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Statistical mechanics of antiparticles

  1. Mar 31, 2010 #1
    I have some questions that my teacher was unable (and unwilling) to answer in class, so I thought I'd ask them here.

    The chemical potential

    [tex]\mu=\left(\frac{\partial U}{\partial N}\right)_{S,V} [/tex]

    is given by the derivative of the energy with respect to particle number at constant volume and entropy. Usually the chemical potential is negative, because if you add a particle, in order to keep entropy constant, then you have to decrease the energy.

    However, because of the exclusion principle, for fermions at zero temperature, you can add particles to the system without changing the arrangements (or entropy), as these particles would just occupy the next highest level. So the chemical potential is positive, and it has to do with the exclusion principle. If you increase the temperature from absolute zero, however, the fermion gas becomes more of an ordinary ideal gas where the spin statistics doesn't matter, and the chemical potential should become negative, because adding more particles to the gas should require that you reduce the energy in order to keep entropy constant. Is there anything special about the temperature at which this transition happens? Does this temperature have a specific name?

    I'm also confused about an ideal gas of anti-fermions. Antiparticles have the opposite chemical potential than particles. So does this mean adding more anti-fermions to the gas would require an increase in the energy to preserve constant entropy? This doesn't seem right physically, but the equations seem to indicate this.

    I'm not sure this is the right place for this post, but the other most logical place, the "classical physics forum" which includes thermodynamics, I'm not sure is the right place to discuss bosons and fermions and antiparticles.
    Last edited: Mar 31, 2010
  2. jcsd
  3. Apr 1, 2010 #2
    I believe that would be Fermi temperature. For metallic electrons, that temperature is between 10,000 and 100,000 K.

    Working solution I came up with is that antiparticle states have negative energy, just like Dirac's equation states. When you sum up all the negative energy and plug it into upper formula, your chemical potential will reverse sign. Anyone else?
    Last edited: Apr 1, 2010
  4. Apr 1, 2010 #3
    I just looked at the equation for a free electron gas:

    [tex]\mu=E_f\left(1-\frac{\pi^2}{12}\left(\frac{kT}{E_f}\right)^2-\frac{\pi^4}{80}\left(\frac{kT}{E_f}\right)^4 \right)+... [/tex]

    and if you set kT=Ef, then it doesn't look like the chemical potential vanishes. Both pi^2/12 and pi^4/80 look like they are about 1. However, this expression is probably only for kT<Ef, so maybe the chemical potential does vanish at kT=Ef.

    That's really interesting. Negative energy states bother me. I'm so used to field theory where antiparticles have positive energy. But in hole theory, antiparticles should still have positive energy, as they are just the holes left behind when negative energy electrons are given enough energy to kick them out of the negative energy sea and into positiveness. I think with states that are arbitrarily negative in energy, the distribution functions will begin to have problems. Maybe it's just semantics: the hole or the antiparticle has positive energy - what had negative energy was still the electron and not the positron.
  5. Apr 1, 2010 #4
    The physics of antiparticles is correctly described only by quantum field theory, where both particles and antiparticles, alike, contribute positive energy to the system. The 1st quantized Dirac equation is inadequate.

    To answer RedX's question, we must be careful about what we mean when we say the chemical potential. Is the chemical potential for the total number of particles in the system? or for the difference in the number of particles and antiparticles, i.e. for the charge?

    In the first, case the chemical potential will contribute the same sign for particles and antiparticles. And, in the second case, the chemical potential will contribute opposite signs for particles and antiparticles. In high energy physics, it is the one for charge that is more commonly used, since it is charge that is conserved.
  6. Apr 1, 2010 #5
    I recognize that, that is Sommerfeld expansion. It is a trick to see how thermodynamical variables change on a finite temperature. Basically, chemical potential will become negative when temperature is so high that you can't say for any state that it will be occupied by electron; that is, when FD distribution totally "melts down".

    Not, this is interesting. First time I ran into this problem was in leptogenesis. Plasma of electrons and positrons at ultra high temperatures was considered by a friend of mine. Only source of particles is pair creation, so I guess the charge is fixed. What it really bothered me is when you plug opposite signed chem. potentials in FD equation, you get significantly different distributions and that can't be right.

    But if, like you said, "... both particles and antiparticles, alike, contribute positive energy to the system.", then chemical potential will contribute the same sign for particles and antiparticles does make sense. Still, tell me if I got this right: opposite signs are considered when you are trying to calculate how much energy does one require to introduce a surplus of charge into ideal gas, right?
  7. Apr 1, 2010 #6
  8. Apr 1, 2010 #7
    So if you have:

    [tex]dS=\frac{1}{T}\left(dU+pdV-\mu dN+\mu d\bar{N}\right) [/tex]

    what you're saying is that this can be written as:

    [tex]dS=\frac{1}{T}\left(dU+pdV-\mu d[N-\bar{N}]\right) [/tex]

    so you can take the conjugate variable to [tex]\mu [/tex] to be [tex]N-\bar{N}[/tex]?

    So this defines the chemical potential as:

    [tex]\mu=\left(\frac{\partial U}{\partial (N-\bar{N})}\right)_{S,V} [/tex]

    where [tex]\mu[/tex] is negative. If the number of particles minus antiparticles is conserved, then does this derivative even make sense? You could create particle/antiparticle pairs, and only put say the antiparticle in the system, so the system gains a net antiparticle. Then you are stuck again with the interpretation that to add an antiparticle to a gas of antiparticles at constant entropy, you must also add energy in order to have a negative chemical potential. This doesn't make physical sense.

    Which case is high energy physics, the 1st or the second?

    Would you regard this point to be a phase transition? Or is there nothing special about going from positive to negative? To me, the physical interpretation is so radical whether [tex]\mu [/tex] is positive or negative, that this ought to be a phase transition.
  9. Apr 2, 2010 #8
    I would say this is not phase transition. Why would it be? Let's say we are looking at hot electron plasma. There is no more FD "step" and even for ground state there is no significant probability to find electron. Adding one electron to the party enlarges entropy so much (because you have no clue where it went - no more bias to set it above Fermi energy )
    that you actually need to cool down plasma to maintain constant entropy. I am sorry, but I don't see why is this so shocking: this is exactly how classical gas works.
  10. Apr 2, 2010 #9
    I guess another way to state it is like this: when you go from a quantum gas to a classical gas by raising the temperature, should the two gases be considered different phases, or is the transition so gradual that you say it's one phase.

    Your answer makes sense, that the gas is fundamentally a quantum mechanical gas, and this quantumness is always there even at high temperatures, but the quantum effects get smaller and smaller. So there is no phase change: it was and always will be a quantum gas since everything is quantum mechanical.
  11. Apr 2, 2010 #10

    Physics Monkey

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    It depends. Nearly free fermions don't experience a phase transition as you raise the temperature. They just slowly lose track of their quantumness amid the thermal noise. Weakly interacting bosons, on the other hand, may undergo an actual phase transition from a superfluid state to a high temperature disordered (classical) phase.
  12. Apr 4, 2010 #11
    So was using opposite signs wrong for the distribution? At equilibrium, the process of annihilation and creation (of pairs) is equal:

    [tex]\mu_{\gamma}=\mu_{+}+\mu_{-} [/tex]

    and since the chemical potential of a photon [tex]\gamma[/tex] is 0, then the chemical potential of the particle and antiparticle are the opposite.

    If you want to introduce a surplus of positive charge, then you inject positrons. If you want to introduce a surplus of negative charge, then you inject electrons. Why would there be opposite signs for the different cases, if you had a neutral plasma to begin with? It looks asymmetrical.
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