this answer has been expanded at the end.
I am convinced that macroscopic wormholes are impossible because they would violate the energy conditions etc. so it is not a top priority to improve the consistency of semi-consistent stories. At the same moment, I also think that any form of time travel is impossible as well, so it's not surprising that one may encounter some puzzles when two probably impossible concepts are combined.
However, it is a genuinely confusing topic. You may pick Leonard Susskind's 2005 papers about wormholes and time travel:
http://arxiv.org/abs/gr-qc/0503097
http://arxiv.org/abs/gr-qc/0504039
Amusingly enough, for a top living theoretical physicist, the first paper has 3 citations now and the second one has 0 citations. The abstract of the second paper by Susskind says the following about the first paper by Susskind:
"In a recent paper on wormholes (gr-qc/0503097), the author of that paper demonstrated that he didn't know what he was talking about. In this paper I correct the author's naive erroneous misconceptions."
Very funny. The first paper, later debunked, claims that the local energy conservation and uncertainty principle for time and energy are violated by time travel via wormholes. The second paper circumvents the contradictions from the first one by some initial states etc. The discussion about the violation of the local energy conservation law in Susskind's paper is relevant for your question.
I think that if you allowed any configurations of the stress-energy tensor - or Einstein's tensor, to express any curvature - it would also be possible for one throat of an initial wormhole to be time-dilated - a gravity field that is only on one side - and such an asymmetry could gradually increase the time delay between the two spacetime points that are connected by the wormhole. For example, you may also move one endpoint of the wormhole along a circle almost by the speed of light. The wormhole itself will probably measure proper time on both sides, but the proper time on the circulating endpoint side is shortened by time dilation, which will allow you to modify the time delay between the two endpoints.
Whatever you try to do, if you get a spacetime that can't be foliated, it de facto proves that the procedure is physically impossible, anyway. Sorry that I don't have a full answer - but that's because I fundamentally believe that the only correct answer is that one can't allow wormholes that would depend on negative energy density, and once one allows them, then he pretty much allows anything and there are many semi-consistent ways to escape from the contradictions.
Expansion
Dear Julian,
I am afraid that you are trying to answer more detailed questions by classical general relativity than what it can answer. It is clearly possible to construct smooth spacetime manifolds such that a wormhole is connecting places X, Y whose time delay is small at the beginning but very large - and possibly, larger than the separation over $c$ - at the end. Just think about it.
You may cut two time-like-oriented solid cylinders from the Minkowski spacetime. Their disk-shaped bases in the past both occur at $t=0$ but their disked-shaped bases in the future appear at $t_1$ and $t_2$, respectively. I can easily take $c|t_1-t_2| > R$ where $R$ is the separation between the cylinders. Now, join the cylinders by a wormholes - a tube that goes in between them. In fact, I can make the wormhole's proper length decreasing as we go into the future. It seems pretty manifest that one may join these cylinders bya tube in such a way that the geometry will be locally smooth and Minkowski.
These manifolds are locally smooth and Minkowski, when it comes to their signature. You can calculate their Einstein's tensor - it will be a function of the manifold. If you allow any negative energy density etc. - and the very existence of wormholes more or less forces you to allow negative energy density - then you may simply postulate that there was an energy density and a stress-energy tensor that, when inserted to Einstein's equations, produced the particular geometry. So you can't possibly avoid the existence of spacetime geometries in which a wormhole produces a time machine sometime in the future just in classical general relativity without any constraints.
The only ways to avoid these - almost certainly pathological - configurations is to
postulate that the spacetime may be sliced in such a way that all separations on the slice are spacelike (or light-like at most) - this clearly rules "time traveling" configurations pretty much from the start
appreciate some kind of energy conditions that prohibits or the negative energy densities
impose other restrictions on the stress-energy tensor, e.g. that it comes from some matter that satisfies some equations of motion with extra properties
take some quantum mechanics - like Susskind - into account
If you don't do either, then wormholes will clearly be able to reconnect spacetime in any way they want. This statement boils down to the fact that the geometry where time-like links don't exist at the beginning but they do exist at the end may be constructed.
All the best
Lubos
Disclaimer: I'm not a GR expert, but this is how this question has been explained to me by other physicists before. If I got something wrong, please correct me.
The traveler does indeed not have to exert as much work to leave the gravity well via the wormhole compared to the normal route. They are not repelled from mouth A nor attracted to mouth B by any effect having to do with the gravity of the planet.
Conservation laws are preserved, however, by interaction with the wormhole mouths themselves. When the traveler enters mouth A and leaves mouth B, no work is required to raise their mass because mouth A appears to gain equal mass to the traveler, and mouth B loses it. As far as conservation laws are concerned, it's as if the traveler crashed into and merged with an asteroid in low orbit (mouth A), and then an identical copy of the traveler got assembled out of the mass of another asteroid (mouth B) and ejected in high orbit.
So, if you try to generate infinite energy by throwing something through the wormhole and then running a generator off it as it falls back down, your plans will be foiled by mouth A becoming steadily more massive while mouth B becomes steadily less massive, until mouth A collapses into a black hole.
Best Answer
It's a bit hard to exactly construct a stress-energy tensor similar to a wormhole in normal space since part of the assumption is that the topology isn't simply connected, but consider the following scenario :
Take a thin-shell stress-energy tensor such that
$$T_{\mu\nu} = \delta(r - a) S_{\mu\nu}$$
with $S_{\mu\nu}$ the Lanczos surface energy tensor, where the Lanczos tensor is similar to a thin-shell wormhole. For a static spherical wormhole, that would be
\begin{eqnarray} S_{tt} &=& 0\\ S_{rr} &=& - \frac{2}{a}\\ S_{\theta\theta} = S_{\varphi\varphi} &=& - \frac{1}{a}\\ \end{eqnarray}
If we did this by the usual cut and paste method (cutting a ball out of spacetime before putting it back in, making no change to the space), the Lanczos tensor would be zero due to the normal vectors being the same (there's no discontinuity in the derivatives). But we're imposing the stress-energy tensor by hand here. This is a static spherically symmetric spacetime, for which we can use the usual metric
$$ds^2 = -f(r) dt^2 + h(r) dr^2 + r^2 d\Omega^2$$
with the usual Ricci tensor results :
\begin{eqnarray} R_{tt} &=& \frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{f'}{rhf}\\ R_{rr} &=& - \frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{h'}{rh^2}\\ R_{\theta\theta} = R_{\varphi\varphi} &=& -\frac{f'}{2rhf} + \frac{h'}{2rh^2} + \frac{1}{r^2} (1 - \frac{1}{h}) \end{eqnarray}
Using $R_{\mu\nu} = T_{\mu\nu} - \frac 12 T$ (this will be less verbose), we get that $T = -\delta(r - a) [2(ah)^{-1} + 2 (ar^2)^{-1}]$, and then
\begin{eqnarray} \frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{f'}{rhf} &=& \delta(r - a) \frac{1}{a} (\frac{1}{h} + \frac{1}{r^2}) \\ - \frac{1}{2 \sqrt{hf}} \frac{d}{dr} \frac{f'}{\sqrt{hf}} + \frac{h'}{rh^2} &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1]\\ -\frac{f'}{2rhf} + \frac{h'}{2rh^2} + \frac{1}{r^2} (1 - \frac{1}{h}) &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1] \end{eqnarray}
This is fairly involved and I'm not gonna solve such a system, so let's make one simplifying assumption : just as for the Ellis wormhole, we'll assume $f = 1$, which simplifies things to
\begin{eqnarray} 0 &=& \delta(r - a) \frac{1}{a} (\frac{1}{h} + \frac{1}{r^2}) \\ \frac{h'}{rh^2} &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1]\\ \frac{h'}{2rh^2} + \frac{1}{r^2} (1 - \frac{1}{h}) &=& \delta(r - a) \frac{1}{a}[\frac{1}{h} + \frac{1}{r^2} - 1] \end{eqnarray}
The only solution for the first line would be $h = - r^2$, but then this would not be a metric of the proper signature. I don't think there is a solution here (or if there is, it will have to involve a fine choice of the redshift function), which I believe stems from the following problem :
From the Raychaudhuri equation, we know that in a spacetime where the null energy condition is violated, there is a divergence of geodesic congruences. This is an important property of wormholes : in the optical approximation, a wormhole is just a divergent lense, taking convergent geodesic congruence and turning them into divergent ones. This is fine if the other side of the wormhole is actually anoter copy of the spacetime, but if this is leading to inside flat space, this might be a problem (once crossing the wormhole mouth, the area should "grow", not shrink as it would do here).
A better example, and keeping in line with the bag of gold spacetime, is to consider a thin-shell wormholes that still has trivial topology. Take the two manifolds $\mathbb R^3$ and $\mathbb S^3$. By the Gauss Bonnet theorem, a sphere must have a part in which it has positive curvature (hence focusing geodesics). Then perform the cut and paste operation so that we have the spacetime
$$\mathcal M = \mathbb R \times (\mathbb R^3 \# S^3)$$
Through some topological magic, this is actually just $\mathbb R^4$. The thin-shell approximation is easily done here, and it will give you the proper behaviour : geodesics converge onto the mouth, diverge upon crossing the mouth, then go around the inside of the sphere for a bit before possibly getting out.
From there, it's possible to take various other variants, such as smoothing out the mouth to make it more realistic (which will indeed give you a bag of gold spacetime), as well as a time dependancy to obtain this spacetime from flat Minkowski space.