# A Bell spaceship paradox quantitatively

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1. Aug 21, 2015

### 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:

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.

Last edited by a moderator: May 7, 2017
2. Aug 21, 2015

### Staff: Mentor

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.

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.

3. Aug 21, 2015

### 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.

4. Aug 21, 2015

### 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.

5. Aug 21, 2015

### Staff: Mentor

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.

6. Aug 21, 2015

### PAllen

Oops, you're right.

7. Aug 21, 2015

### Staff: Mentor

"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.

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).

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.

8. Aug 22, 2015

### Mentz114

These are quotes from Rindlers book ( 1985)

(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)

I should think not.

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

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

9. Aug 22, 2015

### Mentz114

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.

Last edited: Aug 22, 2015
10. Aug 22, 2015

### 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?

11. Aug 22, 2015

### Mentz114

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 ?

12. Aug 22, 2015

### 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.

13. Aug 22, 2015

### 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 any one 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.

Last edited: Aug 22, 2015
14. Aug 22, 2015

### Mentz114

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).

Last edited: Aug 22, 2015
15. Aug 22, 2015

### PAllen

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.

16. Aug 22, 2015

### 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.

Last edited: Aug 22, 2015
17. Aug 22, 2015

### 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).

18. Aug 22, 2015

### vanhees71

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

19. Aug 22, 2015

### Staff: Mentor

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.

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).

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".

Yes, all of this looks correct.

20. Aug 23, 2015

### vanhees71

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 $\frac{c^2}{\alpha}<L_A$:

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.