Physics 540 Fall 2005
Note: midterm exam is Thursday Oct. 20, 8:20-9:40 am. The coverage will include Kittel up through sec. 17.  On this homework set (#6), the first problem is the only one directly relevant to the exam.  Also please note that the policy about computing numbers on the exam will be -- yes, you need to be able to get real numbers to at least 10% precision for which you are allowed to bring a calculator.  You will not be given the values of the fundamental constants so you will need to memorize them.

Homework #6 Due Tuesday Oct. 25

1.  Consider 4He to be a classical ideal gas.

a.  At p=1 atm and T=4.2 K, what is the thermal wavelength l?
b.  Using this picture for the vapor of 4He, what is the chemical potential mL=(dG/dN)PT of the liquid phase, given that it coexists at p=1 atm and T=4.2 K with the vapor (and boils when heat is added)?
c.  The numerical value of mL is probably not very different at (p,T)=(0,0) from the value at (p,T)=(1 atm, 4.2 K).  Explain qualitatively the interpretation or source of this chemical potential at (p,T)=(0,0).  It is OK to neglect such things as quantum physics and the difference between a liquid and a solid.  Later we will discuss why it is a liquid rather than a solid.
2.  Kittel, p. 85 problem 18.1

3.  Kittel, p. 85 problem 18.2

4.  Kittel, p. 86 problem 18.4  As guidance, separate into parts:

a. Show that the moment µz in the direction of the field (z direction) is gLz where Lz is the angular momentum and the gyromagnetic ratio g is Q/M times some constant.  You may assume the charge is uniformly distributed on the surface of the spherical dust particle of mass M.
b. The classical version of statistical mechanics of rotating objects has a partition function which is an integral over the three components of angular momentum and the three angles needed to specify the orientation of the object times the exponential of the classical Hamiltonian (L2/2I -gµzH) divided by kBT.  Compute <µz> and the susceptibility.
c. Explain why the answer appears different from the Debye-Langevin result 18.50, but is actually very much analogous.