# 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 Jan 28 at 15:28
• Is $\tau$ the proper time? Jan 28 at 15:45
• Yes, it is the proper time. Jan 30 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. Jan 30 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) Jan 30 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. Jan 30 at 18:19