My GR book Hartle says the scalar product of four-velocity with itself $-1$?
Consider the definition of four velocity $\mathbf{u} = \frac{dx^{\alpha}}{d\tau}$. Suppose I take the scalar product of four-velocity with itself. I then get
\begin{equation}
\mathbf{u}\cdot \mathbf{u} = \eta_{\alpha\beta}\frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau}\end{equation}
But these aren't numbers. So why would it yield $-1$?
Granted, I suppose I could consider this a unit four-velocity. But I still don't see how that would yield $-1$. It should be
\begin{equation}
\mathbf{u}\cdot \mathbf{u} = -1(1)(1) + (1)(1)(1) + (1)(1)(1) + (1)(1)(1) = 2
\end{equation}
using
\begin{equation}
\eta_{\alpha\beta}=
\begin{bmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
\end{equation}
Best Answer
First I just want to point out that saying that the four velocity $u_\mu$ satisfies $u_\mu u^\mu=-1$ is a convention, it is not a requirement. It amounts to a choice of the parameterization $\tau$. However, it is a very useful parameterization, it's not common to use other choices.
In this parameterization, the four velocity takes the form
\begin{equation} u^\mu = \left(\gamma, \gamma v_x, \gamma v_y, \gamma v_z\right) \end{equation} where $\gamma=(1-v^2)^{-1/2}$. Then it's easy to explicitly check that \begin{equation} u_\mu u^\mu = -\gamma^2 + \gamma^2 \vec{v} \cdot \vec{v} = -\gamma^2\left(1-v^2\right)=-1 \end{equation}
Note that this assumes $|\vec{v}| < 1$, since otherwise $\gamma$ is not a finite real number. So only timelike paths can be paramaterized so that $\eta_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$.