# The metrics to describe a wormhole

PAllen
2019 Award
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?

• Ibix
Dale
Mentor
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
Mentor
2019 Award
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 lets 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##).

• Dale
PeterDonis
Mentor
2019 Award
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
2019 Award
Actually, the wormhole metric itself doesn't require this; the two asymptotically flat exterior regions could be completely disconnected.

Also, of course, this condition lets 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.

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PeterDonis
Mentor
2019 Award
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
2019 Award
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
Staff Emeritus
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
Mentor
2019 Award
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
Staff Emeritus
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.

PeterDonis
Mentor
2019 Award
It's necessary (but not sufficient) for a metric to have a negative determinant for it to be Lorentzian, I believe.
This is correct, and it means the line element as it's written down in the OP can only be valid for ##r > 2M##. But that, in itself, does not show that the spacetime geometry itself does not extend to a region that corresponds to ##r \le 2M##; it just shows that, if there is such a region, it can't be described by a coordinate chart of the form given in the OP.

I'm pretty sure the stress-energy tensor is really a tensor density.
Not if you're referring to the thing that appears on the RHS of the Einstein Field Equation. That's a tensor.

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PeterDonis
Mentor
2019 Award
We might also be interested in whether or not the strong and weak energy conditions are satisfied
The weak energy condition will always be violated close enough to the throat in any wormhole metric (the Morris-Thorne paper discusses this on p. 405). For this particular metric, I think @Ibix is correct in post #14 that the WEC is violated everywhere. I would expect the null energy condition to be violated everywhere as well. Since this particular metric has a traceless stress-energy tensor, the dominant energy condition is the same as the weak; and since it has ##T^t{}_t = 0##, the strong energy condition is obviously violated since the vector ##T^a{}_b X^b## will vanish for the timelike ##X^b = \partial_t##. So it looks like all of the energy conditions are violated by this particular wormhole.

pervect
Staff Emeritus