Proof of $ g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-c^2$ in GR The differential proper time $d\tau$ is defined by
$$
    c^2 d\tau^2 = -g_{\mu\nu}dx^\mu \otimes dx^\nu
$$
So i believe we can think of $d\tau^2$ as 
$$
    d\tau^2 = -\frac{1}{c^2}ds^2
$$
where $ds^2$ is the inner product.
I've read that 
\begin{equation}
    g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-c^2
\end{equation}
How can I (rigorously) derive this last equation from the first? Is there a way? Physics texts say it is, but I have not seen it done (apart from obviously not acceptable "proofs" that simply "divide" the first equation by $d\tau^2$ on both sides).
 A: Proper time for a timelike path $x^\mu(\lambda)$ is defined by
$$
\tau = \frac{1}{c} \int \left( - g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}\right)^{1/2} d\lambda
$$
where we integrate over a part of the path. So differentiating with respect to $\lambda$ and squaring we find 
$$
c^2 \left(\frac{d\tau}{d\lambda} \right)^2 = - g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}.
$$
Using the chain rule we find
$$
g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} = g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} \left(\frac{d\tau}{d\lambda} \right)^2,
$$
which gives the required equation.
In fact you have probably already seen a very similar thing in differential geometry for regular curves in $\mathbb{R}^3$ parameterized by their arc length. Given an interval $I$, a curve $\alpha: I \to \mathbb{R}^3$ the arc length is defined by
$$
s(t) = \int_{t_0}^t \lvert \alpha'(s) \rvert ds = \int_{t_0}^t \left( \alpha'(s) \cdot \alpha'(s) \right)^{1/2} ds.
$$
Then we have $\frac{ds}{dt} = \lvert \alpha'(s) \rvert$ which is equal to $1$ if $\alpha$ is parameterized by its arc length.
