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## Main Question or Discussion Point

Hi,

I've heard it said that for Schwarzschild spacetimes, then the coordinate 'r' is an affine parameter for radial null geodesics in the region exterior to the horizon. It seems weird to me that one of the coords is an affine parameter.

How can this be seen? I know that the radial null geodesics satisfy [tex] \tfrac{dt}{dr}=\pm \tfrac{1}{1-\tfrac{2M}{r} } [/tex], but what does this have to do with r being affine?

As far as I'm aware affine simply means related to the proper time linearly, i.e. [tex] \lambda=a\tau+b [/tex], and for any parameter related this way the geodesic equation (zero on the RHS) will be satisfied.

The only thing I could think of would be to take my [tex] \tfrac{dt}{dr}=\pm \tfrac{1}{1-\tfrac{2M}{r} } [/tex] and differentiate again, to get [tex] \tfrac{d^2t}{dr^2}=\mp\tfrac{2M}{(2M-r)^2 } [/tex]. Then look at the LHS of the geodesic equation [tex] \tfrac{d^2t}{dr^2}+\Gamma^{t}_{\mu\nu}\tfrac{dx^{\mu}}{dr}\tfrac{dx^{\nu}}{dr} [/tex] and calculate the Christoffel symbols and show this equation equates to zero.

For radial geo's this equation reduces to, [tex] \tfrac{d^2t}{dr^2}=\mp\tfrac{2M}{(2M-r)^2 } [/tex]. Then look at the LHS of the geodesic equation [tex] \tfrac{d^2t}{dr^2}+\Gamma^{t}_{tt}\left(\tfrac{dt}{dr}\right)^2+\Gamma^{t}_{rt}\left(\tfrac{dt}{dr}\right) +\Gamma^{t}_{tr}\left(\tfrac{dt}{dr}\right)+\Gamma^{t}_{rr}[/tex] which follows from dr/dr=1.

I find that [tex] \Gamma^{t}_{t r}=\tfrac{m}{((2m-r) r)} [/tex] and [tex] \Gamma^{t}_{r t}=-\tfrac{m}{((2m-r) r)} [/tex] and the other two zero. So LHS doesn't vanish this way, and there goes that idea...

My other thought is possibley to consider the magnitude of the tangent vector, for null geo's this should be zero?

I've heard it said that for Schwarzschild spacetimes, then the coordinate 'r' is an affine parameter for radial null geodesics in the region exterior to the horizon. It seems weird to me that one of the coords is an affine parameter.

How can this be seen? I know that the radial null geodesics satisfy [tex] \tfrac{dt}{dr}=\pm \tfrac{1}{1-\tfrac{2M}{r} } [/tex], but what does this have to do with r being affine?

As far as I'm aware affine simply means related to the proper time linearly, i.e. [tex] \lambda=a\tau+b [/tex], and for any parameter related this way the geodesic equation (zero on the RHS) will be satisfied.

The only thing I could think of would be to take my [tex] \tfrac{dt}{dr}=\pm \tfrac{1}{1-\tfrac{2M}{r} } [/tex] and differentiate again, to get [tex] \tfrac{d^2t}{dr^2}=\mp\tfrac{2M}{(2M-r)^2 } [/tex]. Then look at the LHS of the geodesic equation [tex] \tfrac{d^2t}{dr^2}+\Gamma^{t}_{\mu\nu}\tfrac{dx^{\mu}}{dr}\tfrac{dx^{\nu}}{dr} [/tex] and calculate the Christoffel symbols and show this equation equates to zero.

For radial geo's this equation reduces to, [tex] \tfrac{d^2t}{dr^2}=\mp\tfrac{2M}{(2M-r)^2 } [/tex]. Then look at the LHS of the geodesic equation [tex] \tfrac{d^2t}{dr^2}+\Gamma^{t}_{tt}\left(\tfrac{dt}{dr}\right)^2+\Gamma^{t}_{rt}\left(\tfrac{dt}{dr}\right) +\Gamma^{t}_{tr}\left(\tfrac{dt}{dr}\right)+\Gamma^{t}_{rr}[/tex] which follows from dr/dr=1.

I find that [tex] \Gamma^{t}_{t r}=\tfrac{m}{((2m-r) r)} [/tex] and [tex] \Gamma^{t}_{r t}=-\tfrac{m}{((2m-r) r)} [/tex] and the other two zero. So LHS doesn't vanish this way, and there goes that idea...

My other thought is possibley to consider the magnitude of the tangent vector, for null geo's this should be zero?