# Geodesic equation using 4-momenta

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?

• Probably relevant: physics.stackexchange.com/q/674564/25301 Commented Jan 28, 2022 at 15:28
• Is $\tau$ the proper time? Commented Jan 28, 2022 at 15:45
• Yes, it is the proper time. Commented Jan 30, 2022 at 11:54

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 $$\boxed{A^\mu=0\,.}$$

• Thank you so much for the detailed answer! It is super clear. Commented Jan 30, 2022 at 11:52
• 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) Commented Jan 30, 2022 at 11:53
• 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. Commented Jan 30, 2022 at 18:19