
#1
Feb1812, 08:11 AM

P: 14

Start with
[itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex]=[itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex] Insert [itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex]=[itex]\nabla^{\mu}[/itex][itex]\frac{g_{\mu\nu}g^{\mu\nu}}{4}[/itex]R[itex]_{\mu\nu}[/itex] Contract the Ricci Tensor [itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex] = [itex]\nabla^{\mu}[/itex][itex]\frac{g_{\mu\nu}}{4}[/itex]R Thus [itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex]=[itex]\frac{1}{4}[/itex][itex]\nabla^{\mu}[/itex][itex]{g_{\mu\nu}}[/itex]R But General Relativity says [itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex]=[itex]\frac{1}{2}[/itex][itex]\nabla^{\mu}[/itex][itex]{g_{\mu\nu}}[/itex]R What is wrong here? 



#2
Feb1812, 09:02 AM

Sci Advisor
HW Helper
PF Gold
P: 4,108

[itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex]=[itex]\nabla^{\mu}[/itex][itex]\frac{g_{\sigma\tau}g^{\sigma\tau}}{4}[/itex]R[itex]_{\mu\nu}[/itex] (since [itex]\mu[/itex] and [itex]\nu[/itex] are already "taken"). Note that since your proposed proof makes no use of the unique properties of Ricci, it would seem that your result would work for any symmetric tensor. So, you must look at it with suspicion. 



#3
Feb1812, 10:58 AM

P: 14

I don't understand your declaration that [itex]\mu \nu[/itex] are already taken. Does that mean we can never assume that such a metric as
g[itex]^{\mu\nu}[/itex] exists with out first proving that it is so for the specific case of [itex]\nabla^{\mu}[/itex]R[itex]_{\mu\nu}[/itex]=[itex]\frac{1}{4}[/itex][itex]\nabla^{\mu}[/itex]g[itex]_{\mu\nu}[/itex]R 



#4
Feb1812, 01:21 PM

P: 260

Einstein Tensor; What is wrong here? 



#5
Feb1812, 02:46 PM

P: 476




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