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Variation of system energy in Canonical Ensemble

  1. Sep 22, 2015 #1
    A system is in contact with a reservoir at a specific temperature. The macrostate of the system is specified by the triple (N,V,T) viz., particle number, volume and temperature.

    The canonical ensemble can be used to analyze the situation. In the canonical ensemble, the system can exchange energy with the reservoir, with the system energy varying, in principle, from zero to infinity.

    I don't understand how the system energy can vary once we specify (N,V,T). The internal energy, after all, can be considered to be a function of T and V, U=U(T,V).

    Surely, we're not talking about the fluctuation, i.e. the standard deviation in energy, since that is very small for systems with large N. Where am I going wrong?
  2. jcsd
  3. Sep 22, 2015 #2
    The ##\underline{\text{expected value}}## of the internal energy ##<U>## is the (macroscopical) quantity that depends of ##N##,##T##.
  4. Sep 22, 2015 #3
    In order to find the density matrix, you have to minimize the Gibb's entropy: $$S= -k_B Tr(\hat \rho ln \hat \rho)$$
    With the constraints: $$Tr(\hat \rho)=1$$ and $$ <U>=Tr(\hat \rho \hat H) =E= constant$$
    i.e, you let your system explore all the microstates (which include microstates with different energy) that are compatible with the macroscopic constraint ##<U>=E=constant##
  5. Sep 22, 2015 #4
    Yes, i think i get where i was going wrong. I was equating the thermodynamic notion of internal energy to the internal energy of one individual system, instead of the ensemble average. Thanks for the reply.

    Just to be clear, there are multiple microstates with different U which correspond to the same macrostate defined by (N,V,T), right?
    Also, this exchange of energy between the system and reservoir is basically due to statistical fluctuation, which i was ignoring?
  6. Sep 22, 2015 #5
    Moreover, you can prove that, if we define ##\delta U = \frac{<U>-U}{U}## then the standard deviation of ##\delta U## (the relative fluctuations) decay as:
    $$ <(\delta U)^{2}> \sim \frac{1}{N}$$
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