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Physics 562: Statistical MechanicsSpring 2003, James P. SethnaHomework 1, due Wednesday Jan. 29Latest revision: January 27, 2003, 4:12 pmReadingPathria, chapters 1 and 2.Problems(1.1) Entropy and Hard Spheres.rWe can improve on the realism of the ideal gas by giving the atoms a small radius. Ifwe make the potential energy inﬁnite inside this radius (“hard spheres”), the potentialenergy is simple (zero unless the spheres overlap, which is forbidden). Let’s do this in twodimensions.A two dimensional L × L box with hard walls contains an ideal gas of N hard disks ofradius r L (left ﬁgure). The disks are dilute: the summed area Nπr2 L2.LetA bethe eﬀective volume allowed for the disks in the box: A =(L − 2r)2.(a) The area allowed for the second disk is A − π(2r)2(right ﬁgure), ignoring the smallcorrection when the excluded region around the ﬁrst disk overlaps the excluded regionnear the walls of the box. What is the allowed 2N-dimensional volume in conﬁgurationspace, of allowed zero-energy conﬁgurations of hard disks, in this dilute limit? Ignoresmall corrections when the excluded region around one disk overlaps the excludedregions around other disks, or near the walls of the box. Remember the 1/N ! correctionfor identical particles. Leave your answer as a product of N terms.(b) What is the conﬁgurational entropy for the hard disks? Here, simplify your answer sothat it does not involve a sum over N terms, but valid to ﬁrst order in the area of thedisks πr2. Show, for large N, that it is well approximated by SX= NkB(1+log(A/N −b)), with b representing the eﬀective excluded area due to the other disks. (Youmay want to derive the formulaNn=1log (A − (n − 1))=N log (A − (N − 1)/2) +O(2).) What is the value of b, in terms of the area of the disk?(c) Just as for the ideal gas, the internal energy is purely kinetic, and the kinetic energyand momentum-space entropy depend only on temperature and not on volume. So,if we isothermally expand this hard-disk gas from initial area A1to A2, the heat ﬂowQ = T ∆S is equal to the work done by the pressure. By diﬀerentiating both formulasfor Q with respect to A2, ﬁnd the pressure for the hard-sphere gas in the large Napproximation of part (b). Does it reduce to the ideal gas law for b =0?1(1.2) Thermodynamics Review I: Microcanonical Thermo.Thermodynamics was understood as an almost complete scientiﬁc discipline before statis-tical mechanics was invented. Stat mech can be thought of as the “microscopic” theory,which yields thermo as the “emergent” theory on long length and time scales where theﬂuctuations are unimportant.The microcanonical stat mech distribution introduced in class studies the properties atﬁxed total energy E, volume V , and number of particles N. We derived the microscopicformula S(N,V,E)=−kBlog Ω(N,V,E). The principle that entropy is maximal led usto the conclusion that two weakly-coupled systems in thermal equilibrium would exchangeenergy until their values of∂S∂E|N,Vagreed, leading us to deﬁne the latter as the inverseof the temperature. By an analogous argument we ﬁnd that systems that can exchangevolume (by a thermally insulated movable partition) will shift until∂S∂V|N,Eagrees, andthat systems that can exchange particles (by semipermeable membranes) will shift until∂S∂N|V,Eagrees.How do we connect these statements with the deﬁnitions of pressure and chemical potentialwe get from thermodynamics? In thermo, one deﬁnes the pressure as minus the changein energy with volume P = −∂E∂V|N,S, and the chemical potential as the change in energywith number of particles µ =∂E∂N|V,S; the total internal energy satisﬁesdE = TdS− PdV + µdN. (1.2.1)(a) Show by solving equation (1.2.1) for dS that∂S∂V|N,E= P/T and∂S∂N|V,E= −µ/T(simple algebra).I’ve always been uncomfortable with manipulating dX’s. Let’s do this the hard way. Our“microcanonical” equation of state S(N,V,E) can be thought of as a surface embedded infour dimensions.(b) Show that, if f is a function of x and y, that∂x∂y|f∂y∂f|x∂f∂x|y= −1. (Draw a picture of asurface f(x, y), and draw a path that starts at (x0,y0,f0) and moves along a contourat constant f to y0+∆y. The ﬁnal point will be at (x0+∂x∂y|f∆y, y0+∆y, f0). Drawit at constant x back to y0, and then at constant y back to (x0,y0). How much mustf change to make this a single-valued function?) Applying this formula to S at ﬁxedE, derive the two equations in part (a) again.(c) Ideal Gas Thermodynamics. Using the microscopic formula for the entropy of amonatomic ideal gas we derived in class,S(N, V, E)=52NkB+ NkBlogVNh3 4πmE3N3/2, (1.2.2)calculate T , P ,andµ. The ﬁrst two should look familiar.(d) Maxwell Relations. Imagine solving the microcanonical equation of state of somematerial (not necessarily an ideal gas) for the energy E(S, V, N): it’s the same surface2in four dimensions, but looked at with a diﬀerent direction pointing “up”. One knowsthat the second derivatives of E are symmetric: at ﬁxed N,wegetthesameanswerwhichever order we take derivatives with respect to S and V . Use this to show theMaxwell relation∂T∂VS,N= −∂P∂SV,N. (1.2.3)(This should take about two lines of calculus). Generate two other similar formulasby taking other second partial derivatives of E. There are a bewildering number ofthese relations.(e) Stat Mech check of the Maxwell relation. Using equation (1.2.2) above, writeformulas for E(S, V, N), T (S, V, N )andP (S, V, N) for the ideal gas (non trivial!).(This is diﬀerent from T and P in part (c), which were functions of N, V ,andE.)Show explicitly that the Maxwell relation equation (1.2.3) is satisﬁed.3(1.3) Stirling’s Approximation and Asymptotic Series.See http://www.lassp.cornell.edu/sethna/Cracks/WhatIs Radius of Convergence.html,http://www.lassp.cornell.edu/sethna/Cracks/Stirling.html,http://www.lassp.cornell.edu/sethna/Cracks/QED.html.One important approximation useful in statistical mechanics is Stirling’s approximationfor n!, valid for large n. It’s not a traditional Taylor series: rather, it’s an asymptoticseries. Later on in the course understanding the distinction may be important, so let’sstudy it here.(a) Show, by converting the sum to an integral, that log(n!) ∼ (n +1/2) log(n +1/2) −n − 1/2log(1/2). Show that this is compatible with the more

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