I'm introducing myself to statistical mechanics using two books: Introduction to Statistical Physics by S. Salinas, and Statistical Physics of Particles, by Mehran Kardar. Both textbooks work on an example of a isolated system which is composed of two systems which are allowed to exchange heat with each other only. System $1$ and $2$ have energy $E_1$ and $E_2$ respectively such that the total energy $E_1+ E_2$ is fixed( microcanonical ensemble). The systems have phase space volumes $\Omega_1(E_1)$ and $\Omega_2(E_2=E- E_1)$, respectively. In salinas book, it is assumed that the energy $E_1$ can assume only discrete values, so the phase space volume of the composite system is given by (eq 4.5 of textbook)

$$\Omega(E) = \sum_{E_1=0}^E \Omega_1(E_1)\Omega_2(E- E_1) \tag1$$

However, Mehran Kardar considers that the energy is a non-countable parameter, such that the total phase space volume is (eq. 4.4 of the textbook)

$$ \Omega(E) = \int_0 ^E dE_1\Omega_1(E_1)\Omega_2(E- E_1) \tag2$$

My problem: This seems to be incorrect to me, since Integration over the energy $E_1$ should imply that the product $\Omega_1 \Omega_2$ would be some kind of density of phase space volume, and not the phase space volume by itself. Also we have that the dimension of $\Omega(E)$ in $(1)$ is different from the $\Omega(E)$ from $(2)$. What point am I missing?

Edit: As far as I have read about this topic, some texts define $\Omega(E)$ as the accessible "area" in system's phase space associated with energy $E$, but $\Omega(E)$ in actually the density of states of the system, not an area. Interpret $\Omega(E)$ as a density of microstates is ok to me since eq. $(2)$ means that $\int dE_1 \Omega_1 \Omega_2$ is also a density. But i'm quite confused about the discrete energy spectrum in a classical description, since the Interpretation of $\Omega(E)$ as a density in eq. $(1)$ makes no longer sense to me.

  • 2
    $\begingroup$ I think that in $(2)$, $\Omega(E)$ should be seen as a density rather than a volume, i.e. $\Omega_1,\Omega_2$ are then also densities $\endgroup$ Jan 19, 2022 at 16:30
  • $\begingroup$ @Mathphys meister : I agree, but none of the books/notes about microcanonical ensemble i have read talks about this $\endgroup$ Jan 19, 2022 at 19:08
  • $\begingroup$ You are correct that Eq (2) is dimensionally inconsistent (given that $\Omega$ is unitless and $E$ is an energy. You can correct this by adding in a "small" energy scale $\Delta E$ and writing the integral like: $\int\frac{dE_1}{\Delta E} \Omega_1(E_1)\Omega_2(E - E_1)$ $\endgroup$
    – hft
    Jun 12, 2023 at 16:06

1 Answer 1


When we pass from the discrete to the continuous domain probabilities become probability densities. The same applies to the microcanonical partition function. One often views this is a minor mathematical issue, but the truth of the matter is it throws off students who approach the subject for the first time.

The correct way to define the microcanonical partition function in continuous energy $E$ is this:

  • Define $\Omega(E)$ (dimensionless) to be the number of microstates with energy anywhere between $0$ and $E$. The number of microstates in a narrow energy band $dE$ is $d\Omega(E)$.

  • Define the microcanonical partition function $\omega(E)$ as $$\omega(E) = \frac{d\Omega(E)}{dE}$$

Clearly, $\omega(E)$ is a density, number of microstates per unit energy. The number of microstates in energy band $dE$ is $$d\Omega = \omega dE \Rightarrow \int d\Omega = \int \omega dE$$ Then all integrals in $E$ involve the density $\omega$, not $\Omega$.

I recommend using $\Omega$ when dealing with discrete systems and $\omega$ when dealing with continuous. This notation is not universally accepted but avoids confusion and I have adopted it in my lectures.


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