Weve spent quite a bit of time learning

We’ve spent quite a bit of time learning about how the individual fundamental particles that compose the universe behave. Naturally, I am sure you are wondering: Can we start with that “microscopic” knowledge and learn something about bulk properties of matter such as temperature? Will the peculiar rules governing the behavior of an individual particle or small group of particles (such as the Pauli principle) influence the properties of the matter they compose?

Particles have identical physical properties…but can be distinguished by following their (well defined) classical paths. ASSUMPTIONS In equilibrium, the energy distribution of the particles will converge to the most probable allowed. In principle, there is no limit on the number of particles occupying each state.

Imagine 6 particles with 9 indivisible quanta of energy divided among them. • If the particles are indistinguishable, we only care about how many particles are in each state, and there are 26 unique ways to distribute the energy among them— 26 unique combinations. • If the particles are distinguishable (we make a distinction as to which particle is in which energy state), there are 2002 unique permutations. Remember that for now we have assumed that these particles are distinguishable.

If you are making choices from n objects, then on your first pick you have n choices. On your second pick, you have n-1 choices, n-2 for your third choice and so forth. As illustrated before for 5 objects, the number of ways to pick from 5 objects is 5!. Suppose you are going to pick a subset r out of the total number of objects n, like drawing 5 cards from a deck of 52. For the first pick, you have n choices, then n-1 and so on down to n-r+1 for the last pick. The number of ways you can do it is:

If we care about which particle is in which ‘state’, there are six different states where one particle has all of the energy. Here we have to choose more than one particle for each state, and we can distinguish between different combinations of particles in each state, so the multiplicity gets bigger. If each of these “microstates” is equally likely (and we assumed that we will converge on the most probable), it seems that nature doesn’t favor the situation where one particle has all of the energy.

We assume that each “microstate” (unique permutation) is equally probable. In other words, if there are 180 permutations that will produce a particular energy distribution, then that distribution is more probable than a distribution that can only be produced by six permutations. Energy level Average number 0 2. 143 1 1. 484 2 0. 989 3 0. 629 4 0. 378 5 0. 210 6 0. 105 7 0. 045 8 0. 015 9 0. 003 To find the average number of particles in each state average number of particles in the jth energy level count the number of particles in each state for this distribution multiply by the number of permutations that can produce this distribution divided by the total number of permutation for all distributions There are only 6 (out of 2002) permutations that can produce a situation where one particle has all 9 E. Apparently it’s not very probable.

The Maxwell-Boltzmann distribution can be shown graphically as the plot of the number of molecules traveling at a given speed versus the speed. As the temperature increases, this curve broadens and extends to higher speeds.

Using: follows The equipartition theorem A classical molecule in thermal equilibrium has an average energy of k. T/2 per degree of freedom. But generally, there are more than three degrees of freedom (more than just the translational motion in each of x, y, and z): For molecules that can rotate, you can have a rotational degree of freedom. A one dimensional harmonic oscillator has two degrees of freedom, one corresponding to its potential energy, the other to its kinetic energy. There is an energy k. T/2 associated with each of these degrees of freedom.

The first two pictures give the same outcome. Even though a and b are identical, you can tell them apart by following them along their unique paths. a b q q a b b a a ? ? p q Quantum mechanically, each particle has some probability of being somewhere at a particular time, which overlaps greatly at the collision point. Which particle emerges where? In wave terms, they interfered. b

Since we know that particles are really “wavicles” and Maxwell Boltzmann statistics is only good for distinguishable particles, what good is it? The Maxwell-Boltzmann distribution assumes that the particles they describe are distinguishable. Two particles can be considered distinguishable if the distance separating the particles is large compared to their De. Broglie wavelength. Put in mathematical terms, if their average separation is larger than the uncertainty in their momentum. Using the uncertainty principle: At thermal equilibrium: the particles are moving randomly and the directions will cancel each other but their magnitudes aren’t zero for each degree of freedom Maxwell Boltzmann statistics are valid for low density and high temperature and particle mass

Before, we had distinguishable particles…. assuming they are in two different states, you get two distinct wavefunctions, where r 1 and r 2 are the positions of the toe particles… the probabililty of both particles being in the same state is then… …but indistinguishable particles require a more complicated wavefunction. or If both particles are in the same state these become… “bosons” as they are called, are more likely to be found in the same energy state than apart… “fermions”, however, can never be found in the same state! The uncertainty principle…again.

We are now considering indistinguishable particles we are now only counting how many particles are in each energy state, rather than which particles are in each energy state. However, our assumption that there is no theoretical limit on the number of particles occupying each state still holds. Now each of the 26 energy combinations show occur with equal probability.

Average Energy number Maxwelllevel Boltzmann Average number Bose. Einstein 0 2. 143 2. 269 1 1. 484 1. 538 2 0. 989 0. 885 3 0. 629 0. 538 4 0. 378 0. 269 5 0. 210 0. 192 6 0. 105 0. 115 7 0. 045 0. 077 8 0. 015 0. 038 9 0. 003 0. 038 This probability is now the same for each energy distribution, just 1 divided by the total number of distributions. Not surprisingly, the distribution for bosons show a higher probability at the extremes, where there were fewer permutations producing the observed energy spectrum.

The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. here, U is thermal energy Thermal energy per atom: number of degrees of freedom per dimension (kinetic and potential) in real life, there are three dimensions energy per degree of freedom (from the equipartition theorem) To get the energy per mole, multiply by Avagadro’s number:

Note that the specific heat predicted by the classical equipartition theorem is constant with temperature! This prediction disagrees with the data at low temperature. Einstein’s solution: The energy is then: The specific heat is then: Treat each atom as an independent quantum simple harmonic oscillator, and quantize its energy. this approaches the classical result, k. T, at high temperatures as we know it should (the correspondence principle)

Einstein’s approach worked (mostly). Some fine tuning was required, however. Debye pointed out that the atoms in a solid do not move independently but interacts with its neighbors. The result - continuous vibrational waves-sound.

• Once again we assume that the particles are distinguishable. • Thanks to the Pauli exclusion principle, we must, however, remove assumption that there is no theoretical limit on the number of particles in each state! As in the case for bosons, each allowed energy distribution has an equal probability. Unlike the boson case, only five energy distributions are allowed! All others have more than the permitted two particles in each state in violation of the exclusion principle.

E Average number Maxwell. Boltzmann Average number Bose. Einstein Average number Fermi. Dirac 0 2. 143 2. 269 1. 8 1 1. 484 1. 538 1. 6 2 0. 989 0. 885 1. 2 3 0. 629 0. 538 0. 8 4 0. 378 0. 269 0. 4 5 0. 210 0. 192 0. 2 6 0. 105 0. 115 0 7 0. 045 0. 077 0 8 0. 015 0. 038 0 9 0. 003 0. 038 0 Now there are few allowed states. Since the particles are indistinguishable, each state is again equally likely. Note that the energy distribution will flatten out at higher energies. If one particle has all of the energy state, that would require all of the other particle to be in the same energy state – zero - in violation of the Pauli principle.