Solving the Spaceship Paradox: A New Explanation

[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.

 

[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.