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Homework Help: Dirac delta function / Gibbs entropy

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    This is an issue I'm having with understanding a section of maths rather than a coursework question. I have a stage of the density function on the full phase space ρ(p,x);

    [itex] ρ(p,x) = \frac {1}{\Omega(E)} \delta (\epsilon(p,x) - E) [/itex]

    where [itex]\epsilon(p,x)[/itex] is the Hamiltonian.

    Then the logarithm of that is taken to obtain;

    [itex] ln(ρ(p,x)) = -ln(\Omega (E)) + ln(\delta (\epsilon(p,x) - E)) [/itex]

    And then both sides are multiplied by ρ(p,x) and integrated over the phase space

    [itex] ∫ρ(p,x)ln(ρ(p,x))dpdx = -ln(\Omega (E))∫ρ(p,x)dpdx + ln(\delta (\epsilon(p,x) - E))∫ρ(p,x)dpdx [/itex]

    The term ∫ρ(p,x)dpdx equals one clearly, and the term on the LHS stays as it is. And [itex] ln(\Omega (E)) = S [/itex] (omitting the Boltzmann constant for the moment [S = S/k]).

    However my text book completely negates the [itex] ln(\delta (\epsilon(p,x) - E))[/itex] term, the dirac delta function. As if it goes to zero. Leaving us with;

    [itex] S = -∫ρ(p,x)ln(ρ(p,x))dpdx [/itex]

    Why can we do this? Why is that term equal to zero? I understand the function of the dirac delta function, δ(x) is zero anywhere else except x = 0, where it is equal to infinity. The intergal of the dirac delta function over all space is equal to one. So why then does [itex] ln(\delta (\epsilon(p,x) - E))[/itex] equal zero? That makes little sense to me, where [itex]\epsilon(p,x)[/itex] is not equal to zero, then we're logging the number zero, and when it's equal to E, we're logging infinity. Neither case makes much sense, so why can we omit it/ say it is equal to zero?
  2. jcsd
  3. Oct 7, 2012 #2


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    I'm not sure how to handle the dirac delta function here either.

    Usually the microcanonical distribution is defined in such a way as to allow for a small uncertainty in the total energy E. So, instead of an energy surface in phase space, you deal with a very thin energy “slab” of thickness dE. Then the density of states would be [itex] ρ(p,x) = \frac {1}{\Omega(E,dE)} \Delta (p,x) [/itex] where [itex]\Delta (p,x)[/itex] has the value 1 if ##(p,x)## lies in the slab and 0 otherwise. ##\Omega(E,dE)## is the volume in phase space of the slab. So, [itex]\Delta (p,x)[/itex] would take the place of the dirac delta function.

    Now, [itex] ∫ρ(p,x)ln(ρ(p,x))dpdx [/itex] should give you what you want. But this doesn't really answer your question of how to do it directly with the dirac-delta function.
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