A Where exactly is the wormhole in the Kruskal-Szekeres diagram?

[haushofer]

TL;DR

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.

 

[Ibix]

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.

 

[haushofer]

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 )

 

[romsofia]

I believe the final paragraph (and diagrams) might help: https://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/kruskal.html
(Tevian Dray's book "Differential Forms and the Geometry of General Relativity")

 

[Ibix]

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.

 

[Ibix]

I believe the final paragraph (and diagrams) might help: https://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/kruskal.html
(Tevian Dray's book "Differential Forms and the Geometry of General Relativity")

There are similar, although less graphically sophisticated, diagrams in chapter 7 of Carroll's lecture notes.

 

[Orodruin]

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.

 

[haushofer]

I believe the final paragraph (and diagrams) might help: https://sites.science.oregonstate.edu/physics/coursewikis/GGR/book/ggr/kruskal.html
(Tevian Dray's book "Differential Forms and the Geometry of General Relativity")

Yes, it does, thanks!

 

[haushofer]

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?

 

[Orodruin]

But how can the wormhole close if some particle enters it? Isn't the Schwarzschild solution static?

Only outside the event horizon.

 

[haushofer]

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.

 

[Orodruin]

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.

 

[haushofer]

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

 

[Ibix]

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.

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.

 

[Orodruin]

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]

The Killing field that's timelike outside is null on the horizon and spacelike inside.

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.

 

[Orodruin]

According to Wald the Killing field vanishes on the horizon, not goes null. Sorry!

I guess the zero vector is technically a null vector …

 

[Orodruin]

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.

 

[Orodruin]

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).

 

[PeterDonis]

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).

 

[PeterDonis]

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).

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.

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.

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.

 

[Orodruin]

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.

 

[Ibix]

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.

 

[PeterDonis]

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).

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".

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.

 

[Orodruin]

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]

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.

 

[Ibix]

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"?

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[Orodruin]

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.

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 …