# Proper time of a timelike geodesic

In the contest of the newtonian limit in general relativity, if I consider a timelike geodesic that can represent the motion of a free falling particle under the influence of the gravitational force of the perturbed metric, I can't understand why the proper time of the particle is such that: $$-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2$$ I know that a geodesic is parametrized at proper time if $$g(\dot \gamma, \dot \gamma)=-1$$, but from here I don't know how deduce the expression above. Really it is an expression for the proper time that I have encountered also many other times but I can't understand it:

$$\textbf{EDIT:}$$ I mean why in general can I write $$d\tau=\sqrt{-g_{\alpha\beta}dx^{\alpha}dx^\beta}$$?
I have thoguht that if $$\gamma=(t(\tau), x(\tau), y(\tau), z(\tau))$$ then $$g(\dot \gamma, \dot \gamma)=-1\iff g(\frac{d\gamma}{d\tau}, \frac{d\gamma}{d\tau})=-1\iff -(\frac{dt}{d\tau})^2+(\frac{dx}{d\tau})^2+(\frac{dy}{d\tau})^2+(\frac{dz}{d\tau})^2=-1\iff -dt^2+dx^2+dy^2+dz^2=-d\tau^2$$ where for $$g$$ I have used the perturbed metric at first order that is minkowski metric.

Instead in non relativitusc measure units the condition $$g(\dot \gamma, \dot \gamma)=-1$$ becomes $$g(\dot \gamma, \dot \gamma)=-c^2$$, right?
Can you help me?

The proper time is the time experienced by the particle itself. When you thus choose your reference frame as that of the particle itself, you fix your spatial coordinates at the particle's location. So $$dx^i=0$$.
• Ok thanks! Two clarifications: 1) the expression $-c^2dt^2+dx^2+dy^2+dz^2$ is referred to what? The minkowski metric? 2) where is used in this expression the definition of proper time $g(\dot \gamma, \dot \gamma)-1$? Commented Apr 1, 2021 at 6:15
• 1) $ds^2$ is called the line element (and in your case, it is the line element corresponding to the Minkowski metric, yes). 2) I'm not really sure what you mean by the definition you presented. Proper time is nothing but the (square root of) line element itself, when the reference frame is attached to the particle itself. Commented Apr 1, 2021 at 18:17
• Ok thanks! for the second question: if the proper time is the square root of line how can I obtain $-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2$? this is the key point of my question above! Commented Apr 1, 2021 at 18:22