Solutions to sample quiz problems and assigned problems

1 1 Solutions to sample quiz problems and assigned problemsSample Quiz ProblemsQuiz problem 1. Prove the expression for the Carnot efficiency for a perfectly reversible Carnot cycle using anideal :The ideal Carnot cycle consists of four segments as follows (1) An isothermal expansion during whichheatQHis added to the system at temperatureTH; (2) an adiabatic expansion during which the gas cools fromtemperatureTHtoTC; (3) An isothermal compression during which heatQCis extracted from the system to a coldreservoir at temperatureTC; (4) Adiabatic compression during which the temperature of the gas rises efficiency of the engine is given by =W/QH, whereWis the useful work done by the system andQHis theheat added to the system. This is a thermodynamic cycle provided it is carried out reversibly ( kept very closeto equilibrium), and then the internal energy is the same at the beginning and the end of the cycle. In that case,dU= 0, andWtotal=QH QL, so we can also write the efficiency as = 1 QLQH= 1 TL STH S= 1 TLTH(1)The introduction of Entropy by Clausius greatly simplified understanding of engines.

2 Quiz problem 2. Show that in the eigenfunction basis, the von Neuman entropyS= kBtr( ln( )) reduces tothe Gibbs formS= kB ipiln(pi) where the sum is over all See Eq. (20) of Lecture notes. Quiz problem 3. Use Stirling s approximation to show that,ln((Nn)) N[pln(p) + (1 p)ln(1 p)],wherep=n/N(2)Solution. Using Stirling s approximation we have,ln((Nn)) =ln(N!) ln(n!) ln(N n)! =Nln(N) N nln(n) +n (N n)ln(N n) + (N n)(3)which reduces toln((Nn)) =Nln(N) nln(n) (N n)ln(N n).(4)Definingp=n/Ngives,ln((Nn)) =Nln(N) pNln(pN) N(1 p)ln(N(1 p)) = N[pln(p) + (1 p)ln(1 p)](5) Quiz problem 4. IfS= kBN[pln(p) + (1 p)ln(1 p)], by doing a variation with respect topfind the valueofpthat gives the maximum entropy. Demonstrate that it is a maximum by showing that the second derivative withrespect topis Taking a derivative with respect topgives, S p= kBN[ln(p) ln(1 p)](6)2 The extrema occur at values ofp=p where this expression is zero, which yieldsp = 1/2.

3 To determine whetherthis is a minimum of maximum, we take the second derivative to find that, 2S p2= kBN[1p+11 p]p=p = kBN(7)Since the second derivative is negative the extremum atp = 1/2 is a maximum. We have thus found the value ofthe probability at which the entropy is a maximum. Quiz problem 5. Use the Master equation to prove the second law of thermodynamics, in a closed systemdS/dt see lecture notes page 10. Quiz problem 6. Give three examples of systems where the ergodic hypothesis fails. Explain why it fails in Non-interacting classical gas, as in the absence of interactions the systems is non-chaotic. Coupledharmonic oscillators with no other interactions, as the normal modes do not share energy. Any system with spon-taneous symmetry breaking at low temperature, magnetic systems. Once in the up magnetized state at lowenough temperature the system stays in that state. Similarly once a crystal has formed from the gas phase at lowtemperature it almost never transforms to another crystal phase.

4 Quiz problem 7. Why is the ideal gas law still a good starting point for high temperature gases, even though anon-interacting gas is non-ergodic?Solution. Even very small interactions can make the system chaotic. These interactions have little effect onthe equilibrium thermodynamic behavior though they are critical to the transport properties. Quiz problem 8. Usingpi=e Ei/Zin the Gibbs form for the entropy, show thatF= kBTln(Z), whereF=U TSis the Helmholtz free energy. HereZ= ie Eiis the canonical partition The Gibbs formula for the entropy isS= kB ipiln(pi).(8)Using the Boltzmann probability in the canonical ensemblepi=exp( Ei)/Z, we have,S= kB ipi[ EikBT ln(Z)] =UT+kBln(Z) soU TS= kBTln(Z),(9)where we used ipi= 1, U= ipiEi Quiz problem 9. Ifx=e (Ei Ej), so thatwji f(x). Show that the Metropolis algorithm is given byf(x) =min(x,1).SolutionWith this notation, the Metropolis algorithm is: Ifx >1,wji= 1, and ifx <1,wji=x.

5 This can beput in one equation aswji=f(x) =min(1,x).3 Quiz problem 10. Ifx=e (Ei Ej), so thatwji f(x). Show that detailed balance is satisfied for anyf(x)satisfyingf(x)/f(1/x) =x. Moreover, show thatf(x) =x/(1 +x), the heat bath algorithm, is one such 1/wji, iff(x) =wji, and iff(1/x) =wij, then detailed balance is satisfied providedf(x)/f(1/x) =x. For the heat bath case,f(x)/f(1/x) = (x/1 +x)/(1/x/(1 + 1/x)) =x(10) Quiz problem 11. For a monatomic interacting classical gas, with interactions that only depend on the particleco-ordinates, derive the Maxwell Boltzmann distribution of velocities and show that the average kinetic energy isgiven by< KE >= 3 See page 23 of the notes. Quiz problem 12. Using the fact that E2=< E2> < E >2=kBT2 CVshow that E/Eis proportional 1/N1 See page 28 of the notes. Quiz problem 13. Write down the central difference form of the second derivative. Using this expression for theacceleration, write down the basic form of the Verlet algorithm for Molecular See.

6 Page 21-22 of the notes. Quiz problem that the virial is defined to beG= k~pk ~rk. Explain why the average of its timederivative is expected to be zero at long The average of the time derivative of the virial gives,<dGdt>=1 0dGdtdt= (G( ) G(0))/ where the average is taken over the time interval . SinceGis bounded, as goes to infinity, the average of the timederivative ofGgoes to zero. Quiz problem ground state structure of Argon is a fcc crystal. If a MD simulation is carried out wherea high temperature structure is quenched to a low temperature where the fcc structure is the equilibrium state. Ifthe starting high temperature state in your MD simulation is a liquid phase, do you expect the simulation to find thecorrect fcc state? For most particle systems it is difficult to find the low temperature ground state, so the fcc single crystalstate is usually not found. In experiments a polycrystal is quit often found but it is harder to make single MD simulations, the timescale is very short so unless special conditions are applied, the low temperature state isa frozen glass Quiz problem a MC calculation of the ferromagnetic Ising model on a square lattice, you calculate themagnetization as a function of temperature as accurately as you can.

7 However your simulations remain roundedat the critical point, instead of showing the behavior|Tc T| expected in the thermodynamic limit. Explain thephysical origins of the rounding that you observe. Your experimental colleague does a measurement of a model Isingmagnet that consists of quantum dots with around 1000 spins in each quantum dot. She also observes a roundingof the behavior of the magnetization. Provide some reasons why her magnetization experiment also gives a roundedbehavior near the critical Three reasons for rounding of the transition are finite size effects, sample inhomogeneity and non-equilibrium effects. Non-equilibrium effects occur when the system is not fully equilibrated. In experiments all threeeffects occur, while in simulations the sample inhomogeneity effect is avoided. Quiz problem the relation,PV=NkBT+13internal k~rk ~Fk(11)Explain the physical origin of each of the three terms. This relation is not always true for periodic boundary condi-tions, though rather fortuitiously it is true for periodic boundaries provided the interaction potential is a pair See page 23-24 of the notes.

8 Quiz problem that( E)2=kBT2CV(12)Solution. See page 28 of the to assigned problemsProblem 1. Find the relation between pressure and volume for an ideal gas under going an adiabatic :. In an adiabatic process no heat is added to the system, sodU+dW= 0, wheredW=PdV. Combiningthe ideal gas law and the equipartion theorem, we havedU= NkBdT= d(PV) = (V dP+PdV)(13)where =DOF/2. Using the first law, we then have, PdV= (V dP+PdV)or dPP= ( + 1)dVV(14)Integrating gives,PP0= (V0V)1+1/ (15)which may be written in its most common form,PV =constantwhere = + 1 (16)For a mono-atomic ideal gas in three dimensions there are three translational degrees of freedom, soDOF= 3, = 3/2, and = 5 problem 2:Derive the ideal monatomic gas law using kinetic theory. You may use the equipartition resultU= :The internal energy of an ideal gas is purely kinetic energy, so that,U=32 NkBT=12 im <[(vix)2+ (viy)2+ (viz)2]>=12Nm < ~v2>(17)The pressure is calculated by considering a particle incident normally on a perfectly reflecting wall,Fx=max=m px t=2mvx t(18)The time taken for the particle to strike the wall again is t= 2L/vx, so thatFwall=2mv2x2 Lso thatP= iFixA=m i(vix2)AL(19)Noting thatV=AL, with i(vix)2=13< ~v2>and combining (12) and (14) yields,PV=m i(vix)2=NkBT(20) problem the degeneracy (E,N) forNa spin 1/2 Ising system with HamiltonianH= h :The energy of a spin state depends only on the number of up spinsNu, and the number of down spinsNd, but not their arrangement.

9 We then have,E= h2(Nu Nd),with degeneracy (E) =N!Nu!Nd!(21)We also haveN=Nu+Nd, so thatE= h2(2Nu N) orNu=N2 Eh, Nd=Eh N2(22)Since entropy is thekBln( (E)), we use Stirling s approximation to write,ln( (E)) =NlnN N NulnNu+Nu NdlnNd+Nd=NlnN NulnNu NdlnNd(23)It is useful to introduce the fractionf=Nu/N, so thatNd/N= 1 fandln( ) =NlnN fNln(fN) (1 f)Nln[(1 f)N] =N[fln(f) + (1 f)ln(1 f)](24)which shows that the entropy in this model is extensive. In terms off,E= (Nh/2)(2f 1). We define =E/N, sothatf= 1/2 /h. Finally the entropy is,S= kBln( (E)) = kBN[(1/2 /h)ln(1/2 /h) + ( /h 1/2)ln( /h 1/2)](25) 6 problem a random walk on a one dimensional lattice where the probability of stepping to the rightor to the left is 1/2. Find the probability that afterNsteps the walker is at positionxlattice units from the startingposition. Use Stirling s approximation to show that for largeNthe probability distribution is a Gaussian inx. Findthe average mean square distance the walker is from the origin as a function :Let us define the number of steps that the student takes to beN=t/ t.

10 The number of steps taken inthe positive x-direction,n, and the number taken in the negative x-direction,m, must add up toN, so that,N=n+m(26)Thexposition of the student at timet=N tis given byx=d x, whered= (n m).(27)But what is the probability that the student reaches this position at timet?P(d,N) =12NN!m!n!=12NN!((N+d)/2)!((N d)/2)!(28)Stirling s approximation:Log(n!) nln(n) n(29)We then have,Ln(P(d,N)) NLn(2) +NLn(N) N N+d2Ln((N+d)/2) + (N+d)/2 (N d2)Log((N d)/2) + (N d)/2(30)This reduces to,Ln(P(d,N)) NLn(2) +NLn(N) N+d2[Ln(N) +Ln(1 +d/N) Ln(2)] (N d2)[Ln(N) +Ln(1 d/N) Ln(2)](31)Definingy=d/Nand expanding for smallyyields,ln(P(d,N)) = N2[(1 +y)ln(1 +y) + (1 y)ln(1 y)] N2[(1 +y)(y 12y2) +..) + (1 y)( y 12y2) +..)] (32)To leading order this yieldsln(P(d,N)) Ny2= d2N(33)or,P(d,N) e d2/N e x2 t/( x)2t.(34)Comparing to the Solutions to the continuum diffusion equation in one dimension, this shows that the diffusionconstant is related to the stochastic parameters xand tthrough,D ( x)2/ t.