Why is in statistical thermodynamics $β=\frac{∂\ln\left(Ω\right)}{∂U}=\frac{1}{k_{B}T}$? Why is in statistical thermodynamics $β=\frac{∂\ln\left(Ω\right)}{∂U}=\frac{1}{k_{B}T}$? $β$ appears later during the derivation of the Boltzmann canonical partition function.
$Ω$ is the number of accessible microstates for a canonical ensemble (I think, that it is important here, that it is not a microcanonical ensemble).
$U$ is the internal energy of the system.
$k_{B}$ is the Boltzmann constant and $T$ is the absolute temperature.
Is this relationship related to the fact that the temperature of a system depends only on its translational kinetic energy and not on its potential energy or other forms of kinetic energy?
This equation appeared in the lecture without much explanation, and I want to understand the role of this equation: Is it a definition? An axiom? A theorem (i.e. something that can be proven)? Is there a unique answer or does the answer depend on the context?
I'm particularly interested in knowing if the validity of this equation can be proved or if it is a postulate.
In my own textbook it is said about this equation: "Strictly speaking, we can only conclude here that $β$ must be a monotonically decreasing function of the temperature. [After introducing] the statistical interpretation of the entropy [it will be] clear that $β$ [must be] proportional to the reciprocal temperature [in order to be] phenomenological consistent with the statistical definition of entropy."
So, I understand that only because of the assumption made by Boltzmann that $S=k_{B}\ln\left(Ω\right)$ we can conclude that $β=\frac{∂\ln\left(Ω\right)}{∂U}=\frac{1}{k_{B}T}$, because $\left(\frac{∂S}{∂U}\right)_{V,N}=\frac{1}{T}$. In conclusion, the validity of the relationship $β=\frac{∂\ln\left(Ω\right)}{∂U}=\frac{1}{k_{B}T}$ cannot be demonstrated and is only a consequence of the Boltzmann definition of entropy. Is this correct?
 A: Here is one way to identify which thermodynamic properties correspond to which statistical properties.

*

*Start with the entropy of classical thermodynamics:
$$\frac{dS}{k} = \frac{dU}{kT}-\frac{PdV}{kT} + \frac{\mu dN}{kT}
\Rightarrow 
\boxed{\frac{S}{k} = \frac{U}{kT}-\frac{PV}{kT} + \frac{\mu N}{kT}}
$$


*Now to statistics: postulate the number of microstates $\Omega(E,V,N)$ to have the homogeneity property $$\log\Omega(\lambda E,\lambda V,\lambda N) = \lambda\log\Omega(E,V,N)$$


*Apply Euler's theorem
$$\boxed{\log\Omega 
= \frac{\partial\log\Omega}{\partial E}E
+ \frac{\partial\log\Omega}{\partial V}V
+ \frac{\partial\log\Omega}{\partial N}N
}
$$


*Assuming $E$ and $U$ to be the same thing, postulate the equalities by direct comparison between the boxed results:
$$ \log\Omega = \frac{S}{k}$$
$$ \frac{\partial\log\Omega}{\partial E} = \frac{1}{kT}$$
$$ \frac{\partial\log\Omega}{\partial V} = -\frac{P}{kT}$$
$$ \frac{\partial\log\Omega}{\partial N} = \frac{\mu}{kT}$$


*Accept these as exact equalities until contradicted  by experiment
Why should $\log\Omega$ be homogeneous, as postulated in step #2? Naive lattice calculations and calculations with non interacting particles (classical or quantum) show this to be the case.
