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I know that the Geodesic equation has the form: $$\frac{d^2x^{\mu}}{d\lambda^2}+\Gamma^{\mu}_{\alpha \beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=0\tag{1}$$ where $\lambda$ is the parameter of the curve $\gamma(\lambda)$, which parametrizes the world-line of the particle.

As far as I know I can change the parametrization using a different curve $\gamma'(\lambda')$ (as long as $\gamma$ and $\gamma'$ have the same image, which is the trajectory of the particle) and the geodesic equation doesn't change form. Usually the proper time $\tau$ is used as $\lambda$.

Now I have reasons to believe that you cannot use the time coordinate $t$ to parametrize the curve, but I cannot argue why this is indeed the case.

So my question is: Why can't you use the coordinate time $t$ to parametrize the geodesic and write: $$\frac{d^2x^{\mu}}{dt^2}+\Gamma^{\mu}_{\alpha \beta}\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt}=0~?$$

Qmechanic
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Mathew
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    You can use the time in place of the proper time because $dt/d\tau \neq 0$ since the curve is timelike (if I correctly understand). However, changing parameter, the equation does not preserve the form you wrote. That form is preserved if and only if the new parameter is related to the old one by means of a linear non homogeneous in general transformation. – Valter Moretti Oct 31 '21 at 13:11

2 Answers2

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  1. OP considers the affine geodesic equation (GE) $$ {d^2 x^{\mu} \over d\lambda^2} + \Gamma^{\mu}_{\alpha\beta} {dx^{\alpha} \over d\lambda} {dx^{\beta} \over d\lambda} ~=~ 0.\tag{1}$$ For timelike geodesics (= a massive point particle) the affine GE (1) holds when the parameter $\lambda$ is affinely related to the arc length $s=c\tau=a\lambda+b$ of the geodesic. (Here $\tau$ is the proper time.)

  2. For a generic parametrization $\lambda$ the GE contains an extra term proportional to the velocity: $$ {d^2 x^{\mu} \over d\lambda^2} + \Gamma^{\mu}_{\alpha\beta} {dx^\alpha \over d\lambda} {dx^\beta \over d\lambda} ~\propto~ {d x^{\mu} \over d\lambda},\tag{2}$$ cf e.g. my Phys.SE answer here. To be specific, if $\frac{d\tau}{d\lambda}=f(\lambda)$, then the GE reads $$ {d^2 x^{\mu} \over d\lambda^2} + \Gamma^{\mu}_{\alpha\beta} {dx^\alpha \over d\lambda} {dx^\beta \over d\lambda}~=~ \frac{d\ln |f|}{d\lambda}{d x^{\mu} \over d\lambda}.\tag{2'}$$

  3. Generically, if we choose a given coordinate time $t=\lambda$ as parametrization, we have to use the GE (2) rather than the affine GE (1).

Qmechanic
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$\lambda$ can be any affine parameter. For an interesting discussion of this read the answers to What is the physical meaning of the affine parameter for null geodesic?

However coordinate time is not an affine parameter. You can rewrite the geodesic equation using coordinate time by using the chain rule, but as mentioned in Qmechanic's answer this introduces an extra term:

$$ \frac{d^2x^{\mu}}{dt^2} + \Gamma^{\mu}_{\alpha \beta}\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt} - \Gamma^{0}_{\alpha \beta}\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt}\frac{dx^{\mu}}{dt} = 0 $$

John Rennie
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  • This is clear now. But how do you find that (in this case) the extra term is the one you wrote? I used the chain rule and got an extra term proportional to the velocity, but the factor in front of it has something to do with the derivative of the parameter of the new parametrization with respect to the the parameter of the old one. I don't see how you get that Christoffel symbol with the 0 upper index. – Mathew Nov 01 '21 at 09:55
  • @Mathew to be honest I didn't work through the derivation, I just looked up the result. – John Rennie Nov 01 '21 at 10:18
  • Could you please share the resource where you found that result? – Mathew Nov 01 '21 at 11:42
  • @Mathew It was from Wikipedia. – John Rennie Nov 01 '21 at 16:38