Solving the Spaceship Paradox: A New Explanation

[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.

 

[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?

 

[Mentz114]

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?

From the Wiki page

In Bell's version of the thought experiment, three spaceships A, B and C are initially at rest in a common inertial reference frame, B and C being equidistant to A. Then a signal is sent from A to reach B and C simultaneously, causing B and C starting to accelerate (having been pre-programmed with identical acceleration profiles), while A stays at rest in its original reference frame. According to Bell, this implies that B and C (as seen in A's rest frame) "will have at every moment the same velocity, and so remain displaced one from the other by a fixed distance. Now, if a fragile thread is tied between B and C, it's not long enough anymore due to length contractions, thus it will break.

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.

Ergo, no paradox.

 

[PeterDonis]

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

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

I suggest that there are three cases here, corresponding to

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

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.)

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

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:

According to Bell, this implies that B and C (as seen in A's rest frame) "will have at every moment the same velocity, and so remain displaced one from the other by a fixed distance".

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.

 

[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..

 

[WannabeNewton]

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

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.

 

[PeterDonis]

writing $a_n t$ for $\beta_n$

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.

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

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.

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

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).

 

[Mentz114]

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.

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

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.

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 $a_1=a_2$.

 

[WannabeNewton]

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

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.

 

[Mentz114]

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

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.

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

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.

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.

 

[WannabeNewton]

I have a problem with this.

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?

 

[PeterDonis]

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.

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.

$\bar{\gamma}=-U^\mu V_\mu$

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.

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.

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.

 

[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.

 

[WannabeNewton]

PBut in spite of that I'm still agnostic about '... attributes this to the string resisting length contraction'.

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.

 

[Mentz114]

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.

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.

 

[pervect]

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.

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.

 

[Mentz114]

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.

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.

Your remarks are apposite, thanks.

 

[PeterDonis]

This is why I expected $U^\mu V_\mu$ to be a scalar in flat spacetime.

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).

 

[PeterDonis]

Something along the lines of all objects whose 4-velocity is orthogonal to a time-like Killing vector field are at rest, I think.

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).

 

[WannabeNewton]

If we dangle a thread in a sufficiently strong gravitational field it will break under its own weight.

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.

I'm trying to put some equations together and reasonably optimistic.

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.

 

[georgir]

On a somewhat possibly related note, what are the correct acceleration profiles that the ships must follow in order for the string to not break and the distance between them to appear constant in their own rest frames? Is that even possible for both ships at once?
On a hunch I'd say if it is possible, then the most likely candidate is a setup where the front ship (according the direction of their acceleration) starts when a signal that the back ship sent when it started reaches it. But I'm at work now and would attract too much weird looks if I started to try to calculate this :(

 

[Mentz114]

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.

Yep, the string breaks.

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.

The first quote above means that it makes no difference if the string is tethered to the trailing ship or not, it will still break. The string will experience tension as if in a gravitational field which gets stronger ( linearly in my first estimation) with time, and so must break.

Now the inertial observer who sees the ships maintaining constant separation can ascribe the string breaking to this inertial lag force. The ship observer can ascribe a string breaking to the increasing separation or the inertal lag or both. This probably depends on the composition of the string.

I find this plausible, since it is expressed in Newtonian terms and fairly intuitive.

I have equations but I've run out of time for some hours and they are not done yet.

 

[georgir]

Mentz, a constant proper acceleration means that the experienced "tension as if in a gravitational field" will be constant, not getting stronger with time. It means that the string can always be made strong enough to not break due to this effect alone.

 

[WannabeNewton]

On a somewhat possibly related note, what are the correct acceleration profiles that the ships must follow in order for the string to not break and the distance between them to appear constant in their own rest frames?

Hi georgir. We call this Born rigidity. If we have a line of spaceships and accelerate them simultaneously longitudinally and want the distances between them to remain constant in all of their rest frames then it turns out we have to accelerate them so that in the rest frame of any of the ships, the 4-velocity field of the line of ships is given by $u = \frac{1}{x}(1,0,0,0)$ where $x$ is the location of each ship in this rest frame. The 4-acceleration is then $a = \frac{1}{x}(0,1,0,0)$and so each ship in fact has a different acceleration which in this rest frame varies with the constant spatial position of each ship. As an side, there's a precise mathematical definition of Born rigidity that I will omit, unless you want to see it.

In the inertial frame in which all the ships were simultaneously accelerated each ship will have a different velocity varying in accordance with the different accelerations of each ship and the line of ships as a whole will length contract continuously in time relative to this frame. The fact that the line is length contracting in the inertial frame is, in an intuitive sense, equivalent to demanding Born rigidity because the resistance to length contraction is what led to for example the string breaking in the inertial frame of the Ball spaceship paradox.

The relationship between $x$ for each spaceship and the coordinates $(X,T)$ of the inertial frame is as it turns out $x^2 = X^2 - T^2$. Note then that $\frac{dX}{dT} = \frac{T}{\sqrt{T^2 + x^2}}$ meaning that the 3-velocity of each spaceship in the inertial frame varies with its $x$ value and similarly so does $\frac{d^2 X}{dT^2} = \frac{1}{ \sqrt{T^2 + x^2}}- \frac{T^2}{(T^2 + x^2)^{-3/2}}$. The length contraction formula between the frontmost spaceship and rearmost spaceship is also easy to calculate but rather messy to write down-just subtract the positions of said spaceships in the inertial frame.

EDIT: see here for more details: http://www.mathpages.com/home/kmath422/kmath422.htm

Mentz, a constant proper acceleration means that the experienced "tension as if in a gravitational field" will be constant, not getting stronger with time. It means that the string can always be made strong enough to not break due to this effect alone.

Thanks! You saved me some typing there you did :)

 

[PeterDonis]

The first quote above means that it makes no difference if the string is tethered to the trailing ship or not, it will still break. The string will experience tension as if in a gravitational field which gets stronger ( linearly in my first estimation) with time

No, it won't; it will be constant with time, as georgir said. This is because only one end of the string has a motion that is constrained by being attached to something else; the other end can move freely (as can all the pieces of the string in between). That changes the congruence of worldlines that the individual pieces of the string follow: they follow a congruence with zero expansion in this case. It is still *possible* for the string to break in this case, but only if the proper acceleration of the ship that is pulling on the string is large enough that the string's weight exceeds its tensile strength.

(Technically, there is another condition as well: the string must be short enough that its trailing end is less than a distance $1 / a$ from the ship, in the ship's instantaneous rest frame, where $a$ is the ship's proper acceleration. If the trailing end is further from the ship than that, the string will break because its trailing end would have to move faster than light to keep up. I've been assuming in this entire thread that we are ruling out this kind of thing.)

 

[Mentz114]

Addressing the time dependence issue - how do you model the constant acceleration ? If a rest frame is boosted by $\beta$ then the proper acceleration in the boosted frame is $\dot{\beta}\gamma^3$ which is time dependent. The expansion coefficient is $\partial_t(\gamma)$. This obviously misleads me about the time dependence.

Peter, the length issue could be a cruncher, so it could be back to the pencil and note book for me.

Aside : This chart of the Minkowski metric $ds^2= -x^2dt^2+t^2dx^2+dy^2+dz^2$ gives an acceleration of $1/x$ and expansion scalar $1/xt$. Does this mean that two tethered comoving observers will see the tether get tighter ? Probably doesn't help my case.

[edit] corrected a typo in expansion scalar.

 

[PeterDonis]

Addressing the time dependence issue - how do you model the constant acceleration ?

Constant path curvature of the worldline; i.e., the 4-acceleration $a = u^a \partial_a u^b$ is constant. In the case of the spaceship paradox, $a$ is furthermore the same for *all* the worldlines (both spaceships, and all of the pieces of the string in between).

If a rest frame is boosted by $\beta$ then the proper acceleration in the boosted frame is $\dot{\beta}\gamma^3$ which is time dependent. The expansion coefficient is $\partial_t(\gamma\beta)$. This obviously misleads me about the time dependence.

That's because "time" is frame-dependent. Which means that what looks like "time dependence" in one frame becomes "time and space dependence" in another frame, even when the invariant path curvature is constant! This is a good illustration of how trying to think of things in terms of frames instead of invariants can cause confusion.

In the original rest frame (A's rest frame), $\beta$ is a function of time (i.e., coordinate time $t$), but not space (i.e., $\beta$ is not a function of $x$). However, in any other inertial frame, $\beta$ will be function of both $t'$ and $x'$. The reason A's original rest frame is different in this respect is that in that frame, both spaceships start accelerating (i.e., firing their rockets) simultaneously (at time $t = 0$). In any other frame, because of relativity of simultaneity, the ships start accelerating at different times.

 

[Mentz114]

... $a$ is furthermore the same for *all* the worldlines (both spaceships, and all of the pieces of the string in between).

Are you sure ? The spaceships have synchronised starts but the strings are being towed so they have a lag.

That's because "time" is frame-dependent. Which means that what looks like "time dependence" in one frame becomes "time and space dependence" in another frame, even when the invariant path curvature is constant!

OK.

This is a good illustration of how trying to think of things in terms of frames instead of invariants can cause confusion.

Preachy. I've been trying to stick to invariants and covariant calculations throughout.

In the original rest frame (A's rest frame), $\beta$ is a function of time (i.e., coordinate time $t$), but not space (i.e., $\beta$ is not a function of $x$). However, in any other inertial frame, $\beta$ will be function of both $t'$ and $x'$. The reason A's original rest frame is different in this respect is that in that frame, both spaceships start accelerating (i.e., firing their rockets) simultaneously (at time $t = 0$). In any other frame, because of relativity of simultaneity, the ships start accelerating at different times.

Well, a proper model will take all this into account.

 

[PeterDonis]

The spaceships have synchronised starts but the strings are being towed so they have a lag.

Strictly speaking, yes, all the pieces of the string won't start moving at the same time (in A's original rest frame) as the ships do. The assumption of constant proper acceleration for all the pieces of the string is an idealization. However, dropping the idealization doesn't change the conclusion: it just makes the math more complicated (because you have to show that the expansion scalar I computed is the invariant that describes the *average* motion of the string).

Preachy. I've been trying to stick to invariants and covariant calculations throughout.

I didn't mean to be preachy; but your question about time dependence that I was responding to doesn't make sense if you're only looking at invariant/covariant quantities, since the "time dependence" you were describing is dependence on the coordinate time in a particular frame. The only invariant/covariant time dependence in the problem is dependence on proper time along one of the given worldlines (the worldline of one of the spaceships or a piece of the string).

 

[WannabeNewton]

Peter, the length issue could be a cruncher, so it could be back to the pencil and note book for me.

You keep trying to do calculations but you aren't thinking about the physics. Put down the pencil and note and set aside the blind calculations. Just think about what's going on conceptually in the inertial frame. It's really simple. Length contraction causes the equilibrium (natural) length of the string to continuously decrease in the inertial frame but the length of the string itself in the inertial frame is held constant by being fastened between the two instantaneously equal velocity spaceships. Clearly there comes a point when the stresses are too much to sustain the difference between fixed length in the inertial frame and equilibrium length in the inertial frame and the string breaks.

 

[Mentz114]

You keep trying to do calculations but you aren't thinking about the physics. Put down the pencil and note and set aside the blind calculations. Just think about what's going on conceptually in the inertial frame. It's really simple. Length contraction causes the equilibrium (natural) length of the string to continuously decrease in the inertial frame but the length of the string itself in the inertial frame is held constant by being fastened between the two instantaneously equal velocity spaceships. Clearly there comes a point when the stresses are too much to sustain the difference between fixed length in the inertial frame and equilibrium length in the inertial frame and the string breaks.

I only believe fully covariant calculations. I would prefer being guided to the right way to do the calculation myself.

If I'm not able to play the acceleration gradient card, then there must be a kinematic explanation, viz. one which deends only on the velocities, not accelerations. Perhaps I will end up in the right place.

 

[PAllen]

I only believe fully covariant calculations. I would prefer being guided to the right way to do the calculation myself.

If I'm not able to play the acceleration gradient card, then there must be a kinematic explanation, viz. one which deends only on the velocities, not accelerations. Perhaps I will end up in the right place.

It seems to me that you have one covariant explanation providing minimum 'explanation', and a choice of additional frame dependent explanations - each providing more 'motivation' but at the cost of being frame dependent.

1) The covariant description is simply that by virtue of how the congruence is specified, its expansion tensor is nonzero, thus the string is under growing tension or separation. The force causing this (assuming the string is passive) is the rockets pulling on it. Nothing can be said about distances because that requires specification of simultaneity. Nothing can be said about why the expansion tensor is nonzero except: that is how the congruence was defined. Different congruence, representing a different physical set up, would have no expansion.

2) In (any) inertial frame, the distance between the rockets is constant (once they are both accelerating), and the additional frame dependent explanation is that the reason the rockets cause tension and a nonzero expansion tensor is that the equilibrium length of the string contracts, but the rockets are forcing it to stay the same length.

3) In either rocket frame, realized, for example, as Fermi-Normal coordinates, the additional explanation is that the distance between the rockets grows, causing the tension and expansion of the string.

This is similar to any number of other situations where there is an invariant/covariant description, and multiple frame dependent explanations.

 

[Mentz114]

I have a calculation which may do the trick.

The expansion tensor is defined as $\theta_{\mu\nu}= \Theta h_{\mu\nu}$ where $h=g_{\mu\nu}+U_\mu U_\nu$. For $U_\mu=-\sinh(at) dt + \cosh(at) dx$ I get the components of $\theta_{\mu\nu}$

$\left[\begin{array}{cccc} a\,cosh\left( a\,t\right) \,\left( {sinh\left( a\,t\right) }^{2}-1\right) & -a\,{cosh\left( a\,t\right) }^{2}\,sinh\left( a\,t\right) & 0 & 0\\\ -a\,{cosh\left( a\,t\right) }^{2}\,sinh\left( a\,t\right) & a\,cosh\left( a\,t\right) \,\left( {cosh\left( a\,t\right) }^{2}+1\right) & 0 & 0\\\ 0 & 0 & a\,cosh\left( a\,t\right) & 0\\\ 0 & 0 & 0 & a\,cosh\left( a\,t\right) \end{array}\right]$

and if we boost this tensor by $\beta=at$ the components are.

$\left[\begin{array}{cccc} -a\,cosh\left( a\,t\right) & 0 & 0 & 0\\\ 0 & 2\,a\,cosh\left( a\,t\right) & 0 & 0\\\ 0 & 0 & a\,cosh\left( a\,t\right) & 0\\\ 0 & 0 & 0 & a\,cosh\left( a\,t\right) \end{array}\right]$

I think the answer is right there in the way those components change between the two frames. The $\theta_{tx}$ is crucial.

Sorry about the matrix explosion, it's hot off the presses.

 

[WannabeNewton]

This is similar to any number of other situations where there is an invariant/covariant description, and multiple frame dependent explanations.

Which is exactly what I've been trying to say over and over again in this thread. And yet there has been no progress since the first page of this thread because blind calculations are being used to substitute for conceptual understanding, to no avail unsurprisingly.

I feel that some advice from Feynman is in order:

https://www.youtube.com/watch?v=obCjODeoLVw

tl;dr no calculation you do is useful if you don't know how to actually interpret it in terms of physics.

 

[Mentz114]

tl;dr no calculation you do is useful if you don't know how to actually interpret it in terms of physics.

Please stop the anodyne advice and disparaging remarks about my lack of understanding of physics.

The calculation in my last post explains the situation exactly without videos or handwaving. The expansion tensor shows how time enters the expansion in the inertial frame. Do you understand what that calculation is saying ? I do.

 

[PeterDonis]

This chart of the Minkowski metric $ds^2= -x^2dt^2+t^2dx^2+dy^2+dz^2$

Can you give the coordinate transformation from the standard Minkowski chart to this one? It's not the standard Rindler chart; that line element doesn't have the $t^2$ coefficient in front of $dx^2$. But with that coefficient there I'm not sure what the coordinate transformation is supposed to be.

gives an acceleration of $1/x$ and expansion scalar $1/xt$.

I assume you mean that these are the proper acceleration and expansion scalar for the congruence of worldlines that are at rest (i.e., have constant $x$, $y$, $z$ coordinates) in this chart, correct? Acceleration and expansion are properties of congruences of worldlines, not coordinate charts.

The expansion tensor is defined as $\theta_{\mu\nu}= \Theta h_{\mu\nu}$ where $h=g_{\mu\nu}+U_\mu U_\nu$.

[STRIKE]I'm not sure this is a correct expression for the expansion tensor.[/STRIKE] This expression for the expansion tensor is restricted to the case where the shear is zero (see below). I'm assuming that $\Theta$ is supposed to be the expansion scalar, which is the trace of the expansion tensor; but the expansion tensor itself includes shear as well as expansion (shear is the trace-free part, expansion is the trace). The formula I'm used to seeing for the full expansion tensor (including shear) is

$$
\theta_{\mu \nu} = \frac{1}{2} h^{\alpha}{}_{\mu} h^{\beta}{}_{\nu} \left( \partial_{\beta} U_{\alpha} + \partial_{\alpha} U_{\beta} \right)
$$

[STRIKE]I'm not sure I see how to get from that to the formula you give.[/STRIKE] If the trace-free part of this (i.e., the shear) is zero, then this can be reduced to the expression you give (note that a factor of 1/3 is normally included because there are three spatial dimensions and the expansion tensor is supposed to be purely spatial).

In the particular case we're discussing, the shear *is* zero, so all of the information in the expansion tensor is contained in its trace, the expansion scalar; so I'm not sure why you are computing the full tensor anyway. The expansion scalar, as we've seen, is just $\Theta = \partial_{\mu} U^{\mu}$, which can be computed in any inertial frame; you just need the correct transformed expression for $U^{\mu}$ in that frame.

For $U_\mu=-\sinh(at) dt + \cosh(at) dx$

I'm not sure this is right either. The 4-velocity field in question is $U^{\mu} = \gamma \partial_t + \gamma \beta \partial_x$; lowering an index on this gives $U_{\mu} = \eta_{\mu \nu} U^{\nu} = - \gamma dt + \gamma \beta dx$. Since $\gamma = \cosh (at)$ and $\gamma \beta = \sinh (at)$, this gives $U_{\mu} = - \cosh (at) dt + \sinh (at) dx$.

if we boost this tensor by $\beta = a t$

I'm also not sure what you mean by this. Are you transforming from one inertial frame to another? If so, $\beta$ should be a constant, not a function of $t$, which is what it looks like you're doing here. Unless what you mean is that you are picking some particular value of $t$, and boosting everywhere by the (constant) value of $\beta$ associated with that value of $t$; but if you're doing that, then you are basically boosting into the instantaneous rest frame of one of the spaceships at time $t$, and the expansion tensor you get should be purely spatial in that frame (i.e., it should have no 0-0 component, which yours does).

 

[WannabeNewton]

Please stop the anodyne advice and disparaging remarks about my lack of understanding of physics.

I didn't mean to come off as disparaging and I profusely apologize if I did and I certainly didn't mean to say anything about your understanding of physics. My point was simply that these calculations aren't going to do you any favors. They aren't telling you anything you already don't know and they aren't shedding light on the dynamics leading up to the breaking of the string as observed in the inertial frame. You can do all the calculations you want but in the end all they're telling you is "the string breaks" which is a trivial statement. If you're going to object to the role of length contraction in breaking the string in the inertial frame then you're going to have to come up with an alternative physical explanation for why the string breaks in this inertial frame. You still haven't done that. Calculations are not a substitute for physical explanation.

If I ask you "why are the rest frames of spaceships in uniform circular orbits at the photon radius in Schwarzschild space-time non-rotating for all possible angular velocities?" and you tell me "well because you can write down the tangent Killing field $\eta^{\mu} = \delta^{\mu}_t + \omega \delta^{\mu}_{\phi}$ and show that $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]}|_{R = 3M} = 0$ hence the rest frames of the spaceships following orbits of $\eta^{\mu}$ are non-rotating" well this doesn't really tell me why now does it? It just tells me, using the covariant twist of the Killing field, that it does indeed happen, which I already know. I want to know why it happens.

The same goes for the Bell spaceship paradox but the "why" depends on the frame of reference, that's all.

 

[Mentz114]

Peter, thanks a lot for finding the error in U. I recalculated and it got rid of a problem. So now
$U_\mu=-\cosh(at)\partial_t+\sinh(at)\partial_x$ I get the tensor components

$\theta_{\mu\nu}=\left[\begin{array}{cccc} a\,\left( {\cosh\left( a\,t\right) }^{2}-1\right) \,\sinh\left( a\,t\right) & -a\,\cosh\left( a\,t\right) \,{\sinh\left( a\,t\right) }^{2} & 0 & 0\\\ -a\,\cosh\left( a\,t\right) \,{\sinh\left( a\,t\right) }^{2} & a\,{\cosh\left( a\,t\right) }^{2}\,\sinh\left( a\,t\right) & 0 & 0\\\ 0 & 0 & a\,\sinh\left( a\,t\right) & 0\\\ 0 & 0 & 0 & a\,\sinh\left( a\,t\right) \end{array}\right]$

This is in the coordinate basis of the inertial observer. To transform the tensor to the local basis of the accelerating ship we use the tetrad (vierbein) based on U, which happens to be the LT that connects the bases. So $\theta_{AB} = \Lambda^\mu_A \Lambda_B^\nu \theta_{\mu\nu}$. This gives the components

$\theta_{AB}=\left[\begin{array}{cccc} 0 & 0 & 0 & 0\\\ 0 & a\,\sinh\left( a\,t\right) & 0 & 0\\\ 0 & 0 & a\,\sinh\left( a\,t\right) & 0\\\ 0 & 0 & 0 & a\,\sinh\left( a\,t\right) \end{array}\right]$

This is exactly right. I also checked and found that $\theta_{AB}\theta^{AB}=\theta_{\mu\nu}\theta^{\mu\nu}=3\Theta^2$.

I could do with some help interpreting these. If we take only the spatial parts we can see a factor of $\cosh(at)^2$ between the $xx$ components. So something is different expansion-wise between the frames. I still hope this is the kinematic effect that explains the breaking of the string in the inertial frame.

 

[PeterDonis]

A note: the usual notation for the expansion tensor is $\theta_{\mu \nu}$; the notation $\omega_{\mu \nu}$ is normally used for the vorticity.

If we take only the spatial parts

But you can't just compare spatial parts; that breaks covariance. The expansion tensor is only purely spatial in the instantaneous rest frame of the ship. In fact that's exactly what your computation shows; just run it in the other direction. That is, your computation shows that, if we start with the purely spatial expansion tensor in the instantaneous rest frame of the ship, and then transform to a different inertial frame, the tensor is no longer purely spatial. See further comments below.

something is different expansion-wise between the frames.

No, it isn't, because, as you show, the expansion scalar $\Theta$ is the same in both frames. The scalar is the physical invariant; the different forms of the tensor in different frames are just different ways of representing that physical invariant.

It's true that only one representation--the one in the instantaneous rest frame of the ship--gives a purely spatial tensor, as above; but that's to be expected, because in any other frame the 4-velocity of the ship at that event on its worldline is not "purely temporal" either (i.e., $U^{\mu}$ has nonzero spatial components in any frame other than the instantaneous rest frame). The covariant way of saying that the expansion tensor is purely spatial in the instantaneous rest frame is to say that it always lies in the hypersurface that is orthogonal to the 4-velocity; in other words $U^{\mu} \theta_{\mu \nu} = 0$ must always hold. But that means that if $U^{\mu}$ has nonzero spatial components, $\theta_{\mu \nu}$ must have a nonzero time component for the orthogonality condition to hold.

 

[Mentz114]

I didn't mean to come off as disparaging and I profusely apologize if I did and I certainly didn't mean to say anything about your understanding of physics.

Sure, I was too touchy. No problem.

My point was simply that these calculations aren't going to do you any favors. They aren't telling you anything you already don't know and they aren't shedding light on the dynamics leading up to the breaking of the string as observed in the inertial frame. You can do all the calculations you want but in the end all they're telling you is "the string breaks" which is a trivial statement. If you're going to object to the role of length contraction in breaking the string in the inertial frame then you're going to have to come up with an alternative physical explanation for why the string breaks in this inertial frame. You still haven't done that. Calculations are not a substitute for physical explanation.

I know that the physics - i.e. the string breaks is indisputable. The only reason it is necessary to explain this is to show that SR is consistent. That is the first step in finding a physical interpretation. This would bring in the material properties of the string including its inertia.

I'm not there yet, but I think the kinematic part is done.

...
The same goes for the Bell spaceship paradox but the "why" depends on the frame of reference, that's all.

I would say the 'how' is what I'm after, and I admit that it can be frame dependent like the components of $\omega_{\mu\nu}$.

Anyhow, I appreciate your input.

 

[Mentz114]

A note: the usual notation for the expansion tensor is $\theta_{\mu \nu}$; the notation $\omega_{\mu \nu}$ is normally used for the vorticity.

Yes, sorry. That's an annoying mistake.

something is different expansion-wise between the frames.

No, it isn't, because, as you show, the expansion scalar $\Theta$ is the same in both frames. The scalar is the physical invariant; the different forms of the tensor in different frames are just different ways of representing that physical invariant.

Sorry, I phrased myself badly. What I meant was that the tensors have different components. Right now that is all I'm claiming. (Sheesh, you got to be careful what you say around here ).

 

[PeterDonis]

What I meant was that the tensors have different components. Right now that is all I'm claiming.

Fair enough.

(Sheesh, you got to be careful what you say around here ).

I didn't mean to come across as nitpicking. But the statement I was responding to was immediately after "if we take only the spatial parts" (probably I should have quoted that as well to help clarify where I was coming from), which you can't really do, even if you're just looking for a "kinematic" explanation. Even if you note that the spatial tensor components differ, you also should note that the temporal components differ as well (zero in one frame, nonzero in another), and factor that into your kinematic analysis.

(Personally, I have a hard time doing that for a tensor whose usual physical interpretation depends on it being purely spatial. What does "expansion in the time direction" mean? But I think if you're going to have a complete kinematic interpretation of what's going on that co-varies properly between frames, you have to include the nonzero temporal components somehow. Taking the trace is one obvious way to do that, but you seem to be resisting going that route.)

 

[Mentz114]

Fair enough.



... the statement I was responding to was immediately after "if we take only the spatial parts" (probably I should have quoted that as well to help clarify where I was coming from), which you can't really do, even if you're just looking for a "kinematic" explanation. Even if you note that the spatial tensor components differ, you also should note that the temporal components differ as well (zero in one frame, nonzero in another), and factor that into your kinematic analysis.

I was thinking about projecting into a 3D submainfold to represent the material. As in 'relativistic elastodynamics'. But I'm probably not going to attempt that .

(Personally, I have a hard time doing that for a tensor whose usual physical interpretation depends on it being purely spatial. What does "expansion in the time direction" mean? But I think if you're going to have a complete kinematic interpretation of what's going on that co-varies properly between frames, you have to include the nonzero temporal components somehow. Taking the trace is one obvious way to do that, but you seem to be resisting going that route.)

I agree with this.

Responding in general, if we say that the potential energy that holds the material together depends on time dilation, then a simple (physical ?) explanation is that the string wants to get shorter because of this but is kept at the same length. But the string already length contracted in this frame so it sort of cancels out.

I'm going to look at what $\theta_{\mu\nu}$ means physically. It reminds me of tidal forces.

 

[WannabeNewton]

Responding in general, if we say that the potential energy that holds the material together depends on time dilation, then a simple (physical ?) explanation is that the string wants to get shorter because of this but is kept at the same length. But the string already length contracted in this frame so it sort of cancels out.

The intermolecular potential between elements of the string (Lennard-Jones potential) is time-independent so time dilation plays no role here. Furthermore the string has not length contracted. It's equilibrium length is length contracting but its actual length in the inertial frame is fixed between the spaceships which is precisely why it eventually breaks. The actual length of the string in the inertial frame only length contracts if the spaceships are accelerated Born rigidly.

I'm going to look at what $\theta_{\mu\nu}$ means physically. It reminds me of tidal forces.

See section 2.8 of Malament's text.

 

[stevendaryl]

The intermolecular potential between elements of the string (Lennard-Jones potential) is time-independent so time dilation plays no role here.

It seems to me that if a molecule is attracted to a second molecule, and the second molecule is accelerating, then the force felt by the first molecule is time-dependent.

A "toy" model of a relativistic string that features some of the properties of real strings is this: Imagine that each molecule is equipped with a clock, a radio transmitter and receiver, and a little rocket. Each molecule continuously sends signals to its neighbors, who immediately send a return signal. The clock is used to time the round-trip signal. If the round-trip time for an exchange with a neighbor is longer than some cut-off, then it accelerates toward that neighbor. If it is longer than a second cut-off, then it gives up, and assumes the chain has been broken.

 

[PeterDonis]

I was thinking about projecting into a 3D submainfold to represent the material.

That's what $h_{\mu \nu} = \eta_{\mu \nu} + U_{\mu} U_{\nu}$ does; it's a projection tensor into the 3D hypersurface that's orthogonal to the 4-velocity at a given event. But for an accelerated worldline, two such hypersurfaces at different events are not parallel to each other. That's why a time component appears in the expansion tensor when you transform it from the instantaneous rest frame of the ship to a different frame.

if we say that the potential energy that holds the material together depends on time dilation, then a simple (physical ?) explanation is that the string wants to get shorter because of this but is kept at the same length.

Hm, I hadn't thought of looking at it this way. Interesting.

But the string already length contracted in this frame

In which frame? The original rest frame (i.e., A's rest frame)? In that frame the string is *not* length contracted; only its "natural" or "unstressed" length is. Basically the viewpoint you are suggesting here says that contraction of the string's "unstressed length" is due to the time dilation of the string because of relative velocity, and the effect of that time dilation on potential energy. No cancellation there that I can see.

 

[Mentz114]

The intermolecular potential between elements of the string (Lennard-Jones potential) is time-independent so time dilation plays no role here. Furthermore the string has not length contracted. It's equilibrium length is length contracting but its actual length in the inertial frame is fixed between the spaceships which is precisely why it eventually breaks. The actual length of the string in the inertial frame only length contracts if the spaceships are accelerated Born rigidly.

OK, that's good. It was too facile as I thought.

See section 2.8 of Malament's text.

Loads of interesting stuff there. It will take some time for me absorb all that.

 

[Mentz114]

That's what $h_{\mu \nu} = \eta_{\mu \nu} + U_{\mu} U_{\nu}$ does; it's a projection tensor into the 3D hypersurface that's orthogonal to the 4-velocity at a given event. But for an accelerated worldline, two such hypersurfaces at different events are not parallel to each other. That's why a time component appears in the expansion tensor when you transform it from the instantaneous rest frame of the ship to a different frame.

OK. I'm still trying to understand what this will mean to the string as perceived by the inertial frame. I must study Mallaments stuff on this.

In which frame? The original rest frame (i.e., A's rest frame)? In that frame the string is *not* length contracted; only its "natural" or "unstressed" length is. Basically the viewpoint you are suggesting here says that contraction of the string's "unstressed length" is due to the time dilation of the string because of relative velocity, and the effect of that time dilation on potential energy. No cancellation there that I can see.

Apparently time dilation does not have the necessary effect ( see WNB's post above).

Thanks for your (and others) replies and patience to stick with my ramblings. Maybe it's hard to believe but I have learned a lot from this and I'm pleased I found the mystery component (hidden in clear view), even if it turns out to be irrelevant.

 

[WannabeNewton]

OK, that's good. It was too facile as I thought.

Actually in light of Steven's comment, I'm not so sure about my time dilation comment. I'm not entirely sure in what sense you're using time dilation with regards to the intermolecular potential between elements of the string as far as the Bell spaceship paradox goes. Could you perhaps expand on your thought?

 

[Mentz114]

Actually in light of Steven's comment, I'm not so sure about my time dilation comment. I'm not entirely sure in what sense you're using time dilation with regards to the intermolecular potential between elements of the string as far as the Bell spaceship paradox goes. Could you perhaps expand on your thought?

It seems there's two ways to put tension into a string. Pulling the ends apart or changing the forces (gradients of a potential) that hold it in equilibrium internally while the ends are kept a fixed distance apart. Gravitational time dilation sometimes acts like a potential so maybe an accelerative pseudo-potential can do the same. I don't know enough about solid state physics to judge if this approach could be correct.

But I couldn't accept it if it was not expressed in (covariant) equations.

I have two avenues right now - (1) the effect of the expansion tensor ( or extrinsic curvature tensor of the spatial hyperslices) on the separation vectors or (2) a tidal tensor which could be something like $\dot{\Theta} h_{\mu\nu}$. $\dot{\Theta}$ is the time derivative of a velocity field and could be interpreted as an acceleration field, maybe. All a bit loose, but it's edifying mining Malament for useful nuggets.

( A question about Malament section 2.8- are the separation vectors $\eta^a$ spatial, so they have a direction and a length ?).

Sadly for me I don't have a lot of time to spend on this for a few days.