1. Feb 25, 2014

### Mentz114

I'm sticking my neck out because I just worked this out and may regret this post later.

The spaceship paradox arises because the expansion scalar of the 'Bell' congruence is always positive, implying that the thread will break, even when the ships are not separating.

For this vector field $V=\gamma \partial_t + \gamma\beta \partial_x$ where $\beta$ is a function of $t$ I find

$\Theta=\frac{d\gamma}{dt}=\gamma^3\ B\,\left( \frac{dB}{d\,t}\right)$.

I suggest that there are three cases here, corresponding to

$\Theta<0,\ \Theta=0,\ \Theta>0$.

The 'Bell' congruence is $V$ with $\Theta > 0$, and the case where there is no separation is obviously $\Theta=0$. This ties in with a number of other calculations.

I rest my case.

Last edited: Feb 25, 2014
2. Feb 25, 2014

### dauto

I don't think you rested your case. In fact I don't think you explained your case very well. What in the world are you talking about?

3. Feb 25, 2014

### Mentz114

From the Wiki page

I'm saying that in the case described here, the thread does not break because the B and C are not moving apart.

The only proper justification I have been given for the thread breaking is that the expansion scalar $\partial_\mu V^\mu$ is always positive. I'm saying it can be zero so we cannot use the expansion scalar as a reason why the thread breaks.

4. Feb 25, 2014

### Staff: Mentor

Huh? The positive expansion scalar is the invariant *definition* of the ships "separating". Any other sense of the word "separating" is frame-dependent.

The three cases correspond to three *different* congruences. The $\Theta > 0$ case is the Bell congruence. The $\Theta = 0$ case is just an ordinary inertial congruence (i.e., a family of inertial observers all at rest with respect to each other). I haven't seen any discussion in textbooks or articles of the $\Theta < 0$ case, but I think we can leave it out of discussion for now; see below.

(Note: I'm not sure that your $\Theta$ is the same as the expansion scalar; I'll have to check the math when I get a chance. However, I think the *sign* of $\Theta$ will be the same as the sign of the expansion scalar, which is sufficient for this discussion.)

No, that's not correct; the Wiki page is describing the $\Theta > 0$ case. Both B and C are accelerating relative to a fixed inertial frame (the one in which A remains at rest), so $\gamma$ increases with coordinate time in that frame; i.e., $\Theta = d \gamma / dt > 0$. That means the expansion scalar is positive and the thread breaks.

To see how this is consistent with Bell's description, note the bolded phrase:

Distance relative to a fixed frame is, obviously, frame-dependent, i.e., it's not an invariant, so it's not a good way to describe the actual physics. The actual physics depends on the distance in the instantaneous rest frame of either B or C, and how that distance changes with proper time along either B or C's worldline. That is what the (invariant) expansion scalar captures.

I have a proposed FAQ on this that's been in the works for some time, which addresses the above; I'll work on getting it published and visible in this forum.

5. Feb 25, 2014

### Mentz114

Thanks, Peter. I'll mull that over. In my original post I suggested it was three congruences and then deleted it.

I've done another calculation that has a nice result. To get rocket C's motion relative to B, I boosted a rest frame with $\beta_1$ ( for B) and boosted a rest frame by $\beta_2$ for C. To get C in B's frame I then boost both with $-\beta_1$, which gives for C

$V=\left( 1-\beta_1\,\beta_2\right) \,\gamma_1\,\gamma_2\partial_t+\left( \beta_2-\beta_1\right) \,\gamma_1\,\gamma_2\partial_x$.

Note that in the rest frame the velocity is $\left( \beta_2-\beta_1\right) /\left( 1-\beta_1\,\beta_2\right)$. The expansion scalar for the composite 4-velocity is ( writing $a_nt$ for $\beta_n$)

$\Theta = \nabla_\mu V^\mu = \frac{ t(a_2-a_1)^2 ( 1-a_1a_2t^2 )} {(1-a_1^2t^2)^{3/2}(1-a_2^2t^2)^{3/2} }$

I think this models the three spaceships and shows that the expansion scalar of the case $a_1=a_2$ is zero. However, $\Theta$ is never negative.

My point is that the string will not break when $a_1=a_2$.

I see that you agree with my view that frame-dependent effects cannot do work - like straining a material.

I'm satisfied that there is no paradox and no need to ascribe the cause of the breaking to length contraction..

Last edited: Feb 25, 2014
6. Feb 25, 2014

### WannabeNewton

If so, describe how an observer in the inertial frame in which the rockets are accelerated simultaneously with the same proper acceleration would explain the non-vanishing expansion scalar.

You're still seriously misunderstanding the difference between frame-dependent explanations of the non-vanishing of an invariant and the frame-independent consequences of the non-vanishing of an invariant.

7. Feb 26, 2014

### Staff: Mentor

No, this isn't right, at least not for a constant proper acceleration, which I assume is what you're trying to express here. See below.

If $a_n$ is the proper acceleration of spaceship $n$, then this is not correct; $a_1 = a_2$ means equal proper acceleration, which means the expansion scalar is positive and the string will break.

I've looked at the long-winded derivation of the expansion scalar now, and I agree with your formula $\Theta = \nabla_a u^a$ (i.e., it matches what I get as the general formula for the expansion scalar--short derivation below). Note that, since we're working in an inertial frame, $\nabla_a = \partial_a$, so we just have $\Theta = \partial_a u^a$.

But writing this out, we get

$$\Theta = \partial_a u^a = \partial_t u^t + \partial_x u^x = \partial_t \gamma + \partial_x \left( \gamma v \right)$$

For the simplest case of constant proper acceleration (i.e., $a_1 = a_2$ in your notation), everything is a function of $t$ only, so we have

$$\Theta = \partial_t \gamma = \partial_t \left( 1 - v^2 \right)^{-1/2} = - \frac{1}{2} \left( 1 - v^2 \right)^{-3/2} \left( - 2 v \right) \partial_t v = \gamma^3 v \partial_t v$$

For constant proper acceleration, we have $\partial v / \partial \tau$ constant (where $\tau$ is the proper time along a given worldline in the congruence); but $\partial v / \partial \tau = \left( \partial t / \partial \tau \right) \partial v / \partial t = \gamma \partial v / \partial t$, so we have $\partial_t v = \gamma^{-1} a$ if $a$ is the constant proper acceleration. (Note, btw, that we are assuming here that both $a$ and $v$ are positive--we are starting from rest and accelerating in the positive $x$ direction. A more sophisticated analysis would account for the other possibilities for the relative signs, to show that $\Theta$ always comes out positive, but I won't go into that detail here.) So we have

$$\Theta = \gamma^3 v \gamma^{-1} a = \gamma^2 v a$$

I haven't tried to solve the more general case of letting the proper acceleration vary from worldline to worldline; this would take some time for me to model because in the inertial frame, the proper acceleration itself must be a function of both $t$ and $x$, with the constraint that it must be constant along a given worldline, i.e., $u^a \nabla_a a = 0$ if $a$ is the proper acceleration. But the above is sufficient to show that, for the case of constant proper acceleration (your $a_1 = a_2$), the expansion is positive.

Short derivation of the expansion scalar: for a timelike congruence defined by a vector field $u^a$, the expansion scalar is the trace of the expansion tensor $\theta_{ab}$, which is given by

$$\theta_{ab} = h^m{}_a h^n{}_b \frac{1}{2} \left( \nabla_n u_m + \nabla_m u_n \right)$$

where $h_{ab} = g_{ab} + u_a u_b$ is the projection tensor orthogonal to $u_a$. The expansion scalar is just the trace of this tensor, which is

$$\theta = \theta^a{}_a = h^{ma} h^n{}_a \frac{1}{2} \left( \nabla_n u_m + \nabla_m u_n \right) = \frac{1}{2}\left( g^{ma} + u^m u^a \right) \left( g^n{}_a + u^n u_a \right) \left( \nabla_n u_m + \nabla_m u_n \right)$$

$$\ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \left( g^{mn} + u^m u^n \right) \left( \nabla_n u_m + \nabla_m u_n \right) = \nabla_m u^m + u^m u^n \nabla_n u_m = \nabla_m u^m$$

where in the last equality we have used the fact that the 4-acceleration $u^a \nabla_a u_b$ is orthogonal to the 4-velocity.

Not just in virtue of being frame-dependent effects, no. Any real work done must always correspond to some invariant that is not frame-dependent (such as the expansion scalar).

However, it seems like a lot of people use frame-dependent effects to formulate a physical "interpretation" of what is going on. Bell's discussion of the spaceship paradox, where he says that "length contraction" is what causes the string to break, is an example. The FAQ entry I mentioned in my previous post addresses that point (in fact a previous thread in which the point came up is what prompted me to write the FAQ entry).

Last edited: Feb 26, 2014
8. Feb 26, 2014

### Mentz114

The expansion scalar vanishes with equal acceleration as I show in my previous post.

The invariant vanishes.
My calculation is covariant and the final result is a scalar. The way I set up the frames ensures that B and C have the same proper acceleration if $a_1=a_2$.

9. Feb 26, 2014

### WannabeNewton

It doesn't vanish. If the proper accelerations of the spaceships are equal and simultaneous in the inertial frame then the string will break meaning the expansion scalar must be positive. The observer in the inertial frame attributes this to the string resisting length contraction, simple as that.

10. Feb 26, 2014

### Mentz114

Thanks for that. We seem to be in agreement with the numbers I left in the quote. The way I'm thinking now is that the paradox arises is because the comoving ship frames see the ships B,C separating while A sees them comoving. I have a problem with this.

This $\bar{\gamma}=-U^\mu V_\mu$ is an invariant in SR and has the value 1 if U and V are comoving. Under the initial conditions specified in the Wiki quote of Bell's position, $\bar{\gamma}$ will be 1, since B,C have the same worldline. So this will still be 1 if we transform to either B or C's frame. I could be missing something here but I don't see *how* the ship observer can think the other ship is separating and A thinks they are comoving.

I don't find it simple. At least the expansion scalar is not frame dependent, but I still think the thread breaks only if the ships separate in all frames.

Anyhow, I'm still mulling all this. Thanks for the input.

Last edited: Feb 26, 2014
11. Feb 26, 2014

### WannabeNewton

Motion and (spatial) distance are relative so I'm not seeing why exactly it's bothering you. Take a rod and Born rigidly accelerate it along its length. Then in the rest frame of any point of the rod the distance to neighboring points is always constant but in the inertial frame through which the rod accelerates the distance between any two points is always decreasing due to "length contraction" (I put it in quotes because it differs from the usual gamma factor length contraction). Is this also troubling to you?

12. Feb 26, 2014

### Staff: Mentor

With this interpretation of "comoving", "comoving" is frame-dependent. You have to find a sense of "comoving" which is invariant in order to use it in a physical argument, since we've already agreed that only invariants can be so used. The usual invariant sense of "comoving" is that the congruence is rigid, which means both the expansion and the shear are zero. But the expansion is not zero for the Bell congruence, as I've already shown.

That's not what I was using in my calculation; in my calculation, $\gamma (t) = 1 / \sqrt{ 1 - v(t)^2 }$, where $v(t)$ is the velocity of either spaceship, B or C, in A's rest frame, as a function of coordinate time in that frame. By hypothesis, $v$ is a function of $t$ only *if* we do the calculation in A's frame; but in any other frame, $v$ will be a function of $x'$ as well as $t'$ (I'm using primes for the coordinates in the other frame). So the analysis becomes more complex in any frame other than A's rest frame.

Also, this definition of $\bar{\gamma}$ is not actually invariant, because it is taking the inner product of vectors at different events. See below.

No, they don't have "the same worldline". They have worldlines which have the same path curvature (i.e., the same proper acceleration), but the worldlines are spatially separated, so they're not "the same". This is a critical fact that you have left out of your analysis of what is "invariant".

Your definition of $\bar{\gamma}$ above implicitly evaluates $U^a$ at some time $t$ on B's worldline, and $V^a$ at the same time $t$ on C's worldline. But "at the same time" is frame-dependent. In A's rest frame, at any time $t$, we will indeed have $U^a = V^a$. But if you evaluate $U^a$ and $V^a$ at the same time $t'$ in any other frame, they will *not* be equal; doing that is equivalent to evaluating $U^a$ at some time $t_B$ in A's rest frame and $V^a$ at some different time $t_C$ in A's rest frame, because B and C are spatially separated so relativity of simultaneity comes into play. So in any frame other than A's rest frame, $\bar{\gamma} \neq 1$, and therefore $\bar{\gamma}$ is not an invariant.

13. Feb 26, 2014

### Mentz114

PD and WNB, thanks to both of you for the posts. I have to admit that there may be no invariant definition of relative velocity and that relative velocity ( like the spatial components of a 4-velocity) transforms like a tensor component. In which case the conclusions I thought I drew are not true. But in spite of that I'm still agnostic about '... attributes this to the string resisting length contraction'. I'm also unsure about the causality.

It's necessary to work out how relative velocity transforms. I feel a calculation coming on and I'll be back.

14. Feb 26, 2014

### WannabeNewton

Why? You can't just reject the explanation because it doesn't appeal to your intuitions. It's not that hard to see intuitively how the Lorentz contraction of the string in the inertial frame contributes to the macroscopic and microscopic dynamics in this frame that causes the string to break.

15. Feb 26, 2014

### Mentz114

My current line is to use the equivalence principle. If we dangle a thread in a sufficiently strong gravitational field it will break under its own weight. I believe what the inertial observer sees is the same, with the acceleration gradient playing the part of the gravity. This means the thread breaks because of its inertia and the differential acceleration. Nothing to do ( directly) with time dilation. I'm trying to put some equations together and reasonably optimistic.

16. Feb 26, 2014

### pervect

Staff Emeritus
Parallel transport of a velocity vector is coordinate independent, and in a flat space-time gives the right result for relative velocity (when you compare the parallel-transported vectors).

Unfortunately, it's path dependent in curved space-time, so you aren't guaranteed a unique answer unless you specify a unique path.

For the purposes of Bell's spaceship, I've found that an adequate substitute for "relatively at rest" is having a constant two-way propagation delay for light signals. A static metric is sufficient to cause the two-way light propagation to be independent of time.

So in coordinate dependent terms, if objects have constant spatial coordinates, and none of the metric coefficients is a function of time, we can say they are at rest.

I'm not sure how to express this in coordinate independent language. Something along the lines of all objects whose 4-velocity is orthogonal to a time-like Killing vector field are at rest, I think.

17. Feb 26, 2014

### Mentz114

This is why I expected $U^\mu V_\mu$ to be a scalar in flat spacetime. I know in curved space it is only a 'local' tensor in a vicinity of a point of coincidence of U and V. But I think PeterDonis revoked my expectation.

18. Feb 26, 2014

### Staff: Mentor

Parallel transport between two fixed events in flat spacetime is path-independent, yes. But changing frames changes which two fixed events you are parallel transporting between to compare the vectors $U$ and $V$ (because it changes the surfaces of simultaneity, which define at what points on the worldlines of B and C you evaluate $U$ and $V$). So changing frames changes $U^\mu V_\mu$ because it changes the endpoints, not because it changes the path taken between the same endpoints (the latter would indeed not change $U^\mu V_\mu$ in flat spacetime).

19. Feb 26, 2014

### Staff: Mentor

This defines "at rest" in an invariant way, yes, but it has to be the *same* timelike KVF for all the objects. In Minkowski spacetime, there are two infinite families of timelike KVFs--one for each possible inertial frame, and one for each possible Rindler coordinate chart. So, for example, two inertial observers in relative motion are both following orbits of a timelike KVF, but it's a different KVF for each of them, so they're not both at rest relative to the same definition of "at rest".

Similarly, the two spaceships in the Bell Spaceship Paradox are both following orbits of a timelike KVF, but it's a different timelike KVF for each of them (this time because they are following orbits of different "Rindler" KVFs, i.e., worldlines of constant proper acceleration that asymptote to different Rindler horizons).

20. Feb 26, 2014

### WannabeNewton

This is because the tension in the string due to your grip on the near end of the string isn't enough to balance the gravitational force acting on each infinitesimal element of the string.

In other words the string is brought beyond its equilibrium (natural) length to the point of overwhelming elastic stresses.

In the same spirit, in the inertial frame of the Bell spaceship paradox setup there is an interplay between length contraction and equilibrium length when the string is fastened between the spaceships maintaining constant distance between them in said inertial frame.

It's definitely instructive to try and mess around with this stuff on your own so go for it but I guarantee you that there won't be any way to avoid length contraction when explaining why the string breaks in the inertial frame. If you want a more detailed physical explanation then feel free to ask.

Last edited: Feb 26, 2014