I What is the Strange Solution to the EFE with a Born Frame and Rotating System?

[Mentz114]

TL;DR Summary

Starting with a rotating frame field (spherical Born coordinates) and setting $\omega\equiv \omega(r)$ then solving the differential equation $\vec{a}=0$ , $\vec{a}$ being the proper acceleration gives the frame field of a circular geodesic.

The Born frame field (see ref below) describes a rotating system and the proper acceleration $\vec{a}=\nabla _{{{\vec {p}}_{0}}}\,{\vec {p}}_{0}={\frac {-\omega ^{2}\,r}{1-\omega ^{2}\,r^{2}}}\,{\vec {p}}_{2}$. If $\omega$ depends on coordinate $r$ then $\vec{a}=\frac{{r}^{2}\,w\,\left( \frac{d}{d\,r}\,w\right) +r\,{w}^{2}}{{r}^{2}\,{w}^{2}-1}$ and solving the ODE $\vec{a}=0$ gives $\omega(r)=M/r$ where $M>0$ is a constant.

Obviously there must be a source now and sure enough the Ricci and Einstein tensors are not zero. The metric is transformed to
<br /> g_{\mu\nu}=\begin{pmatrix}<br /> -1 &amp; 0 &amp; 0 &amp; -\frac{M}{r}\\<br /> 0 &amp; 1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; {r}^{2} &amp; 0\\<br /> -\frac{M}{r} &amp; 0 &amp; 0 &amp; -\frac{{M}^{2}-1}{{r}^{2}}<br /> \end{pmatrix}<br />
and clearly $M<1$ is a constraint.

The Einstein tensor in the local frame is
<br /> E_{mn}=\begin{pmatrix}<br /> -\frac{r\,{M}^{2}-4\,{M}^{2}+4\,r}{4\,{r}^{3}} &amp; 0 &amp; 0 &amp; \frac{M\,\left( 3\,{M}^{2}-4\,r\right) }{4\,{r}^{3}}\\<br /> 0 &amp; \frac{\left( M-2\right) \,\left( M+2\right) }{4\,{r}^{2}} &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -\frac{{M}^{2}-8}{4} &amp; 0\\<br /> \frac{M\,\left( 3\,{M}^{2}-4\,r\right) }{4\,{r}^{3}} &amp; 0 &amp; 0 &amp; \frac{{M}^{2}\,\left( 3\,{M}^{2}-4\,r-3\right) }{4\,{r}^{4}}<br /> \end{pmatrix}<br />

I don't know what to make of this so any comments welcomed.https://en.wikipedia.org/wiki/Born_coordinates

 

[PeterDonis]

If $\omega$ depends on coordinate $r$

In the Born chart, it doesn't; $\omega$ is a constant, the "angular velocity" of the rotating frame relative to an inertial observer at rest relative to the center of rotation.

I don't know what to make of this

I'm not sure what you are trying to do. Are you trying to derive an analogue of the Born chart for Schwarzschild spacetime?

 

[Mentz114]

It looks as if the result of $\omega\rightarrow M/r$ ( a potential) has resulted in an EMT that looks like a thin rotating disc with all the mattter in geodesic motion.

 

[PeterDonis]

the result of $\omega\rightarrow M/r$

First you have to explain why you are letting $\omega$ be a function of $r$ at all. The frame field that is used as the basis for Born coordinates does not have this property; for that frame field, $\omega$ is a constant. So if you want $\omega$ to be a function of $r$, you are talking about a different frame field, and I'm not sure what the motivation for considering that frame field is.

 

[Mentz114]

First you have to explain why you are letting $\omega$ be a function of $r$ at all. The frame field that is used as the basis for Born coordinates does not have this property; for that frame field, $\omega$ is a constant. So if you want $\omega$ to be a function of $r$, you are talking about a different frame field, and I'm not sure what the motivation for considering that frame field is.

Why do I have to justify it ? The motivation is curiosity.
Naturally it is a different space-time - but what is it ?
I'm having a problem converting the cylindrical chart to spherical polar so the Einstein tensor could be wrong - lots of fun trying to fix it.

Thanks for your responses.

 

[PeterDonis]

Why do I have to justify it ?

Perhaps "justify" was the wrong word. My point is, what physical congruence of worldlines (or family of observers) are you trying to describe? It can't be the same congruence as the one that's used to derive the Born chart, since that congruence has a constant $\omega$: it describes a family of observers that are in circular orbits around some common center in flat spacetime, with different orbital radius but the same angular velocity $\omega$.

Mathematically, what you've apparently done is take the description of the frame field of the family of observers at rest in the Born chart, and say "let's let $\omega$ be a function of $r$". But doing that invalidates the whole construction of the Born chart, so it's not clear what you're trying to describe or even if what you're trying to describe is consistent.

If the basic idea you're trying to pursue is "what happens if we let $\omega$ be a function of $r$", I think a better approach would be to start with a chart in which you can unambiguously define a frame field with that property without invalidating the chart itself. So I would take the cylindrical chart on Minkowski spacetime (the one in which the Minkowski metric is expressed at the very start of the "Langevin observers in the cylindrical chart" section) and the Langevin frame field as expressed in that chart, and then let $\omega$ be a function of $R$ (note the capital $R$ since this is the Minkowski cylindrical chart, not the Born chart) and compute the proper acceleration. The formal expression of the frame field should remain the same as long as $\omega$ is only a function of $R$; but the formal expression for the proper acceleration will change.

Naturally it is a different space-time - but what is it ?

I don't know, because I don't even know what family of observers you're trying to describe. See above.

I'm having a problem converting the cylindrical chart to spherical polar

Why would you need to? You can compute tensors just fine in the cylindrical chart. Furthermore, since your family of observers is not spherically symmetric, but only axially symmetric, a cylindrical chart seems like a better choice for describing it.

 

[PeterDonis]

The metric is transformed to

How are you obtaining this? You're supposed to already know the metric in order to compute the proper acceleration.

 

[Mentz114]

Perhaps "justify" was the wrong word. My point is, what physical congruence of worldlines (or family of observers) are you trying to describe? It can't be the same congruence as the one that's used to derive the Born chart, since that congruence has a constant $\omega$: it describes a family of observers that are in circular orbits around some common center in flat spacetime, with different orbital radius but the same angular velocity $\omega$.

I know this !

Mathematically, what you've apparently done is take the description of the frame field of the family of observers at rest in the Born chart, and say "let's let $\omega$ be a function of $r$". But doing that invalidates the whole construction of the Born chart, so it's not clear what you're trying to describe or even if what you're trying to describe is consistent.

Yes , agreed.

But, I have a new metric. It has a circular geodesic and matter so it no longer matters how it was Born (pun intended). I'll take it from there. I will explore cylindrical coords as well.

 

[PeterDonis]

I have a metric. It has a circular geodesic and matter so it no longer matters how it was Born (pun intended).

In the sense that you can write down whatever symmetric 4 x 4 matrix you like and call it a metric, yes, that's true. But I don't see any way to give that 4 x 4 matrix a physical interpretation without being able to link it to something physically understandable. In that sense, how it was "Born" does matter.

 

[Mentz114]

How are you obtaining this? You're supposed to already know the metric in order to compute the proper acceleration.

Ricci rotation coefficients are defined by the frame field. The frame covariant derivative can then be used to get the proper acceleration.

 

[PeterDonis]

Ricci rotation coefficients are defined by the frame field

Yes, but that means you need a consistent definition of a frame field in the chart you are using. I'm not convinced you have that for the Born chart. But I can see an obvious consistent way to define one using the Minkowski cylindrical chart, which I described previously; and I think that will end up giving you the same sort of metric you have now.

 

[PeterDonis]

Ricci rotation coefficients are defined by the frame field

And the covariant derivative, which requires knowledge of the metric.

 

[Mentz114]

And the covariant derivative, which requires knowledge of the metric.

If the local metric $\eta_{ab}$ is known and the frame field is known, then the metric is just a transformation of $\eta$.

The result for the Born chart in cylindrical coordinates gives a metric which does not cover the whole space $0<r<\infty$ because $r<1/M$ and $0<M<1$ is required. So it could be a very thin disc of something. I do not think the parameter $M$ can represent matter or energy. A small puzzle but not worth any more time spent.

<br /> g_{\mu\nu}=\begin{pmatrix} <br /> \frac{1-{r}^{2}\,{M}^{2}}{{M}^{2}-1} &amp; 0 &amp; 0 &amp; r\,M\\<br /> 0 &amp; 1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> r\,M &amp; 0 &amp; 0 &amp; 1-{M}^{2}<br /> \end{pmatrix}<br />

 

[PeterDonis]

If the local metric $\eta_{ab}$ is known and the frame field is known, then the metric is just a transformation of $\eta$.

You also need to know the connection, because you need to know the covariant derivatives at your chosen event. Just knowing that the metric at that event is $\eta_{ab}$ is not enough, because you can always find a chart in which the metric is $\eta_{ab}$ at that event. But without knowing the connection, you don't know how to parallel transport vectors from the tangent space at one event to the tangent space at another event, so you have no way of comparing frame field vectors at different events.

 

[PeterDonis]

The result for the Born chart in cylindrical coordinates

Is based on inconsistent assumptions, as far as I can see; you started with a chart that is defined in a way that requires $\omega$ to be constant, but you're treating the frame field as though $\omega$ can vary with $r$. So even if we just say you wrote down the metric you give arbitrarily, without deriving it from anything, I still don't think you have a consistent model, so there's no point in asking what it might represent physically.

 

[Mentz114]

You also need to know the connection, because you need to know the covariant derivatives at your chosen event. Just knowing that the metric at that event is $\eta_{ab}$ is not enough, because you can always find a chart in which the metric is $\eta_{ab}$ at that event. But without knowing the connection, you don't know how to parallel transport vectors from the tangent space at one event to the tangent space at another event, so you have no way of comparing frame field vectors at different events.

The frame covariant derivative allows comparison between points on the curve. I don't understand your objection. The curvature tensors as experienced by the transported frame can be calculated from the first derivatives of the tetrad. No need for Christoffel connections.

Is based on inconsistent assumptions, as far as I can see; you started with a chart that is defined in a way that requires ω to be constant, but you're treating the frame field as though ω can vary with r. So even if we just say you wrote down the metric you give arbitrarily, without deriving it from anything, I still don't think you have a consistent model, so there's no point in asking what it might represent physically.

I think you've made this point already. Again, feel free not to consider what it may mean.

 

[pervect]

Why is having circular geodesics surprising? I don't understand the physical significance of what you're doing at all, unfortunately.

But it's easy to construct a very simple example with circular geodesics. Consider a flat plane, 2-space + 1 time. A point remaining at rest in said flat plane follows a geodesics. Now, adopt a rotating frame. The geodesics in the rotating frame , of the points that used to be stationary, are now circular.

If we assume your calculations are correct, can you explain why having circular geodesics is necessarily surprising, in light of this example?

 

[PeterDonis]

The frame covariant derivative allows comparison between points on the curve.

What do you mean by "the frame covariant derivative"?

The curvature tensors as experienced by the transported frame can be calculated from the first derivatives of the tetrad.

First derivatives of the tetrad with respect to what?

 

[Mentz114]

What do you mean by "the frame covariant derivative"?

See https://jila.colorado.edu/~ajsh/astr3740_17/grbook.pdf

First derivatives of the tetrad with respect to what?

coordinates.

 

[Mentz114]

Why is having circular geodesics surprising? I don't understand the physical significance of what you're doing at all, unfortunately.

But it's easy to construct a very simple example with circular geodesics. Consider a flat plane, 2-space + 1 time. A point remaining at rest in said flat plane follows a geodesics. Now, adopt a rotating frame. The geodesics in the rotating frame , of the points that used to be stationary, are now circular.

If we assume your calculations are correct, can you explain why having circular geodesics is necessarily surprising, in light of this example?

The Born observer experiences proper acceleration. This is removed by a space-time in which the same frame field is a geodesic.

However, my latest calculation hints that $G_{00} < 0$ so it is unrealistic.

 

[PeterDonis]

See https://jila.colorado.edu/~ajsh/astr3740_17/grbook.pdf

I see the term "frame covariant derivative" in section 11.16.1. I don't see how it allows you to take derivatives without knowing the metric. Using tetrads doesn't eliminate the metric, it just re-expresses it in a different form that's often more useful for calculations.

coordinates

Which, as I said, I suspect your modified Born chart does not form a consistent set of. But re-doing your computations in the standard cylindrical chart would keep much of what you've done formally the same; what I don't know is whether it would lead to an Einstein tensor that is formally the same. At some point if I have time I'll fire up Maxima and see what it outputs for that formulation.

 

[Mentz114]

I see the term "frame covariant derivative" in section 11.16.1. I don't see how it allows you to take derivatives without knowing the metric. Using tetrads doesn't eliminate the metric, it just re-expresses it in a different form that's often more useful for calculations.

This is what I've been saying all along ! Why are you so argumentative about this .
You keep saying I've 'pulled a metric out of a hat' - but it comes from the tetrad and the local Minkwoski metric - uniquely defined (I think).

Which, as I said, I suspect your modified Born chart does not form a consistent set of. But re-doing your computations in the standard cylindrical chart would keep much of what you've done formally the same; what I don't know is whether it would lead to an Einstein tensor that is formally the same. At some point if I have time I'll fire up Maxima and see what it outputs for that formulation.

Doing the calculation in the cylindrical chart simplifies the expressions and shows clearly the '00' component of the Einstein tensor is negative.

That explains why it so weird ! Whatever produces the curvature is certainly not matter (' as we know it, Jim').

(I have Maxima batch scripts to do the calculations and I have high confidence in them but it could all be a down to a miscalculation).

 

[PeterDonis]

This is what I've been saying all along !

I'm not sure it is. You're saying that the metric...

...comes from the tetrad and the local Minkwoski metric

...which I don't agree with; if you don't already know the metric, you can't figure it out just from knowing some frame field and that the metric is locally Minkowski (the latter is true of any spacetime so I don't see how it helps any).

For example: suppose all you know is the tetrad (frame field) of the Langevin observers, expressed in the Minkowski cylindrical chart (the one with the capital $R$), as shown in the Wikipedia article. And you know that at any event, the metric is locally Minkowski. But you don't know the global metric, and you don't know anything else about the frame field. How do you get from that to the line element shown for the Minkowski cylindrical chart?

 

[Mentz114]

I'm not sure it is. You're saying that the metric...
...which I don't agree with; if you don't already know the metric, you can't figure it out just from knowing some frame field and that the metric is locally Minkowski (the latter is true of any spacetime so I don't see how it helps any).

For example: suppose all you know is the tetrad (frame field) of the Langevin observers, expressed in the Minkowski cylindrical chart (the one with the capital $R$), as shown in the Wikipedia article. And you know that at any event, the metric is locally Minkowski. But you don't know the global metric, and you don't know anything else about the frame field. How do you get from that to the line element shown for the Minkowski cylindrical chart?

OK.

Is the Ricci tensor below showing negative curvature ? $0<L<1$ is a constant.

<br /> \pmatrix{-\frac{{L}^{2}}{2\,{r}^{4}} &amp; 0 &amp; 0 &amp; \frac{3\,L}{2\,{r}^{2}}\cr 0 &amp; 0 &amp; 0 &amp; 0\cr 0 &amp; 0 &amp; \frac{{L}^{2}}{2\,{r}^{4}} &amp; 0\cr \frac{3\,L}{2\,{r}^{2}} &amp; 0 &amp; 0 &amp; -\frac{{L}^{2}}{2\,{r}^{2}}}<br />

 

[PeterDonis]

Is the Ricci tensor below showing negative curvature ?

I can't tell just from the components. To compute the Ricci scalar, which is the simplest invariant, and see if it's negative I would need to know the metric.

 

[Mentz114]

I can't tell just from the components. To compute the Ricci scalar, which is the simplest invariant, and see if it's negative I would need to know the metric.

The curvature scalar is $\frac{{L}^{2}}{2\,{r}^{4}}$. But $G_{00}$ is definitely negative.
This is all calculated in the Born frame with $\omega$ depending on $r$.

From this angle it is not physical.

If we start with the metric however, then everything is OK. The Einstein tensor I think is OK and I'm going on to do a full analysis.

 

[PeterDonis]

The curvature scalar is $\frac{{L}^{2}}{2\,{r}^{4}}$.

Ok, that would normally be termed positive curvature.

But $G_{00}$ is definitely negative.

That's not necessarily an issue. The energy density measured by a timelike observer is $T_{ab} u^a u^b = \frac{1}{8 \pi} G_{ab} u^a u^b$, where $u$ is the observer's 4-velocity. If that is negative, that would indeed be considered unphysical (at least in the sense that it violates energy conditions).

If we start with the metric however, then everything is OK.

Start with what metric?

 

[Mentz114]

Ok, that would normally be termed positive curvature.

That's not necessarily an issue. The energy density measured by a timelike observer is $T_{ab} u^a u^b = \frac{1}{8 \pi} G_{ab} u^a u^b$, where $u$ is the observer's 4-velocity. If that is negative, that would indeed be considered unphysical (at least in the sense that it violates energy conditions).

In the frame calculation $\frac{1}{8 \pi} \hat{G}_{ab} u^a u^b=\frac{1}{8 \pi} \hat{G}_{00} <0$ so it is a problem for sure.

Start with what metric?

I attach a Maxima batch file that does the calculation for 'the metric' in the holonomic basis.
In this case $\frac{1}{8 \pi} {G}_{ab} u^a u^b$ is less than zero. If you run it rename to .mac, if necessary.

 

[PeterDonis]

I attach a Maxima batch file

If I'm reading this correctly, the metric you are using is

$$
ds^2 = - \frac{r^2 L^2 - 1}{L^2 - 1} dt^2 + dr^2 + dz^2 + 2 r L dt d\varphi + \left( 1 - L^2 \right) d \varphi^2
$$

What are the units of $L$ supposed to be? If $\varphi$ is an angular coordinate, the $dt d \varphi$ term indicates that $L$ should be dimensionless (since the term as a whole should have dimensions of length squared, and $r$ and $dt$ both have dimensions of length); but the $r^2 L^2$ in the $dt^2$ term indicates that $L$ should have dimensions of inverse length.

 

[Mentz114]

If I'm reading this correctly, the metric you are using is

$$
ds^2 = - \frac{r^2 L^2 - 1}{L^2 - 1} dt^2 + dr^2 + dz^2 + 2 r L dt d\varphi + \left( 1 - L^2 \right) d \varphi^2
$$

What are the units of $L$ supposed to be? If $\varphi$ is an angular coordinate, the $dt d \varphi$ term indicates that $L$ should be dimensionless (since the term as a whole should have dimensions of length squared, and $r$ and $dt$ both have dimensions of length); but the $r^2 L^2$ in the $dt^2$ term indicates that $L$ should have dimensions of inverse length.

I think that finishes the 'strange geodesic' - whose existence troubled me from its (accidental) birth.
It was digression in my rotation agenda which I return to gratefully with order restored.

Thanks for all the help.

 

[PeterDonis]

I think that finishes the 'strange geodesic'

Are you saying that I got the metric wrong? Or that I got the metric right, and it is inconsistent (because the units of $L$ aren't consistent)?

 

[Mentz114]

Are you saying that I got the metric wrong? Or that I got the metric right, and it is inconsistent (because the units of $L$ aren't consistent)?

You got the metric right and it is rubbish because of the units .
I should have said 'finished off' as in 'coup de grace'.

 

[Mentz114]

I spotted a mistake in my Born cylindrical chart script and as a result the strange circle is back. The line element is now
$${dt}^{2}\,\left( {L}^{2}-1\right) +2\ r\,L\,d\phi\,dt\,+\,{r}^{2}\ {d\phi}^{2}+{dz}^{2}+{dr}^{2}$$
and the earlier analyses still apply. Viz., the scalar curvature is positive $\frac{{L}^{2}}{2\,{r}^{2}}$ but $\frac{1}{8 \pi} \hat{G}_{ab} u^a u^b=\frac{1}{8 \pi} \hat{G}_{00} <0$ so the stationary observer in this frame 'sees' negative energy density.

Cool or what ?

 

[PeterDonis]

The line element is now

Is there a sign error in the $dt^2$ term? Or did you just prefer to write $(L^2 - 1) dt^2$ instead of $- (1 - L^2) dt^2$?

 

[PeterDonis]

Also, is $L$ a constant? Or a function of $r$?

 

[PeterDonis]

the scalar curvature is positive $\frac{{L}^{2}}{2\,{r}^{2}}$ but $\frac{1}{8 \pi} \hat{G}_{ab} u^a u^b=\frac{1}{8 \pi} \hat{G}_{00} < 0$

I'm not sure that's true. The result I get from Maxima (assuming that I posted the metric correctly before and that $L$ is a constant) for $G_{00}$ is (giving it just as Maxima gives it)

$$
- \frac{L^2 \left( L^2 - 1 \right)}{4r^2} - \frac{\left( L - 1 \right) L \left( 1 + L \right)}{2r^2}
$$

On its face this looks negative, but it's not. Consider: for a worldline stationary in this metric (i.e., all coordinates except $t$ constant) to be timelike (so it can be the worldline of an observer), we must have $g_{00} < 0$ (since the way the metric is written makes it evident that the $-+++$ signature convention is being used). That means (assuming there is no sign error in the $dt^2$ term as I posted it) $L^2 - 1 < 0$, or $1 - L^2 > 0$. So we can rewrite the above in a form that makes it manifestly positive:

$$
\frac{L^2 \left( 1 - L^2 \right)}{4r^2} + \frac{\left( 1 - L \right) L \left( 1 + L \right)}{2r^2} = \left( 1 - L^2 \right) \left( \frac{L^2}{4r^2} + \frac{L}{2r^2} \right)
$$

The actually measured energy density is the same as this except that the $1 - L^2$ factor is gone (it's canceled by the two factors of $u^0$).

 

[Mentz114]

I'm not sure that's true. The result I get from Maxima (assuming that I posted the metric ...
[]
The actually measured energy density is the same as this except that the $1 - L^2$ factor is gone (it's canceled by the two factors of $u^0$).

I've attached a short script that does the calculation and it looks correct. Saves a lot of explaining. Note that in this $U^\mu$ has only a $\vec{\partial_t}$ component because it is at rest.

I must apologise for using the symbols $U$ and $V$ for the same vector. An old bad habit to remind me if something is covariant or contavariant.

 

[Mentz114]

I'm not sure that's true. The result I get from Maxima (assuming that I posted the metric correctly before and that $L$ is a constant) for $G_{00}$ is (giving it just as Maxima gives it)

$$
- \frac{L^2 \left( L^2 - 1 \right)}{4r^2} - \frac{\left( L - 1 \right) L \left( 1 + L \right)}{2r^2}
$$

On its face this looks negative, but it's not.

It is positive because $0<L<1$.

Working in the holonomic coordinates I get $G_{00}=\frac{3\,\left( 1-L\right) \,{L}^{2}\,\left( L+1\right) }{4\,{r}^{2}}$, $G_{14}=G_{41}=-\frac{L\,\left( 3\,{L}^{2}-2\right) }{4\,r}$, $G_{22}=-G_{33}=\frac{{L}^{2}}{4\,{r}^{2}}$ and $G_{44}=-\frac{3\,{L}^{2}}{4}$. I think the two negative terms ('pressures') are unphysical.

However $G_{\mu\nu}U^\mu U^\nu = \frac{3\,{L}^{2}}{4\,{r}^{2}}$ which is regular.

It seems to be the description of a non-rigidly rotating disc, which is unusual. The great puzzle is still - where does the energy/curvature come from ? There is a spin-singularity at $r=0$ because the vorticity is $L/2r$ around the z-axis. My pet (naive) theory is that this is dragging the space-time so that anything at rest will rotate and feel no force. The negative pressure in the Φ-direction and r-direction could account for the geodesic motion. The observer with $U_\mu=\sqrt{1-{L}^{2}}\ \vec{\partial_t}$ feels an acceleration in the r-direction but the worldline $U_\mu=\sqrt{1-{L}^{2}}\vec{\partial_t}-\vec{\partial_\phi}\ \frac{r\,L}{\sqrt{1-{L}^{2}}}$ does not. Whatever is there has reversed the usual situation !

The Petrov classification is I i.e. four principal null directions.

I don't think there is much more to say about this.

Thanks again for your feedback.

 

[PeterDonis]

It is positive because 0<L<10

Yes, that's what I pointed out in post #36 after the part you quoted.

It seems to be the description of a non-rigidly rotating disc

It can't be a description of anything rotating that is made of ordinary matter, since no Einstein tensor components would be negative for that case.

There is a spin-singularity at $r=0$

Not just a spin singularity, a curvature singularity; the Ricci scalar is singular at $r = 0$.

The observer with $U_\mu=\sqrt{1-{L}^{2}}\ \vec{\partial_t}$ feels an acceleration in the r-direction

Are you sure? The Christoffel symbol $\Gamma^r{}_{tt}$ vanishes, and that's the only one that would produce a proper acceleration in the $r$ direction for this observer. In fact, there are no non-vanishing Christoffel symbols that would give this observer a proper acceleration in any direction.

the worldline $U_\mu=\sqrt{1-{L}^{2}}\vec{\partial_t}-\vec{\partial_\phi}\ \frac{r\,L}{\sqrt{1-{L}^{2}}}$ does not

Again, are you sure? The only relevant non-vanishing Christoffel symbol is $\Gamma^r{}_{t \phi}$, which is $L / 2$. This gives a nonzero proper acceleration for any 4-velocity with non-vanishing $t$ and $\phi$ components. For your particular case, I get a radial proper acceleration of $- r L^2$, i.e., inward, as you would expect for an observer riding along with some kind of rotating object.

 

[Mentz114]

Yes, that's what I pointed out in post #36 after the part you quoted.
It can't be a description of anything rotating that is made of ordinary matter, since no Einstein tensor components would be negative for that case.

Not just a spin singularity, a curvature singularity; the Ricci scalar is singular at $r = 0$.

Agreed. It certainly can't be anything recognisable.

...
Are you sure?
...
Again, are you sure?

I will check the accelerations again as soon as I get time. The script has never got one wrong !

 

[PeterDonis]

I will check the accelerations again as soon as I get time.

You should also check your 4-velocity vectors; they don't look like they're normalized correctly. For a stationary observer, for example, I get

$$
U = \frac{1}{\sqrt{1 - L^2}} \partial_t
$$

in order to have $g_{ab} U^a U^b = -1$.

 

[Mentz114]

You should also check your 4-velocity vectors; they don't look like they're normalized correctly. For a stationary observer, for example, I get

$$
U = \frac{1}{\sqrt{1 - L^2}} \partial_t
$$

in order to have $g_{ab} U^a U^b = -1$.

Do you mean $U_\mu = \frac{1}{\sqrt{1 - L^2}} \partial_t$ ?
That is what I use for the stationary (covariant) vector and I get the a different acceleration than you do, viz $\frac{{L}^{2}}{2\,r}$. So we are not doing the same calculation it appears.
But for this one $V_\mu = \frac{1}{\sqrt{1 - L^2}} \partial_t -\frac{r\,L}{\sqrt{1-{L}^{2}}}\partial_\phi$ I get no acceleration (yet ?).
Still checking.

 

[PeterDonis]

Do you mean $U_\mu = \frac{1}{\sqrt{1 - L^2}} \partial_t$ ?

No, the index should be upper, not lower. Again, we should have $g_{ab} U^a U^b = -1$ for correct normalization. If the only nonzero component of $U^a$ is the $t$ or $0$ component, then we must have $U^0 = 1 / \sqrt{g_{00}}$ for normalization to be satisfied.

That is what I use for the stationary (covariant) vector

You don't use covectors to assess 4-acceleration. The tangent vector to an observer's worldline is a vector (upper index), not a covector (lower index). So, therefore, is the derivative of that tangent vector with respect to proper time.

 

[PeterDonis]

The tangent vector to an observer's worldline is a vector (upper index), not a covector (lower index). So, therefore, is the derivative of that tangent vector with respect to proper time.

To write this out explicitly, we have

$$
A^a = \frac{d}{d\tau} U^a = U^b \nabla_b U^a = U^b \left( \partial_b + \Gamma^a{}_{bc} U^c \right) U^a
$$

 

[PeterDonis]

Btw, if you are treating partial derivatives with respect to the coordinates as vectors (more precisely, coordinate basis vectors in the tangent space), then they are upper index vectors. The notation $\partial_t$ for partial derivatives is very unfortunate in this respect, as it invites the incorrect inference that these are covectors, not vectors. The correct way to interpret what is going on is that there is a 1-to-1 correspondence between tangent vectors, i.e., upper index vectors in the tangent space, and directional derivatives, and $\partial_t$, for example, is just the directional derivative along a curve where $t$ is the only coordinate that is changing. I don't know if all GR textbooks go into this in detail, but MTW certainly does.

 

[Mentz114]

Btw, if you are treating partial derivatives with respect to the coordinates as vectors (more precisely, coordinate basis vectors in the tangent space), then they are upper index vectors. The notation $\partial_t$ for partial derivatives is very unfortunate in this respect, as it invites the incorrect inference that these are covectors, not vectors. The correct way to interpret what is going on is that there is a 1-to-1 correspondence between tangent vectors, i.e., upper index vectors in the tangent space, and directional derivatives, and $\partial_t$, for example, is just the directional derivative along a curve where $t$ is the only coordinate that is changing. I don't know if all GR textbooks go into this in detail, but MTW certainly does.

Yes, I'm sorry about the abuse of notation.

You are calculating $A^a = \frac{d}{d\tau} U^a = U^b \nabla_b U^a = U^b \left( \partial_b + \Gamma^a{}_{bc} U^c \right) U^a$ but I'm using this $A_i=\nabla_n U_i U^n$ which is the "covariant derivative of $U_\mu$ projected in the $U^\mu$ direction". You can see why I need the vector and covector explicitly. The quote and the expression are from Stephani's book 'General Relativity' (about page 175).

So we are not calculating the same thing and I may have spread confusion in other ways. I need time to sort this out but for now I am not completely confident about these acceleration results.

 

[PeterDonis]

I'm using this $A_i=\nabla_n U_i U^n$ which is the "covariant derivative of $U_\mu$ projected in the $U^\mu$ direction".

I would write this as $A_i = U^n \nabla_n U_i$ to avoid ambiguity in what the covariant derivative is operating on. This is just lowering an index on what I wrote:

$$
A_i = g_{ai} A^a = g_{ai} U^n \nabla_n U^a = U^n \nabla_n g_{ai} U^a = U^n \nabla_n U_i
$$

So if $A^a$ vanishes, $A_i$ must vanish as well.

 

[Mentz114]

I would write this as $A_i = U^n \nabla_n U_i$ to avoid ambiguity in what the covariant derivative is operating on. This is just lowering an index on what I wrote:

$$
A_i = g_{ai} A^a = g_{ai} U^n \nabla_n U^a = U^n \nabla_n g_{ai} U^a = U^n \nabla_n U_i
$$

So if $A^a$ vanishes, $A_i$ must vanish as well.

I have worked it by hand and my calculation is correct. I will post the details later.

 

[Mentz114]

I would write this as $A_i = U^n \nabla_n U_i$ to avoid ambiguity in what the covariant derivative is operating on. This is just lowering an index on what I wrote:

$$
A_i = g_{ai} A^a = g_{ai} U^n \nabla_n U^a = U^n \nabla_n g_{ai} U^a = U^n \nabla_n U_i
$$

So if $A^a$ vanishes, $A_i$ must vanish as well.

It does by my calculation.

The proper acceleration of the vector
$$U_\mu=\sqrt{1-{L}^{2}}\ \vec{\partial_t}-rL\vec{\partial_\phi}/\sqrt{1-{L}^{2}}$$
is $\dot{U}_\nu=(\nabla_\mu U_\nu) U^\mu$, the covariant derivative of $U_\nu$ projected in the direction $U^\mu$.
The covector $g^{\mu\nu}U_\nu$ has only a time (first) component so the only terms of $D_{\mu\nu}=\nabla_\mu U_\nu$ which can be selected by the contraction of $D_{\mu\nu}$ with $U^\nu$ are in the first column, $D_{1,k}$.

Using $\nabla_\mu U_\nu = \partial_\mu U_\nu - {\Gamma_{\mu\nu}}^\alpha U_\alpha$ there are eight instances where ${\Gamma_{\mu\nu}}^\alpha U_\alpha$ is non-zero listed below in the 'Trace output section' in the pdf.

The first two lines are equivalent to
<br /> \begin{align*}<br /> D_{12} &amp;= -{ \Gamma_{12}}^1 U_1 - { \Gamma_{12}}^4 U_4\\<br /> &amp;=\frac{L\,\left( {L}^{3}-L\right) }{2\,r\,\sqrt{1-{L}^{2}}}-\frac{\left( L-1\right) \,{L}^{2}\,\left( L+1\right) }{2\,r\,\sqrt{1-{L}^{2}}}\\<br /> &amp;= 0<br /> \end{align*}<br />
and since there are no other candidates in the first column of $D_{\mu\nu}$ this is sufficient to eliminate any proper acceleration.

Calculating the mixed covariant derivative $D^\mu_\nu$ and contracting with $U^\nu$ also gives a zero proper acceleration ( naturaly this uses a different form of the definition of $D_{\mu\nu}$ above).

 

[Mentz114]

Solution ?

(The details of the frame and metric are given in the posts above)

In the static frame $T_{ab} = R_{ab}-R\eta_{ab}/2$ has components
$$T_{tt}=\frac{3\,{L}^{2}}{4\,{r}^{2}},\quad T_{zz}=T_{\phi\phi}=-T_{rr}=\frac{{L}^{2}}{4\,{r}^{2}},\quad T_{t\phi}=T_{\phi t}=\frac{L}{2\,{r}^{2}}$$

An electromagnetic field with Faraday tensor $F^{\mu \nu}$ has stress-energy tensor

$E^{\mu\nu} = \frac{1}{4\pi} \left[ F^{\mu \alpha}F^\nu{}_{\alpha} - \frac{1}{4} \eta^{\mu\nu}F_{\alpha\beta} F^{\alpha\beta}\right]$

and with electric field $F^{0r}=P/r,\ F^{r0}=-P/r$ in the r-direction this evaluates to a diagonal tensor with components $\frac{1}{4\pi}P^2/2r^2$ except for $E_{rr}$ which is the negative of the other components.

Calculating $M_{\mu\nu}=8\pi T_{\mu\nu}-4\pi E_{\mu\nu}$ and setting $P=L$ gets components $M_{00}={4\,\pi\,{L}^{2}}/{{r}^{2}}$ and $M_{0\phi}=M_{\phi 0}={4\,\pi\,L}/{{r}^{2}}$.

So $T_{\mu\nu}$ is the sum of the emt for an electric field and what looks like rotating matter or energy.
The $8\pi$ is necessary to bring the components to the same units. However there is latitude in setting the value of $P(L)$ so those constants can be adjusted.

This looks like a physically plausible solution of the EFE - except for point source which always causes problems.

It is not so strange after all !

I will write this up sometime.