# I Strange circular geodesic

#### Mentz114

Gold Member
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

Mentor
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

#### Mentz114

Gold Member
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.

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#### PeterDonis

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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

Gold Member
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

Mentor
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

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

#### PeterDonis

Mentor
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

Gold Member
I'm not sure that's true. The result I get from Maxima (assuming that I posted the metric ...
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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.

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#### Mentz114

Gold Member
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.

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#### PeterDonis

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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

Mentor
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

Mentor
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

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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

Gold Member
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
\begin{align*} D_{12} &= -{ \Gamma_{12}}^1 U_1 - { \Gamma_{12}}^4 U_4\\ &=\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}}}\\ &= 0 \end{align*}
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).

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