Maxwell–Boltzmann distribution derivation using only thermodynamic equations

Question

I want to derive the Maxwell–Boltzmann distribution using only the fundamental relation and the Boltzmann's entropy formula.

$dU=TdS-pdV\tag1$ $S=k_b ln(\Omega)\tag2$

So using only the two equations I get:

$$dS=k_b\frac{1}{\Omega} d\Omega$$

$$dS=\frac{1}{T}dU$$

$$\frac{1}{T}dU=k_b\frac{1}{\Omega} d\Omega$$ $$\frac{1}{Tk_b}dU=\frac{1}{\Omega} d\Omega$$ $$\frac{U}{Tk_b}=ln(\Omega)$$

$U=\frac{mv^2}{2}$ I also substitute the kinetic energy formula.

$$\frac{mv^2}{2Tk_b}=ln(\Omega)$$

$$\Omega=exp\left(\frac{mv^2}{2Tk_b}\right)$$

$$\frac{1}{\Omega}=exp\left(\frac{-mv^2}{2Tk_b}\right)$$

$$P(v)=\frac{exp\left(\frac{-mv^2}{2Tk_b}\right)}{\sum exp\left(\frac{-mv^2}{2Tk_b}\right)}$$

$$P(v)=\frac{exp\left(\frac{-mv^2}{2Tk_b}\right)}{\displaystyle Z}$$

where ${\displaystyle Z=\sum exp\left(\frac{-mv^2}{2Tk_b}\right)}={\displaystyle Z=\sum _{i}e^{-\beta E_{i}}}$ is the canonical partition function (whatever that means).

and so

$${\displaystyle f(\mathbf {v} )~d^{3}\mathbf {v} =A\exp \left(-{\frac {mv^{2}}{2k_{\text{B}}T}}\right)~d^{3}\mathbf {v} }$$

where $A$ is the normalization constant I obtain by solving the Gaussian integral.

I want to derive it like this, because I am not yet familiar so much with statistical mechanics, but I know classical thermodynamics and I wonder if this derivation is "legal" (whatever that means).

Answer 1

Probably not legal, for a few reasons. The thermal energy $U$ tends to depend on $T$, so that side $dU/T$ needs to be integrated properly. Additionally, the probability should be proportional to $\Omega$, not $1/\Omega$.

Thermodynamic quantities typically deal with macroscopic numbers of particles, so the thermal energy $U$ refers to $N$ particles. The equipartition theorem for a monatomic ideal gas, which perhaps you want to avoid, equates $U=3NkT/2$ with $U=\sum_i \frac{m v_i^2}{2}=Nm\bar{v^2}/2$ for root-mean-square velocity $v_{RMS}=\sqrt{\bar{v^2}}$. This makes manifest the requisite temperature dependence and particle-number dependence that make the derivation in question "illegal."

Answer 2

Your approach is doomed to fail.

The classical ideal gas obeying Maxwell-Boltzmann distribution rather famously has the quantum-defying Sackur-Tetrode entropy. It is unbounded below as $T\to0$, and so then $\Omega$ is undefine-able.

Boltzmann's formula should be paired with quantum theory, i.e. you should derive either Bose-Einstein distribution or Fermi-Dirac distribution, or both, and then take the limits to Maxwell-Boltzmann if you want that.