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Studied physics and maths at school many years ago.


Aug
29
accepted Derivation of geodesic deviation equation from two neighbouring geodesics
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.
Aug
28
asked Derivation of geodesic deviation equation from two neighbouring geodesics
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 $d\tau$.