Working Content > Thermodynamics and statistical physics > Boltzmann distribution

Prerequisites: 

• Boltzmann distribution
• How energy is distributed: Fluctuations 

In our previous readings we have read how a degree of freedom interacting with a thermal bath will be continually exchanging energy with the elements of the bath and, as a result, its energy will fluctuate. At thermal equilibrium, the average energy in each degree of freedom will be the same and the fluctuations will be governed by the Boltzmann factor, e^-E/k_BT. To see how this works, let's do an example.

The Boltzmann factor and a degree of freedom

Let's consider a gas of diatomic molecules where the interaction of the atoms in the molecule looks like a Lennard-Jones potential. Potential energies like the L-J potential look a lot like the potential energy of a spring. Give them a little energy and the atoms will vibrate back and forth. The coordinate in the graph of the atom-atom PE at the right represents the separation of the two atoms (r) and the vertical coordinate represents the interaction potential energy of the two atoms. If the total energy of the atoms were given by the red line, E1, the atoms would oscillate: that their separation would go back and forth along the red line to the walls of the PE well.   But atoms and molecules are quantum systems. They behave like an oscillator (as if connected by a spring), but their energy can not be equal to any value; they can only take on specific (quantized) values -- for example the values E0 and E1 shown. (See Quantum oscillators - discrete states.)

If one of these molecules is in a thermal bath, we would expect that it most likely would start out in the state E₀, its ground state (the lowest energy possible). But through collisions with other molecules, it would sometimes get bumped up into its excited state, E₁. The probability that it would get excited is given by the Boltzmann factor, e^-ΔE/k_BT, where ΔE is the excitation energy, E₁ - E₀. Here's a problem that might involve this calculation.

A. Suppose that a two-atom molecule interacts by an attractive potential that has a number of excited states. The ground state has an energy E₀. If the molecule gets excited into a state E_n, there is an enhanced likelihood that it will undergo a chemical reaction. If the system is in a thermal bath at temperature T, what is the probability that the molecule will be found in the excited state E_n compared to the probability that it will be found in its ground state? If the system is at room temperature (so k_BT ~ 250 meV) and E_n - E₀ = 1.65 eV, what is the relative probability of finding the excited state compared to the ground state?

Since the probability of finding the molecule is an energy state E is proportional to e^-E/k_BT, the ratio of the probabilities of finding the molecule in the state E_n compared to being in the state E₀ is

P_n/P₀ = e^-E_n/k_BT/e^-E₀/k_BT

Since e^a/e^b = e^(a-b) our ratio is

P_n/P₀ = e^{-(E_n - E₀)/k_BT} = e^-(ΔE)/k_BT

For the numbers we are given, ΔE = 1650 meV (milli-eV) and ΔE/k_BT = (1650/250) = 6.6. Our result is therefore e^-6.6 = 1.4 x 10^-3. About 0.1% of the time.

Joe Redish 2/15/16

 Workout