# Deriving the equation relating the metric and the coordinates to the proper time in general relativity

If $$x^a = x^a(\tau)$$ is the worldline of a particle in general motion, then

$$V^a = \frac{dx^a}{d\tau}$$

is a four-vector field along the worldline. If

$$g_{ab}V^aV^b = g_{ab}\dot{x}^a\dot{x}^b = 1$$

then $$V$$ is called the four-velocity and $$\tau$$ is called the proper time. When

$$g_{ab}V^aV^b = 1$$

the increment in $$\tau$$ between the events on the worldline with coordinates $$x^a$$ and $$x^a + dx^a$$ is

$$d\tau = \sqrt{g_{ab}dx^ady^b}.$$

How do we get the equals 1 bit?

Also has the square root equation been derived?

Both seem to have plucked out of mid air.

• Possible connection to my other question, physics.stackexchange.com/q/449170 – Permian Dec 20 '18 at 17:15
• $\sqrt{g_{ab}V^aV^b}$ is the norm of the four velocity, which is always $1$ (or $c$ if you're not using $c=1$). Are you asking why the norm of the four velocity is always $1$? – John Rennie Dec 20 '18 at 17:24
• Note that the last equation should be $d \tau = \sqrt{ g_{ab} dx^a dx^b}$, not $d \tau = \sqrt{ g_{ab} V^a V^b}$. – Michael Seifert Dec 20 '18 at 19:16
The square of the spacetime interval is $$ds^2 = g_{\mu \nu} dx^\mu dx^\nu$$ and is an invariant.
In the rest frame we have $$dt = d\tau$$, where $$\tau$$ is the proper time, $$dx = dy = dz = 0$$ and the metric is the Minkowski (flat) metric $$\eta = diag(-1, 1, 1, 1)$$. So $$ds^2 = \eta_{\mu \nu} dx^\mu dx^\nu = -d\tau^2$$. The square of the four velocity $$U^\mu = dx^\mu / d\tau$$ is $$-1$$.
The invariance allows to state $$g_{\mu \nu} U^\mu U^\nu = -1$$ in any reference frame.