Sunday, September 17, 2017

general relativity - Form of the geodesic equation


I have been studying the basics of general relativity with Hartle's Gravity. He presents the geodesic equation as


d2xμds2=Γμαβdxαdsdxβds


However, in reading Padmanabhan's Gravitation, he says that the equation


0=kααkβ


is the geodesic equation for a wave vector k defined as kα=αψ, where ψ is just a scalar function.


How do these two definitions of the geodesic equation represent the same thing? They do not look at all alike. In fact, if I try to work it out, I get


0=kα(kβxαΓδβγkδ)


0=kαkβxαΓδβγkδkα



For one thing, the xα is not supposed to be on the bottom of the derivative! If kα is supposed to satisfy the geodesic equation, I expected this to look like the first equation I wrote.



Answer



In short, the first equation you wrote is in component form where curves are parametrized as xμ(s), while the second equation is more general. Now let's flesh out the details.


What does it mean for a curve to be a geodesic? Intuitively it has to be straightest curve possible in curved spacetime. The way you do that is to propagate the tangent vector Tα of a curve C along itself! That yields the coordinate independent geodesic equation: TααTβ=0


Then you can write this tensor equation in component form, by choosing a coordinate system and its associated Christoffel symbol. The covariant derivative is expressed as: αTβ=αTβ+ΓβαγTγ

Now if you plug this equation into the coordinate independent geodesic equation, you get: TααTβ+TαΓβαγTγ=0
Now if you parametrize the curve C as xμ(s), then the tangent vector field becomes:Tμ=dxμds, and Tαα represents the derivative operator dds. Therefore, the previous equation turns into: ddsdxβds+Γβαγdxαdsdxγds=0
d2xβds2+Γβαγdxαdsdxγds=0
Which is precisely the equation that you started with.


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