Physics 562: Statistical MechanicsSpring 2002, James P. SethnaHomework 5, due Wednesday, April 17Latest revision: April 17, 2002, 13:16ReadingPathria Chapter 8. Also look at sections 10.1, 10.4-10.6.“Order Parameters, Broken Symmetry, and Topology”,http://www.lassp.cornell.edu/sethna/OrderParameters/, James P. Sethna, 1991 Lecturesin Complex Systems, eds. L. Nadel and D. Stein, SFI Studies in the Sciences of Complexity,Proc. Vol. XV, Addison Wesley, 1992, p. 243.Yeomans 6.1-6.4.Feynman “Lectures on Physics”, volume III, chapter 21.Problems(5.1) White Dwarves, Neutron Stars, and Black Holes: Fermi Pressure.As the energy sources in large stars are consumed, and the temperature approaches zero,the final state is determined by the competition between gravity and the chemical ornuclear energy needed to compress the material.A simple model of ordinary stellar matter is a Fermi sea of non-interacting electrons, withenough nuclei to balance the charge. Let’s model a white dwarf (or black dwarf, since weassume zero temperature) as a uniform density of He4nuclei and a compensating uniformdensity of electrons. Assume Newtonian gravity. Assume the chemical energy is givensolely by the energy of a gas of non-interacting electrons (filling the levels to the Fermienergy).(a) Assuming non-relativistic electrons, calculate the energy of a sphere with N zero-temperature non-interacting electrons and radius R. Calculate the Newtonian grav-itational energy of a sphere of He4nuclei of equal and opposite charge density. Atwhat radius is the total energy minimized? (This should agree with Pathria 8.4(21),up to his fudge factor α.)A more detailed version of this model was studied by Chandrasekhar and others as a modelfor white dwarf stars.(b) Using the non-relativistic model in part (a) (or Pathria’s formula if you got lost inthe algebra), calculate the Fermi energy of the electrons in a white dwarf star of themass of the Sun, 2 × 1033gm. (i) Compare it to a typical chemical binding energyof an atom. Are we justified in ignoring the electron-electron and electron-nuclearinteractions (i.e., chemistry)? (ii) Compare it to the temperature inside the star, say107K. Are we justified in assuming that the electron gas is degenerate (roughly zerotemperature)? (iii) Compare it to the mass of the electron. Are we justified in using a1non-relativistic theory? (iv) Compare it to the mass difference between a proton andaneutron.The electrons in large white dwarf stars are relativistic. This leads to an energy whichgrows more slowly with radius, and eventually to an upper bound on their mass.(c) Assuming extremely relativistic electrons with  = pc, calculate the energy of a sphereof non-interacting electrons. Notice that the sum of this energy plus the gravitationalenergy of the nuclei doesn’t have an extremum except with a special value of the mass,M0.CalculateM0. (This should agree with Pathria 8.4(23), up to his fudge factor.)How does your M0compare with the mass of the Sun?A star with mass larger than M0continues to shrink as it cools. The electrons (note (a.iv)above) combine with the nuclei, staying at a constant density as the star shrinks into aball of almost pure neutrons (a neutron star, often forming a pulsar because of trappedmagnetic flux). Recent speculations by Rajagopal and Wilczek (Phys. Rev. Lett. 86,3492 (2001)) suggests that the “neutronium” will further transform into a kind of quarksoup with many strange quarks, forming a transparent insulating material.For an even higher mass, the Fermi repulsion between quarks can’t survive the gravitationalpressure, and the star collapses into a black hole. At these masses, general relativity isimportant, going beyond the purview of this course.2(5.2) Fermions in Semiconductors.Perhaps useful: Ashcroft and Mermin, Solid State Physics, Chapter 28.Let’s consider a simple model of a doped semiconductor. Consider a crystal of phosphorous-doped silicon, with N − M atoms of silicon and M atoms of phosphorous. Each siliconatom contributes one electron to the system, and has two states at energies ±∆/2, where∆=1.16eV is the energy gap. Each phosphorous atom contributes two electrons andtwo states, one at −∆/2 and the other at ∆/2 − ,where =0.044eV is much smallerthan the gap.∗(Our model ignores the quantum mechanical hopping between atoms thatbroadens the levels at ±∆/2 into the conduction band and the valence band. It also ignoresspin and chemistry: each silicon really contributes four electrons and four levels, and eachphosphorous five electrons and four levels.) To summarize, our system has N + M spinlesselectrons (maximum of one electron per state), N states at energy −∆/2, M states atenergy ∆/2 − ,andN − M states at energy ∆/2.(a) Derive a formula for the number of electrons as a function of temperature T andchemical potential µ for the energy levels of our system.(b) Find the asymptotic form of µ(T ) in the limit T →∞. What is the limiting occupationprobability for the states?(c) Draw an energy level diagram showing the filled and empty states at T =0. Findthechemical potential µ(T ) in the low temperature limit T → 0. (Hint: Because thereis a gap between the filled and empty states, you can’t just set T =0. Onelevelisbasically completely filled.) Your answer should not involve the variable T.(d) In a one centimeter cubed sample, there are M =1016phosphorous atoms; siliconhas about N =5×1022atoms per cubic centimeter. At room temperature (1/40 eV),what fraction of the phosphorous atoms are ionized (have their upper energy stateempty)? What is the density of holes (empty states at energy −∆/2?Phosphorous is an electron donor, and our sample is doped n-type, since the dominantcarriers are electrons: p-type semiconductors are doped with holes.∗The phosphorous atom is neutral when both of its states are filled: the upper statecan be thought of as an electron bound to a phosphorous positive ion. The energy shift represents the Coulomb attraction of the electron to the phosphorous ion: it’s smallbecause the dielectric constant is large (see A&M above).3(5.3) Bloch walls in Magnets.If needed: Matthews and Walker, Mathematical Methods of Physics, Chapter 12 (Calculusof Variations).The free energy density of an Ising magnet below Tccan be roughly approximated as adouble well potential, with two minima at ±M0:F[M]=(K/2)(∇M)2+(g/4)(M2− M20)2. (5.3.1)This problem