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I am a bit confused by the following problem I am asked to solve:

Show that the geodesic equation is consistent with the normalization of the 4-velocity $v^\mu$, $v_\mu v^\mu = -1$ or $0$ for particles with or without mass. As expression for the equation in terms of the tangent vector $u^\mu$, it is given that:

$$u^\mu\nabla_\mu u^\nu = 0$$

Now I guess this is quite simple, but I just got stuck doing it.

First of all: do I even have to show it? Because basically, as $u^\mu = \frac{dx^\mu}{d\tau}$, I could apply the same proof I made to show the normalization in the first place, which would make it automatically normalized.

I am not sure about that though, since there is another line that says that from this statement it follows that if $u^\mu$ is a tangent vector that $u^\mu/|u|$ is also a solution to the equation. For me that suggests that it cannot automatically be normalized then, as this statement would be pointless then.

Now my question is: What part am I missing, that makes this question not as trivial as it appears to me now? I would love to solve it myself, so a push in the direction is appreciated just as much.

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The question is asking you to show that $u_\mu u^\mu$ won't change as you move along a geodesic. Whether this is trivial or not depends on how exactly you proved that $\frac{dx^\mu}{d\tau}$, and be careful because the geodesic equation isn't invariant under arbitrary reparametrization. In other words, if instead of proper time you feel like using a weird function $\lambda(\tau)$ as a parameter, $v^\mu = \frac{dx^\mu}{d\lambda}$ won't satisfy the same geodesic equation. It will still be parallel transported, but with a covariant derivative equal to a multiple of itself:

$$v^\mu \nabla_\mu v^\nu = \kappa(\lambda)v^\nu$$

Again, the details depend on how exactly you proved everything. The standard way I know is to define a geodesic as a curve that parallel transports its tangent vector, i.e. it satisfies the above equation for $v^\mu$. You then show that you can reparametrize so that the right hand side becomes zero, and then show what your question is asking: if the RHS is zero, the square of the tangent vector is constant. This lets you choose the parameter so that $u_\mu u^\mu = -1$, i.e. your curve is now parametrized by arclength, i.e. proper time.

By the way, be careful with the distinction between tangent vector and four velocity. You can define a tangent vector for any curve using any parameter, but the four velocity is only defined for timelike curves and using proper time as a parameter so that it is normalized. This is relevant to the second part of your equation: If your curve is a geodesic and $u^\mu$ is any tangent vector, $u^\mu/|u|$ satisfies the geodesic equation with the RHS equal to zero but $u^\mu$ is general satisfies the geodesic equation that I wrote above.

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