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Average Energy and the Equipartition Theorem

  1. Sep 15, 2013 #1
    The Equipartition Theorem states that each quadratic degree of freedom contributes 1/2 kT of energy. This can be derived for the translational degrees by integrating the average kinetic energy multiplied by the Maxwell velocity distribution:

    [itex]\int_{-\infty }^{\infty } \frac{m v^2}{2} \sqrt{\frac{m}{2 \pi k T}} e^{-\frac{m v^2}{2 k T}}dv=\frac{1}{2}k T[/itex]

    Doing this integral in 3-dimensions gives [itex]\frac{3}{2}k T[/itex]

    However the average energy from the Boltzmann energy distribution, which is used in the derivation for energy density per wavelength in blackbody radiation, is

    [itex]\int_{0 }^{\infty } E \sqrt{\frac{1}{k T}} e^{-\frac{E}{k T}}dE=k T[/itex]

    I'm having difficulty reconciling these two results. I understand we are integrating over 2 different variables (velocity for the first and energy for the second), but both results are average energy.

    The 1st is a very specific statement on quadratic degrees of freedom. The 2nd is a general result from the Boltzmann energy distribution, which is sometimes smaller than the 1st.

    Any thoughts?
     
  2. jcsd
  3. Sep 16, 2013 #2

    DrClaude

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    Staff: Mentor

    When considering blackbody radiation, you are dealing with oscillatory degrees of freedom. Each corresponds to two quadratic degrees of freedom (kinetic and potential energy, think of the harmonic oscillator), hence ##2 \times \frac{1}{2}kT = k T##.
     
  4. Sep 16, 2013 #3
    Thank you for the response.

    Isn't that integral far more general though? I guess what I'm asking is, where in that integral do we specify 2 quadratic degrees of freedom?
     
  5. Sep 18, 2013 #4

    jasonRF

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    Science Advisor
    Gold Member

    You are on the right track, in that the mean energy is simply
    [tex]
    \bar{E} = \int E\, p(E) \, dE
    [/tex]
    where p(E) is the probability the system (single particle in this case) has energy between E and E+dE. However, your second equation is assuming 2 degrees of freedom. Here is why. The probability that a system has energy between E and E+dE is
    [tex]
    p(E) = \sum_n P_n
    [/tex]
    where [itex]P_n[/itex] is the probability of state n, and the sum is only over those states with energy between E and E+dE. Assume dE is small, so [itex]P_n \propto e^{-E/kT}[/itex] for all n. Then
    [tex]
    p(E) \propto \Omega(E) e^{-E/kT}
    [/tex]
    where [itex]\Omega(E)[/itex] is the number of states with energy between E and E+dE. [itex]\Omega(E)[/itex] isn't so easy to calculute - it is usually easier to calculate [itex]\Phi(E)[/itex], the number of states with energy less than E, and then find
    [tex]
    \Omega(E) = \frac{d \Phi(E)}{dE}
    [/tex]

    For a particle with energy E, the classical approach tells us
    [tex]
    \Phi(E) \propto \int dx\, \int dp \,
    [/tex]
    where the integral is over the portion of position-momentum space corresponding to energies less than E.

    For a particle in a mono atomic gas, [itex]E = p^2/2m[/itex], which does not depend upon x. So the x integral is just a constant. Hence
    [tex]
    \Phi(E) \propto \int_0^{\sqrt{2mE}} dp \propto \sqrt{E}.
    [/tex]
    Hence [itex]\Omega(E) \propto E^{-1/2}[/itex] and
    [tex]
    P(E) = C_1 E^{-1/2} e^{-E/kT}.
    [/tex]
    If you find the normalization constant [itex]C_1[/itex], you should find that the mean energy is kT/2.

    For a harmonic oscillator, [itex]E = k x^2 /2 + p^2 / 2m[/itex], so the integral for [itex]\Phi(E)[/itex] is over an ellipse in position-momentum space. The semi-major and semi-minor axes of this ellipse are [itex]\sqrt{2E/k}[/itex] and [itex]\sqrt{2Em}[/itex]. Hence the integral, which is just the area of the ellipse, is proportional to E. That is, [itex]\Phi(E) \propto E[/itex] so [itex]\Omega(E)\propto 1[/itex]. So for this two degree of freedome harmonic oscillator we find
    [tex]
    P(E) = C_2 e^{-E/kT}.
    [/tex]
    This is the expression you used in your second case (although you should check your normalization factor - it has the wrong units!). Anyway, find the normalization constant [itex]C_2[/itex] and you should find that the mean energy is kT, just as you found before.

    jason
     
  6. Sep 18, 2013 #5

    jasonRF

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    Science Advisor
    Gold Member

    Just thought I would point out a couple more things

    1) If what I wrote is too complicated, a simpler thing you can do is take your velocity space integral and change variables to [itex]E=\frac{1}{2} m v^2 [/itex]. Note that you can use the fact that the integrand is even, so the integral from -infinity to infinity can be replace by twice the integral from 0 to infinity. Anyway, you should find a square root of energy shows up just as it did in my calculation for a single degree of freedom; you have already done the integral so you know it is 1/2 kT. Since your second integral doesn't have the square root, it cannot be for a single degree of freedom.

    2) The calculation I did for one and two quadratic degrees of freedom can easily be generalized to n quadratic degrees of freedom. In that case [itex]\Phi(E)[/itex] is a volume in n-dimensional space, where the size in each dimension is proportional to [itex]\sqrt{E}[/itex]. Hence [itex]\Phi(E) \propto E^{n/2} [/itex] and we find
    [tex]
    p(E) = C_n E^{n/2-1} e^{-E/kT}
    [/tex]

    jason
     
  7. Sep 21, 2013 #6
    Wow, thank you very much Jason. That was incredibly helpful. I very much appreciate the time and effort it took to write that out.

    Thanks again!
     
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