Solutions to practice problems today

Hi class-

If you were are still feeling a bit unsure about the practice problems today, here is a written explanation.

Problem 1:

In this problem there are M things from which you are choosing: the sites to which you will assign particles. Assume you take the following procedure. You start with particle 1 and assign it to site 1 that you choose, then particle 2 to site 2, and so on and so forth.

(a) Since the particles are all identical, the order in which we choose the sites does not matter, since they will all eventually be assigned to a blue particle. Therefore, we use the combination formula, MCN = M!/N!(M-N)!.

(b) Here the order that we pick sites *does* matter because the particles are all colored. Therefore, we use the permutation formula, NPM = M!/(M-N)!.

(c) Here, we are picking sites with replacement; that is, it doesn't matter whether or not we have chosen a site before or not. Thus, for distinguishable, we have that there are M options for us to place each one. In total, there are M^N options.

(d) Now we have to consider that this is the same case as (c), but if we swap the locations of two particles, then we don't have a new configuration. In the case where M >> N, for the vast majority of configurations, we never have more than one molecule on a site. In this case, we have over-counted configurations using the M^N rule by a factor that depends on the number of ways there are to swap around particle identities, which is N!. Thus we have that the number of conformations is M^N/N!. On the other hand if M is of the same order as N, then we will have lots of configurations in which there is more than one particle in each lattice spot. In such cases, we have *not* overcounted by N! but by some number that is less that that factor. For example, consider the case when all particles are on the same lattice site. In the M^N counting of all configurations, this one is included only once. Therefore, we have not overcounted it at all. This is a subtlety to this problem that we did not discuss in class.

Problem 2:

Essentially, to count classical microstates (systems with continuous degrees of freedom), we have to introduce some discretization into space, which is what we did in this problem. Ultimately we will let this discretization be extremely small. Here, we want to count the number of microstates (combinations of x,y values) that lie witin a certain energy range U to U+dU. The potential energy function is u=a r^2.

To start, we realize that the number of combinations of x and y in the perscribed energy range is related to the area enclosed by two energy contours (U and U+dU) in the x,y plane: it is given by the number of grid points of our discretized x and y axes that lie in this area. For sufficiently small discretization dx, the number of points is just the area A divided by the size of a grid square, dx^2. Now the problem of finding the number of microstates is the same thing as finding the area between these energy contours.

Each contour is a circle, since u=a(x^2+y^2). Therefore, we can find the squared radii of the two circles as, r^2 = U/a and (U+dU)/a, respectively. The area of each circle is pi times these values, and the intermittant area is the difference between the two:

A = pi*((U+dU)/a - U/a)

The number of microstates is A/dx^2, such that our final answer simplifies to

Omega = pi * dU/ (a * dx^2).

We could have also determined this answer using a different approach: first finding the total number of microstates with energy less than or equal to U. This is simply the area of a circle of radius r^2 = U/a, divided by dx^2:

Xi = pi*U/(a* dx^2)

The number of microstates that lie within an energy range dU centered around an energy U is given by the derivative of this function

dXi/dU = # of microstates per energy U

(dXi/dU) * dU = # of microstates in a window of width dU centered around U

Therefore,

Omega = (dXi/dU)*dU = pi * dU / (a * dx^2)

And this is the same answer as before.

-MSS

Office hours

Dear 210A,

The office hours for our course are as follows:

me - Wednesdays 10:30am-12pm, 3321 Engineering II
Dong Hun - Thursdays 2-3:30pm, 3315 Engineering II

And of course, feel free to contact either of us if you would like to meet individually at another time, or stop by, if my office door is open.

Cheers,
MSS

Final exams

Class,

If you did not get back your final exams, you may stop by my office this week to get them.

In addition, if you have subscribed to this blog by email, you may want to remove yourself at this point, as it will be used for future courses that I teach.

Good luck in your future studies,

MSS

Final exam and course grades

Dear class,

Summaries of grade distributions for the final exam and the course grades can be found at the course website.

Also, I will be handing out the graded exams the first day of the 110B course next week. We will also discuss the solutions then.

MSS

Leftover HW9 sets

If you didn't get your graded HW9 back, you can pick it up at my office tomorrow morning.

MSS

HW9

HW9 solutions have been posted and the graded problem sets will be distributed at the review session.

MSS

Review Session

Badri and I will be holding a Review Session on Thursday Evening at 7 pm in Engr. II 3361. We will mainly be answering questions so bring any questions that you have.

I am not planning on holding any official office hours today or tomorrow, so if you have questions you want answered before the review session (or if you can't make the review session), feel free to email me (mblack@engr.ucsb.edu) with questions or to set up a meeting.

-Matt