Four Velocity Sign of Time: \dot t>0?

[Onyx]

TL;DR

Is it generally the case even with light like paths that $\dot t>0$?

Is it generally the case even with light like paths that $\dot t>0$?

 

[Sagittarius A-Star]

A four velocity is not defined for a light like path.

 

[Ibix]

Up to you, really. It is true that all future-pointing vectors will have the same sign in their time component, assuming your time coordinate is reasonably named and the spacetime has a global distinction between past and future. But there's nothing to stop you having your time coordinate increase towards the past, in which case all future-pointing four vectors would have negative time components.

As @Sagittarius A-Star points out, four velocity is not defined for null paths. However, you can define other four vectors tangent to null curves, such as the four momentum.

 

[vanhees71]

It's convenient to have the world-line parameter defined such that $\dot{t}>0$. For massive particles you have time-like worldlines, and you can choose the proper time, $\tau$ as a natural world-line parameter. Then the four-velocity is "normalized": $u_{\mu} u^{\mu}=c^2$.

For massless particles ("naive photons") of course you cannot choose proper time, because it's not defined but you can choose any affine parameter you like. Then you have $\dot{x}^{\mu} \dot{x}_{\mu}=0$, i.e., light-like worldlines.

In both cases it is natural to choose $\dot{t}>0$, where $t$ is the time-like coordinate since then with increasing world-line parameter you describe a motion into the future.