1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equality of expectation value integral over coordinate space and over energy

  1. Aug 3, 2012 #1
    Dear all,

    I'm wondering, how one could justify mathematically the equality
    [itex]\int O(E(\vec{x}_1,...\vec{x}_N)) exp(-\beta E(\vec{x}_1,...,\vec{x}_N)) d\vec{x}_1...d\vec{x}_N[/itex] = [itex]\int g(E) O(E) exp(-\beta E) dE[/itex]

    where O(E(x)) is an observable and g(E) the density of states.

    Is there a mathematical justification for the equality?

  2. jcsd
  3. Aug 3, 2012 #2
    This is a non-trivial question. There IS a mathematical justification (that you'd have to be really clever to figure out), but I think it's much easier to think about the PHYSICAL justification.

    Regardless of whatever variable you choose to use in your integral, the probability of occupation of a state is dictated by Maxwell-Boltzmann statistics. Whether you express the Boltzmann factor as a function of E or a function of x and x's time derivative is irrelevant, since the Boltzmann factor only depends on the energy of that state (and the energy is typically a function of x and x's time derivatives). Here's a mathematical way of stating that:

    P(E)=\frac{e^{\frac{-E}{kT}}}{\int _Ee^{\frac{-E}{kT}} dE}\Leftrightarrow P(x_1, x_2, ..., x_n, \dot{x}_1, \dot{x}_2, ..., \dot{x}_n)=\frac{e^{-\frac{E(x_1, x_2, ..., x_n, \dot{x}_1, \dot{x}_2, ..., \dot{x}_n)}{kT}}}{\int_{x_1}\int_{x_2}...\int_{x_n}\int_{\dot{x}_1}\int_{\dot{x}_2}...\int_{\dot{x}_n} e^{-\frac{E(x_1, x_2, ..., x_n, \dot{x}_1, \dot{x}_2, ..., \dot{x}_n)}{kT}}dx_1 dx_2 ... dx_n d\dot{x}_1d\dot{x}_2...d\dot{x}_n}
    where I have assumed the energy is only a function of the generalized coordinates and their time derivatives. (Or more precisely, the x dots should be generalized momenta.)

    The reason for this is physically obvious: whether you label a state based on its energy or its coordinates/momenta, you still have the same Boltzmann factor to dictate the occupation of that state.
    Last edited: Aug 3, 2012
  4. Aug 3, 2012 #3
    Thanks for your answer.

    Well, I should have mentioned that the physical argumentation is clear to me. I'm really interested in the pure mathematical justification.

    (By the way: the (unnormalized) probability density, when expressed as a function of the energy, is given by [itex]g(E) e^{\frac{-E}{kT}}[/itex], the density of states is somewhat like the Jacobian determinant that emerges when performing a variable-transformation on probability densities)
  5. Aug 3, 2012 #4
    Well, in thermodynamics, a routine way of proving the equivalence of statements like the big equation I just posted is by simply evaluating the two integrals and showing they give the same probabilities. It's fairly easy to do this with the standard kinds of energy functions, which are quadratic in x's and xdot's. I'm fairly certain it holds for any power of x and xdots too. I think proving it for an arbitrary E(x's, xdot's) would be quite challenging, if not impossible.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook