Einstein Tensor; What is wrong here?

In summary, the conversation discusses the insertion of a factor in the equation \nabla^{\mu}R_{\mu\nu}=\nabla^{\mu}\frac{g_{\mu\nu}g^{\mu\nu}}{4}R_{\mu\nu} and the question of whether it is valid to assume the existence of a metric g^{\mu\nu} without first proving it for the specific case. The expert summarizer points out that overloading indices and a double sum in the inserted term raise doubts about the validity of the proof.
  • #1
nobraner
13
0
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?
 
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  • #2
nobraner said:
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.
 
  • #3
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 without 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
nobraner said:
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 without 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.
 
  • #5
nobraner said:
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 without 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]
 

1. What is the Einstein tensor?

The Einstein tensor, also known as the Einstein-Hilbert tensor, is a mathematical object used in the theory of general relativity to describe the curvature of spacetime.

2. Why is the Einstein tensor important?

The Einstein tensor is important because it is used in Einstein's field equations, which relate the curvature of spacetime to the distribution of matter and energy. This allows us to understand how matter and energy affect the curvature of spacetime and vice versa.

3. How is the Einstein tensor calculated?

The Einstein tensor is calculated by taking the Ricci curvature tensor and subtracting half of its trace. This results in a symmetric tensor that describes the curvature of spacetime in a way that is consistent with the principles of general relativity.

4. What does it mean if there is something wrong with the Einstein tensor?

If there is something wrong with the Einstein tensor, it could mean that there is a mistake in the calculations or assumptions made in the theory of general relativity. It could also indicate the presence of new physics that is not accounted for in the current understanding of the universe.

5. What are some potential implications of an error in the Einstein tensor?

An error in the Einstein tensor could have significant implications for our understanding of the universe and our ability to make accurate predictions about its behavior. It could also lead to the development of new theories or modifications to existing ones in order to better explain the observed phenomena.

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