Is Proper Time Only Perceived by External Observers?

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Dale said:
Excellent, thanks. I guess I just got in the habit of always getting an affine parameter such that I have completely forgotten when it is necessary.
Of course, working with affine parameters is of great advantage. That's why it's better to use the "square form" of the Lagrangian for the geodesics, i.e.,
$$L=-\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where the dot again means the derivative wrt. an arbitrary parameter ##\lambda##, but thanks to Noether's theorem, since the Lagrangian is quadratic in the ##\dot{x}## and since it doesn't explicitly depend on ##\lambda## the "Hamilton-like" conserved quantity ##H=L## is conserved along the solutions of the equations of motion (which are just the geodesic equation). This means that for the solution ##\lambda## is automatically an affine parameter along the trajectories of the particle.

Another advantage is that this works without trouble for both light-like as well as time-like geodesics. In the latter case you simply choose the conserved quantity ##g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=c^2##. Then you have ##\lambda=\tau##, with ##\tau## the proper time along the geodesic. For the light-like case you have to set ##g_{\mu \nu} =g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=0##, and of course there's no proper time, but ##\lambda## is still some arbitrary affine parameter. The physics of course doesn't depend on the choice of this parameter.
 
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