I know the geodesic equation in the form $$\frac{d^2X^\mu}{d\lambda^2} + \Gamma^\mu_{\nu\alpha}\frac{dX^\nu}{d\lambda} \frac{dX^\alpha}{d\lambda} = 0,$$ where $\lambda$ is a parametrization of the curve $X^\mu(\lambda)$.

However, in the textbook I also see a formulation of the geodesic equation using the 4-momentum $P^\mu$, $$ \frac{d X^{\nu}}{d \tau} \nabla_{\nu} \frac{d X^{\mu}}{d \tau}=P^{\nu} \nabla_{\nu} P^{\mu}=0. $$

It is not clear to me how we arrive at this formulation. It seems that we simply switched out $\lambda$ to $\tau$ — but why are we able to do that?


1 Answer 1


The geodesic equation can be written equivalently when $X$ is parametrized with proper time: $$ \frac{d^2X^\mu}{d\tau^2} + \Gamma^\mu_{\nu\alpha}\frac{dX^\nu}{d\tau} \frac{dX^\alpha}{d\tau} = 0\,. $$ Here, $\tau\mapsto X^\mu(\tau)$ is a world line whose four velocity is $\frac{dX^\mu}{d\tau}$ and whose acceleration is $\frac{d^2X^\mu}{d\tau^2}\,.$

Now suppose we have a four-velocity field $U^\mu$ at every point of space time (or at least around the world line $X^\mu$) such that $$ \frac{dX^\mu(\tau)}{d\tau}=U^\mu(X^\mu(\tau))\,. $$ By the chain rule $$\tag{1} \frac{d^2X^\mu(\tau)}{d\tau^2}=\frac{dX^\mu(\tau)}{d\tau}\partial_\nu U^\mu(X^\mu(\tau))=(U^\nu\partial_\nu U^\mu)(X^\mu(\tau))\,. $$ Dropping the clumsy argument $X^\mu(\tau)$ the geodesic equation then implies $$ U^\nu\partial_\nu U^\mu+\Gamma^\mu_{\nu\alpha}U^\nu U^\alpha=U^\nu\nabla_\nu U^\mu=0. $$ Four momentum $P^\mu$ is just rest mass $m_0$ times four velocity $U^\mu\,.$ This does not alter the geodesic equation.

Few remarks.

  • In curved space time $\frac{d^2X^\mu}{d\tau^2}$ is not the four acceleration. Since by (1) it is $U^\nu\partial_\nu U^\mu$ which does not transform as a tensor due to the partial derivative there.

  • The correct way of defining four acceleration is $$\tag{2} A^\mu\equiv\frac{d^2X^\mu}{d\tau^2}+\Gamma^\mu_{\nu\alpha}\frac{dX^\nu}{d\tau}\frac{dX^\alpha}{d\tau}\,. $$ As above this is seen to be equal to $U^\nu\nabla_\nu U^\mu\,.$ This is the covariant derivative of $U$ in the direction of $U\,.$ Some authors also write $\dot U^\mu$ for $A^\mu$ and the shortest form of the geodesic equation is $$ \boxed{A^\mu=0\,.} $$

  • $\begingroup$ Thank you so much for the detailed answer! It is super clear. $\endgroup$
    – Loika
    Jan 30 at 11:52
  • $\begingroup$ One a second note - can this equation be used for photons? Or is it only valid for particles with mass? (since the proper time is zero along the photon worldline) $\endgroup$
    – Loika
    Jan 30 at 11:53
  • $\begingroup$ Indeed, the geodesic equations hold only for timelike world lines, exactly because these are the world lines that have a non zero proper time. This does however not mean that only timelike geodesics exist. Another related post is this one. In short: if photons are only subject to gravity and otherwise unperturbed they follow lightlike geodesics. $\endgroup$
    – Kurt G.
    Jan 30 at 18:19

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