User:Yen851115/CHEME5740(I)

User:Yen851115/CHEME5740(I)

CHEME5740 project I: Maxwell-Boltzmann statistics
>>> link to the target webpage: Maxwell-Boltzmann statistics

>>> Target Paragraph

Derivations
Maxwell–Boltzmann statistics can be derived in various statistical mechanical thermodynamic ensembles: In each case it is necessary to assume that the particles are non-interacting, and that multiple particles can occupy the same state and do so independently.
 * The grand canonical ensemble, exactly.
 * The canonical ensemble, but only in the thermodynamic limit.
 * The microcanonical ensemble, exactly

Derivation from microcanonical ensemble
Suppose we have a container with a huge number of very small particles all with identical physical characteristics (such as mass, charge, etc.). Let's refer to this as the system. Assume that though the particles have identical properties, they are distinguishable. For example, we might identify each particle by continually observing their trajectories, or by placing a marking on each one, e.g., drawing a different number on each one as is done with lottery balls.

The particles are moving inside that container in all directions with great speed. Because the particles are speeding around, they possess some energy. The Maxwell–Boltzmann distribution is a mathematical function that speaks about how many particles in the container have a certain energy. More precisely, the Maxwell–Boltzmann distribution gives the non-normalized probability that the state corresponding to a particular energy is occupied.

In general, there may be many particles with the same amount of energy $$\varepsilon$$. Let the number of particles with the same energy $$\varepsilon_1$$ be $$N_1$$, the number of particles possessing another energy $$\varepsilon_2$$ be $$N_2$$, and so forth for all the possible energies {$$\varepsilon_i$$ | i=1,2,3,...}. To describe this situation, we say that $$N_i$$ is the occupation number of the energy level $$i.$$ If we know all the occupation numbers {$$N_i$$ | i=1,2,3,...}, then we know the total energy of the system. However, because we can distinguish between which particles are occupying each energy level, the set of occupation numbers {$$N_i$$ | i=1,2,3,...} does not completely describe the state of the system. To completely describe the state of the system, or the microstate, we must specify exactly which particles are in each energy level. Thus when we count the number of possible states of the system, we must count each and every microstate, and not just the possible sets of occupation numbers.

'The number of different ways of performing an ordered selection of one single object from N objects is obviously N. The number of different ways of selecting two objects from N objects, in a particular order, is thus N(N &minus; 1) and that of selecting n objects in a particular order is seen to be N! / (N &minus; n) ! . It is divided by the number of permutations, n!, if order does not matter. The binomial coefficient, N!/(n!(N &minus; n)!), is, thus, the number of ways to pick n objects from $$N$$. If we now have a set of boxes labelled a, b, c, d, e, ..., k, then the number of ways of selecting Na objects from a total of N objects and placing them in box a, then selecting Nb objects from the remaining N &minus; Na objects and placing them in box b, then selecting Nc objects from the remaining N &minus; Na &minus; Nb objects and placing them in box c, and continuing until no object is left outside is'

To begin with, let's ignore the degeneracy problem: assume that there is only one way to put $$N_l$$ particles into the energy level $$l$$. What follows next is a bit of combinatorial thinking which has little to do in accurately describing the reservoir of particles. 'For instance, let's say there is a set of boxes labelled $$a$$, $$b$$, ..., $$k$$. With the concept of combination, we could calculate how many ways to arrange $$N$$ balls into respective boxes in which there would be $$N_l$$ balls without an order. Then we select $$N_a$$ balls from a total of $$N$$ balls, placing them in box $$a$$, and continuing on selection from the remaining until no ball is left outside is '



\begin{align} W & = \frac{N!}{N_a!(N-N_a)!} \times \frac{(N-N_a)!}{N_b!(N-N_a-N_b)!} ~ \times \frac{(N-N_a-N_b)!}{N_c!(N-N_a-N_b-N_c)!} \times \ldots \times \frac{(N-\ldots-N_l)!}{N_k!(N-\ldots-N_l-N_k)!} = \\ \\ & = \frac{N!}{N_a!N_b!N_c!\ldots N_k!(N-\ldots-N_l-N_k)!} \end{align} $$

>>>and because not even a single ball is to be left outside the boxes (all balls should be put in boxes), which implies that the sum made of the terms $$N_a$$, $$N_b$$, ..., $$N_k$$ must equal $$N$$; thus the term ($$N$$ - $$N_a$$ - $$N_b$$ - ... - $$N_k$$)! in the relation above evaluates to 0! (0!=1), and we simplify the relation as



\begin{align} W & = N!\prod_{l=a,b,...}^k \frac{1}{N_l!} \end{align} $$. This is just the multinomial coefficient. The number of ways of arranging $$N$$ balls into $$k$$ boxes, the $$l$$-th box holding $$N_l$$ items and ignoring the permutation of items in each box.

Now going back to the degeneracy problem which characterizes the reservoir of particles. If the i-th box has a "degeneracy" of $$g_i$$; that is, it has $$g_i$$ "sub-boxes", such that any way of filling the i-th box where the number in the sub-boxes is changed is a distinct way of filling the box, then the number of ways of filling the i-th box must be increased by the number of ways of distributing the $$N_i$$ objects in the $$g_i$$ "sub-boxes". The number of ways of placing $$N_i$$ distinguishable objects in $$g_i$$ "sub-boxes" is $$g_i^{N_i}$$ (the first object can go into any of the $$g_i$$ boxes, the second object can also go into any of the $$g_i$$ boxes, and so on). Thus the number of ways $$W$$ that a total of $$N$$ particles can be classified into energy levels according to their energies, while each level $$i$$ having $$g_i$$ distinct states such that the i-th level accommodates $$N_i$$ particles is:


 * $$W=N!\prod_{i} \frac{g_i^{N_i}}{N_i!}$$

This is the form for W first derived by Boltzmann. Boltzmann's fundamental equation $$S=k\,\ln W$$ relates the thermodynamic entropy S to the number of microstates W, where k is the Boltzmann constant. It was pointed out by Gibbs; however, the above expression for W does not yield an extensive entropy, and is therefore faulty. This problem is known as the Gibbs paradox. The problem is that the particles considered by the above equation are not indistinguishable. In other words, for two particles (A and B) in two energy sub-levels, the population represented by [A,B] is considered distinct from the population [B,A]; while for indistinguishable particles, they are not. If we carry out the argument for indistinguishable particles, we are led to the Bose–Einstein expression for W:


 * $$W=\prod_i \frac{(N_i+g_i-1)!}{N_i!(g_i-1)!}$$

The Maxwell–Boltzmann distribution follows from this Bose–Einstein distribution for temperatures well above absolute zero, implying that $$g_i\gg 1$$. The Maxwell–Boltzmann distribution also requires low density, implying that $$g_i\gg N_i$$. Under these conditions, we may use Stirling's approximation for the factorial:



N! \approx N^N e^{-N}, $$

to write:


 * $$W\approx\prod_i \frac{(N_i+g_i)^{N_i+g_i}}{N_i^{N_i}g_i^{g_i}}\approx\prod_i \frac{g_i^{N_i}(1+N_i/g_i)^{g_i}}{N_i^{N_i}}$$

Using the fact that $$(1+N_i/g_i)^{g_i}\approx e^{N_i}$$ for $$g_i\gg N_i$$ we can again use Stirling's approximation to write:


 * $$W\approx\prod_i \frac{g_i^{N_i}}{N_i!}$$

This is essentially a division by N! of Boltzmann's original expression for W, and this correction is referred to as .