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!

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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dirac delta function / Gibbs entropy
  1. Dirac Delta Function (Replies: 1)

Loading...