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

[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.

 

[WannabeNewton]

Certainly in a general sense, the components of $\sigma$ in the Lorentz frame don't have to be the same as the components of $\sigma$ in the coordinate basis; in that sense yes they are all frame dependent. But in the special case wherein the components of $\sigma$ vanish identically in say the Lorentz frame, they must vanish identically in the coordinate basis as well because $\sigma_{\mu\nu} = e_{\mu}{}{}^{a}e_{\nu}{}{}^{b}\sigma_{ab}$ where latin indices are the Lorentz frame indices and greek indices are the coordinate indices.

Yes.

Cool, thanks!

 

[Mentz114]

My earlier statement about the shear being zero in the frame basis and non-zero in the coordinate basis is confusing because I've only been looking at the spatial components of $\nabla_{(a} u_{b)}$. If all components are taken into account, then it is non-zero in both, but has spatial components in one but not the other. The kinematic quantities should be projected in the local 3-space using the spatial projection tensor and this disappears the timelike components. Whoops.

I hope this clears up the issue and we can agree that it is possible to have shear ( projected into 3D) in one frame and not another since that depends only on some components.

 

[PeterDonis]

it is possible to have shear ( projected into 3D) in one frame and not another since that depends only on some components.

Yes, I agree that if you project into different 3-surfaces you can get different projections.

Btw, I found a bunch of errors in my previous post of the shear computation using Rindler coordinates, so since the post was still within the edit window I went back and corrected them, since it was easier than reposting.

Also, the corrected result for the shear scalar invariant can be cast in an interesting form:

$$
\sigma = - \frac{\gamma^2}{z^2} - r^2 \omega^4 \gamma^4 = - \frac{\gamma^4}{z^2} \left[ 1 - r^2 \omega^2 \left( 1 - z^2 \omega^2 \right) \right]
$$

The obvious next step is to re-do my computation in the Rindler chart but allowing $\omega$ to be a function of $z$, to see if a solution with zero shear is possible. I think this will leave the tensor traceless, but it will change the specific values of at least one component, $\sigma_{tz}$, and will also add one more nonzero component, $\sigma_{z \phi}$. I'm working that through now.

 

[PeterDonis]

I'm working that through now.

Here's a quick post of my results so far (which could stand checking given that I had to go back and correct my last post of results ). We have the same congruence as before except that now we are letting $\omega$ be a function of $z$ instead of being a constant. This means we now have two more nonzero partial derivatives to consider: $\partial_z \omega$, and $\partial_z \gamma = \gamma^2 r^2 \omega \partial_z \omega$.

These partial derivatives come into play in two of the components of the shear tensor (the expansion still appears to be zero):

$$
\sigma_{tz} = \partial_z ( - z \gamma ) + 2 \gamma = \gamma - \gamma^3 r^2 z \omega \partial_z \omega = \gamma \left( 1 - \gamma^2 r^2 z \omega \partial_z \omega \right)
$$

$$
\sigma_{z \phi} = \partial_z ( \gamma \omega r^2 ) = \omega r^2 \partial_z \gamma + \gamma r^2 \partial_z \omega = \gamma^3 r^2 \partial_z \omega
$$

The other two nonzero components of the shear remain the same: $\sigma_{tr} = - z r \omega^3 \gamma^3$ and $\sigma_{r \phi} = r^3 \omega^3 \gamma^3$.

The shear scalar now gets some extra terms:

$$
\sigma = - \frac{\gamma^2}{z^2} \left( 1 - \gamma^2 r^2 z \omega \partial_z \omega \right)^2 - r^2 \omega^4 \gamma^4 + \gamma^6 r^2 \left( \partial_z \omega \right)^2
$$

This can be reworked into a form that makes the issue with trying to set it to zero more evident (and also reduces to the previous result, as desired, if $\partial_z \omega = 0$):

$$
\sigma = - \frac{\gamma^4}{z^2} \left[ z^2 r^2 ( \partial_z \omega )^2 + 2 z r^2 \omega \partial_z \omega + 1 - r^2 \omega^2 \left( 1 - z^2 \omega^2 \right) \right]
$$

I don't see any simple solution for making this zero, which leads me to believe that there is no simple congruence of the form we've been considering (i.e., a Rindler congruence boosted only tangentially, with the angular velocity allowed to vary with $z$) that can be shown to be rigid.

 

[Mentz114]

Yes, I agree that if you project into different 3-surfaces you can get different projections.

Btw, I found a bunch of errors in my previous post of the shear computation using Rindler coordinates, so since the post was still within the edit window I went back and corrected them, since it was easier than reposting.

Also, the corrected result for the shear scalar invariant can be cast in an interesting form:

$$
\sigma = - \frac{\gamma^2}{z^2} - r^2 \omega^4 \gamma^4 = - \frac{\gamma^4}{z^2} \left[ 1 - r^2 \omega^2 \left( 1 - z^2 \omega^2 \right) \right]
$$

The obvious next step is to re-do my computation in the Rindler chart but allowing $\omega$ to be a function of $z$, to see if a solution with zero shear is possible. I think this will leave the tensor traceless, but it will change the specific values of at least one component, $\sigma_{tz}$, and will also add one more nonzero component, $\sigma_{z \phi}$. I'm working that through now.

Looking back to your earlier post where you started the Rindler calculation I see you worked out the shear from $\sigma_{ab} = \nabla_{(a}u_{b)}$, there being no trace term to subtract. However, according to Stephani, this is valid only for geodesics. For non-geodesics there is an additional term ${\dot{u}}_a u_b$ where ${\dot{u}}_a = u^b\nabla_b u_a$ i.e the acceleration. This term is orthogonal to $u^a$.

The reason is that we need to decompose the the parts of the covariant derivative that are orthogonal to the velocity. This is explained in Stephani's book* in section 17.2 "Timelike vector fields".

It makes a big difference as one might expect. For instance, including the extra term in the Born chart calculation gives zero shear, leaving out the acceleration term gives non-zero shear. The vorticity is not affected because ${\dot{u}}_a u_b$ is fully symmetric.

Anyhow, for non geodesics, it seems that $\sigma_{ab} = \nabla_{(a}u_{b)}+{\dot{u}}_{(a} u_{b)}$, assuming no trace term.

I'm fairly confused now. The extra term makes sense but Stephani's defintions of $\sigma_{ab}$ and $\omega_{ab}$ do not agree ( for instance ) with those in this paper http://arxiv.org/abs/1012.4806 by Abreu and Visser.

I'm sticking with Stephani because it gives the correct answers in cases I've been able to verify.

* "General Relativity", 2nd Edition. Cambridge, 1990.

 

[WannabeNewton]

Yes that is correct. The shear is defined as $\sigma_{ab} = \theta_{ab} - \frac{1}{3}h_{ab}\theta$ where $\theta_{ab}$ is the expansion tensor; the expansion tensor and rotation tensor are defined as $\theta_{ab} = h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{(c}u_{d)}$ and $\omega_{ab} = h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{[c}u_{d]}$ respectively. One can show that $\nabla_a u_b = \omega_{ab} + \theta_{ab} - u_a a_b$ where $u^a$ is a time-like congruence and $a_b = u^c \nabla_c u_b$ is its acceleration.

This is because $\theta_{ab} + \omega_{ab} \\= h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{(c}u_{d)} + h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{[c}u_{d]} \\= (g_{a}{}{}^{c} + u_a u^c )(g_{b}{}{}^{d} + u_b u^d)\nabla_{c}u_{d} \\= \nabla_a u_b + u_a u^c \nabla_c u_b$
since $u^a \nabla_b u_a = 0$.

The last term of course vanishes if $u^a$ is geodesic in which case $\sigma_{ab} = \nabla_{(a} u_{b)} - \frac{1}{3}h_{ab}\theta$. If $u^a$ is not geodesic then we will have $\sigma_{ab} = \nabla_{(a}u_{b)} + u_{(a}a_{b)} - \frac{1}{3}h_{ab}\theta$

 

[PeterDonis]

If $u^a$ is not geodesic then we will have $\sigma_{ab} = \nabla_{(a}u_{b)} + u_{(a}a_{b)} - \frac{1}{3}h_{ab}\theta$

And I take it the vorticity for a non-geodesic congruence would then be $\omega_{ab} = \nabla_{[a}u_{b]} + u_{[a}a_{b]}$, correct? [Edit: This appears to work for the Rindler congruence, i.e., straight Rindler observers with no rotation.]

It looks like I'll need to go back and rework my computations.

 

[Mentz114]

Yes that is correct.
..
..

Excellent. That clears up any issues I had. Thanks.

And I take it the vorticity for a non-geodesic congruence would then be $\omega_{ab} = \nabla_{[a}u_{b]} + u_{[a}a_{b]}$, correct?

It looks like I'll need to go back and rework my computations.

Sorry about that, I wondered why we were getting such different results.

(deleted my stupid remark. Tsk).

 

[WannabeNewton]

And I take it the vorticity for a non-geodesic congruence would then be $\omega_{ab} = \nabla_{[a}u_{b]} + u_{[a}a_{b]}$, correct?

Yeah it should be since $\theta_{ba} = h_{b}{}{}^{c}h_{a}{}{}^{d}\nabla_{(c}u_{d)} = h_{b}{}{}^{d}h_{a}{}{}^{c}\nabla_{(d}u_{c)} = h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{(c}u_{d)} = \theta_{ab}$ so $\theta_{[ab]} = 0$ and similarly $\omega_{ab} = \omega_{[ab]}$, giving us $\nabla_{[a}u_{b]} = \omega_{ab} - u_{[a}a_{b]}$.

 

[PeterDonis]

$u_{[a}u_{b]}$ is zero, I think because U_aU_b is a symmetric tensor (?)

Yes, but $u_{[a} a_{b]}$ is not.

Also, the antisymmetrized covariant derivative is the same as the antisymmetrized partial derivative (because the connection coefficients are symmetric in their lower indexes), so the vorticity expands to $\omega_{ab} = \partial_a u_b - \partial_b u_a + u_a a_b - u_b a_a$.

 

[Mentz114]

Yes, but $u_{[a} a_{b]}$ is not.

Also, the antisymmetrized covariant derivative is the same as the antisymmetrized partial derivative (because the connection coefficients are symmetric in their lower indexes), so the vorticity expands to $\omega_{ab} = \partial_a u_b - \partial_b u_a + u_a a_b - u_b a_a$.

You're right. I removed that remark from my post probably while you were writing. Apologies. It is in the Born case but not generally.

 

[PeterDonis]

It looks like I'll need to go back and rework my computations.

I have reworked them. I'll summarize what I've got for the case where we allow $\omega$ to be a function of $z$; the case of constant $\omega$ can be obtained by just setting $\partial_z \omega = 0$.

First, to summarize my results for $\nabla_{(a} u_{b)}$:

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

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

$$
\nabla_{(z} u_{\phi)} = \gamma^3 r^2 \partial_z \omega
$$

$$
\nabla_{(r} u_{\phi)} = \gamma^3 r^3 \omega^3
$$

Now for the proper acceleration covector $a_a$:

$$
a_a = u^b \nabla_b u_a = u^b \partial_b u^a - u^b \Gamma^c{}_{ba} u_c
$$

The contraction $u^b \partial_b$ vanishes because nothing depends on $t$ or $\phi$, so we're left with the connection coefficient term, which gives two nonzero components of $a_a$:

$$
a_z = - u^t \Gamma^t{}_{tz} u_t = \frac{\gamma^2}{z}
$$

$$
a_r = - u^{\phi} \Gamma^{\phi}{}_{\phi r} u_{\phi} = - \gamma^2 \omega^2 r
$$

Both of these make obvious sense physically. We now want the symmetric bivector $u_{(a} a_{b)}$, which will have four nonzero components that match up nicely with the four nonzero components of $\nabla_{(a} u_{b)}$:

$$
u_{(t} a_{z)} = - \gamma^3
$$

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

$$
u_{(z} a_{\phi)} = \gamma^3 \frac{\omega r^2}{z}
$$

$$
u_{(r} a_{\phi)} = - \gamma^3 \omega^3 r^3
$$

We now just match up corresponding components to obtain $\sigma_{ab} = \nabla_{(a} u_{b)} + u_{(a} a_{b)}$. We find that $\sigma_{tr}$ and $\sigma_{r \phi}$ vanish; but the other two components do not:

$$
\sigma_{tz} = \gamma \left[ 1 - \gamma^2 \left( 1 + r^2 z \omega \partial_z \omega \right) \right]
$$

$$
\sigma_{z \phi} = \gamma^3 r^2 \left( \frac{\omega}{z} + \partial_z \omega \right)
$$

The shear scalar is $\sigma = \left( \sigma_{ab} \right)^2 g^{aa} g^{bb}$, which gives, after simplifying as much as possible (this could stand checking as the algebra gets rather tedious for me):

$$
\sigma^2 = \gamma^6 r^2 \left[ \left( 1 - \frac{r^2 \omega^2}{\gamma^4} \right) \left( \partial_z \omega + \frac{\omega}{z} \right)^2 - \frac{1}{z^2 \gamma^4} \right]
$$

An obvious ansatz is to set $\partial_z \omega = - \omega / z$, which gives $\omega = 1 / z$; this is what we would "naively" guess if we were trying to ensure that all of the stack of disks were rotating "in sync" by making $\omega$ decrease in step with the time dilation factor. This makes $\sigma_{z \phi} = 0$, but it leaves $\sigma_{tz} = \gamma$ and therefore $\sigma^2 = - \gamma^2 / z^2$. So this does zero out the purely spacelike components of the shear in this chart; but it does *not* zero out the shear completely. I'll save further comment on the physical meaning of this for a follow-up post.

 

[yuiop]

$$
\sigma_{tz} = \gamma \left[ 1 - \gamma^2 \left( 1 + r^2 z \omega \partial_z \omega \right) \right]
$$

Assuming v=0 and $\gamma^2 = 1/(1 - r^2\omega^2)$ and $\partial_z \omega = - \omega / z$, I get
$$
\sigma_{tz} = \gamma \left[ 1 - \gamma^2 \left( 1 + r^2 z \omega (-\omega/z) \right) \right] = \gamma \left[ 1 - \gamma^2 \left( 1 - r^2\omega^2 \right) \right] = \gamma \left[ 1 - \gamma^2 \left( 1/\gamma^2 \right) \right] = 0$$

The shear scalar is $\sigma = \left( \sigma_{ab} \right)^2 g^{aa} g^{bb}$, which gives, after simplifying as much as possible (this could stand checking as the algebra gets rather tedious for me):
$$
\sigma^2 = \gamma^6 r^2 \left[ \left( 1 - \frac{r^2 \omega^2}{\gamma^4} \right) \left( \partial_z \omega + \frac{\omega}{z} \right)^2 - \frac{1}{z^2 \gamma^4} \right]
$$

An obvious ansatz is to set $\partial_z \omega = - \omega / z$, which gives $\omega = 1 / z$; this is what we would "naively" guess if we were trying to ensure that all of the stack of disks were rotating "in sync" by making $\omega$ decrease in step with the time dilation factor. This makes $\sigma_{z \phi} = 0$, but it leaves $\sigma_{tz} = \gamma$ and therefore $\sigma^2 = - \gamma^2 / z^2$. So this does zero out the purely spacelike components of the shear in this chart; but it does *not* zero out the shear completely. I'll save further comment on the physical meaning of this for a follow-up post.

I work that out to $\sigma^2 = - \gamma^2 r^2/ z^2 = - \gamma^2 r^2 \omega^2$

when $\partial_z \omega = - \omega / z$ and assuming your original equation ...

$$
\sigma^2 = \gamma^6 r^2 \left[ \left( 1 - \frac{r^2 \omega^2}{\gamma^4} \right) \left( \partial_z \omega + \frac{\omega}{z} \right)^2 - \frac{1}{z^2 \gamma^4} \right]
$$

.. is correct.

 

[Mentz114]

I have reworked them.
..
..
$$
a_z = - u^t \Gamma^t{}_{tz} u_t = \frac{\gamma^2}{z}
$$

$$
a_r = - u^{\phi} \Gamma^{\phi}{}_{\phi r} u_{\phi} = - \gamma^2 \omega^2 r
$$

Both of these make obvious sense physically.

I agree with those components. Yippee.

$$
\sigma_{tz} = \gamma \left[ 1 - \gamma^2 \left( 1 + r^2 z \omega \partial_z \omega \right) \right]
$$

$$
\sigma_{z \phi} = \gamma^3 r^2 \left( \frac{\omega}{z} + \partial_z \omega \right)
$$

I get
$\sigma_{z \phi}= \frac{ 1}{2z}\gamma^\frac{3}{2}{r}^{2}\,\left( \partial_z\,w \,z+w\right)$ and $\sigma_{z t}= \frac{ 1}{2}\gamma^\frac{3}{2}{r}^{2}\,\left( \partial_z\,w \,z+w\right)$. My $\sigma_{z \phi}$ is different from yours by a factor.

I'll save further comment on the physical meaning of this for a follow-up post.

What do the time components of the shear signify ? I think the purely spatial component is physical and can be removed by solving the differential equation. With my solution all the components go. I'm out of time and dashed this off, but I think we're close enough to hope for exact agreement eventually.

I attach some notes I've been scribbling which show some intermediate results which I'll check with yours when I can.

 

[PeterDonis]

$$
\sigma_{tz} = \gamma \left[ 1 - \gamma^2 \left( 1 + r^2 z \omega (-\omega/z) \right) \right] = \gamma \left[ 1 - \gamma^2 \left( 1 - r^2\omega^2 \right) \right] = \gamma \left[ 1 - \gamma^2 \left( 1/\gamma^2 \right) \right] = 0
$$

Hm, yes, this looks right. But that means $\sigma^2$ should be zero as well (since all components are zero), so I'll need to check the equation for that again; it's quite possible I made an algebra error. (Or I made one in calculating $\sigma_{tz}$, but I'm more confident of the individual component calculations than I am of the one for $\sigma^2$.)

This also takes us back to the question about the Herglotz-Noether theorem, which appears to claim that the shear should *not* be zero for this congruence. I'm not sure what the resolution to that question is.

 

[WannabeNewton]

What do the time components of the shear signify ?

In the coordinates comoving with any given observer represented by an integral curve of $u^a$, the time components of the shear will always vanish because $u^a \sigma_{ab} = u^ah_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{(c}u_{d)} - \frac{1}{3}\theta u^a h_{ab} = 0$ and similarly $u^b \sigma_{ab} = 0$. In other words, $\sigma_{ab}$ is a purely spatial tensor, and this is manifest in the coordinates comoving with said observer. And since the physical interpretation of $\sigma_{ab}$ is described using said comoving coordinates, the time components don't mean anything.

 

[PeterDonis]

My $\sigma_{z \phi}$ is different from yours by a factor.

I get

$$
\sigma_{z \phi} = \nabla_{(z} u_{\phi)} + u_{(z} a_{\phi)} = \partial_z u_{\phi} + u_{\phi} a_z = \partial_z ( \gamma \omega r^2 ) + ( \gamma \omega r^2 ) \frac{\gamma^2}{z} = \omega r^2 \partial_z \gamma + \gamma r^2 \partial_z \omega + \gamma^3 \frac{\omega r^2}{z}
$$

$$
\ \ \ \ \ = \left( \omega r^2 \gamma^3 r^2 \omega + \gamma r^2 \right) \partial_z \omega + \gamma^3 r^2 \frac{\omega}{z} = \gamma^3 r^2 \left[ \left( \omega^2 r^2 + \frac{1}{\gamma^2} \right) \partial_z \omega + \frac{\omega}{z} \right] = \gamma^3 r^2 \left( \partial_z \omega + \frac{\omega}{z} \right)
$$

You might want to check how you're computing $\partial_z \gamma$; it comes out with a factor of $\left( 1 - r^2 \omega^2 \right)^{- 3/2}$ in front, but that's $\gamma^3$, not $\gamma^{3/2}$. Also, the factor of $\frac{1}{2}$ from the derivative is canceled by a factor of $2$ when you take $\partial_z \omega^2 = 2 \omega \partial_z \omega$. So we have $\partial_z \gamma = \gamma^3 r^2 \omega \partial_z \omega$.

 

[Mentz114]

In the coordinates comoving with any given observer represented by an integral curve of $u^a$, the time components of the shear will always vanish because $u^a \sigma_{ab} = u^ah_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{(c}u_{d)} - \frac{1}{3}\theta u^a h_{ab} = 0$ and similarly $u^b \sigma_{ab} = 0$. In other words, $\sigma_{ab}$ is a purely spatial tensor, and this is manifest in the coordinates comoving with said observer. And since the physical interpretation of $\sigma_{ab}$ is described using said comoving coordinates, the time components don't mean anything.

That $\sigma_{ab}$ is a purely spatial tensor is made clear in my textbooks and articles I've read. This is why I happily declared that they went to zero under transformations and caused a fuss.

But $\sigma= \sigma^{ab}\sigma_{ab}$ is a Lorentz scalar so I assume includes four dimensions, yes ?

 

[PeterDonis]

In the coordinates comoving with any given observer represented by an integral curve of $u^a$, the time components of the shear will always vanish

Yes, but the chart I was using is not comoving except with the observer at the center of the disk, who is not rotating. So it's possible to have meaningful time components of shear at points other than the center of the disk in this chart; they signify that the observers at those points are not at rest in the chart we're using. Transforming to locally comoving coordinates at an event that's not at the center of the disk would make the time components of the shear vanish at that point, but might also make spatial components appear that weren't there in the original chart.

(It looks like that isn't actually the case here if we set $\omega = 1 / z$, but it would be the case if, for example, we made $\omega$ a constant, not dependent on $z$.)

 

[PeterDonis]

But $\sigma= \sigma^{ab}\sigma_{ab}$ is a Lorentz scalar so I assume includes four dimensions, yes ?

If you're not using a comoving chart (as I wasn't), then yes, you have to evaluate $\sigma^2$ (I should have used that notation before since it's actually a quadratic invariant) as a Lorentz scalar, using all four dimensions.

If you're in a chart that's comoving at a particular event, you can evaluate $\sigma^2$ in that chart, at that event, as if it were a spatial 3-tensor, since the time components must vanish in that particular chart. But it will still be a Lorentz scalar--you'll get the same number as you would get if you evaluated $\sigma^2$ at the same event in a different, non-comoving chart, where there might be nonzero timelike components.

 

[Mentz114]

I get

$$
\sigma_{z \phi} = \nabla_{(z} u_{\phi)} + u_{(z} a_{\phi)} = \partial_z u_{\phi} + u_{\phi} a_z = \partial_z ( \gamma \omega r^2 ) + ( \gamma \omega r^2 ) \frac{\gamma^2}{z} = \omega r^2 \partial_z \gamma + \gamma r^2 \partial_z \omega + \gamma^3 \frac{\omega r^2}{z}
$$

$$
\ \ \ \ \ = \left( \omega r^2 \gamma^3 r^2 \omega + \gamma r^2 \right) \partial_z \omega + \gamma^3 r^2 \frac{\omega}{z} = \gamma^3 r^2 \left[ \left( \omega^2 r^2 + \frac{1}{\gamma^2} \right) \partial_z \omega + \frac{\omega}{z} \right] = \gamma^3 r^2 \left( \partial_z \omega + \frac{\omega}{z} \right)
$$

You might want to check how you're computing $\partial_z \gamma$; it comes out with a factor of $\left( 1 - r^2 \omega^2 \right)^{- 3/2}$ in front, but that's $\gamma^3$, not $\gamma^{3/2}$. Also, the factor of $\frac{1}{2}$ from the derivative is canceled by a factor of $2$ when you take $\partial_z \omega^2 = 2 \omega \partial_z \omega$. So we have $\partial_z \gamma = \gamma^3 r^2 \omega \partial_z \omega$.

The problem was not in the calculation, but with me ( as always). I translated by eye and made that mistake ( I don't have $\gamma$ explicitly in my workings). So I actually get $\sigma_{z \phi}= \frac{ 1}{2z}\gamma^3{r}^{2}\,\left( \partial_z\,w \,z+w\right)$. The factor of two comes from symmetrization, I think. I'm using $T_{(ab)}= (1/2)(T_{ab}+T_{ba})$. Should I drop the factor of 1/2 ?

Thanks for pointing that out the power of $\gamma$ mistake.

 

[PeterDonis]

The factor of two comes from symmetrization, I think, but I'll check.

If so, it may just be a difference in how we're defining "symmetrization"; I didn't include the factor of 1/2, not because I don't think it belongs, but because I was being lazy , since the main question was whether it was possible for the shear to be zero.

 

[Mentz114]

If so, it may just be a difference in how we're defining "symmetrization"; I didn't include the factor of 1/2, not because I don't think it belongs, but because I was being lazy , since the main question was whether it was possible for the shear to be zero.

OK, that explains the 1/2 so the only visible difference now is with $\sigma_{tz}$.

I think you've found the zero-shear congruence, but I find all this rotational stuff a bit hard to interpret ...

Thanks for confirming some of my results - as a side effect of your calculations, of course. It's a rare thing.

 

[PeterDonis]

the only visible difference now is with $\sigma_{tz}$

Just to check that, I get:

$$
\sigma_{tz} = \nabla_{(t} u_{z)} + u_{(t} a_{z)} = \partial_z u_t - 2 \Gamma^{t}{}_{tz} u_t + u_t a_z = \partial_z ( - z \gamma ) - 2 \frac{1}{z} ( - z \gamma ) + ( - z \gamma ) \frac{\gamma^2}{z}
$$

$$
\ \ \ \ \ = - \gamma - z \partial_z \gamma + 2 \gamma - \gamma^3 = \gamma - \gamma^3 - z \gamma^3 r^2 \omega \partial_z \omega = \gamma \left[ 1 - \gamma^2 \left( 1 + z r^2 \omega \partial_z \omega \right) \right]
$$

This vanishes if $\partial_z \omega = - \omega / z$, so that ansatz does indeed make the shear vanish (since it also obviously makes $\sigma_{z \phi}$ vanish).

 

[Mentz114]

Just to check that, I get:

$$
\sigma_{tz} = \nabla_{(t} u_{z)} + u_{(t} a_{z)} = \partial_z u_t - 2 \Gamma^{t}{}_{tz} u_t + u_t a_z = \partial_z ( - z \gamma ) - 2 \frac{1}{z} ( - z \gamma ) + ( - z \gamma ) \frac{\gamma^2}{z}
$$

$$
\ \ \ \ \ = - \gamma - z \partial_z \gamma + 2 \gamma - \gamma^3 = \gamma - \gamma^3 - z \gamma^3 r^2 \omega \partial_z \omega = \gamma \left[ 1 - \gamma^2 \left( 1 + z r^2 \omega \partial_z \omega \right) \right]
$$

This vanishes if $\partial_z \omega = - \omega / z$, so that ansatz does indeed make the shear vanish (since it also obviously makes $\sigma_{z \phi}$ vanish).

Ha. Our results are identical. I get ( without the factor of 1/2)

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

which is just $\gamma - \gamma^3 - z \gamma^3 r^2 \omega \partial_z \omega$.

The expression above simplifies to $\frac{{r}^{2}\,w\,\left( \frac{d}{d\,z}\,w\right) \,z+{r}^{2}\,{w}^{2}}{\sqrt{1-{r}^{2}\,{w}^{2}}\,\left( {r}^{2}\,{w}^{2}-1\right) }$.

High fives all around.:thumbs: