Parametrization of the Geodesic Equation in GR

Question

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

Answer 1

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).

Answer 2

$\lambda$ 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:

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