EFE: Why is there a curvature tensor and curvature scalar?by PerpStudent Tags: einstein field equat 

#1
May912, 03:21 PM

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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?




#2
May912, 03:53 PM

PF Gold
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Are you aware that the curvature scalar is the contraction of the Ricci tensor, so R = R^{μ}_{μ}. 



#3
May912, 03:55 PM

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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(and it turns out the constants have to be 8π and 8π, to give the Newtonian inversesquare law in the lowfield limit) 



#4
May912, 07:02 PM

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EFE: Why is there a curvature tensor and curvature scalar?One can write down the EFE in terms of the RiemannChristoffel 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. 



#5
May912, 07:13 PM

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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] 



#6
May1012, 03:46 AM

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Continuiity conditions on the stressenergy 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. 



#7
May1012, 03:45 PM

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pervect, ben, as a matter of interest, do you know any easytounderstand reason why it's 8π ?
4π i'm more or less used to (and even 4π 10^{7} ) … but why 8 ? 



#8
May1012, 06:15 PM

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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.




#9
May1012, 07:30 PM

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#10
May1112, 10:48 AM

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#11
May1112, 12:03 PM

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from http://en.wikipedia.org/wiki/Stress–...nservation_law …
The stress–energy tensor is the conserved Noether current associated with spacetime translations. 



#12
May1112, 01:07 PM

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#13
May1112, 01:10 PM

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#14
May1112, 02:00 PM

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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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