Skip to main content
Chemistry LibreTexts

Particle number fluctuations

[ "article:topic", "Author tag:Tuckerman", "showtoc:no" ]
  • Page ID
    5211
  • 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.