Affine parameter Schwazschild

  • Thread starter LAHLH
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  • #1
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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?
 

Answers and Replies

  • #2
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In this case proper time is not a well defined concept, since you are considering a null curve. I cannot give you a formal definition, but I think of an affine parameter as something that parametrizes the curve under consideration in a good way - i.e. every point on the curve is in one-to-one correspondence with a single value of the parameter. Hope this helps..
 
  • #3
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If you compute it carefully you'll get:
[tex] \frac{d^2 t}{d r^2} = \frac{\frac{2m}{r^2}}{\left( 1-\frac{2m}{r} \right)^2}[/tex]
[tex] \Gamma^{t}_{tt} =0[/tex],
[tex] \Gamma^{t}_{rt} = \Gamma^{t}_{tr} =
\frac{\frac{m}{r^2}}{\left( 1-\frac{2m}{r} \right)}[/tex],
[tex] \Gamma^{t}_{rr} =0[/tex]
so we have for [tex] \frac{d t}{d r} = - \frac{1}{1-\frac{2m}{r}} [/tex]:
[tex]
\frac{d^2 t}{d r^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} = 0
[/tex]
 
  • #4
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Oh, I must have dropped a sign (should have realised from torsion free condition anyway), so is this sufficient to say r is affine? I mean we also would have to show the r comp of geodesic is zero too I guess, but providing this is true is this all we need?

EDIT:

I used the package GRTensorII on Maple just to check these christoffel symbols, and it gave my original answers i.e. [tex] \Gamma^{t}_{tr}=-\Gamma^{t}_{rt} [/tex] instead of [tex] \Gamma^{t}_{tr}=+\Gamma^{t}_{rt} [/tex]. I find this strange now as Carroll shows the answer you stated, and obviously I now expect the symbol to be symmetric on lower indices not anti sym. strange
 
Last edited:
  • #5
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Probably you have to show that it holds also for r component of geodesic (I don't
see the reason why proving this only for t component would be sufficient). It's
very simple and you can easily check it.
 
  • #6
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I haven't had chance to check the r comp of geodesic yet, but just finding it odd that the GRTensor package on Maple seems to get such a simple christoffel symbol for schwarzschild slightly wrong.

Anyhow, so what I meant to say was, is it sufficient to show that the geodesic equation is satisfied with a zero LHS to prove something is an affine parameter, is this the way people would normally go about it?
 
  • #7
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We call the curve [tex] t \rightarrow x^\mu(\tau) [/tex] a geodesic iff
[tex] \frac{d^2 x^\mu}{d \tau^2} +
\Gamma^\mu_{\alpha\beta}\frac{d x^\alpha}{d \tau}\frac{d x^\beta}{d \tau}
= \alpha \frac{d x^\mu}{d \tau}[/tex] for some function [tex] \alpha(\tau) [/tex].
If this equation if fulfilled with [tex]\alpha \equiv 0 [/tex] then we say that
geodesic is in affine parametrisation (or [tex]\tau [/tex] is a affine parameter
of this geodesic). So the answer to your question is yes.
 
  • #8
pervect
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GRTensor uses a rather unusual notation for the Christofffel symbols. To get the "standard" christoffel symbols, use the following defintion:


grdef(`CC{ ^a b c} := Chr{b c ^a}`);

and calculate the Christoffel symbols as CC, not Gamma.

See https://www.physicsforums.com/showpost.php?p=375565&postcount=6 (or the earlier posts in the same thread where I was once upon a time confused by the same issue).
 
  • #9
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GRTensor uses a rather unusual notation for the Christofffel symbols. To get the "standard" christoffel symbols, use the following defintion:


grdef(`CC{ ^a b c} := Chr{b c ^a}`);

and calculate the Christoffel symbols as CC, not Gamma.

See https://www.physicsforums.com/showpost.php?p=375565&postcount=6 (or the earlier posts in the same thread where I was once upon a time confused by the same issue).

Thanks alot for that, glad I found it here for Schw as I prob wouldn't have noticed for more complicated solns
 

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