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location United Kingdom
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visits member for 3 years, 6 months
seen Dec 7 at 18:05

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.


2d
awarded  Popular Question
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.