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Particle number fluctuations

  • Page ID
    5211
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    In the grand canonical ensemble, the particle number \(N\) is not constant. It is, therefore, instructive to calculate the fluctuation in this quantity. As usual, this is defined to be

    \[\Delta N = \sqrt{\langle N^2 \rangle - \langle N \rangle^2}\]

    Note that

    \( \zeta{\partial \over \partial \zeta}\zeta {\partial \over \partial \zeta}\ln {\cal Z}(\zeta,V,T)\)

    $\textstyle =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$

    \( {1 \over {\cal Z}}\sum_{N=0}^{\infty}N^2 \zeta^N Q(N,V,T) -{1 \over {\cal Z}^2} \left[\sum_{N=0}^{\infty} N \zeta^N Q(N,V,T)\right]^2\)

    $\textstyle =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$

    \( \langle N^2 \rangle - \langle N \rangle^2 \)


    Thus,

    \[\left(\Delta N\right)^2 =\zeta{\partial \over \partial \zeta} \zeta {\partial \over \partial \zeta} \ln {\cal Z} (\zeta, V, T) = ({KT}^2){\partial^2 \over \partial \mu^2}\ln {\cal Z}(\mu,V,T) = kTV{\partial^2 P \over \partial \mu^2}\]

    In order to calculate this derivative, it is useful to introduce the Helmholtz free energy per particle defined as follows:

    \[a(v,T) = {1 \over N}A(N,V,T)\]

    where \(v={V \over N} = {1 \over \rho}\) is the volume per particle.
    The chemical potential is defined by
    \(\mu\) $\textstyle =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$

    \( {\partial A \over \partial N} =a(v,T) + N{\partial a \over \partial v}{\partial v \over \partial N}\)

    $\textstyle =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$ data-cke-saved-style =$

    \( a(v,T) - v{\partial a \over \partial v}\)


    Similarly, the pressure is given by

    \[P = -{\partial A \over \partial V} = -N{\partial a \over \partial v}{\partial v \over \partial V} = -{\partial a \over \partial v}\]

    Also,

    \[ {\partial \mu \over \partial v} = -v{\partial^2 a \over \partial v^2}\]

    Therefore,

    \[ {\partial P \over \partial \mu} = {\partial P \over \partial v}{\partial v \over \partial \mu} = {\partial^2 a \over \partial v^2} \left[v{\partial^2 a \over \partial v^2}\right]^{-1} = {1 \over v}\]

    and

    \[{\partial^2 P \over \partial \mu^2} = {\partial \over \partial v}{\partial P \over \partial \mu}{\partial v \over \partial \mu} = {1 \over v^2} \left[ v {\partial^2 a \over \partial v^2}\right]^{-1}= -{1 \over v^3 \partial P/\partial v}\]

    But recall the definition of the isothermal compressibility:

    \[\kappa_T = -{1 \over V}{\partial V \over \partial P}=-{1 \over v \partial p/\partial v}\]


    Thus,

    \[{\partial^2 P \over \partial \mu^2} = {1 \over v^2}\kappa_T\]


    and

    \[\Delta N = \sqrt{\frac{\langle N \rangle kT \kappa_T}{v}}\]


    and the relative fluctuation is given by

    \[{\Delta N \over N} = {1 \over \langle N \rangle}\sqrt{\frac{\langle N \rangle kT \kappa_T}{v}} \sim {1 \over \sqrt {\langle N \rangle }}\rightarrow 0\;\;{as}\langle N \rangle \rightarrow \infty\]


    Therefore, in the thermodynamic limit, the particle number fluctuations vanish, and the grand canonical ensemble is equivalent to the canonical ensemble.

    Contributors and Attributions

    Mark Tuckerman (New York University)


    Particle number fluctuations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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