for the non-physicists, all you need to know to answer my question is that I'm talking about a $6N$ dimensional space of the coordinates $\{\vec{q}_i,\vec{p}_i\}_{i=1} ^{N}$ which I call the phase space and I define a probability distribution function on that space with a normlization constat $\Omega$.

I have a problem with the definition of $\Omega(E,V,N)$ — the number of microstates with $V$, $N$ and energy $E$. It starts with the definition of the PDF. If for a given Hamiltonian $H$ defined on phase space, one defines (for a fixed $E$) the PDF as follows:

$P_E(\{q_i,p_i\})=\dfrac{1}{\Omega(E,V,N)}\cdot \begin{cases}1\ , \ H(\{q_i,p_i\})=E \\0 \ ,\ else \end{cases}$

then, In this case it follows that in order for $P$ to be normalized we should demand that: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{H=E} d\Gamma$$

where I use the notation $d\Gamma =\prod _i d^3q_id^3p_i$.

but this integral is zero because the domain of integration is a set of measure zero. So as I saw, there are two options. the first is to redefine the PDF as: $$P(\{q_i,p_i\})=\frac{1}{\Omega(E,V,N)}\cdot \delta(H(\{q_i,p_i\})-E)$$ and then we get: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int \delta(H(\{q_i,p_i\})-E) d\Gamma$$ which solves the problem of zero integral but now it has the dimension of $\dfrac{1}{[E]}$ which is problematic if I want to consider $\log \Omega(E,V,N)$.

The second option is to redefine the PDF as:

$P_E(\{q_i,p_i\})=\dfrac{1}{\Omega(E,V,N)}\cdot \begin{cases}1\ , \ H(\{q_i,p_i\})\in [E,E+dE] \\0 \ ,\ else \end{cases}$

and then we get: $$\Omega(E,V,N)=\frac{1}{h^{3N}}\cdot\int_{E<H<E+dE} d\Gamma$$ which also solves the problem of zero integral and the dimensionality issue (it is now a dimensionless quantity) but the downside now is that it feels to me as if it is not well defined in this fashion because I defined the PDF by some infinitesimal qunatity. I know that there is a formal consturction of the infinitesimals but I'm not familiar with it.

can someone please shed some light on which of these 2 definitions are "better"?

I asked this question on physics stack exchange here but didn't get any good answer, so I guess it's more of a mathematical issue.


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