# 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?

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 Be aware that the normalization depends of the trace of the metric chosen--there are weirdos who use the $\mathop{Tr}(g) = 1$ metric, for instance. – dmckee♦ Nov 11 '11 at 23:21 I'm using the signature (-+++). But my question still applies, changing -1 to +1, in the signature (+---). (Not sure what you mean by $Tr(g)=1$ though, since in either of these two signatures $g^\mu_\mu = \delta^\mu_\mu = 4$) – Joss L Nov 11 '11 at 23:39 @Joss L: dmckee is referring to that some authors use a normalized trace ${\rm Tr}(M)=\frac{1}{n}\sum_{i=1}^n M^i{}_i$. – Qmechanic♦ Nov 12 '11 at 8:36 Oh, okay. Ususlly in general relativity when people say trace I think they mean $M^\mu_{\phantom{\mu}\mu}=g^{\mu\nu}M_{\mu\nu}$. But this is all besides the point. – Joss L Nov 12 '11 at 17:53