# Dirac delta function / Gibbs entropy

1. Oct 6, 2012

### Silversonic

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);

$ρ(p,x) = \frac {1}{\Omega(E)} \delta (\epsilon(p,x) - E)$

where $\epsilon(p,x)$ is the Hamiltonian.

Then the logarithm of that is taken to obtain;

$ln(ρ(p,x)) = -ln(\Omega (E)) + ln(\delta (\epsilon(p,x) - E))$

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

$∫ρ(p,x)ln(ρ(p,x))dpdx = -ln(\Omega (E))∫ρ(p,x)dpdx + ln(\delta (\epsilon(p,x) - E))∫ρ(p,x)dpdx$

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

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

$S = -∫ρ(p,x)ln(ρ(p,x))dpdx$

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 $ln(\delta (\epsilon(p,x) - E))$ equal zero? That makes little sense to me, where $\epsilon(p,x)$ 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. Oct 7, 2012

### TSny

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 $ρ(p,x) = \frac {1}{\Omega(E,dE)} \Delta (p,x)$ where $\Delta (p,x)$ 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, $\Delta (p,x)$ would take the place of the dirac delta function.

Now, $∫ρ(p,x)ln(ρ(p,x))dpdx$ 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.