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I Q re equilibrium energy relationships for a mixture

  1. Feb 25, 2017 #1
    The question I am asking comes from a side issue in another thread. I am seeking an explanation that will explain my wrong understanding of the physics.
    In post #19 of the above thread I describe my uncertainty about my understanding.
    I have been assuming that the concept of equilibrium at a given temperature, with respect to a mixture of particles, including photons and matter particles, means that for each kind of matter particle, the average kinetic energy of those particles is the same as the average photon energy. Is this wrong? If this is wrong, can you suggest where I can find out what the correct relationship is?​
    In post #20, mfb responded.
    It is wrong. Classically (and the protons behave like classical particles here), you have the same average energy for every degree of freedom. The degrees of freedom of the electromagnetic field are a bit more complex, they do not correspond to photons moving around. In addition, you get a Bose-Einstein statistics due to quantum mechanics, which differs from the classical statistics.​

    So I now know I was wrong, but I still don't understand what the correct relationship is. I would much appreciate someone (1) posting an explanation of the correct relationship of the average kinetic energy among the particles of a particular matter species, e.g., protons, and the average kinetic energy of the photons in the mixture, OR (2) citing a reference that has such an explanation.
     
  2. jcsd
  3. Feb 26, 2017 #2

    PeterDonis

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    Staff: Mentor

    Your question makes a very important wrong assumption: it assumes that the number of photons in the mixture is conserved. But photon number is not conserved: photons can be created and destroyed in arbitrary amounts due to interactions with matter. That is a key reason why you can't think of the temperature of radiation as the "average energy of the photons", the way you can think of the temperature of a gas of massive particles as the average kinetic energy of the particles.

    You can still do thermodynamics with photons in a mixture, you just have to be careful to use the correct state functions and to not try to interpret the thermodynamic parameters too literally in terms of a similar microphysics to that of a gas of massive particles. The Wikipedia article on photon gas actually gives a decent quick summary:

    https://en.wikipedia.org/wiki/Photon_gas
     
  4. Feb 26, 2017 #3
    Hi %PeterDonis:

    Perhaps you can help me using a specific example.

    Consider a plasma consisting of protons, electrons and photons. (Protons and electrons in equal numbers.) I would like to be able to calculate the average kinetic energy of the protons for a specific temperature, e.g., 120 GK.

    Previously I calculated the "average" photon energy and then used this average energy as the average kinetic energy of the protons. I then learned that this is not the right way to calculate the average kinetic energy of the protons. However, when I was taught that fact, I did not learn the right way. I would like to learn the right way to do this, and also to understand why it is the right way. The Wikipedia article provides the proper (and intuitively logical) way to calculate the average energy of the photons (U/N), but I can find no guidance there about the protons.

    I did find an equation applicable to the protons but not an explanation.
    The average translational kinetic energy of particles in a plasma is 3kT/2, i.e. the equation for kinetic energy of plasma particles is the same as any other form of matter.
    http://www.answers.com/Q/What_is_the_kinetic_energy_level_of_particles_in_plasma
    I have unsuccessfully tried to search the Internet to find the explanation.

    ADDED
    I found a Wikipedia article that gives the same answer as answers.com, but it seems to be limited to cases where speeds are far from relativistic.
    Therefore the explanation seems to me to be incomplete, and this makes me uncertain about the equation
    E = (3/2) kT .​

    Regards,
    Buzz
     
    Last edited: Feb 26, 2017
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