Birkhoff's Theorem: Schwarzschild Metric Unique Solution?

[La Guinee]

Hi. Birkhoff's Theorem says that the Schwarzschild metric is the unique spherically symmetric vacuum solution. Isn't the Robertson-Walker metric spherically symmetric?

 

[atyy]

The Schwarzschild solution is a vacuum solution. The Robertson-Walker metric may include matter.

 

[La Guinee]

Why can't the Robertson Walker metric be a vacuum solution? The Ricci tensor for the Robertson Walker metric can be easily calculated. Choosing a linear scale factor with appropriate coefficients satisfies R_mu nu = 0 (or am I doing something wrong)?

 

[MeJennifer]

The Schwarzschild solution is a vacuum solution. The Robertson-Walker metric may include matter.

May? The FRW solution contains only matter there is no vacuum in this solution.

Interesting things to consider in the light of this fact:
- Why does the majority of cosmologists consider the FRW solution a rough aproximation of our universe if the FRW has no vacuum.
- How can they possibly make statements about the distance between objects if the FRW solution has no distance between objects.

Beware, it might be similar to asking a catholic priest too many questions about hell, he may answer it is the place whe you go if you keep asking those questions.

 

[La Guinee]

Sorry, I still don't understand. You have the Robertson Walker metric. From this you can construct the Ricci tensor. Now set the components of the Ricci tensor equal to 0. This gives you a condition on the scale factor, namely that it's linear in time. What's wrong with this?

 

[atyy]

Beware, it might be similar to asking a catholic priest too many questions about hell, he may answer it is the place whe you go if you keep asking those questions.

So if I keep asking the "majority of cosmologists" too many questions, where do I end up - the singularity?

 

[atyy]

Sorry, I still don't understand. You have the Robertson Walker metric. From this you can construct the Ricci tensor. Now set the components of the Ricci tensor equal to 0. This gives you a condition on the scale factor, namely that it's linear in time. What's wrong with this?

Is there any way to change coordinates to get it into the Schwarzschild form?

 

[Jonathan Scott]

Is there any way to change coordinates to get it into the Schwarzschild form?

I may have missed something here, but as far as I know:

The RW metric describes a whole universe with the same space-time curvature and energy density everywhere, such as a hypersphere.

The Schwarzschild solution describes the shape of space-time around a single central mass where mass is assumed to be distributed in a spherically symmetrical way about the central point and space-time is assumed to be flat at sufficient distance from that mass.

I find it difficult to see anything in common between these two cases.

 

[Ich]

La Guinee,

you have an interesting point here.
FRW and Schwarzschild are (in the empty case) in fact the same metric, expressed in different coordinates.
http://arxiv.org/ftp/astro-ph/papers/0602/0602102.pdf" addresses the issue.
My interpretation: You can arrange receding test particles in flat spacetime such, that the overall picture becomes isotropic if you associate an appropriate boost with each space translation, fitting your velocity to that of the test particle at the new location.
FRW space is a slice through Minkowski space that makes the distribution of test particles additionally homogeneous, as necessary for cosmologigal models.

 

[George Jones]

Why can't the Robertson Walker metric be a vacuum solution? The Ricci tensor for the Robertson Walker metric can be easily calculated. Choosing a linear scale factor with appropriate coefficients satisfies R_mu nu = 0 (or am I doing something wrong)?

Have you calculated the Riemann tensor in this case? I think

https://www.physicsforums.com/showthread.php?t=234224

is relevant.

 

[La Guinee]

Thank you for the replies. I don't see how the FRW can reduce to the schwarzschild metric with a suitable coordinate change because the FRW is time dependent.

 

[George Jones]

Thank you for the replies. I don't see how the FRW can reduce to the schwarzschild metric with a suitable coordinate change because the FRW is time dependent.

Careful. Massless FRW reduces to a patch of flat Minkowski spacetime in unusual coordinates. Minkowski spacetime is isotropic.

 

[La Guinee]

Careful. Massless FRW reduces to a patch of flat Minkowski spacetime in unusual coordinates. Minkowski spacetime is isotropic.

Consider the following FRW metric:

ds² = -dt² + 9t² [ dr² / (1+9r²) + r² dOmega² ]

This satisfies einsteins equations in vacuum. So are you're saying under suitable coordinate change this reduces to Minkowski? This seems weird because one is time dependent and the other isn't.

 

[George Jones]

Consider the following FRW metric:

ds² = -dt² + 9t² [ dr² / (1+9r²) + r² dOmega² ]

This satisfies einsteins equations in vacuum. So are you're saying under suitable coordinate change this reduces to Minkowski? This seems weird because one is time dependent and the other isn't.

Look at post #30 in the thread to which I previously gave a link.

 

[La Guinee]

I see. Thank you.

 

[atyy]

Birkhoff and Minkowski
http://scienceworld.wolfram.com/physics/BirkhoffsTheorem.html

Burko et al
http://arxiv.org/abs/gr-qc/0008065
"The spacetime is Schwarzschild outside the shell and Minkowski inside the shell. Note that g_tt → −1 as r → ∞, but g_tt != −1 inside the shell, although spacetime is (locally) Minkowski."

Edit: Heuristically, it seems the Minkowski metric is obtained by setting both the Schwarzschild radial coordinate and mass parameter to zero. I have no idea if this can be rigorously justified.

Edit: I can just set the mass parameter to zero without touching the radial coordinate.