EFE: Why is there a curvature tensor and curvature scalar?

PerpStudent

In the Einstein tensor equation for general relativity, why are there two terms for curvature: specifically the curvature tensor and the curvature scalar multiplied by the metric tensor?

Mentz114

When the Einstein-Hilbert action is extremized wrt the inverse metric, that is what emerges. See here http://en.wikipedia.org/wiki/Einstei...Hilbert_action.

Are you aware that the curvature scalar is the contraction of the Ricci tensor, so R = R^μ_μ.

tiny-tim

Hi PerpStudent!

Because the traceless symmetric part of A_ij is A_ij - 1/4 tr(A) g_ij.

Any tensor equation can be "traced" and "tracelesssed".

ie the trace of the equation is true, and the traceless symmetric part of the equation is true.

So in the Einstein field equations we expect …

R = constant*T

R_ij - 1/4 R g_ij = constant* (T_ij - 1/4 T g_ij)

(and it turns out the constants have to be -8π and 8π, to give the Newtonian inverse-square law in the low-field limit)

dextercioby

There's only one independent / fundamental curvature, namely the Riemann-Christoffel curvature tensor. The so-called Ricci curvature and the curvature scalar are simply contractions of the 4th rank tensor with respect with the metric once and twice, respectively. They are susequently derived concepts.

One can write down the EFE in terms of the Riemann-Christoffel curvature tensor only (in the absence of matter) as:

[tex] g^{\mu \alpha}R_{\mu \nu|\alpha \beta}-\frac{1}{2}g_{\nu \beta}g^{\mu \alpha}g^{\lambda \sigma}R_{\mu \lambda|\alpha \sigma} = 0 [/tex]

but it won't look pretty, that's why the Ricci curvature tensor and the Ricci scalar are put into GR.

Ben Niehoff

The "reason" the particular combination

[tex]G^{\mu\nu} \equiv R^{\mu\nu} - \frac12 R g^{\mu\nu}[/tex]
appears is because is this the unique combination of curvatures that satisfies

[tex]\nabla_\mu G^{\mu\nu} = 0.[/tex]

pervect

To expand on this , Einstein's equation says that G_uv = 8 pi T_uv, where T_uv is the stress-energy tensor.

Continuiity conditions on the stress-energy tensor, T_uv require that
[tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] i.e. that the tensor be divergence free.

So since T_uv, the rhs is divergence free, the lhs has to be divergence free as well.

tiny-tim

pervect, ben, as a matter of interest, do you know any easy-to-understand reason why it's 8π ?

4π i'm more or less used to (and even 4π 10^-7 ) …

but why 8 ?

dextercioby

My 2ç. There's no <physical> reason, the "extra" 2 comes from the 1/2 of the Christoffel symbols which has to do with the metric being assumed torsionless and symmetric.

Matterwave

The constant comes from matching the Einstein Field Equations to the Newtonian equation for gravitation in the low speed, low gravity case (makes sure that General Relativity gives the same predictions in this case as Newtonian gravity).

PerpStudent

Is the requirement that [tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] due to energy and momentum conservation?

tiny-tim

from http://en.wikipedia.org/wiki/Stress–...nservation_law …

The stress–energy tensor is the conserved Noether current associated with spacetime translations.

dextercioby

Yes, but that works only in flat space-time. ∇^μT_μν=0 comes from the EFE and the ∇^μG_μν=0, with the G the Einstein tensor in the LHS of the EFE.

dextercioby

No, see the reason in the my post above this one.

pervect

It's very closely related.

You can think of it as being due to energy and momentum conservation in a local sense, i.e. at a point.

See for instance http://en.wikipedia.org/wiki/Continuity_equation

I'm pretty sure Wald and MTW discuss this with more rigor - I'd have to look stuff up to refresh my recollection to give any real detail, at least if I wanted to avoid misleading anyone.