Assignment 6. 1 Kadar Ch. 5 Problem 6. Tyler Shendruk April 5, Part a)

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1 Assignment 6 yler Shendruk April 5, 00 Kdr Ch. 5 Problem 6 Model section of polyelectrolyte s chrged cylinder surrounded by ions who re dispersed in solvent. he totl system chrge is neutrl. here re N ions. If we ignore the Coulomb interctions between ions nd sy they ech only interct with the chrged rod by the potentil then the Hmiltonin is H Ur Ne L N i ln r L [ p i m + ri ] e n ln L where n N/L is the number density, L is the height of the cylinder, r the rdil coordinte, R the rdius of the continer, the rdius of the rod nd e unit chrge.. Prt o nd the cnonicl prtition function, we consider the momentum of the ions to be n idel but we cn not pull it out of the rdil/coordinte-spce integrl since it hs n r dependence. I repet Z Z 0 Z int if Z is the totl prtition function, Z 0 the idel prtition funtion nd Z i nt the interction term to the prtition function. With tht sid Z N d 3 p i d 3 q i N! h 3 exp [ βh] i N d 3 p i d 3 [ q i N [ ] ] p N! h 3 exp β i m + U i i N d 3 p i d 3 [ ] [ ] q i N p N i N! h 3 exp β exp β U m i i i [ ] N N {d N! λ 3 exp β U 3 q } N i

2 where we've used the idel gs result but without the volume term V N becuse we hven't integrted over sptil coordintes yet. Notice lso tht U relly only depends on the rdil position not the index so Z N! N! N! N! N! N exp [ βnu] { d 3 q } N λ 3 [ N L π R λ 3 rdrdθdl exp [ βu] l0 θ0 r [ N ] N R λ 3 πl rdr exp [ βu] r [ N πl R λ 3 [ N πl R λ 3 rdr exp r rdr r ] N [ r βne ln L] ] N r L [ βne ] ] N [ N πl R N! λ 3 L Nβne r βne dr r [ N πl r βne + N! λ 3 L Nβne βne ] N ] R r N ] N πl [R βne βne N L Nβne. 3 Z N! λ 3 βne. Prt b Find the probbility of being t certin rdil position r. Notice this is not the probbility of being in stte pr, v but just of being t rdil position, i.e. we don't cre bout the momentum. pr pq i p i pq i, p i pp i e βh /Z e βh0 /Z 0 where H 0 nd Z 0 re respectively the Hmiltonin nd prtition function for n idel gs. pr e βh0+u /Z e βh0 /Z 0 Z 0 Z e βu.

3 Remember Z Z 0 Z U in this prticulr cse. So the Z 0 will cncel out ll the pesky coecients in Z nd ll we will hve is the integrnd tht we hd in Z divided by the integrl: r exp [ βu] pr R r exp [ βu] dr r r βne R r r βne dr [ R βne βne r βne βne ] pr βne r βne R βne βne. 4 he verge rdil position is found in the stndrd wy < r > R r R rprdr r βne r βne r R βne dr βne βne R R βne r βne dr βne r βne R 3 βne 3 βne R βne βne 3 βne βne R 3 βne 3 βne 3 βne R βne βne βne R 3 βne 3 βne < r > 3. 5 βne R βne βne.3 Prt c Here we investigte the limits of high nd low tempertures. For this exercise we ssume R. We hve only two types of energy here:. therml energy. electrosttic energy ne. We see the rtio of the two s βne in every result we found in Sections. to.. So very qulittively, if the electrosttic energy is greter thn the therml energy, we expect the ions to be ttrcted to the chrged rod nd thus condense on it. But on the other hnd, if the therml energy is much greter thn the 3

4 electrosttic ttrction, the motion of the ions will be rndomized nd they will ct like n idel gs. When the two energies re comprble the behviour will be the most complex becuse the ions' behviour will be trnsitioning between the two limiting cses. he trnsition temperture is then when the two energies blnce or Let's look t the two limits: Low empertures: In this cse, c ne k B. 6 ne > nd remembering R such tht R ne k B ne k B rdil probbility distribution from Eq. 4 becomes pr ne ne R ne r ne ne ne k r B ne which mens the ne r ne pr for ne > nd R 7 With this limit of the probbility density, the men rdil position is < r > R R prdr ne r ne R ne ne ne ne ne ne rdr r ne dr ne ne R ne k r B dr [ ne 3 k r B ne [ ne 3 k B ] R ne [ ] ne 3 k B 0 3 ne ne k B ne 3 k R B ] 4

5 < r > ne for ne > nd R 8 he key is < r > nd doesn't go o to innity. he ions hve condensed on to the polyelectrolyte. High empertures: Now ne < which mens the rdil prob- nd still R so now R ne k B ne k B bility distribution Eq. 4 becomes pr ne ne R ne r ne R ne ne k r B ne pr ne r ne R R for ne < nd R 9 he verge rdil position is esy to nd just like in the low temperture limit: < r > R R prdr ne ne R ne R ne r R R ne ne R3 ne ne k R B ne rdr R ne k r B dr [ ne 3 k r B ] R < r > R ne for ne < nd R. 0 Here < r > R, the continer size, s expected for gs. 5

6 .4 Prt d he counter-ions exert pressure on the continer wlls. Wht is it? he pressure is the chnge in free energy with volume. P F V ln Z V. In this question, we hven't been vrying nything but the rdil component of the volume so then in cylindricl coordintes V πlr R. hen the pressure is P ln Z V R ln Z V R ln Z πlr R πlr R ln ] N πl [R βne βne N N! λ 3 L Nβne βne N [ πlr R ln R βne βne ] 0 Nk B πlr R βne βne βne R βne Nk B πlr R βne βne R Nk B βne V /R. βne Notice wht we hve there: the idel gs lw with two correction terms: P V N [ ne ] ne k B. R If we continue to ssume tht R then the second correction term pproches one nd we hve nice clen P V N ne but Eq. llows for the fct tht even if the polyelectrolyte is quite wek but in very conned spce the ions will hve correction term of pproximtely R P V N R 6

7 .5 Prt e o try to ccount for the interction between counter-ions previously considered non-intercting we simply sy tht the number of ions N cn be broken up into two popultions: N ions re condensed on the rod nd the other N N N ions re free out in the solvent. We sy tht the N free ions re still nonintercting but the N do some shielding of the chrged rod nd lower it's electrosttic potentil mking the modied Hmiltonin N H i [ p i m + e n ln ri ]. L Argue without clculting nything wht the equlibrium number of N should be or equivlently n. We know t high enough tempertures N 0 nd N N mening ll the counter-ions re free. At very low tempertures, they ll condense nd N 0 nd N N. he equlibrium vlue of n will be the vlue tht blnces the electrosttic energy with the therml energy so n k B e. 3 When we hppen to hve n < n i.e. mny condensed ions nd few free ions then too much of the electrosttic potentil is shielded nd less ions condense nd more re ble to leve the re round the rod mening n will be lowered towrds n. On the other hnd, if we hppen to hve n > n i.e. mny free ions nd few condensed ions the electrosttic energy is momentrily higher thn the therml energy nd more ions re ttrcted to the rod mening n will be incresed towrds n. he system is lwys driven towrds n. Kdr Ch. 5 Problem 7 Consider n ensemble of hrd rods of length l. Account for their interctions by mking n excluded volume. his question is very similr to the hrd sphere problems lredy considered.. Prt If ech rod hs some excluded volume ω you cn thnk me for my better nottion thn the book's Ω next time you see me thn we expect the number of sttes Ω to go like Ω V ωn N. in the exct sme wy s the hrd spheres did. We lso expect n indistinguishblity term Ω N!. 7

8 And nlly unlike sphere which hs no rottionl degrees of freedom, we expect there to be some number of sttes ssocited with ech rod's bility to tke some number of orienttions. Kdr clls this Ω A N but we know tht A should be proportionl to the rch length formed from the restricting ngle θ we rbitrrily cll the proportionlity constnt Ω A N lθ N. 4 So putting these three together Ω N! V ωn N lθ N 5 nd so the entropy s lwys dened only up to n rbitrry constnt is S k B ln Ω k B ln N! k B [ln A N + ln V ωn V ωn N A N N N ln N + N] S Nk B [ln A + ln n ω ] + 6 where n N/V is the number density of rods nd we hve used Stirling's pproximtion.. Prt c I'm doing the rst prt of Prt c before Prt b becuse it mkes the most sense tht wy. he rottionl motion is restricted to n ngle θ so s shown in Kdr's gure, the re is mde up of two equlterl tringle nd two pizz slices. he re of ech slice is α θ l 0 0 rdr θl / nd the re of ech tringle is α l sin θ/. his mens the excluded re is ω l θ + sin θ 7 8

9 .3 Prt b In equlibrium the entropy must be t n extrem. Use this fct to nd the density s function of restriction ngle. [ S θ Nk B ln A + ln θ n ω ] + [ A θ Nk B A + ] ω θ n ω + 0 [ ] A Nk B A ω n ω 0 where denotes the derivtive with respect to θ. herefore, A A ω n ω n ω A A + ω A ω A + ωa. We've lredy sid A lθ nd ω l θ + sin θ so then A n ω A + ωa l l + cos θ lθ + l θ + sin θ l n l θ + θ cos θ + sin θ 8 which is plotted in Fig...4 Prt d Wht's the equlibrium stte t high densities? Referring to Fig., we see high density only occurs t smll θ which is to sy, ll the rods re ligned. If we consider smll θ then the density goes like n l θ + θ cos θ + sin θ l θ + θ θ +... l θ s the rods pproch close-pcking. 9

10 Density..0 n l θ Figure : Density is function of excluded volume nd number of orienttionl sttes both of which re functions of θ. Notice: solutions of θ π re nonsensicl. If you llow θ π you will nd second extrem locl mxim. Is there phse trnsition? here is density minimum which gives sort of criticl ngle θ c. All other densities re higher thn n c nθ c. he minimum is t but tht's pin to work with. denomintor is mximized or dn dθ 0 d θ + θ cos θ + sin θ 0 dθ + cos θ c θ c sin θ c + cos θ c 0 he minimum is lso when the θ c sin θ c + cos θ c. 9 his isn't n expression for θ c but θ c must stisfy the bove eqution. Numericlly, θ c

11 3 Kdr Ch. 5 Problem 9 Sy we hve eqution of stte P n b n + c 6 n3 where n N/V is the density nd b nd c re positive constnts. 3. Prt o nd the criticl temperture, density nd pressure, we require he two derivtives re P n 0 P 0. b n P n n k B n b n + c 6 n3 P n herefore, from the second derivtive bn + cn 3 n k B bn + cn b + cn 3b n c b c 4 nd then from the rst derivtive c bn c + cn c 0 b c b + c b 0 c c c b c 0 c b c 5 nd then from the eqution of stte itself P c c n c b n c + c 6 n3 c b b b b c c c + c 6 3 b c

12 P c b3 6c. 6 hese criticl vlues men c n c P c b b c c b 3 6c b b6c ccb 3 c n c P c Prt b Clculte the isotherml compressibility. he isotherml compressibility is κ V V P which we wnt in terms of density. So we write κ V V P N/n N/n P n N N n P n n n P P n n [ bn + cn n 8 ] κ For n n c, the compressibility κ c b n bn + cn3 n c bn c + cn3 c b c k B b b c b c + cb3 c 3 9

13 κ t k B n c c 30 diverges t c 3.3 Prt c Find P P c long the criticl isotherm. P P c c n b n + c 6 n3 P c b c n b n + c 6 n3 b3 6c c 6 n3 b n + b c n b3 6c c [n 3 3 bc ] 6 n + 3 b c n b3 c 3 c [ n 3 3n c n + 3n 6 cn n 3 c] P P c c 6 [n n c] Prt d he liquid nd the gs phse seprte into two dierent densities n + nd n respectively. For tempertures close to c, the densities of both should be close to n c. We expect the dierences to be symmetric nd so sy n ± n c ± δ. 3 Kdr gives hint. Where does it come from? Recll the Gibbs-Duhem reltion: Sd V dp + Ndµ 0. Along n isotherm d 0 so this reduces to 0 V dp + Ndµ dµ V N dp n dp dn P n n dn n [ bn + cn ]. 33 Do we hve ny expecttions bout chemicl potentil? Yes. Becuse the two phse coexist, the chemicl potentils must be equl µn + µn 0. But 3

14 µn + µn is just n + n dµ. So 0 µn + µn n+ dµ n ] dn [ bn + cn n kb dn n+ n n+ n n b + cn [ ln n bn + cn 4 ] n+ n ln n + bn + + cn + 4 k B ln n + bn cn 4 n+ ln b n + n + c n 4 + n n but we sid n ± n c ± δ, n c b/c nd c b c so nc + δ 0 ln b n c + δ n c δ + c n c + δ n c δ n c δ 4 + δ ln bn c δ + cn c + δ δ δ 4 + δ ln bn c δ + cn c δ 4 4δ + δ ln δ b δ c + δ b c + δ ln c δ δ Since δ is pretty smll, let's expnd the logrithm to δ δ δ 3... c c δ 4 Hrden δ + δ ln. 34 c δ δ c Explore the Ising model for D. Sy we hve dimensionless mgnetiztion 35 m N + N N 36 4

15 where N ± re the number of up nd down spins so N N + + N. We cn sy 4. Prt N + N + m 37 N N m. 37b If the spins re independent see the next section then we cn sy tht the number of sttes re N N! Ω N+ N +m! N m. 38! Using Stirling's pproximtion, the entropy is S k B ln Ω k B [ln N! ln N + m k B [N ln N N N + m N m k B [N ln N N k B N! ln ln N m ln N + m [ ln N + m ln ln N + m N m ]! N + m + N + m + N m m ln N + m ] N m N m ln ] N m ] + N S Nk B [ ] + m m + m ln + m ln Prt b Wht's the Hmiltonin? H µ B B i σ i J σ i σ j 40 i,j NN where N N denotes summing only over nerest neighbours, B is the externl mgnetic eld, J is the exchnge interction, µ B is the Bohr mgneton nd σ i szj ± is the spin. We wnt to let there be some men eld tht represents the stte of the entire ensemble of spins. In this wy, we'll hve non-intercting spins but in n eective eld tht they ll creted. So let's cll σ the verge spin. Let's dd 5

16 it to σ i σ j in wy tht we cn get n estimte σ i σ j σ i + 0 σ j + 0 σ i σ + σ σ j σ + σ σ i σ j σ + σ σ σ j σ + σ + σ σ j σ + σ σ i σ j σ + σ i σ σ σ j σ σ + σ σ j σ + σ σ i σ j σ + σ σ i σ σ σ j σ + σ σ j σ + σ [ σ σ i σ + σ σ j σ + σ ] + [σ i σ j σ σ σ j σ] [ σ σ i σ + σ σ j σ + σ ] + [σ i σ σ j σ]. }{{}}{{} No i, j terms Correltionterm Since we re only considering the men, we ignore ny cross terms nd so neglect the correltion term on the fr right, sying σ i,j σ σ i + σ j σ 4 which mens tht if we dene z to be the number of nerest neighbours, the Hmiltonin is pproximtely H µ B B i µ B B i µ B B i µ B B i σ i J σ i σ j σ i Jσ σ i Jσ σ i Jσ i,j NN σ i + σ j σ i,j NN i NN z i σ i + j NN σ i zσ σ j σ NN µ B B + zjσ i σ i zjσ H µ B B e σ i zjσ 4 where B e B + zjσ/µ B is the eective eld seen by the now non-intercting spins. In this men eld ech spin cn tke n energy ±B e i.e. it's clssic two level system now. At this point we cn clculte the prtition function Z e NzJσ / e µ BB e / + e µ N BB e / nd then the free energy F ln Z i 6

17 if we wnt. Or we cn tke step bck nd do this following the ssignment instructions. Setting the eective externl eld to B e 0 i.e. the externl eld to B 0 nd mke n even worse pproximtion for the men eld: i,j NN σ i σ j σ, the number of nerest neighbours to z 4, σ Nm nd using the nottion H E. E JNm. 43 Using this pproximtion for E nd our previous result for S, the free energy F E S is F JNm + Nk [ ] B + m m + m ln + m ln Prt c Find n implicit eqution for the mgnetiztion m s function of the derivtive of the free energy h f where f is the free energy per spin f F/N. he derivtive is h f m m m { Jm + k B 4Jm + k B 4Jm + k B 4Jm + k B ln Rerrnging the bove we get [ ]} + m m + m ln + m ln [ + m ln + + m + m ] m ln + m m [ ] + m m ln + ln + m. m h + 4Jm ln + m. m For convenience, let's momentrily dene χ h+4jm e χ + m m e χ me χ + m m eχ e χ + tnh χ h + 4Jm tnh. nd see tht 7

18 If we dene c zj/k B nd h h/k B then we re left with h + m c m tnh Prt d For h 0, wht is good pproximtion to the mgnetiztion? Setting h 0 nd expnding the hyperbolic tngent, we see 0 + mc m tnh m c m 3 c c c m 3 3 m c 3 c 3 m ± 3 c ± 3 c c c ± c 3 c c. If ssume tht c i.e. c but lso < c so tht the root is rel then / c nd the mgnetiztion is m ± c 3 c c. 46 8

MATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The half-angle formula cos 2 θ = 1 2

MATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The half-angle formula cos 2 θ = 1 2 MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xy-plne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points