Peter4075
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 Dec 9 awarded Popular Question Nov 22 comment Derivation of the Riemann tensor confusion After spending ages carefully typing it out in LyX, I couldn't resist showing it off here! Nov 22 comment Derivation of the Riemann tensor confusion I've just found this for the covariant derivative for a general tensor:$\nabla_{k}T_{j_{1}\ldots j_{s}}^{i_{1}\ldots i_{r}}=\partial_{k}T_{j_{1}\ldots j_{s}}^{i_{1}\ldots i_{r}}+\Gamma_{lk}^{i_{1}}T_{j_{1}\ldots j_{s}}^{li_{2}\ldots i_{r}}+\ldots+\Gamma_{lk}^{i_{r}}T_{j_{1}\ldots j_{s}}^{i_{1}\ldots i_{r-1}l}-\Gamma_{j_{1}k}^{l}T_{lj_{2}\ldots j_{s}}^{i_{1}\ldots i_{r}}\ldots-\Gamma_{j_{s}k}^{l}T_{j_{1}\ldots j_{s-1}l}^{i_{1}\ldots i_{r}}.$ Nov 22 comment Derivation of the Riemann tensor confusion Carroll gives it on p58 here: arxiv.org/pdf/gr-qc/9712019.pdf. I notice that the upper index on the negative Gammas and the lower indices on the positive Gammas are constant, whilst the lower indices on the negative Gammas and the upper index on the positive Gammas are not. Nov 22 accepted Derivation of the Riemann tensor confusion Nov 21 comment Derivation of the Riemann tensor confusion OK. It was that particular detailed equation for a general tensor I was interested in, but I'll take a look around. Nov 21 comment Derivation of the Riemann tensor confusion Thanks. Does $T^{d...a_{n}}$ mean $T^{da_{2}...a_{n}}$ etc? Do you have a reference for the derivation of this equation? Nov 21 comment Derivation of the Riemann tensor confusion Right. So the mechanism is that I start with a four component vector $\vec{V}$. Take the covariant derivative of this using $\frac{\partial V^{\alpha}}{\partial x^{\beta}}+V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}$. Summing over the repeated $\gamma$ index then gives me a $4\times4$ matrix aka a rank-2 tensor $V_{;\beta}^{\alpha}$. Is that correct? Nov 21 comment Derivation of the Riemann tensor confusion The semi-colon threw me. Nov 21 comment Derivation of the Riemann tensor confusion Oh. I didn't realise you treat $\lambda_{a;b}$ as having two lower indices. That does tend to put things in a new light. Nov 21 asked Derivation of the Riemann tensor confusion Nov 17 accepted Is this covariant derivative identity true? Nov 17 comment Is this covariant derivative identity true? Excellent answer. I can see it now. Thanks. Nov 17 comment Is this covariant derivative identity true? @ValterMoretti Any hints as to how I can show it's true? Nov 17 asked Is this covariant derivative identity true? Nov 4 awarded Popular Question Oct 26 awarded Benefactor Oct 24 accepted Geodesic deviation equation - why does the ordinary second derivative give the correct answer? Oct 20 comment Geodesic deviation equation - why does the ordinary second derivative give the correct answer? So, can I assume that as only one of the coordinate curves (hope that's the correct term) is a geodesic (ie the $\phi=constant$ one), then the answer has nothing to do with Riemann normal coordinates? Also, I didn't realise that if $\frac{D\zeta^{\mu}}{d\lambda}=0$ then $\frac{D\zeta_{\mu}}{d\lambda} =0$. I've seen a similar problem to this online but with one of the particles moving along the equator ($\theta =\pi/2$), meaning the connection coefficients “naturally” disappear and the absolute 2nd derivative equals the ordinary 2nd derivative, which makes things a lot simpler. Oct 19 revised Geodesic deviation equation - why does the ordinary second derivative give the correct answer? added 159 characters in body