Deriving Boltzmann's Distribution

[harbour]

I am sure this has a simple answer, but I don't seem to get it at the moment. I am going through a derivation of the Boltzmann's distribution by maximising the entropy with the constraints that the sum of the probabilities add to 1 and the average energy is some constant value. My question is really why must we maximise the entropy in order to get Boltzmann's distribution; thus does that mean that a system under this distribution have maximum entropy as a result?

 

[AngeSurTerre]

The second principle of thermodynamics says that the entropy of an isolated system can only increase. This is justified by the fact that the probability that a system evolves from a given microscopic configuration to a configuration with higher disorder is greater than the probability that it evolves to a configuration with less disorder. Consider for example a pool game with many balls that are initially put with an arbitrary ordering. Then strike a ball. This will break the ordered pattern. Strike again... and again, this will add more and more disorder. At this point, it is hard to believe that a random strike on a ball will put all balls back to the initial configuration. Of course, in principle, this can happen. But you will have to strike the ball during the age of the Universe for that special event to happen. And if it ever happens, the next striking will destroy again the ordered pattern. This is why the probability to evolve to a disordered configuration is dominant, and finding the equilibrium state is given by maximizing the entropy, which is just a measure of the disorder.

 

[Fantasist]

Ithus does that mean that a system under this distribution have maximum entropy as a result?

Yes, the Maxwell-Boltzmann distribution has maximum entropy. Any closed system of a large number of particles will approach this state after a sufficiently long time.

 

[Jano L.]

Yes, the Maxwell-Boltzmann distribution has maximum entropy. Any closed system of a large number of particles will approach this state after a sufficiently long time.

Why do you think that?

 

[Fantasist]

Why do you think that?

It is a well accepted fact that can be proven in various ways. It is evident from the fact that the Maxwell-Boltzmann distribution depends only on one parameter, namely the the temperature. No other information can be obtained from it. This means entropy is maximized.

 

[Jano L.]

It is a well accepted fact that can be proven in various ways. It is evident from the fact that the Maxwell-Boltzmann distribution depends only on one parameter, namely the the temperature. No other information can be obtained from it. This means entropy is maximized.

You said any system of particles will evolve towards the Maxwell-Boltzmann distribution (energies or velocities, it does not matter). This is not evident and it it is hardly true. For example, system of gravitationally interacting point particles may not reach Maxwell-Boltzmann distribution because the attractive force does not decrease fast enough with distance of particles and because the phase space volume is infinite due to singular potential energy (gravothermal catastrophe).

Another example: a system of solid spheres with equal masses and only binary collisions maintains the same distribution of speeds indefinitely.

 

[Fantasist]

You said any system of particles will evolve towards the Maxwell-Boltzmann distribution (energies or velocities, it does not matter). This is not evident and it it is hardly true. For example, system of gravitationally interacting point particles may not reach Maxwell-Boltzmann distribution because the attractive force does not decrease fast enough with distance of particles and because the phase space volume is infinite due to singular potential energy (gravothermal catastrophe).

If the system is closed (i.e. no particles can escape), and there is a random energy and momentum exchange between the particles, then the distribution function will eventually become isotropic and approach the Maxwell-Boltzmann distribution.

Another example: a system of solid spheres with equal masses and only binary collisions maintains the same distribution of speeds indefinitely.

That would only be true if you have exclusively head-on collisions, which practically is only possible for collisions in 1D space.

 

[dextercioby]

Why do you think that?

Boltzmann's kinetic equation (molecular chaos hypothesis, non-correlation of multiparticle distribution function) leads to the H-theorem (principle of entropy maximization for closed systems) -> Maxwell-Boltzmann distribution,

 

[Jano L.]

If the system is closed (i.e. no particles can escape), and there is a random energy and momentum exchange between the particles, then the distribution function will eventually become isotropic and approach the Maxwell-Boltzmann distribution.That would only be true if you have exclusively head-on collisions, which practically is only possible for collisions in 1D space.

You stated any system will end up with the Maxwell-Boltzmann distribution. Now you say only those systems where the exchange of energy and momentum is random will end up with the Maxwell-Boltzmann distribution?

What do you mean by random here? The systems considered in classical mechanics are not random. They obey equations of motion.

That would only be true if you have exclusively head-on collisions, which practically is only possible for collisions in 1D space.

No, it is valid also for system of spheres moving in 2D and 3D space. In any binary collision of two equally massive spheres, spheres exchange velocities which leaves the distribution unchanged.

 

[Jano L.]

Boltzmann's kinetic equation (molecular chaos hypothesis, non-correlation of multiparticle distribution function) leads to the H-theorem (principle of entropy maximization for closed systems) -> Maxwell-Boltzmann distribution,

Boltzmann's assumptions on the behaviour of many-body system that lead to his H-theorem are reasonable only for some models of mechanical systems such as dilute gases.

According to Edwin Jaynes, systems with initially Maxwellian distribution may actually violate behaviour stated by the H-theorem:

http://bayes.wustl.edu/etj/articles/violation.pdf

There is a class of initial conditions for which interparticle forces automatically produce and maintain H-theorem-violating states, with the result that $\dot{H}$ remains positive, on the average, throughout the approach to equilibrium.

If the initial distribution of particles is spatially homogeneous and Maxwellian, the approach to equilibrium takes place through an increase (decrease) in the Boltzmann H, according as the initial potential energy is less (greater) than the equilibrium value.

 

[dextercioby]

Thanks for the reference, but I also recommend: