No. of field equations and components or Riemann tensor?

damnedcat
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no. of field equations and components or Riemann tensor??

Someone was trying to explain to me about curvature in space. From what I got from what they were saying doesn't make sense to me. I'm not sure I understand what the number of components, N, of R\alpha,\beta,\gamma,\delta when compared to
the number of field equations, M, implies about curvature (when i say compare i mean: N>M, N<M, N=M). I thought looking at just the Riemann tensor tells u about curvature. mYBE i Just didn't get what he was saying. Any help?
 
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Was your friend referring to something like this? http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html See section 8.1.2, first para.

-Ben
 
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Something like that. He was saying that certain property can only occur for space times of more than 3 spatial dimensions. Since for instance the einstein eqn for vacuum in 1,2 and 3 dimensions have the number of field equations being less than or equal to the number of components of the Riemann tensor. I think he was talking about curvature of the space.
 


Sorry, but can you clarify by what you mean by alpha/beta/gamma/delta ?
 


nicksauce said:
Sorry, but can you clarify by what you mean by alpha/beta/gamma/delta ?

I think he means the indicies of the Riemann tensor?
 


damnedcat said:
Something like that. He was saying that certain property can only occur for space times of more than 3 spatial dimensions. Since for instance the einstein eqn for vacuum in 1,2 and 3 dimensions have the number of field equations being less than or equal to the number of components of the Riemann tensor. I think he was talking about curvature of the space.

I think this must be somewhat garbled. In any number of dimensions, the Riemann tensor, which has 4 indices, has at least as many independent components as the Ricci tensor, which has two indices and is constructed from it. The Einstein tensor has the same number of components as the Ricci tensor. The field equations have the Einstein tensor on one side. So the number of field equations will always be equal to the number of components in the Ricci and Einstein tensors, and less than or equal to the number of components in the Riemann tensor.
 
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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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