# How does that Boltzmann distribution interact with entropy?

In an ideal gas, the Boltzmann distribution predicts a distribution of particle energies $E_i$ proportional to $ge^{-E_i/k_bT}$.

But, doesn't entropy dictate that the system will always progress towards a state of maximum disorder? In other words the system evolves towards a macro-state which contains the maximum possible number of indistinguishable micro-states. This happens when all particles have the same energy, which seems to contradict the Boltzmann distribution.

I'm pretty sure I've misinterpreted entropy here, but I'd be please if someone could explain how!

• "This happens when all particles have the same energy" ... I think your confusion is here. The individual particles energies are secondary. – Nikolaj-K Nov 4 '13 at 14:44
• "...This happens when all particles have the same energy,....". No, having the same energy means a very ordered system, so with a very low entropy. At the contrary, a maximum entropy means a repartition between all possible energy levels, compatible with the external constraints. The exact form of the repartition depends on the external constraints (temperature, etc...) – Trimok Nov 4 '13 at 18:02
• Surely all particles having the same energy is the maximum possible as for (say) n particles and n quanta of energy there are $n!$ ways the quanta could be assigned to the particles, that all look like all particles having the same energy. – Sideshow Bob Nov 5 '13 at 10:58

In any system in equilibrium, the entropy of such system is the maximum given a set of constrains. If you think of a microcanonical ensemble, the total energy is fixed while in an canonical ensemble of particles the temperature is the one being held constant.

This distribution probability you mention, is for a canonical situation. Given that the temperature is being held fixed, the many different microstates available for such macrostate are given by that exponential function which depends of the energy of the particles and the temperature.

• Could you elaborate on the beginning of the first sentence. What works in the microcanonical ensamble, which doesn't in the canonical one. – Nikolaj-K Nov 4 '13 at 14:45
• @NickKidman thanks, think I overstated it. As you point in the other comment, the energy of each particle is a secondary issue. I'll rephrase. – Ignacio Vergara Kausel Nov 4 '13 at 14:51

What you say is not true. Macrovariables of a system will evolve towards and fluctuate around equilibrium values which maximize/minimize the thermodynamic potential corresponding to the constraint imposed on your system.

For an isolated system, this corresponds to maximum entropy states while for a system in contact with thermostat it corresponds to minimizing the free energy.

It is easy to see from the Boltzmann distribution you wrote:

$p(E) = \frac{ge^{-\beta E}}{Q}$

Where $g$ can be written as $g=e^{S(E)/k_B}$.

This gives in the end

$p(E) = \frac{e^{-\beta( E - T S(E))}}{Q}$

and hence the most probable energy state is the one that minimizes the free energy.

For a finite system, the energy per particle fluctuates around the most probable value until, eventually, in the thermodynamic limit, the magnitude of these fluctuations vanishes.

• Hmm. So does the Boltzmann distribution assume a system in contact with a reservoir? I thought it would work in isolation. – Sideshow Bob Nov 5 '13 at 11:00
• yes the Boltzmann distribution assumes that the system is in contact with a reservoir. – gatsu Nov 6 '13 at 17:40

@SideshowBob, for two particles, all distributions have equal multiplicity. For three particles, you can parameterize the multiplicity by the energy of the first particle $E_0$. The multiplicity depends on the ways of distributing $E_{tot}-E_0$ among two remaining particles, which is proportional to $E_{tot}-E_0$. Thus $E_0=0$ maximizes the total multiplicity, which is to say that $E=0$ is the most probable energy for each particle to have, and not $\frac{E_{tot}}{3}$ as you suspected. Be careful: intuition and statistics rarely go well together!