Question 34.2: Show that 〈∆S〉 = −mkB/2, explain the sign of the answer, a......

\left\langle ∆S\right\rangle =\left\langle -\frac{1}{2}\sum\limits_{ ij}{g_{ij} } α_{i} α_{j} \right\rangle =-\frac{1}{2}\sum\limits_{ ij}{g_{ij} }\left\langle α_{i} α_{j} \right\rangle =-\frac{k_{B} }{2}\sum\limits_{i=1}^{m}{δ_{ii} } =-\frac{mk_{B} }{2}.           (34.25)

The equilibrium configuration, α = 0, corresponds to maximum entropy, so 〈∆S〉 should be negative; a fluctuation corresponds to a statistically less likely state. If the system has m degrees of freedom, then its mean thermal energy is mk_{B}T/2, which is equal to −T〈∆S〉.