Einstein Tensor; What is wrong here?


by nobraner
Tags: einstein, tensor
nobraner
nobraner is offline
#1
Feb18-12, 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?
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
robphy
robphy is offline
#2
Feb18-12, 09:02 AM
Sci Advisor
HW Helper
PF Gold
robphy's Avatar
P: 4,108
Quote Quote by nobraner View Post
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]

[snip]

What is wrong here?
Your inserted factor should be
[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.
nobraner
nobraner is offline
#3
Feb18-12, 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

elfmotat
elfmotat is offline
#4
Feb18-12, 01:21 PM
elfmotat's Avatar
P: 260

Einstein Tensor; What is wrong here?


Quote Quote by nobraner View Post
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
You're overloading your indices. An index shouldn't appear more than twice in any term.
juanrga
juanrga is offline
#5
Feb18-12, 02:46 PM
P: 476
Quote Quote by nobraner View Post
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
There is a double sum in the term that you inserted. Therefore you cannot contract the Ricci leaving out the g[itex]_{\mu\nu}[/itex]


Register to reply

Related Discussions
Einstein Tensor; super simple derivation; where did I go wrong? Special & General Relativity 8
General Tensor contraction: Trace of Energy-Momentum Tensor (Einstein metric) Special & General Relativity 5
Einstein Tensor... Special & General Relativity 15
Einstein Tensor Special & General Relativity 5
Einstein Tensor Calculus & Beyond Homework 12