Partition Function

Boltzmann Factor

Consider a system connected to an isolated reservoir, in which the system and the reservoir as a whole is isolated. The multiplicity of the system at state \(s\) is \(\Omega_S = 1\) so the total multiplicity is \(\Omega_S \cdot \Omega_R = \Omega_R\). The probability ratio of the system between two state \(s_1\) and \(s_2\) is \begin{align} \frac{P(s_1)}{P(s_2)} &= \frac{\Omega_R(s_1)}{\Omega_R(s_2)} \\ &= \frac{e^{S_R(s_1)/k}}{e^{S_R(s_2)/k}} \\ &= e^{(S_R(s_1) - S_R(s_2))/k} \end{align} where \(\Omega_R(s)\) and \(S_R(s)\) are the multiplicity and entropy of the reservoir at state \(s\). If the reservoir volume and particles are fixed but the system is allowed to exchange energy with the reservoir $$ dS = \frac{1}{T}(dU + PdV -\mu dN) = \frac{dU}{T}$$ So, $$\frac{P(s_1)}{P(s_2)} = e^{(U_R(s_1) - U_R(s_2))/kT}$$ Let \(U_{tot}\) be the total energy, i.e. energy of the reservoir plus energy of the system. As the reservoir and the system as a whole is isolated, \(U_{tot}\) is a constant and $$U_R = U_{tot} - E$$ where \(E\) is the energy of the system. Hence, $$\frac{P(s_1)}{P(s_2)} = e^{-(E(s_1) - E(s_2))/kT}$$ And the probability at a state \(s\) is proportional to the term $$ \bbox[5px,border:2px solid #666] { P(s) \propto e^{-E(s)/kT} }$$ which is called the Boltzmann factor.

Partition Function

Let \(C\) be a constant such that $$P(s) = C e^{-E(s)/kT}$$ As the sum of the probability of all states must be equal to 1, $$\sum_sP(s) = C \sum_s e^{-E(s)/kT} = 1$$ Define $$ \bbox[5px,border:2px solid #666] { Z = \sum_s e^{-E(s)/kT} }$$ Then, substitute \(C = \frac{1}{Z}\) into probability equation, $$P(s) = \frac{1}{Z}e^{-E(s)/kT}$$ Let \(\beta = 1/kT\). The average energy of the system is \begin{align} \bar{E} &= \sum_s P(s)E(s) = \sum_s \frac{1}{Z}e^{-E(s)\beta} E(s)\\ &= \frac{1}{Z} \left(-\frac{\partial}{\partial \beta} \right)\sum_s e^{-E(s)\beta} \\ &= -\frac{1}{Z} \frac{\partial Z}{\partial \beta} \end{align} Thus, the average energy in terms of the partition function is $$ \bbox[5px,border:2px solid #666] { \bar{E} = -\frac{1}{Z} \frac{\partial Z}{\partial \beta} }$$

Helmholz Free Energy

\begin{align} \bar{E} &= F + TS = F - T\frac{\partial F}{\partial T} \\ &= T^2\left(\frac{F}{T^2} - \frac{1}{T}\frac{\partial F}{\partial T} \right) \\ &= kT^2 \frac{\partial}{\partial T}\left(-\frac{F}{kT}\right) \\ \end{align} As the average energy is equal to $$\bar{E} = -\frac{1}{Z}\frac{\partial Z}{\partial \beta}$$ we have $$ \bbox[5px,border:2px solid #666] { Z = e^{-\beta F} }$$

Equipartition Theorem

When

• the temperature is high such that number of energy levels is much smaller than \(kT\)
• the energy of each degree of freedom \(E(q) \propto q^2 \), where \(q\) can be considered as continuous

then, the average energy of each degree of freedom at temoerature \(T\) is given by \(\frac{1}{2}kT\). This is called the equipartition theorem.
Proof
Consider a system with \(f\) degrees of freedom, then the energy of the system is $$E = \sum_i^f c_i q_i^2$$ The partition function is $$Z = e^{-\sum_i^f \beta c_i q_i^2} = \frac{1}{\Delta q_1}\frac{1}{\Delta q_2}...\frac{1}{\Delta q_f} \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}...\int^{\infty}_{-\infty}e^{- \sum_i^f\beta c_i q_i^2} dq_1 dq_2 ... dq_f$$ Let \(x = q\sqrt{\beta c}\) $$\int^{\infty}_{-\infty} e^{-\beta cq^2} dq = \frac{1}{\sqrt{\beta c}}\int^{\infty}_{\infty} e^{-x^2}dx = \sqrt{\frac{\pi}{\beta c}}$$ Let \(C_i = \sqrt{\frac{\pi}{c}}\), the partition function becomes $$Z = C_1C_2...C_f \beta^{-f/2}$$ Let \(C = C_1C_2...C_f\), $$Z = C\beta^{-f/2}$$ The average energy is \begin{align} \bar{E} &= \frac{1}{Z}\frac{\partial Z}{\partial \beta} \\ &= \frac{\beta^{f/2}}{C}\left(-\frac{fC}{2}\beta^{-3f/2}\right) \\ &= \frac{f}{2}\beta^{-1}\\ &= \frac{f}{2}kT \end{align}

Maxwell's Speed Distribution

For a system with particles that carry only kinetic energy, i.e. $$E = \frac{1}{2}mv^2 $$ then the probability of a particle with speed \(v\) is $$P(v) = C e^{-mv^2/2kT}$$ where \(C\) is a constant. Let \(D(v)\) be the speed distribution function. Since sum of all probability must be 1, we have $$ \int^{\infty}_{0} D(v)dv = 4\pi C\int^{\infty}_{0} e^{-mv^2/2kT}dv = 1 $$ Let \(x = v\sqrt{m/2kT}\) \begin{align} \int^{\infty}_{0} e^{-mv^2/2kT}dv &= \int^{\infty}_{0} \frac{2kT}{m} x^2e^{-x^2}\sqrt{\frac{2kT}{m}}dx \\ &= \left(\frac{2kT}{m} \right)^{3/2}\frac{\sqrt{\pi}}{4} \end{align} So $$ C = \left(\frac{m}{2kT}\right)^{3/2}\frac{4}{\sqrt{\pi}}\frac{\pi}{\pi} = 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2}$$ Thus, the speed distribution function is $$ \bbox[5px,border:2px solid #666] { D(v) = \left(\frac{m}{2\pi kT}\right)^{3/2}4 \pi v^2 e^{-mv^2/2kT} }$$

Multiple Particles

Suppose there are two particles in a system. When in isolated system, they each carry energy \(E_1\) and \(E_2\). If the two particles are not interacting, i.e. no potential energy due to each other, the total energy is $$ E_{tot} = E_1 + E_2 $$ Then the partition function is $$ Z = \sum_s e^{-\beta E_{tot}} = \sum_s e^{-\beta(E_1 + E_2)}$$ If the particles are distinguishable, for each microstates of particle 1 forms a microstate with each microstates of particle 2, the partition function becomes \begin{align} Z &= \sum_{s_1}\sum_{s_2} e^{-\beta(E_1 + E_2)}\\ &= \sum_{s_1}e^{-\beta E_1} \sum_{s_2} e^{-\beta E_2}\\ &= Z_1 Z_2 \end{align} This can be generalized to \(N\) particles $$Z_{tot} = Z_1Z_2...Z_N$$ If the particles are indistinguishable, when \(N\) is large enough, the double counted microstates can be eliminated by dividing \(N!\) $$Z_{tot} = \frac{1}{N!}Z_1Z_2...Z_N$$