# 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
• Ill have a think about this, this isnt made clear where I read it (GR by Woodhouse) – Permian Dec 20 '18 at 17:45
• This is all just taken directly from differential geometry. In particular, the last two definitions arise from the fact that you can parameterize paths by the arc length traced out by those paths. – Jerry Schirmer Dec 20 '18 at 18:25
• 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.