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location United Kingdom
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visits member for 2 years, 10 months
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Manual worker. Live in England. Formal scientific education "peaked" many years ago when I failed maths and chemistry A-level and scraped a bare pass in physics. Still enjoy pottering about in maths and physics.


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
Nov
7
awarded  Popular Question
Sep
18
revised Getting started general relativity
additional information
Sep
10
awarded  Informed
Sep
9
comment Index raising and lowering - how does it work?
I've no idea what this means.
Sep
9
accepted Index raising and lowering - how does it work?
Sep
9
comment Index raising and lowering - how does it work?
I'm afraid I don't understand where the “integrability condition” comes from. However, I can live with that as (I think) I get the gist that all gradients are covectors but not all covectors are gradients, and therefore $g_{\mu\nu}V^{\nu}=\partial_{\mu}\phi$ is only sometimes true. How about going the other way? Does $g_{\mu\nu}V^{\nu}$ always give a tangent vector to a parametrised curve, and how would you show that? Apologies if you've already shown this and I've just got lost in the maths.
Sep
9
asked Index raising and lowering - how does it work?
Aug
28
awarded  Popular Question
Aug
5
answered Understanding Tensors
Jul
13
accepted Derivation of freely falling frame in Schwarzschild spacetime
Jul
13
comment Derivation of freely falling frame in Schwarzschild spacetime
“The transformation to $\bar{r}$ then allows to write the metric directly in the standard form, in a mathematically rigorous way.” Yes, I can see that now. Is there a specific name for the above transformation? One more thing. I confess I could look at $r=\bar{r}\left(1+\frac{r_{\text{s}}}{4\bar{r}}\right)^{2}$ for many years and not think, “ah yes, that describes a local, freely falling frame”. Is there an obvious way of showing that the radial coordinate $\bar{r}$ describes such a frame? Thank you for your patience.