Gravitational Time Dilation: A Thought Experiment

[Chris Hillman]

Rindler versus Bell congruences

Of course I don't: they aren't the same problem. They aren't even analogous problems.

View attachment 8788

In this picture, I've drawn three problems.

Problem 1:
On the red coordinates, I've drawn the one-rocket-with-two-clocks problem. The two black lines are the worldlines of the head and tail. The gray area is the worldsheet traced out by the rocket.

Problem 2:
On the green coordinates, I've drawn the two-rockets-and-string problem. The two black lines are the worldlines of the two rockets (assumed to be point-particles). The blue area is the worldsheet tracet out by the string.

Just wanted to point out that these are respectively what I called the Rindler and Bell congruences. As Hurkyl says, these are distinct congruences, which is rather Bell's point. If you know that the Rindler observers are rigid, it follows at once that the Bell congruence must not be rigid! And it is not; the string must eventually break, as Bell said. See the version of the WP article "Bell's spaceship paradox" listed at http://en.wikipedia.org/wiki/User:Hillman/Archive (more recent versions might be better, or might be much worse; several physics Ph.D.s plus myself spent months unsuccessfully trying to persuade a dissident WP editor not to munge the version written mostly by myself and Peter Jacobi).

 

[Chris Hillman]

The actual mistake?

Hi, RandallB,

You haven’t recognized the string with the front attached to a clock in one rocket the back attached to a clock on a following rocket is essentially the same problem as your two clocks in one rocket problem?

Hurkyl is correct; these are two distinct congruences and that is the point of the "paradox". The Rindler congruence is the rigid one; it is the pseudoeuclidean analog of a family of concentric circles, so of course the inner hyperbolic arcs in the Rindler congruence have larger path curvature. The Bell congruence is not rigid; in the version of the WP article "Bell's spaceship paradox" which was written mostly by myself and Peter Jacobi (see http://en.wikipedia.org/wiki/User:Hillman/Archive) I computed the expansion tensor of the Bell congruence, which clearly shows why the string must break.

As to “point at an actual mistake”,

Looks like I may have found the source of the disagreement.

The most straightforward way of disambiguating what we are talking about would be to write down the congruences in question. From Hurkyl's post and diagrams it is clear that he is talking about the Rindler versus Bell congruences, as discussed by Bell. Note that it is mathematically incorrect to claim that the Bell congruence is rigid; it is not rigid, and this can be confirmed by a simple computation of a standard quantity in differential geometry, the expansion tensor of the congruence. So there is no question about what must eventually happen to a string stretched between two Bell observers!

 

[Chris Hillman]

Debunking an urban legend

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.

I'd like to point out that cranks often claim that "the CERN theory group disagrees with Bell", or some such nonsense. What Bell actually said was that after he (re)-discovered the "spaceship and string paradox", he approached various of his CERN colleagues in the lunchroom, and they mostly initially disagreed with his claim that the string would break. But of course, the issue is resolved by computing the expansion tensor of the Bell congruence (and in any case, once you realize that the Bell congruence is obviously distinct from the Rindler congruence, which is rigid, it is not surprising that the Bell congruence exhibits a nonzero expansion tensor or that the string will break). And because the issue is unambiguously resolved by a standard computation (or even by trigonometric reasoning!), it would be absurd to claim that "the CERN theory group" somehow "eternally disagrees" with Bell's conclusion.

(In fact, over at Wikipedia, User:SCZenz happens to be a current member of the CERN theory group, and he was one of the WikiProject Physics members who attempted to assist myself and Peter Jacobi in trying to persuade the dissident to stop "correcting" [sic] the article by replacing correct statements with incorrect ones!)

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.

Exactly. It is really ridiculous that some people refuse to accept that there is no question whatever that Bell's answer is correct: the string must break.

I hope we can begin to wrap up this thread now, since it is beginning to look like a repetition of the apparently endless argument at WP between the dissident editor and the members of WikiProject Physics.

[EDIT: after writing that I read some more of the posts in more detail and noticed more possible misunderstandings, which I described in a few more posts below. These misunderstandings turned out to be of the kind I feared I would find; none of them are new in discussions of the so-called "spaceship and string paradox".]

 

[Chris Hillman]

Uh oh!

Because there is gravitational time dilation between the "far away removed" and the surface of the shell. The calculations are elementary, didn't you know that? Try googling Pound-Rebka.

Maybe nakurusil was just being careless here, but just to be clear: consider an isolated massive thin nonrotating spherical shell with vacuum inside and out. Then the the vacuum interior is locally flat and the vacuum exterior is locally isometric to the Schwarzschild vacuum. So signals from observers on the shell to distant static observers are redshifted. OTH, signals from a shell observer to a static observer in the interior (static wrt the shell) are not redshifted.

 

[Chris Hillman]

Maybe no real disagreement here?

Hi again, RandallB,

Using my approach and seeing the distance between the attachment points as also shorter, you do expected it to break or have slack at one or the other observation point.

Well, if you mean that a string stretched between two Bell observers will eventually break, maybe there is no disagreement between yourself and Hurkyl?

Either I missed something or you didn't write down what congruence you are discussing, so I can only guess what you and Hurkyl are talking about here...

 

[Chris Hillman]

Another possible source of confusion

I worry that your approach is fundamentally misguided -- you seem to be extremely insistant that one chooses their coordinate frame properly, and keeps switching between frames to analyze different parts of the problem. But a choice of coordinate chart is completely aphysical

Here is another possible source of confusion: the modern literature tends to use "frame" to mean "frame field" in the sense of the WP article "Frame fields in general relativity", in the version listed at http://en.wikipedia.org/wiki/User:Hillman/Archive, which is a very different notion from "coordinate chart". A frame field is a "geometric object" which can be specified by writing it down in any coordinate chart; it consists of a set of vector fields which are orthonormal at each event.

It is probably obvious why I resisted being drawn into this: outdated terminology and inappropriate techniques bedevil this kind of discussion. I would repeat that the notion of the expansion tensor is standard in (semi)-Riemannian geometry, and it is exactly what is needed to tell whether something is being pulled apart or not. So it is really quite silly to try to avoid, as some do, dealing with congruences and their kinematical decomposition here (acceleration, expansion, vorticity). This would be analogous to insisting on computing volumes by tricky simplicial approximations--- if you use inappropriate techniques, it is little wonder if you get snarled up in struggling to explain your computations or even get apparently conflicting results. In this case, as far as I can tell, Hurkyl and RandallB seem to agree that a string stretched between two Bell observers must eventually break (which it will), so the apparent disagreement may well be spurious, in which case this thread is a waste of time.

 

[RandallB]

Hi again, RandallB,
Well, if you mean that a string stretched between two Bell observers will eventually break, maybe there is no disagreement between yourself and Hurkyl?
Either I missed something or you didn't write down what congruence you are discussing, so I can only guess what you and Hurkyl are talking about here...

Wow you are having trouble following the problem.
If the distance between the two Bell observer clocks as call them, is viewed as shorter, and the length of the string between them is also shorter, why would you think I'd ever suggest the string would break?

You are correct though, I should have edited that post better – should have had a guest ion there … do you expect it to break or ... using one or the other observation point? But I’m sure Hurkle can tell where I disagree.

As to “congruence” apply it to an F0 observer’s view of the two Bell observer clocks with a string attached between them; but with the F0 frame accelerating and the two clocks remaining stationary. You will still see the string break, which is why I think Bell got this one wrong.

 

[Chris Hillman]

My last attempt to clarify possible misunderstanding between myself and RandallB

Wow you are having trouble following the problem.

IMO, you have been consistently rather rude in this thread, and you have ignored my gentle hints to avoid outbursts of this kind.

As to “congruence” apply it to an F0 observer’s view of the two Bell observer clocks with a string attached between them; but with the F0 frame accelerating and the two clocks remaining stationary.

I don't know what you mean by "FO observer", and because you neglected to write down any congruences, I still don't know if you understand what I mean by "Bell observer" and "Rindler observer" (see Hurkyl's diagram for the gist).

You will still see the string break, which is why I think Bell got this one wrong.

The standard "spaceship and string paradox" involves a string stretched betwen two spacecraft whose worlds lines belong to the Bell congruence, not to the Rindler congruence. In both congruences, the world lines are hyperbolic arcs, i.e. have constant path curvature along each arc, just as circular arcs have constant path curvature in euclidean geometry, but in the Rindler congruence, not all the arcs have the same path curvature, just in the analogous euclidean congruence, a family of concentric circles, the inner circles bend faster than the outer ones.

The standard way to study congruences in (semi)-Riemannian geometry is to compute their kinematic decomposition (acceleration, expansion, vorticity). This shows that the Rindler congruence has zero expansion (we say it is a "rigid" congruence), but the Bell congruence has nonzero expansion.

Physically, this means that a string stretched between two Rindler observers will not be stretched by the relative motion of these observers, which initially strikes many as odd because the trailing Rindler observer is accelerating harder than the leading Rindler observer!

OTH, a string stretched between two Bell observers must eventually break, because the expansion tensor of the Bell congruence shows that positive expansion along the direction in question. IOW, because the trailing Bell observer is accelerating with the same magnitude and direction of acceleration as the leading Bell observer, comparing with the Rindler congruence, we see at once that the string must suffer unbounded stretching, so it must eventually break. This is just Minkowski geometry; only the physical interpretation of Minkowski geometry (due to Minkowski) involves physics.

So this is why it is crucial to understand whether you are claiming that a string stretched between two Bell observers must eventually break (correct), or that a string stretched between two Rindler observers must eventually break (incorrect). Bell was talking about the first scenario, and he correctly concluded that the string will break.

 

[Chris Hillman]

Constant distance?

Hi, MeJennifer

In flat space-time, two completely identical ideal clocks sepearated by a distance l accelerate with a constant proper acceleration a for an identical proper time interval t.

Sounds like you might be trying to describe the Bell congruence, but if so, do you see what is wrong here? (See my immediately previous post, reply to RandallB.)

Spoiler:

Recall that the Rindler congruence is the Minkowski analogue of a family of nested circles. The trailing Rindler observer is accelerating harder, just as an interior circle is bending faster. Concentric circles maintain constant distance along orthogonal geodesic arcs (radii); in the same way, the Rindler observers maintain constant distance along the spatial hyperslices of the Rindler congruence.

But the Bell congruence is analogous to a family of circular arcs, all having the same path curvature, which are all orthogonal to a particular line. You can see that if you form curves everywhere orthogonal to these circular arcs, they do not maintain constant distance. In the same way, if you form the spatial hyperslices for the Bell congruence, the Bell observers do not maintain constant distance along these surfaces.

(I am referring to "pedometer distance" here, the distance computed by integrating arc length along spacelike geodesics. The expansion tensor computation refers to nearby pairs of observers, so that all reasonable notions of distance agree.)

So the situation you described cannot arise in Minkowski geometry.

 

[Chris Hillman]

Bell versus Rindler congruence again

Well if the smart money is against me that must mean I'd get odds on a c-note, I could use an extra grand on getting published! Anyway for this part of the argument, the claim Hurkyl makes is the front "higher" clock in a rocket ship experiences a different rate of time than a clock in the back "lower" end as the ship is accelerating. His "proof" is your post #5...Which is ridiculous, The ship clocks both have a common identical acceleration and will see time pass at the same rate.

Since Hurkyl wrote down the Rindler metric, it seems reasonable to assume he was talking about Rindler observers. If we consider two Rindler observers aligned with the direction of acceleration common to all the Rindler observers, the trailing Rindler observer is accelerating harder.

As far as I can see, the dispute may arise entirely from confusion about whether the standard form of the Bell paradox (aka "spaceship and string paradox") concerns Rindler observers or Bell observers. It is useful to discuss both, but FWIW the "classic" form of the Bell paradox concerns Bell observers, as one would expect from the name. A taut string held between a pair of Rindler observers will remain taut (but will not be stretched by the relative motion of these observers), even though the trailing Rindler observer is accelerating harder. OTH, a taut string held between a pair of Bell observers is stretched at a constant rate by the relative motion of these observers (even though they are accelerating in the same direction and with the same magnitude of acceleration), so it must eventually break, as Bell stated.

I certainly hope this clears up this "dispute" once and for all!

 

[MeJennifer]

Hi, MeJennifer
Sounds like you might be trying to describe the Bell congruence, but if so, do you see what is wrong here? (See my immediately previous post, reply to RandallB.)

Spoiler:

Recall that the Rindler congruence is the Minkowski analogue of a family of nested circles. The trailing Rindler observer is accelerating harder, just as an interior circle is bending faster. Concentric circles maintain constant distance along orthogonal geodesic arcs (radii); in the same way, the Rindler observers maintain constant distance along the spatial hyperslices of the Rindler congruence.

But the Bell congruence is analogous to a family of circular arcs, all having the same path curvature, which are all orthogonal to a particular line. You can see that if you form curves everywhere orthogonal to these circular arcs, they do not maintain constant distance. In the same way, if you form the spatial hyperslices for the Bell congruence, the Bell observers do not maintain constant distance along these surfaces.

(I am referring to "pedometer distance" here, the distance computed by integrating arc length along spacelike geodesics. The expansion tensor computation refers to nearby pairs of observers, so that all reasonable notions of distance agree.)

So the situation you described cannot arise in Minkowski geometry.

Sorry Chris but I have no idea what your objection to this scenario is.
I am simply asking a yes or no answer to a simple scenario. Why you say that such a scenario "cannot arise" is honestly beyond me.

In flat space-time, two completely identical ideal clocks separated by an initial distance l accelerate with a constant proper acceleration a for a proper time interval t. After this time interval each clock stops counting but leaving the final time on their displays.

An observer fetches both clocks and compares the time as displayed on their displays.

Are the readings identical or not?

To me this seems to be a simple yes or no answer.
What am I missing here?

Furthermore there is another scenario that also should have a simple yes or no answer:

In flat space-time, an arbitrary end of a ridgid rod of a length l is accelerated in the direction of the other side with a constant proper acceleration a for a proper time interval t. Two completely identical ideal clocks were placed at each end of this rod with a built in accelerometer. Each clock is individually programmed to start counting as soon as the acceleration starts and to stop counting as soon as the acceleration stops. Once the clock stops it leaves the final time on its display.
An observer compares the time as displayed on their displays.

Are the readings identical or not?

 

[Chris Hillman]

A yes or no answer? That's not possible here...

Sorry Chris but I have no idea what your objection to this scenario is.
I am simply asking a yes or no answer to a simple scenario. Why you say that such a scenario "cannot arise" is honestly beyond me.

To me this seems to be a simple yes or no answer.
What am I missing here?

Try to draw the two euclidean congruences I described, and then try to draw their two Minkowski analogues (the Bell and Rindler congruences) as I described them. Note that circular arcs are the euclidean analog of hyperbolic arcs in Minkowski geometry (that is, these are the constant path curvature curves in these two geometries.)

Your description above assumes a situation which would be impossible in Minkowski geometry. Thus, your question lacks a "yes or no answer" because it posits a situation which cannot in fact arise in Minkowski geometry.

If this is still confusing, try to very carefully and precisely translate what you wrote into euclidean geometry so that you can see that the euclidean analog would describe a situation which cannot arise in euclidean geometry!

In both cases (euclidean and Minkowksi geometry), the scenarios are indeed simple. Nonetheless they are impossible in those geometries, so as you can see, this stuff is just a bit tricky until you have mastered hyperbolic trignometry (the kind appropriate for two-dimensional Minkowski geometry) versus trigonometric trigonometry (the latter is ordinary high school trig, the trig associated with two-dimensional euclidean geometry).

 

[MeJennifer]

Try to draw the two euclidean congruences I described, and then try to draw their two Minkowski analogues (the Bell and Rindler congruences) as I described them. Note that circular arcs are the euclidean analog of hyperbolic arcs in Minkowski geometry (that is, these are the constant path curvature curves in these two geometries.)

Your description above assumes a situation which would be impossible in Minkowski geometry. Thus, your question lacks an answer because it describes an impossible situation. If this is still confusing, try to very carefully and precisely translate what you wrote into euclidean geometry so that you can see that the euclidean analog would describe a situation which cannot arise in euclidean geometry.

I still don´t get it, it is simply an experiment which could be performed. What in this experiment woud be impossible in your opinion.

Say I manufacture two clocks with a simple rocket engine and some automated navigation software, what exactly is `impossible`here?

Also, since we cross posted, what do you think of the second scenario?

 

[Chris Hillman]

I still don´t get it, it is simply an experiment which could be performed.

Actually no, it couldn't---not if you assume that str is correct!

What in this experiment woud be impossible in your opinion.

You wrote

In flat space-time, two completely identical ideal clocks separated by an initial distance l accelerate with a constant proper acceleration a for a proper time interval t.

The problem is that this statement is geometrically self-contradictory. This is not a matter of opinion--- it is provable, at least once we all know what we mean by "distance", "accelerate" and so on.

You can have a congruence in which pairs of timelike curves from the congruence (pairs of world lines of observers) exhibit "acceleration in the same direction with the same constant magnitude, b"; this describes the Bell congruence. Or you can have a congruence in which the timelike curves all have constant path curvature (form hyperbolic arcs), but in which a pair of leading and trailing observer maintain constant (pedometer) distance d throughout; this describes the Rindler congruence. These are two distinct congruences, and the conditions I just noted cannot be simultaneously realized in Minkowski geometry.

To answer your question about two spacecraft : these can either behave like Rindler observers, or they can behave like Bell observers, but they cannot possibly do both! That is why your question has no answer-- the statement of your question assumed they can do both. But they can't.

To repeat: both the Bell and Rindler congruences are made up of hyperbolic arcs (constant path curvature curves). But in the Rindler congruence, trailing observers have larger path curvature (large magnitude of acceleration) than leading observers. The Rindler congruence is "rigid" (vanishing expansion tensor) but the Bell congruence is nonrigid (nonvanishing expansion tensor). A string which is initially stretched taut between two Bell observers will be stretched more and more, until at some point it breaks. OTH, a string which is stretched taut betwen two Rindler observers is neither stretched nor becomes slack; it maintains constant pedometer distance along its length.

I suggested that you look at the euclidean analogues first and make sure you understand why the analogue of the statement you wrote would likewise be impossible in euclidean geometry! This should make it easier to appreciate that this "paradox" is geometrically a triviality. No doubt this accounts for why people often get upset when trying to explain it--- once you "see" the resolution, it can get frustrating when others insist on their incorrect "resolution".

Hope this helps!

 

[MeJennifer]

I must be missing something, please bear with me.

Suppose I have one single clock with a simple rocket and some computerized navigation system.
I programmed this system in such a way that the clock will accelerate with a proper constant acceleration and proper time interval and after that time interval the clock will stop counting and the last recorded time will be shown.
Now would that be possible?

Assumming it is, suppose I build two of those.

Now I place one five meters to the left of me and another one five meters to the right and make sure I stand in the middle and launch them at the same time.

Then afterwards I fetch each clock and check their times.

What exactly is impossible here?

Also, what about my second scenario, is that impossible as well?

Sorry for being slow here but I really do not see the problem.

 

[Chris Hillman]

I must be missing something, please bear with me.

Suppose I have one single clock with a simple rocket and some computerized navigation system.
I programmed this system in such a way that the clock will accelerate with a proper constant acceleration and proper time interval and afterwards it will record the time.

Now would that be possible?

Yes, of course.

Assumming it is, suppose I build two of those.

Also possible, in principle!

Now I place one to the left of me and one to the right and make sure I stand in the middle and launch them at the same time.

"In the middle" could be a problem--- let's say the two craft are initially comoving inertial and then all notions of "distance in the large" should agree, and also "at the same time" should be unambiguous.

Then afterwards I fetch each clock and check their times.

Uh oh! First, you didn't specify how the craft accelerate after you begin the experiment. Second, you didn't specify how you try to compare the elapsed times after the craft have executed specified motions. One direct method would be to make them move some more until they are once again comoving inertial, whereupon we can try to check them via Einstein's synchronization procedure, and will presumably find that they are no longer synchronized, according to Einstein's procedure for comparing ideal clocks carried by comoving inertial observers.

Recall what we said about multiple operationally significant notions of distance. Likewise, there are multiple operationally significant notions of comparing times of clocks carried by distant observers.

What exactly is impossible here?

Please reread what I wrote in my earlier two posts, including the sentence you originally wrote, which I quoted and stated was trying to posit an situation which cannot arise in Minkowski geometry. Please try to very carefully draw the Bell and Rindler congruences and their euclidean analogs. I think you will see what I mean if you keep trying.

 

[MeJennifer]

Uh oh! First, you didn't specify how the craft accelerate after you begin the experiment.

I said constant proper acceleration, see the satement below.
What else would you like to know?

Second, you didn't specify how you try to compare the elapsed times after the craft have executed specified motions.

I did. I explained that after a proper time interval the clocks stop counting and that the last recorded time was visible on the display. See the statement below.


Here is what I wrote so you can acknowledge.

Scenario 1

In flat space-time, two completely identical ideal clocks separated by an initial distance l accelerate with a constant proper acceleration a for a proper time interval t. After this time interval each clock stops counting but leaving the final time on their displays.

An observer fetches both clocks and compares the time as displayed on their displays.

Are the readings identical or not?

Scenario 2.

In flat space-time, an arbitrary end of a ridgid rod of a length l is accelerated in the direction of the other side with a constant proper acceleration a for a proper time interval t. Two completely identical ideal clocks were placed at each end of this rod with a built in accelerometer. Each clock is individually programmed to start counting as soon as the acceleration starts and to stop counting as soon as the acceleration stops. Once the clock stops it leaves the final time on its display.
An observer compares the time as displayed on their displays.

Are the readings identical or not?

 

[Chris Hillman]

Sorry, MeJennifer!

I said constant proper acceleration, see the satement below.

Oh no! I see now that I did misread the sentence I quoted in my previous post (italics added and notation slightly modified):

In flat space-time, two completely identical ideal clocks separated by an initial distance d accelerate with a constant proper acceleration a for a proper time interval t.

I thought you wrote:

"In flat space-time, two completely identical ideal clocks separated by constant distance d accelerate with a constant proper acceleration a for a proper time interval t."

Sorry, sorry, sorry! The latter statement, which conflates the Bell and Rindler congruences, is exactly the mistake apparently made by RandallB in arguing with Hurkyl. Because I was trying hard to explain this to RandallB, I misread the crucial word "initial" in your own post. I apologize profusely--- this thread has already seen more than its fair share of misunderstandings, and I regret having unintentionally added to the obscuring smoke when I meant to increase the level of illumination...sigh...

But there is a larger point here. For those who haven't yet grasped the distinction between the Bell and Rindler congruence:

When one says that two craft "accelerate with a constant proper acceleration a for a proper time interval t", one is saying they belong to the Bell congruence (for acceleration of magnitude a in a certain direction). When one says "constant distance l", one is saying they belong to the Rindler congruence (for acceleration in a certain direction while maintaining rigidity). But a pair of leading and trailing observers (with no separation transverse to the common direction of acceleration) can belong to the Bell congruence, or they can belong to the Rindler congruence, but they can't belong to both.

Be this as it may, I have some remaining objections to your scenarios:

1. In your "Scenario 1" you wrote that a third observer "fetches both clocks and compares the time as displayed on their displays", which I understand to mean that a third craft rendevous's with the trailing and then the leading observer, fetching their clocks, and then someone compares them directly inside the cabin of the third craft. If so, you would need to describe exactly how the clocks are moved to bring them together inside the spaceship of the third observer in order to compare them, before this thought experiment would be well defined.

2. In your "Scenario 2" you wrote "In flat space-time, an arbitrary end of a ridgid rod of a length l is accelerated in the direction", but this is also impossible, since "rigid" bodies cannot be accelerated in str, without becoming nonrigid. More precisely, as the Rindler congruence shows, we cannot "rigidly accelerate" a body by pushing at one end; we need to accelerate different bits by different amounts, so that each bit accelerates like a Rindler observer. This would be a highly artificial situation.

 

[RandallB]

OK, maybe the term uniform gravitational field was a bad choice of words. Let's replace uniform gravitational field with uniform gravitational potential gradient in a specific direction. Now gravitational potential linearly change in that direction, while gravitational acceleration remains constant.

This, IMO, represents the situation of linear acceleration where the "g-meters' on the two clocks, line astern, read the same acceleration, while the gravitational redshift differs. The front clock gains time on the rear clock.

Jorrie

This is the "equivalence" argument Hurkyl referred to in the early posts, I hope he is not still holding to that.
Think through your description here.
Earth has such a "gradient" that is the gravitational potential at the surface is stronger than the potential say a quarter way to the moon. If you had a mountain that tall you would weigh much less there. But you mass has not changed, if you jump down 16 feet can it be done in just one second? How is that smaller force still going to accelerate you at same 32 ft/sec^2 you would have at the surface?

 

[MeJennifer]

1. In your "Scenario 1" you wrote that a third observer "fetches both clocks and compares the time as displayed on their displays", which I understand to mean that a third craft rendevous's with the trailing and then the leading observer, fetching their clocks, and then someone compares them directly inside the cabin of the third craft. If so, you would need to describe exactly how the clocks are moved to bring them together inside the spaceship of the third observer in order to compare them, before this thought experiment would be well defined.

That is not neccesary since the clocks stop as soon as the acceleration is done. So we can simply look at the display to verify the recorded time. Just check it I wrote it in the description.

2. In your "Scenario 2" you wrote "In flat space-time, an arbitrary end of a ridgid rod of a length l is accelerated in the direction", but this is also impossible, since "rigid" bodies cannot be accelerated in str, without becoming nonrigid. More precisely, as the Rindler congruence shows, we cannot "rigidly accelerate" a body by pushing at one end; we need to accelerate different bits by different amounts, so that each bit accelerates like a Rindler observer. This would be a highly artificial situation.

I thought the argument was that in Rindler congruence there would be no stress on the rod while in Bell congruence we would see that the rod tears apart. No?
So you are saying that for instance the commonly used accelerated elevator in space does not show Rindler congruence? Then what does it show, certainly not Bell congruence right?

 

[MeJennifer]

The latter statement, which conflates the Bell and Rindler congruences, is exactly the mistake apparently made by RandallB in arguing with Hurkyl.

Amusing! The reason I presented the two scenarios in the first place was to an attempt to clear up the confusion, something in which I obviously and miserabily failed.

 

[Chris Hillman]

Oh the humanity! The humanity!

Truly, this is like watching an airship wreck!

the situation of linear acceleration where the "g-meters' on the two clocks, line astern, read the same acceleration, while the gravitational redshift differs. The front clock gains time on the rear clock.

The first sentence clearly refers to the Bell congruence (since the leading and trailing observers are said to have the same path curvature). Without knowing exactly how Jorrie intended to compare the time kept by these two observers I can't say whether the second sentence is correct.

That is not neccesary since the clocks stop as soon as the acceleration is done. So we can simply look at the display to verify the recorded time. Just check it I wrote it in the description.

The observer riding in spaceship A can check the time kept by the ideal clock carried in ship A, at any event on A's world line. Likewise, the observer riding in spaceship B can check the time kept by the ideal clock carried in ship B, at any event on B's world line. But when one wants to compare the time kept by these two clocks after the two ships have performed some motions, one needs to specify a procedure. Generally speaking, either the two ships will need to "rendevous" so that they are once again comoving inertial (ideally, even share the same world line!), or else they will need to signal to each other somehow. In both cases, the details of the rendevous manuevering or signalsing (respectively) are absolutely crucial.

I thought the argument was that in Rindler congruence there would be no stress on the rod while in Bell congruence we would see that the rod tears apart.

Yes, although I was using taut strings and trying as far as possible to avoid elastodynamics, the non-Lorentz covariance of Hooke's law and Hookean constitutive relationships, and so on.

So you are saying that for instance the commonly used accelerated elevator in space does not show Rindler congruence? Then what does it show, certainly not Bell congruence right?

Sigh... I confess to tiring of this. I hesitate to say anything until I know exactly what we mean by "the commonly used accelerated elevator in space".

It is probably true that most authors use "uniform field" to mean Rindler congruence, which as we have seen in this thread is potentially confusing. However, it would also be potentially confusing (no pun intended) to use the Bell congruence. A third candidate would be a certain nontrivial Weyl vacuum which happens to be locally isometric to the Minkowski vacuum (take the Newtonian gravitational potential of a uniform mass density ray; the equipotentials look like nested parabolas in a parabolic chart; this generates a Weyl vacuum solution which turns out to be essentially the Rindler vacuum written in a parabolic chart). A fourth candidate would be the Weyl vacuum generated by the Newtonian gravitational potential of a uniform density thin plate, which is not locally flat, but does capture the idea that equipots should be "parallel planes" (because this exact vacuum solution possesses a Lie algebra of Killing vector fields which is isomorphic to e(2), the Lie algebra of the group of planar euclidean isometries).

All of these have multiple properties which are quite different from what one would expect from Newtonian physics. I have written voluminously in the past about the trickiness of "uniform gravitational field" in gtr.

Amusing! The reason I presented the two scenarios in the first place was to an attempt to clear up the confusion, something in which I obviously and miserabily failed.

Yes, I surely do know that sinking feeling (no pun intended)...

 

[RandallB]

IMO, you have been consistently rather rude in this thread,
I don't know what you mean by "FO observer",

I'm RUDE!
I've been told to ignore my own thoughts as I'm not the "Science Adviser", even you pass judgments while explaining you just skimmed the material. No one acknowledges that an Earth style gravitational field is not equivalent to a constant acceleration. And when I suggest accelerating the F0 point, defined by Hurkyl as the observation point and frame from which the graphs are constructed, to explain how that would not produce the same exact graphs as moving the clocks. Not only does no one even look at it, you don't know what I'm talking about because, I don't know, I guess you don't like reading his posts.

I see no point in rewriting and posting what is already being ignored, and I can do without the abuse.
So I'll leave this thread and wish Hurkyl good luck. He at least seemed to put some sincere effort and thought into it, maybe he'll get around to consider all angles and either change his mind or find a more complete solution than these two issues have seen to support Bell's ideas. Either way, by that I mean do better than Bell did. I love his Bell Theorem but I just don't see his work here on these two issues as complete.

I'm Done Here.

 

[Chris Hillman]

I've been told to ignore my own thoughts as I'm not the "Science Adviser", even you pass judgments while explaining you just skimmed the material.

Regarding being "told to ignore my own thoughts", I think you might be misattributing a remark by pervect to myself. I did say that I considered your comment "Wow you are having trouble following the problem" to be a bit rude, as was a remark by another poster (who I won't name), who demanded (of a fourth poster), "Can you read?"

I try to avoid the appearance of making "personal attacks", although this can sometimes be hard to do; no doubt I could have done a bit better myself in this thread, and I apologize if I have offended anyone.

One principle which can be useful: try to choose words which distinguish between criticizing a particular claim made by X, and attacking or being rude to X himself/herself.

Not only does no one even look at it, you don't know what I'm talking about because, I don't know, I guess you don't like reading his posts.

For the record, I have no problem reading Hurkyl's posts!

I think you misunderstood why (much earlier) I emphasized that I was only skimming the thread. In the end I did wind up reading a number of posts more carefully than I intended to, but my desire to avoid spending much time on reading more carefully was, perhaps, understandable if you know a bit of background.

I have explained elsewhere (so you might not have seen these remarks) why I am reluctant to discuss the so-called "spaceship and string paradox" any more: I feel that I have paid my dues amply in another forum. (I don't feel that this would really be a very enlightening exercise as far as physics goes, but anyone wishing to verify said payment of dues may skim the archives of http://en.wikipedia.org/wiki/Talk:Bell's_spaceship_paradox and peruse http://en.wikipedia.org/wiki/Wikipedia:Requests_for_comment/Rod_Ball) .

No one acknowledges that an Earth style gravitational field is not equivalent to a constant acceleration.

Er... I just mentioned that very issue! Hurkyl also alluded to it, if I am not mistaken.

I see no point in rewriting and posting what is already being ignored, and I can do without the abuse.

My own take is that I tried rather hard to play peacemaker here, and I don't think I "abused" anyone, but I regret that you feel put upon, particularly since, as Chekhov put it, "ninety percent of human misery is based upon simple misunderstanding".

Oh well... let's wrap this thread up by saying that it's terribly important to try hard to make sure everyone is talking about exactly the same thing, which generally requires typing in a lot more mathematics than anyone (including myself) was willing to do here.

 

[Hurkyl]

I am getting interested in accelerating clocks.

Let me get this absolutely right:
...
Is that what is claimed?

What you stated is not the problem I was stating about the clocks.

The two-clocks-on-a-uniformly-accelerating-rocket could be explicitly stated in different ways. Probably the simplest is:

(1) We're working in Minkowski space. (which is flat)
(2) It looks the same in any inertial reference frame.
(3) The tail never passes the head.
(4) A light signal emitted from the tail will always reach the head.

From these assumptions, you get a diagram that looks like the first diagram I drew here on the red coordinates.

It also turns out to be true that in this situation:
(5) The proper acceleration of the tail has a constant magnitude.
(6) The proper acceleration of the head has a constant magnitude, which is less than that of the tail.
(7) In any inertial frame, there is an instant when the entire rocket is at rest. At that instant, the coordinate length of the rocket is L. (All inertial frames agree on the value of L)


If we just want to uniformly accelerate for a bit, then stop accelerating, we get a diagram like this:

I've marked proper time on the diagram. (Sorry, I was lazy and stopped marking after the acceleration stopped) The blue line is the extent of the rocket (as measured in what is then the rockets rest frame) that the acceleration stops, and the distance along that blue line is exactly the original length of the rocket.

(Just to emphasize, it is somewhat remarkable that it works out that there exists an inertial frame in which every point of the rocket stops accelerating simultaneously)

As you can see, the rear clock has ticked fewer times than the front clock during the acceleration phase. Additionally, the rear clock has dilated more than the front clock with respect to the red inertial frame, as you can see from the red time coordinate of the third tick of each clock.


The situation you were describing sounds more like you were trying to describe this:

In this picture, both rockets start simultaneously (as measured by the red frame). They both have an acceleration of constant magnitude, and those magnitudes are the same for each. Both rockets shut off after an equal amount of proper time.

As you expected, each rocket reads 4 ticks when it stops. (Of course, in most frames, they did not stop simultaneously) The blue line indicates the instant the tail rocket stops, as measured in the frame in which the rockets eventually come to rest. As you can see, in this frame, the rockets are then very far apart, and that the head rocket reads a much larger time than the tail rocket.

 

[Hurkyl]

OH, but wait you did give three diagrams from the view of an accelerating point without the clocks or string moving at all.
The same three diagrams you gave for holding the observation point stationary. This must prove that string tied to fixed points will break spontaneously! I’ve put up a sting and am waiting for it to break, but no luck so far. Maybe it means that it won’t break unless some accelerating observer goes by and actually looks at the string and its attachment points! But that would mean the moon is not really there unless someone looks at it, and I don’t buy that view either.

Well, maybe I shouldn't reply since you've said you're done with the thread, but I still think it's worth saying.

The equivalence principle only applies to inertial frames -- the laws of physics in a noninertial frame have a different form than they do in an inertial frame. This is a terrific example of that.

(And, once again, measurements are made relative to a coordinate chart... not relative to a point)

 

[pervect]

I'm RUDE!
I've been told to ignore my own thoughts as I'm not the "Science Adviser", even you pass judgments while explaining you just skimmed the material.

Interesting that Randall describes himself as rude - I would say that he has a 'tude (slang for attitude), but he seems to be reasonably polite, he just doesn't listen very well. (I might even add "if at all".)

Probably my main concern is that third parties recognize that RandallB is "following his own thoughts", as he puts it, as opposed to to, for example, doing his best to represent some "standard view" taken from the literature and the textbooks, with references to sources wherever possible.

I wouldn't particularly suggest that RandallB do or not do anything, except perhaps to abide by PF guidelines (not that I have any complaints in that department). He's simply far too independent for me to wish to offer him any such suggestions.

 

[Jorrie]

Potential gradient

Hi Randall:

This is the "equivalence" argument Hurkyl referred to in the early posts, I hope he is not still holding to that.

From my side, I sincerely hope that Hurkyl is holding on to that!

Think through your description here.
Earth has such a "gradient" that is the gravitational potential at the surface is stronger than the potential say a quarter way to the moon. If you had a mountain that tall you would weigh much less there. But you mass has not changed, if you jump down 16 feet can it be done in just one second? How is that smaller force still going to accelerate you at same 32 ft/sec^2 you would have at the surface?

Earth doesn't have a uniform potential gradient, while the hypothetical gravity field with uniform potential gradient in one direction has. So what you said above is not relevant.

Jorrie