Solving the Spaceship Paradox: A New Explanation

[A.T.]

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.

I think the assumption in Bell's spaceship scenario is that the acceleration is small, so that the speed of sound in the rope is not a limiting factor. The point of the scenario is that no matter how cautiously you accelerate, the rope will break anyway.

 

[WannabeNewton]

Gravitational time dilation sometimes acts like a potential so maybe an accelerative pseudo-potential can do the same.

Gravitational (resp. pseudo gravitational) time dilation indeed arises from a gravitational (resp. pseudo-gravitational) potential in stationary space-times but there is only a gradient if the potential is varying with height or equivalently if the orbits of the stationary Killing field are observers of different proper accelerations. The observers in the Bell spaceship paradox all have the same proper acceleration and they don't even follow orbits of the same time-like Killing field so no that won't lead to the string breaking because there won't be such a gradient.

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

Only at the initial event considered. At the very next event they will no longer be orthogonal to the tangent field they are being Lie transported along unless the tangent field happens to be a Killing field or a geodesic field.

 

[Mentz114]

Gravitational (resp. pseudo gravitational) time dilation indeed arises from a gravitational (resp. pseudo-gravitational) potential in stationary space-times but there is only a gradient if the potential is varying with height or equivalently if the orbits of the stationary Killing field are observers of different proper accelerations. The observers in the Bell spaceship paradox all have the same proper acceleration and they don't even follow orbits of the same time-like Killing field so no that won't lead to the string breaking because there won't be such a gradient.

Yes I think this was established earlier. I'm was surprised you asked.

Only at the initial event considered. At the very next event they will no longer be orthogonal to the tangent field they are being Lie transported along unless the tangent field happens to be a Killing field or a geodesic field.

OK, as expected.

I found that $\dot{\Theta}<0$ for the boosted congruence. Still looking for a negative x-component in something like a tidal tensor. We know they can do work and deform stuff.

 

[PAllen]

I don't know if this will help any, but one way state this issue is:

Independent of frame or coordinates, the reason the string breaks is that the rockets are pulling on it, and in the local frame of any piece of the string, its neighbors are moving away (due to rocket pull). This is exactly what the expansion tensor is saying.

The only frame dependent description is what the length of the overall string is (equivalently, the distance between the rockets). It happens that there exists a frame where the frame dependent length contraction exactly balances the local (invariant) expansion such that the rocket distance remains constant and the string length remains constant until it breaks.

 

[Mentz114]

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.

That is interesting. I've been trying to imagine a lot of scenarios similar to this. Would this model work if the an element used Doppler to estimate the velocity of its neighbours and acted to reduce the relative velocity ?

 

[Mentz114]

I don't know if this will help any, but one way state this issue is:

Independent of frame or coordinates, the reason the string breaks is that the rockets are pulling on it, and in the local frame of any piece of the string, its neighbors are moving away (due to rocket pull). This is exactly what the expansion tensor is saying.

OK, I think that is clear.

The only frame dependent description is what the length of the overall string is (equivalently, the distance between the rockets). It happens that there exists a frame where the frame dependent length contraction exactly balances the local (invariant) expansion such that the rocket distance remains constant and the string length remains constant until it breaks.

Ok. I'm having trouble getting equations for this but I'll try harder.

 

[PeterDonis]

Since it's relevant to the topic of this thread, here's a link to the new FAQ entry that has been posted in this forum on the Bell Spaceship Paradox:

https://www.physicsforums.com/showthread.php?t=742729

 

[Mentz114]

Since it's relevant to the topic of this thread, here's a link to the new FAQ entry that has been posted in this forum on the Bell Spaceship Paradox:

https://www.physicsforums.com/showthread.php?t=742729

Having had time to read and think more about this I am pretty sure the correct explanation of why the string breaks in the inertial frame is the change in the components of the expansion tensor $\theta_{\mu\nu}$. According to Malament the expansion tensor acts like a tidal tensor and can produce volume and shape changes in the distribution of nearby members of the congruence.

The difference in the expansion tensor in the 3D space carried along the congruence and the 3D space of the inertial observer is a factor of $\gamma^2$ which arises from $U_\mu U_\nu$. This is a stretching force which grows indefinately and is bound to break the thread eventually.

This explanation fits in with the way that tidal forces work and is the result of a covariant calculation and to me is much more convincing than the one in the FAQ.

 

[WannabeNewton]

And we're back to square one...

I refer you again to my comment: https://www.physicsforums.com/showpost.php?p=4675992&postcount=37

$\theta \neq 0$ is not an explanation. It's simply an observation. Why is $\theta \neq 0$? I've already explained this above and you still have not substantiated it.

 

[Mentz114]

And we're back to square one...

I refer you again to my comment: https://www.physicsforums.com/showpost.php?p=4675992&postcount=37

$\theta \neq 0$ is not an explanation. It's simply an observation. Why is $\theta \neq 0$? I've already explained this above and you still have not substantiated it.

The expansion tensor in the local spaceship basis is $diag(\Theta,\Theta,\Theta)$ and in a colocated inertial frame it is $diag(\gamma^2\Theta,\Theta,\Theta)$. Are we going to ignore this ? The first is shape but not volume preserving, the second is neither. The material (in the POV of) the inertial frame is undergoing an increasing tidal force in the direction of motion. Tidal forces break things.

 

[WannabeNewton]

You haven't answered my question. What is the physical reason for why $\theta \neq 0$? I've already explained to you why simply saying $\theta \neq 0$ is not a valid explanation. It's simply the frame-invariant statement that the string breaks, nothing more and nothing less. It doesn't tell you why. You're constantly ignoring this.

 

[PeterDonis]

The difference in the expansion tensor in the 3D space carried along the congruence and the 3D space of the inertial observer is a factor of $\gamma^2$ which arises from $U_\mu U_\nu$. This is a stretching force which grows indefinately and is bound to break the thread eventually.

The problem with this argument is that it appears to say that the string should break in the 3D space of the inertial observer, but *not* in the 3D space carried along the congruence, because you are attributing the breaking of the string to the *change* in the expansion (because of the change in $\gamma$), but in the 3D space carried along the congruence the expansion does not change (but it is nonzero). This can't be right, because the string breaking is an invariant; any anaylsis done in any frame, including the frame carried along the congruence, has to give the same answer.

If you are going to look at the expansion, you have to explain why the expansion being nonzero, whether or not its value changes, is sufficient for the string to stretch and break.

 

[PeterDonis]

I've already explained to you why simply saying $\theta \neq 0$ is not a valid explanation.

As I'm reading his argument, he isn't saying the string breaks because $\theta \neq 0$. He's saying it breaks because $\theta$ is *increasing*. That can't be right; see my previous post.

 

[Mentz114]

I seem to remember that $\dot{\Theta}\neq 0$ (Raychaudhuri's number) but I'm not sure if that is relevant.

Thanks to both of you. I'll read your posts and think about it. I'm also getting a bit fatigued with this topic, so maybe it's time for me to move on.

 

[PeterDonis]

I seem to remember that $\dot{\Theta}\neq 0$ (Raychaudhuri's number) but I'm not sure if that is relevant.

For the case under discussion, yes, I believe Raychaudhuri's equation gives $\dot{\Theta} \neq 0$. But the invariant that corresponds to the string breaking is $\Theta \neq 0$, not $\dot{\Theta} = 0$.

 

[PeterDonis]

For the case under discussion, yes, I believe Raychaudhuri's equation gives $\dot{\Theta} \neq 0$.

Just to confirm this, here's a quick computation of the expansion scalar $\theta$ (I'll stick to the lower-case Greek letter here since that's the standard symbol) and the Raychaudhuri equation for the Bell congruence.

We'll work in the inertial frame in which the ships are initially at rest. In that frame we have the following for the coordinates $(t, x)$ of a given ship [Edit: I've normalized so that the proper acceleration of each ship is $1$; that means $t$ and $x$ are essentially dimensionless coordinates, in ordinary units what I'm writing as $t$ and $x$ would be $at$ and $ax$, where $a$ is the proper acceleration]:

$$
(t, x) = (\gamma v, x_0 - 1 + \gamma)
$$

where $\gamma = \cosh \tau$ and $\gamma v = \sinh \tau$ ($\tau$ is the proper time along any ship's worldline [Edit: again, this is normalized, in ordinary units what I'm writing as $\tau$ would be $a \tau$]), and $x_0$ is the $x$ coordinate at which the ship starts (i.e., its $x$ coordinate before it turns on its engines and begins accelerating). Taking derivatives with respect to $\tau$ gives us the 4-velocity $(u^t, u^x$ and the 4-acceleration $(a^t, a^x)$:

$$
(u^t, u^x) = (\gamma, \gamma v)
$$

$$
(a^t, a^x) = (\gamma v, \gamma)
$$

It's straightforward to eliminate $\tau$, by observing that $\tau = \sinh^{-1} t$ and working through the math accordingly; we find that $\gamma$ and $\gamma v$ are functions of $t$ only, such that $\partial_t \gamma = v$ and $\partial_t \left( \gamma v \right) = 1$. (It's actually evident that these relations must hold from looking at the $(t, x)$ vector above and realizing that $v = dx / dt$.)

Armed with all this, computing the expansion scalar is simple, because we're in a global inertial frame so all the connection coefficients are zero; we get

$$
\theta = \partial_a u^a = \partial_t u^t = v
$$

So $\theta > 0$ as soon as the ships start moving. The Raychaudhuri equation (with terms that are identically zero not shown) is:

$$
\dot{\theta} = - \frac{\theta^2}{3} + \partial_a a^a = - \frac{v^2}{3} + \partial_t a^t = 1 - \frac{v^2}{3} = \frac{1}{\gamma^2} + \frac{2}{3} v^2
$$

If this is correct, then $\dot{\theta}$ starts at $1$ but approaches a limit of $2/3$ as $t \rightarrow \infty$.

The only thing I'm not sure about is the factor of $1/3$ in the first term; it's possible that this is due to a different definition of $\theta^2$ than just the square of the scalar $\theta$ that I calculated above. It would be nice and neat if that factor of $1/3$ were not there, since that would make $\dot{\theta} = 1 / \gamma^2$, which looks nicer (and would make it approach $0$ as $t \rightarrow \infty$, which is also neater). But as far as I can tell the factor of $1/3$ should be there given the definitions I'm using.

 

[Mentz114]

You haven't answered my question. What is the physical reason for why $\theta \neq 0$? I've already explained to you why simply saying $\theta \neq 0$ is not a valid explanation. It's simply the frame-invariant statement that the string breaks, nothing more and nothing less. It doesn't tell you why. You're constantly ignoring this.

I'm not ignoring it. I'm trying to find a physical explanation - what you call the 'why' but I would say is the 'how'. That is whole point of the exercise. It may not be an appropriate analogy but tidal forces in curved spacetime are frame dependent. I'm hoping to find a model where the spaceships attribute the breaking to the separation they experience ( and maybe something else as well) but the inertial observer attributes the breaking to something other than separation.

I have calculated that the distance between the ships increases as $\gamma$, i.e. $x'_2-x'_1 = \gamma(x_2-x_1)$ which is like the effect a fixed separation and a shrinkage of $1/\gamma$.

The main problem with the effect I found is that it goes as $\gamma^2$, so I need to compensate for some time dilation to get a rudimentary match up.

But my main non-undertanding is the expansion scalar itself. Stephani defines it as 'an isotropic velocity field orthogonal to the congruence' and Wald calls it 'an average velocity field'. Why does it produce no shape-change in the rocket-frame ? They are experiencing proper acceleration in the x direction, so we do have a special direction.

It's very hard to get my head around all this.

 

[PeterDonis]

The only thing I'm not sure about is the factor of $1/3$ in the first term

I'm now even more not sure about it, because there's an obvious computation we can do as a check; just compute $\dot{\theta} = d \theta / d \tau = dv / d \tau$, since we know $\theta = v$. That's simple; we have $v = dx / dt = u^x / u^t = \tanh \tau$, so:

$$
\frac{dv}{d\tau} = \frac{d}{d\tau} \left( \tanh \tau \right) = 1 - \tanh^2 \tau = 1 - v^2 = \frac{1}{\gamma^2}
$$

This strongly suggests that the factor of $1/3$ should not be there in the Raychaudhuri equation given the definitions I'm using; it should just be

$$
\dot{\theta} = - \theta^2 + \partial_a a^a
$$

 

[WannabeNewton]

II'm hoping to find a model where the spaceships attribute the breaking to the separation they experience ( and maybe something else as well) but the inertial observer attributes the breaking to something other than separation.

The spaceships indeed attribute the breaking of the string to the continuously increasing separation they undergo in their instantaneous rest frames and the resulting stretching of the proper length of the string in their instantaneous rest frames. This comes directly from the physical interpretation of $\nabla_{\mu}u^{\mu}$. The inertial observer on the other hand attributes the string breaking to the continuous contraction of the equilibrium length of the string in conjunction with the constant length of the string in the inertial frame. You can use instantaneous Lorentz transformations to go from one explanation to the other.

Why does it produce no shape-change in the rocket-frame ?

You need to apply forces in directions non-parallel to the string in order for its shape to change. This is what the shear tensor codifies. The expansion scalar on the other hand represents radial forces parallel to the string so you cannot change the shape of the string, you can only change its length. The existence of a preferred direction doesn't really change anything. Take for example a rotating spherical shell. The velocity field of the shell has a non-zero vorticity whose axis of circulation defines a preferred direction. Imagine also that there is a gas inside the shell which on average provides an outwards isotropic (radial) force to the inner surface of the shell. The rotation is unaffected since there is no torque and the shell will have a non-vanishing expansion scalar driving it to increase in volume whilst retaining its spherical shape so the existence of a preferred direction doesn't change anything.

 

[Mentz114]

I'm now even more not sure about it, because there's an obvious computation we can do as a check; just compute $\dot{\theta} = d \theta / d \tau = dv / d \tau$, since we know $\theta = v$. That's simple; we have $v = dx / dt = u^x / u^t = \tanh \tau$, so:

$$
\frac{dv}{d\tau} = \frac{d}{d\tau} \left( \tanh \tau \right) = 1 - \tanh^2 \tau = 1 - v^2 = \frac{1}{\gamma^2}
$$

This strongly suggests that the factor of $1/3$ should not be there in the Raychaudhuri equation given the definitions I'm using; it should just be

$$
\dot{\theta} = - \theta^2 + \partial_a a^a
$$

I'm still trying to understand your earlier post. Have you seen this http://arxiv.org/pdf/gr-qc/0511123.pdf or http://arxiv.org/abs/1012.4806. Also section 9.2 in Wald.

However, $\dot{\theta}$ is not vital to my argument. Thanks for the calculations in any case.

 

[Mentz114]

The spaceships indeed attribute the breaking of the string to the continuously increasing separation they undergo in their instantaneous rest frames and the resulting stretching of the proper length of the string in their instantaneous rest frames. This comes directly from the physical interpretation of $\nabla_{\mu}u^{\mu}$. The inertial observer on the other hand attributes the string breaking to the continuous contraction of the equilibrium length of the string in conjunction with the constant length of the string in the inertial frame. You can use instantaneous Lorentz transformations to go from one explanation to the other.



You need to apply forces in directions non-parallel to the string in order for its shape to change. This is what the shear tensor codifies. The expansion scalar on the other hand represents radial forces parallel to the string so you cannot change the shape of the string, you can only change its length. The existence of a preferred direction doesn't really change anything. Take for example a rotating spherical shell. The velocity field of the shell has a non-zero vorticity whose axis of circulation defines a preferred direction. Imagine also that there is a gas inside the shell which on average provides an outwards isotropic (radial) force to the inner surface of the shell. The rotation is unaffected since there is no torque and the shell will have a non-vanishing expansion scalar driving it to increase in volume whilst retaining its spherical shape so the existence of a preferred direction doesn't change anything.

Thanks. I'm looking at "The expansion scalar on the other hand represents radial forces parallel to the string so you cannot change the shape of the string, you can only change its length. "

Does this mean we cannot interpret the $\theta_{ab}$ as a deformation tensor because it predicts expansion in every direction around a point on the congruence ?

Since I wrote the above, I have found that Malament analyses this in great detail before equation 2.8.15 where he gets the directional change of volume. I'll get into this over the weekend.

 

[PeterDonis]

I'm still trying to understand your earlier post.

If it helps, think of what I was doing in that post as follows: I was taking the expansion scalar $\theta$ (and its proper rate of change $\dot{\theta}$ as well, though as you say that's not actually needed for the argument) as known, and figuring out what, if any, quantities in the original inertial frame correspond to them. As it turns out, $\theta = v$ in that frame, i.e., the numerical value of the expansion scalar (which is an invariant, it is the same at a given event in every frame) is the same as the numerical value of the velocity of the spaceships in that frame.

Have you seen this http://arxiv.org/pdf/gr-qc/0511123.pdf or http://arxiv.org/abs/1012.4806.

Thanks for the links, I hadn't seen these before but they look interesting.

However, $\dot{\theta}$ is not vital to my argument.

Ok, good. I wanted to calculate $\dot{\theta}$ as well as $\theta$ in any case, to make sure I had a complete picture of what was happening; I expected $\dot{\theta} > 0$ on physical grounds, but I wanted to confirm it.

 

[WannabeNewton]

Does this mean we cannot interpret the $\theta_{ab}$ as a deformation tensor because it predicts expansion in every direction around a point on the congruence ?

$\theta_{ab}$ contains both $\theta$ and $\sigma_{ab}$ so it generates both isotropic (radial) volume change, which is what $\theta$ codifies, and shape deformation, which is what $\sigma_{ab}$ codifies.

In fact using an instantaneously comoving local inertial frame of an infinitesimal volume element $\mathcal{V}$ being carried i.e. Lie transported by the flow generated by $u^{\mu}$ it's easy to show that $\theta = \nabla_{\mu}u^{\mu} = \frac{1}{\mathcal{V}}u^{\mu}\nabla_{\mu}\mathcal{V}$. The effect(s) of $\sigma_{ab}$ can be determined by diagonalizing it through its eigenbasis (spectral theorem) and looking at how the eigenvalues affect an initially spherical ball Lie transported by the flow generated by $u^{\mu}$. Clearly the eigenbasis will constitute the principal axes of the deformation and the eigenvalues will determine the rate of deformation-in the absence of expansion the spherical ball will be deformed into an ellipsoid with major and minor axes determined by the deformation rate of the principal axes (the eigenbasis).

 

[PeterDonis]

To expand on (pun intended ) my previous computation of the expansion, I found an archived version of the Wikipedia page on the spaceship paradox, in which there is a computation (by Chris Hillman, I believe) of the full frame field and the expansion tensor:

http://en.wikipedia.org/w/index.php?title=Bell's_spaceship_paradox&oldid=57888610

His result for $\theta$ is basically the same as mine, but his result is not normalized (i.e., he allows arbitrary proper acceleration instead of normalizing it to $1$), so it actually shows something that my results didn't show; if we de-normalize so that the proper acceleration is $a$, then in the original inertial frame, we have $\theta = a v$ (instead of $\theta = v$). He does not compute $\dot{\theta}$, but it's straightforward to show that $\dot{\theta} = a^2 / \gamma^2$.

These results are important because they show the correct units for the expansion: $\theta$ has units of acceleration, and $\dot{\theta}$ has units of acceleration squared (assuming "natural" units where $c = 1$, so velocity is dimensionless). That may help with physical interpretation.

One other thing from Hillman's computation is notable: the expansion tensor $\theta_{ab}$ has only one nonzero component [STRIKE]in the original inertial coordinate chart (which is the chart Hillman uses)[/STRIKE] [Edit: he gives the expansion tensor in the frame comoving with a particular ship], $\theta_{xx} = \theta$ (i.e., this component is the same as the trace, the expansion scalar). In other words, the expansion is only in the $x$ direction (the direction of motion of the ships), not in the other two directions. This means that, contrary to what I said (I think) in an earlier post in this thread, the shear of this congruence is *not* zero. [Edit: I had originally tried to compute the shear, but that was assuming that the expansion components were in the original inertial coordinate chart, which they aren't, see above.]

 

[PeterDonis]

This means that, contrary to what I said (I think) in an earlier post in this thread, the shear of this congruence is *not* zero.

To follow on from this, here is a quick computation of the shear in the frame momentarily comoving with a particular ship. We have $\theta_{\hat{x} \hat{x}} = \theta$ and all other components zero (note that in this frame we can treat all of these tensors as spatial 3-tensors). In the comoving frame, the projection tensor $h_{\hat{a} \hat{b}}$ is just the spatial identity tensor, diag(1, 1, 1). Then we have $\sigma_{\hat{a} \hat{b}} = \theta_{\hat{a} \hat{b}} - \frac{1}{3} \theta h_{\hat{a} \hat{b}}$, which gives:

$$
\sigma_{\hat{x} \hat{x}} = \theta - \frac{1}{3} \theta = \frac{2}{3} \theta
$$

$$
\sigma_{\hat{y} \hat{y}} = \sigma_{\hat{z} \hat{z}} = - \frac{1}{3} \theta
$$

Note the similarity to, for example, the tidal tensor surrounding a spherically symmetric body.

 

[PeterDonis]

Does this mean we cannot interpret the $\theta_{ab}$ as a deformation tensor because it predicts expansion in every direction around a point on the congruence?

Remember that the tensor $\theta_{ab}$ contains both expansion and shear; it can be decomposed as $\theta_{ab} = \sigma_{ab} + \frac{1}{3} \theta h_{ab}$, where the projection tensor $h_{ab}$ can be thought of as the spatial metric of the spacelike hypersurface that is orthogonal to the spaceship's worldline (at the event at which we are evaluating all these tensors). Since the shear is nonzero (see my previous posts), the tensor $\theta_{ab}$ is describing both volume expansion and deformation; essentially, the combination of expansion and shear in this case results in stretching the string in the direction of motion of the ships, with no change in its size in the other two directions, which changes both its volume and its shape.

 

[PeterDonis]

To follow on from this, here is a quick computation of the shear in the frame momentarily comoving with a particular ship.

And following on from *that*, a corrected computation of the Raychaudhuri equation, which fixes the issue I had noted before, about why my original computation did not match the obvious check of computing $d \theta / d \tau$. Since the shear is nonzero, there is an extra term in the equation that I left out before. The correct equation is:

$$
\dot{\theta} = - \frac{1}{3} \theta^2 - 2 \sigma^2 + \partial_a a^a
$$

We have $\theta = a v$ (I'll de-normalize as I did in the last couple of posts) and $2 \sigma^2 = \sigma_{ab} \sigma^{ab} = \theta^2 \left( \frac{4}{9} + \frac{1}{9} + \frac{1}{9} \right) = \frac{2}{3} \theta^2$ (note that since this is a scalar invariant, we can compute it in any frame we wish as long as we express the result in terms of scalar invariants--which $\theta$ is--so I've done it in the comoving frame to make it simple). Plugging in gives

$$
\dot{\theta} = - \theta^2 + \partial_a a^a = a^2 \left( 1 - v^2 \right) = \frac{a^2}{\gamma^2}
$$

which now matches what I got from the check computation when de-normalized (and explains why the factor of $1/3$ is there in the $\theta^2$ term). Sweet.

 

[Mentz114]

Thanks for the replies. I have not had time to absorb them entirely. What I did do yesterday was to calculate the directional derivative $\xi^b\partial_b V = V\theta$ (Malament 2.8.15). The volume function $V$ turns out to be $1/\sqrt{-g}$ which is 1 in our case, and obviously frame independent.

Peter, your post #75 is interesting because I got zero shear and did not look further. There's a lot of stuff here I struggling with. WBN's post#73 is a bit mind blowing.

 

[Mentz114]

To expand on (pun intended ) my previous computation of the expansion, I found an archived version of the Wikipedia page on the spaceship paradox, in which there is a computation (by Chris Hillman, I believe) of the full frame field and the expansion tensor:

http://en.wikipedia.org/w/index.php?title=Bell's_spaceship_paradox&oldid=57888610

His result for $\theta$ is basically the same as mine, but his result is not normalized (i.e., he allows arbitrary proper acceleration instead of normalizing it to $1$), so it actually shows something that my results didn't show; if we de-normalize so that the proper acceleration is $a$, then in the original inertial frame, we have $\theta = a v$ (instead of $\theta = v$). He does not compute $\dot{\theta}$, but it's straightforward to show that $\dot{\theta} = a^2 / \gamma^2$.

These results are important because they show the correct units for the expansion: $\theta$ has units of acceleration, and $\dot{\theta}$ has units of acceleration squared (assuming "natural" units where $c = 1$, so velocity is dimensionless). That may help with physical interpretation.

One other thing from Hillman's computation is notable: the expansion tensor $\theta_{ab}$ has only one nonzero component [STRIKE]in the original inertial coordinate chart (which is the chart Hillman uses)[/STRIKE] [Edit: he gives the expansion tensor in the frame comoving with a particular ship], $\theta_{xx} = \theta$ (i.e., this component is the same as the trace, the expansion scalar). In other words, the expansion is only in the $x$ direction (the direction of motion of the ships), not in the other two directions. This means that, contrary to what I said (I think) in an earlier post in this thread, the shear of this congruence is *not* zero. [Edit: I had originally tried to compute the shear, but that was assuming that the expansion components were in the original inertial coordinate chart, which they aren't, see above.]

Using the frame field in the Wiki archive above I have agreement with your calculations that follow the post above. What I did was to use the congruence $\vec{u}=\vec{e}_0$ to get the shear tensor $\sigma_{\mu\nu}= \nabla_{(\nu}u_{\mu)}+ \dot{u}_{(\mu}u_{\nu)}-\theta h_{\mu\nu}/3$ in the global inertial frame. This gives $\sigma^{\mu\nu}\sigma_{\mu\nu}=3\theta^2/2$. Now using the tertrad and inverse tetrad to transform $\sigma_{\mu\nu}$ to the local frame, the components of $\sigma_{\hat{\mu}\hat{\nu}}$ are the same as those you quote for the comoving inertial frame, viz

$\sigma_{\hat{x}\hat{x}}=2\theta/3,\ \ \sigma_{\hat{y}\hat{y}}=\sigma_{\hat{z}\hat{z}}=-\theta/3$ and of course $\sigma^{\hat{\mu}\hat{\nu}}\sigma_{\hat{\mu}\hat{\nu}}=3\theta^2/2$.

I've had a misunderstanding about the shear tensor which is sorted thanks to the example you gave.

What remains is whether $\sigma_{\mu\nu}$ has any physical significance. Can this be interpreted as the local shear tensor as 'seen' from the inertial frame ? There are problems with this because the trace is not zero, whether using all components or the projection into 3-space.

 

[WannabeNewton]

What remains is whether $\sigma_{\mu\nu}$ has any physical significance.

$\sigma_{\mu\nu}$ really only has a direct, subtlety-free, simple physical interpretation in the instantaneously comoving (local) inertial frames of observers following orbits of the 4-velocity field $u^{\mu}$ that $\sigma_{\mu\nu}$ is expressed in terms of. In such a case, we can interpret $\sigma_{\mu\nu}$ physically as follows. First, for simplicity, assume $\nabla_{\mu}u^{\mu} = 0$ and $\nabla_{[\mu}u_{\nu]} = 0$ so that we only have to deal with physical effects of $\sigma_{\mu\nu}$; also, let's take $u^{\mu}$ to be geodesic so that $\sigma_{\mu\nu} = \nabla_{(\mu}u_{\nu)}$

Consider now an observer following an orbit of $u^{\mu}$ and an instantaneously comoving (local) inertial frame of this observer. In this frame we have $\Gamma^{\mu}_{\nu\gamma}= 0$ at this instant and $u^{\mu} = \delta^{\mu}_0$ at this instant. Then since $u^{\mu}\sigma_{\mu\nu} = u^{\nu}\sigma_{\mu\nu} = 0$ only $\sigma_{ij} = \partial_i u^j + \partial_j u^i$ are non-vanishing. This is, up to some $\gamma$ factors, just the Newtonian expression for shear and hence carries the same interpretation: http://en.wikipedia.org/wiki/Deformation_(mechanics)

 

[PeterDonis]

the trace is not zero, whether using all components or the projection into 3-space.

No, the trace is zero. The trace is $\sigma^{\mu}{}_{\mu}$, not $\sigma^{\mu \nu} \sigma_{\mu \nu}$. The shear tensor is always trace-free. In the comoving frame, $\sigma^{\mu}{}_{\mu} = \sigma^{\hat{x}}{}_{\hat{x}} + \sigma^{\hat{y}}{}_{\hat{y}} + \sigma^{\hat{z}}{}_{\hat{z}} = \theta \left( \frac{2}{3} - \frac{1}{3} - \frac{1}{3} \right) = 0$.

 

[Mentz114]

No, the trace is zero. The trace is $\sigma^{\mu}{}_{\mu}$, not $\sigma^{\mu \nu} \sigma_{\mu \nu}$. The shear tensor is always trace-free. In the comoving frame, $\sigma^{\mu}{}_{\mu} = \sigma^{\hat{x}}{}_{\hat{x}} + \sigma^{\hat{y}}{}_{\hat{y}} + \sigma^{\hat{z}}{}_{\hat{z}} = \theta \left( \frac{2}{3} - \frac{1}{3} - \frac{1}{3} \right) = 0$.

I don't understand why you wrote this. First of all - I know what a trace is ! Where do I call $\sigma^{\mu \nu} \sigma_{\mu \nu}$ the trace ? Please re-read my post.

I was trying to say that the trace-less shear tensor in the local basis, when transformed to the inertial coordinates is no longer traceless. Therefore undermining any claim that it is still a shear tensor.

 

[Mentz114]

$\sigma_{\mu\nu}$ really only has a direct, subtlety-free, simple physical interpretation in the instantaneously comoving (local) inertial frames of observers following orbits of the 4-velocity field $u^{\mu}$ that $\sigma_{\mu\nu}$ is expressed in terms of. In such a case, we can interpret $\sigma_{\mu\nu}$ physically as follows. First, for simplicity, assume $\nabla_{\mu}u^{\mu} = 0$ and $\nabla_{[\mu}u_{\nu]} = 0$ so that we only have to deal with physical effects of $\sigma_{\mu\nu}$; also, let's take $u^{\mu}$ to be geodesic so that $\sigma_{\mu\nu} = \nabla_{(\mu}u_{\nu)}$

Consider now an observer following an orbit of $u^{\mu}$ and an instantaneously comoving (local) inertial frame of this observer. In this frame we have $\Gamma^{\mu}_{\nu\gamma}= 0$ at this instant and $u^{\mu} = \delta^{\mu}_0$ at this instant. Then since $u^{\mu}\sigma_{\mu\nu} = u^{\nu}\sigma_{\mu\nu} = 0$ only $\sigma_{ij} = \partial_i u^j + \partial_j u^i$ are non-vanishing. This is, up to some $\gamma$ factors, just the Newtonian expression for shear and hence carries the same interpretation: http://en.wikipedia.org/wiki/Deformation_(mechanics)

Thanks. But I'm not dealing with a geodesic which might make a difference. I'll check the link.

 

[WannabeNewton]

Thanks. But I'm not dealing with a geodesic which might make a difference. I'll check the link.

It won't make a difference. I was just using a geodesic to make the calculation trivial. If you use an arbitrary time-like congruence the physical interpretation of $\sigma_{\mu\nu}$ will still be the same in a momentarily comoving local inertial frame.

 

[PeterDonis]

I was trying to say that the trace-less shear tensor in the local basis, when transformed to the inertial coordinates is no longer traceless.

But that can't be right. The trace is a scalar invariant; if it's zero in one frame, it's zero in every frame. So you must have made an error in your computation somewhere. I was mistaken to think the error was that you were computing $\sigma^{\mu \nu} \sigma_{\mu \nu}$; sorry about that. But there must be an error somewhere if you are not getting $\sigma^{\mu}{}_{\mu} = 0$ in every frame.

 

[WannabeNewton]

I was trying to say that the trace-less shear tensor in the local basis, when transformed to the inertial coordinates is no longer traceless. Therefore undermining any claim that it is still a shear tensor.

You must have made a computational error because the trace-free condition is coordinate-invariant and frame-invariant.

 

[PeterDonis]

The trace is a scalar invariant; if it's zero in one frame, it's zero in every frame.

To illustrate how this works, I'll compute the trace in the original inertial frame, which is moving at $- v$ in the $x$ direction relative to the comoving frame. So the Lorentz transformation we need is ("hatted" indexes are in the comoving frame, non-hatted indexes are in the original inertial frame):

$$
\Lambda^{\hat{t}}{}_t = \Lambda^{\hat{x}}{}_x = \gamma
$$

$$
\Lambda^{\hat{t}}{}_x = \Lambda^{\hat{x}}{}_t = \gamma v
$$

$$
\Lambda^{\hat{y}}{}_y = \Lambda^{\hat{z}}{}_z = 1
$$

We only need the diagonal components of the shear tensor in the new frame; they are

$$
\sigma_{tt} = \Lambda^{\hat{\mu}}{}_t \Lambda^{\hat{\nu}}{}_t \sigma_{\hat{\mu} \hat{\nu}} = \gamma^2 v^2 \sigma_{\hat{x} \hat{x}}
$$

$$
\sigma_{xx} = \Lambda^{\hat{\mu}}{}_x \Lambda^{\hat{\nu}}{}_x \sigma_{\hat{\mu} \hat{\nu}} = \gamma^2 \sigma_{\hat{x} \hat{x}}
$$

$$
\sigma_{yy} = \sigma_{\hat{y} \hat{y}}
$$

$$
\sigma_{zz} = \sigma_{\hat{z} \hat{z}}
$$

The trace is then:

$$
\eta^{\mu \nu} \sigma_{\mu \nu} = - \sigma_{tt} + \sigma_{xx} + \sigma_{yy} + \sigma_{zz} = \left( 1 - v^2 \right) \gamma^2 \sigma_{\hat{x} \hat{x}} + \sigma_{\hat{y} \hat{y}} + \sigma_{\hat{z} \hat{z}} = 0
$$

 

[Mentz114]

You must have made a computational error because the trace-free condition is coordinate-invariant and frame-invariant.

Yep, that is a possibility. I remember, I did make a mistake once ...

To illustrate how this works, I'll compute the trace in the original inertial frame, which is moving at $- v$ in the $x$ direction relative to the comoving frame. So the Lorentz transformation we need is ("hatted" indexes are in the comoving frame, non-hatted indexes are in the original inertial frame):

Thanks a million. I'll check this all out later and compare with my calculation which is using the rocket coords in the archived Wiki article. I enjoyed the article which poss some interesting questions.

 

[Mentz114]

Shear in the global frame basis

I found my error - I didn't raise an index of $\sigma_{\mu\nu}$ before adding the diagonal terms. Now everything is consistent and ${\sigma^\mu}_\mu={\sigma^{\hat{\mu}}}_{\hat{\mu}}=0$.

With expansion scalar $\theta=\frac{{k}^{2}\,t}{\sqrt{{k}^{2}\,{t}^{2}+1}}$ the components of $\bar{\sigma}_{\mu\nu}=\sigma_{\mu\nu}/\theta$ are

$\bar{\sigma}_{tt} =\frac{2\,{k}^{2}\,{t}^{2}}{3}$, $\bar{\sigma}_{tx} =\bar{\sigma}_{xt} =-\frac{2\,k\,t\,\sqrt{{k}^{2}\,{t}^{2}+1}}{3}$, $\bar{\sigma}_{xx} =\frac{2\,\left( {k}^{2}\,{t}^{2}+1\right) }{3}$, $\bar{\sigma}_{yy} =\bar{\sigma}_{zz} =-\frac{1}{3}$.

The congruence is $\vec{u}=\sqrt{{k}^{2}\,{t}^{2}+1}\ \partial_t+ kt\ \partial_x$ which can be written $\vec{u}=\gamma\ \partial_t+ \gamma\beta\ \partial_x$. The proper acceleration in the local frame field has one component $k$ in the $d\hat{x}$ direction.

The question is - what does $\sigma_{\mu\nu}$ mean ? The time components seem to have information but $\sigma_{\mu\nu}$ is a rank-3 tensor, despite appearances. So I'm somewhat puzzled.

If you could indulge me one more time - any thoughts ?

 

[PeterDonis]

The time components seem to have information but $\sigma_{\mu\nu}$ is a rank-3 tensor, despite appearances.

No, it isn't. It's a rank-4 tensor, just like any other tensor in spacetime. It just happens to be a rank-4 tensor that, in a particular frame (the comoving frame), has only "space-space" components nonzero, so it can be treated, *within that frame*, as a purely spatial rank-3 tensor.

There's another way to put this: the rank-3 tensor, which is what has the direct physical interpretation, is really the rank-4 tensor projected into the spacelike 3-surface that is orthogonal to the 4-velocity of the spaceship. That projection is done using the projection tensor $h_{\mu \nu} = g_{\mu \nu} + u_{\mu} u_{\nu}$. So what you should be looking at for physical interpretation is not $\sigma_{\mu \nu}$; it's $\sigma_{ab} = h^{\mu}{}_a h^{\nu}{}_b \sigma_{\mu \nu}$.

It just so happens that, in the comoving frame, $h_{\mu \nu}$ is the purely spatial identity tensor, so $\sigma_{ab} = \sigma_{\hat{\mu} \hat{\nu}}$ in that frame (the "hats" on the indexes are to emphasize that we are looking at the shear tensor components in the comoving frame, where we know the tensor is purely spatial). But in any other frame, you have to actually compute $h_{\mu \nu}$ and contract it with $\sigma_{\mu \nu}$ as above. If you do this, you should end up with the same rank-3 tensor $\sigma_{\hat{\mu} \hat{\nu}}$ as in the comoving frame (by a process similar to the one by which we showed that the trace is frame-invariant).

In short, the rank-3 tensor in the comoving frame is the one that has the direct physical interpretation. In any other frame, just looking at the tensor components isn't a good way to think about physical interpretation, because what is physically meaningful, in frame-invariant terms, is the contraction of the shear tensor with the projection tensor. The comoving frame is just the one in which that contraction works out to just give the tensor components in that frame directly.

 

[Mentz114]

No, it isn't. It's a rank-4 tensor, just like any other tensor in spacetime. It just happens to be a rank-4 tensor that, in a particular frame (the comoving frame), has only "space-space" components nonzero, so it can be treated, *within that frame*, as a purely spatial rank-3 tensor.

OK, but $|\sigma_{\mu\nu}|=0$.

There's another way to put this: the rank-3 tensor, which is what has the direct physical interpretation, is really the rank-4 tensor projected into the spacelike 3-surface that is orthogonal to the 4-velocity of the spaceship. That projection is done using the projection tensor $h_{\mu \nu} = g_{\mu \nu} + u_{\mu} u_{\nu}$. So what you should be looking at for physical interpretation is not $\sigma_{\mu \nu}$; it's $\sigma_{ab} = h^{\mu}{}_a h^{\nu}{}_b \sigma_{\mu \nu}$
..
...
you have to actually compute $h_{\mu \nu}$ and contract it with $\sigma_{\mu \nu}$ as above. If you do this, you should end up with the same rank-3 tensor $\sigma_{\hat{\mu} \hat{\nu}}$ as in the comoving frame (by a process similar to the one by which we showed that the trace is frame-invariant)..

In response to your post I calculated $\sigma_{ab} = h^{\mu}{}_a h^{\nu}{}_b \sigma_{\mu \nu}$ with the components of $\sigma_{\mu\nu}$ in my last post, and found $\sigma_{ab}=\sigma_{\mu\nu}$. I was surprised too. Maybe I misinterpret what you're saying.

In short, the rank-3 tensor in the comoving frame is the one that has the direct physical interpretation. In any other frame, just looking at the tensor components isn't a good way to think about physical interpretation, because what is physically meaningful, in frame-invariant terms, is the contraction of the shear tensor with the projection tensor. The comoving frame is just the one in which that contraction works out to just give the tensor components in that frame directly.

Firstly, I understand that the physics is in the invariants. That is why in my calculation no invariant has been hurt in any way. This is what I take to be the meaning of 'covariant calculation'.

If two bodies collide and make a big explosion - the physics is in the details of the explosion and is independent of which frame we choose. But each body could claim that the energy for the explosion came from the kinetic energy of the *other* body. So components can tell us how differnt observers slice and dice (attribute) the components of the physics.

 

[PeterDonis]

OK, but $|\sigma_{\mu\nu}|=0$.

I don't understand; what is $|\sigma_{\mu\nu}|$? And which frame are we talking about?

In response to your post I calculated $\sigma_{ab} = h^{\mu}{}_a h^{\nu}{}_b \sigma_{\mu \nu}$ with the components of $\sigma_{\mu\nu}$ in my last post, and found $\sigma_{ab}=\sigma_{\mu\nu}$. I was surprised too. Maybe I misinterpret what you're saying.

Again, which frame are we talking about? [STRIKE]What you should find is that $\sigma_{ab} = \sigma_{\hat{\mu} \hat{\nu}}$; note carefully the hats on the indexes, indicating that projecting the shear tensor, in any frame, into the spacelike hypersurface orthogonal to the spaceship's 4-velocity, as expressed in that frame, will give you back the shear tensor components in the comoving frame.[/STRIKE] [Edit: actually, not; see follow-up post.] But I'm not sure you're saying that's what you got.

each body could claim that the energy for the explosion came from the kinetic energy of the *other* body.

Kinetic energy by itself isn't really a "component"; but that's a minor point. See below.

So components can tell us how differnt observers slice and dice (attribute) the components of the physics.

Not in themselves, no. To give physical meaning to any "component", you have to express it as an invariant. For example, if two bodies collide, assuming they have equal rest mass $m$, the kinetic energy each one sees the other to have is $K = E - m = u^a p_a - \sqrt{ p^a p_a } = \left( \gamma - 1 \right) m$, where $u^a$ is the "observing" body's 4-velocity and $p^a$ is the "observed" body's 4-momentum. We can view the total energy $E$ as a "component" of the observed body's 4-momentum in the observing body's frame, but that only has physical meaning because we can express it as the invariant $u^a p_a$.

In other words, what tells us how different observers slice and dice the physics are the invariants we form by contracting a particular vector, tensor, or whatever with the vectors that describe the observer (his 4-velocity and the spatial vectors of his frame). So if you want to try to give a physical interpretation to shear tensor components in some frame other than the comoving frame, you'll need to figure out what observer, other than the comoving observer, you're interested in, and what it means, physically, to contract that observer's frame vectors with the shear tensor. In general I'm not sure this will be very useful for the shear tensor, because the whole point of that tensor is to capture properties measured by the comoving observer.

 

[PeterDonis]

What you should find is that $\sigma_{ab} = \sigma_{\hat{\mu} \hat{\nu}}$; note carefully the hats on the indexes, indicating that projecting the shear tensor, in any frame, into the spacelike hypersurface orthogonal to the spaceship's 4-velocity, as expressed in that frame, will give you back the shear tensor components in the comoving frame.

I just realized that this can't be right as it stands, because the shear tensor components in the original inertial frame, $\sigma_{\mu \nu}$ (no hats on the indexes) are obtained by applying the Lorentz transform to the components in the comoving frame, $\sigma_{\hat{\mu} \hat{\nu}}$ (with hats on the indexes). But applying the (inverse, to go from original frame to comoving frame) Lorentz transform is not the same as contracting with the projection tensor. So I need to think about this some more.

 

[PeterDonis]

So I need to think about this some more.

Ok, I was confusing myself about the projection tensor. Now I think I'm un-confused.

Contracting with the projection tensor $h_{\mu \nu}$ gives the "transverse" part of a tensor, i.e., the part that is orthogonal to the 4-velocity $u^{\mu}$. But the shear tensor is already transverse; i.e., it is already orthogonal to $u^{\mu}$. So contracting with the projection tensor should leave the shear tensor unchanged. That is, in any frame, if you express the shear tensor in that frame, and contract it with the projection tensor expressed in that frame, the shear tensor components should be unchanged. Sorry for my previous confusion about this.

 

[Mentz114]

Peter, thanks again.

Just to clear up the notation - the un-hatted indexes are for the coordinate basis, and the hats are the local frame basis of the accelerating congruence.

My point about the determinant of $\sigma_{\mu\nu}$ being zero, is that as a bilinear transformation it is 3-dimensional ( there are 3 non-zero eigenvalues). I would expect this after some thought because the transformation with the vierbien into the local frame basis gives a rank-3 tensor.

Regarding the possible physical meaning of $\sigma_{\mu\nu}$, I have some ideas but no time to explicate now.

 

[PeterDonis]

Just to clear up the notation - the un-hatted indexes are for the coordinate basis, and the hats are the local frame basis of the accelerating congruence.

Yes.

My point about the determinant of $\sigma_{\mu\nu}$ being zero, is that as a bilinear transformation it is 3-dimensional ( there are 3 non-zero eigenvalues).

Ah, ok. Yes, if you view $\sigma_{\mu \nu}$ as a 4x4 matrix, it has zero determinant because only three eigenvalues are nonzero. That will be generally true of any transverse tensor, since evaluating the eigenvalues basically amounts to transforming into the comoving frame, where only the space-space components can be nonzero.

 

[WannabeNewton]

Regarding the possible physical meaning of $\sigma_{\mu\nu}$...

Wasn't this already answered?

 

[Mentz114]

Wasn't this already answered?

Not in my understanding. In the local frame the meaning is clear, but what does it mean when written in the coordinate basis ? I've been doing a lot of thinking and calculation with a Born congruence ( ie $\dot{u}_{\hat{\mu}}= k\ dx$, $k$ constant ) where the velocity depends on $t$ and $x$. Some interesting things have emerged.

 

[WannabeNewton]

In the local frame the meaning is clear, but what does it mean when written in the coordinate basis ?

It has no meaningful physical interpretation in that case. The problem is that $\sigma_{\mu\nu}$ is defined in terms of $h_{\mu\nu}$ which physically measures spatial distances (between events) relative to the instantaneously comoving inertial frames of the spaceships i.e. it is only defined relative to the spaceship frames. It has no direct relationship to other frames, not even the background global inertial frame in which the spaceships are accelerated simultaneously. As such, while the components of $\sigma_{\mu\nu}$ relative to a given arbitrary frame (field) are measurable quantities in the mathematical sense, they do not necessarily entail a physical interpretation unless the frame field consists of the local Lorentz frames attached to the orbits of the Bell congruence.

This isn't peculiar to the shear tensor. A lot of quantities in the kinematic decomposition only have meaningful physical interpretation relative to the instantaneous (local) inertial frames of observers in the congruence. Take for example the vorticity $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\partial_{\alpha}\xi_{\beta}$. In an instantaneously comoving local inertial frame, we can use the fact that $\epsilon^{ijk} \Leftrightarrow \epsilon^{\mu \nu\alpha\beta}\xi_{\mu}$ and $\xi^{\mu} = \delta^{\mu}_0$ to see that $\omega^{\mu}\xi_{\mu} = 0$ (purely spatial vector in the local inertial frame) and $\vec{\omega} = \vec{\nabla}\times \vec{\xi}$ which means the vorticity, in the local inertial frame, is just a measure of local rotation relative to local gyroscopes if the congruence is Born rigid. But what is the interpretation of $\omega^{\mu}$ in other local Lorentz frames? Well if we boost from the instantaneously comoving local inertial frame of an observer in the congruence to the rest frame of a coincident observer (say along the $x$-axis of the local inertial frame) then the boost will rotate the local $t-x$ plane which means that $\omega^{\mu}$ relative to the new frame will pick up a time-like component and hence will no longer be a purely spatial vector in this frame: a measure of local rotation with a time-like component...yeah we can't interpret that physically.

 

[Mentz114]

OK, that makes sense.

I got to thinking about the Doppler shift between the spaceships. Obviously there will be a kinematic frequency shift between the ships because they are not comoving. What causes the shift in the global inertial frame, where the ships are comoving ?

The only way I can make sense of this is to introduce a gravitational field which brings the spaceships to rest wrt each other in all frames. With $\vec{u}= \sqrt{u^2+1} \ \partial_t + u\ \partial_x$, if we impose the condition that the local acceleration vector is $\dot{u}_{\hat{\mu}}= k\ d\hat{x}$, then the acceleration vector in the coordinate basis is $\dot{u}_\mu = -u\ k \ dt + \sqrt{1+u^2}\ k\ dx$. $u$ can be a function of $t$, $x$ or both.

So the gravitational field (pulling in the -x direction) must create acceleration $\sqrt{1+u^2}k$. The acceleration (force?) ratio between two points $P_1$ $P_2$ is then $\sqrt{1+u^2}|_{P_1}/\sqrt{1+u^2}|_{P_2}$ and from the 4-velocity the ratio of clock rates $d\tau_1/d\tau_2$ is the reciprocal of the acceleration ratio. Given that $P_1$ $P_2$ are the end points of a null geodesic connecting two spaceship worldlines, I suspect that the frequency shift due to the field is identical to the kinematical shift when the field is not present.

If it is true then I don't see immediately how to apply it to the strings breaking question, although a growing weight could be relevant.