Wormholes are unphysical?

stevendaryl
Staff Emeritus
Physically, a wormhole spacetime would have to be a solution of the EFE, and the only solution of the EFE with zero curvature everywhere is flat Minkowski spacetime. So any physical spacetime with wormholes present would have to have nonzero curvature.

Are you sure about that? I would think that a cylindrical (or toroidal) universe would be consistent with the EFE. After all, as far as the field equations are concerned, there is no difference between a toroidal universe and one whose mass/energy density just happens to be periodic in space.

PeterDonis
Mentor
2020 Award
Are you sure about that? I would think that a cylindrical (or toroidal) universe would be consistent with the EFE.

Hm, yes, you're right, since the EFE is local, it can't by itself constrain the global topology.

After all, as far as the field equations are concerned, there is no difference between a toroidal universe and one whose mass/energy density just happens to be periodic in space.

If the mass/energy density is nonzero, however, there must be curvature; that *is* required by the EFE. So yes, a given solution of the EFE could describe a toroidal (but non-flat) universe, or an infinite universe with periodic mass/energy density; but a toroidal vacuum universe (like a square region of Minkowski spacetime with boundaries identified) would still require zero stress-energy for the EFE to be satisfied.

pervect
Staff Emeritus
Do you have any reference where I could see the geodesic equations for the Kruskal metric?

You can write down the geodesic equations directly from the metric by calculating the Christoffel symbols.

See for instance the general geodesic equation as described in http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

This doesn't usually give a lot of insight though, the equations are messy. What gives more insight with less calculation is finding the Killing vectors.

See for instance http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html

The dot product of the any killing vector and the tangent vector of a geodesic is constant. So instead of a very messy nonlinear differential equation, you have a set of simple linear differential equation.

You can also add the requirement that the norm of the tangent vector be unity if the geodesic is parameterized by proper time (which is usually what you want0.

The only Killing vector that has a different form in the KS coordinates is ##\partial / \partial t##. You can find some discussion of what this Killing vectors look like expressed in Kruskal coordinates at.

So if you follow the discussion of how to get the motion from the Killing vectors, and how to transform the Killing vectors to the new coordinates, you'll have the math to work out motions in KS coordinates.

However, the discussion already given to your problem is much easier to understand than the detailed math you'll need to work out motions in KS coordinates.

I'm not sure that the more advanced answer to your question will help you understand more, but just in case it does I'm giving it.

I think that re-reading the less technical answers already would be a much easier approach to understanding what's happening

1 person
WannabeNewton
So getting back to wormholes. A wormhole is a topological feature of spacetime; it means that spacetime is not simply connected. What does that imply about curvature?

Space-time having a non-trivial fundamental group does not imply the existence of a wormhole. It is a sufficient but not necessary condition. Furthermore the Einstein-Hilbert action does not contain any topological terms in it so it does not codify the topology in itself, an initial Cauchy surface must be provided and if the space-time is hyperbolic the Cauchy data can be evolved uniquely under Einstein's equations (think of it as a boundary value problem). That the Einstein-Hilbert action contains no topological terms is a feature of it being four dimensional; in two dimensions the same integral is entirely topological thanks to the Gauss-Bonnet theorem. One cannot easily relate the existence of non-trivial fundamental groups in Lorentzian 4-manifolds constrained to satisfy Einstein's equations to the Riemann or Ricci curvature of that manifold. .

Bill_K
Space-time having a non-trivial fundamental group does not imply the existence of a wormhole. It is a sufficient but not necessary condition.
Don't you mean necessary but not sufficient?

an initial Cauchy surface must be provided and if the space-time is hyperbolic the Cauchy data can be evolved uniquely under Einstein's equations (think of it as a boundary value problem).
What about a periodic spacetime, in which the hypersurface t = t0 has been identified with t = t0 + C? There's no guarantee that the data evolved from t0 will match the data that has already been determined at t0 + C.

WannabeNewton
Don't you mean necessary but not sufficient?

Sorry, yes.

What about a periodic spacetime, in which the hypersurface t = t0 has been identified with t = t0 + C? There's no guarantee that the data evolved from t0 will match the data that has already been determined at t0 + C.

My apologies but, while I don't disagree that such a situation can arise, I'm not seeing what point is being made exactly with regards to the quoted statement.

Bill_K
if the space-time is hyperbolic the Cauchy data can be evolved uniquely under Einstein's equations (think of it as a boundary value problem).
What about a periodic spacetime, in which the hypersurface t = t0 has been identified with t = t0 + C? There's no guarantee that the data evolved from t0 will match the data that has already been determined at t0 + C.
My apologies but, while I don't disagree that such a situation can arise, I'm not seeing what point is being made exactly with regards to the quoted statement.
For such a spacetime, the general Cauchy initial value problem fails to have a solution. Only for very specific values of the data will it be self-consistent. This is a mathematical statement, but with important physical consequences! It's exactly the reason that I can't kill my grandfather.

1 person
WannabeNewton
For such a spacetime, the general Cauchy initial value problem fails to have a solution. Only for very specific values of the data will it be self-consistent. This is a mathematical statement, but with important physical consequences!

Oh I see. I'll try to read more on that, thank you for pointing that out :)

It's exactly the reason that I can't kill my grandfather.

Well we wouldn't want that now would we!

Gold Member
Don't you mean necessary but not sufficient?

What about a periodic spacetime, in which the hypersurface t = t0 has been identified with t = t0 + C? There's no guarantee that the data evolved from t0 will match the data that has already been determined at t0 + C.

would that mean you compactify time in a circle? then I guess due to periodicity you're driven to making your last statement true...

PAllen
For such a spacetime, the general Cauchy initial value problem fails to have a solution. Only for very specific values of the data will it be self-consistent. This is a mathematical statement, but with important physical consequences! It's exactly the reason that I can't kill my grandfather.

I thought the initial value formulation only produced globally hyperbolic manifolds, which would exclude such a solution. To get such a solution, I would think you need to pose additional topological constraints, which may or may not be consistent with initial conditions.

stevendaryl
Staff Emeritus
Space-time having a non-trivial fundamental group does not imply the existence of a wormhole. It is a sufficient but not necessary condition.

So what is the definition of a "wormhole"? The descriptions I've seen (in Misner, Thorne and Wheeler, for example), demonstrate by reducing space to two-dimensions, a flat sheet. A wormhole is depicted as a "tunnel" connecting two distant points in space. For example:
http://www-tc.pbs.org/wnet/hawking/strange/assets/images/ss.wormholes.jpg [Broken]

Are you saying that having that topology (or the 3D analog, rather) is not sufficient to have a wormhole?

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WannabeNewton
The definition of a wormhole is actually more restrictive than that of simply having a non-trivial fundamental group. Wikipedia has a good description of the restriction: http://en.wikipedia.org/wiki/Wormhole

For a more detailed and more mathematical treatment of wormholes, see section 10.7.2. of "Introduction to Black Hole Physics"-Frolov and Zelnikov

stevendaryl
Staff Emeritus
The definition of a wormhole is actually more restrictive than that of simply having a non-trivial fundamental group. Wikipedia has a good description of the restriction: http://en.wikipedia.org/wiki/Wormhole

For a more detailed and more mathematical treatment of wormholes, see section 10.7.2. of "Introduction to Black Hole Physics"-Frolov and Zelnikov

Hmm. The definition in Wikipedia is this:
If a Minkowski spacetime contains a compact region Ω, and if the topology of Ω is of the form Ω ~ R x Σ, where Σ is a three-manifold of the nontrivial topology, whose boundary has topology of the form ∂Σ ~ S2, and if, furthermore, the hypersurfaces Σ are all spacelike, then the region Ω contains a quasipermanent intra-universe wormhole.

That's a little hard for me to parse, but taking it one step at a time:

• "the topology of Ω is of the form Ω ~ R x Σ". I think that's just saying that the region can be split into space + time, with time being represented by the real numbers, as usual.
• "Σ is a three-manifold of the nontrivial topology". The spatial part of the wormhole is an ordinary chunk of 3-space.
• "whose boundary has topology of the form ∂Σ ~ S2". Here's where my understanding of higher-dimensional geometry gets a little fuzzy. S2 is just a sphere. So I think this is just saying that the boundary of the wormhole is a sphere. The analogy in 2-dimensions is that the boundary is a circle. It seems to me that in the 2-D case, the wormhole is a cylinder, so the boundary would be a PAIR of circles, one on each end of the wormhole. So I would think that in the 3-D case, the boundary would be a pair of spheres, one on each end.
• "the hypersurfaces Σ are all spacelike". That's just saying that the spatial part of the wormhole really is spatial.

So I'm not sure exactly what this definition is saying, nor am I sure of how it differs from the intuitive idea of the 3-D analog of the 2-D case of two distant circles joined by a cylinder.

Hmm. The definition in Wikipedia is this:

That's a little hard for me to parse, but taking it one step at a time:

• "the topology of Ω is of the form Ω ~ R x Σ". I think that's just saying that the region can be split into space + time, with time being represented by the real numbers, as usual.
• "Σ is a three-manifold of the nontrivial topology". The spatial part of the wormhole is an ordinary chunk of 3-space.
• "whose boundary has topology of the form ∂Σ ~ S2". Here's where my understanding of higher-dimensional geometry gets a little fuzzy. S2 is just a sphere. So I think this is just saying that the boundary of the wormhole is a sphere. The analogy in 2-dimensions is that the boundary is a circle. It seems to me that in the 2-D case, the wormhole is a cylinder, so the boundary would be a PAIR of circles, one on each end of the wormhole. So I would think that in the 3-D case, the boundary would be a pair of spheres, one on each end.
• "the hypersurfaces Σ are all spacelike". That's just saying that the spatial part of the wormhole really is spatial.

So I'm not sure exactly what this definition is saying, nor am I sure of how it differs from the intuitive idea of the 3-D analog of the 2-D case of two distant circles joined by a cylinder.

"The spatial part of the wormhole is an ordinary chunk of 3-space."
No. "Nontrivial topology" essentially means that the space contains a hole (whose boundary is ∂Σ).

"So I think this is just saying that the boundary of the wormhole is a sphere."
Not quite. It's saying the boundary has the topology of a sphere (in the same sense that oranges and bananas both have the topology of a sphere in that their surfaces can be incrementally deformed without surgery into a sphere).

"It seems to me that in the 2-D case, the wormhole is a cylinder, so the boundary would be a PAIR of circles, one on each end of the wormhole. So I would think that in the 3-D case, the boundary would be a pair of spheres, one on each end."
No. A wormhole is defined here as a spacetime in which the spatial part has a hole in it. To obtain what we normally consider to be a wormhole (with its two mouths) we must mathematically identify that hole with another hole elsewhere in the space.

The problem with this (Visser's) definition, as mentioned in the Wikipedia article, is that "nontrivial topology" could be interpreted to mean more than merely possessing a hole. It could be interpreted to mean that the space possesses a "handle" in the Euler sense. In this case, a connection between two universes, which we would normally consider to be a wormhole, would not satisfy the definition.

The proper fix, as implied in the article, is to dispense with the term "nontrivial topology" and describe the space as possessing a hole. This is equivalent to describing the space as having the property of preventing the incremental shrinking of some enclosed surfaces to a point (an enclosed surfaces of arbitrarily small surface area).

stevendaryl
Staff Emeritus