Monday, 23 May 2016

homework and exercises - Carroll's derivation of the geodesic equations



In Carroll's derivation of the geodesic equations (page 69, http://preposterousuniverse.com/grnotes/grnotes-three.pdf), he starts with τ=(gμνdxμdλdxνdλ)1/2dλ

and arrives atδτ=(gμνdxμdλdxνdλ)1/2(12σgμνdxμdλdxνdλδxσgμνdxμdλd(δxν)dλ)dλ.
He then changes the curve parametrization from arbitrary λ to proper time τ by plugging dλ=(gμνdxμdλdxνdλ)1/2dτ
into the above to obtain


δτ=(12σgμνdxμdτdxνdτδxσgμνdxμdτd(δxν)dτ)dτ.


I cannot see how that substitution works. I've been told it uses the chain rule, but I just can't see it. Can anyone help? Thanks.



Answer



Basically think of it this way. Take the original equation τ=f(x)dλ

which in differential form becomes


dτ=f(x)dλ



after a little rearranging gives


dλdτ = (f(x))1--------(3)


with the function f(x) in this case being equal to


f(x) = (gμνdxμdλdxνdλ)1/2--------(4)


as was demonstrated


EDIT:


Using eq (3)


dλdτ = (f(x))1 = (gμνdxμdλdxνdλ)1/2


Substitute into


δτ= (gμνdxμdλdxνdλ)1/2 (12gμν,σdxμdλdxνdλδxσgμνdxμdλd(δxν)dλ) dλ



gives


δτ= dλdτ (12gμν,σdxμdλdxνdλδxσgμνdxμdλd(δxν)dλ) dλ


δτ= dλdτ (12gμν,σdxμdτdxνdτδxσgμνdxμdτd(δxν)dτ) dτdλdτdλ dλ


Use chain rule to get


δτ= (12gμν,σdxμdτdxνdτδxσgμνdxμdτd(δxν)dτ) dτ


Here I use , to represent the partial derivative with respect to xσ.


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