Why is the scalar product of four-velocity with itself -1? 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}
 A: The derivatives with respect to $\tau$ very much are numbers, but they are not all $1$.
Consider your worldline as a curve $\gamma$ parameterized by $\lambda$. We have
\begin{align}
\gamma : \mathbb{R} & \to \mathbb{R}^4 \\
\lambda & \mapsto (x^0, x^1, x^2, x^3).
\end{align}
At any point in your worldline you have a position $(x^0, x^1, x^2, x^3)$, where all components are scalar functions of $\lambda$. You can differentiate these to find your velocity components:
$$ \dot{x}^\alpha \equiv \frac{\mathrm{d}x^\alpha}{\mathrm{d}\lambda}. $$
Again, these are functions of $\lambda$, and each of the four evaluates to a real number at every point on the line.
In the special case where your parameter $\lambda$ is the proper time, then we call $\dot{x}^\alpha$ instead $u^\alpha$, the $\alpha$ component of $4$-velocity.

For an example, if you were not moving with respect to the coordinates, then each spatial $x^i$ would be constant as $\lambda$ varied, so we would have $\dot{x}^i = 0$. We also know
$$ \mathrm{d}\tau = \sqrt{-\mathrm{d}s^2} = \sqrt{-g_{\alpha\beta} \mathrm{d}x^\alpha \mathrm{d}x^\beta} = \sqrt{-g_{\alpha\beta} \dot{x}^\alpha \dot{x}^\beta} \mathrm{d}\lambda = \sqrt{-g_{00}} \, \dot{x}^0 \, \mathrm{d}\lambda, $$
so
$$ u^\alpha \equiv \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} = \frac{\mathrm{d}x^\alpha}{\mathrm{d}\lambda} \frac{\mathrm{d}\lambda}{\mathrm{d}\tau} = \dot{x}^\alpha \frac{1}{\sqrt{-g_{00}} \dot{x}^0} = \big((-g_{00})^{-1/2}, 0, 0, 0\big). $$
You can easily check that $g_{\alpha\beta} u^\alpha u^\beta = -1$ in this case.

1 In general relativiy, replace $\mathbb{R}^4$ with some open set $U \subseteq \mathbb{R}^4$ covering the appropriate part of the manifold. 
A: 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 $\eta_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$.
A: 
Why is the scalar product of four-velocity with itself -1



*

*The scalar product is invariant

*In the coordinate system in which the object is (momentarily) at rest, the only
non-zero component is the temporal component.


See that, in the rest frame, $\gamma = 1$ thus $d\tau = dt$.  Then, (setting $c = 1$) we have
$$\frac{dx^0}{dt} = 1,\,\frac{dx^i}{dt}=0 $$
Thus
$$\eta_{\alpha \beta}\frac{dx^\alpha}{dt}\frac{dx^\beta}{dt} = -1 $$
A: I will consider $c=1$.
We define the interval by $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$
and the proper time by
$$d\tau^2=-ds^2=-g_{\mu\nu}dx^\mu dx^\nu$$
From here, it's straightforward to show that
$$g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$$
Now, let's take the inner product of the four-velocity $U^\mu=dx^\mu/d\tau$,
$$U=g_{\mu\nu}U^\mu U^\nu=g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$$
for what was said before.
In conclusion, the four-velocity always will be normalized $-1$.
A: From the definition of the spacetime invariant and a 4-displacement vector, we know:
$$ \mathrm ds^2 = \mathrm d\mathbf x\cdot \mathrm d\mathbf x $$
We also know proper time is defined as:
$$ \mathrm d\tau = \sqrt{-\mathrm ds^2} $$
Thus:
$$ \mathrm d\tau^2 = -\mathrm ds^2 $$
Now:
$$ \mathrm d\tau^2 = - \mathrm d\mathbf x\cdot \mathrm d\mathbf x $$
$$ \frac{\mathrm d\mathbf x}{\mathrm d\tau} \cdot \frac{\mathrm d\mathbf x}{\mathrm d\tau}  = - \mathbf U\cdot \mathbf U = -1 $$
