There are two ways to derive Boltzmann exponential probability distribution of ensemble:
1) Microcanonical Ensemble: We assume a system S(E,V,N)
E= internal Energy, V=volume, N=number of molecules or entities.
We have different energy states that the molecules can take, but the total energy E of the system is fixed. So whatever be the distribution of molecules in different energy levels, the energy of the over all system is fixed. Then we find the maxima of Entropy of the system to find out the equilibrium probability distribution of molecules in energy levels. We introduce two Lagrange multipliers for two constraints: total probability is unity and total energy is constant E. What we get is an exponential distribution.
2) Canonical Ensemble: We have a system with N molecules. The Helmholtz energy is defined as F=F(T,V,N). So this time energy is not fixed but the temperature is. Instead of different energy states for the molecules, now we have different energy levels of the entire system to be. So by minimization of F we get the equilibrium probability distribution of the system to be in different energy levels. This time the constraint is total probability is unity. The distribution we get is an exponential one.
Now the question is:
How can the probability distribution of the canonical ensemble can give population distribution of molecules in different energy states which is rather found from micro-canonical ensemble?
In the book Molecular driving forces (Ken A Dill) Chapter 10. Equation 10.11 says something similar.
The probabilities you get from the different ensembles (microcanonical, canonical) are probabilities for finding the entire N-particle system in a particular microstate (not to be confused with the probability of finding constituent molecules in particular states. There is a simple relation for interaction free system though). For the microcanonical ensemble in equilibrium, each microstate has the same probability. For the canonical ensemble, where the system is in contact with a heat reservoir, the probability is given by a Boltzmann factor. There is no contradiction. One can show ensemble equivalence in the thermodynamic limit of very large systems. The reason is the strong peaking of the distributions around the average energy. In principle the system can have all energies when in contact with a heat reservoir but the vast majority is near the average therefore the canonical ensemble behaves essentially as is if the energy were fixed that means like a microcanonical ensemble. This also means that fluctuation away from the average are increasingly insignificant the lager the system gets. In the thermodynamic limit they are completely insignificant.
Thank You Prof. Binek.. Now I understand the statistical equivalence between Canonical and Micro-canonical Ensemble.. So will the probability distribution for canonical ensemble which is P(Ej)= Exp(-Ej/KT)/Q where Q is the partition function and Ej represents energy "states" of the system, be equivalent to the probability distribution of micro-canonical ensemble, which gives distribution of particles in different energy "levels"? For this to be true, Ej will now be equivalent to ej (energy levels within the system for the particles).
It is very instructive to look into the example of a monatomic ideal gas. The probability of finding the entrire N-particle system at an energy between [E,E+dE] is given by P(E)dE proportional to exp(-beta E) X E^(3/2 N-1). Armed with this probability distribution you can calculate the internal energy U = which of course yields U=3/2 N k_B T . The width of P(E) is given by sigma=(3/2N)^(1/2) k_BT. The interesting part happens when you calculate sigma/ which obviously is proportional to 1/N^(1/2) . This tells you that in the limit of large N the distribution function becomes narrower and narrower resembling asymptotically a delta function. This is the origin of the ensemble equivalence as discussed above. This simple example can be generalized to show that sigma/ in general decreases with 1/N^(1/2). For the internal energy you can show that deviations from the average become completely irrelevant U=+/- sigma =3/2Nk_BT (1+/-(2/(3N))^(1/2)). Plug in N=1mol and you see what happens.
I would like to add a single comment: the equivalence between these ensembles requires that energy is additive. For small systems, with surface and bulk energy comparable, or long range interacting systems the additive criteria is broken. In that case ensembles are inequivalent
Dear Prof. Rituraj Borah, in addition to all interesting answers to this thread, I would like to add that the microcanonical ensemble yields entropy as a function of energy and volume S(U, V) (at fixed particle number N).
Now for the canonical distribution when there is an exchange of energy with the heat bath, the following average thermodynamics equations can be used to describe the system: Delta F = -T Delta S - P Delta V = T Delta S - P Delta V = Delta where F = and P =
.
The average comes from the equation
= - delta_F/ delta_T where U is the internal free energy.
It means that in the canonical ensemble by using these relations S(, V), it also yields entropy as a function of energy and volume, as in the microcanonical assemble when the probability distribution is used.
For instance, for the whole derivation see: pp. 41 & 42 of L. Landau and E. Lifshitz, Vol. 6, Statistical Physics, Pergamon 1980, Part I. Chapter II.
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