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Maxwell-Boltzmann accounts for identical particles?

  1. Dec 26, 2015 #1
    The probability of finding one single particle at energy ##\epsilon## in a system of such distinguishable particles at thermal equilibrium in the thermodynamic limit is given by the Boltzmann distribution that is: (up to a constant factor in front)

    ##e^{-\frac{\epsilon}{kT}}##

    To find this, one does combinatorics of distinguishable particles, it's not correct for a gas! Even in a classical treatment ignoring any QM effects one has to do different combinatorics for a gas.

    Now if I'm not mistaken this result is used in the derivation of the Maxwell-Boltzmann velocity distribution ##n(v)##.

    Why is not accounting for the identical particles combinatorics OK here?
     
  2. jcsd
  3. Dec 26, 2015 #2
    Because they are farther away from eachother than their de-Broglie wavelength.
     
  4. Dec 26, 2015 #3

    DrClaude

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    That result is exactly the same for indistinguishable classical particles. The difference only appears in the partition function, which you have omitted here.

    The details of the partition function do not enter in the derivation of the Maxwell-Boltzmann velocity distribution. If I'm not mistaken, you would get the same result for distinguishable and indistinguishable particles.
     
  5. Dec 26, 2015 #4
    What do you mean by "indistinguishable classical particles"? The statistics for bosons or for fermions give probability distributions that both differ from that for distinguishable particles. The difference is negligible if the distances between them are large in comparison with their de-Broglie wavelength.

    The velocity distribution of electrons in ordinary metals is quite different from the Maxwell-Boltzmann distribution (Fermi sphere). But in a semiconductor with a low carrier concentration, the MB distribution is ok.
     
  6. Dec 26, 2015 #5

    DrClaude

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    I meant identical classical particles.
     
  7. Dec 27, 2015 #6

    Jano L.

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    What different combinatorics do you mean? If you're thinking of diminishing the number of accessible states by the number of permutations of all identical particles, this has trivial effect - the corresponding phase volume gets divided by ##N!##, the probability density gets multiplied by ##N!## so all the values of measurable quantities come out the same.

    If you mean discrete states instead of continuum of states, this has impact on the whole calculation procedure and on the values of probabilities, but for gas in common conditions (rarified, hot) the results for measurable physical quantities are almost the same as for continuum of states. The classical Boltzmann formula often works well enough for gas so people do not bother with using the discretized version.
     
  8. Dec 27, 2015 #7
    Thanks for all the answers, what I meant was the following. To find the probability of a single particle having an energy ##E## which is mentioned in my original post one has to maximize the logarithm of the following expression under conservation of total energy and number of particles

    ##\frac{N!}{n_1!n_2!...}##

    where N is the total number of particles and ##n_i## is the number of particles in energy level ##E_i##.

    This kind of counting is only OK for distinguishable particles, in a crystal for example.

    Now the Maxwell Boltzmann distribution still uses this ##exp(-E/kT)## probability to find one single particle at a certain energy.

    It does also not seem okay to me to JUST divide by ##N!## in the above expression to get the combinatorics for the gas since dividing by ##N!## assumes that there are much more states than particles which is obviously not self consistent with all these ##n_i!## terms in the denominator.
     
  9. Dec 27, 2015 #8
    The Boltzmann factor itself does not depend on this. It is just proportional to the multiplicity of the heat bath with temperature T that the smaller system is in contact with.

    But yes, you are right about the counting statistics for indistinguishable particles. I do not know what the expressions are, but they are different for fermions and bosons, both very different from the expression above. It is clear they cannot both be obtained by dividing by N!.
     
  10. Dec 29, 2015 #9

    vanhees71

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