# I Strange circular geodesic

#### Mentz114

Gold Member
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
$$g_{\mu\nu}=\begin{pmatrix} -1 & 0 & 0 & -\frac{M}{r}\\ 0 & 1 & 0 & 0\\ 0 & 0 & {r}^{2} & 0\\ -\frac{M}{r} & 0 & 0 & -\frac{{M}^{2}-1}{{r}^{2}} \end{pmatrix}$$
and clearly $M<1$ is a constraint.

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

I don't know what to make of this so any comments welcomed.

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#### PeterDonis

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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.

#### PeterDonis

Mentor
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

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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

Mentor
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

Mentor
Ricci rotation coefficients are defined by the frame field
And the covariant derivative, which requires knowledge of the metric.

#### Mentz114

Gold Member
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.

$$g_{\mu\nu}=\begin{pmatrix} \frac{1-{r}^{2}\,{M}^{2}}{{M}^{2}-1} & 0 & 0 & r\,M\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ r\,M & 0 & 0 & 1-{M}^{2} \end{pmatrix}$$

#### PeterDonis

Mentor
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

Mentor
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

Gold Member
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

Staff Emeritus
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

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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.

$$\pmatrix{-\frac{{L}^{2}}{2\,{r}^{4}} & 0 & 0 & \frac{3\,L}{2\,{r}^{2}}\cr 0 & 0 & 0 & 0\cr 0 & 0 & \frac{{L}^{2}}{2\,{r}^{4}} & 0\cr \frac{3\,L}{2\,{r}^{2}} & 0 & 0 & -\frac{{L}^{2}}{2\,{r}^{2}}}$$

#### PeterDonis

Mentor
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.

"Strange circular geodesic"

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