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visits member for 3 years, 11 months
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Studied physics and maths at school many years ago.


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 $d\lambda$?
May
14
asked Carroll's derivation of the geodesic equations
May
14
accepted Why two different Lagrangians to derive geodesic equations?
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
12
asked Why two different Lagrangians to derive geodesic equations?
May
10
comment Differentiating the Lagrangian to find geodesic equations?
Yes, the Kronecker delta renames the index. I see it now. Thanks.
May
10
accepted Differentiating the Lagrangian to find geodesic equations?
May
10
asked Differentiating the Lagrangian to find geodesic equations?
Mar
4
accepted How to tell if a star is in a galaxy?
Mar
4
asked How to tell if a star is in a galaxy?
Dec
15
awarded  Popular Question
Dec
9
awarded  Popular Question