# Static point in de Sitter-Schwarzschild spacetime

• I
Gold Member

## Main Question or Discussion Point

In de Sitter-Schwarzschild spacetime things close to the black hole are falling towards it whereas in greater distance they are receding. So there should be a certain (unstable) ##r##-coordinate, where things are static. The de Sitter-Schwarzschild metric has according to Wikipedia
https://en.wikipedia.org/wiki/De_Sitter–Schwarzschild_metric
2 solutions:

##f(r)=1-2M/r##

##f(r)=1-\Lambda*r^2##

Equating yields ##r=(2M/\Lambda)^{1/3}##

Is this the wanted static r-coordinate? Heuristically it seems to make sense, r increases with increasing ##M## and decreases with increasing ##\Lambda##.

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PeterDonis
Mentor
2019 Award
2 solutions
No, one solution with two extra terms in ##f(r)##, i.e., the superposition of the two individual solutions. The correct formula is ##f(r) = 1 - 2M / r - \Lambda r^2 / 3##. (Note the factor of ##3## in the ##\Lambda## term; the Wikipedia article is a little misleading since it fails to include that.)

Equating
That's not how you find the static ##r## coordinate. You find it by looking for the value of ##r## for which the proper acceleration of a worldline with constant ##r## is zero. That gives a value that is close to, but not the same as, the one you derived. See here:

https://www.physicsforums.com/threa...ween-two-spherical-shells.989533/post-6347791

Note that the ##A## in that post is ##\Lambda / 3##.

• timmdeeg
Gold Member
No, one solution with two extra terms in ##f(r)##, i.e., the superposition of the two individual solutions. The correct formula is ##f(r) = 1 - 2M / r - \Lambda r^2 / 3##. (Note the factor of ##3## in the ##\Lambda## term; the Wikipedia article is a little misleading since it fails to include that.)

That's not how you find the static ##r## coordinate. You find it by looking for the value of ##r## for which the proper acceleration of a worldline with constant ##r## is zero. That gives a value that is close to, but not the same as, the one you derived. See here:

https://www.physicsforums.com/threa...ween-two-spherical-shells.989533/post-6347791

Note that the ##A## in that post is ##\Lambda / 3##.
Ok, got it, so zero proper acceleration of this timelike geodesic is the correct criterion.
Then with ##A=\Lambda/3## the correct result is

##r=(3M/\Lambda)^\frac{1}{3}##

Its by a factor 2 different from what results by equating the two extra terms. As always I appreciate your explanation, thanks.

Last edited:
PeterDonis
Mentor
2019 Award
the correct result is
With the exponent fixed (it should be ##\frac{1}{3}##, not ##\frac{1}{1/3}##), yes.

Its by a factor 2 different from what results by equating the two extra terms.
A factor of ##2^\frac{1}{3}##.

• timmdeeg
Gold Member
With the exponent fixed (it should be ##\frac{1}{3}##, not ##\frac{1}{1/3}##), yes.
Indeed, fixed.

A factor of ##2^\frac{1}{3}##.
Yes, my fault.