In special relativity (SR), momentum is defined as  

$\frac{mv}{\sqrt {1 - \frac{v^2}{c^2}}}$

While one way of interpreting the relativistic momentum is to consider the mass to be the "relativistic mass" $m\_r = m/\sqrt {1 - v^2/c^2}$, the other way is to consider the velocity as a "relativistic velocity":  

$v\_r = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}$  

or, $v\_r = \frac{v}{\sqrt{1 - \beta^2}}$, where $\beta = v/c$.

This can be rewritten as:  

$v\_r = \frac{dx}{dt}(\frac{1}{\sqrt{1 - \beta^2}}) = \frac{dx}{dt\sqrt{1 - \beta^2}} = \frac{dx}{d\tau}$, where $\tau$ is the proper time.

What is the physical interpretation of this relativistic velocity? If we visualize this on a space-time diagram, an observer on the rest frame, observing the moving body, appears to be taking the ratio of a space interval $dx$ of the moving body as measured from his rest frame, with a proper time interval $d\tau$ that has meaning only on the frame of the moving body (i.e. the proper time measured by a clock attached to the moving body). Essentially, it appears to be a velocity greater than the observed velocity $dx/dt$ of the body, but how do we give it a physical meaning?