In special theory of relativity, why is 4-velocity defined as:
$$ u^\mu = \frac{dx^\mu}{d\tau} $$
and not as $$ u^\mu = \frac{dx^\mu}{dt} $$
where ${\tau}$ is proper-time and t is time in some other observers' frame of reference. What does derivation with respect to invariant (like proper-time here) have to do with contravariant components of the vector? I'd prefer intuitive explanation rather than strict mathematical.
Proper time is the reference time all observers can agree on. Makes for calculating things much much easier.
Since tau is attached to the particle the space differentials are zero
Imagine we used some other reference time (which you are more than welcome to do). Then the spacetime line element would have dx, dy, and dz not equal to zero.
Furthermore using a different frame would mean we would have to transform everything from one frame to another. Yikes!
Proper time is simply chosen to avoid all these extra nastynesses.
Hope that helps
4-vectors have invariant length as defined by $$\vec{v}\cdot \vec{v} = g_{ab}v^a v^b.$$
Coordinate velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}$ does not have this property. Proper velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau}$ does. Coordinate velocity is defined. It's not a 4-vector, so it's not that useful.
Given a path $x(s)$ on a manifold, the velocity with respect to that path is defined taking derivatives with respect to the invariant parameter the path is described with, on the manifold (the arc length). In general relativity such parameter is the proper length $s$ (or proper time $\tau = \gamma s$).
