dx, dy, dz

(These differentials are one forms because they associate a scalar, i.e Lorentz interval, with a vector).

the dual of these one-forms gives an ONB of vectors that constitutes a frame field

[tex]\left[ \frac{1}{A z c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right] = \left[ \hat{t}, \hat{x}, \hat{y}, \hat{z} \right][/tex]

and we can compute in this frame-field (a non-coordinate but orthonormal basis) that

[tex]\Gamma^{\hat{z}}{}_{\hat{t}\hat{t}} = -1/z[/tex]

This connection coefficient is what I'm talking about when I say "what an accelerometer will measure".

In a coordinate basis (which is not normal) for comparison we get

[tex]\Gamma^z{}_{tt} = A^2 z[/tex]
 
Last edited:
  • #35
hover
343
0
Hey everyone,
speaking of time dilation what equation can predict how slow a clock will run when it is accelerated??:confused:

thanks!!
 

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