2
$\begingroup$

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 ?

$\endgroup$
  • 3
    $\begingroup$ Possible duplicate of number of states in microcanonical ensemble $\endgroup$ – valerio Jan 31 '18 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 Jan 31 '18 at 14:12
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.