# boltzmann weight factor and statistical ensembles

i am working on a project about in-equivalence between statistical ensembles ( micro-canonical and canonical to be more precise ).

how can we show that the in the canonical ensemble the system is weighed by $e^{-\beta x}$ ?

Also can any one suggest me some interesting articles on the subject.

• Is the project about lack of equivalence between statistical ensembles? The word in-equivalent is a little confusing. May 9, 2016 at 16:31
• for me as long as they dont give exactly the same thermodynamic limit they are in-equivalent. May 9, 2016 at 16:41
• "inequivalence" is clearer (even if spellcheck doesn't recognize the word). May 9, 2016 at 17:09
• yes exactly, i first wrote it inequivalence it's funny that you said that. so any answer ?? May 9, 2016 at 17:18
• Well, you need to have non-convex thermodynamic potentials, which happens, for example, in mean-field models or other models with very long-range interactions. (For models with short-range interactions, there are general theorems stating that different statistical ensembles are always equivalent.) One possible review paper discussing some of these issues is arxiv.org/abs/1403.6608 . May 9, 2016 at 17:39

There is a really neat way to prove this.

The equilibrium probability distribution $P_{e}$ for the canonical ensemble is that which minimizes the free energy functional $F[P]$, i.e.

$$F[P] \geq F[P_e]$$

using the method of lagrangian multipliers, we impose that the functional derivative of the free energy with the constraint of normalization

$$\int P(C) dC =1$$

is zero ($C$=a configuration of the system -we integrate in configuration space-). So, we have to take the functional derivative of

$$F_{\lambda}[P]=E[P]-TS[P]=$$ $$=\int P(C) H(C) dC+ k T \int P(C) \ln P(C) dC+\lambda \left(\int P(C) dC -1\right)$$

We obtain

$$\frac{\delta F_{\lambda}}{\delta P}[P_e]=H(C)+kT(\ln P_e+1)+\lambda=0$$

from which

$$\ln P_e=-\frac{H(C)}{kT}-\frac{\lambda}{kT}-1$$

That is to say,

$$P_e (C) = e^{\frac{-H(C)}{kT}} e^{-1-\frac{\lambda}{kT}}$$

Normalizing we finally obtain the canonical distribution:

$$P_e (C) = \frac{e^{-\beta H(C)}}{\int e^{-\beta H(C)} dC}$$

For an alternative demonstration, try K.Huang: Statistical Mechanics. It goes like this:

Consider a system composed of two sub-systems of hamiltonians $H_1$, $H_2$ and number of particles $N_1$, $N_2$, with $N_1>>N_2$. Consider a microcanonical ensamble of the composite system with energy between $E$ and $E+2\Delta$. This includes a large set of energies, but only one set of values $\bar E_1$, $\bar E_2$ will be important (the values which maximize the entropy). We assume $\bar E_2 >> \bar E_1$. If we are only interested in the state of system 1, we can integrate over the degrees of freedom of system 2, so that the probability density of system 1 verifies

$$P_1(p_1, q_1) \propto \Gamma_2 (E_2) = \Gamma_2 (E-E_1)$$

where $\Gamma(E) = \exp \left(\frac {S(E)} k\right)$. Since only the values $\bar E_1$, $\bar E_2$ are important, we expand $k \ln \Gamma_2 (E_2)$:

$$k \ln \Gamma_2 (E-E_1) = S_2(E-E_1) \simeq S_2(E)-E_1 \left( \frac{\partial S_2(E_2)}{\partial E_2} \right)_{E_2=E} = S_2(E)-\frac{E_1}{T}$$

where $T$ is the temperature of system 2 (it is now clear that system 2 is a heat reservoir). Hence

$$P_1(p_1,q_1) \propto \Gamma_2 (E_2) \propto \exp \left(-\frac{E_1}{kT}\right)$$

Since $E_1 = H_1(p_1,q_1)$, the demonstration is complete.

You can find a slightly different demonstration on M.E.Tuckerman: Statistical Mechanics Theory and Molecular Simulation

• I've seen this one before, i was actually wondering if it's possible to derive it from the micro-canonical ensemble ? i know a demonstration that starts from micro-canonical ensemble but in there reasoning they suppose that the energie is small with respect to the energie of the reservoir which isn't not true since the energie is then integrated from 0 to infinity. May 9, 2016 at 19:58
• There is no integration from $0$ to $\infty$ of the energy. I'll update my answer to include that demonstration. May 10, 2016 at 0:02
• Note that, as pointed out by Yvan Velenik in a comment, $\Gamma_2(E_2) = \Gamma_2(E-E_1)$ only if the various constituents of these systems interact with short range interactions otherwise there should also be the interaction energy $E_{12}$ between the two subsystems. May 10, 2016 at 10:10
• Valerio92 in your demonstration you used the fact that $E_1<<E$ in order to do the expansion, but in the canonical-ensemble we allow the energie to go to $\infty$ . (unless what you mean is that for high energies it's zero so the partition function isn't sensible to the high energie region ) May 10, 2016 at 12:56
• Yes, the energy can formally be between $0$ and $\infty$ but if you calculate the probability density function of the energy you will discover that it is a gaussian centered at a certain value $U$ whose width is $\Delta E=\sqrt{ 2 k T^2 C_v}$. Since $U \propto N$ and $C_v \propto N$, $\Delta E / U \propto 1/\sqrt{N}$ and will go to $0$ in the thermodynamic limit $N \rightarrow \infty$, i.e.: fixing the temperature of the system, for $N \rightarrow \infty$, is the same as fixing the energy. May 10, 2016 at 13:50