A Riemann Tensor Formula in Terms of Metric & Derivatives

Jogging-Joe
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Could someone please write out or post a link to the Riemann Tensor written out solely in terms of the metric and its first and second derivatives--i.e. with the Christoffel symbol gammas and their first derivatives not explicitly appearing in the formula.
Could someone please write out or post a link to the Riemann Tensor written out solely in terms of the metric and its first and second derivatives--i.e. with the Christoffel symbol gammas and their first derivatives not explicitly appearing in the formula.

Thanks.
 
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This link has both the Riemann tensor and the Christoffel symbols. I am not inclined to retype everything just to substitute one into the other. Where are you headed with this? Even if you could find such an expression prechurned on the internet, it is one of those things I would not trust to be error free.
 
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Thanks.

I know that a person can get the Riemann Tensor in the form I want by substituting the metric and its first derivatives into the Christofell gammas in the standard Riemann Tensor formula.. As you noted, it is easy to make errors and that is why I would not trust me doing it, and would want a second opinion calculation. Someone must have put it in that form, somewhere.
 
Jogging-Joe said:
Thanks.

I know that a person can get the Riemann Tensor in the form I want by substituting the metric and its first derivatives into the Christofell gammas in the standard Riemann Tensor formula.. As you noted, it is easy to make errors and that is why I would not trust me doing it, and would want a second opinion calculation. Someone must have put it in that form, somewhere.
But why do you want it in that form?
 
I can show that Birkhoff's Theorem is wrong, and I think I might be able to prove it in a second independent way, which would require me to use that particular form of the Riemann Tensor.
 
Jogging-Joe said:
I can show that Birkhoff's Theorem is wrong

No, you can't. You may think you can, but you've made an error somewhere.

Jogging-Joe said:
I think I might be able to prove it in a second independent way, which would require me to use that particular form of the Riemann Tensor.

Sorry, we don't help people with personal theories on this site, particularly personal theories which we already know are wrong (see above).

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I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
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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