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

by PerpStudent
Tags: einstein field equat
 P: 27 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?
PF Gold
P: 4,087
 Quote by 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?
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 = Rμμ.
 Sci Advisor HW Helper Thanks P: 26,148 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)
HW Helper
P: 11,947
EFE: Why is there a curvature tensor and curvature scalar?

 Quote by 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?
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.
 Sci Advisor P: 1,594 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 $$\nabla_\mu G^{\mu\nu} = 0.$$
Emeritus
P: 7,659
 Quote by Ben Niehoff 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.
 Sci Advisor HW Helper Thanks P: 26,148 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 ?
 Sci Advisor HW Helper P: 11,947 My 2ç. There's no 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.
P: 2,950
 Quote by 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 ?
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).
P: 27
 Quote by 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 $$\nabla_\mu T^{\mu\nu} = 0.$$ 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?
 Sci Advisor HW Helper Thanks P: 26,148 from http://en.wikipedia.org/wiki/Stress–...nservation_law … The stress–energy tensor is the conserved Noether current associated with spacetime translations.
HW Helper
P: 11,947
 Quote by tiny-tim from http://en.wikipedia.org/wiki/Stress–...nservation_law … The stress–energy tensor is the conserved Noether current associated with spacetime translations.
Yes, but that works only in flat space-time. ∇μTμν=0 comes from the EFE and the ∇μGμν=0, with the G the Einstein tensor in the LHS of the EFE.
HW Helper
P: 11,947
 Quote by PerpStudent 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.
Emeritus
 Quote by PerpStudent Is the requirement that $$\nabla_\mu T^{\mu\nu} = 0.$$ due to energy and momentum conservation?