Static Point in de Sitter-Schwarzschild Spacetime

[timmdeeg]

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 $\Lambda$.

But is this correct? Please advise.

 

[PeterDonis]

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]

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.

 

[PeterDonis]

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]

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.