The metrics to describe a wormhole

  • #1
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Main Question or Discussion Point

Lets consider some kind of metrics:
\begin{equation}
ds^2 = dt^2 - \frac{dr^2}{1-\frac{2M}{r}} - r^2(d\theta^2 + \sin^2\theta d\phi^2).
\end{equation}
here ##r = l/2\pi## is the radial coordinte like in Schwarzschild metrics.
As far as I know, this metrics is describe the whormhole.

Lets put ##\theta=\pi/2##.

First, I can find the velocities:
\begin{equation}
\frac{dr}{ds} = \sqrt{1- \frac{2M}{r}} \left[ (E^2 - 1) - \frac{L^2}{r^2}\right].
\end{equation}
whrere the conserved quantities are ##E = \frac{dt}{ds} = \frac{1}{\sqrt{1-v^2/c^2}} = \gamma## and ##L = r^2\frac{d\phi}{ds} ##.
Also, the velocity of any particle measured by stationary observer ##v = \frac{proper\, distance = \frac{dr}{\sqrt{1-\frac{2M}{r}}}}{proper\, time = dt}##is constant:
\begin{equation}
\frac{dl}{dt} = \sqrt{\frac{E^2 - 1}{E^2}} = v.
\end{equation}

But, If I have not read a literature, how do I know is it a wormhole metric? What properties should lead me to the conclusion that it's a wormhole?
 
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Answers and Replies

  • #2
Nugatory
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Is that metric even possible? That is, is it a solution of the Einstein Field Equations?
 
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  • #3
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Is that metric even possible? That is, is it a solution of the Einstein Field Equations?
Why not? One can come up with an exotic matter that generates such a metric according with Einstein equations.

The invention of different metrics is a kind of kinematics, when we are not interested in the distribution of matter that generates a metrics.
 
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  • #4
PeterDonis
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Lets consider some kind of metrics
Where did you get this metric from? Please give a reference.

is it a solution of the Einstein Field Equations?
@sergiokapone is correct that any metric (provided it meets some minimal requirements like being locally Lorentzian) can "solve" the Einstein Field Equations, since you can always just turn the crank to compute its Einstein tensor, multiply it by ##8 \pi##, and call that the "stress-energy tensor" of the spacetime. The question is whether that SET is physically reasonable.

If I have not read a literature, how do I know is it a wormhole metric?
If you have not read the literature, where did you get this metric from? And why did you think it was a wormhole metric in the first place?
 
  • #5
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Where did you get this metric from?
Firstly, I just come up it for looking geodesics. I was wondering whether the non-flatness in spatial coordinates affects the motion of bodies.
Then I found it in "Wormholes in spacetime and their use for interstellar travel: A tool for teaching general relativity" by Michael S. Morris and Kip S. Thorne
(1987)
 
  • #6
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If any interested:
Non zero components Christoffel symbols in such metrics:
\begin{align}
\Gamma^r_{rr} &= -\frac{M}{r^2}\frac{1}{1-\frac{2M}{r}},\\
\Gamma^r_{\theta\theta} &= 2M - r\\
\Gamma^r_{\phi\phi} &= (2M - r)\sin^2\theta\\
\Gamma^{\theta}_{r\theta} &= \lambda^{\phi}_{r\phi} = \frac1r\\
\Gamma^{\theta}_{\phi\phi} &= -\sin\theta\cos\theta \\
\Gamma^{\phi}_{\theta\phi} &= \cot\theta.
\end{align}

Non zero components of curvature tensor:
\begin{align}
R_{r\theta r\theta} & = \frac{M}{r}\frac{1}{1-\frac{2M}{r}} \\
R_{r\phi r\phi} & = \frac{M}{r}\frac{\sin^2\theta}{1-\frac{2M}{r}} \\
R_{\theta\phi\theta\phi} &= - 2Mr\sin^2\theta.
\end{align}
Scalar curvature ##R = 0##.
 
  • #7
PeterDonis
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  • #8
PeterDonis
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Scalar curvature ##R = 0##.
If that's true (I haven't checked it), it's an indication that the metric you wrote in your OP is not a wormhole metric (or at least not one with a traversable wormhole). Those have a nonzero Ricci scalar. See, for example, post #6 in the thread I linked to above.
 
  • #9
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If that's true (I haven't checked it), it's an indication that the metric you wrote in your OP is not a wormhole metric (or at least not one with a traversable wormhole). Those have a nonzero Ricci scalar. See, for example, post #6 in the thread I linked to above.
But now I'm not interested in whether this wormhole is traversible or not. I want to understand how one can guess that this wormhole?
 
  • #10
PeterDonis
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I want to understand how one can guess that this wormhole?
First we have to establish that it is, in fact, a wormhole. I'm not convinced of that.

If the metric you had written down were the standard Schwarzschild metric (with ##\left( 1 - 2M / r \right)## in the ##dt^2## term), then there would be a non-traversable "wormhole" in it, yes. But the metric you wrote down isn't the standard Schwarzschild metric.

Can you give a specific chapter/section/page reference in Morris & Thorne where the metric you wrote down in the OP appears?
 
  • #11
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First we have to establish that it is, in fact, a wormhole. I'm not convinced of that.
Ok, let's find out if this is a wormhole?
 
  • #12
PeterDonis
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Ok, let's find out if this is a wormhole?
No, let's find out if the metric you posted in the OP is even relevant to a discussion of wormholes at all. I don't see it anywhere in the Morris & Thorne paper. So where did you get it from?

If you have read the Morris & Thorne paper, it discusses the general characteristics of wormholes. Is there something in particular in that discussion that you have questions about?

If you are asking how you can "read off" from a particular metric that it is a wormhole, you can't. It's more complicated than that. Too complicated in its entirety for a PF thread; you would be asking for a course in wormhole physics. That's why we need to focus on something more specific, like particular things you have questions about from the Morris & Thorne paper.
 
  • #13
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If you are asking how you can "read off" from a particular metric that it is a wormhole, you can't.
Sad, but that's what I wanted to know.

About article:
Paper, p. 402 C. Spatial geometry of whormhole
just put ##b = 2M##
 
  • #14
Ibix
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A couple of points.

The Einstein tensor is diagonal with elements ##(0,-2m/r^3,m/r^3,m/r^3)##. This is exotic everywhere, I think.

The Kretschman scalar is ##24m^2/r^6##, which doesn't immediately look to my (non expert) eyes like a wormhole. There don't seem to be two asymptotically flat regions unless ##r## can go negative. But the metric isn't Lorentzian for negative r - it has a --++ signature.
 
  • #15
PeterDonis
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just put b=2M
This option is not discussed explicitly in the paper; I suspect it's because it does not meet one or more of the constraints given in section II. An obvious issue is the coordinate singularity at ##r = 2M##, and the fact that ##g_{rr}## becomes positive instead of negative at ##r < 2M##, whereas there is no similar transition in ##g_{tt}## at ##r = 2M## as there is in the standard Schwarzschild metric (##g_{tt} = 1## everywhere).
 
  • #16
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There don't seem to be two asymptotically flat regions unless ##r## can go negative. But the metric isn't Lorentzian for negative r - it has a --++ signature.
As I think, the body never go under the ##r<2M##. If we look at the equation (derived from metric, ##E = \frac{dt}{ds} = \mathrm{const} \ge 1##, ##r^2\frac{d\phi}{ds} =L = \mathrm{const} \ge 0## are constants of motion).
\begin{equation}
\frac{dr^2}{ds^2} = (E^2 - 1) - \left( \frac{2M}{r} (E^2 - 1) + \left(1 - \frac{2M}{r}\right)\frac{L^2}{r^2}\right)
\end{equation}
We can get the effective potential:
\begin{equation}
U _{eff} = \frac{2M}{r} (E^2 - 1) + \left(1 - \frac{2M}{r}\right)\frac{L^2}{r^2}
\end{equation}
The Turning Point never be ##r < 2M##. So, metrics allways be a physical and ##+\,-\,-\,-##
 
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  • #17
Ibix
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Yes. So it looks like the metric doesn't look like what I understand a wormhole to look like, based on the curvature.
 
  • #18
PeterDonis
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There don't seem to be two asymptotically flat regions unless rr can go negative.
That's not what you need to look for. Since ##r## is the "areal radius" (i.e., ##4 \pi r^2## is the surface area of the 2-sphere labeled by ##r##), it can never be negative; but that's just as true of "wormhole" spacetimes as any others.

What we would need to look for is the possibility of a transformation to a different "radial" coordinate (such as the ##l## in the wormhole metric discussed in the thread I linked to in post #7) which does allow for the full range ##- \infty < l < \infty## while still giving positive ##r## (areal radius) everywhere. In other words, in any wormhole metric, there must be two regions that both cover the same ranges of "areal radius" ##r##, from its minimum value (which is called ##b_0## in the Morris-Thorne paper) all the way out to infinity.

As I think, the body never go under the r<2M
There will be a minimum (positive) value of ##r## in any wormhole metric, yes. However, you can't just declare by fiat that your metric does not allow ##r < 2M##. You have to show that it doesn't--i.e., that it is not possible for any geodesic to start with ##r > 2M## and eventually reach ##r \le 2M##.

Also, while a complete discussion of how to check whether a given metric describes a wormhole is complicated, as I said, there is one obvious check that can be made. As I noted above, any wormhole spacetime (traversable or not) must have two regions that both cover the range ##b_0 \le r < \infty##. Does yours?
 
  • #19
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Also, while a complete discussion of how to check whether a given metric describes a wormhole is complicated, as I said, there is one obvious check that can be made. As I noted above, any wormhole spacetime (traversable or not) must have two regions that both cover the range b0≤r<∞b0≤r<∞b_0 \le r < \infty. Does yours?
If we plot embedding diagram, it will be with a throat (equations from (24) to (27) in K. Thorn paper). Is it enough to consider it as a wormhole? Also, all the equations (31) - (36) 3. General whormhole will be valid for my metrics.
 
  • #20
Ibix
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That's not what you need to look for.
Right. A lesson I should have absorbed from the Schwarzschild to Kruskal-Wallace coordinate transform.

Aside: my new phone still thinks Schwarzenegger is a more likely autocomplete than Schwarzschild. It'll learn... :cool:
 
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  • #21
PeterDonis
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We can get the effective potential:
No, this isn't correct. The effective potential can't contain ##E##; the whole point is to find an equation that separates out ##E## and the effective potential. In other words, you need to find an equation that looks like this:

$$
\left( \frac{dr}{d\tau} \right)^2 = \left( E^2 - 1 \right) + U_{\text{eff}}
$$

where ##U_{\text{eff}}## does not contain ##E## at all (just like Newtonian gravity).

I'm not even sure that's possible for the metric you wrote down in the OP; but certainly the equation in your post #16 isn't it.
 
  • #23
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No, this isn't correct. The effective potential can't contain ##E##; the whole point is to find an equation that separates out ##E## and the effective potential. In other words, you need to find an equation that looks like this:

$$
\left( \frac{dr}{d\tau} \right)^2 = \left( E^2 - 1 \right) + U_{\text{eff}}
$$

where ##U_{\text{eff}}## does not contain ##E## at all (just like Newtonian gravity).

I'm not even sure that's possible for the metric you wrote down in the OP; but certainly the equation in your post #16 isn't it.
I do not know how to exclude ##E ## from the eff-potential. Therefore, I thought that it is logical to consider the eff-potential as everything that depends on ##r##. Yes, it depend on particle velocity, but why not, it is just a constant?
 
  • #25
PeterDonis
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I do not know how to exclude ##E## from the eff-potential.
Do the corresponding derivation for the standard Schwarzschild metric. You will find that you can obtain an equation of the form I gave, with no dependence on ##E## in ##U_{\text{eff}}##.

If you can't do that for your metric, that just means there is no well-defined "effective potential" for your metric.

I thought that it is logical to consider the eff-potential as everything that depends on ##r##.
It's not. The term "effective potential" has a particular meaning which requires that it appears in an equation of the form I gave, where it does not depend on ##E##. As above, if you can't find such an equation for your metric, that means the concept of "effective potential" can't be used with your metric.
 

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