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Energy fluctuations in the canonical ensemble

  • Page ID
    5231
  • 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 \(\langle H \rangle \):

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

     

     

    \(\langle H \rangle \)

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    \( - \frac {\partial}{\partial \beta} \ln Q (N,V,T) \)

     

    \(\langle H^2 \rangle \)

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    \(\frac {1}{Q} C_N \int dx H^2 (x) e^{- \beta H (x)} \) 

     
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    \( \frac{1}{Q} C_N \int dx \frac{\partial^2}{\partial \beta^2}e^{-\beta H(x)}\)

     
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    \( \frac{1}{Q} \frac {\partial^2}{\partial \beta^2}Q\)

     
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    \(\frac{\partial^2}{\partial \beta^2}\ln Q + \frac {1}{Q^2} \left( \frac {\partial Q}{\partial \beta}\right)^2\)

     
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    \(\frac{\partial^2}{\partial \beta^2}\ln Q +\left[\frac{1}{Q} \frac{\partial Q}{\partial \beta}\right]^2\)

     
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    \(\frac{\partial^2}{\partial \beta^2}\ln Q + \left[ \frac {\partial}{\partial \beta}\ln Q\right]^2\)

     


    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.