# Basic GR questions

1. Apr 5, 2015

### kau

I have a problem in understanding krushkal extension of Schwarzchild metric. So i solved the Einstein equation for the vacuum part of a spherically symmetric matter distribution and solved it in the inside region too. they are fine and compatible at schwarzchild radius. But the problem i am having is how come one is using the schwarzchild metric (solution of vacuum) and changing coordinates for few time and then finally in krushkal coordinates describing the spacetime all the way down to r=0 (but this metric solution we started with is a vacuum solution). How is this justified???
My second confusion is that what will happen if someone is close to horizon.. Does he feel sudden pull of infinite strength or is it very smooth as was elsewhere outside the mass.??? I think he should not feel any difference at least locally (as far as equivalence principle is concerned)..

2. Apr 5, 2015

### wabbit

For the first question, the Kruskal metric does have a singularity at r=0, as you can see from the expression for the line element which becomes inifinite there. But the Kruskal map is equivalent to the Scharzchild maps, they descripe the same spacetime.

As to the horizon you are right, everything is smooth across it, and this is a nice thing about the Kruskal map : it makes this immediately clear. The only thing that happens is that he might not know it but there's no going back once he has crossed it (actually I think this could be a danger for a specraft hovering above the horizon: how do you know exactly were you stand, and whether you might be getting too close?) .

3. Apr 5, 2015

### kau

i can see the singularity at r=0 and nowhere else in kruskal extension. but the confusion is something else.i do agree that schwarzchild metric and krushkal explain the same spacetime (i.e. metric outside the mass)..then why we are using it to describe physics inside the schwarzchild radius.\???? why not we start with the metric solution that stands for the inside one and then doing these transformations???

4. Apr 5, 2015

### pervect

Staff Emeritus
I'm not sure what your question/confusion is about. What you should find is that that both the kruskal solution and the Schwarzschld solution solve Einstein's equation, that you can double-cover the Kruskal solution with two Schwarzschild solutions pasted together, making the Kruskal solution more complete than the later (so that the Schwarzschild solution is only part of the complete solution), and also that the Kruskal solution does NOT have a singularity at the event horizon r=2M/c^2, in contrast to the Schwarzschild solution which does have a singularity at the event horizon.

Then the reason for preferring the Kruskal solution are twofold - it is more complete than the Schwarzschild solution, and it doesn't have the problem with an avoidable sigularity at the event horizon. The existence of the Kruskal solution shows that the singularity at the event horizon is "removable" by a proper choice of coordinates, thus the singularity in Schwarzchild coordinates is called a "coordinate singularity".

5. Apr 5, 2015

### kau

Yes. I agree with each of your statements. let me tell you how did i arrive at kruskal extension. i took the the schwarzchild metric (solution of vacuum einstein equation) then said that let's not worry about angular coordinates and take one slice of them and think about radial and time direction only. then we changed the coordinates for two three times and finally we reaches to the one so called kruskal coordinates. ok everything is fine. i can see no singularity at the horizon. but now the problem is that why i can infer that r=0 is a definite singularity since my starting solution is not valid inside,it's valid only upto the horizon. if someone wants to say something about r=0 then we should take the interior solution instead of starting with schwarzchild solution. maybe you are trying to say that kruskal cover the entire space. but there is my objection since we arrived at this changing coordinates and writing the schwrzchild solution in that changed coordinate which is valid for vacuum only not at the interior.

6. Apr 5, 2015

### wabbit

Ah I think I get your meaning : why do we study Scharzchild first instead of going to Kruskal right away since it's simpler ?
If so I don't really know except that the purpose initially is to describe a spherically symmetrical spacetime, so the natural way to do this is with something like Schwarzchild - the fact that Kruskal actually describes this too is not obvious, to me at least, from just looking at the metric.

Also, Kruskal is great, but each coordinate system has its advantages and in practice, I don't think Scharwzchild and other maps are made obsolete by Kruskal.

7. Apr 5, 2015

### kau

Ok. fine. You are right but the thing is how do you ensure the fact that kruskal extension is a solution of einstein equation in presence of matter as well. since you went to this coordinate just starting with schwarzchild solution (vacuum solution) and then doing few coordinate transformations. so nowhere it's obvious why kruskal extension would describe the interior solution ...it certainly describe the vacuum part because you started with schwarzchild solution and this coordinate tell you that horizon is free from singularity ..all good. but why you can push it to describe interior is not obvious to me. may be i am missing some physical notion here that making these troubles. Thanks anyway and if you are not tired then can certainly reply to my query.

8. Apr 5, 2015

### wabbit

But it's not. It's a vacuum solution too, or I really missed something. (Quite possible too, I'm no expert in any way). And so is Schwarzchild, in or out. This is all vacuum. The only mass is in the singularity, which isn't part of the solution.

And I enjoy these questions, I'm learning too and it's good to think about these things. Just hope I dont lead you astray with too many of my misunderstandings.

Last edited: Apr 5, 2015
9. Apr 5, 2015

### kau

Let me describe my problem again. How do we derive the kruskal extension. we start with Schwarzchild solution (it's a vacuum solution) and then we do few coordinate transformations (namely 4 or 5) and we arrive at a metric which is so called kruskal extension. so what we did to arrive at this is just coordinate transformation. and this tells us that there is no singularity at r=2GM (schwarzchild radius). and this describe the vacuum since it's the same metric (namely schwarzchild metric) so it describe vacuum very well. But why are we using it to say that there is a singularity at r=0 since it does not describe the interior. So my confusion is related to the fact that kruskal extension is derived from vacuum solution. it might happen that even if i start with interior solution i can transform it to kruskal extension as well,but that part of derivation i didn't see. so if you have seen it you can refer me to that. thanks.

10. Apr 5, 2015

### kau

I didn't get this part of your answer when i replied. i really didn't know that the mass is only at r=0.. i guess that's not the case. mass can be spread till anywhere at any radius but have to be smaller than schwarzchild radius.... this information that mass is at r=0 is not known and also not clear to me..

11. Apr 5, 2015

### wabbit

Ah OK maybe I get it now (sorry I'm slow here...)
You're saying : starting from the exterior solution alone, we find that it can be extendee using Kruskal to one that has a singularity. But what tells us this is the only way ? Could there be another extension that doesn't have a singularity at all, or maybe two, or something else ? (Presumably we re constraining this solution to also be shperically symmetrical since that was the original purpose.)

Hah. I don't know. But don't we have a unicity theorem here we can draw on ? Hmm... Need to think about this, or maybe an expert will step in.

12. Apr 5, 2015

### wabbit

Well I'm quite sure that if you allow mass inside, away from the singularity, then you get a different solution. It won't be Kruskal, nor interior Schwarzchild (though exterior will be the same), just a different spacetime. The initial prescription is for a vacuum solution that's all.

A solution with mass is for instance the model used for a star. Within the mass distribution it differs from the Schwarzchild interior or Kruskal solution.

13. Apr 5, 2015

### kau

yeah i looked at that solution where we took a model of perfect liquid inside and got some solution. but you are saying that in case of kruskal we are taking a model where all the mass is at r=0 and otherwise vacuum. ok if i consider this then everything is fine at its place. let me think a little more and go through a bit more and then ci will comeback. i have a confusion regarding penrose diagram for a collapsing spherical shell of photons... but i will post about it next time if i dont find answer...thanks by the way..

14. Apr 5, 2015

### wabbit

No problem, and again this is just how I understand it, definitely not an expert answer.

Just wanted to add : it's not really that we re looking for a solution that has no mass except at r=0, it's a solution with no mass at all within the solution. r=0 happens to be where a singularity forms, i.e. the solution doesn't extend to that point (or surface)

Last edited: Apr 5, 2015
15. Apr 5, 2015

### pervect

Staff Emeritus
It is possible (though tedious) to demonstrte that the Kruskal metric satisfies the vacuum Einstein field equatons without any reference to the Schwarzschild solution at all. Using the direct route, there is then no issue with crossing the horizon because there is no singularity present at the horizon.

As far as demonstrating that r=0 is a true singularity that can't be removed, the usual approach is to look at curvature invariants, such as the http://en.wikipedia.org/wiki/Kretschmann_scalar. An infinite curvature scalar is a sign of bad behavior that cannot be removed by a coordinate change as we did for the coordinate singularity at the event horizon.

16. Apr 5, 2015

### Staff: Mentor

It does describe the interior; there are two ways of showing this. One way is to simply re-do the derivation for the interior (see below). The other way is to analytically extend the exterior solution into the interior region. The way we do that is to look at geodesics that hit the "edge" of the exterior solution (i.e., they reach the horizon, where the exterior Schwarzschild line element becomes singular) and note that they are extendible beyond that point once we have switched to the Kruskal line element. Then we just extend the geodesics as far as we can, and find that we now have a valid solution in a new region of spacetime, the interior region.

(Actually, when we do the maximal analytic extension of the Kruskal line element, we find four regions total in the spacetime: the original exterior region, the future interior region--the one we usually think of as the "interior", also called the "black hole" region--plus a second exterior region and a second, "past" interior region, also called the "white hole" region. These last two regions are only present in the idealized case where the solution is vacuum everywhere; in any real black hole formed by gravitational collapse, those regions will not be present.)

Yes, it does. The derivation goes through the same way, except for a few sign changes to account for the fact that $r < 2M$ instead of $r > 2M$.

17. Apr 6, 2015

### stevendaryl

Staff Emeritus
As I understood the question, it was not whether the Kruskal solution for the interior of a black hole is A solution, but whether it is unique. Can there be more than one way to extend the Schwarzschild exterior solution across the event horizon? Obviously, there can be, in at least two ways:
1. Specifying that the stress-energy tensor $T_{\mu \nu}$ is zero outside the event horizon doesn't say anything about $T_{\mu \nu}$ inside the horizon.
2. Even if we insist that $T_{\mu \nu}$ is zero everywhere, it doesn't uniquely determine the metric. For example, as I understand it, the Weyl part of the curvature tensor is not determined by the stress-energy.
So I think it's probably true that knowing the metric outside the event horizon doesn't uniquely determine it inside the event horizon. So then the question is: Is the singularity present in EVERY extension to the Schwarzschild exterior?

18. Apr 6, 2015

### Staff: Mentor

Birkhoff's Theorem proves that it is if it's a vacuum solution. More precisely, Birkhoff's Theorem proves that the only vacuum, spherically symmetric solution of the EFE is the Schwarzschild geometry, and the Kruskal line element describes that geometry.

Correct. But it's worth noting that if we know the SET vanishes outside the horizon, then the singularity theorems show that whatever stress-energy is inside the horizon must collapse and form a singularity, leaving behind an interior vacuum region that must be described by the Schwarzschild geometry (assuming spherical symmetry).

It does if the spacetime is spherically symmetric; see above. If the spacetime is not spherically symmetric, then the Kruskal line element doesn't apply anyway, even outside the horizon, since that line element describes a spherically symmetric spacetime.
In the general case, yes, this is true. But if we impose additional symmetry constraints, those plus a specification of the SET may be sufficient to determine the Weyl tensor. In the particular case under discussion--vacuum plus spherical symmetry--it is.

19. Apr 6, 2015

### pervect

Staff Emeritus
While I would agree that knowing that $T_{\mu\nu}$ doesn't uniquely specify the metric, my understanding, based on http://math.ucr.edu/home/baez/gr/ricci.weyl.html is that the situation is rather similar to electromagnetism, where knowing the charge distribution everywhere in space does not specify the electromagnetic field. The situation appears to be analogous in that you need to specify boundary conditions. In the EM case, you can add an electromagnetic wave to any solution of Maxwell's equations and get a different field solution which is also valid. It's a generic property of differential equations, you have to specify the boundary conditions to get a unique solution to either Maxwell's equations or Einstein's equations - or to any differential equation. If you really want mathematical precision, I'd suggest Wald's section on why GR is a well-posed initial value problem, but I suspect this is not what the OP was after.

Going with the gravitational waves as the reason for multiple solutions, if the exterior region (which would be two regions in the Kruskal solution, the black hole exterior region and the white hole exterior region) contain no gravitational waves at any time, the interior region won't either. And you get a unique solution. But - you have made some seemingly innocuous assumptions to get this far, namely the notion that the solution is spherically symmetrical.

However, it's rather doubtful that an actual black hole caused by gravitational collapse is spherically symmetric inside the event horizon, it's believed to be unstable, it's rather like a mathematical solution of a pin standing on its point. It's prefectly valid mathematically, but not likely on physical grounds, a real pin will fall over. If you have an almost perfectly spherically symmetric shell of matter that collapses, small irregularities in the matter distribution may and probably will start to grow when the shell collapses inside its event horizon. There are also some wrinkles due to the fact that the angular momentum of the shell is unlikely to be zero, giving rise to a rotating singularity rather than the zero angular momentum Schwarzschild / Krusal solution.

So the ultra short version is that the Kruskal solution really is just an extended version of the Schwarzschild solution (which is what I think the OP is struggling with), but neither solution should be taken to describe the interior region due to gravitational collapse, which is a much more complex problem that's not totally well understood, though there is a fair amount of modern literature on the topic so there is not a state of total ignorance either. It is, however, much more complex than dealing with the Kruskal solution, which is well understood.