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

[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:

 

[yuiop]

As I understand it, if all the components of $\sigma$ are zero (which is what PeterDonnis and Mentz114 appear to have found), then $\sigma$ itself should also be zero. It would be nice to find the correct expression for $\sigma$ just to round off the conclusions of this thread and see if there is really is a problem with the accepted understanding of the Herglotz-Noether theorem.

By the way guys, thanks for all the hard and sometimes tedious work you have put into this investigation. Impressive stuff on a non trivial subject.

 

[WannabeNewton]

If by $\sigma$ you mean $\sigma = \sigma^{ab}\sigma_{ab}$ then yes it vanishes if $\sigma_{ab} = 0$ identically.

 

[Mentz114]

As I understand it, if all the components of $\sigma$ are zero (which is what PeterDonnis and Mentz114 appear to have found), then $\sigma$ itself should also be zero. It would be nice to find the correct expression for $\sigma$ just to round off the conclusions of this thread and see if there is really is a problem with the accepted understanding of the Herglotz-Noether theorem.

By the way guys, thanks for all the hard and sometimes tedious work you have put into this investigation. Impressive stuff on a non trivial subject.

It was fun, some of the time.

I'm sure PeterDonis will be back to discuss the physics but I'll give my (speculative ) thoughts now.

It looks as if we have a spinning disc which experiences an acceleration in the z-direction ( the axis of rotation) but is not moving translationally. Two disks close together will have angular velocity $\omega_1=\omega_0/z$ and $\omega_2=\omega_0/(z+\delta z)$ and the shear is zero as is the expansion scalar. It looks as if matching the angular velocity with the time dilation factor eliminates the shear so locally every disc appears to have the same $\omega$. Something that is unexpected is that the proper z-acceleration ( in the coordinate basis) is modified from $1/z$ to $\gamma^2/z$. So the discs feel heavier because they are spinning ? Is that possible ?

Also, the relationship between $z$ and $\omega$ means the shear-free disc stack cannot be cylindrical, but is constrained to be a cone.

 

[PeterDonis]

It looks as if we have a spinning disc which experiences an acceleration in the z-direction ( the axis of rotation) but is not moving translationally.

Just to clarify, it's not moving translationally because I used a comoving chart (comoving with the observer at the center of the disk), which is non-inertial. In an inertial chart the disk would of course be moving translationally in the $z$ direction (since it's accelerating in that direction).

Two disks close together will have angular velocity $\omega_1=\omega_0/z$ and $\omega_2=\omega_0/(z+\delta z)$ and the shear is zero as is the expansion scalar. It looks as if matching the angular velocity with the time dilation factor eliminates the shear so locally every disc appears to have the same $\omega$.

It depends on what "the same $\omega$" means, which is why it's important to specify that. Observers in the center of disks at different $z$ coordinates will observe their local disks as rotating with different angular velocities; that's what $\omega_1$ and $\omega_2$ are. But an observer at the center of a particular disk will observe *all* the disks to be rotating at the same angular velocity as his local disk, because of the variation of $\omega$ (the number that appears in the 4-velocity) with $z$, which just cancels the variation of the time dilation factor with $z$.

Something that is unexpected is that the proper z-acceleration ( in the coordinate basis) is modified from $1/z$ to $\gamma^2/z$. So the discs feel heavier because they are spinning ?

Basically, yes. The acceleration an observer rotating with one of the disks will experience is $a = \sqrt{a^a a_a} = \sqrt{g^{aa} (a_a) ^2}$, which includes both the $z$ and $r$ components (the formula works out that way because the metric is diagonal). This gives

$$
a = \sqrt{ g^{zz} ( a_z )^2 + g^{rr} ( a_r )^2} = \sqrt{\frac{\gamma^4}{z^2} + \gamma^4 \omega^4 r^2} = \gamma^2 \sqrt{\left( \frac{1}{z} \right)^2 + \left( \omega^2 r \right)^2 }
$$

So for an observer at the center of a disk, $a = 1 / z$ (because $\gamma = 1$ when $r = 0$), which is expected since this observer is not rotating. For an observer not at the center of a disk, yes, the proper acceleration is larger than it would be if the disk were moving inertially; it is the vector sum of the $r$ component due to the proper acceleration, which is the same as if the disk were moving inertially (i.e., the factor of $\gamma^2$ would still be present anyway), and the extra $z$ component due to the acceleration, which is indeed increased by the factor $\gamma^2$.

I think this goes back to what pervect posted in one of these threads, about a surface of constant $z$ appearing curved, not flat, to an observer moving transversely across it, if the surface is being accelerated in the $z$ direction.

Also, the relationship between $z$ and $\omega$ means the shear-free disc stack cannot be cylindrical, but is constrained to be a cone.

I assume you mean that, since $\omega$ varies with $z$, so will the tangential velocity at the edge of the disk? This makes the maximum $r$ coordinate that the chart I'm using can cover vary with $z$, since the tangential velocity at the maximum $r$ must be less than 1 (the speed of light); but as long as the radial coordinate of all the disk edges is less than the maximum allowed $r$ coordinate for the disk with the largest $z$ (which will have the largest tangential velocity at the edge for a given $r$), all the physical disks can have the same radius. Disks with smaller $z$ will just occupy a smaller portion of the spatial region that can be covered by the chart I used at their value of $z$.

 

[Mentz114]

Just to clarify, it's not moving translationally because I used a comoving chart (comoving with the observer at the center of the disk), which is non-inertial. In an inertial chart the disk would of course be moving translationally in the z direction (since it's accelerating in that direction).

I have to disagree. The congruence has $dz/d\tau=0$ so I think it is not moving, but is at a fixed z. The proper acceleration is $\gamma^2/z$ so something is holding it. Different z positions means different time dilation factors.

I like your acceleration calculation.

My remark about the cone shape comes from putting $\omega=\omega_0/z$ into $1/\sqrt{1-\omega^2 r^2}$ so that $\omega_0 r/z < 1 \rightarrow r< z/\omega_0$. ( for r,z > 0 )

 

[PAllen]

I have to disagree. The congruence has $dz/d\tau=0$ so I think it is not moving, but is at a fixed z. The proper acceleration is $\gamma^2/z$ so something is holding it. Different z positions means different time dilation factors.

In SR, how can you have proper acceleration without having motion (except for an instant) in an inertial frame?

 

[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.

I found the tetrad (for the case we have been discussing ) which does this ${\Lambda_A}^a {\Lambda_B}^b g_{ab} = \eta_{AB}$, and also ${\Lambda_A}^a u_a = -\partial_t$. So the tetrad takes covariant tensors from the coordinate basis to the local Minkowski frame basis of $u_a$. I then calculated the transform of $\nabla_a u_b + \dot{u}_a u_b$ and all the temporal components went to zero when it is symmetrized. I am amazed. Can this be shown in general from the first two equations ?

 

[Mentz114]

In SR, how can you have proper acceleration without having motion (except for an instant) in an inertial frame?

I don't know. But how can you have movement without $dz/d\tau \neq 0$ ?

 

[PAllen]

I don't know. But how can you have movement without $dz/d\tau \neq 0$ ?

Because the z is not a coordinate in and inertial frame. It is comoving with the center of a disk.

 

[Mentz114]

Because the z is not a coordinate in and inertial frame. It is comoving with the center of a disk.

What is the coordinate velocity ? Ie $dz/dt$. We've been working in the coordinate basis not a local frame.

 

[PAllen]

What is the coordinate velocity ? Ie $dz/dt$. We've been working in the coordinate basis not a local frame.

The calculations, started from adapting a Rindler metric. The z in that metric is comoving with a disk. So in an inertial frame in which a disk is momentarily at rest, the growth in z would be hyperbolic - just as Rindler to Minkowski coordinates.

 

[PeterDonis]

The calculations, started from adapting a Rindler metric. The z in that metric is comoving with a disk. So in an inertial frame in which a disk is momentarily at rest, the growth in z would be hyperbolic - just as Rindler to Minkowski coordinates.

Exactly.

 

[PeterDonis]

The congruence has $dz/d\tau=0$ so I think it is not moving, but is at a fixed z.

Motion is relative; "not moving" is not an absolute statement. An observer at the center of a disk is not moving relative to himself, obviously; and since $z$ is a coordinate comoving with him, his $dz / d \tau$ will be zero. So will his $dz / dt$ in this chart, because the $t$ of this chart is the Rindler $t$ coordinate, which is just his $\tau$ corrected by a time dilation factor that depends on $z$. (I normalized the chart so the observer at the center of the disk at $z = 1$ has his $\tau$ equal to the coordinate $t$.)

The proper acceleration is $\gamma^2/z$ so something is holding it.

Sure, whatever is accelerating all the disks (presumably they have rockets attached to them or something) will appear to be "holding them in place" in the chart I used. (But as PAllen and I have posted, this chart is not an inertial chart, so objects at rest are not moving inertially, they are accelerated.)

My remark about the cone shape comes from putting $\omega=\omega_0/z$ into $1/\sqrt{1-\omega^2 r^2}$ so that $\omega_0 r/z < 1 \rightarrow r< z/\omega_0$. ( for r,z > 0 )

Yes, this is the same thing I was saying in my previous post (but your way of expressing it is much clearer). But this is a restriction on the region that can be covered by the chart I used; there is no reason why every disk has to have the maximum possible radius it can have while staying within the chart, for its value of $z$. If each disk has a radius $R$ at its rim that satisfies $R < z_{max} / \omega_0$, where $z_{max}$ is the largest value of $z$ occupied by any disk, then all the disks can have the same radius while remaining within the limitation you give.

(Btw, I was also implicitly normalizing the chart I used so that $\omega_0 = 1$, and that value of $\omega$ corresponds to $z = 1$.)

 

[PAllen]

So, assuming these calculations checked by a few people are correct, it seems either:

1) We have an exception to Herglotz-Noether.
or
2) The particular congruence analyzed here, involving uniform acceleration along the spin axis plus just the right rotation, is actually a killing motion(!?).

 

[Mentz114]

So, assuming these calculations checked by a few people are correct, it seems either:

1) We have an exception to Herglotz-Noether.
or
2) The particular congruence analyzed here, involving uniform acceleration along the spin axis plus just the right rotation, is actually a killing motion(!?).

It is true that $\nabla_{(a} u_{b)} = 0$ with $\omega=1/z$ which suggests that $u$ is a Killing motion.

There is a stationary KVF, $\xi^a = \partial_t$, which gives $E=-\xi^\mu u_\mu = \frac{z}{\sqrt{1-{r}^{2}\,{\omega}^{2}}}$ . Substituting 1/z for ω gives $E=\frac{{z}^{2}}{\sqrt{{z}^{2}-{r}^{2}}}$ which is only constant if z is constant.

 

[PeterDonis]

(Btw, I was also implicitly normalizing the chart I used so that $\omega_0 = 1$, and that value of $\omega$ corresponds to $z = 1$.)

On further consideration, I realized that this remark was wrong; $\omega_0$ is a free parameter that is not constrained by my analysis, even with the normalization of coordinates that I did. I mention it because it will come into play below.

1) We have an exception to Herglotz-Noether.

This is what I thought at first, but now I'm not so sure. See below.

2) The particular congruence analyzed here, involving uniform acceleration along the spin axis plus just the right rotation, is actually a killing motion(!?).

I think it's possible that it actually is, despite various statements in the literature that seem to imply the contrary.

Looking at the proof of the H-N theorem in the Giulini paper, it depends on three key properties that a rigid congruence with non-zero vorticity must have. These are given by Lemmas 19, 20, and 21 in the paper, and they are (I'll be using my notation, not the notation of the paper):

(1) The vorticity must be constant along any worldline in the congruence; i.e., we must have $u^c \nabla_c \omega_{ab} = 0$. (This is from Lemma 19.)

(2) The proper acceleration must be constant along any worldline in the congruence; i.e., we must have $u^b \nabla_b a_a = 0$. (This is from Lemma 21.)

(3) The proper acceleration must be an exact 1-form; that is, we must have $a_a = \partial_a f$ for some scalar function $f$. (This is from Lemma 20, with the results for the other two lemmas plugged into the equation that Lemma 20 gives.) [Edit: Note that since $f$ is a scalar, partial derivatives are equivalent to covariant derivatives, which is why I wrote $\partial_a$ above instead of $\nabla_a$; i.e., the gradient of a scalar never has any connection coefficient terms.]

The proof of the H-N theorem then amounts to saying that the three properties above, combined, guarantee that the motion is a Killing motion.

Looking at the congruence we've come up with, it seems evident that it satisfies properties #1 and #2, because everything depends only on $z$ and $r$, and every worldline in the congruence is a curve of constant $z$ and $r$. That leaves property #3, which leads us to look for some scalar function $f$ such that:

$$
a_z = \frac{\gamma^2}{z} = \partial_z f
$$

$$
a_r = - \gamma^2 \omega^2 r = - \frac{\gamma^2 \kappa^2 r}{z^2} = \partial_r f
$$

where we have substituted $\omega = \kappa / z$, and $\kappa$ is what Mentz114 was calling $\omega_0$ before.

Trying the ansatz $f = \ln ( z / \gamma )$, we find that it satisfies both of the above equations:

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

$$
\partial_r f = - \frac{\gamma}{z} \frac{z}{\gamma^2} \partial_r \gamma = - \frac{\gamma^2 \kappa^2 r}{z^2} = - \gamma^2 \omega^2 r
$$

So it looks like this congruence actually does satisfy the conditions of the H-N theorem, despite the various statements in the literature that appear to say it shouldn't.

 

[PeterDonis]

It is true that $\nabla_{(a} u_{b)} = 0$ with $\omega=1/z$

No, that's not true; we spent a bunch of posts showing that. What is true is that $\nabla_{(a} u_{b)} + u_{(a} a_{b)} = 0$.

However, given the results of my last post, we should be able to find a KVF that generates the worldlines of the congruence; I think it will look something like $\partial_t + \kappa \partial_{\phi}$ (where $\kappa$ is what you were calling $\omega_0$, per my last post).

 

[Mentz114]

No, that's not true; we spent a bunch of posts showing that. What is true is that $\nabla_{(a} u_{b)} + u_{(a} a_{b)} = 0$.

However, given the results of my last post, we should be able to find a KVF that generates the worldlines of the congruence; I think it will look something like $\partial_t + \kappa \partial_{\phi}$ (where $\kappa$ is what you were calling $\omega_0$, per my last post).

Yes, I forgot that only geodesics can be KMs. So $u^a$ is not a Killing motion and the choice offered by PAllen is a problem.

I checked and found that $K^a=\partial_t+k\ \partial_\phi$ is a KVF with k constant. But $-K^\mu u_\mu = z^2$.

I'll stop banging on about this, but I still don't see the motion in the z-direction.

 

[PeterDonis]

I forgot that only geodesics can be KMs.

That's not true either. The standard Rindler congruence (i.e., linear acceleration without rotation) is a Killing motion, but it's not geodesic.

So $u^a$ is not a Killing motion

I think it is. For a 4-velocity field to be a Killing motion, it's not required that $u^a$ itself satisfy Killing's equation; in fact in general one wouldn't expect it to, since $u^a$ is normalized so it's always a unit vector, whereas KVFs are not.

What is required is that there is *some* KVF $\xi^a$ whose integral curves are the same as the curves generated by $u^a$. For example, in Schwarzschild spacetime, the 4-velocity field $u^a = (1 / \sqrt{1 - 2M/ r}) \partial_t$ is a Killing motion, because, even though $u^a$ doesn't satisfy Killing's equation, $\xi^a = \partial_t$ does, and its integral curves are the same as the curves generated by $u^a$ (namely, curves of constant $r$, $\theta$, $\phi$ in the standard Schwarzschild chart).

In our case, we have a 4-velocity field $u^a = ( \gamma / z ) \partial_t + ( \gamma \kappa / z ) \partial_{\phi}$, which should generate the same family of curves as $\xi^a = \partial_t + \kappa \partial_{\phi}$, since the ratios of the two terms in both cases are the same. (This goes against the intuition I originally had about this, btw, since it clearly shows that the ratio between the $\partial_t$ and $\partial_{\phi}$ terms is *not* a function of $z$; $\gamma$ is, but $\gamma / z$ appears in both terms, so it cancels out.)

I checked and found that $K^a=\partial_t+k\ \partial_\phi$ is a KVF with k constant.

Ok, good, that's what I would have expected based on the results from my recent posts and the above.

But $-K^\mu u_\mu = z^2$.

Yes, that's saying that as you increase $z$, you are increasing your "potential energy" in the apparent "gravitational field" along the $z$ dimension.

I still don't see the motion in the z-direction.

There isn't any in the chart we're using; but there would be in an inertial chart. Put another way, there is no motion in the $z$ direction relative to observers who are comoving with the disks; but there would be motion in the $z$ direction relative to inertial observers. I don't understand why that is hard to grasp; surely the fact that "motion" is relative is one of the basic ideas of relativity.

 

[Mentz114]

Thanks for the explanation of Killing motion. I understand I have some mis-understandings.

There isn't any in the chart we're using; but there would be in an inertial chart. Put another way, there is no motion in the z direction relative to observers who are comoving with the disks; but there would be motion in the z direction relative to inertial observers.

We're working in the coordinate basis. I'm not insisting that I'm right, but I really don't understand.

I don't understand why that is hard to grasp; surely the fact that "motion" is relative is one of the basic ideas of relativity.

I find that patronising, but I forgive you.

I'll stop posting here now and think, rather than calculating anything that stands still long enough.

 

[PeterDonis]

We're working in the coordinate basis.

Which chart or basis we're using is immaterial, at least for the definition of "motion" that I've been using. That's why I made a point of expressing the same thing in a way that is obviously independent of basis or chart, in terms of motion relative to actual observers. If an observer is comoving with the disks, the fact that the disks are not moving in the $z$ direction relative to him is independent of basis or chart. Similarly, if an observer is inertial, the fact that the disks *are* moving in the $z$ direction relative to him is independent of basis or chart.

I find that patronising

I'm sorry, I didn't mean to be, but I honestly don't understand where the miscommunication is; I suspect we're talking past each other. Again, that's why I'm trying to express things in a way that is clearly independent of which chart or basis we are using.

Put another way: by "there is no motion in the $z$ direction" did you mean "the disks are not moving in the $z$ direction, relative to observers who are comoving with the centers of the disks"? If so, of course I agree; but I think the qualification "relative to observers who are comoving with the centers of the disks" is necessary, and you didn't put it in, so I'm not sure if that was what you actually meant.

 

[PeterDonis]

I checked and found that $K^a=\partial_t+k\ \partial_\phi$ is a KVF with k constant.

One other thing that's important to note here: the reason this is a KVF is that $\partial_t$ and $\partial_{\phi}$ are both KVFs, and they are being combined with constant coefficients. But that means that, in order for the congruence we've discovered to be a Killing motion, and hence to be rigid, the centers of the disks must be following curves in the Rindler congruence; i.e., their proper acceleration has to be constant. Put another way, what allows this motion to be rigid is the fact that the Rindler congruence in Minkowski spacetime is a Killing congruence; if the centers of the disks follow worldlines that are not Rindler worldlines (i.e., if they move in the $z$ direction in the Rindler chart I have been using), the motion is no longer a Killing motion.

(This is evident from the properties cited in the Giulini paper, that I noted earlier, one of which is that the proper acceleration must be constant along each worldline in the congruence.)

Similarly, if we transfer all this to Schwarzschild spacetime (which was, after all, the original point of the analysis ), a stack of disks all "hovering" at constant $r$ can be rotating rigidly, as long as the angular velocity of the disks, as seen by local observers at their centers, varies with $r$ in the right way. (Note that this way will be *different* than the simple $1 / z$ dependence we found in flat spacetime, since the time dilation factor varies differently as a function of $r$. If I have time I'll post the actual calculation.) However, this conclusion depends on the fact that the family of worldlines described by the centers of the disks, i.e., the family of "hovering" worldlines of constant $r$, $\theta$, $\phi$, is a Killing congruence, because $\partial_t$ in Schwarzschild coordinates is a KVF; then the Killing congruence that describes the disks' full motion, including rotation, will be something like $\partial_t + \kappa \partial_{\phi}$, with $\kappa$ constant, just as for flat spacetime. If the disks are moving radially, that is no longer true.

 

[Mentz114]

Peter, thanks for the explanations. I've been very short of time for this but I'll be able to read back and do some thinking over the weekend.

[later]
I'm back and I've had a big 'aha !' moment. Again, I think we are both right because there are two interpretations of $u^\mu$. If the time dilation factor 1/z is attributed to a gravitational field ( however unlikely) then $u^\mu$ is the worldline of a hovering disc. However if the time dilation factor is caused by rocket motors (say) then $u^\mu$ is the worldline of a disc comoving with a rocket in a Rindler rocket-chain with proper distance between rockets preserved. This is a notable instance of the equivalence of the gravitational field and proper acceleration.

If someone has already pointed this out, I apologise. I'm very relieved to have resolved my problem with this.

I'll read what you've written about KMs when I've recovered from this ...