1. The Maxwell-Boltzmann Distribution Function at energy ε & temperature T: Boltzmann developed statistical mechanics & was a pioneer of quantum mechanics. His work contained elements of relativity & quantum mechanics, including discrete atomic energy levels. Maxwell-Boltzmann Statistics He “introduced the theory of probability into a fundamental law of physics & thus in his statistical interpretation of the 2 nd Law of Thermodynamics he broke with the classical prejudice, that fundamental laws have to be strictly deterministic.” “With his pioneering work, the probabilistic interpretation of quantum mechanics had already a precedent.” He constantly battled for acceptance of his work. He also struggled with depression & poor health. He committed suicide in 1906. Many believe that thermodynamics was the cause.

2. Paul Eherenfest was Boltzmann’s student. He delivered Boltzmann’s eulogy & carried on (among other things) the development of statistical thermodynamics for nearly 3 decades. Not unlike his mentor Boltzmann, he was filled with self-doubt & was deeply troubled by the disagreements between his friends (Bohr, Einstein, etc.) which arose during the development of quantum mechanics. Eherenfest shot himself to death in 1933!!

3. US physicist Percy Bridgmann (man on right in photo), the 1946 Nobel Physics Prize winner, was also a thermal & statistical physics pioneer. He studied the physics of matter under high pressure. Bridgman committed suicide in 1961. There’s no need to worry! I’ve never known a student who didn’t survive a course in thermal & statistical physics!!

4. Number of Particles at Energy ε, Temperature T: A is a Normalization Constant Integrate n(ε) over all ε to get N = Total Number of Particles ε = Particle Energy, k = Boltzmann's constant T = Temperature in Kelvin. g(ε) = Number of States With Energy ε. g(ε) depends on the problem. Maxwell-Boltzmann Distribution Function:

5. Assume a continuous distribution of energies & calculate g(ε)  Number states with energy ε in the range ε to ε + dε. g(ε)  The “Density of States.” To do this calculation, we first need to make some assumptions. Specifically, we need to assume that we’ve calculated the energies ε of the particles. Consistent with the assumption of an ideal gas, we assume that each particle is “free” so that ε = [(p 2 /(2m)] = (½)mv 2 (1) To calculate g(ε), it turns out to be easier to find the number of momentum states corresponding to a momentum p, & then use (1) to change variables from p to ε. Corresponding to every value of momentum p is a value of energy ε.

6. ε = [(p 2 /(2m)] = (½)mv 2 (1) Momentum is a 3-dimensional vector & every point (p x,p y,p z ) in 3-D momentum (p) space corresponds to some energy ε. Think of (p x,p y,p z ) as forming a 3-D grid in p space. “Free” Particles Now, count how many momentum states there are in a region of space (density of momentum states) & use (1) to find the density of energy states: g(ε)  Number states with energy ε in the range ε to ε + dε First, rewrite (1) in the form: The number of p states in a spherical shell from p to p + dp is proportional to 4πp 2 dp = volume of the shell. ε = a sphere in p space.

7. So, the number of states g(p) with momentum between p & p + dp has the form: B is a proportionality constant, which we’ll calculate later. Each p corresponds to a single ε, so Also, ε = [(p 2 /(2m)] = (½)mv 2 so that This givesSo, We had So, this now has the form: Constant C contains B & all the other proportionality constants together.

8. To find the constant C, evaluate N  Total number of particles in the system. Look the integral up in a table & find: The result is So that n(ε)  Number of molecules with energy between ε & ε + dε in a sample containing N molecules at temperature T.

9. Notice that “no” molecules have E = 0, few molecules have high energy (a few kT or greater), & there is no maximum of molecular energy. Plot of the Distribution:

10. This is how the distribution changes with temperature (each vertical grid line corresponds to kT). Notice how the probability of a particle having energy greater than 3kT (in this example) increases as T increases. Notice that the distribution for higher temperatures is skewed towards higher energies but all three curves have the same total area (makes sense).

11. The Total Energy of the System is Evaluation of the Integral Gives This is the total energy for the N molecules, so the average energy per molecule is This is exactly the result obtained from the elementary kinetic theory of gases.

12. Some things to note about the ideal gas energy: 1. The energy is independent of the molecular mass. 2. Which gas molecules will move faster at a given temperature: lighter or heavier ones? Why? 3. The average energy at room temperature, kT is about 40 meV, or (1/25) eV. This is not a large energy. 4. (½)(kT) of energy "goes with" each degree of freedom.

13. Because ε = (½)mv 2, the number of molecules having speeds between v and v + dv can also be calculated. The result is n(v) v

14. The speed of a molecule having the average energy comes from solving for v. The result is v rms is the speed of a molecule having the average energy It is an rms speed because we took the square root of the square of an average quantity.

15. The average speed is calculated from: The result is Comparing this with v rms : Because the velocity distribution curve is skewed towards high energies.

16. Find the most probable speed by setting (dn(v)/dv) = 0. The result is The subscript “p” means “most probable.” Summary of the different velocity results: