The metrics to describe a wormhole

[sergiokapone]

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?

 

[Nugatory]

Is that metric even possible? That is, is it a solution of the Einstein Field Equations?

 

[sergiokapone]

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.

 

[PeterDonis]

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?

 

[sergiokapone]

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)

 

[sergiokapone]

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 $R = 0$.

 

[PeterDonis]

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)

The metric you gave in your OP is not the Morris-Thorne wormhole metric I'm familiar with, which has been discussed previously on PF. See, for example, this thread:

https://www.physicsforums.com/threads/morris-thorne-wormhole-metric-terms.877834/

Are you sure you're writing down the metric correctly?

 

[PeterDonis]

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.

 

[sergiokapone]

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?

 

[PeterDonis]

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?

 

[sergiokapone]

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?

 

[PeterDonis]

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.

 

[sergiokapone]

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 $b = 2M$

 

[Ibix]

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.

 

[PeterDonis]

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).

 

[sergiokapone]

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 $+\,-\,-\,-$

 

[Ibix]

Yes. So it looks like the metric doesn't look like what I understand a wormhole to look like, based on the curvature.

 

[PeterDonis]

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?

 

[sergiokapone]

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.

 

[Ibix]

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...

 

[PeterDonis]

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.

 

[PeterDonis]

Kruskal-Wallace

I think you mean Kruskal-Szekeres? Another auto-complete glitch, perhaps?

 

[sergiokapone]

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?

 

[Ibix]

I think you mean Kruskal-Szekeres? Another auto-complete glitch, perhaps?

Yes, but this time in my brain. And I even mis-spelled Wallis (Kruskal-Wallis test)

 

[PeterDonis]

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.

 

[PAllen]

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...

Would that be a good metric for testing the very strong equivalence principle?

 

[Dale]

What properties should lead me to the conclusion that it's a wormhole?

There should be points that are connected by multiple timelike geodesics, some of which are very short. I am not sure if closed timelike curves are essential

 

[PeterDonis]

There should be points that are connected by multiple timelike geodesics

Actually, the wormhole metric itself doesn't require this; the two asymptotically flat exterior regions could be completely disconnected.

Also, of course, this condition let's in a lot of cases that aren't wormholes: for example, consider a circular orbit in the Schwarzschild vacuum region around a planet (I put in the latter to ensure that the spacetime as a whole is not a wormhole spacetime) and a radial geodesic with just the right initial upward velocity to come back down to the same altitude in the same time as it takes to complete one circular orbit.

One definition that I have seen that seems to me to be useful is: a wormhole is present if there is a compact region with a simple boundary (such as a 2-sphere, or rather a "3-cylinder" when the time dimension is included) but non-trivial topology inside (i.e., something other than $R^4$).

 

[PeterDonis]

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?

Those equations only work if all of the other requirements are met. Yes, if all of the requirements in the paper are met, the spacetime describes a wormhole; that's what the requirements are for.

all the equations (31) - (36) 3. General whormhole will be valid for my metrics.

Yes, agreed. A spacelike slice of constant coordinate time in your metric is identical to a spacelike slice of constant coordinate time in the Schwarzschild metric (more precisely, in the exterior Schwarzschild metric, for $r > 2M$), and we already know such a spacelike slice, when analytically extended, contains a wormhole.

The difference in your metric, as compared to the standard Schwarzschild metric, is that $g_{tt} = 1$ everywhere. Obviously that can't affect anything that doesn't involve the $t$ coordinate, which equations (31) - (36) do not. But other equations given in the paper do. Those need to be checked as well.

 

[PAllen]

Actually, the wormhole metric itself doesn't require this; the two asymptotically flat exterior regions could be completely disconnected.

Also, of course, this condition let's in a lot of cases that aren't wormholes: for example, consider a circular orbit in the Schwarzschild vacuum region around a planet (I put in the latter to ensure that the spacetime as a whole is not a wormhole spacetime) and a radial geodesic with just the right initial upward velocity to come back down to the same altitude in the same time as it takes to complete one circular orbit.

One definition that I have seen that seems to me to be useful is: a wormhole is present if there is a compact region with a simple boundary (such as a 2-sphere, or rather a "3-cylinder" when the time dimension is included) but non-trivial topology inside (i.e., something other than $R^4$).

Of course wormholes between disconnected exteriors aren’t good for sci fi travel purposes.

For the orbital multiple geodesic case to have a large multiple in the proper time, the orbit would need to be close to a BH.

 

[PeterDonis]

For the orbital multiple geodesic case to have a large multiple in the proper time, the orbit would need to be close to a BH.

Or an object sufficiently compact that circular orbits near $r = 3M$ were possible. The minimum radius for a stable object in equilibrium is $r = (9/4) M$, so it would be possible for such a compact object to exist (and such a spacetime would not raise the wormhole issues that a black hole spacetime would). I don't think any neutron stars are close to this limit, though; AFAIK even the most compact ones known have a surface radius of something like $10 M$ or larger.

 

[PAllen]

Or an object sufficiently compact that circular orbits near $r = 3M$ were possible. The minimum radius for a stable object in equilibrium is $r = (9/4) M$, so it would be possible for such a compact object to exist (and such a spacetime would not raise the wormhole issues that a black hole spacetime would). I don't think any neutron stars are close to this limit, though; AFAIK even the most compact ones known have a surface radius of something like $10 M$ or larger.

I wasn’t thinking of the BH raising any topology issues, I was thinking of a BH from collapse, which is believed to have none. Just noting how it would be possible to get a large ratio for the orbit versus slingshot geodesics.

 

[stevendaryl]

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.

On the other hand, the major reason that it's nice to have an effective potential is so that you can reduce the equations of motion to something that's effectively a one-dimensional problem (after taking advantage of all the constants of motion). If you can rewrite the equations of motion in the form:

$(\dot{r})^2 = F(r, c_1, c_2, ...)$

where $c_1, c_2, ...$ are constants of the motion, then you can figure out a lot about the orbits. Whether or not the constants involve the energy doesn't seem all that relevant. Is this just a matter of terminology, or are there specific things you want to calculate where it's important that the right-hand side be expressible as $F(r,c_1, c_2, ...) = E_{eff} - V_{eff}$?

 

[PeterDonis]

are there specific things you want to calculate where it's important that the right-hand side be expressible as $F(r,c_1, c_2, ...) = E_{eff} - V_{eff}$?

The standard usage of the effective potential depends on $V_{\text{eff}}$ being independent of $E$, since you are using the value of $E$ as compared with $V_{\text{eff}}$ to determine what kind of orbit it is (elliptic, parabolic, or hyperbolic). You can't use that reasoning if $V_{\text{eff}}$ depends on $E$. The term "effective potential" comes from the fact that this is what is done in Newtonian mechanics; the only thing that changes in, for example, Schwarzschild spacetime is the presence of an extra term, independent of $E$, in the effective potential.

If you have enough constants of the motion, I agree that you can reduce the problem to a one-dimensional one regardless of the form of the RHS of the resulting equation in $\dot{r}$. However, I would not use the term "effective potential" if the RHS cannot be put in the specific form $E - V_{\text{eff}}$, with $V_{\text{eff}}$ independent of $E$. So in that sense it would be a matter of terminology.

However, I would also say you should look at the specific solutions you obtain for a case like that of the OP where $V_{\text{eff}}$ is not independent of $E$, to see what they are like; I don't think, for example, that you can assume they will take the same form as they do in the cases where $V_{\text{eff}}$ is independent of $E$.

 

[pervect]

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.

We recently had a very similar question, I would guess the answer from that question may not have gotten through since this one is so similar.

I'll try a different and more concise answer. Look at the determinant of the above metric. It changes sign. This isn't good, it's why people question if the above line element is even Lorentzian. It's necessary (but not sufficient) for a metric to have a negative determinant for it to be Lorentzian, I believe. I'm working from memory here, which isn't as good as it used to be, so feel free to double or triple check this remark, as elementary as it seems. Elementary errors are the worst :(.

Note the last similar example ALSO had this "feature" of the determinant of the metric changing sign. I'm not sure of the origin of said post, though I'm guessing a shared origin.

$\sqrt{-g}$ is basically a 4-volume element. If we consider what happens in the region where we have a Lorentzian metric, i.e. g<0, we note that the volume element in these coordinates is going to zero. So we have at least a coordinate singularity at r=2m, where one of the metric coeffficients vanish which, along with the diagonal nature of the metric, makes the deteriminant g vanish. We can calculate the stress-energy tensor and/or the Lagrangian density associated with the metric above to look at it's physical reasonableness. I believe people have calculated the stress-energy tensor in this thread via calculating the Einstein tensor G and using Einstein's field equations. I don't think anyone has gone further to look at the associated Lagrangian density.

BTW, I'm pretty sure the stress-energy tensor is really a tensor density. But I'm working from memory again.

Going into the details of the calculation won't help if one doesn't already have some familiarty with Lagrangian densities. I'd have to review Wald a bit myself, though I recently glanced at one of the sections that talks about this. On the plus side, looking at the Lagrangian density means we look at only one number, which is simpler to ask questions about than a tensor density. The obvious question is to ask if the Lagrangian density is finite - or not.

We might also be interested in whether or not the strong and weak energy conditions are satifisfied, this may require us to look at the stress-energy tensor rather than just the Lagrangian density anyway.