PHYSICS DEPARTMENT, PRINCETON UNIVERSITYPHYSICS 301 MIDTERM EXAMINATIONOctober 26, 2005, 10:00–10:50 am, Jadwin A06SOLUTIONSThis exam contains two problems. Work both problems. The problems count equallyalthough one might be harder than the other. Do all t he work you want graded in theseparate exam books.Write legibly. If I can’t read it, it doesn’t count!Put your name on all exam books that you hand in. (Only one should be necessary!!!)On the first exam boo k, rewrite and sign the honor pledge: I pledge my honor that I havenot violated the Honor Code during this examination.Copyrightc 2005, Princeton University Physics Department, Edward J. GrothPhysics 301 Midterm Exam Solutions 26-Oct-2005 Page 21. Consider atomic hydrogen in equilibrium at temperature τ. The temperature is lowenough that the el ectron is in its ground state a s far as its orbital motion is concerned. Be-cause the proton ( the nucleus of the hydrogen atom) and electron each have spin 1/2, thereare four possible states. There are very weak interactions (called hyperfine interactions)between the magnetic dipole moments of the proton and t he electron. This produces anenergy difference between the singlet and triplet states. This i s called hyperfine splitting.The singlet state occurs when the proton and electron spins anti-align to form a spin0 a tom.The three triplet stat es occur when the proton and electron spins are aligned forminga spin 1 atom.Our model is a four state system with one state (the singlet) with energy 0 and theother three (the triplet states) with energy ǫ > 0.We will be concerned with the modifications this internal hyperfine interaction makesto the statistical mechanics of a gas of N atoms. (In o ther words, we’re goi ng to ignore thestates of the atoms that correspond to the center of mass mo tion and focus on the internalstates resulting from the hyperfine splitting.)(a) Without elaborate calculation, what are the internal entropy and energy as τ → 0(that is, τ ≪ ǫ)?SolutionAt very low temperatures all atoms will be in the singlet state. So the internal entropyis σint= 0 and the internal energy is Uint= 0 .End Solution(b) Again without elaborate calculation, what are the internal entropy and energy ifτ →≫ ǫ?SolutionIn this case each state is equally likely, so σint= N log 4 and Uint= 3Nǫ/4 .End Solution(c) What is the partition function for these i nternal states at arbitrary temperature?(Don’t forget that there are N independent atoms).SolutionSince the atoms are independent, we can compute the partition function for one atomand take the product of N partition functions to get the partition function for N atoms.There is no N ! correction because that occurs in the parti tion function for the center ofmass states. In other words, the atoms are regarded as located at fixed positions, so wecan identify (in principle) which of the four possible states each atom is in. The partitionCopyrightc 2 005, Princeton University Physics Department, Edward J. GrothPhysics 301 Midterm Exam Solutions 26-Oct-2005 Page 3function is just the sum of Bo ltzmann factors. Each atom has one state at energy 0 and 3states at energy ǫ. So,Zint=1 + 3e−ǫ/τN.End Solution(d) What are the free energy, entropy, and energy of the internal states of the N atomsat arbitrary temperature?SolutionHere, one has to remember a formula or two:Fint= −τ log Zint=−Nτ log1 + 3e−ǫ/τ.σint= −∂Fint∂τ=N log1 + 3e−ǫ/τ+3N(ǫ/τ)e−ǫ/τ1 + 3e−ǫ/τ.Uint= Fint+ τσint=3Nǫe−ǫ/τ1 + 3e−ǫ/τ.Note that the expressions for the entropy and energy go to the answers to parts (a) and (b)in the limits of low and high temperatures. The transition i s the“21 cm line” of neutralhydrogen. This is the wavelength ( in the radio band) of the electromagnetic radiationemitted or absorbed by the transition between the single and tri plet stat es. You might beinterested to calculate the temperat ure that separates low and high temperatures for thistransition. You should come up with a very low temperature!End SolutionCopyrightc 2 005, Princeton University Physics Department, Edward J. GrothPhysics 301 Midterm Exam Solutions 26-Oct-2005 Page 42. You may have read Abbott’s boo k Flatland which besides having some political andcultural themes is also an introduction to the “physics of other dimensions.” An omissionin the book is that cavity radiation in Flatland is not discussed. It turns out physics inFlatland is very simila r to physics in our own world. The speed of light is still c. Quan-tum mechanics still applies, so there is still an ¯h, and harmonic oscillators are quantumoscillators. The laws of thermodynamics are obeyed! There is a small difference: there isonly one polari zation of light.(a) In Flatland, a square box has area (volume) L × L = A. In such a box, how manymodes of electromagnetic radiation with frequency ω are there in the frequency intervaldω?SolutionJust as in the three dimensional case, the modes are standing waves in the box soω2= (πc)2nxL2+nyL2,orn2= n2x+ n2y=ωLπc2.The number of modes with frequency less t han ω must be the same as the area of aquadrant of a disk of radius n ( since there is a mode at each la ttice point (nx, ny) withinthe disk). T he area isN(ω) = πn2/4 =Aω24πc2.Differentiating,n(ω) dω =Aω dω2πc2,where n(ω) is now the density of modes—the number of modes per unit frequency interval.End Solution(b) What is the average energy of a mode of frequency ω when it is in equilibrium attemperature τ? Ignore the zero po int energy of the oscillators.SolutionIf you don’t rememb er t he average energy of a harmonic oscillator i t’s easy to workout. The partition function isZ = 1 + e−¯hω/τ+ e−2¯hω/ τ+ · · · =11 − e−¯hω/τ.The energy isu = τ2∂ log Z∂τ=¯hωe¯hω/τ− 1.Copyrightc 2 005, Princeton University Physics Department, Edward J. GrothPhysics 301 Midterm Exam Solutions 26-Oct-2005 Page 5End Solution(c) What is the total energy in a Flatland cavity of area A in equilibrium at temperat ureτ? How does this total energy depend on temperature? (That is, what power oftemperature?) If you come across an integral you can’t do, be sure to put it indimensionless form so that it’s clearly just a number.SolutionWe know the density of modes and the average energy per mode, so we just add it allup:U(τ) =Z∞0Aω2πc2¯hωe¯hω/τ− 1dω .Let x =