I The metrics to describe a wormhole

[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.

 

[PeterDonis]

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.

 

[PeterDonis]

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]

Oh, there's something else I wanted to say. I'm guessing that it's possible that the motivation of the OP might be to understand how the Schwarzschild geometry can be viewed as an Einstein-Rosen bridge, a sort of non-traversable wormhole.

MTW has a discussion of this view of the Schwarzschild geometry in their textbook "Gravitation". And I suppose there's also the Einstein-Rosen paper.

If this was the intent, then the discussion of the underlying issue is getting confused by introducing alternative metrics that aren't the Schwarzschild metric and turn out to have some fundamental issues such as not being Lorentzian manifolds in the interior region. But perhaps understanding the wormhole nature of the Schwarzschild geometery wasn't the original intent. More could be said on this, but it may be a digression so I won't belabor it anymore.