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## 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?

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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.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?

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

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tiny-tim

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Because the traceless symmetric part of A

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})

R

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

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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.

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Ben Niehoff

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[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]

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

[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]

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.

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tiny-tim

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but why *8* ?

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Matterwave

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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).

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

but why8?

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

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tiny-tim

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

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Yes, but that works only in flat space-time. ∇

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

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

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It's very closely related.Is the requirement that [tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] due to energy and momentum conservation?

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.

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