Building a wormhole We regularly get questions about wormholes on this site. See for example Negative Energy and Wormholes and How would you connect a destination to a wormhole from your starting point to travel through it?. Various wormhole solutions are known, of which my favourite is Matt Visser's wormhole because it's closest to what every schoolboy (including myself many decades ago) thinks of as the archetypal wormhole.
The trouble is that Visser has pulled the same trick as Alcubierre of starting with the required (local) geometry and working out what stress-energy tensor is required to create it. So Visser can tell us that if we arrange exotic string along the edges of a cube the spacetime geometry will locally look like a wormhole, but we know nothing about what two regions of spacetime are connected.
My question is this: suppose I construct a Visser wormhole by starting in Minkowksi spacetime with arbitrarily low densities of exotic matter and gradually assembling them into the edges of a cube, how would the spacetime curvature evolve as I did so?
I'm guessing that I would end up with something like Wheeler's bag of gold spacetime. So even though I would locally have something that looked like a wormhole it wouldn't lead anywhere interesting - just to the inside of the bag. I'm also guessing that my question has no answer because it's too hard to do any remotely rigorous calculation. Still, if anyone does know of such calculations or can point me to references I would be most interested.
 A: 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. 
A: This article deals with passable wormhole 
http://scitation.aip.org/content/aapt/journal/ajp/56/5/10.1119/1.15620 
Wormhole Trafficable properties 
As we have seen, there are several objections to the possibility of 
come true interstellar travel through holes 
black or wormholes from Schwarzschild. To make 
passable a wormhole should have the following 
properties: 


*

*spherically symmetric and static geometry. It is a 
condition imposed to simplify the calculations. 

*Being solution of Einstein's equations. 

*contain a throat (a narrow fragment 
space-time, highly curved) connecting two 
asymptotically flat regions of spacetime. 

*Absence of horizons to allow the trip in 
two directions. 

*small tidal forces, not to destroy possible 
travelers. 

*Allow a traveler can cross the wormhole 
in a suitable time and at a time coordinated 
reasonable. The latter is measured by an observer 
far away from the sources of the gravity field. 

*Matter and fields that generate the curvature of 
spacetime is described by a tensor of energy-
-momento with physical meaning. 

*The solution must be stable for small perturbations 
during the passage of the traveler. 

*Finally, the wormhole must be built with a 
finite amount of material, certainly less than the 
the material content of the universe, and an interval of 
finite time, clearly less than the age of the universe.

