# Dude! Where's my wormhole?

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#### haushofer

TL;DR Summary
Confusion about the location of the wormhole in a conformal diagram of the Schwarzschild black hole.
Dear all,

recently I was brushing up my knowledge of black holes with (among others) Zee's "Einstein gravity in a Nutshell" and encountered the analytical continuation of the Schwarzschild black hole in the famous Kruskal-Szekeres coordinates (Zee: chapter VII.2). The corresponding diagram can be translated into a Penrose diagram, which I attached for reference. My question is quite simple: I'm confused about the exact location of the wormhole in this diagram (or the Kruskal-Szekeres diagram for that matter). Should I identify both singularities of the black and white hole in this diagram such that topologically the diagram becomes a cilinder? What happens exactly with the physical singularity? Don't you eventually hit it when you enter the horizon? How can one reconcile the idea that geodesics terminate at the singularity with the idea of a wormhole bringing you to region III or IV in the Penrose diagram?

Many thanks!
continuation of the Schwarzschild solution.

X marks the spot - it's in the middle, both on Kruskal and Penrose diagrams. Note that both lines of the X represent spherical surfaces of radius 2GM, so it's not actually a point. Passing through that X in any direction is where the nested spherically symmetric surfaces stop getting smaller and start getting larger again, but since the radius isn't zero it's not just like passing through the center of a polar coordinate system on flat space.

Timelike worldlines may pass through it from the white hole to the black hole. Ingoing null worldlines that lie on the white hole horizon pass through it and become outgoing (outward into the other exterior) worldlines on the black hole event horizon.

Note that literature is not consistent in the numbering scheme. I and II are usually consistent, but III and IV can be named either way around. This has confused me before. I suggest referring to them by names - something like black and white hole interiors, and exterior and mirror exterior.

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haushofer
Ibix said:
X marks the spot - it's in the middle, both on Kruskal and Penrose diagrams. Note that both lines of the X represent spherical surfaces of radius 2GM, so it's not actually a point. Passing through that X in any direction is where the nested spherically symmetric surfaces stop getting saller and start getting larger again, but since the radius isn't zero it's not just like passing through the center of a polar coordinate system on flat space.

Timelike worldlines may pass through it from the white hole to the black hole. Ingoing null worldlines that lie on the white hole horizon pass through it and become outgoing (outward into the other exterior) worldlines on the black hole event horizon.

Note that literature is not consistent in the numbering scheme. I and II are usually consistent, but III and IV can be named either way around. This has confused me before. I suggest referring to them by names - something like black and white hole interiors, and exteriorand mirror exterior.
Ok, so observers in region I cannot enter the wormhole then, only observers in region III? In this case my confusion about "hitting the singularity" is also removed (I guess it was a coordinate confusion instead of a physical confusion :P )

haushofer said:
Ok, so observers in region I cannot enter the wormhole then, only observers in region III?
Correct. Only spacelike paths can reach the wormhole from either exterior region.

haushofer
dextercioby and romsofia
Ibix said:
Correct. Only spacelike paths can reach the wormhole from either exterior region.
This is why it is called a non-transversable wormhole. It collapses faster than anything can get from I to IV or vice versa. The Flamm paraboloid (which is the typical throat-like structure you often see in images - such as the cover of my problem book) is the spatial structure of the fully extended Schwarzschild solution for fixed t. If you translate a surface of fixed t upwards (it will no longer be a surface of fixed t) or downward, the wormhole starts closing. It does so fast enough for nothing to get through.

Orodruin said:
This is why it is called a non-transversable wormhole. It collapses faster than anything can get from I to IV or vice versa. The Flamm paraboloid (which is the typical throat-like structure you often see in images - such as the cover of my problem book) is the spatial structure of the fully extended Schwarzschild solution for fixed t. If you translate a surface of fixed t upwards (it will no longer be a surface of fixed t) or downward, the wormhole starts closing. It does so fast enough for nothing to get through.
But how can the wormhole close if some particle enters it? Isn't the Schwarzschild solution static?

haushofer said:
But how can the wormhole close if some particle enters it? Isn't the Schwarzschild solution static?
Only outside the event horizon.

Orodruin said:
Only outside the event horizon.
So inside the horizon it becomes time-dependent? And it only pinches off when a particle enters the throat?

I guess I should recheck the math again.

haushofer said:
So inside the horizon it becomes time-dependent? And it only pinches off when a particle enters the throat?

I guess I should recheck the math again.
It is only outside the horizon that ##\partial_t## is a time-like Killing field. The existence of a time-like Killing field is the definition of a stationary spacetime.

The wormhole is non-transversable regardless of a particle enters or not. The point is that no timelike path exists going between regions I and IV.

Sidequestion: can such analytic extensions of static solutions introduce time-dependency in general?

haushofer said:
So inside the horizon it becomes time-dependent?
Yes. The Killing field that's timelike outside is null on the horizon and spacelike inside. The other Killing fields remain associatedwith the spherical symmetry.
haushofer said:
And it only pinches off when a particle enters the throat?
No, it's just that there's no timelike path that starts in either exterior region and can reach the wormhole. On the diagrams, null paths are 45° slopes everywhere, so you can see this easily. You can draw a spacelike line through it from anywhere in the exterior, you just can't reach it on a timelike path.

Ibix said:
Yes. The Killing field that's timelike outside is null on the horizon and spacelike inside. The other Killing fields remain associatedwith the spherical symmetry.
It is not directly obvious to me that ##\partial_t## can be extended to a smooth field across the horizon. I would have to do the maths to check this. Probably the easiest would be to write it down in Kruskal coordinates…

Ibix said:
The Killing field that's timelike outside is null on the horizon and spacelike inside.
Orodruin said:
It is not directly obvious to me that ##\partial_t## can be extended to a smooth field across the horizon. I would have to do the maths to check this. Probably the easiest would be to write it down in Kruskal coordinates…
According to Wald the Killing field vanishes on the horizon, not goes null. Sorry! Edit: nope! See #23, below.

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Ibix said:
According to Wald the Killing field vanishes on the horizon, not goes null. Sorry!
I guess the zero vector is technically a null vector …

Ibix
Ibix said:
According to Wald the Killing field vanishes on the horizon, not goes null. Sorry!
I thought a bit about it and it intuitively makes sense that it is zero on the horizon. The surfaces of constant t get closer and closer together when you look at the Kruskal diagram and so it is kind of intuitive that the field goes to zero. That it is null (assuming it exists) is of course a necessity for it to be smooth across the horizon since the norm square changes sign from outside to inside.

Ibix
Ibix said:
According to Wald the Killing field vanishes on the horizon, not goes null. Sorry!
Ok, here is the math.

The KS coordinates are on the form
$$T = f(r) \sinh(t/2R_S), \qquad X = f(r) \cosh(t/2R_S)$$
with ##f(R_S) = 0## (outside the horizon but that suffices).
This means that
$$\partial_t = \frac{ f(r)}{2R_S} [ \cosh(t/2R_S)\partial_T + \sinh(t/2R_S)\partial_X] = \frac{X \partial_T + T\partial_X}{2R_S}.$$
This would seem to vanish only at the origin, but is clearly null on the horizon (as the metric is proportional to ##dT^2 - dX^2##, the squared norm is proportional to ##X^2 - T^2 = 0## on the horizon).

Ibix
Orodruin said:
This would seem to vanish only at the origin, but is clearly null on the horizon
Yes. This is easy to check by, for example, using Painleve or Eddington-Finkelstein coordinates, where the components become simply ##(1, 0, 0, 0)## and it is clear that the vector does not vanish on the horizon (although to check it on all four branches of the bifurcate Killing horizon you have to use four different patches of such coordinates--Kruskal makes it easier by having one patch that covers the entire maximal analytic extension of the manifold, and by covering the bifurcation point, which none of the other coordinate patches do).

Ibix said:
X marks the spot
Actually it's not just at that spot; that spot is simply where the throat of the wormhole is at it maximum size. At earlier times the wormhole throat is expanding to that size from zero; at later times the wormhole throat is contracting from that size to zero. The reason it is impossible to traverse the wormhole is not that only spacelike paths go into it, but that any timelike path that goes into it gets trapped by the contraction of the throat and forced into the singularity instead of escaping out the other side. This is somewhat easier to see in the Kruskal chart than the Penrose chart because of the way the latter distorts the spacetime inside the horizon (pulling the singularity "up" to a horizontal line instead of being a hyperbola).

Ibix said:
Only spacelike paths can reach the wormhole from either exterior region.
Only spacelike paths can reach the point of maximal size of the wormhole throat (from the exterior region--as noted in another post, timelike paths from the white hole region can reach this point, but they also are forced to go into the black hole region from there). But that does not mean only spacelike paths can enter the wormhole. See above.

haushofer said:
observers in region I cannot enter the wormhole then, only observers in region III?
No. Observers from both regions can enter the wormhole; they just cannot enter it soon enough to escape out the other side. See above.

haushofer said:
how can the wormhole close if some particle enters it? Isn't the Schwarzschild solution static?
"Static" just means there is a Killing vector field that is timelike, at least on some region of the spacetime (in this case the region outside the horizon). It does not mean that a particular observer's worldline must be an integral curve of such a KVF. Only if the latter is true will the observer see an unchanging spacetime along his worldline. But we are talking about observers who are trying to go through the wormhole; such observers must have decreasing ##r## along their worldlines as they enter the wormhole, so they will not see the spacetime geometry around them as unchanging. Only observers who "hover" at a constant ##r## forever see that.

haushofer
PeterDonis said:
Yes. This is easy to check by, for example, using Painleve or Eddington-Finkelstein coordinates
Why go to EF coordinates when the KS coordinates cover everything and we just concluded that the Killing field is ##X\partial_T + T\partial_X##? That seems pretty definitive to me. (Yes, I just showed it for the exterior, but it is pretty clear that it is what will be in the interior as well given how the coordinate lines look.)

Bottom line is: Wald seems to be wrong here in claiming that the Killing field actually vanishes at the horizon. It vanishes as ##r\to R_S## with constant ##t##, but that is just the limit approaching the maximal size wormhole throat - which is invariant under the symmetry and therefore should have a vanishing Killing field.

In fact, the Killing field in KS coordinates just so happens to be expressed in exactly the same way as the Killing field of Minkowski space that generates Lorentz transformations. The difference being that the KS metric has an additional r-dependent factor in front of the Minkowski metric. However, as lines of constant r are the invariant hyperbolae, this still leaves the metric invariant.

Orodruin said:
Ok, here is the math.

The KS coordinates are on the form
$$T = f(r) \sinh(t/2R_S), \qquad X = f(r) \cosh(t/2R_S)$$
with ##f(R_S) = 0## (outside the horizon but that suffices).
This means that
$$\partial_t = \frac{ f(r)}{2R_S} [ \cosh(t/2R_S)\partial_T + \sinh(t/2R_S)\partial_X] = \frac{X \partial_T + T\partial_X}{2R_S}.$$
This would seem to vanish only at the origin, but is clearly null on the horizon (as the metric is proportional to ##dT^2 - dX^2##, the squared norm is proportional to ##X^2 - T^2 = 0## on the horizon).
Apparently I just can't read, which fits with the way the rest of the day has gone. What Wald actually says (p154) is ...we see that ##\nabla_ar=0## at ##X=T=0##, and it is not difficult to verify that the static Killing field ##\xi^a## vanishes there also. Note also that ##\nabla_ar## and ##\xi^a## become colinear along the null lines ##X=\pm T##.

Ugh. So I was right in the first place.

Orodruin said:
Why go to EF coordinates
Of course you don't have to. I am just saying that if you want an easy way to verify that the KVF does not vanish at the horizon (but just becomes null), the fact that the KVF is simply the vector ##(1, 0, 0, 0)## in EF coordinates (or Painleve coordinates) tells you that without having to do any further math at all. The only caveat is that no patch of EF coordinates (or Painleve coordinates) covers the bifurcation point, so you can't use this method to evaluate the KVF there (and of course the KVF does vanish there).

Orodruin said:
Wald seems to be wrong here in claiming that the Killing field actually vanishes at the horizon.
I would want a specific reference to exactly where Wald is claimed to say this. I don't see one in the thread; I just see "Wald".

Orodruin said:
the Killing field in KS coordinates just so happens to be expressed in exactly the same way as the Killing field of Minkowski space that generates Lorentz transformations.
Yes, that's one of the reasons KS coordinates are very useful for understanding the causal structure of the spacetime.

Ibix said:
Apparently I just can't read, which fits with the way the rest of the day has gone. What Wald actually says (p154) is ...we see that ##\nabla_ar=0## at ##X=T=0##, and it is not difficult to verify that the static Killing field ##\xi^a## vanishes there also. Note also that ##\nabla_ar## and ##\xi^a## become colinear along the null lines ##X=\pm T##.

Ugh. So I was right in the first place.
Good, good. We all agree. Including Wald in absentia.

PeterDonis said:
I would want a specific reference to exactly where Wald is claimed to say this.
And we now have the specific passage (that post "crossed in the mail" with mine) and the question is resolved, good.

PeterDonis said:
Actually it's not just at that spot; that spot is simply where the throat of the wormhole is at it maximum size. At earlier times the wormhole throat is expanding to that size from zero; at later times the wormhole throat is contracting from that size to zero.
...wait, so the entire interior of the black and white holes counts as "wormhole"? And that's why I can see into both exterior regions from inside the black hole? Or is there some specific part of the diagram that's considered to be "the wormhole"?

Ibix said:
so the entire interior of the black and white holes counts as "wormhole"?
Any spacelike 3-surface which connects the two exterior regions counts as a constant time slice of the wormhole (at least for some time slicing), because any such 3-surface will have the same topology as the ##T = 0## surface in Kruskal coordinates, ##S^2 \times R##, which is the "wormhole" topology. It is a matter of convention what portion of each of these surfaces to call "the wormhole" as opposed to "the exterior regions". To say that the wormhole is "closed", you need to have two disconnected spacelike 2-surfaces, each of which only includes one of the two exterior regions.

In the Penrose chart, all of the spacelike 3-surfaces of constant "Penrose time" connect the two exterior regions and have the topology described above, so all of them are "part of the wormhole". That is why I said that the Penrose chart is not the best chart to use for the "wormhole" description. In the Kruskal chart, if we normalize the coordinates in units of ##2M##, then we can say that for ##T \le -1##, there is no wormhole and the two exterior regions are disconnected; for ##-1 < T < 1##, the wormhole starts growing from zero throat size to maximum throat size, then shrinks back down to zero size; and for ##T \ge 1##, again there is no wormhole and the two exterior regions are disconnected.

If we consider a timelike worldline that tries the best it can, so to speak, to go through the wormhole, it would go as close as possible to the antihorizon in Region I (the right exterior region), would cross the horizon as close as possible to the bifurcation point, and would continue to try to stay as close as possible to the antihorizon after that. To such an observer, it would seem like they entered the wormhole but found it closing too fast for them to exit out the other side, and they would end up hitting the singularity instead.

Ibix said:
I can see into both exterior regions from inside the black hole?
You can see light coming from both exterior regions inside the black hole region, yes. You will, of course, only see a portion of what happens in those regions before you hit the singularity.

PeterDonis said:
To such an observer, it would seem like they entered the wormhole but found it closing too fast for them to exit out the other side, and they would end up hitting the singularity instead.
Can you qualify this? What do you mean by ”seem like they entered the wormhole”? Any timelike path in the black hole region will have ever decreasing r-coordinate and will eventually hit the singularity. Where it hits the singularity will always be a Kruskal boost away from being very much to the right in the black hole region of the Kruskal diagram.

PeterDonis said:
You can see light coming from both exterior regions inside the black hole region, yes. You will, of course, only see a portion of what happens in those regions before you hit the singularity.
Although it should be noted that first one would need to find an actual maximally extended Schwarzschild black hole …

Orodruin said:
What do you mean by ”seem like they entered the wormhole”?
Any such notion must rely on a choice of coordinates (for time slicing) and a convention for what counts as "inside the wormhole". The notion I implicitly used in my post is that we are using the Kruskal time slicing (and normalizing the coordinates in units of ##2M##), so that the wormhole exists for the range ##-1 < T < 1##, and that you are "inside the wormhole" at any event on your worldline that is in that time range in Kruskal coordinates and is at a value of ##r## that is less than or equal to ##2M##. The reason I think this convention is reasonable is that, as I pointed out in an earlier post, all of the surfaces of constant Kruskal coordinate time in the range given above have the same "wormhole" topology as the ##T = 0## surface that contains the bifurcation 2-surface; the only difference between them is the size of the wormhole "throat" (i.e., the area of the 2-sphere of minimum area within the spacelike 3-surface).

One could of course adopt other conventions; for example, the convention implicitly used in post #2 and some of its follow-ups is that you are only "inside the wormhole" if you are at an event on the bifurcation 2-surface at the origin of Kruskal coordinates. Which means that no timelike path from either exterior region can ever be inside the wormhole by this convention (only timelike paths coming from the white hole region can reach the bifurcation surface).

Orodruin said:
Although it should be noted that first one would need to find an actual maximally extended Schwarzschild black hole
Yes, of course, which means that this thread in its entirety is almost certainly irrelevant to our actual universe, which almost certainly contains no such objects.

PeterDonis said:
Any such notion must rely on a choice of coordinates (for time slicing) and a convention for what counts as "inside the wormhole".
Hence it is not really something the observer actually sees. Things like curvature invariants increase monotonically etc. Just checking.

Orodruin said:
Hence it is not really something the observer actually sees.
If we want to take that point of view, then the observer never sees themselves inside the wormhole. Unless we want to say that being able to see incoming light from both exterior regions (which is an invariant) counts as being inside the wormhole--but then that would be true anywhere inside the black hole region.

PeterDonis said:
Any spacelike 3-surface which connects the two exterior regions counts as a constant time slice of the wormhole (at least for some time slicing)
Got it. Here's the Kruskal diagram from Wikipedia (originally by @DrGreg, modified by a user called Vilnius - note that regions III and IV are labelled the other way around to the version in the OP):

If I draw a horizontal line across the diagram, each point is representative of a sphere. As I move my finger along the line, the radius of the associated sphere falls until it reaches a minimum on the ##T## axis, then starts to grow again. But each sphere is "inside" the previous one in the sense that I have to pass through the sphere I'm on if I want to get from a previous one to one I haven't touched yet, so if I try to draw an embedding of this it forces me to draw a classic wormhole neck.

You can see from the diagram that the sphere radius on axis depends on where I draw my horizontal line. As Peter said, it is maximised at ##2GM## at the origin (the diagram uses units of ##2GM##, so labels this ##r=1##) and falls to zero at either singularity. So, as Peter said, the "neck" of the wormhole is narrower the further away from the X I draw my horizontal line - and it, in fact, falls to zero radius if I draw the line grazing either singularity. Any further away, and my line passes through two disjoint regions of spacetime and a region in the diagram that doesn't represent anywhere in the spacetime between them. That is, the wormhole collapsed (or hasn't formed yet if I'm at the bottom of the diagram).

The diagram shows a black line starting on the ##X## axios and curving into the upper singularity. This is the worldline of a test body falling into the black hole and striking the singularity. This one doesn't make it through the wormhole. The interesting thing about Kruskal diagrams, though, is that you can pick a different time origin and the whole thing rearranges rather like a boosted Minkowski diagram - the spokes "rotate" and the hyperbolae stay put. That means that by a different choice of origin I can make that test particle pass through the neck or not - in fact, there's no invariant sense in which the particle does or does not pass through the wormhole. It's kind of a moot point since it hits the singularity.

The other interesting thing is that if you pick interior Schwarzschild coordinates, the spatial plane become the hyperbolae of constant ##r## - that is, the spheres I encounter as I move along those lines have equal radii - there is no wormhole in these coordinates! That's actually a cheat. Because those coordinates only cover the upper and lower wedges, I'm able to shuffle all my wormholiness off into the regions the coordinates don't cover - in particular the origin. There's no way to wish the wormhole away completely - it's a fact of the geometry, so any coordinate system that covers the whole diagram must include lines that extend to ##X=\pm\infty## and therefore necessarily have smaller radius spheres in the middle than either end. The very best you can do, I think, is Schwarzschild coordinates, where the planes with wormhole-like topology are the spacelike straight lines that pass through the origin, where the coordinates are singular and I can't describe it.

I hadn't quite grasped that before. It's fascinating how much complexity lurks inside the simplest non-trivial solution to the EFEs.

Ibix said:
there's no invariant sense in which the particle does or does not pass through the wormhole
Yes: you have to pick an arbitrary boundary for what counts as "through the wormhole", and a Kruskal boost can change whether a particular worldline does or does not pass inside the boundary.

Ibix said:
you can pick a different time origin and the whole thing rearranges rather like a boosted Minkowski diagram - the spokes "rotate" and the hyperbolae stay put
I'm not sure "pick a different time origin" is the correct way to describe what is being done. What you are doing is a "time translation" along the integral curves of the Killing vector field that is timelike in the exterior regions. ("Time translation" is also something of a misnomer because the KVF is not timelike everywhere; it's null on the two horizons and spacelike in the black hole and white hole regions.) In Schwarzschild coordinates this KVF is ##\partial_t##. This KVF indeed works like the boost KVF in Minkowski spacetime, except that its norm varies with ##r## in a different way than the boost KVF in Minkowski spacetime varies with ##x##. But the integral curves and boost action look the same.

Also, though, you can't shift the spacetime origin of this KVF the way you can shift the spacetime origin of the boost KVF in Minkowski spacetime--or, to put it in what might be a better way, there is not just one "boost KVF" in Minkowski spacetime, there are an infinite number of them, depending on which point in spacetime you pick as the "bifurcation point" of the boost (the point where the two Killing horizons generated by the boost meet). In Schwarzschild spacetime, you can't do that; there is only one possible spacetime origin of the boost, and that origin is not a spacetime point but a 2-sphere, the one represented by the origin of the Kruskal diagram. All this is why I'm not sure "pick a different time origin" is a good description for the Schwarzschild case.