Peter4075
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 Aug 27 revised Textbook disagreement on geodesic deviation on a 2-sphere added 776 characters in body Aug 27 comment Textbook disagreement on geodesic deviation on a 2-sphere Is my question controversial? It's been tagged as "Homework", but I'm in my sixth decade and haven't formally studied physics for forty years. I'm just curious as to whether I'm on the right track. Thanks. Aug 27 revised Textbook disagreement on geodesic deviation on a 2-sphere edited tags Aug 27 asked Textbook disagreement on geodesic deviation on a 2-sphere Jul 2 awarded Inquisitive Jul 2 awarded Curious Jun 19 awarded Necromancer Jun 17 awarded Yearling Jun 5 comment Minkowski metric and definition of coordinate differentials? Yes, I can see it now. Thanks. Jun 5 comment Minkowski metric and definition of coordinate differentials? Of course! I should have realised. Jun 5 accepted Minkowski metric and definition of coordinate differentials? Jun 5 asked Minkowski metric and definition of coordinate differentials? Mar 27 awarded Popular Question Mar 12 awarded Popular Question Mar 10 awarded Notable Question Jan 28 awarded Popular Question Jan 8 awarded Nice Question Dec 19 accepted Riemann curvature tensor symmetries confusion Dec 19 comment Riemann curvature tensor symmetries confusion @StanLiou: Thanks. Final question. Can the Ricci tensor therefore be defined using other index permutations that don't involve the Riemann tensor having the same 1 and 2 or 3 and 4 indices as the metric, ie $R_{\mu\nu}=g^{\sigma\rho}R_{\sigma\mu\nu\rho}$ , $R_{\mu\nu}=g^{\sigma\rho}R_{\mu\sigma\rho\nu}$ , $R_{\mu\nu}=g^{\sigma\rho}R_{\mu\sigma\nu\rho}$ ? Dec 18 comment Riemann curvature tensor symmetries confusion Oh, showing my huge ignorance, I thought you could just sort of “cancel” any upper and lower index. I didn't know about symmetric/anti-symmetric. Does that mean that $g^{\alpha\beta}R_{\alpha\beta\gamma\mu}=g^{\alpha\beta}R_{\gamma\mu\alpha\beta}‌​=0$ ? Would that also mean (swapping the metric tensor indices) that $g^{\beta\alpha}R_{\alpha\beta\gamma\mu}=g^{\beta\alpha}R_{\gamma\mu\alpha\beta}‌​=0$ ? So, if I avoid the Riemann tensor having the same 1 and 2 or 3 and 4 indices as the metric, can I state the Ricci tensor as (for example) $R_{\mu\nu}=g^{\sigma\rho}R_{\sigma\mu\rho\nu}$ ?