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

## Main Question or Discussion Point

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?

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

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

tiny-tim
Homework Helper
Hi PerpStudent!

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

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

Rij - 1/4 R gij = constant* (Tij - 1/4 T gij)​

(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
Homework Helper
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?
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:

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

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

Ben Niehoff
Gold Member
The "reason" the particular combination

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

$$\nabla_\mu G^{\mu\nu} = 0.$$

pervect
Staff Emeritus
The "reason" the particular combination

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

$$\nabla_\mu G^{\mu\nu} = 0.$$
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
$$\nabla_\mu T^{\mu\nu} = 0.$$ 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.

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tiny-tim
Homework Helper
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
Homework Helper
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
Gold Member
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 ?
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).

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

dextercioby
Homework Helper
dextercioby
Homework Helper
Is the requirement that $$\nabla_\mu T^{\mu\nu} = 0.$$ due to energy and momentum conservation?
No, see the reason in the my post above this one.

pervect
Staff Emeritus
Is the requirement that $$\nabla_\mu T^{\mu\nu} = 0.$$ due to energy and momentum conservation?