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visits member for 4 years, 2 months
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Studied physics and maths at school many years ago.


Aug
29
comment Derivation of geodesic deviation equation from two neighbouring geodesics
Thanks. I would never have seen that.
Aug
28
comment Derivation of geodesic deviation equation from two neighbouring geodesics
Thanks, but I still don't see why something multiplied by the derivative of that something should be second order. How do we know that $\frac{d\xi^{c}}{du}$ is a small number? I'm assuming second order in this derivation means something squared, not a second order differential equation.
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
2
comment How are these two Riemann tensor equations equivalent?
Yes, I see it now. Thanks.
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.
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
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 $d\lambda$?
May
13
comment Why two different Lagrangians to derive geodesic equations?
Any chance of a hint as to what that chain rule looks like? I cannot see how he substitutes (3.52) into (3.51) to get the equation at the bottom of page 69. Thank you for your patience.
May
13
comment Why two different Lagrangians to derive geodesic equations?
Thanks, but I just cannot see how he gets $d\tau$ as the denominator after making the affine parameter substitution (Equation 3.52).
May
13
comment Why two different Lagrangians to derive geodesic equations?
Thanks. I will try to derive the non-affine geodesic equations you give. In the meantime do you have a link to this derivation? I'm still puzzled as to how Moore starts with the square root Lagrangian and ends up with the affine geodesic equations.
May
13
comment Why two different Lagrangians to derive geodesic equations?
@Qmechanic - thanks. How come Valter Moretti's derivation here (which is over my head) starts with the square root Lagrangian and ends with (what I assume is) the affinely parametrized geodesic equations, ie my first equation?
May
10
comment Differentiating the Lagrangian to find geodesic equations?
Yes, the Kronecker delta renames the index. I see it now. Thanks.
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.