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Why are 3 states sufficient to describe a statistical ensemble?

  1. Feb 2, 2008 #1
    Hello!

    I have a question arising from my course in statistical physics: Describing microcanonical, canonical and grandcanonical ensemble you take three variables (like E,V,N for microncanonical) and get the "rest" from derivatives of the free energies. Is there a deeper meaning why there are only three variables to be fixed and not four or five?

    Another question: Is there any deeper meaning that all free energies transform into each other by Legendre transformations? There should be. Any references on these topics that could be helpful?

    Thanks for all answers and help!

    Blue2script
     
  2. jcsd
  3. Feb 2, 2008 #2

    Mapes

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    There are three variables because you are considering increases in system energy via (1) heating, (2) PV work, and (3) adding material. If you also considered adding energy through forming a surface ([itex]\gamma\,dA[/itex]), applying stress ([itex]\sigma V\,d\epsilon[/itex]), etc., you would use more variables (in these cases, [itex]A[/itex] and [tex]\inline{\epsilon}[/tex]) and you'd have a wider number of possible ensembles to try.

    As for the potentials question, we always want to use a potential that makes calculations easiest. We know that energy U is minimized for all spontaneous processes, so that [itex]dU=T\,dS-P\,dV+\mu\,dN=0[/itex]. Since it's easy to perform experiments at constant pressure and temperature, let's transform to [itex]G=U-TS+PV[/itex] so that now [itex]dG=-S\,dT+V\,dP+\mu\,dN=0[/itex]. From this we get, for example, that when two phases are in equilibrium the chemical potential [itex]\mu_i[/itex] of material [itex]i[/itex] must be equal in each phase.

    A excellent reference is Callen's Thermodynamics.
     
    Last edited: Feb 3, 2008
  4. Feb 2, 2008 #3
    Hi Mapes,

    thanks for your answer, it really makes sense! On the second question: As far as I knew are the free energies sort of the energies we can gain when we variate the systems state variables. E.g. G(T,P,N), change T and the difference of both Gibbs energies is the energy you can get out of the system. Is that right? And how is that related to Legendre transforms?

    Again, thanks for your answers!!

    Blue2script
     
  5. Feb 2, 2008 #4

    Mapes

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    The reduction in the Gibbs free energy G is the driving force on any process at constant T and P. We can indeed completely capture and store this [itex]\Delta G[/itex] as energy by using a non-heat engine, for example an electrochemical cell, whose efficiency can in theory reach 100% (recall that the efficiency of a heat engine must always be less than 100%).

    The substitutions [itex]G=U-TS+PV[/itex], [itex]F=U-TS[/itex], [itex]H=U+PV[/itex], etc. are the Legendre transforms. Their primary use is to transform the original [itex]dU=T\,dS-P\,dV+\mu\,dN+\dots[/itex] into expressions of variables we can control, like T and P. We can extend this idea: if we need to incorporate stress and strain, then we have

    [tex]dU=T\,dS-P\,dV+\sigma V\,d\epsilon[/tex]

    where I have assumed a closed system so that [itex]dN=0[/itex]. Now let's say I can control temperature, pressure, and stress. Then I define a potential [itex]\phi[/itex] using the Legendre transform

    [tex]\phi=U-TS+PV-\sigma V\epsilon[/tex]

    which is now minimized when the system is at equilibrium: [itex]d\phi=-S\,dt+V\,dP-\epsilon \,d(\sigma V)=0[/itex]. (If you are familiar with the Clausius-Clapeyron relation, you can probably see how I might change the temperature of a phase transformation by altering the stress on a solid, rather than the pressure which is usually considered. We can calculate things like that with our new potential [itex]\phi[/itex].)
     
    Last edited: Feb 3, 2008
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