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825
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location United Kingdom
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visits member for 4 years, 1 month
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Studied physics and maths at school many years ago.


Jul
19
awarded  Disciplined
Jun
25
awarded  Popular Question
Jun
24
awarded  Enthusiast
Jun
17
awarded  Yearling
Jun
16
comment Nabla or semicolon notation for covariant derivative?
I guess saying it's just a matter of opinion is an answer of sorts.
Jun
16
asked Nabla or semicolon notation for covariant derivative?
Jun
6
awarded  Nice Question
Jun
2
comment How are these two Riemann tensor equations equivalent?
Yes, I see it now. Thanks.
Jun
2
accepted How are these two Riemann tensor equations equivalent?
Jun
2
comment How are these two Riemann tensor equations equivalent?
OK, I multiply by $g^{ax}$ (?) to give$$\lambda_{;bc}^{x}-\lambda_{;cb}^{x}=R_{\phantom{\mu}bc}^{dx}\lambda_{d}.$$ Then what?
Jun
2
comment How are these two Riemann tensor equations equivalent?
OK, so what metric do I multiply by? If I try $g^{ax}$ it does weird things to the rhs (ie I have two upper indices in the Riemann tensor). And I'm still left with $\lambda^{a}$.
Jun
2
comment How are these two Riemann tensor equations equivalent?
So I multiply by $g^{ad}$ to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}=R_{\phantom{\mu}abc}^{d}\lambda^{a}.$$ Then use the Riemann tensor symmetries to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}= -R_{\phantom{\mu}dbc}^{a}\lambda^{a}.$$ But it's still not the same as Poisson's equation.
Jun
2
revised How are these two Riemann tensor equations equivalent?
edited tags
Jun
2
asked How are these two Riemann tensor equations equivalent?
May
15
comment Why two different Lagrangians to derive geodesic equations?
I couldn't figure it out so I did ask a separate question, here physics.stackexchange.com/questions/183970/…. It was answered very nicely by Horus. The penny has now dropped. Thanks.
May
14
comment Carroll's derivation of the geodesic equations
@tok3rat0r - hasn't Horus multiplied through twice by $\frac{d\lambda}{d\tau}\frac{d\tau}{d\lambda}$? That gives the $\frac{d\lambda}{d\tau}\frac{d\lambda}{d\tau}d\lambda$ term at the end, which cancels to give $d\tau$.
May
14
accepted Carroll's derivation of the geodesic equations
May
14
comment Carroll's derivation of the geodesic equations
Well, I think I get it now. Very well explained. Thanks.
May
14
comment Carroll's derivation of the geodesic equations
Although my question has been flagged as being asked before and already having an answer, it wasn't answered at the sub-graduate level I could understand. Horus's answer, on the other hand, I could follow, and many thanks to him for persevering.
May
14
comment Carroll's derivation of the geodesic equations
Sorry, I realise I'm missing the blindingly obvious here (and being close voted to oblivion), but I can't still can't see how all those denominator $d\lambda$s become $d\tau$s. Do I substitute $d\lambda=\frac{d\lambda}{d\tau}d\tau$ for every single $d\lambda$?