# Geodesic equation in terms of four velocity

I am trying to show that for timelike paths, we can write the geodesic equation in terms of the four-velocity $$U^\mu=\frac{dx^\mu}{d\tau}$$ as $$U^\lambda\nabla_\lambda U^\mu=0.$$

In other words, substituting $$U^ \mu=\frac{dx^\mu}{d\tau}$$ into the above equation should produce the affinely parameterised geodesic equation

$$\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\rho\lambda}\frac{dx^\rho}{d\tau}\frac{dx^\lambda}{d\tau}=0.$$

Performing the substitution, I got instead

$$U^\lambda(\partial_\lambda U^\mu +\Gamma^\mu_{\rho\lambda}U^\rho)=0$$ $$U^\lambda\partial_\lambda U^\mu +\Gamma^\mu_{\rho\lambda}U^\rho U^\lambda=0$$ $$\frac{dx^\lambda}{d\tau} \frac{\partial}{\partial x^\lambda} \frac{dx^\mu}{d\tau}+\Gamma^\mu_{\rho\lambda}\frac{dx^\rho}{d\tau}\frac{dx^\lambda}{d\tau}=0$$ $$\frac{dx^\lambda}{d\tau}\frac{d}{d\tau} \frac{\partial x^\mu}{\partial x^\lambda}+\Gamma^\mu_{\rho\lambda}\frac{dx^\rho}{d\tau}\frac{dx^\lambda}{d\tau}=0$$

$$\frac{dx^\lambda}{d\tau}\frac{d}{d\tau} \delta^\mu_\lambda+\Gamma^\mu_{\rho\lambda}\frac{dx^\rho}{d\tau}\frac{dx^\lambda}{d\tau}=0$$

$$\Gamma^\mu_{\rho\lambda}\frac{dx^\rho}{d\tau}\frac{dx^\lambda}{d\tau}=0$$

where I used $$\frac{d}{d\tau} \delta^\mu_\lambda=0$$ in the last line since $$\delta^\mu_\lambda$$ is a constant.

What am I doing wrong?

• You are confusing your self, because you use $x^\mu$ to represent both your path, and as a general coordinate. Try introducing a different symbol for your path, e.g. $\gamma^\mu$. Mar 2 '21 at 7:30
• @mmeent That makes sense, I am indeed confusing those two variables.. Mar 2 '21 at 7:43

$$$$\frac{\mathrm d x^\lambda}{\mathrm d\tau} \left[\frac{\partial}{\partial x^\lambda} \frac{\mathrm d x^\mu}{\mathrm d\tau}\right]=\frac{\mathrm d x^\lambda}{\mathrm d\tau}\left[\frac{\partial }{\partial\tau}\left(\frac{\mathrm d x^\mu}{\mathrm d\tau}\right) \frac{\partial\tau}{\partial x^\lambda}\right]=\frac{\partial }{\partial\tau}\left(\frac{\mathrm d x^\mu}{\mathrm d\tau}\right) \underbrace{\frac{\partial\tau}{\partial x^\lambda}\frac{\mathrm d x^\lambda}{\mathrm d\tau}}_{1}=\frac{\mathrm d^2 x^\mu}{\mathrm d\tau^2} \tag{01}\label{01}$$$$