Why does my relativistic gas pressure formula include v/c?

  • Context: Graduate 
  • Thread starter Thread starter Jeremy Wittkopp
  • Start date Start date
  • Tags Tags
    Molecules Relativistic
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
3 replies · 1K views
Jeremy Wittkopp
Messages
4
Reaction score
0
Suppose the molecules of an ideal gas move with a speed comparable to the speed of light. I am trying to adapt the kinetic theory to express the pressure of the gas in terms of m and the relativistic energy, but each time I try to derive the expression, I get: P = (1/3)(N/V)(v/c)√(E2 - m2c2).

N/V - number density
v - velocity
c - speed of light
E - relativistic energy
m - rest mass

There shouldn't be the term for velocity. Idk what I am doing wrong.
 

Discussion

Physics news on Phys.org
What is v/c for an ultrarelativistic particle?
 
Wouldn't it be: pc/E? But in my derivation, I already inputted momentum, so I'm still a bit confused.
 
I tried this derivation again, I'll show my work this time:

Classical Ideal Gas: P = (1/3)(N/V)mv2 = (2/3) * (1/2)mv2 * (N/V) = (2/3)K(N/V) (assuming only 3 degrees of freedom)
P - pressure
N - # of particles
V - volume
m - mass
v - velocity
K - kinetic energy

Krel = (ϒ - 1)mc2
where
c - speed of light
ϒ - Lorentz factor

So then: P = (2/3)(N/V)(ϒ - 1)mc2

Tell me if I am just crazy, relativity isn't my strongest area.