Spherically symmetric metric form

1. Aug 22, 2010

mersecske

spherically symmetric metric used to write in he following form:

ds^2 = -h(r,t}^2 * dt^2 + f(r,t)^2 * dr^2 + r^2 * d_omega^2

ds^2 = -f(r,t}^2 * dt^2 + f(r,t)^(-1) * dr^2 + r^2 * d_omega^2

and

ds^2 = -f(r}^2 * dt^2 + f(r)^(-1) * dr^2 + r^2 * d_omega^2

how can we interpret these forms?

This is the general spherically symmetric vacuum solution, or what?

2. Aug 22, 2010

Petr Mugver

Since in your first equation f and h can be whatever functions you like (yes, in this way you get the most general spherically symmetric metric) the last two equations are just special cases of the first one (i.e. with particular choices of f and h).

3. Aug 22, 2010

bcrowell

Staff Emeritus
For any spherically symmetric spactime, even one that is not a vacuum solution, coordinates exist such that its metric can be expressed in your h(r,t), f(r,t) form.

In the case of a vacuum solution, Birkhoff's theorem says that the unique solution (up to a choice of coordinates) is the Schwarzschild spacetime. For the Schwarzschild spacetime, it's possible to choose coordinates such that you get the f(r), 1/f(r) form. See http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.3 [Broken] , subsection 7.3.7. If f(r) is chosen arbitrarily, then your f(r), 1/f(r) form will not in general be a vacuum solution; only if f(r) is chosen to be of the Schwarzschild form 1-2m/r is it a vacuum solution.

Last edited by a moderator: May 4, 2017
4. Aug 23, 2010

mersecske

Yes I know that spherically symmetric metric has the first form in general,
however the precise statement is "spherically symmetric and STATIC" as I know!

But my question is:
Can we say something similar about the second and third one in general?
I mean: All "..." spacetimes has the form...
and the form... always describes "..."

Not only the Schwartzschild but de Sitter or charged BH has the same form with f(r).

And I have another question:
most textbooks said that spherically symmetric and STATIC spacetimes
can be described with a metric in the first form.
However this is not true because the Schwarzschild metric has a singularirty at the event horizon! What is the precise statement?

5. Aug 23, 2010

mersecske

but at section: 7.3.6 Birkhoff's theorem
the author(s) write "time" when t is used,
however t is not time in general,
because t is not time-like coordinate below the horizon!
So I think this is not a precise textbook.

6. Aug 23, 2010

George Jones

Staff Emeritus
At and below the event horizon, Schwarzschild spacetime is not static.

7. Aug 23, 2010

mersecske

A spacetime is static or not static isn't it?
How can define "static" locally?

So what is the precise statement for the first metric form?
general spherically symmetric metric
or
general spherically symmetric and static metric?

8. Aug 23, 2010

George Jones

Staff Emeritus
No.
If U is a connected open subset of a spacetime M, then U is itself a spacetime. U could be static, while M\U is not.

9. Aug 23, 2010

bcrowell

Staff Emeritus
I'm pretty sure you're incorrect about this. First, there are definitely non-static spacetimes, such as FRW cosmological solutions, that can be put in the first form. Also, I believe it is true that all spherically symmetric spacetimes can be put in this form. If you have a copy of Hawking and Ellis, take a look at their proof of Birkhoff's theorem. They pretty clearly state that the metric can be put in this form in general, and then they write down the non-vacuum field equations, including source terms. Only later do they specialize to the case of a vacuum solution, and then the final result of the proof is that the spacetime is of the Schwarzschild form, which is static outside the horizon. If they had to assume staticity in order to write down the metric in that original form, then their proof would be logically erroneous, because they'd be assuming what is essentially the main result of the proof: staticity.

Is the claim that (a) all spherically symmetric and static spacetimes can be put in that form, or (b) that any spacetime that can be put in that form is spherically symmetric and static? Claim b is clearly false, since cosmological solutions are a counterexample. Claim a is true, but unnecessarily weak, since the assumption of staticity is unnecessary.

10. Aug 23, 2010

mersecske

But what about the metric at the horizon?
The event horizon of a Schwarzschild BH is also spherically symmetric,
but you cannot put the metric in the first form!

11. Aug 23, 2010

bcrowell

Staff Emeritus
One thing I hadn't noticed before is that you were talking about the form

$$ds^2 = -h(r,t)^2 dt^2 + f(r,t)^2 dr^2 + r^2 d\Omega^2$$ ,

whereas Hawking and Ellis are using the more general form

$$ds^2 = -h(r,t)^2 dt^2 + f(r,t)^2 dr^2 + g(r,t) d\Omega^2$$ .

I think their form can be used to cover the entire Schwarzschild metric, including the horizon, since the metric has this form when expressed in Kruskal–Szekeres coordinates (with t->V, r->U, g->r).

I don't know whether it is possible to cover the Schwarzschild spacetime with coordinate patches, each of which gives a metric in your form, without any coordinate singularities. Do you have an argument to show that this is impossible? You can do any coordinate transformation of the form $t\rightarrow t'=f(t,r)$ while still keeping the metric in your form.

I'm glad you pointed this out, because it is definitely a possible weak point in my own version of the proof, which I linked to in #3. I'll have to think about it more. I'm not sure that it's a terribly big issue to have a coordinate singularity on a particular surface.

Another thing to be careful about is that many authors, such as Ludvigsen, use the term "stationary" to mean asympotically stationary, so some statements they make may seem incorrect unless you take into account their definition of the term.

[EDIT] I found a discussion of this point in Misner, Thorne, and Wheeler, section 32.2. On p. 843 they introduce the form with $r^2 d\Omega^2$, then note that it may fail to apply on certain 3-surfaces, and suggest in an exercise on p. 845 how to remove these coordinate singularities.

Last edited: Aug 23, 2010
12. Aug 30, 2010

mersecske

Ok, thanks! And Kruskal-Szekeres or Eddington-Finkelstein like coodinates can be used in general to any spherically symmetric metric? What is the form of a general spherically symmetric metric, which is not singular anywhere?

13. Aug 30, 2010

bcrowell

Staff Emeritus
I general you can't cover a manifold with a single coordinate patch; if you could, then the definition of a manifold would be a lot simpler! Therefore I don't think there is any such general form.

14. Aug 31, 2010

mersecske

I dont say single just non-singular.
For example Schwarzschid coordinates is good below and above the horizon,
but not on the horizon.

15. Sep 2, 2010

bcrowell

Staff Emeritus
I'd suggest you study the proof in Misner, Thorne, and Wheeler, section 32.2, and figure out what goes wrong in the places where $\nabla r=0$. I suspect that the form you're talking about can't be applied at all in a place like this, but I haven't thought it through in enough detail. For the proof of Birkhoff's theorem, all that matters is that the places where it fails to have that form are 3-dimensional rather than 4-dimensional. It would surprise me very much if you could cover the entire Schwarzschild spacetim4e with patches that all had the form you're talking about, and that were all smooth and well-behaved.