# Kinetic Theory

1. Mar 27, 2009

### 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?

2. Mar 28, 2009

### 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 $$p = mv$$

3. Mar 28, 2009

### 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

4. Mar 28, 2009

### 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)

5. Mar 28, 2009

### Pinu7

Also worth noting in understanding kinetic energy

is that

KE= dp/dv

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