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The definition of the entropy is :

$$S=k_b \ln(\Omega(E))$$ for a system that has energy $E$ fixed.

But when we look at the definition of the number of accessible microstates, we have :

$$ \Omega(E) = \int \frac{dp dq}{h}\delta(E-H)$$ that has the unit of the inverse of an energy.

But we write it inside of a logarithm in the definition of the entropy, thus it should be unitless.

Thus, there is something I don't totally understand.

Can we define $\Omega(E)$ with a Dirac delta like this?

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    $\begingroup$ Possible duplicate of number of states in microcanonical ensemble $\endgroup$
    – valerio
    Commented Jan 31, 2018 at 11:51
  • $\begingroup$ I think you are missing something because the delta function has units of inverse the argument. If $q$ and $p$ are the canonical position and momentum, then your result has units of inverse energy. I am guessing the arguments in the delta function are normalized (i.e., unitless)? $\endgroup$
    – honeste_vivere
    Commented Jan 31, 2018 at 14:12

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This question is almost an exact duplicate of this one.

However, long story short is that the correct definition is

$$ \Omega(E) = E_0 \int \frac{d^{3N}p d^{3N}q}{h^{3N} N!}\delta(E-H)$$

where $E_0$ is an arbitrary constant with the dimensions of energy whose value has no influence on the thermodynamic quantities. Also notice that I included the factor $N!$ for the correct Boltzmann counting.

For more details, see the linked question.

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  • $\begingroup$ This definition is arbitrary, because $E_0$ is arbitrary. The very reason for dividing by $h^{3N}$ is to get a value of number of states that does not depend on choice of units like classical phase volume does. So if there is $E_0$, it should have a definite value, not arbitrary value. $\endgroup$
    – Ján Lalinský
    Commented Nov 14, 2023 at 2:18
  • $\begingroup$ Maybe $\Omega(E) = E \int \frac{d^{3N}p ~d^{3N}q}{h^{3N}N!} \delta(E-H)$. $\endgroup$
    – Ján Lalinský
    Commented Nov 14, 2023 at 2:20
  • $\begingroup$ @JánLalinský In the continuum, the "number of states" doesn't really make sense. So you either define a "number density of states" (with dimensions $1/E$) or you choose some way to normalize the density to get a dimensionless number so you can take a logarithm. This normalization shouldn't matter in the end since we're always interested in changes in the entropy in thermodynamics. $\endgroup$
    – Andrew
    Commented Nov 14, 2023 at 2:36
  • $\begingroup$ I always thought the "integration over phase space" was a bit of a hack to try to get a statistical-mechanics-like counting argument in a continuous classical theory. If you do a counting of states quantum mechanically in a situation where the number of states is actually finite (like a particle in a box) you can actually calculate the normalization. I'm not sure offhand what happens if you try to take a limit of a particle in a box where the box becomes infinite (getting back to a continuum of states), maybe you get some kind of infrared divergence. $\endgroup$
    – Andrew
    Commented Nov 14, 2023 at 2:41
  • $\begingroup$ @Andrew Actually, there is a way to define number of states even for continuum of possible states, e.g. via restriction to psi functions that have nodes on the box walls; the possible psi functions are from a continuous space and there is uncountable infinity of them, but the number of states is a different concept, because only eigenfunctions count. $\endgroup$
    – Ján Lalinský
    Commented Nov 14, 2023 at 4:27

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