Rotating Disk Spinoff: Is 3D Timelike Congruence Born Rigid?

[pervect]

Let me apologize in advance for not reading the entire rotating disk thread. I think that the following question is closely related, but if it was answered in that thread, I didn't spot it.

Let us consider the following timelike congruences, which maps congruence parameters t,r,theta and z to points in a minkowskii space (t', x', y', z'). g is a "free parameter" (which could even be set to 1 to simplify the problem)

(t,r,theta, z) -> t' = (1/g+z) sinh(g t) , x'=r cos (w t + theta), y' = r sin(w t + theta),
z' = (1/g + z) cosh(g t)

The point (r=0, z=0), which one might describe as "the center of the disk" is undergoing hyperbolic motion in the z direction, i.e. it's accelerating at a constant rate, as can be seen by comparing it to the equations in MTW, pg 173, for hyperbolic motion. The parameter "g" represents the rate of said acceleration, and can be set to 1 if desired.

The points at r>0, theta = constant are "rotating around the center of the disk" with an angular velocity w.

The point of writing out the congruence is to make the notion of a "disk" mathematically exact and allow us to calculate things about it. I can't think of any clearer way to define a disk precisely than to write down the congruence of worldlines which make up the disk.

Now consider the case z=0 (a "thin" disk). Then we have:

(t, r, theta) ->t' = (1/g) sinh(gt), x' = r cos(w t + theta), y' = r sin(wt + theta) z' = (1/g)cosh(gt)

The question is - is this 3-parameter set of congruences Born rigid, in the appropriate reduced dimensional manner? (It's only got 1 time and 2 space dimensions).

Looking at Wald, the first step to confirm this would be to express the congruences as a function of proper time. However, the congruence parameter t is proper time for the particle at "the center of the disk", but not elsewhere.

My feeling is that this congruence "should be" Born rigid, because if we look at an instantaneously co-moving inertial frame around the point (r=0, z=0) it describes a rotating plane.

The 4-velocity of every point in the plane is the same as the 4-velocity of a rotating disk.

Further, the rotation rate (measured by clocks in the co-moving inertial frame above) doesn't change with time.

I'm hoping someone will have some insight as to how to compute the expansion tensor to confirm this idea that 2-space 1-time congruence is Born rigid WITHOUT re-parameterizing everything in terms of proper time.

The inuitive point of this (assuming the math works out) is that the thin disks for z=constant are each Born rigid, but rotate with respect to each other, so that the 3-space + 1 time congruence is not Born rigid.

 

[PeterDonis]

The inuitive point of this (assuming the math works out) is that the thin disks for z=constant are each Born rigid, but rotate with respect to each other, so that the 3-space + 1 time congruence is not Born rigid.

I think this is right; I believe the 3+1 congruence has nonzero shear. I thought I had computed that in some recent thread or other, but I haven't been able to find the relevant post (the rotating disk thread that this one spun off from was itself spun off from yet another thread, but neither of those seems to have the post I thought I had written).

 

[PeterDonis]

Further, the rotation rate (measured by clocks in the co-moving inertial frame above) doesn't change with time.

But it *does* vary with z--at least, IIRC that was the conclusion we came to in the thread that spun off the thread that spun off this one...

 

[PAllen]

Page 2 of:

http://arxiv.org/abs/0810.0072

gives a complete enough description of computing the expansion tensor that I could follow it in principle (but for me, it would take at least a whole weekend, and I'd make some stupid mistakes). It does start right off with expressing the tangent unit vector field, but you don't have to explicitly write the congruence in terms of proper time. Also, particular to your case, where after setting z=0, you have 2X1 subspace, you have to induce Lorentzian 3-metric on this. The procedure to do this is described. If you were dealing with a 3 parameter family of world lines (have 3x1 world tube) there would be no need to induce a metric. All the steps described seem fully defined to me, but very tedious to carry out.

 

[WannabeNewton]

With regards to the thin rotating disk accelerating along the direction perpendicular to the plane of rotation, why not just take $u^{\alpha} = (\gamma \cosh g\tau, 0 ,\omega\gamma,\gamma \sinh g\tau) = (\gamma gz, 0 ,\omega\gamma,\gamma gt)$, where the 4-velocity field $u^{\alpha}$ is expressed in cylindrical coordinates $(t,r,\varphi,z)$ relative to a global inertial frame, $\omega$ is the angular velocity, and $\gamma = \frac{1}{\sqrt{1 - \omega^{2}r^{2}}}$, and directly compute $\nabla_{\alpha}u_{\beta}$. From this one can determine $\nabla_{\mu}u^{\mu}$ and $\nabla_{(\alpha}u_{\beta)}$ and hence $\sigma_{ab} = \nabla_{(\alpha}u_{\beta)} - \frac{1}{3}\nabla_{\mu}u^{\mu}h_{\alpha\beta}$ where $h_{\alpha\beta} = g_{\alpha\beta} + u_{\alpha}u_{\beta}$.

 

[yuiop]

The point (r=0, z=0), which one might describe as "the center of the disk" is undergoing hyperbolic motion in the z direction, i.e. it's accelerating at a constant rate, as can be seen by comparing it to the equations in MTW, pg 173, for hyperbolic motion. The parameter "g" represents the rate of said acceleration, and can be set to 1 if desired.

OK, so g is the constant proper acceleration as measured in S by the accelerated observer O that remains at rest wrt the centre of the disc. The hyperbolic motion (non constant acceleration) is measured by the inertial observer O' at rest in S'.

The points at r>0, theta = constant are "rotating around the center of the disk" with an angular velocity w.

Constant in S and "rotating around the center of the disc" as measured in S'. Assuming the disc acts as a perfect flywheel, angular velocity w remains constant in S and slows down in S' such that the instantaneous angular velocity is given by $w' = w/\gamma= w/\sqrt{1 + (gt')^2}$ or alternatively $w' = w/\gamma = w/\cosh(gt)$. We know w is constant in S, because a perfect flywheel is effectively a clock and its rotation rate will exactly match any clock at rest in S that measures the rotation rate.

Now consider the case z=0 (a "thin" disk). Then we have:

(t, r, theta) ->t' = (1/g) sinh(gt), x' = r cos(w t + theta), y' = r sin(wt + theta) z' = (1/g)cosh(gt)

From the above:

$r' =\sqrt{(x')^2 +(y')^2}$

$r' =\sqrt{r^2 * [\cos^2(wt +\theta) +\sin^2(wt + \theta)]}$

Using the Pythagorean trigometric identity we get:

$r' = r$

For the $\theta$ relationship we obtain:

$tan(\theta') = \frac{y'}{x'} = \frac{r \sin(wt + \theta)}{r \cos(wt + \theta)} = \tan(wt + \theta)$

$\theta = \theta' - wt$

To a make a meaningful comparison of angles in S and S' we should make the measurements of 2 points on the perimeter of the rotating disc simultaneously in S' and transform the measurements to S where where they are stationary.

$\theta = \theta' - w * \sinh^{-1}(t'g)/g$

$\Delta\theta = \Delta\theta' - w\left(\frac{\sinh^{-1}(t'_2g)}{g} - \frac{\sinh^{-1}(t'_1g)}{g}\right)$

Since the measurement is simultaneous in S', $t'_2 = t'_1$, so it follows that:

$\Delta\theta = \Delta\theta'$

If my reasoning above is correct then the transformations of $\Delta\theta$ and r are completely independent of the acceleration orthogonal to the plane of the disc and independent of the time parameter, so the geometry of the disc (and size) as measured by S and S' is equal. If the shape and size remains constant according to S, then they remain constant to S' and vice versa.

P.S. There is wiggle room to claim I have not directly shown that the circumference remains equal and constant in S and S'. I have done that in my head and will write it up later.

 

[yuiop]

...
P.S. There is wiggle room to claim I have not directly shown that the circumference remains equal and constant in S and S'. I have done that in my head and will write it up later.

The observer in S calculates the circumference to be C = vt where v is the tangential velocity of the disc. This equates to C = r*w*t in terms of angular velocity. We already know r=r' and $w = w'*\gamma$. t is simply $t'/\gamma$. This means that $C = r'*(w'*\gamma)*(t'/\gamma) = r'*w'*t' = C'$.

So now we have:

r=r', $\Delta\theta = \Delta\theta'$ and C=C' and all these measurements are independent of the acceleration g and the time t or t'.

 

[pervect]

With regards to the thin rotating disk accelerating along the direction perpendicular to the plane of rotation, why not just take $u^{\alpha} = (\gamma \cosh g\tau, 0 ,\omega\gamma,\gamma \sinh g\tau) = (\gamma gz, 0 ,\omega\gamma,\gamma gt)$, where the 4-velocity field $u^{\alpha}$ is expressed in cylindrical coordinates $(t,r,\varphi,z)$ relative to a global inertial frame, $\omega$ is the angular velocity, and $\gamma = \frac{1}{\sqrt{1 - \omega^{2}r^{2}}}$, and directly compute $\nabla_{\alpha}u_{\beta}$. From this one can determine $\nabla_{\mu}u^{\mu}$ and $\nabla_{(\alpha}u_{\beta)}$ and hence $\sigma_{ab} = \nabla_{(\alpha}u_{\beta)} - \frac{1}{3}\nabla_{\mu}u^{\mu}h_{\alpha\beta}$ where $h_{\alpha\beta} = g_{\alpha\beta} + u_{\alpha}u_{\beta}$.

I'm pretty sure that that the 4-velocity field you wrote down isn't equivalent to the one that would result from I wrote down. - at least if I understand the coordinates. In particular, yours has a component which is zero, while mine doesn't

Relative to some global inertial frame, the 4 velocity in the direction of acceleration should't be zero, because of the acceleration. Getting rid of it gets rid of the acceleration and it's resulting effects.

But your post was helpful, I see that I don't actually need to re-parameterize all the geodesics, I just need to normalize the 4-velocity field to unit legnth.

 

[WannabeNewton]

I have the acceleration along the $z$-axis and the rotation in the $x$-$y$ plane. The above 4-velocity was obtained by boosting in the $\varphi$ direction to get the rotation and then boosting in the $z$ direction to get the acceleration, all in cylindrical coordinates.

 

[yuiop]

I have the acceleration along the $z$-axis and the rotation in the $x$-$y$ plane. The above 4-velocity was obtained by boosting in the $\varphi$ direction to get the rotation and then boosting in the $z$ direction to get the acceleration, all in cylindrical coordinates.

What conclusions can be drawn from your 4 velocity field?

 

[PeterDonis]

$\sigma_{ab} = \nabla_{(\alpha}u_{\beta)} - \frac{1}{3}\nabla_{\mu}u^{\mu}h_{\alpha\beta}$ where $h_{\alpha\beta} = g_{\alpha\beta} + u_{\alpha}u_{\beta}$.

For what we're looking for here, I think there's an even simpler approach: just compute $\nabla_{(a} u_{b)}$, since that must be zero for the expansion tensor as a whole to be zero (i.e., to have zero expansion and shear, which is what's required for Born rigid motion). Here's what I get when I try to do that for a 3+1 congruence in Rindler coordinates; the metric is

$$
ds^2 = - z^2 dt^2 + dz^2 + dr^2 + r^2 d\phi^2
$$

where I have normalized so the proper acceleration of the "fiducial" observer, the one at $r = 0$ whose proper time is given by the coordinate time $t$, is 1.

The congruence I'll consider has 4-velocity components $u^a = ( \gamma / z, 0, \gamma v, \gamma \omega)$, where $\gamma = 1 / \sqrt{1 - v^2 - \omega^2 r^2}$. That is, $v$ is ordinary velocity in the radial direction (relative to an observer at $r = 0$ but with the same $z, \phi$), and $\omega$ is the angular velocity (again relative to an observer at $r = 0$ but with the same $z$). What the following calculation will show (assuming it's correct--checking is welcome!) is that it is impossible for any congruence of this form to have a zero expansion tensor, by showing that it's impossible for all components of $\nabla_{(a} u_{b)}$ to be zero.

Start by re-expressing $\nabla_{(a} u_{b)}$ in a form we can work with:

$$
\nabla_{(a} u_{b)} = \nabla_a u_b + \nabla_b u_a = \partial_a u_b + \partial_b u_a - 2 \Gamma^c{}_{ab} u_c
$$

where the only nonzero Christoffel symbols are $\Gamma^z{}_{tt} = z$ and $\Gamma^t{}_{tz} = 1 / z$ (we can deal with only one ordering of coordinates in cases like $\Gamma^t{}_{tz}$ because everything is manifestly symmetric in the lower indexes). We also need the expression for $u_a$, because one key term in it does differ significantly from $u^a$:

$$
u_a = g_{ab} u^b = ( - z \gamma, 0, \gamma v, \gamma \omega)
$$

Now we just turn the crank. First, from the $\nabla_{(r} u_{r)}$ equation, we find something interesting; we must have $v = 0$:

$$
\nabla_{(r} u_{r)} = 2 \partial_r u_r = 2 \partial_r v = 0
$$

This tells us that $\partial_r v = 0$; but we also know that $v = 0$ at $r = 0$, so we must have $v = 0$ everywhere. So there can't be any radial motion in our congruence.

Continuing on, most of what we find is that certain partial derivatives have to be zero; for example:

$$
\nabla_{(t} u_{t)} = 2 \partial_t u_t = \frac{2}{z} \partial_t \gamma = 0
$$

which tells us that $\partial_t \gamma = 0$, which of course makes intuitive sense. We can similarly conclude that $\partial_r \gamma = 0$ from the $\nabla_{(t} u_{r)}$ equation. From the $\nabla_{(\phi} u_{\phi)}$ equation, we find that $\partial_{\phi} ( \gamma \omega) = 0$, which, after noting that $\omega$ itself can't depend on $\phi$ by definition, tells us that $\partial_{\phi} \gamma = 0$. The $\nabla_{(t} u_{\phi)}$ equation then tells us that $\partial_t \omega = 0$, and the $\nabla_{(r} u_{\phi)}$ equation tells us that $\partial_r \omega = 0$.

So we're left with two equations, and here's where we'll find the impossibility. First:

$$
\nabla_{(t} u_{z)} = \partial_z u_t - 2 \Gamma^t{}_{tz} u_t = \partial_z (- z \gamma ) - \frac{2}{z} ( - z \gamma ) = - \gamma - z \partial_z \gamma + 2 \gamma = 0
$$

This tells us that $\partial_z \gamma = \gamma / z$. But we can expand this out (remember that we found $v = 0$):

$$
\partial_z \gamma = \partial_z \left( 1 - \omega^2 r^2 \right)^{- 1/2} = - \frac{1}{2} \left( 1 - \omega^2 r^2 \right)^{- 3/2} \partial_z \left( 1 - \omega^2 r^2 \right) = \gamma^3 \omega r^2 \partial_z \omega
$$

So $\partial_z \gamma = \gamma / z$ gives us

$$
\partial_z \omega = \frac{1}{\gamma^2 \omega r^2 z}
$$

Now for the second equation that remains:

$$
\nabla_{(z} u_{\phi)} = \partial_z u_{\phi} = \partial_z (\gamma \omega) = \gamma \partial_z \omega + \omega \partial_z \gamma = 0
$$

We substitute $\partial_z \gamma = \gamma / z$ and divide through by $\gamma$ to get

$$
\partial_z \omega = - \frac{\omega}{z}
$$

which cannot be consistent with the other equation for $\partial_z \omega$.

So we have found that no congruence of the type given here (i.e., with no $z$ motion) can have a zero expansion tensor. So *if* it's possible to have a rigid congruence at all, it will have to have motion in the $z$ direction at least somewhere. I'm working through that calculation now.

 

[PAllen]

In response to Peter's last post:

This calculation is all based on the 4-metric, and expansion tensor as a 4-tensor. The 'trick' pulled by Epp at all, which I reference in my post #4, is to allow 2-surface (like our example) to arbitrarily change its embedding, in ways not consistent with rigidity in 3X1 space. Thus, they first derive 2X1 subspace, with induced Lorentzian 3-metric, and only require rigidity here (using an expansion 3-tensor). This reduces the number of constraints. We have to decide which cases we're interested in.

 

[PeterDonis]

Well, it didn't take too long for me to spot a mistake once I posted. Posting things appears to make errors jump out more easily. However, this one might not change the conclusion.

The mistake was in the $\nabla_{(r} u_{r)}$ equation; I should have remembered that $u_r = \gamma v$, which gives

$$
\nabla_{(r} u_{r)} = \partial_r (\gamma v) = v \partial_r \gamma + \gamma \partial_r v = 0
$$

Expanding out $\partial_r \gamma$ similarly to the way I expanded out $\partial_z \gamma$ (but now not assuming $v = 0$) gives:

$$
\partial_r \gamma = \gamma^3 \left( v \partial_r v + r \omega^2 \right)
$$

Substituting this back into the previous equation gives

$$
\gamma \partial_r v + v \gamma^3 \left( v \partial_r v + r \omega^2 \right) = 0
$$

This simplifies to

$$
\left(1 + \gamma^2 v^2 \right) \partial_r v + \gamma^2 v \omega^2 r = 0
$$

Noting that $1 + \gamma^2 v^2 = \gamma^2$, we find that $\partial_r v = - v \omega^2 r$.

The other equation we need to rework is the $t-z$ equation that gave us $\partial_z \gamma = \gamma / z$. That equation is still OK, but now we need to re-expand $\partial_z \gamma$ without assuming that $v = 0$. The expansion goes similarly to what we just did for $\partial_r \gamma$, and we get

$$
\partial_z \gamma = \gamma^3 \left( v \partial_z v + r^2 \omega \partial_z \omega \right)
$$

Setting this equal to $\gamma / z$, and using the fact that $\partial_z \omega = - \omega / z$, which still holds (since that comes from the $z-\phi$ equation, which hasn't changed), we obtain, after rearranging factors:

$$
\partial_z v = \frac{1}{z v} \left( \frac{1}{\gamma^2} + \frac{\omega^2 r^2}{z} \right) = \frac{1}{zv} \left( 1 - v^2 \right)
$$

So $v$ as a function of $r$ and $z$ must satisfy both of the above partial differential equations. My suspicion is that there is no function $v(r, z)$ that will satisfy them both, but I haven't been able to get a definite answer yet one way or the other. If I'm right that there is no such function, then the conclusion from my previous post would still hold.

 

[PeterDonis]

I found another error (actually two of them) in the $\phi - \phi$ equation and the $r - \phi$ equation; I had forgotten that there are nonzero Christoffel symbols $\Gamma^r{}_{\phi \phi} = - r$ and $\Gamma^{\phi}{}_{r \phi} = 1 / r$ that add a term to each of these equations. The first one should read

$$
\nabla_{(\phi} u_{\phi)} = 2 \partial_{\phi} \left( \gamma \omega \right) - 2 ( - r ) \gamma v = 0
$$

We must have $\partial_{\phi} \omega = 0$ by definition, so this leaves us with

$$
\partial_{\phi} \gamma = - \frac{2 v \gamma r}{\omega}
$$

The interesting thing about this is, if we assume that nothing can vary with $\phi$, which seems intuitively reasonable, then this equation requires $v = 0$, which puts us right back to the same conclusion I made a couple of posts ago (see below for a check that that conclusion is still valid after correcting the errors I've found in these last two posts). But for the sake of argument, let's go forward assuming that the above equation does not just work out to $0 = 0$.

The second equation should read

$$
\nabla_{(r} u_{\phi)} = \partial_r ( \gamma \omega ) + \partial_{\phi} ( \gamma v ) - \frac{2}{r} \gamma \omega = 0
$$

I'm still working on expanding all this out to see where it ends up, but I wanted to get this posted since I may not have a chance to go further with it for a while.

 

[PeterDonis]

Expanding out $\partial_r \gamma$ similarly to the way I expanded out $\partial_z \gamma$ (but now not assuming $v = 0$) gives:

$$
\partial_r \gamma = \gamma^3 \left( v \partial_r v + r \omega^2 \right)
$$

Just a quick note, this is only correct if we assume $\partial_r \omega = 0$, which seems intuitively reasonable, but which also needs to be looked at in the light of my other corrections.

 

[Mentz114]

To find out what happens when a spinning disc moves or is accelerated in the axial direction I started with the Born coordinates in flat spacetime, found the frame field for the 'static' observer, and boosted it in the axial direction by $\beta(t)$.
The metric is
$ds^2=-(1-\omega r^2)dt^2+2\,\omega{r}^{2}\,d\phi\,dt\,+dz^2+dr^2+\,{r}^{2}{d\phi}^{2}$, the static worldline is $u^\mu=\frac{1}{\sqrt{1-{r}^{2}\,{w}^{2}}}\partial_t$ and the frame basis is
$e_0=\frac{1}{\sqrt{1-{r}^{2}\,{w}^{2}}}\partial_t,\ e_1=\partial_z,\ \ e_2=\partial_r,\ \ e_3= \frac{r\,w}{\sqrt{1-{r}^{2}\,{w}^{2}}}\partial_t+\frac{\sqrt{1-{r}^{2}\,{w}^{2}}}{r}\partial_\phi$

After boosting this frame field the results are

Expansion scalar
$\Theta=\frac{\beta\,\left( \frac{d\beta}{d\,t}\,\right) }{\sqrt{1-{r}^{2}\,{w}^{2}}\,\sqrt{1-{\beta}^{2}}\,\left( 1-{\beta}^{2}\right) }$

Acceleration
$\dot{u}=u^\nu\nabla_\nu u^\mu = \frac{\Theta}{\beta}\partial_z - \frac{\omega^2 r}{(1-\omega^2 r^2)(1-\beta^2)} \partial_r$

Vorticity
$\omega^\mu = \frac{\omega}{(1-\omega^2 r^2)\sqrt{1-\beta^2}}\partial_z+\frac{\beta \omega^2 r}{(1-\omega^2 r^2)(1-\beta^2)}\partial_\phi$

Shear is zero.

The effects of acceleration and constant velocity can be separated easily. I'm not sure what the physical interpretation of these numbers could be, yet. An interesting thing is that the analog of the centripetal acceleration has a correction for the velocity in the z-direction.

 

[PeterDonis]

boosted it in the axial direction by $\beta(t)$.

So $\beta$ is a boost in the $z$ direction, but its magnitude is a function of $t$ only? It's not a function of any of the other coordinates? This looks to me like a rotating version of the Bell congruence (the one that appears in the Bell Spaceship Paradox); the disks that are "in front" (i.e., at a larger $z$) will pull away because all of the disks are being subjected to the same proper acceleration in the $z$ direction. That's why the expansion is nonzero.

I think a better congruence for what we're discussing would be obtained by allowing $\beta$ to vary with $z$ as well as $t$; basically, along any slice of constant $t$, $d \beta / dt$ would get smaller as $z$ got larger (though the exact numerical values of $d \beta / dt$ would vary with $t$ as well). I suspect that, if the $z$ variation of $d \beta / dt$ were chosen properly, it would make the expansion scalar zero, but at the expense of creating nonzero shear. However, I'd be curious to see the actual computation.

 

[Mentz114]

So $\beta$ is a boost in the $z$ direction, but its magnitude is a function of $t$ only? It's not a function of any of the other coordinates?

Yes, that's it.

This looks to me like a rotating version of the Bell congruence (the one that appears in the Bell Spaceship Paradox); the disks that are "in front" (i.e., at a larger $z$) will pull away because all of the disks are being subjected to the same proper acceleration in the $z$ direction. That's why the expansion is nonzero.

Yes, if $\beta=at$ and the 'spaceshps' are rotating disks that sounds right.

I think a better congruence for what we're discussing would be obtained by allowing $\beta$ to vary with $z$ as well as $t$; basically, along any slice of constant $t$, $d \beta / dt$ would get smaller as $z$ got larger (though the exact numerical values of $d \beta / dt$ would vary with $t$ as well). I suspect that, if the $z$ variation of $d \beta / dt$ were chosen properly, it would make the expansion scalar zero, but at the expense of creating nonzero shear. However, I'd be curious to see the actual computation.

Again you are right. If we let the boost velocity ( z-direction) be a function of t and z i.e. $\beta \equiv \beta(t,z)$ the expansion scalar becomes

$\Theta=\frac{\sqrt{1-{\beta}^{2}}\,\left( \left( 1-{r}^{2}\,{w}^{2}\right) \,\left( \frac{d\beta}{d\,z}\,\right) +\sqrt{1-{r}^{2}\,{w}^{2}}\,\beta\,\left( \frac{d\beta}{d\,t}\,\right) \right) }{(1-\beta^2)^2(1-\omega^2 r^2)}$

clearly this is zero if

$\frac{d}{dz}\beta=-\frac{\beta\,\left( \frac{d}{dt}\beta\right) }{\sqrt{1-{r}^{2}\,{w}^{2}}}$. There is no shear in any case.

Something that bothers me about my calculation is that there is no shear in this frame basis, but in the coordinate basis there is shear. The expansion scalar comes out the same which is a sort of sanity check, so my unease is small.

 

[WannabeNewton]

That's not possible. If the shear vanishes in one coordinate system it must vanish in all coordinate systems.

 

[Mentz114]

That's not possible. If the shear vanishes in one coordinate system it must vanish in all coordinate systems.

If we calculate $\nabla_a u_b$ then transform this with the tetrad into the new basis and then symmetrize, could the transformed tensor lose (or gain) a symmetric part ?

 

[PeterDonis]

there is no shear in this frame basis, but in the coordinate basis there is shear.

Are you taking all the nonzero connection coefficients into account? There are a number of them in this chart.

 

[PeterDonis]

if $\beta=at$

It won't be, at least not if the "ships" (disks) experience constant proper acceleration (more precisely, if an observer at the center of each disk, who is not rotating, experiences constant proper acceleration). If that's the case, we will have $\beta = \tanh at$ in general; $\beta = at$ will only be true for small $t$. But this will still results in the disks pulling away from each other.

 

[Mentz114]

Are you taking all the nonzero connection coefficients into account? There are a number of them in this chart.

Everything is taken into account. I just did a check and I find that in the coordinate basis $\nabla_a u_b$ is completely symmetric ( shear but no vorticity) but $\lambda^a_A \lambda^b_B \nabla_a u_b$ is completely anti-symmetric, so we get vorticity but no shear in the frame basis. I believe this result.

A change of basis can change the components of tensors like $\nabla_a u_b$ so no rules are being broken if the symmetry properties change.

If $\nabla_a u_b$ was zero, then obviously the transformed tensor $\nabla_A u_B$ will also be so.

It won't be, at least not if the "ships" (disks) experience constant proper acceleration (more precisely, if an observer at the center of each disk, who is not rotating, experiences constant proper acceleration). If that's the case, we will have $\beta = \tanh at$ in general; $\beta = at$ will only be true for small $t$. But this will still results in the disks pulling away from each other.

OK. A matter of detail.

 

[PeterDonis]

Everything is taken into account. I just did a check and I find that in the coordinate basis $\nabla_a u_b$ is completely symmetric ( shear but no vorticity) but $\lambda^a_A \lambda^b_B \nabla_a u_b$ is completely anti-symmetric, so we get vorticity but no shear in the frame basis. I believe this result.

This doesn't make sense. The kinematic decomposition is an invariant property of a frame field; the same frame field must have the same expansion, shear, and vorticity (meaning the scalars derived from those tensors--the tensor components themselves will of course transform covariantly) in any basis. I suspect that you are (inadvertently) transforming the frame field itself when you transform the basis.

My own computations are using a different chart (the Rindler chart instead of the Born chart), so I can't compare my results directly with yours; but I am getting zero expansion and nonzero shear in the Rindler coordinate basis. (I haven't--so far--tried transforming to a frame basis using the Rindler chart.)

 

[Mentz114]

This doesn't make sense. The kinematic decomposition is an invariant property of a frame field; the same frame field must have the same expansion, shear, and vorticity (meaning the scalars derived from those tensors--the tensor components themselves will of course transform covariantly) in any basis. I suspect that you are (inadvertently) transforming the frame field itself when you transform the basis.

My own computations are using a different chart (the Rindler chart instead of the Born chart), so I can't compare my results directly with yours; but I am getting zero expansion and nonzero shear in the Rindler coordinate basis. (I haven't--so far--tried transforming to a frame basis using the Rindler chart.)

It makes complete sense. The kinematic quantities are all found by decomposing the tensor $\nabla_a u_b$ into the trace, the symmetric part and the anti-symmetric part. It is the scalar contractions formed from it that remain invariant, the values of the decomposed parts can easily change while not affecting the contractions.

I'm very sad that you don't believe this and I'll start a new topic if you think it's worth it.

 

[PeterDonis]

The kinematic quantities are all found by decomposing the tensor $\nabla_a u_b$ into the trace, the symmetric part and the anti-symmetric part.

Agreed.

It is the scalar contractions formed from it that remain invariant

But if the shear is nonzero in some basis, then the scalar contraction $\sigma_{ab} \sigma^{ab}$ will also be nonzero, correct? But if you claim the shear is zero in some other basis, then that same scalar invariant must be zero in that basis, which contradicts the statement that scalar contractions remain invariant.

I'm very sad that you don't believe this

I suspect we're talking at cross purposes; see above for how I am interpreting your statement about the shear.

 

[WannabeNewton]

If $\sigma_{ab} = 0$ in some coordinate system then $\sigma_{ab} = 0$ in all coordinate systems; this is very trivial. The components of a tensor cannot identically vanish in one coordinate system but not identically vanish in another; this is one of the most basic (and important) facts about calculations in GR.

 

[Mentz114]

But if the shear is nonzero in some basis, then the scalar contraction $\sigma_{ab} \sigma^{ab}$ will also be nonzero, correct? But if you claim the shear is zero in some other basis, then that same scalar invariant must be zero in that basis, which contradicts the statement that scalar contractions remain invariant.

With $\nabla_b u_a = \nabla_{(b} u_{a)} + \nabla_{[b} u_{a]}$ and the transformed equation $\nabla_B u_A = \nabla_{(B} u_{A)} + \nabla_{[B} u_{A]}$, then

$\nabla^b u^a \nabla_b u_a = \nabla^B u^A \nabla_B u_A$ does not imply that $\nabla_{(b} u_{a)} =\nabla_{(B} u_{A)}$ nor $\nabla_{[b} u_{a]}=\nabla_{[B} u_{A]}$

Is this not true ?

 

[WannabeNewton]

That's not what is being said at all. If $T_{\alpha \beta} = 0$ in some coordinate system $(U,x^{\mu})$ and we have an overlapping coordinate system $(V,x^{\mu'})$ then clearly $T_{\alpha'\beta'} = 0$ as well because $T_{\alpha'\beta'} = \frac{\partial x^{\mu}}{\partial x^{\alpha'}}\frac{\partial x^{\nu}}{\partial x^{\beta'}}T_{\mu\nu} = 0$. So it's nonsensical to have $\sigma_{\mu\nu} = 0$ in one coordinate system (the one formed by the frame field) but $\sigma_{\mu'\nu'} \neq 0$ in another coordinate system (the one formed by the coordinate basis field).

 

[Mentz114]

That's not what is being said at all. If $T_{\alpha \beta} = 0$ in some coordinate system $(U,x^{\mu})$ and we have an overlapping coordinate system $(V,x^{\mu'})$ then clearly $T_{\alpha'\beta'} = 0$ as well because $T_{\alpha'\beta'} = \frac{\partial x^{\mu}}{\partial x^{\alpha'}}\frac{\partial x^{\nu}}{\partial x^{\beta'}}T_{\mu\nu} = 0$. So it's nonsensical to have $\sigma_{\mu\nu} = 0$ in one coordinate system (the one formed by the frame field) but $\sigma_{\mu'\nu'} = 0$ in another coordinate system (the one formed by the coordinate basis field).

Changing to a frame field basis is not a holonomic transformation so $T_{\alpha'\beta'} \ne \frac{\partial x^{\mu}}{\partial x^{\alpha'}}\frac{\partial x^{\nu}}{\partial x^{\beta'}}T_{\mu\nu}$

Is my previous post mathematically true or not ?

 

[WannabeNewton]

See appendix J of Carroll; non-coordinate bases don't change the issue with regards to $\sigma= 0$. The representation of $\sigma$ in the Lorentz frame is related to its representation in the coordinate frame via $\sigma_{\mu\nu} = e_{\mu}{}{}^{a}e_{\mu}{}{}^{b}\sigma_{ab}$ so $\sigma$ vanishing identically in the Lorentz frame implies it vanishes identically in the coordinate frame.

 

[PeterDonis]

I am getting zero expansion and nonzero shear in the Rindler coordinate basis.

I'll go ahead and post these computations [edit: corrected [STRIKE]some[/STRIKE] a bunch of errors--I am not very reliable at these kinds of computations ] since they're pretty straightforward [or so I thought ] for the congruence Mentz114 is considering (which is simpler than the more general family of congruences I've been trying to investigate).

The Rindler chart in cylindrical coordinates is:

$$
ds^2 = - z^2 dt^2 + dz^2 + dr^2 + r^2 d\phi^2
$$

with nonzero connection coefficients $\Gamma^z{}_{tt} = z$, $\Gamma^{t}{}_{tz} = 1 / z$, $\Gamma^{r}{}_{\phi \phi} = -r$, $\Gamma^{\phi}{}_{r \phi} = 1/r$.

The congruence in question has the 4-velocity field (I won't write down the whole frame field because all we need is the 4-velocity field to compute the expansion tensor):

$$
u^a = \frac{1}{z \sqrt{1 - r^2 \omega^2}} \partial_t + \frac{\omega}{\sqrt{1 - r^2 \omega^2}} \partial_{\phi}
$$

Lowering the index gives the corresponding covector field:

$$
u_a = - \frac{z}{\sqrt{1 - r^2 \omega^2}} dt + \frac{\omega r^2}{\sqrt{1 - r^2 \omega^2}} d \phi
$$

We define $\gamma = 1 / \sqrt{1 - r^2 \omega^2}$ for ease of writing; and we note that the only nonzero partial derivative of $\gamma$ is $\partial_r \gamma = r \omega^2\gamma^3$. This will be useful in what follows.

We want to calculate $\nabla_{(a} u_{b)} = \partial_a u_b + \partial_b u_a - 2 \Gamma^c{}_{ab} u_c$. The nonzero components turn out to be:

$$
\nabla_{(t} u_{z)} = \partial_z ( - z \gamma ) - 2 \frac{1}{z} ( - z \gamma ) = \gamma
$$

$$
\nabla_{(t} u_{r)} = \partial_r ( - z \gamma ) = - z \partial_r \gamma = - z r \omega^2 \gamma^3
$$

$$
\nabla_{(r} u_{\phi)} = \partial_r ( \gamma \omega r^2 ) - 2 \frac{1}{r} \gamma \omega r^2 = r^3 \omega^3 \gamma^3
$$

The expansion, which is the trace of $\nabla_{(a} u_{b)}$, is obviously zero. The shear tensor is nonzero, and its scalar invariant is $\sigma = \sigma_{ab} \sigma^{ab} = ( \sigma_{ab} )^2 g^{aa} g^{bb}$ since the metric is diagonal in this chart. So we have

$$
\sigma = ( \sigma_{tz} )^2 g^{tt} g^{zz} + ( \sigma_{tr} )^2 g^{tt} g^{rr} + ( \sigma_{r \phi} )^2 g^{rr} g^{\phi \phi}
$$

We have $g^{tt} = - 1 / z^2$, $g^{zz} = g^{rr} = 1$, and $g^{\phi \phi} = 1 / r^2$, so we obtain

$$
\sigma = - \frac{\gamma^2}{z^2} - \frac{z^2 r^2 \omega^4 \gamma^6}{z^2} + \frac{r^6 \omega^6 \gamma^6}{r^2} = - \frac{\gamma^2}{z^2} - r^2 \omega^4 \gamma^4
$$

 

[WannabeNewton]

This is the frame field obtained by boosting the static observers in the Rindler chart in the tangential ($\phi$) direction yes?

 

[PeterDonis]

This is the frame field obtained by boosting the static observers in the Rindler chart in the tangential ($\phi$) direction yes?

Yes.

 

[Mentz114]

See appendix J of Carroll; non-coordinate bases don't change the issue with regards to $\sigma= 0$. The representation of $\sigma$ in the Lorentz frame is related to its representation in the coordinate frame via $\sigma_{\mu\nu} = e_{\mu}{}{}^{a}e_{\mu}{}{}^{b}\sigma_{ab}$ so $\sigma$ vanishing identically in the Lorentz frame implies it vanishes identically in the coordinate frame.

The acceleration can go from a non-zero value to zero we change from a non-geodesic to a geodesic basis. So I don't see why shear cannot depend on the frame.

I always understood that all the kinematic quantities are frame dependent but I'll think about it. For now I'll stick with my current POV.

@Peter, I'll check your Rindler calculation when I get time. But I'm away now. Thanks for input.Your input appreciated as always.