Solving the Spaceship Paradox: A New Explanation

[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.