Hi all, is there in special relativity any particular interpretation or meaning of the relativistic momentum divided by the rest mass?

(as in contrast to rel. momentum divided by mass which is the velocity).

@RaphaelJ.F.Berger The four-momentum divided by the rest mass is the four-velocity. The spatial momentum is not a covariant quantity and therefore has no special meaning in SR.

@RaphaelJ.F.Berger note that "relativist mass" is an outdated concept, and most physicists only think of rest mass when you say "mass" physics.stackexchange.com/questions/133376/…

incidentally, your question is an example of why invariant mass is a more useful concept: as John said, it relates four-momentum and four-velocity, which are two useful four-vectors.

What is then the four momentum. divided by mass times gamma.?

a not very meaningful quantity

$\gamma$ is not a Lorentz scalar, so dividing a 4-vector by it doesn't give you a 4-vector, it just gives you a frame-dependent mess

"a frame dependent mess" is a nice punny way to sum up the situation on $\gamma m$ :D

I have obtained some expressen in rel qm which is like beta c alpha, where c alpha is the velocity operator, its anti hermitian and it seems to me it corresponds to the frame dependent mess. Possibly the name will stick to it.

note that if the $\gamma$ is the Lorentz factor for the object moving with that 4-velocity (and not that of another observer), the "mess" is just $(c,v^i)$, where $v^i$ is the non-relativistic 3-velocity.

OK! Thats in interesting. So its about an electron moving around a atomic nucleus. I suppose the non-relativistic velocity has no particular physical meaning except that its equal to the 4-velocity for slow motions, right?

it has the ordinary physical meaning of the 3-velocity in that it is the time-derivative of the 3-position $x^i(t)$

but whats the time?

the $x^0$ coordinate of your current frame

I see!

or $x^0 /c$ if you're using annoying units where $c\neq 1$ :P

its 137

Wiki shows pretty clearly how to arrive at this interpretation of the spatial part of the 4-velocity

Thank you!

What does it mean in non relativistic fluid mechanics for

$\nabla _ r$

??

is there some part of that formula missing? :P

or are you asking what the gradient $\nabla_r$ means?

It's in a mass conservation eon

i.e. $\partial_t \rho + \nabla_r (\boldsymbol{u}\rho) = 0$

Even so I'm not sure what it means in that equation

As in, how that's different form the usual equation without the subscript