Recent content by nobraner

Bill, Finally found the time to fix this. One of my biggest weaknesses is ignoring the upper/lower covariant derivative convention. I guess I always think of covariant derivatives as always being covariant.

The full derivation of the covariant derivative of the Ricci Tensor as Einstein did it, is now available on line at https://sites.google.com/site/generalrelativity101/appendix-c-the-covariant-derivative-of-the-ricci-tensor For those who wish to study it.

Thanks for the advice, but anyone can do it that way. I'm trying to do it the way Einstein did it; the hard way. Einstein didn't know about the Bianchi Identities.

Then, how did Einstein get [SIZE="5"]\nabla_{μ}R_{αβ}=\frac{1}{2}g_{αβ}\nabla_{μ}R

[SIZE="5"]I am trying to calculate the covariant derivative of the Ricci Tensor the way Einstein did it, but I keep coming up with \nabla_{μ}R_{αβ}=\frac{∂}{∂x^{μ}}R_{αβ}-2\Gamma^{α}_{μ\gamma}R_{αβ} or...

I don't understand your declaration that \mu \nu are already taken. Does that mean we can never assume that such a metric as g^{\mu\nu} exists without first proving that it is so for the specific case of \nabla^{\mu}R_{\mu\nu}=\frac{1}{4}\nabla^{\mu}g_{\mu\nu}R

Start with \nabla^{\mu}R_{\mu\nu}=\nabla^{\mu}R_{\mu\nu} Insert \nabla^{\mu}R_{\mu\nu}=\nabla^{\mu}\frac{g_{\mu\nu}g^{\mu\nu}}{4}R_{\mu\nu} Contract the Ricci Tensor \nabla^{\mu}R_{\mu\nu} = \nabla^{\mu}\frac{g_{\mu\nu}}{4}R Thus...

If as you say, Professor Susskind is wrong, then I feel betrayed that a physicist of his stature would teach error.

Professor Lenard Susskind explicitly states in his YouTube videos that g^{\mu\nu}g_{\mu\nu}=\delta^{\mu}_{\nu} The product of the covariant and contravariant metric is the kroniker delta (the multiplicative identity matrix). Although, he does say somewhere that if you have \delta^{a}_{\nu}...

Start with \nabla_{μ}R^{\mu\nu}=\nabla_{μ}R^{\mu\nu} insert the multiplicative identity, expressed as the product of the covariant and contravariant metric \nabla_{μ}R^{\mu\nu}=\nabla_{μ}(g^{\mu \nu}g_{\mu\nu})R^{\mu\nu} contract the indices of the Ricci Tensor, to get...