Bell spaceship paradox quantitatively

[vanhees71]

Yesterday, I found the time to write a bit further on my SRT FAQ and wanted to give a quantitative analysis of the Bell space-ship paradox on the example of the two rockets accelerating with constant proper acceleration, and I found a problem, I cannot solve. So I took this section out from my FAQ for the time being.

The problem occurs, when one looks at the situation from the point of view of the heading rocket C and if $\alpha L_A>c^2$, where $\alpha$ is the constant proper acceleration and $L_A$ the constant distance of the space ships in A's reference frame, where the space ships start accelerating from rest simultaneously at time $t=0$. Perhaps you can help me out here. The problem finally seems to be to find a proper, i.e., invariant definition of the distance of the spaceships. I still have to understand the physical meaning of bcrowell's analysis using the time-like congruence of space-like separated hyperbolic-motion hyperbola. This seems to be the only analysis in terms of a frame-independent quantity in the literature.

I've put my analysis here, because for some reason I cannot upload it as an attachment here:

http://fias.uni-frankfurt.de/~hees/tmp/bell-paradox.pdf

Note that the figure in bcrowell's Insights article is depicted in other form as the left-hand panel of the figure in the text. For this case (point of view of the rear space-ship B) no problem occurs. The problem occurs in the situation depicted in the right panel, but that's explained in detail in the text too.

 

[PeterDonis]

The problem occurs, when one looks at the situation from the point of view of the heading rocket C and if $\alpha L_A>c^2$, where $\alpha$ is the constant proper acceleration and $L_A$ the constant distance of the space ships in A's reference frame, where the space ships start accelerating from rest simultaneously at time $t=0$.

This means the rear spaceship starts out behind the front spaceship's Rindler horizon. Note that this will always happen eventually if the ships continue accelerating indefinitely; but the condition you give means it is true from the very start, as soon as the ships start accelerating.

The problem finally seems to be to find a proper, i.e., invariant definition of the distance of the spaceships.

There isn't one. That's true regardless of whether the issue referred to above (rear ship behind front ship's Rindler horizon) is taken into account. We went over this in the previous thread on this topic. Once the ships start accelerating, neither one is at rest in the instantaneous rest frame of the other, so there is no invariant way to define the distance between them.

The only difference when the rear ship is behind the front ship's Rindler horizon is that the rear ship can no longer send light signals to the front ship (it can send light signals towards the front ship, but they will never catch up with the front ship). That means there is no way for the rear ship to use the "radar" method (send a light signal and time how long it takes to return) to determine the distance to the front ship.

 

[vanhees71]

Ok, that explains why in the calculation from the point of view of C the "distance" to B has a limiting value $c^2/\alpha$, but it shows that this is not the right quantity to decide whether the original string breaks, because this is just an artifact of the fact that I cannot measure the distance to B from C, because no light signal can reach C from B at the instant, where I choose the instantaneous restframe or C. So there is no way to measure the distance to B by sending a light signal from C to B and let it reflect such that I can infer the space ships' distance from the point of view of C. Is the correct conclusion than that there is simply no "proper" distance to C that can be defined? But what's then the meaning of the distance $L_C$ I calculated in my writeup in Eqs. (18) and (22)? And is then the distance measure from the point of view of B, which seems to be better behaved (it's continuously increasing during the entire acceleration and always $>L_A$, cf. (14)), a true distance of C from B as measured by B? And, if so, is then the conclusion that the rope must break justified by the fact that this "distance" $L_B$ is increasing and $>L_A$?

So can one solve the paradox as follows: Since at one instant (or even from the very beginning of the acceleration of both ships in frame A) B is behind C's Rindler horizon, one cannot connect the space ships with a rope.

Last but not least is the very short "resolution" of the paradox in Bell's original article in his famous book correct? I ask, because it seems to me that when arguing like just written above, nowhere occurs the notion of length contraction of a "quasi-rigid" rod in this argument.

On the other hand, that would be a much simpler argument and that was the argument I brought forward in this other thread in a wrong say. With the insights I gained now from our discussion, I'd reformulate this argument as follows: Suppose the space ships are connected by a rigid rod before acceleration (which is trivially possible, because the space ships are at rest wrt. the inertial frame A) and suppose then that both space ships starts at the same time precisely accelerating in an equal way such that their distance (as measured at A) stay always the same, can this rod stay intact? The answer is of course no, because as argued in my very first draft of the Bell paradox then I had to assume that in the space time diagram the rod is depicted by the horizontal line, i.e., it connects the two points of the space ships worldlines which are at equal times wrt. A, and then the proper distance of these points would be the length of the rod, assumed to be quasi-rigid, but this is not true, because this "distance" is larger then $L_A$, and becomes arbitrarily large, but this is a contradiction to the assumption that the rod is quasi-rigid and thus no quasi-rigid rod can connect the space-ships. For me, in fact, this seems to be the most unambigous explanation, and it coincides with Bell's resolution of the paradox, and thus Bells' argument is a well-founded analysis of the paradox: $L_A$ simply is not the proper length of a supposed-to-exist quasi-rigid rod connecting the space ships but the length-contracted value. So one has $L_{\text{proper}}=\gamma L_A$, and $\gamma$ becomes arbitrarily large because of the acceleration of the space ships. So a "real rod" must stretch and thus breaks at some time in A.

 

[PAllen]

I'll repeat my take on all this.

1) Proper distance between a pair of events (in SR) is invariant, and there are no ambiguities or conventional aspects.
2) Rest length (often called proper length) of an object is an invariant quantity if you have a well defined object - its constituent parts are in mutual rest. The Herglotz-Noether theorem puts great restrictions on when this is possible for general motion in SR. If you can't achieve mutual rest between constituent world lines, then you don't really have an object and any overall dimension is a conventional quantity dependent on how you choose to define simultaneity, with no preferred option.
3) The behavior of material bodies under general motion is kinematically described by the expansion tensor of the congruence of world lines defining the body. Decomposition of this tensor describes local expansion, shear, and vorticity.
4) Distance between world lines is inherently purely conventional unless the world lines remain at mutual rest. This is simply because there is no unambiguous simultaneity applying to both world lines. An example of such a convention that works for arbitrary motion as long as the world lines are inertial before some event, and after another event (anything can happen in between) is radar simultaneity. This will lead to a consistent pairing of events on world lines where momentarily comoving frames fail for both world lines. However, without the restriction about pre and post inertial motion, even this can fail to provide consistent simultaneity.

You are seeking to reject (4), and you can't succeed in this. Meanwhile, the physics of a string, which can certainly be laid out between rockets that will later lose two way causal connection, is truly defined by (3) - the global relation between the rockets is irrelevant.

 

[PeterDonis]

The behavior of material bodies under general motion is kinematically described by the expansion tensor of the congruence of world lines defining the body. Decomposition of this tensor describes local expansion, shear, and vorticity.

Actually, the expansion tensor decomposes into the expansion scalar (the trace of the expansion tensor) and the shear (the traceless part of the expansion tensor). The vorticity is a separate tensor.

 

[PAllen]

Actually, the expansion tensor decomposes into the expansion scalar (the trace of the expansion tensor) and the shear (the traceless part of the expansion tensor). The vorticity is a separate tensor.

Oops, you're right.

 

[PeterDonis]

Is the correct conclusion than that there is simply no "proper" distance to C that can be defined?

"Proper" distance, as the term is usually used, requires the objects between which the distance is being measured to be at mutual rest, as PAllen pointed out. If they are not, there is no such thing as "proper distance" between them.

Since at one instant (or even from the very beginning of the acceleration of both ships in frame A) B is behind C's Rindler horizon, one cannot connect the space ships with a rope.

Sure you can. The rope will break, but it won't break "instantaneously". It will take time, because, heuristically, it takes time for the information that B is behind C's Rindler horizon, once the ships start accelerating, to propagate through the string (since information can't propagate faster than the speed of light).

Bells' argument is a well-founded analysis of the paradox: $L_A$ simply is not the proper length of a supposed-to-exist quasi-rigid rod connecting the space ships but the length-contracted value.

Yes, it is, as long as you are willing to accept "the length-contracted value" as a "real" thing that can affect the stresses in the rod, even though it is the "length" of the rod in a frame in which it is not at rest. A more comprehensive analysis would justify this by actually computing the stresses in the rod, based on the forces between its atoms.

 

[Mentz114]

Last but not least is the very short "resolution" of the paradox in Bell's original article in his famous book correct? I ask, because it seems to me that when arguing like just written above, nowhere occurs the notion of length contraction of a "quasi-rigid" rod in this argument.

On the other hand, that would be a much simpler argument and that was the argument I brought forward in this other thread in a wrong say. With the insights I gained now from our discussion, I'd reformulate this argument as follows: Suppose the space ships are connected by a rigid rod before acceleration (which is trivially possible, because the space ships are at rest wrt. the inertial frame A) and suppose then that both space ships starts at the same time precisely accelerating in an equal way such that their distance (as measured at A) stay always the same, can this rod stay intact? The answer is of course no, because as argued in my very first draft of the Bell paradox then I had to assume that in the space time diagram the rod is depicted by the horizontal line, i.e., it connects the two points of the space ships worldlines which are at equal times wrt. A, and then the proper distance of these points would be the length of the rod, assumed to be quasi-rigid, but this is not true, because this "distance" is larger then $L_A$, and becomes arbitrarily large, but this is a contradiction to the assumption that the rod is quasi-rigid and thus no quasi-rigid rod can connect the space-ships. For me, in fact, this seems to be the most unambigous explanation, and it coincides with Bell's resolution of the paradox, and thus Bells' argument is a well-founded analysis of the paradox: $L_A$ simply is not the proper length of a supposed-to-exist quasi-rigid rod connecting the space ships but the length-contracted value. So one has $L_{\text{proper}}=\gamma L_A$, and $\gamma$ becomes arbitrarily large because of the acceleration of the space ships. So a "real rod" must stretch and thus breaks at some time in A.

These are quotes from Rindlers book ( 1985)

In relativity, length contraction is a kind of velocity perspective effect ( in Weyl's phrase) - analogous to the visual foreshortening of a stationary rod that is viewed from the back rather than the top. For our world-map of of the moving rod we pick what in the rod's frame are later events at its back than at its front, so that the back appears closer to the front.

(that description sounds like a radar distance measurement. If we reverse 'front' and 'back' we measure the back first then the result is a larger measurement for a rod moving away. Rindler thinks the two are the same but they are not)

But of course nothing at all has happened to the rod itself.

I should think not.

Nevertheless, relativistic length contraction is no 'illusion': it is real in every sense.

Gulp. So its not like perspective at all. Perspective is what makes the distant cows look smaller but they are not.

For example, just as a rotated rod can be placed in the space between two parallel planes which it would not fit perpendicularly, so a moving rod can be momentarily confined in a space it would not fit when at rest.

This seems to mean that the rod must be rotated - but 'nothing at all has happened to the rod itself'.

 

[Mentz114]

I'll repeat my take on all this.

[agree with the snipped items]
4) Distance between world lines is inherently purely conventional unless the world lines remain at mutual rest. This is simply because there is no unambiguous simultaneity applying to both world lines. An example of such a convention that works for arbitrary motion as long as the world lines are inertial before some event, and after another event (anything can happen in between) is radar simultaneity. This will lead to a consistent pairing of events on world lines where momentarily comoving frames fail for both world lines. However, without the restriction about pre and post inertial motion, even this can fail to provide consistent simultaneity.

You are seeking to reject (4), and you can't succeed in this. Meanwhile, the physics of a string, which can certainly be laid out between rockets that will later lose two way causal connection, is truly defined by (3) - the global relation between the rockets is irrelevant.

I agree (emphatically) with all the points you make.

@vanhees71

To reinforce point (4). Measuring the length of an object in the rest frame of the object is equivalent to $L=\int_0^Ldx$. A ruler is placed along the object and the marks on the ruler are added up. If the object is moving wrt to the ruler ( or whatever) then this becomes $L'=\int_0^{L/\gamma}dx'=\gamma \int_0^{L/\gamma} dx + \beta\gamma \int_{t_1}^{t_2}dt$. Notice that $L/\gamma$ appears in the limit of the new spatial part because it is $L$ projected into $dx'$. Obviously $L'$ depends now on the choice of $t_1,t_2$, which are times in the apparatus frame. If these are made equal in the object frame, the result is the rest length $L$.

The number one gets depends on how one sets up the apparatus.

 

[vanhees71]

Thanks a lot for all your answers. Particularly PAllen's four points are very convincing, and I agree with all of them (particularly after my blunder at the last weekend ;-(). Nevertheless, for me the paradox is not satisfactorily solved, because it is a welldefined problem: How does the rope or "rigid" rod connecting the spaceships before they accelerate behave. As PAllen's point 2 tells me that there is no clear definition of a distance between the space ships, because there's no frame of reference where both are at rest after they started accelerating. This means that it is not clear, a priori how to describe this connecting rod. Obviously no quasi-rigid rod can connect the two space ships, because of point 2, and thus a real rod must break, but how do I make this quantitative at this apparently simple example for constant proper acceleration of the two rockets.

Ok, so one needs a model for the concrete realization of the connection of the space ships. What I don't get concerning the ansatz with the congruences, which seems to define strain in a covariant way and thus is so far the best ansatz to clearly resolve the paradox, is, how to choose them to describe this rod. Of course, the endpoints are given by the worldlines of the two spaceships, which should be among the timelike worldlines defining the congruence, because this defines the rod connecting the spaceships, but to describe the kinematics of the rod I need a concrete realization of this rod, i.e., the world tube between the space ships in terms of this congruence, and there seems to be a lot of arbitrariness in this choice. In BCrowell's great SRT book, he chooses simply the spatially translated hyperbola of the proper acceleration, but what's the physical picture about the connecting rod for this particular choice?

 

[Mentz114]

What I don't get concerning the ansatz with the congruences, which seems to define strain in a covariant way and thus is so far the best ansatz to clearly resolve the paradox,

I don't see how something that depends on a choice of measurement apparatus can be a physical effect.

Will you build into your model the details of the measurement apparatus ?

 

[vanhees71]

What do you mean by measurement apparatus? This is just the description of the string or rod connecting the space ship. And you are absolutely right: There must be a covariant description to decide whether this connection breaks of not, but this obviously depends on the concrete thing you use for this connection, i.e., on the elasticity/stability properties of the material. I don't plan to make such a detailed model, which would be an overkill just to resolve such a paradox. I'm content with a purely geometric description, i.e., the kinematics of some simple model, and bcrowell's model seems to be the most natural one: One just assumes the spaceships are somehow connected by a rod described by a horizontal worldline in A's reference frame at each instant of A's time. Then you get bcrowell's choice of the congruences, and the strain of the connection is a well-defined covariant quantity. I just have to work out the details for myself. Then I'll put it in my little SRT writeup.

 

[PAllen]

First, the choice of timelike congruence is constrained to include the rocket world lines, and that it be a congruence - continuous, filling the space-time between the end worldlines, and non-intersecting. Then, you can get a key argument from the Herglotz-Noether theorem itself - born rigidity is the criteria for no expansion, shear, or vorticity along the string. Herglotz-Noether establishes that given one end point that is accelerating and definition of the congruence bounds in anyone 3-slice, the rest of such a congruence is uniquely determined. So then you exhibit (since it is easy to guess) the congruence with either rocket world line as one bound, and that fills the distance between them in the starting inertial frame, and that meets born rigidity. You know that this is unique, once you've found it (by Herglotz-Noether). Then you note, that in any inertial frame, the distance between the unconstrained end and the other rocket grows without bound. This amounts to a rigorous demonstration that the string must eventually break, without any unnecessary assumptions.

Of course, it is not a very elementary argument. However, the two string variant of the Insight Article is an accessible presentation of the gist of this method.

 

[Mentz114]

What do you mean by measurement apparatus? This is just the description of the string or rod connecting the space ship. And you are absolutely right: There must be a covariant description to decide whether this connection breaks of not, but this obviously depends on the concrete thing you use for this connection, i.e., on the elasticity/stability properties of the material. I don't plan to make such a detailed model, which would be an overkill just to resolve such a paradox. I'm content with a purely geometric description, i.e., the kinematics of some simple model, and bcrowell's model seems to be the most natural one: One just assumes the spaceships are somehow connected by a rod described by a horizontal worldline in A's reference frame at each instant of A's time. Then you get bcrowell's choice of the congruences, and the strain of the connection is a well-defined covariant quantity. I just have to work out the details for myself. Then I'll put it in my little SRT writeup.

You seem to be basing the paradox on the 'length contraction' so we can forget about actual measurements. The quantity $L/\gamma$ is a frame dependent geometrical projection - it is not a physical effect. In the words of W. Rindler 'nothing at all happens to the rod itself'.

We already have a covariant expression for the separation of all elements of the congruence including the string.

1. We agree that the ships are moving apart
2. Therefore any logic that deduces they are not moving apart is flawed.
3. Any conclusions based on the flawed logic is also flawed ( and unecessary).

 

[PAllen]

You seem to be basing the paradox on the 'length contraction' so we can forget about actual measurements. The quantity $L/\gamma$ is a frame dependent geometrical projection - it is not a physical effect. In the words of W. Rindler 'nothing at all happens to the rod itself'.

We already have a covariant expression for the separation of all elements of the congruence including the string.

1. We agree that the ships are moving apart
2. Therefore any logic that deduces they are not moving apart is flawed.
3. Any conclusions based on the flawed logic is also flawed ( and unecessary).

well I disagree in that I don't see any way (1) is an invariant statement. Clearly, there is an inertial frame where this is false. Congruence argument applies to the string, but does not say anything about the distance between two world lines each with proper acceleration. I claim there is no way to make an invariant statement about the distance between such world lines. Nor do you need to. What you need to show for the string to break is that any timelike congruence that spans the spactime between these world lines has expansion that grows without bound. That is not the same as establishing an invariant distance between two such world lines.

 

[PAllen]

One consequence of the approach of asking about the expansion free congruence for certain boundary conditions is that if pose the problem as:

Find an expansion free congruence where, in some inertial frame, all lines of the congruence are stationary before some time 0 in that frame, the congruence spans length L in this frame up to t=0, and rightmost world line of the congruence has constant proper acceleration g (to the right) after t=0.

Then there is a maximum L for which this is possible at all. Physically, this means that an unconstrained object must break if it is long enough and one end is accelerated at some constant proper acceleration. This follows from SR without recourse to any theory of the matter of the object. But it is hardly really surprising if you think about it. And this in principle limit is much larger than any practical limit, e.g. for g = acceleration of gravity at Earth's surface, the maximum object length is about 1 light year.

 

[vanhees71]

I'm obviously too stupid to understand this explanation only given in words. Could you please give equations?

My analysis shows that it is not clear from the point of view of the front ship C that rear ship B stays behind in all cases. If we have $L_A \alpha/c^2 >1$ (as PeterDonis pointed out in #2 of this thread, that means that the front ship is beyond B's Rindler horizon), to the contrary the distance of B decreases. So far I have not understood why this is not a correct definition of distance as seen from C, but no matter, it is obviously not a distance stating that the ships drift apart. Contrary to that, from the point of view of the front ship B using the usual definition of his distance of the front ship C, no such riddle occurs, and the ships are indeed shifting apart. So these are all frame-dependent arguments and thus it's not clearly defining a solution to the paradox. In bcrowells book, p 58, only this latter argument is brought to state that the string must break. The riddles when using an observer at instantaneous rest with the front ship is not discussed there. Here's the link to the book:

http://www.lightandmatter.com/sr

As has also been pointed out several times in this thread, and this I understand, there is after (in the sense of the inertial observer A) the acceleration of B and C starts no instantaneous rest frame, where both ships are at rest, and thus you cannot define a proper distance. In other words, the space ships are not in "rigid motion" relative to each other. I come to this later.

Finally, I have tried to understand the argument in Crowell's book using the notion of a timelike congruence (p. 179). I translate this from maxima into math and introduce my notation. The congruence for a proper acceleration $\alpha$ is given in parametric form by
$$x^{\mu}(\tau,l)=\begin{pmatrix}c^2/\alpha \sinh(\alpha \tau/c) \\ c^2/
\alpha [\cosh(\alpha \tau/c)-1]+l \end{pmatrix}.$$
The timelike world lines are parametrized by $\tau$ and the parameter $l \in [0,L_A]$ sweeps out the world tube between B and C. The four-velocity $u$ is
$$u^{\mu}=\frac{1}{c} \partial_{\tau} x^{\mu} = \begin{pmatrix} \cosh(\alpha \tau/c) \\ \sinh(\alpha \tau/c) \end{pmatrix}.$$
For our purposes we have to express this in terms of $t=x_0/c$ and $x=x^1$ (as a field):
$$u^{\mu}=\begin{pmatrix} \sqrt{1+(\alpha t/c)^2} \\ \alpha t/c \end{pmatrix}.$$
Then the expansion scalar
$$\Theta = \partial_{\mu} u^{\mu}=1/c \frac{\partial u^0}{\partial t}=\frac{\alpha^2 t}{c^3 \sqrt{1+\alpha^2 t^2/c^2}}.$$
Since $\beta=u^1/u^0=\frac{\alpha t}{c \sqrt{1+\alpha^2 t^2/c^2}}$ this is the same what Crowell finds in his book, and it's indeed $>0$, which means that the the space ships whose world lines are given by $l=0$ and $l=L_A$ in the congruence. So some "elastic rope" connecting the space ships stretches, and if stretched beyond the limit of stability it breaks. That's the only covariant resolution of the Born paradox I could find googling.

For PAllen's idea in #13, one doesn't need to calculate anything. The Born-rigid rod has length $L_A$ in its rest frame (note that it is defined by Born such that there's always a instantaneous reference frame, where the entire rod is at rest). In A's frame, the rod is moving, and here the usual length-contraction argument holds, so we are back at Bell's original argument (see also bcrowell's Insights article).

Thanks for all your help!

 

[vanhees71]

One consequence of the approach of asking about the expansion free congruence for certain boundary conditions is that if pose the problem as:

Find an expansion free congruence where, in some inertial frame, all lines of the congruence are stationary before some time 0 in that frame, the congruence spans length L in this frame up to t=0, and rightmost world line of the congruence has constant proper acceleration g after t=0.

Then there is a maximum L for which this is possible at all. Physically, this means that an unconstrained object must break if it is long enough and one end is accelerated at some constant proper acceleration. This follows from SR without recourse to any theory of the matter of the object. But it is hardly really surprising if you think about it. And this in principle limit is much larger than any practical limit, e.g. for g = acceleration of gravity at Earth's surface, the maximum object length is about 1 light year.

Yes, and I think this is the relevant conclusion from the calculation in my writeup.

 

[PeterDonis]

I disagree in that I don't see any way (1) is an invariant statement.

It is if by "the ships are moving apart" we mean "the congruence describing the ships and the string between them has a positive expansion scalar". That's the problem with using ordinary language instead of math: there is ambiguity in interpretation.

If we have $L_A \alpha/c^2 >1$ (as PeterDonis pointed out in #2 of this thread, that means that the front ship is beyond B's Rindler horizon

No, it means the rear ship, B, is behind the front ship's Rindler horizon. The Rindler horizon of an accelerating object is always behind it (i.e., in the direction opposite to the direction of acceleration).

, to the contrary the distance of B decreases.

How are you defining "distance"? If you're defining it as "coordinate distance in A's rest frame", then by hypothesis that distance is always constant; that's how you've defined the scenario. If you mean the coordinate distance in either B's or C's instantaneous rest frame, both of those are increasing; that's true regardless of how far apart the ships are, and the fact that B is behind C's Rindler horizon doesn't change it. If you mean "distance" as defined by the expansion scalar, that is positive, and is an invariant (as has already been pointed out several times, and as you have now calculated as well). So I don't know what definition of "distance" you're using that makes you think "the distance of B decreases".

this is the same what Crowell finds in his book, and it's indeed >0>0,

Yes, all of this looks correct.

 

[vanhees71]

How are you defining "distance"? If you're defining it as "coordinate distance in A's rest frame", then by hypothesis that distance is always constant; that's how you've defined the scenario. If you mean the coordinate distance in either B's or C's instantaneous rest frame, both of those are increasing; that's true regardless of how far apart the ships are, and the fact that B is behind C's Rindler horizon doesn't change it. If you mean "distance" as defined by the expansion scalar, that is positive, and is an invariant (as has already been pointed out several times, and as you have now calculated as well). So I don't know what definition of "distance" you're using that makes you think "the distance of B decreases".

I define "distance" in the same way as when I take the argument with the instantaneous rest frame for the rear ship, which was used in the Insight article by bcrowell: Instead I take the front ship's instantaneous rest frame. The math and the Minkowski diagram is in my writeup. I really checked this several times and can't find a mistake, but I then get that this "distance" decreases mononously from $L_A$ (when the acceleration starts) to $\frac{c^2}{\alpha}<L_A$:

http://fias.uni-frankfurt.de/~hees/tmp/bell-paradox.pdf

So I come to the conclusion that the use of this definition of distances is not the appropriate way to explain the space-ship paradox but that it is in fact Bell's original explanation, which is mathematically nicely expressed by the world-line congruence argument in bcrowell's book.

I think it's clearer to follow PAllen's advice and calculate the trajectory of the endpoint of a Born-rigid rod pointing along the $x$ axis with proper length $L_A$ with the other end tight to one of the space ships. It shouldn't matter whether it's tight to B or C, the conclusion must always be that it doesn't fit in reference frame A, because there its length appears Lorentz contracted and thus cannot connect B and C.

One last thing: I think it's misleading to argue that Lorentz contraction causes stresses, because if I have some medium somewhere at rest (be it a Born-rigid body or a realistic medium or even a gas), the Lorentz contraction is purely kinematical effect. This is clear from the fact that it occurs just from measuring distances from another (inertial) reference frame, i.e., nothing changes on the medium but only another observer is measuring its extension. I think there's a lot of confusion in the literature concerning this point.

BTW the best discussion on the Born-rigidity can be found in Pauli's famous relativity encyclopedia article. It's amazing that he could write such a clear monograph on relativity when he was only 21 and just an undergrad student with Sommerfeld!

 

[PAllen]

It is if by "the ships are moving apart" we mean "the congruence describing the ships and the string between them has a positive expansion scalar". That's the problem with using ordinary language instead of math: there is ambiguity in interpretation.

Of course, I explained all about the expansion scalar right after the statement you clipped. I will add more to why I don't see the invariant statement about expansion scalar leading to any invariant statement about growth of distance between the ships. Note the following points:

1) There is no unique congruence filling the spacetime between the rocket world lines. Different choices will have different expansions at different events. Even trying pick a best by seeking one with the minimum of maximum positive expansion leads to the problem of 'over what slice', and you are back to the chicken and egg as to what the 'right' slice is.

2) Suppose you are able to convince yourself that there is an 'optimum' congruence over the whole bounding world tube. Then, if there is no special slice (e.g. one orthogonal to all members of the congruence), then there is still no invariant distance definition anywhere.

3) In particular, for the obvious congruence for the bell spaceship paradox, there is no preferred slicing, and there exist slicings such the distance measured over the congruence is constant; there are even slicings where the distance decreases even though there is growing expansion. If there is no invariant distance, I don't know how to give invariant meaning to the statement that the ships are moving apart.

So, I stick to my view that once you no longer have any common rest frame, you no longer have any invariant special distance or length (Also, I prefer, rather than over-use proper distance since that is also used for arbitrary event pairs, to speak of rest length or rest distance. That makes the meaning and limitation explicit - when there is no common rest frame, there is no rest length or rest distance).

 

[PeterDonis]

I take the front ship's instantaneous rest frame. The math and the Minkowski diagram is in my writeup. I really checked this several times and can't find a mistake, but I then get that this "distance" decreases mononously from $L_A$ (when the acceleration starts) to $\frac{c^2}{\alpha}$

I haven't gone into your paper in detail, but on a quick skim, it looks to me like this result is occurring because the line of simultaneity in C's instantaneous rest frame intersects B's worldline before B starts accelerating. This is mathematically correct, but physically it is very misleading. This is a good illustration of why it's better to look at invariants like the expansion scalar.

 

[PeterDonis]

There is no unique congruence filling the spacetime between the rocket world lines.

Depends on what you mean by "unique". There is a unique congruence that is "uniform", in the sense that all of the intermediate worldlines (describing points on the string) have the same proper acceleration as the ships at the two ends. That congruence is the Bell congruence, the one Mentz computed the expansion scalar for.

It is true that, in any real string, not all of the points on the string will be likely to follow worldlines in this congruence exactly. What exact worldlines they follow will depend on many things and would require a detailed microphysical model of the string to compute. But on average, they must follow the Bell congruence worldlines, at least as long as the string has not yet broken, because (a) the endpoints of the string (the ships) are specified, by hypothesis, to move on worldlines in this congruence, and (b) there is no "preferred" point in the string that should move, on average, any differently than any other point. In other words, the Bell congruence is the best simple model we have of how the points in the string will move, so its expansion scalar is the relevant one for making physical predictions about the scenario.

I will add more to why I don't see the invariant statement about expansion scalar leading to any invariant statement about growth of distance between the ships.

My statement about the expansion scalar was meant as a suggested definition for an invariant meaning of "growth of distance between the ships". Obviously we can choose other definitions; but the expansion scalar definition has the advantage of neatly bypassing all of the issues you raise, which are genuine issues for any other possible meaning of "growth of distance between the ships". To me, once again, that's an argument for choosing a good definition, one that focuses on the physics rather than on things like reference frames and definitions of "distance" which are irrelevant to the physics--they just distract one from the obvious physical thing going on, which is that the string gets stretched until it breaks.

 

[PAllen]

Depends on what you mean by "unique". There is a unique congruence that is "uniform", in the sense that all of the intermediate worldlines (describing points on the string) have the same proper acceleration as the ships at the two ends. That congruence is the Bell congruence, the one Mentz computed the expansion scalar for.

It is true that, in any real string, not all of the points on the string will be likely to follow worldlines in this congruence exactly. What exact worldlines they follow will depend on many things and would require a detailed microphysical model of the string to compute. But on average, they must follow the Bell congruence worldlines, at least as long as the string has not yet broken, because (a) the endpoints of the string (the ships) are specified, by hypothesis, to move on worldlines in this congruence, and (b) there is no "preferred" point in the string that should move, on average, any differently than any other point. In other words, the Bell congruence is the best simple model we have of how the points in the string will move, so its expansion scalar is the relevant one for making physical predictions about the scenario.

Perhaps in this special case, but how do you generalize to time varying proper acceleration for each ship? Even for the Bell scenario, the substantive statement is that there is no non-expanding congruence. Whether you can find a 'maximally uniform' expansion by some definition, is not relevant to the physics, as it would be exceedingly unlikely that the most uniform congruence (if you can even define it, in general) would be the one followed in any physical system.

My statement about the expansion scalar was meant as a suggested definition for an invariant meaning of "growth of distance between the ships". Obviously we can choose other definitions; but the expansion scalar definition has the advantage of neatly bypassing all of the issues you raise, which are genuine issues for any other possible meaning of "growth of distance between the ships". To me, once again, that's an argument for choosing a good definition, one that focuses on the physics rather than on things like reference frames and definitions of "distance" which are irrelevant to the physics--they just distract one from the obvious physical thing going on, which is that the string gets stretched until it breaks.

So we have an invariant definition of "growth of distance", but no invariant definition of the distance that grows. For the invariant thing we agree on, it boils down to: I want to attach only the words "positive expansion overall is required"; you want to attach (in addition to these words) "growth of distance", while agreeing there is no invariant definition of the distance that grows. Obviously this is purely a matter of taste. I can see the intuition behind wanting to call expansion a growth of distance. I hope you can see how I might be troubled with growth of something that is not well defined.

 

[PeterDonis]

Perhaps in this special case, but how do you generalize to time varying proper acceleration for each ship?

It would depend on the details. But see below for some general comments.

Even for the Bell scenario, the substantive statement is that there is no non-expanding congruence.

No, the substantive statement is that the points in the string are described by a congruence with positive expansion. That is not at all the same as saying that every possible congruence that could be constructed to fill that region of spacetime must have positive expansion; the latter is a much stronger statement.

it would be exceedingly unlikely that the most uniform congruence (if you can even define it, in general) would be the one followed in any physical system.

That isn't the claim I made; again, it's a much stronger statement than the one I made. The statement I made was that, in this case, I would expect the Bell congruence to represent the "average" motion of the points in the string, even if it didn't exactly represent all the details of the motion of all the points. The physical rationale for that statement is not that "the most uniform congruence will be the one followed, on average" in every case; it's that, in the case under discussion, since the front and rear ships have the same proper acceleration, on average we would expect the intermediate points in the string to also have the same proper acceleration, since there is nothing that physically singles out any particular point in the string.

For the case of time-varying proper acceleration, obviously there is now, at a minimum, something physically different between the front and rear ships; and a fortiori we would also expect there to be something physically different between different points in the string. So the whole justification for expecting "uniformity" breaks down; obviously the situation is not uniform. In the general time-varying case, I don't think there is a known formula for constructing a suitable congruence to describe the motion of the string, even "on average".

So we have an invariant definition of "growth of distance", but no invariant definition of the distance that grows.

Yes.

Obviously this is purely a matter of taste.

Yes, that and a pedagogical judgment call. People tend to get hung up on "distance" instead of looking at invariants, so I proposed a way to redirect their attention back to invariants. It could backfire, as you say: the person could ask, "but how can distance be growing in an invariant sense when we don't have an invariant definition of the distance that's growing?" But I could then respond, "yes, that's because the concept of distance is not invariant, but the concept of growth of distance is".

Your alternative is basically to always say "expansion" instead of "growth of distance", and when the person responds, "but isn't expansion the same as growth of distance?", to answer "no, it isn't, because distance isn't an invariant but expansion is." That might work too; or it might backfire because the person can't grasp how "expansion" and "growth of distance" can be different things. Or which strategy works better might depend on the person.

 

[PAllen]

No, the substantive statement is that the points in the string are described by a congruence with positive expansion. That is not at all the same as saying that every possible congruence that could be constructed to fill that region of spacetime must have positive expansion; the latter is a much stronger statement.

Actually, this is not hard to show, even in more general cases that Bell scenario. The key is that an everywhere zero expansion congruence is tightly constrained. Establish it based on one ship's world line and the given initial information along one t=0 slice in a 'problem specifying frame', and if it can't reach the other world line after moment, then any congruence that does must have expansion.

Without the stronger statement, I dont' think you've shown much: suppose there were a non-expanding congruence for the Bell scenario, but it wasn't the uniform one. I could just say your model of the string is wrong, and that a real string would tend to find the non-expanding congruence (and then, not break). To say the string must break, it seems to me you really do need to show no model of string meeting the problem conditions is non-expanding.

 

[PAllen]

Yes, that and a pedagogical judgment call. People tend to get hung up on "distance" instead of looking at invariants, so I proposed a way to redirect their attention back to invariants. It could backfire, as you say: the person could ask, "but how can distance be growing in an invariant sense when we don't have an invariant definition of the distance that's growing?" But I could then respond, "yes, that's because the concept of distance is not invariant, but the concept of growth of distance is".

Your alternative is basically to always say "expansion" instead of "growth of distance", and when the person responds, "but isn't expansion the same as growth of distance?", to answer "no, it isn't, because distance isn't an invariant but expansion is." That might work too; or it might backfire because the person can't grasp how "expansion" and "growth of distance" can be different things. Or which strategy works better might depend on the person.

Another thing besides lack of invariant distance that I think could cause pedagogical issues is the idea that if not invariant, then any frame's definition is as good as another. Then growth of distance runs up against: whys isn't it valid to use the distance in an inertial frame, and the distance doesn't grow in that frame? This is another aspect of why I prefer to separate expansion (local growth of separation between members of the congruence) from any global, coordinate dependent, statements about distance. We establish local expansion must occur, even if overall distance does not change (for some frames). I would choose not to say the frame dependent definition of distance is 'wrong', because it is, in fact, the most common one used in SR. But if it is not wrong, then how can one say 'distance must grow' when it doesn't?

 

[PeterDonis]

Actually, this is not hard to show, even in more general cases that Bell scenario. The key is that an everywhere zero expansion congruence is tightly constrained.

If we assume that the expansion scalar is constant everywhere in the congruence, yes. But there is no reason why that must be the case in general. The kinematic decomposition of a congruence results in geometric objects (one scalar, one traceless symmetric tensor, and one antisymmetric tensor) at each point in the region of spacetime covered by the congruence; those objects can change from point to point (and in the general case, they will). So there could certainly be a congruence covering the same region of spacetime between the two ships as the Bell congruence, with an expansion scalar varying from point to point, that did not have positive expansion everywhere.

Establish it based on one ship's world line and the given initial information along one t=0 slice in a 'problem specifying frame', and if it can't reach the other world line after moment, then any congruence that does must have expansion.

Any congruence that does must have on average positive expansion, yes. But that's not at all the same as saying it has positive expansion at every single point; the latter is a much stronger statement. See above.

This is another aspect of why I prefer to separate expansion (local growth of separation between members of the congruence) from any global, coordinate dependent, statements about distance.

But you still used the word "separation", which is just another term for "distance". It's hard to avoid "distance" or any of its synonyms altogether.

To someone who insists on having some interpretation of "distance", I would say that what the expansion scalar does is tell you how the separation between neighboring pieces of the string (in this example) is changing, with respect to the proper time along the worldline of one of the pieces. The "neighboring" part is what let's you avoid all the stuff about frame-dependence; the relevant "frame" is highly localized (a full unpacking would end up defining Fermi normal coordinates in a small world-tube around the chosen worldline), so you're not committing yourself to any global notion of "distance", just a local one. The reason there can't be a global one is that, in any case of non-rigid motion, there is no way to "knit together" the different local notions of distance into a single consistent global one.

 

[PAllen]

Any congruence that does must have on average positive expansion, yes. But that's not at all the same as saying it has positive expansion at every single point; the latter is a much stronger statement. See above.

I never thought I said everywhere positive expansion. By expanding I meant: not everywhere non-expanding.

But you still used the word "separation", which is just another term for "distance". It's hard to avoid "distance" or any of its synonyms altogether.

To someone who insists on having some interpretation of "distance", I would say that what the expansion scalar does is tell you how the separation between neighboring pieces of the string (in this example) is changing, with respect to the proper time along the worldline of one of the pieces. The "neighboring" part is what let's you avoid all the stuff about frame-dependence; the relevant "frame" is highly localized (a full unpacking would end up defining Fermi normal coordinates in a small world-tube around the chosen worldline), so you're not committing yourself to any global notion of "distance", just a local one. The reason there can't be a global one is that, in any case of non-rigid motion, there is no way to "knit together" the different local notions of distance into a single consistent global one.

I agree with all of this, except the idea that there is no consistent global distance. What I would say is there is no invariant global distance, nor is there any 'rest length'. However, there are perfectly consistent, frame dependent, global distances.

 

[vanhees71]

Ok, I followed now the idea by PAllen and attached a rigid rod to B. With the Lorentz transform to the instantaneous restframe of B at proper time $\tau$

$\hat{\Lambda}(\tau)=\begin{pmatrix} \cosh(\alpha \tau) & -\sinh(\alpha \tau) \\-\sinh(\alpha \tau) & \cosh(\alpha \tau)\end{pmatrix},$

setting $c=1$ for brevity, we get the world line of the end of the rigid rod of proper length $L_A$ (in coordinates of the A frame)

$x_{\text{rod}}(\tau)=x_B(\tau)+\hat{\Lambda}^{-1}(\tau) \begin{pmatrix} 0 \\ L_A \end{pmatrix}=\begin{pmatrix} (1+\alpha L_A)/\alpha \sinh(\alpha \tau) \\ (1+\alpha L_A)/\alpha \cosh(\alpha \tau)-1/\alpha. \end{pmatrix}$

Plotting this together with the world lines of ship B and C, one sees that at any instant of time in A the rod's endpoint is behind of C (after the acceleration started).

When I do the same construction with the rod fixed at C I get the strange result that for $\alpha L_A>1$ at any instant of time in A after the acceleration started the end of the rod is at negative $x$, i.e., the rod appears much longer than $L_A$ when measured in A. So this doesn't make any sense again, or does it?

There seems not to be a clear solution of the paradox in elementary terms, and one must argue with the expansion scalar as in bcrowell's book, but as I said in a previous posting in this thread, it's not clear to me, why choosing this particular world-line congruence to fill the world-line tube between the ships. So this is also arbitrary.

 

[loislane]

There seems not to be a clear solution of the paradox in elementary terms, and one must argue with the expansion scalar as in bcrowell's book, but as I said in a previous posting in this thread, it's not clear to me, why choosing this particular world-line congruence to fill the world-line tube between the ships. So this is also arbitrary.

This might be because the expansion scalar in the book's example is restricted to one spatial dimension. As is well known a 3-surface has vanishing expansion in a static spacetime.

 

[PeterDonis]

By expanding I meant: not everywhere non-expanding.

But this is a much weaker statement than my "on average positive expansion".

What I would say is there is no invariant global distance, nor is there any 'rest length'. However, there are perfectly consistent, frame dependent, global distances.

Agreed, this is a better way of saying it.

 

[vanhees71]

Just to make my problem in #32 clear again, here's the explicit calculation if one considers a Born-rigid rod with one end fixed at the leading spaceship. With the same notation as in #32 the other end of the rigid rod is given by the world line (written in coordinates in from A)
$$x_{\text{rod}}=x_C(\tau)-\hat{\Lambda}^{-1}(\tau)=\begin{pmatrix}
(1-\alpha L_A)/\alpha \sinh(\alpha \tau) \\ (1-\alpha L_A)/\alpha \cosh(\alpha \tau)
\end{pmatrix}.$$
As long as $\alpha L_A<1$ there's no problem. Plotting the world lines one sees that at any instant of time $t>0$ in A the rod is too short to reach to B, i.e., tighing the end of the rod to B would need to stretch it more and more, and thus a "real" (non-rigid) rod would eventually break.

But now, if $1-\alpha L_A<0$, a problem occurs. Then for $t>0$ in A one has $x_{\text{rod}}^1(\tau)<0$ (note that $\tau<0$ to have $x^0(\tau)=c t(\tau)>0$). This doesn't make sense to me. That would mean in this case the rigid rod at any instant $t>0$ is longer (sic!) than $L_A$. So something must be wrong in this argument either, but I still cannot figure out what that might be.

Another solution to the problem is, when one slightly changes the conditions by assuming that after some finite time $t_{\text{end}}$ the two spaceships stop accelerating (e.g., because their fuel is over ;-)) and then drift both with the same constant velocity. Then you have a common rest frame of the spaceships, and the distance in A is still $L_A$ but this is the Lorentz contracted proper distance (which now is well defined as the measured distance in the common restframe) which is thus longer by the $\gamma$ factor $L_{\text{proper}}=\gamma L_A$.

So, maybe the entire trouble is the somewhat artificial example of constant proper acceleration (also known as hyperbolic motion for obvious reasons). This example seems to be a notorious trouble maker as there is the (for me also quite paradoxical) calculation that the Lienard Wiechert retarded solutions of the Maxwell equations for a point charge in hyperbolic motion seem to indicate that there is no radiation field at all, while if stopping the acceleration after some finite time, one finds radiation as expected (see, e.g., Pauli's famous monography as a reference).

 

[PeterDonis]

if $1-\alpha L_A < 0$, a problem occurs

Yes, this is because it's physically impossible to have a rigid rod trailing the leading spaceship with a length longer than $1 / \alpha$ (or $c^2 / \alpha$ in conventional units), since that would put the trailing edge of the rod behind the leading spaceship's Rindler horizon. If you try to base an analysis on physically impossible assumptions, naturally you're going to get incorrect results.

maybe the entire trouble is the somewhat artificial example of constant proper acceleration

No, the trouble is that you are trying to construct a model that's physically impossible; see above. Having a rigid rod "pushed" ahead of the trailing spaceship doesn't have this problem because the Rindler horizon is always behind an accelerating object.

 

[PAllen]

Yes, this is because it's physically impossible to have a rigid rod trailing the leading spaceship with a length longer than $1 / \alpha$ (or $c^2 / \alpha$ in conventional units), since that would put the trailing edge of the rod behind the leading spaceship's Rindler horizon. If you try to base an analysis on physically impossible assumptions, naturally you're going to get incorrect results.

The way this would play out is that if you tried to specify the rigid congruence, you wouldn't be able to extend it as a timelike congruence beyond this limit. Physically, as I explained in an earlier post, this means that an unconstrained object longer than this must break if it is pulled with the stated proper acceleration, irrespective of any theory of matter. In practice, it would break at a much shorter length assuming you are able to generate the stated proper acceleration of the leading edge.