Why are 3 states sufficient to describe a statistical ensemble?

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Three state variables are sufficient to describe a statistical ensemble because they account for energy changes through heating, work, and material addition. Introducing additional variables, like surface area or stress, would expand the number of possible ensembles. Free energies are interconnected through Legendre transformations, which facilitate calculations by expressing energy in terms of controllable variables like temperature and pressure. The reduction in Gibbs free energy serves as the driving force for processes at constant temperature and pressure, and its transformations allow for a comprehensive understanding of system behavior. Understanding these concepts is crucial for advanced studies in statistical physics and thermodynamics.
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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
 
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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 (\gamma\,dA), applying stress (\sigma V\,d\epsilon), etc., you would use more variables (in these cases, A and \inline{\epsilon}) 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 dU=T\,dS-P\,dV+\mu\,dN=0. Since it's easy to perform experiments at constant pressure and temperature, let's transform to G=U-TS+PV so that now dG=-S\,dT+V\,dP+\mu\,dN=0. From this we get, for example, that when two phases are in equilibrium the chemical potential \mu_i of material i must be equal in each phase.

A excellent reference is Callen's Thermodynamics.
 
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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
 
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 \Delta G 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 G=U-TS+PV, F=U-TS, H=U+PV, etc. are the Legendre transforms. Their primary use is to transform the original dU=T\,dS-P\,dV+\mu\,dN+\dots 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

dU=T\,dS-P\,dV+\sigma V\,d\epsilon

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

\phi=U-TS+PV-\sigma V\epsilon

which is now minimized when the system is at equilibrium: d\phi=-S\,dt+V\,dP-\epsilon \,d(\sigma V)=0. (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 \phi.)
 
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