Is 4-velocity normalized to -1 even for non-geodesic timelike curves? In Hartle's General Relativity book ("Gravity"), one of the problems (chapter 8 problem 6) is to prove that $g_{\mu\nu}u^\mu u^\nu$ is conserved along geodesics (really not hard to show), where $u^\mu$ is the 4-velocity. My question is: Isn't it true that $g_{\mu\nu}u^\mu u^\nu$ is equal to $-1$ for any timelike curve whether it is a geodesic or not? This follows (I think) from
$$
g_{\mu\nu}u^\mu u^\nu = g_{\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = \frac{g_{\mu\nu}dx^\mu dx^\nu}{d\tau^2}=\frac{ds^2}{d\tau^2} = \frac{-d\tau^2}{d\tau^2} = -1.
$$
Am I wrong about this? Why should we need the geodesic equation to prove this if it's true for any timelike curve?
 A: I emailed my TA and here was his answer, which I think makes sense:
While it is true that a curve which is everywhere timelike can be parametrized so that its tangent vector has unit norm, it is also possible to draw a curve which starts out timelike and then becomes null or spacelike, so its norm won't be the same everywhere. The problem in Hartle simply shows that a geodesic which starts out timelike will always remain timelike (same holds for null or spacelike geodesics). 
A: Indeed, you are absolutely right, and this is a terrible exercise. The quantity v dot v (using the metric) is the length of the tangent to a curve, and this depends on the parametrization. In an arclength parametrization of timelike curves, as you use to derive the geodesic equation, it is always minus one (in your convention). For null curves, it is zero, for spacelike curves +1. An equivalent statement is that v dot a is zero, the second derivative is relativistically perpendicular to the velocity, and this is just as true in geometry, where in an arc-length parametrization, the second derivative is perpendicular to the first derivative. For null curves, the condition that a is orthogonal to v defines affine parameter, since arclength is zero.
