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Relativistic Molecules

  1. Jan 18, 2015 #1
    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.
     
  2. jcsd
  3. Jan 18, 2015 #2

    Orodruin

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    What is v/c for an ultrarelativistic particle?
     
  4. Jan 18, 2015 #3
    Wouldn't it be: pc/E? But in my derivation, I already inputted momentum, so I'm still a bit confused.
     
  5. Jan 18, 2015 #4
    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.
     
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