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

[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 ...