# Energy fluctuations in the canonical ensemble

In the canonical ensemble, the total energy is not conserved. ( $$H (x) \ne \text {const}$$ ). What are the fluctuations in the energy? The energy fluctuations are given by the root mean square deviation of the Hamiltonian from its average $$\langle H \rangle$$:

$\Delta E = \sqrt{\langle\left(H-\langle H\rangle\right)^2\rangle} =\sqrt{\langle H^2 \rangle - \langle H \rangle^2}$

 $$\langle H \rangle$$ $$- \frac {\partial}{\partial \beta} \ln Q (N,V,T)$$ $$\langle H^2 \rangle$$ $$\frac {1}{Q} C_N \int dx H^2 (x) e^{- \beta H (x)}$$ ﻿}} $$\frac{1}{Q} C_N \int dx \frac{\partial^2}{\partial \beta^2}e^{-\beta H(x)}$$ $$\frac{1}{Q} \frac {\partial^2}{\partial \beta^2}Q$$ $$\frac{\partial^2}{\partial \beta^2}\ln Q + \frac {1}{Q^2} \left( \frac {\partial Q}{\partial \beta}\right)^2$$ $$\frac{\partial^2}{\partial \beta^2}\ln Q +\left[\frac{1}{Q} \frac{\partial Q}{\partial \beta}\right]^2$$ $$\frac{\partial^2}{\partial \beta^2}\ln Q + \left[ \frac {\partial}{\partial \beta}\ln Q\right]^2$$

Therefore

$\langle H^2 \rangle - \langle H\rangle^2 =\frac{\partial^2}{\partial \beta^2}\ln Q$

But

$\frac{\partial^2}{\partial \beta^2}\ln Q = kT^2 C_V$

Thus,

$\Delta E = \sqrt{kT^2 C_V}$

Therefore, the relative energy fluctuation $$\frac {\Delta E}{E}$$ is given by

$\frac{\Delta E}{E} = \frac{\sqrt{kT^2 C_V}}{E}$

Now consider what happens when the system is taken to be very large. In fact, we will define a formal limit called the thermodynamic limit, in which $$N\longrightarrow\infty$$ and $$V\longrightarrow\infty$$ such that $$\frac {N}{V}$$ remains constant.

Since $$C_V$$ and $$E$$ are both extensive variables, $$C_V\sim N$$ and $$E \sim N$$,

$\frac {\Delta E}{E} \sim \frac{1}{\sqrt{N}} \longrightarrow 0\;\;\;{as}\;\;\;N\rightarrow \infty$

But $$\frac {\Delta E}{E}$$ would be exactly 0 in the microcanonical ensemble. Thus, in the thermodynamic limit, the canonical and microcanonical ensembles are equivalent, since the energy fluctuations become vanishingly small.