A Non static and isotropic solution for Einstein Field Eq

[Leonardo Machado]

Hello dear friends, today's question is:

In a non static and spherically simetric solution for Einstein field equation, will i get a non diagonal term on Ricci tensor ? A R[r][/t] term ?

I'm getting it, but not sure if it is right.

Thanks.

 

[PAllen]

Note that spherically symmetric is not the same as isotropic. Your title says one, your post says another. A spherically symmetric solution is anisotropic.

Isoptropic, non-static solutions include all the FLRW solutions except where cosmological constant exactly balances the expansion.

The spherically symmetric non-static vacuum solution is unique (without cosmological constant, at least). It is the interior of an eternal Schwarzschild BH. The Ricci tensor is identically zero because it is vacuum. For non-vacuum solutions, you will, indeed get an (r,t) Ricci term (given normal meaning of such coordinates). This should be the only non diagonal term [ noting (r,t) obviously = (t,r) term].

 

[PeterDonis]

A spherically symmetric solution is anisotropic.

This is not quite correct. FRW spacetime is spherically symmetric and also isotropic. The key is that FRW spacetime is spherically symmetric about every point. A spacetime like Schwarzschild spacetime, which is not isotropic, is not.

More technically: "spherically symmetric" means there is a 3-parameter family of spacelike Killing vector fields with the commutation relations of SO(3). But that does not preclude there being more than one such family. In FRW spacetime, there is an infinite number of such families (one centered on each spatial point--and to be really technical, there is such an infinite family in each spacelike hypersurface, instead of just one as there is in Schwarzschild spacetime).

 

[PeterDonis]

I'm getting it, but not sure if it is right.

Can you post your actual math?

 

[PAllen]

Can you post your actual math?

Is there some reason you want this? For non-vacuum case, the existence of this term is well known.

 

[PeterDonis]

Is there some reason you want this?

I'm interested in how the "non-static" assumption is formulated.

 

[PAllen]

I'm interested in how the "non-static" assumption is formulated.

You just use a metric ansatz with angular coordinates, and two unknown functions of r and t. Different formulations of the ansatz lead to different styles of coordinates. In all such set ups, you find the Ricci tensor is diagonal except for the r,t components. One reference for this is chapter 7, section 3, of Synge's book.

 

[PeterDonis]

You just use a metric ansatz with angular coordinates, and two unknown functions of r and t.

Are you referring to equations (70) and (71) of Synge? The first is a general ansatz with three unknown functions, and the second gives different specializations that determine one of the functions in terms of the other two.

 

[vanhees71]

Yes, that's the way Birkhoff's theorem is proven, i.e., you make a ansatz for a spherically symmetric solution of the free Einstein equation. Then you'll get out that in fact the metric components are necessarily time-independent, i.e., any spherically symmetric solution of the vacuum Einstein equation is static.

 

[PAllen]

Are you referring to equations (70) and (71) of Synge? The first is a general ansatz with three unknown functions, and the second gives different specializations that determine one of the functions in terms of the other two.

Yes. Or, as is done earlier in that chapter, simpler forms of the ansatz for different coordinate styles are given. Those equations (70, 71) use a most general form to derive universal results.

 

[PAllen]

Yes, that's the way Birkhoff's theorem is proven, i.e., you make a ansatz for a spherically symmetric solution of the free Einstein equation. Then you'll get out that in fact the metric components are necessarily time-independent, i.e., any spherically symmetric solution of the vacuum Einstein equation is static.

Note, Birkhoff really states the existence of an extra killing field. Only if it is time like do you then get staticity. Thus, the BH exterior is static, while the interior is not because dt is space like in the interior, so you have an axial extra killing field.

 

[Leonardo Machado]

Is there some reason you want this? For non-vacuum case, the existence of this term is well known.

I'm coursing general relativity for the first time, first contact with these strange spaces.