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.

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    $\begingroup$ Putting on my Cynic Hat for a moment, I'd suggest the difficulty is that the mere existence of wormholes, let alone the ability to generate one, is way over in the "computational wanking" corner of physics. It's tough enough to accept a theory that some 'cosmic cataclysm' poked a hole in spacetime from herenow to therewhen. $\endgroup$ Apr 23, 2014 at 12:01
  • $\begingroup$ Very interesting, but my layman guess is that it would be very complicated. Even Minkowski spacetime with very low density matter that collapses to a black hole seems very complicated in general. But I would love to see something like Openheimer-Snyder for forming a wormhole. $\endgroup$
    – MBN
    Apr 23, 2014 at 12:26
  • $\begingroup$ Seems that a reasonable approach would be to first try the spherically symmetrical case. Start with bag of gold, whose associated wormhole throat is a thin shell of exotic dust. Numerically iterate the Einstein equations to obtain the evolution of the geometry (whose denouement is presumably Minkowski-like spacetime containing a dispersed quantity of exotic matter). Now you can run time backward to see the desired formation of the bag of gold induced by the exotic dust gathering to form a spherical shell. $\endgroup$
    – Belizean
    May 2, 2014 at 5:06
  • $\begingroup$ You could try treating the problem perturbatively, write out the metric as a background Minkowski with an additional piece that resembles the matter you want to glue to the space-time to make the wormhole. Generally speaking, the perturbation scheme maps one space-time to another via a diffeomorphism $\varphi$: $(M_{0},\mathbf{\eta}) \mapsto^{\varphi} (M,\mathbf{\eta + h})$. You can now say that the physical manifold $M$ is a connected sum of three pieces, $M = \cup_{i=1}^{3} M_{i}$, with $M_{1}$ the first region, $M_{2}$ the second and $M_{3}$ is the wormhole throat. $\endgroup$ May 26, 2014 at 6:52
  • $\begingroup$ Proceeding in this way you can analyse the topological properties of the three connected peices by assuming the structure of $M_{3}$ using, say, Visser's model. The various curvature forms will be analytic across the surfaces; and I see no reason why $M_{1}$ and $M_{2}$ could not have more or less arbitrary topology, depending on what kind of matter you allow to live in this 'major' part of the universe. Just some thoughts. $\endgroup$ May 26, 2014 at 6:54

2 Answers 2


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.


This article deals with passable wormhole


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:

  1. spherically symmetric and static geometry. It is a condition imposed to simplify the calculations.
  2. Being solution of Einstein's equations.
  3. contain a throat (a narrow fragment space-time, highly curved) connecting two asymptotically flat regions of spacetime.
  4. Absence of horizons to allow the trip in two directions.
  5. small tidal forces, not to destroy possible travelers.
  6. 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.
  7. Matter and fields that generate the curvature of spacetime is described by a tensor of energy- -momento with physical meaning.
  8. The solution must be stable for small perturbations during the passage of the traveler.
  9. 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.
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    $\begingroup$ While a nice resource, I do not think that this really answers the question (which is about the evolution of space-time after/during the construction of the wormhole). $\endgroup$
    – Kyle Kanos
    Oct 15, 2014 at 18:56
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    $\begingroup$ Thanks, but there are lots of articles on analysing the spacetime around existing wormholes. My question is what would happen when you started in flat spacetime and assembled the material to construct a wormhole. To create the usual Morris–Thorne wormhole would appear to require a topology change, and this is what I find odd. $\endgroup$ Oct 16, 2014 at 5:32
  • $\begingroup$ @JohnRennie I have heard a lot of this holes and everyone talks how it can be stabilised but no one care how it's formed,How actually are wormhole formed? $\endgroup$
    – nihaljp
    Sep 14, 2016 at 16:37
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    $\begingroup$ @NihalJalaluddinP: no-one knows. $\endgroup$ Sep 14, 2016 at 16:45
  • $\begingroup$ @JohnRennie Oh come on,Why people don't research how it's formed rather than researching how to stabilise it without even creating? $\endgroup$
    – nihaljp
    Sep 14, 2016 at 16:47

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