Energy Fluctuation in Canonical Ensemble

Show that for a canonical ensemble the fluctuation of energy in a system of constant volume is related to the specific heat and, hence, deduce that the specific heat at constant volume is nonnegative.

SOLUTION

First solution: For a canonical ensemble:

(1)

where β = 1/τ. On the other hand,

(2)

Differentiating (2), we obtain

(3)

By inspecting (1)–( 3), we find that

(4)

Now, the heat capacity at constant volume, C_v, is given by

(5)

Therefore, comparing (4) and (5), we deduce that, at constant volume,

(6)

or in standard units

(7)

Since

then

Second solution: A more general approach may be followed which is applicable to other problems. Because the probability ω of finding that the value of a certain quantity X deviates from its average value ⟨X⟩ is proportional to e^{S(X - ⟨X⟩)} and denoting x = X - ⟨X⟩, we can write

(8)

Note that ⟨X⟩ = 0. The entropy has a maximum at x = 0. Expanding S(x), we obtain

where

so λ > 0. The probability distribution

(9)

⟨x²⟩ = λ, so

(10)

If we have several variables,

(11)

If the fluctuations of two variables x_i, x_k are statistically independent,

The converse is also true: If ⟨x_ix_k⟩ = 0, the variables x_i and x_k are statistically independent. Now for a closed system we can write

(12)

where S₀ is the total entropy of the system and ∆S₀ is the entropy change due to the fluctuation. On the other hand,

(13)

where W_min is the minimum work to change reversibly the thermodynamic variables of a small part of a system (the rest of the system works as a heat bath), and τ is the average temperature of the system (and therefore the temperature of the heat bath). Hence,

However,

(14)

where ∆E, ∆S, and ∆V are changes of a small part of a system due to fluctuations and τ, P are the average temperature and pressure. So,

(15)

Expanding ∆E (for small fluctuations) gives

(16)

Substituting (16) into (14), we obtain

(17)

where we used

So, finally

(18)

Using V and τ as independent variables we have

(19)

Substituting (19) into (18), we see that the cross terms with ∆V ∆τ cancel (which means that the fluctuations of volume and temperature are statistically independent, ⟨∆V. ∆τ ⟩ = 0):

(20)

Comparing (20) with (10), we find that the fluctuations of volume and temperature are given by

(21)

To find the energy fluctuation, we can expand ∆E:

(22)

where we used ⟨∆V. ∆τ ⟩ = 0. Substituting  and from (21), we obtain a more general formula for

(23)

At constant volume (23) becomes

the same as before.