Boltzmann distribution

1. Mar 14, 2012

aaaa202

Why is the boltzmann distribution for a collection of atoms independent of their total energy? (it only depends on their temperature)

One would assume that if the energy is high there'd be a greater tendency to be in excited states or am I wrong?

2. Mar 14, 2012

vanhees71

The Boltzmann distribution is valid for a subsystem of a larger closed system in thermal equilibrium, which allows the exchange of energy but not the exchange of particles. Energy is thus conserved on average, and the total value of energy is fixed by the temperature, which is the Lagrange parameter introduced into the problem to maximize entropy under the constraint that the mean energy has this given value. Thus, there is a one-to-one correspondence between the average energy and temperature. E.g. for an ideal monatomic gas non-degenerate gas, the total internal energy is given by

$$U=\langle E \rangle=\frac{3}{2} N k T,$$

where $T$ is the absolute temperature, $N$ the particle number contained in the subsystem, and $k$ Boltzmann's constant.

3. Mar 14, 2012

aaaa202

hmm okay, but consider this:

If you try to find the probability of an atom in hot resevoir of other atoms appearing in an excited state at T=300K you find that the probability of finding the atom in the ground state is overwhelmingly more probable.

But what if the energy of the system of N particles is so big, that E/N is far larger than the energy of the ground state. How can the ground state then still be so probable? We still assume that T=300K (temperature isn't directly the same as energy, so isn't this possible?)