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

[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.