Subsections

• Gibbs formula
• Averaging

Canonical Ensemble

Gibbs formula

Let us take a part of Microcanonical Ensemble M. This part is described by canonical ensemble , if the size of the rest (thermal bath ) tends to infinity.

What is the probability for A to be in the microscopical state i with energy E_A=E_i?

• All states in the microcanonical ensemble with the same energy E₀ are equally probable. If A_i is fixed, only B can change  

  $$\Prob(A_i) = \frac{\text{\char93 of states with $A$\space in $i$}}{\text{Total \char93 of states of $M$}}$$ (1)

• Total energy:

  $$E_0 = E_i + E_B, \quad E_B\gg E_i$$

• Probability for A_i

  $$\Prob(A_i) = \frac{W_B(E_0-E_i)}{\text{Total \char93 of states of $M$}}$$

We want to use thermodynamic limit: $W_B\to\infty$, $E_0\to\infty$. We hope that the answer does not depend on the nature of the thermal bath.

Naïve approach:

Let us expand W_B:

$$W_B(E_0-E_i)\approx W_B(E_0) - \frac{\partial W_B}{\partial E_0} E_i$$

Why it does not work:

Taylor expansion works only if the next term is smaller than the previous.

$$W_B = e^{S_B(E)/k}, \quad \frac{\partial W_B}{\partial E_0} = e^{S_B(E)/k}\frac{1}{k}\frac{\partial S_B}{\partial E_0}$$

We need  

$$\frac{1}{k}\frac{\partial S_B}{\partial E_0}E_i \ll 1$$ (2)

But S_B and E₀ are extensive variables. The derivative is intensive and does not blow up. We are not guaranteed that (2) works!

A better way:

Let us expand S_B(E) instead:

$$S_B(E_0-E_i) = S_B(E_0) - \frac{\partial S_B}{\partial E_0} ... ... \frac12 \frac{\partial^2 S_B}{\partial E_0^2} E_i^2 + \dots$$

Thermodynamic limit ($V\to\infty$)

$$\begin{aligned} S_B(E_0) &\sim V\\ \frac{\partial S_B}{\pa... ...artial^2 S_B}{\partial E_0^2} &\sim \frac{1}{V} \end{aligned}$$

Large term S_B(E₀) is canceled by denominator in (1). We are left with the second term.

Answer:

For the canonical ensemble  

$$\begin{gathered} \Prob(A_i) = \frac{1}{Q}\exp(-E_i/kT)\\ Q = \sum_i \exp(-E_i/kT) \end{gathered}$$ (3)

This is Gibbs formula .

In microcanonical ensemble all states are equal. In canonical ensemble states with lower energy are more equal than others!

Dependence on the thermal bath: only through T. All thermal baths with the same T are equivalent.

Averaging

$$\left\langle f \right\rangle = \frac{1}{Q}\sum_i f_i \exp(-E_i/kT)$$

Particular case:

$$E=\left\langle E \right\rangle = \frac{1}{Q}\sum_i E_i \exp(-E_i/kT)$$