Gravitational Time Dilation: A Thought Experiment

[Chris Hillman]

Would it help calm things down at all if I said that even professional physicists tend to get confused by the problem of physically interpreting the Lense-Thirring frame-dragging phenomenon? That there are various good reasons for preferring, in this context or that, one of a set of mutually inconsistent analogies between gtr and other classical field theories?

 

[nakurusil]

I

I am not talking about the surface, I am talking about the inside of the shell. Anyway,... how come you say that inside the shell is no gravitational field?

I said gravitational potential, not field. Do you know the difference?
The difference in gravitational potential is what causes the redshift.

In one sentence you say that the gravitational potential inside the shell is null and in another sentence you imply there is a difference in gravitational potential.
Doesn´t add up to me, does it to you?

Can you read? I said that the observed time dilation is due to the potential difference between the surface and the far away source.. So, if you live inside an empty shell and you observe a source up in the sky, you will see its frequency gravitationally shifted.
Your distance from the center of your sphere $$d<R$$ has no influence on the amount of the shift, the altitude of the source $$h$$ above the ground is all that counts.

Somewhow there seems to be a disconnect between what I think and what you think I think.

No question. As a third reminder, this is what you asked:

Do you think there is a gravitational field in the middle?
And if no, how would you explain the time dilation there (relative to someone far away removed from this shell).

And this is precisely what I answered.

 

[MeJennifer]

Ok, well then, let's agree to disagree.

 

[Hurkyl]

I don’t disagree with that, just the way you’ve analyze the problem incompletely.

Point at an actual mistake. I've worked the problem, and got results. If you think the result is wrong, that means I've made a mistake in my work. Show me that mistake.


You're switching to the string problem now, and not the two-clocks-on-one-rocket problem? Are we considering a version wher the rockets eventually stop accelerating, or one where they accelerate indefinitely? I'm going to assume indefinitely.

I can't really make heads or tails of your post. Too many things are simply nonsensical.

Let F0 denote the reference frame in which the rockets are initally at rest.

Impose instead a choice of coordinates that is moving at 0.5c along our flight path.

What does that mean? How can a coordinate chart be "along a flight path", let alone "moving along a flight path"?

I'm going to assume you intend for there to be a relative velocity of 0.5c between F0 and F1.

I would have guessed that you intended for the origin to lie along the path of one of the rockets, but that contradicts what you say next.

At some instant after t = 0 the speed of the rocket, clocks, and string between the clocks will be in that reference frame and speed.

I assume you mean t=0 as measured by F1. The rocket, clocks, and string are always in F1. Did you mean when they are at rest relative to F1?

You speak of some "instant" -- you seem to be assuming that, somewhere, both rockets, the string, and clocks (what clocks? There aren't any clocks in the string problem) are simultaneously at rest, as measred by F1.

That doesn't happen. Why would you think that it does?

(When I work the two-clocks-on-one-rocket problem, I never assumed that the two clocks are ever simultaneously at rest in any frame -- that is a fact I can prove as I work through the problem)

Using the string example and your approach looking back from that frame you would find that the string must be shorter in length

What does it mean to "look back from F1"?

Of course the string was shorter in the past -- that's what we've been trying to tell you: the string can't break in the future unless it was shorter in the past than it was in the future.

and therefore must necessarily have broken before the trip even started!

And this is exactly the opposite of what you just said!

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.

How do I see that? You have neither computed anything, nor suggested how you might have computed something.

Note also that the simultaneous start at t=0 for the two points (x=0) and (x=+rocket length) is no longer simultaneous from this new frame.

Of course. For exactly the same reason, in F1, the two rockets never have the same velocity simultaneously. In particular, they can never be simultaneously at rest. (as measured by F1)

Likewise if you change your choice of coordinates to one where (x=0) is used for the front clock and the back clock is placed at (x=-rocket length) you will get inconsistent results, because your graph and your approach, as most, does not take simultaneity into account. And I see no reason to “favor” anyone of these reference frame choices.

Show me the actual mistake. Point to something and demonstrate that it's mathematically in error.

The problem looks exactly the same in any reference frame. Lorentz transforms are symmetries of Minkowski space -- switching between different inertial frames when studying SR is no different than, for example, rotating a drawing when doing Euclidean geometry.

However, the two-clocks-on-one-rocket has an additional interesting (but aphysical) feature. Drawing the space-time diagram requires one to break the symmetry (since we're drawing the picture in Euclidean space), but the rocket problem is neat beacuse it looks exactly the same no matter how you draw it. (That is, if you use the related problem where the rocket has been accelerating for all times future and past)

To put it a different way, if you went back to my post and everywhere swapped "red" and "green", my analysis is still correct.


Aha! I have a drawing of the string problem. I didn't know where it would be reasonable to draw the string, so I just drew the rockets.

The red frame is the frame in which the rockets are initially at rest. The green frame is some other frame -- it's not precisely 0.5c: I picked a velocity in which it was convenient to program.

The origin of the green frame occurs where the leftmost rocket happens to be at rest in that frame. (plus or minus a pixel or two)

The black lines are the two rockets.


For fun, I've drawn the exact same picture twice. They represent exactly the same problem -- I just chose two different methods of drawing them in Euclidean space.

(I've gotten the drawing slightly off -- but it looks qualitatively the same as the true picture. I can fix it if you really disbelieve)

 

[Chris Hillman]

Point at an actual mistake. I've worked the problem, and got results. If you think the result is wrong, that means I've made a mistake in my work. Show me that mistake.

Uh oh, Bell's "paradox" may be claiming its latest victims! (That is, I quit WP last year because the rules then more or less required me and others to waste endless time arguing with someone who didn't accept standard computations of standard quantities. So I am sympathetic to Hurkyl's frustration here.)

RandallB, I haven't read the posts in question in any detail, in part because I figure I've enormously overpaid my dues in this topic, but I just wanted to suggest that you and Hurkyl might be talking about different things, so the alleged disagreement might be more apparent than real.

In particular, note that there are multiple distinct operationally significant definitions of "distance in the large" which Bell or Rindler observers can use, and also, of course, the Bell and Rindler congruences are significantly different (this is in fact Bell's point). But I think I'll leave it to others to sort out the details of this alleged disagreement.

Oh, and about the issue I like to call "will the real gravitational field please stand up?", this is not easy to explain, much less resolve, in a short space of time, in fact several thoughtful papers in the past few years have focused on this issue. As is so often the case in our subject, it turns out there are various mutually inconsistent analogies which are valuable for different purposes, so it is probably wisest to resist the temptation to claim that my challenge could ever result in a unique quantity stepping forward from the crowd.

 

[RandallB]

Originally Posted by RandallB
Well sure your formula here is nice but your description of "constant" is wrong. The gravitational field here is only constant in the x & y directions (zero), not z.

Right, but Hurkyl did say "oriented along z", so while he could have written a bit less ambiguously, I think he appreciates this point.

What does that have to do with not using a "constant" value in that direction but claiming it is constant. Gravity on Earth does not remain constant with hieght z?

Also, we call them Lorentzian for the formulas that started from his efforts to explain failing to detect the aether due to physical things changing length in the direction they move through it. Long before using them with Minkowski.

not with "an aether in mind".

 

[RandallB]

Point at an actual mistake. I've worked the problem, and got results. If you think the result is wrong, that means I've made a mistake in my work. Show me that mistake.

You're switching to the string problem now, and not the two-clocks-on-one-rocket problem? Are we considering a version wher the rockets eventually stop accelerating, or one where they accelerate indefinitely? I'm going to assume indefinitely.

I can't really make heads or tails of your post. Too many things are simply nonsensical.

Let F0 denote the reference frame in which the rockets are initially at rest.

?? 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? If you don’t get that no wonder I cannot make heads or tails of your complaints. Neither problem calls for stopping, and I didn’t ask for them to stop, just a revelation of an instant in the future when the two were in the frame that moves at 0.5c with respect to your F0 frame.

As to “point at an actual mistake”, here’s the best I can do for you.
Instead of having the two rockets take off at the same time and acceleration from F0; take the point you’re making your measurements from in F0 and accelerate that in opposite direction you were going to send the rocket(s). When you work this problem and graphs your results from that point you will still get the same results of clocks running at different rates or the string breaking (or getting slack). Even though in this example the rocket(s) haven’t even moved, but do appear to be accelerating from the relative measure of your observation point. The clocks, or string, in fact could not do that as they didn’t even move.

IMO the reason that such effects can look correct, come from the same errors that are often made in thinking “The Twins” don’t age differently, that was my first mistake I held with conviction for awhile when starting out in understanding relativity. That taught me to very careful about simultaneity issues. The approaches I’ve seen here do not do that IMO. If you do not agree with that, then there is little chance we will ever agree with each other on this paradox, so we should just end this.

 

[Chris Hillman]

I can't tell from a quick glance if either RandallB or Hurkyl believes the other is arguing that a string stretched between two observers in the Bell congruence would not break (in fact it would), much less whether either is in fact making that claim.

Still trying to spread oil on the troubled waters: it is quite possible you two don't actually disagree about anything, just that some misunderstanding arose, probably from being insufficient precision in describing a thought experiment.

 

[RandallB]

I can't tell from a quick glance if either RandallB or Hurkyl believes the other is arguing that a string stretched between two observers in the Bell congruence would not break (in fact it would),

That just says you need more than a quick glance.
Hurkle is actually talking about a single rocket with two clocks; I see that as including the problem of the string. Provided the whole rocket accelerates uniformly; so that we do not have to deal with "jerk" (acceleration of acceleration) compression oscillations going up and down the material the rocket is made of.
I assume you agree with him that two clocks on one rocket would run at different rates of time; along with somehow defining gravity on Earth as constant as altitude changes and that you both think the string breaks.
As I just said in my prior post if you use an accelerating observation point and frame give you the same results even though the rocket or string doesn’t actually move.
With just a quick glance, you are only contributing or voting your preconceived notions, not putting oil on the water.

As I’ve said, if you guys do not think a more rigorous treatment of simultaneity is called for on these issues we will never agree. If so I’m willing to leave this and simply say that there is a chance that Hurkle is as right as I am, and I might be as wrong as Hurkle is.

 

[Hurkyl]

?? 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?

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

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.


Of course, if you so desired, you could combine the two problems into one grand problem:
Problem 3:
On the blue coordinates, I've drawn the two-rockets-and-string problem where I've treated the rockets as extended objects, rather than point particles.

just a revelation of an instant in the future when the two were in the frame that moves at 0.5c with respect to your F0 frame.

Why do you think such a thing exists in the two-rockets-and-string problem?

take the point you’re making your measurements from in F0

I don't follow. F0 is a frame. Frames assign coordinates to every events. My measurements are the coordinates assigned by F0.

and accelerate that in opposite direction you were going to send the rocket(s).

Although I'm not sure just what you mean, it sounds like you are telling me to measure things with respect to some noninertial "frame" F2. It's not clear to me exactly what noninertial frame: I think you're telling me that the t-axis should trace out the path of a hypothetical uniformly accelerated particle that started at rest at the origin of F0... presumably so that F2's time coordinate coincides with the proper time experienced by the hypothetical particle. But I don't know how you want to assign coordinates to the rest of Minkowski space.

That taught me to very careful about simultaneity issues. The approaches I’ve seen here do not do that IMO. If you do not agree with that, then there is little chance we will ever agree with each other on this paradox, so we should just end this.

Care is good. But I don't understand what you find lacking in my treatment, nor do I see rigor in yours.

Look at this picture again. My biggest "Eureka!" moment for SR is essentially that you don't have to Lorentz transform into the green coordinates in order to figure out the length of the rocket at t=0. Length, as measured by the green frame, is nothing more than the proper length of a segment of a line of simultaneity. That segment is indicated in the middle of the picture. I know how to compute that proper length using the red coordinates, so I never have to bother with a Lorentz transform.

(as always, green text indicates things relative to the green coordinate chart)

This led me to realize that SR is a geometric theory, and it eventually sunk into my head that only Lorentz invariant quantities matter -- everything else is just an artifact of how you're analyzing the problem.

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: it has absolutely no bearing on the problem. If you get right answers by doing it, then good -- but I fear you are completely missing what's happening in SR!

 

[MeJennifer]

I am getting interested in accelerating clocks.

Let me get this absolutely right:

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.
After completion, each clock sends a digitally encoded signal of its value to a remote observer.
After the observer receives both encoded signals he decodes them and concludes that the sent clock readings are not identical?

Is that what is claimed?

 

[RandallB]

My biggest "Eureka!" moment

Well maybe you need another one.
I note you did not draw the diagrams as seen from a single point on F0 from the view of accelerating that point (and its reference frame view of F0) in the opposite direction without moving the clocks or string at all.
Also the fact that you can fit the clocks and the string onto a single diagram as you did is proof enough that the two are related problems.

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.

As to the frame for speed 0.5c of course both clocks must eventually be in that frame, what are you thinking. The only question is at what point and time in that frame is clock A at a speed of zero in relation to that frame, it can only be a single point and time and what time is on clock A as well. Likewise for clock B, it cannot be the same point in the 0.5c frame. So now how do the numbers compare; is there a difference in time in that frame for those two points? What is the distance between the two points in that frame? And how does the time on clock A, compare to clock B?
It takes detail like this to understand the twin’s paradox, and you simply haven’t gone though enough detail to come to a convincing final conclusion on this problem.
You are just repeating the same pattern and answer as before, which IMO and a moving F0 doesn’t work.

 

[Jorrie]

I am getting interested in accelerating clocks.

Let me get this absolutely right:

In flat space-time, two completely identical ideal clocks separated by a distance l accelerate with a constant proper acceleration a for an identical proper time interval t.
After completion, each clock sends a digitally encoded signal of its value to a remote observer.
After the observer receives both encoded signals he decodes them and concludes that the sent clock readings are not identical?

Is that what is claimed?

I have also puzzled about this fact and came to the following conclusion, right or wrong (to be determined ).

What you described is equivalent to two static clocks sitting at different gravitational potentials in a hypothetical uniform gravitation field. They suffer identical gravitational accelerations (due to the uniform field), yet they suffer different gravitational time dilations (or gravitational redshift as measured by a distant static observer), due to their different potentials.

Herein lurks the problem: their identical proper-time intervals (t, that you specified) per definition means that they send the exact same time stamp to the remote observer - yet they did not send it simultaneously... simultaneous in whose reference frame?

Hope it helps, otherwise ignore!

Jorrie

 

[pervect]

I am getting interested in accelerating clocks.

Let me get this absolutely right:

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.
After completion, each clock sends a digitally encoded signal of its value to a remote observer.
After the observer receives both encoded signals he decodes them and concludes that the sent clock readings are not identical?

Is that what is claimed?

Clarify something, please. Is the constant proper acceleration a identical for both clocks? Or do both clocks maintain a constant distance L? The two conditions are, as I have mentioned once or twice :-), different and incompatible.

 

[RandallB]

equivalent to two static clocks sitting at different gravitational potentials in a hypothetical uniform gravitation field. They suffer identical gravitational accelerations

If they are at different gravitational potentials they cannot produce identical gravitational accelerations. You need to construct a constant gravitation field as noted earlier (that means not a round mass like Earth).

 

[MeJennifer]

Ok, let me try to make it as unambigious as possible:

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 completion, each clock sends a digitally encoded signal of its value to a remote observer.
After the observer receives both encoded signals, which he may or may not receive at the same time, he decodes them and concludes that the sent clock readings are not identical?

Is that what is claimed?
And is this the situation of the Bell spaceship paradox?

 

[Jorrie]

Uniform gravitational field

If they are at different gravitational potentials they cannot produce identical gravitational accelerations. You need to construct a constant gravitation field as noted earlier (that means not a round mass like Earth).

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

 

[pervect]

Ok, let me try to make it as unambigious as possible:

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 completion, each clock sends a digitally encoded signal of its value to a remote observer.
After the observer receives both encoded signals, which he may or may not receive at the same time, he decodes them and concludes that the sent clock readings are not identical?

Is that what is claimed?
And is this the situation of the Bell spaceship paradox?

Yes, to the BSP paradox question, this is one common form of the BSP. But onto the original question.

Let's make the comparison process a bit more specific. It's rather vague right now.

An observer, exactly midway between the two spaceships, receives their time readings. Does he receive the same value?

This is more specific, but still not specific enough. While we've specified the position of the observer (he's midway between the two clocks) we haven't yet specified his velocity. This is a key omission, as differently moving observers will answer the question differently.

If the observer is at rest relative to the initial velocity of the clocks, before they started accelerating, he will find that he receives the same encoded signal values.

IF the observer is at rest relative to the _final_ velocity of the clocks, after they are through accelerating, he will not receive the same encoded signal values.

A space-time diagram would be helpful in understanding these results, you might also want to look at

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

 

[MeJennifer]

Let's make the comparison process a bit more specific. It's rather vague right now.

Ok, I apologize for being vague, so let´s pin this down a bit more, I am sure you agree that being vague or beating around the bush does not serve anything.

So,

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?

 

[Jorrie]

... IF the observer is at rest relative to the _final_ velocity of the clocks, after they are through accelerating, he will not receive the same encoded signal values.

Hi Pervect, as you usually say: Huh?

OK, a misunderstanding. MeJennifer specified (paraphrased) "after an identical proper-time t, the two clocks send encoded signals..."

The time stamps on these signals are the same, by definition. As I tried to point out in post #83, they were just not sent out simultaneously in the frame of the observer.

Jorrie

 

[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.