The Boltzmann entropy equation is commonly used in statistical interpretation of entropy to relate entropy $S$ with the number of microstates $\Omega$:

$$S=k\ln(\Omega) \, .$$

The classical thermodynamic entropy $S$ is related to heat $Q$ and absolute temperature $T$:

$$dS=\frac{\delta Q}{T} \, .$$

Now, the number of microstates $\Omega$ for the distribution of energy to the particles in a system happens to depend on temperature $T$, the partition function $P$, and the number of particles $n$ (assuming constant volume):

$$\ln(\Omega_{\mathrm{thermal}}) = n\ln(P) + \frac{U}{kT}$$

$$P=\sum{\exp \left( {-\frac{E_i}{kT}} \right)} \, .$$

So it seems that the dependence of $\Omega_{\mathrm{thermal}}$ on temperature in the above equation links the classical thermal entropy and the statistical thermal entropy.

However, I'm not aware of $\Omega_\text{config}$, the number of spatial configuration microstates, as having a similar dependence on temperature $T$ or internal energy $U$.

So why is the Boltzmann entropy equation is also used for the case of configuration entropy?

Doing so would seem to imply that there is heat transfer involved in changing the configuration entropy.


Boltzmann entropy equation IS configurational entropy.

Also, the 3rd Eq. you have written is wrong, or at least P isn't pressure. P has to be dimensionless. If I had to guess, I'd say P is probability and that expression comes from considering the microcanonical ensemble, i.e., an ensemble in which you have the energy, the volume and the number of particles fixed.

The connection between Boltzmann entropy and thermodynamic entropy, to my knowledge, comes when you take the entropy,

$S=-k\int dqdp P(q,p)\log P(q,p)+\text{constraints}$,

and maximize it then you obtain an equation which are very similar to the thermodynamics Eqs. However, this is on a mathematical standpoint.

There is an interesting discussion about this here: Proving that the Boltzmann entropy is equal to the thermodynamic entropy

  • $\begingroup$ Thanks Henry for the correction on "P" and the phase space integral. Unfortunately, I will need to study up on generalized position and momentum in Hamiltonian mechanics to make sense of that. Also, would you clarify your statement "Boltzmann entropy IS configurational entropy"? Boltzmann entropy depends on the number of microstates, but there are two different ways to count the microstates. One way is for how energy is distributed. The other way is for the spatial positions that particles occupy. $\endgroup$ – kfd182 Aug 5 '16 at 6:06
  • $\begingroup$ I think maybe your mixing the microcanonical ensemble, which is with fixed volume, particles and energy and you count the microstates by counting the spatial arrangement and orientation of the particles (depending on the specific problem your dealing with), with the canonical ensemble which is with fixed volume particles and the constriction $\int dqdp P(q,p)H(q,p)=E$. In the later case, you weigh the microstate with the Boltzmann factor, i.e., $\exp(-\beta H)$, since you are considering rare events (rare microstates). Is this close to what you had in mind? $\endgroup$ – Henry Kel Aug 6 '16 at 17:19

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