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Studied physics and maths at school many years ago.


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) $R_{\mu\nu}=g^{\sigma\rho}R_{\sigma\mu\rho\nu}$ ?
Dec
18
asked Riemann curvature tensor symmetries confusion