Canonical Ensemble

This approach is suitable for a system which is mechanically isolated but is in contact with a heat bath. Therefore such a system is defined by a fixed number of particles and a fixed volume. And the system is kept at a constant temperature through its interaction with the environment. Therefore, the energy of the system is variable.

Consider a mechanically and thermally isolated system (can be treated using the Micro-canonical Ensemble Approach) defined by $N, V, E$. Consider two subsystems in thermal contact defined by $N_1, V_1, H_1(p_1, q_1)$ and $N_2, V_2, H_2(p_2, q_2)$. Then, if each of the subsystem has a small variation in energy given by $\delta$, $E < (E_1 + E_2) < E + 2 \delta$. Although this includes a range of values for $E_1$ and $E_2$, we have seen in the previous section that only one set of values $\bar E_1, \bar E_2$ are important. Assume $E_1 \ll E_2$. Let $\Gamma(E_2)$ be the phase-space volume corresponding to the subsystem 2. The probability of finding system 1 within $dp_1 dq_1$ of $(p_1, q_1)$ regardless of system 2, is proportional to $\Gamma(E_2) dp_1 dq_1$, where $E_2 = E - E_1$. Therefore, upto a proportionality constant,

$$\rho (p_1, q_1) \propto \Gamma_2(E - E_1).$$ (38)

Now, since, $E_1 \ll E$

$$k_{\rm B} \ln \Gamma_2(E - E_1) = S_2(E- E_1)$$

$$\simeq S_2(E) - E_1 \frac{\partial S_2 (E_2)}{\partial E_2} \left. \right\vert _{E_2 = E}$$

$$= S_2 ( E) - E_1/T.$$ (39)

Since, $T$ is the equilibrium temperature for both the subsystems. Therefore,

$$\Gamma _2 ( E- E_1 ) \sim e^{S_2 ( E)/k } e^{- E_1 / kT}.$$ (40)

Therefore, for the smaller subsystem, we have,

$$\rho(p, q) = e^{- H(p, q)/k_{\rm B}T}$$ (41)

where $\rho$ is normalised. Now, this smaller subsystem can be treated as a system in mechanical isolation but in thermal contact with the environment. Therefore, the phase-space density of such a system is given by the above relation. So, the Canonical Ensemble approach is defined by this phase-space density, where $T$ is defined through the contact with a heat bath.

The phase-space volume, known as the Partition Function is given by,

$$Q_N(V, T) = \frac{1}{h^{3 N} N!} \int dp dq e^{- \beta H (p,q)},$$ (42)

where, $\beta = 1/k_{\rm B}T$. It should be remembered that the integration over $p$ is such that all possible values of $E$ are allowed. However, only one value of $E$ contributes to the integral overwhelmingly (this is the mean or average energy). The connection with thermodynamics is established through the relation,

$$Q_N(V, T) = e^{- \beta A( V, T)},$$ (43)

where, $A$ is the Helmholtz free energy.

Once again, we need to identify $A$ defined thus with its thermodynamics properties. Therefore, we must prove that,

1. $A$ is extensive - If the interaction between the two subsystems (mutually exclusive) is negligible compared to the total energy of either of them, then evidently $A = A_1 + A_2$.
2. $A$ is related to $U = \langle H \rangle$ and $S \equiv - \frac{\partial A }{\partial T} \large \vert _V$ through the relation $A = U - TS$.

For the second condition, let us start with the definition of $A$ through the partition function.

$$\frac{\partial}{\partial \beta} \frac{1}{h^{3 N}} \int dp dq e^{\beta [ A(V, T) - H (p, q)]} = 0$$

$$\frac{1}{N!h^{3N}} \int dp dq e^{\beta [A( v, T ) - H (p, q)]... ...T) - H(p, q) + \beta \left(\frac{\partial A }{\partial \beta} \right)_V \right] = 0$$

$$\Rightarrow A(V, T) - U(V, T) - T \left(\frac{\partial A}{\partial T}\right)_V = 0.$$ (44)

Therefore, the recipe for thermodynamics using Canonical Ensemble Approach is given by,

$$Q_N(V,T) \quad \mbox{from} \quad H(p,q)$$

$$A(V,T) = - k_{\rm B} T \ln Q_N(V, T).$$

The other thermodynamic variables like pressure or entropy are obtained the usual way, like, $T = - (\frac{\partial A}{\partial V })_T$, $S = - (\frac{\partial A}{\partial T})_V$ and so on. The ensemble average of energy is given by,

$$\langle H \rangle = \frac{\int dp dq H e^{-\beta H}}{... ...^{-\beta H}} = - \frac{\partial \ln Q_N(V,T)}{\partial \beta}.$$ (45)

Similarly, the mean square fluctuation of energy is given by,

$$\langle H^2 \rangle - \langle H \rangle^2 = \frac{\partial^2 \ln Q_N(V,T)}{\partial \beta^2}.$$ (46)