How is this relation between energy, temperature and number of microstates derived?

Question

In "Concepts in Thermal Physics" (second edition) by Blundell and Blundell, temperature is defined using the following relation: $\frac{1}{k_BT}=\frac{\mathrm{d}\ln(\Omega)}{\mathrm{d}E}$. I am wondering how this relationship between T, E, and $\Omega$ came to be and how it fits in with the other definitions of temperature.

Answer 1

It's the usual definition. The Boltzmann entropy, valid when microstates have equiprobability, is: $S = k_B \ln\Omega $. From thermodynamics we know that $\frac{1}{T} = \frac{dS}{dE}$, using the Boltzmann entropy we have: $\frac{1}{k_B T} = \frac{d\ln\Omega}{dE}$

Answer 2

This connection is linked from both the thermal dynamics and the statistical microcanonocal ensemble.

From thermal dynamics, the entropy is defined as $dS =\frac{dQ}{T}$. Using the first law of thermal dynamics: $$ \tag{1} dS = \frac{dQ}{T} = \frac{1}{T} \left[ dU + PdV -\mu dN \right] $$ Then, treat entropy as a atate function $S(U,V,N)$, the chain-rule of differentiation: $$\tag{2} dS = \frac{\partial S}{\partial U}\Big\vert_{N,V} dU + \frac{\partial S}{\partial V}\Big\vert_{U,N}dV +\frac{\partial S}{\partial N}\Big\vert_{U,V} dN. $$

Compare Eq.(1) and Eq.(2), we have Maxwell's relations: \begin{align} \frac{1}{T} =& \frac{\partial S}{\partial U}\Big\vert_{N,V}\tag{3}\\ \frac{P}{T} =&\frac{\partial S}{\partial V}\Big\vert_{U,N} \\ \frac{\mu}{T} =& -\frac{\partial S}{\partial N}\Big\vert_{U,V} \\ \end{align}

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Then, from the statistical microcanonical ensemble scheme, the total number of micro-states $\Omega$ is calculated as function of $\Omega(E, V, N)$, where $E = \sum_i^N \epsilon_i$. The equilibrium condition is assume by maximum $\Omega(E, V, N)$ with a constrain that $E = \sum_i^N \epsilon_i = U$, where $U$ is the given internal energy of the thermal dynamics.

Under equilibrium condition, the entropy is defined by Boltzmann formula: $$ \tag{4} S = S(U, V, N) = K_b \ln \Omega(E=U, V, N) $$

Therefore, we maximumize the $\ln \Omega(E, V, N)$ with constrian $E = U$ \begin{align} \frac{\partial }{\partial E} &\left\{ \ln\Omega(E, V, N) -\lambda (E-U) \right\}=0\\ \Longrightarrow &\,\,\,\frac{\partial \ln\Omega(E, V, N) }{\partial E}\big\vert_{E=U} - \lambda = 0 \tag{5} \end{align}

Eq.(5) is assumed to be in thermal equilibrium ($E$ is now replaced by $U$), $S = K_b \ln \Omega$, and We may now compare equation (5) with Maxwell relation in thermal dynamics, Eq.(3).

$$ \lambda = \frac{1}{K_b T}. $$

And rewrite Eq.(5) as: $$ \frac{\partial \ln\Omega(E, V, N) }{\partial E}\big\vert_{E=U}=\frac{1}{K_b} \frac{\partial S }{\partial U} = \frac{1}{K_b T} $$

Answer 3

Here is the argument why statistical entropy has such form. In thermodynamics temperature is defined as $\frac{1}{T}=\frac{\partial S}{\partial E}$, where $E$ is the "energy" of the system, and $S$ is the "entropy". The partial derivative is taken under the condition that all other extensive quantities, like volume and the number of particles, are held constant. The "entropy" must reach maximum at the equilibrium state of the system (given energy, volume and the number of particles are held constant).

Statistical mechanic tries to fit into the definitions of thermodynamics, and one can indeed show that a classical system, described by the probability distribution over the microscopic states, in the limit of vanishingly small (and energy preserving) external noise will distribute uniformly over all available states with the same energy.

For something to be a thermodynamic system, entropy must reach maximum over the equilibrium distribution, and since we know that equilibrium distribution is the "maximally spread distribution" (i.e. uniform on the all available phase space), entropy should be "something that increases as the probability distribution takes larger part of the phase space". What fixes $S=\ln(\Omega)$ is the requirement of the total entropy of non-interacting systems to be additive (for two non-interacting systems $\Omega_{total}=\Omega_1\Omega_2$, and therefore $S_{tot}=S_1+S_2$). Plugging the entropy into the definition of temperature, we obtain the desired connection of statistical mechanics of classical systems and thermodynamics: $$ \frac{1}{T}=\frac{\partial \ln{\Omega}}{\partial E} $$