# Unit of number of microstates $\Omega(E)$

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 ?

• Possible duplicate of number of states in microcanonical ensemble – valerio Jan 31 '18 at 11:51
• 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)? – honeste_vivere Jan 31 '18 at 14:12

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