Why use velocity instead of momentum in kinetic theory

[TriTertButoxy]

Why is it in kinetic theory one uses the velocity variable, instead of the momentum variable? Wouldn't this cause problems when trying to generalize to relativistic systems?

 

[Archosaur]

I think I understand your question.

Momentum is in the equation. It's just hiding.
You could think of the equation for kinetic energy as $$KE = \frac{1}{2}pv$$

because $$p = mv$$

I don't know if this answers your question.

 

[CFDFEAGURU]

Kinetic Theory distribution function (numbrt density) is

N(x,p,t)=dN/d^2V=dN/dVxdVp

N is the number density and it is a function of the position vector, x, the momentum vector, p, and time, t.

In a relativistic setting the number density is the same except now

p=mv/sqrt(1-v^2)

The vector space x and the momentum space p, define a 6 dimensional phase space.

Hope that helps.

Matt

 

[CFDFEAGURU]

The above equation for p in a relativistic setting only holds for a particle with zero rest mass. (travels at the speed of light)

 

[Pinu7]

I think I understand your question.

Momentum is in the equation. It's just hiding.
You could think of the equation for kinetic energy as $$KE = \frac{1}{2}pv$$

because $$p = mv$$

I don't know if this answers your question.

Also worth noting in understanding kinetic energy

is that

KE= dp/dv

or the rate of change of momentum with respect to velocity.