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Harmonic oscillator in a heat bath

  1. Dec 9, 2008 #1
    How to describe a harmonic oscillator defined by

    H(q,p) = \frac{p^2}{2m} + \frac{1}{2}kq^2

    in a heat bath with some fixed temperature [itex]T[/itex]?

    I suppose this question alone is not quite well defined, because it mixes classical and statistical mechanics in confusing manner, but I thought that one could make the question more rigor by assuming, that the oscillator frequency is notably larger than the frequency of instants when the oscillator interacts with the heat bath. In this case we could identify the ellipse trajectories [itex]q^2 + \frac{1}{mk}p^2 = r^2[/itex] with the indexes of the energy states used in statistical treatment. Right? The trajectories, on the other hand, can be identified with the radius [itex]r\in [0,\infty[[/itex]. The energy corresponding to each radius is [itex]E(r) = \frac{1}{2}kr^2[/itex].

    The problem is that if we set the Boltzmann probability measure to be proportional to

    \exp\Big(-\frac{kr^2}{2k_{\textrm{B}} T}\Big) dr,

    it would not give the correct energy distribution. This is, because this formula doesn't correctly take into account the density of the ellipse trajectories. The correct probability measure should be something like this:

    \exp\Big(-\frac{kr^2}{2k_{\textrm{B}} T}\Big) \rho(r) dr,

    but I've been unable to figure out what [itex]\rho(r)[/itex] should be.

    I thought that this could be a good postulate to start with: When the oscillator interacts with the heat bath, the position [itex]q[/itex] remains unchanged, but the momentum [itex]p[/itex] becomes thrown to some arbitrary new value, so that the probability of the new momentum would not be weighted by any particular density function. This would be a model of a collision with some particle from the heat bath. So if we assume that the new momentum has a probability measure proportional to [itex]dp[/itex] (this is not normalizable really, but it shouldn't be a problem, because normalizable factors arise later), it should be possible to solve what measure [itex]\rho(r)dr[/itex] follows for the trajectories.

    Unfortunately I found this task too difficult. Anyone having any comments to this? Did I start into wrong direction with this problem, or can the [itex]\rho(r)[/itex] be solved from what I started?
    Last edited: Dec 9, 2008
  2. jcsd
  3. Dec 10, 2008 #2


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    Homework Helper

    Maybe you could start by calculating the canonical partition function
    Z(T)= \frac{1}{{(2\pi\hbar)}^3} \int d^3 p d^3 r e^{(-H/T)}\;.
  4. Dec 11, 2008 #3
    I suppose that would be

    Z(T) = \int\limits_{\mathbb{R}^2} dp\; dq\; \exp\Big(-\frac{H(q,p)}{k_{\textrm{B}} T}\Big)

    with the notation I was using?

    That doesn't look right. It assumes that the points in the phase space would be the indexes of the energy states, and ignores the assumption I made in the beginning, that the indexes of the energy states should be the ellipse trajectories.

    Once the [itex]\rho(r)[/itex], I talked about in the first post, becomes known, then the partition function would be given by formula

    Z(T) = \int\limits_0^{\infty} dr\; \rho(r) \exp\Big(-\frac{kr^2}{2k_{\textrm{B}} T}\Big)
  5. Dec 11, 2008 #4
    Having written down those two equations, it seems immediately clear, that there exists such [itex]\rho(r)[/itex] that the [itex]Z(T)[/itex] is identical from the both formulas, but I'm not convinced that such [itex]\rho(r)[/itex] would be consistent with the earlier assumption

  6. Dec 13, 2008 #5
    Problem of underlying measure space

    I've understood that if [itex]\Omega[/itex] is some finite index set of the energy states, then the Boltzmann probability measure is given by formula

    P_T(\{\sigma\}) = \frac{\exp\big(-\frac{H(\sigma)}{k_{\textrm{B}} T}\big)}
    {\underset{\sigma'\in\Omega}{\sum} \exp\big(-\frac{H(\sigma')}{k_{\textrm{B}} T}\big)}.

    If, instead of a finite index set, we have some arbitrary measure space [itex](\Omega,\mu)[/itex] as the index set, then the formula becomes naturally generalized as

    P_T(X)= \frac{\underset{X}{\int} d\mu(\sigma) \exp\big(-\frac{H(\sigma)}{k_{\textrm{B}} T}\big)}
    {\underset{\Omega}{\int} d\mu(\sigma') \exp\big(-\frac{H(\sigma')}{k_{\textrm{B}} T}\big)}.

    The problem is that usually physical arguments will only tell what [itex]\Omega[/itex] is, and [itex]\mu[/itex] is left as a mystery. I'm curious to know if there exists some established theory on how [itex]\mu[/itex] should be solved in general.

    I've noticed that some times [itex]\mu[/itex] can be solved by taking the most natural measure and not worrying about it more. For example with a gas molecule the index set is [itex]\mathbb{R}^3[/itex], and indexes are the possible velocities [itex]v[/itex] of the molecule. It turns out that the correct Maxwell-Boltzmann velocity distribution follows if one chooses to use the standard Lebesgue measure [itex]m_3[/itex] (or [itex]d^3x[/itex]).

    Is there enlightening explanations out there, about why precisely the standard measure [itex]m_3[/itex] works so well?

    The example I gave in the first post is an example of a more difficult situation, where the measure is not so obviously guessable.
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