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


Performing the substitution, I got instead

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

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


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?

  • 1
    $\begingroup$ 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. $\gamma^\mu$. $\endgroup$
    – TimRias
    Commented Mar 2, 2021 at 7:30
  • $\begingroup$ @mmeent That makes sense, I am indeed confusing those two variables.. $\endgroup$
    – TaeNyFan
    Commented Mar 2, 2021 at 7:43
  • $\begingroup$ Said differently, a derivative wrt $\tau$ does not commute with a spatial derivative. This is because d/dt commutes with d/dx, but $\tau$ is related to t by the Lorentz factor $\gamma$ which is a function of velocities. $\endgroup$
    – Sidd
    Commented Oct 15, 2023 at 16:35

1 Answer 1


\begin{equation} \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} \end{equation}


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