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# Statistical Mechanics PHY 831

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This 46 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 831 at Michigan State University taught by Scott Pratt in Fall. Since its upload, it has received 22 views. For similar materials see /class/207609/phy-831-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15

CHAPTER 1 11 The function and its constraint are fzy 312 74zyy2 l Czy 31y0i 2 According to the method of Lagrange multipliers we have gx aweAcaw m44y7ampw a g nwiACWMD A4ziAu gvmwemmwgtcmwu Combining 3 4 with 5 we nd the values ofz and y which minimize the function as follows I y 0 6 12 First we enumerate the states one state with both two bosons in the lower level one state with both two bosons in the higher level and one state with one boson in the lower level and one boson in the higher level a The partition function is Z0 16725Te 5Ti 7 b The average energy is I ieT 2 72eT ieT ltEgt4L444 Z0 1672eTeieT AtT0ltEgt0andatTooltEgtei c The entropy is given by S anc lt E gt T 1n1e2 T e T AtT050andatTooSln3i cl For the lower energy level the partition function is 1 707 T 7207 T Z11 M 6 M quot39W7 10 and for the higher energy level the partition function is l 5 T 7257 T Z21e W 6 W quot39W 11 6 2672eTeieT T1 gagT eieT 9 Therefore the grandcanonical partition function is 1 1 chwT Z1Z2 WW 1 The average number of particles lt N gt is N T a 1 Z EMT 676710T lt gt a n 307176HT 176767MT1 13 Because of the chemical potential M lt 0 for boson systems at T 0 ltNgt0andatTooltNgtooi 113 Similarly we enumerate the states the only state With one fermion in the lower level and one fermion in the higher level a The partition function is Z0 ei Ti 14 b The average energy is lt E gt 61 15 AtbothT0andTooltEgtei c The entropy is given by SancltEgtTlne T01 16 AtbothT0andTooS01 cl For the lower energy level the partition function is Z1 1 HOWT 17 and for the higher energy level the partition function is Z2 1 HEWT1 18 Therefore the grandcanonical partition function is 2mg T 2122 1 EMT 1 HEWT 19 The average number of particles lt N gt is a eu 676710T ltNgtTanGCWW1 20 Because of the arbitrary chemical potential M for fermion systems at T 0 lt N gt2 1911 19M 7 6 Where 191 is the step function and atTooltNgtli 14 Let us begin with F E 7 TS dF dE 7 SdT 7 TdSi Because TdS dE PdV 7 MdQ dF 7SdT 7 PdV 7 LLin Therefore 7 BF 7P 7 7 i 8V QT QED 21 22 23 24 15 Start with the standard relation dE TdS7 PdVMdNi However since we want a relation where S P and M are held constant we should consider dE PV 7 MQ TdS VdP 7 Ndu Since V 6E PV 7 MQ 8P SM N 7 6E PV 7 MQ 6 SP we can get the following identity a BM SP 6P 5 QiEiDA 16 Our starting point is the partition function ln ZGC a Since 6 6 PV ltEgt ltgtV1 ZG0 ltgtVT E 6 PV ltNgt 7 anGc 7 lt67gtVT The expressions are lt E gt 7 3 P V 7 66 T 7 lt N gt 7 if V 76a T 25 26 27 28 b We know a lt E gt 6 lt E gt 6ltEgt Ta Tl 6a 34 6ltNgt mmw 6a0i 35 B a 8a 3 From the equation 35 we get agy 6a iWQ w 36 3a 3 Substitution of equation 36 into equation 34 we get 3N 6ltEgt 6ltEgt Egg 6ltEgt 65 7 37 a a 6a gaset Therefore BN 0 7 T26ltEgt7 T2 6ltEgt 6ltEgt lt8 gta 38 V 66 66 0 6a iagy 17 Now we want a relation where T N and P are held constant we should consider G E 7 TS PV and we can get dG iSdT VdP MdNi 39 We nd a 7 40gt 8T PN V BC i 41 BP TN Therefore 85 82G Op T 7 7T 7 42 BT MP 8T2 MP So 3 2 a2 T2 3K A 3 BP TN BTQBP 8T2 PN QiEiDi 18 a See problem 7 b Consider everything to be functions of P7 N and T then x N since both CV and Op assume xed Ni as as P W T Here7 the last term can be identi ed at CpTi Now7 to x the volume7 6639 av av 6V 0 aP T Which means that to x V7 avam gt 6P W m or av6m BSBPl gt CV6T T65Ny 6T Op 7 TW Finally7 beginning With dG iSdTJr VdP MdN7 one can read off the Maxwell relation BS 7 8V w T W P7 to get BVBTlP2 C C iT i P V av61 CHAPTER 2 21 Beginning With E 7 7 V 3 713mm 7213mm T 7 anGc72wh3231dplnlt1e 5 V 3 1 MW 28 l 1 d 101 1 7 yaw 44 one get T 2 1 P 707mg 28 1 d1047r10 1n 1 7 67m 45 Integrating 45 by parts7 one obtain T p3 ei iu d6 27113393 1 47po E1 7 waft 5d 46 Because the relativistic energy is e p2 72127 ClE p 67 7 47gt Therefore 7 1 3 p2 MeiL P 271mg 23 1 d 1 E1 7 we 48 22 Beginning With PV V T ln ch Wes1d3plnltle gt 49 one get T 2 7M7 gt Wlt281gt dp47rp ln 16 1 50 7r Similar to problem 17 one obtain 1 3 p3 MeiL 27113393 1ld 1 361 www 51 213 In 2dim for boson and fermion7 Beginning With PA 7 7 A 2 713mm i1 T 7 anGc 7 ONE 231d pln lt1 ie gt 752 N 01 216 one get T 7amp7 gt i1 W231dp2wpln ltlie M 1 53 Similar to problem 1 one obtain 1 2 p2 MeiL 2715 23 1 d p E 1 i we 54 The pressure is the momentum transfer per particle per unit length per unit time on the side of a rectangle of dimensions L Ly1 A 277w Of fp p X P d2pf p X momentum transfer on side of length Lymultiplied by collision rate CllVlCle 2pyvy 2L1Ly 7 Where the collision rate is vyQLgc and the momentum transfer per collision is 2py1 After stating vy pye and A Lchy one nds 1 2 1 2 P W d2pfltpgtpf W d2pfltpgt Where the last step used that fact that 12ltp2gt1 The radially symmetric potential is WT 7 VeerN 55 By equi partition theorem 8V 8V p 7 ltIEgt 610 ltgt T 56 At the same time 8V 1 8T 12 lt5Egt 7 ltnggt 7 ltEvgt 7 T1 57 Therefore T ff ltXvgt 7 ltTvgt 7 3T 58 3 8H 7 8V 7 2 2 7 7 ltpaipgt 7 7mm 1 7 lt2Vgt 7 T1 59 7 Therefore 2 2 2 ltp a m H gtlt5gtT1 60 817 e b 2 75 1i2 fdpe 5T 7 7T1dppdipe T 61 6 fdpeieT fdpeieT Integrating the numerator by parts7 we get p2 7 fdpeieT 7 lt 7 TW 7 T1 62 217 For massless boson7 e pr 63 7 N5 3 p2 eipT P 7 273 d 1031017 e PT N5 47f 7 7 7 WEdPPSG pTe QPTe SpT 1ltgt4lt1gt4lt1gt4gt N57T2T4 64 9053 So 2 7T AW1 65 E 7 L d3 i V 7 27rh3 pplie PT 4 N5 7 7 ger dpfe pTe QPTe SpT SN 1 4 1 4 1 4 1213T41lt gt ltggt lt1gt N57T2T4 66 gohs So 2 7T BW1 67 28 For massless fermion N5 4 P WENChaps eipT 7 72pT 73pT 7 ei4pT I I 1ltgt4ltgt1ltgt4 lt1lt gt4lt gt4ltigt4gt2ltltgt4ltgt4lt2gt4gtl T4 54 7 T4 W2 7 N 5W2T4 7 68 8 90h3 So 7 Afermion gAbosonA 69 Similarly we can prove 7 Bfermion ngoson 70 N to For either photons or bosons N5 3 6435 E 7 VWd P Pm7 where N5 is the number of polarizations For photons e hm hkc pc whereas for phonons e hm cs p05 where cs is the speed of sound Including the speed of soundlight the energy for bosons then becomes 3 N5 3 e p E 7 l27rfi3d101061764317C T4 N e71 iVis d3 7 03 27rh3 I 176 where c A cs for phonons The speci c heat is dEdT which only lowers the power of T by one and adds a factor of 4 Thus the ratio of speci c heats is CVphonons 7 303 7 7 3 2 1015 CVphotons 20 X 210 If you change the chemical potential the density changes by an amount 1 6p VD6f6M 9 Combining this With the change of density from raising the temperature 1 7T2T2 dB 63 ZDM6M 6 g 0 gives 7T2TQ dDde 5 TT Where in 1D 7 23l dpi 23l dD De Dlt6gt7lt NFL gt d6 7lt NFL gt m267 d6 26 i This gives 6 7 7T2TQ 7 TerT2 M 7 125 7 6p and since the density number per unit length is 7 pi P 7 28 1 7m one gets 2 2 6M mT 228 1 6h p2 Begin With expressions for 6p and 6EV at xed M 7r2T2 dD 6PltM7Tli 6V E7 E 7T2TQ 6ltAgt M M6 6V D00 E 7T2TQ 6 7 Do A EA 6V First calculate the density of states for two dimensions A dp Am De 7 2727rh2 QWPE 7 W 71 Which is independent of pfi So 0 Therefore 5le 0 Since XEAN XE14 the two quantities are equal Do W m 2 6EA 7 77 7 WT 72gt and i E m dT M dT p 3amp2 10 2112 a The number of ways are as follows 0 1 2 3 4 5 6 A 0 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 0 1 1 2 2 3 3 3 0 0 0 1 1 2 3 b Because N22 N12N11N107N027N002 z31 N071 1N171N170121 74 The 3 ways for N63 as follows 0 1 5 0 2 4 1 2 3 2113 a A L MT ZmT HT anGC 6 27rh2dp27rpe AmT uT 75 e 27rh27 so A T 2mm exp W 76gt b l Z0 dOe lNeZGd 29 1 7i i AmT dt k Neexp eeg ff 7 1 4N9 179 AmT j 2wdge j m2 N L7 L1 77 27rh N CHAPTER 3 311 a Calculate pg 1 2 1 0 2 3 3 7 2mT i ipzZmT 2703 d 1 avde gt mT 32 lt27rh2gt b 7 4W 2 MeiL p 7 27rh3 p dp lie me 2403 p2dp e ueipzWLT e2 ueip2WLT 7T e upo EQ MpO 232 C 2 2g 2 P 6 02 01 PV 7 7 47f 2 1 T 7 27rh3p dp1nlt1ie43lt5 gtgt P iTefrig fdp 1nlt17 eime v Trip P 14 W 12gte2 e p2quot T1 7r e upOT EMMMT 2527 P 7 pT m fem ipoT 2752 e 2 PipT m 72752 P0 A2 7 72421 312 Use reduced variables 7 t 1 p 7 v71 v27 p 2 3 T UKamp FL 1727 up 7V5 7 iaJSN a Solve Maxwell relation d5 dEPTdV7 MTdN M 7 BE 7 7w PTdV 7 MW 7 8PT av T 7 66 V b e evoomdv v 71 32t 7 m Edi 1 32t 7 0 d3 de ptdv7 1 1 1 t17dvU717tljgtdv7 7 vb 1 7 vb 71 3573a 7 Au dvv717lnlt7va71gt 01 Mb 7 Ma 5b 7 6a 101012 7 Wu 7 5b 7 8a 11571 1 1 vb va 1 1 7tln 7 777t 7 777 va71 vb va 11571 va71 vb va 2 2 vb va 11571 77 7 t777 7t1 7 vbva ltvb71 va71gt nltva71gt e pbpa says As t 7gt 07 vb 7gt 007 va 15v7 so the rst term above can be thrown away7 and in the second term va can be set to unity7 thus we can set the lihisi to zero and set the second term on the rihisi to unity and get t va 7 1 51L Plugging the pa p5 expression into the requirement that 5 Ma 1 1 7 1 07777tlnltLgt va 7 1 3901 M As t A 07 vb A 007 and va 1 61 so the rst term above can be thrown away7 and in the second term va can be set to unity7 thus 1 t1nvb6v tlnvbt7 telts 3901 f 7 1 L tlnltL 1gttlnltite gt17 vail t L 76t0vva g d Lt 1t 1 7 7 7 dtivbivaitel170as1t 00 CHAPTER 4 411 a EN 32TEimN EimN ea lnmm ea 1n13e x 3Xe X EN 32TW b ZN S anc E7 Zc SN1lnzcN ENzc Zfree39zinh 1n201nzfreelnzinte 1 mT 32 35X 7 43X 7 Plt27rh2gt HERE 13 SN 52 111 35Xe431X 1 mT 32 1134M 171i f 1134 ne 13E zx npf 2M2 ne 1 mT 32 1n 7 Pi 27rh Tf 2 13e fX SBer fX SBiXe IX InWpi In 1 13ei zX 136 fX 13ei eX 3 2 7 g 13e W garbage W m Ti 136 2X b 7 SBer fX SBiXe IX gar age 13e fX 13ei zX 412 a 1 3 T 3 ltegtv EN 32T 3T 92T could just use equipartion theorem b SN 11nzCN EN 1 mT 32 112 ln lt27rh2gt 1n 31 Si Sf 1n TE2 ilnpi 1n T ilnpf Tf 92 Pf Pi A 43 a Wab PdVNTdVVNTln2i b Q AUWab AU0 Q NT1n2i 0 PW Pbequot Wbc b PdVPbe dVV Y Pbe Y 177 177 W71 Vb gt Pbe V771 1707 VC2V v71lt v1H b NTH 7 17217 i WM d POV PM Vb 1 Pch 7 Pbe lt70 7 1 1 Tc Tb 7 TC 1 1 e 7 17E717lt gt 44 a p p0eix22R2 S constAdz plnltT32pgt 32 I2 const Adz p ltlnT R 7 Since the integral of the second term Will give the number of particles N it can also be absorbed into the constant7 and S const N 1nTS2R7 17 23 T32R Tog2R0 T To ltgt b 1 AI 8 8p 81 7 3 a PM A R 12R 1 t i quot Arli l PA R A 7 C 81 81 716P8P ivi 777 at 81 mp 8p 81 T I 2 7 if AIA z 7 mR2 T 1 2 7 if AA 7 mRQi Summarizing the three equations for the evolution of A R and T T1 2 7 if AA 7 mm B A 7 R 7 R0 23 T W Substituting to eliminate T and A n 23 mR To looks likeF ma With PE given by 3 R3 PE E Solving for evolution R dR 7 E0 7 32T0 R0 2mE0 7 PE j d 7 3T0 17 RDR23 45 a 6P 6P 6P a5MEyH BS 6639 SS 7 78p T 7BT p6T707 7 BP BS9p 6P 6P ap aS6T6Tgt NOW7 for the Maxwell relation dE 7 TS 7SdT 7 PdV7 BS 8P 2 W 7 W and 88V771Np 8p 285N 7 7E 3 8p aT After using the fact that CV lNT dSdT7 2 m2 E E E T 5 8p 5 8p T 8T p p2CV b For Van der VVaals7 8P 7 7 7p 6T 7 17 pps CV 32 see HW solution to 2b from Chapi 37 8P T 2 7 7 7 up 93 T 17 pm 5 T me 7 7 Qapi 3 1 7 W35 RW7mM 72152 E 7 Lapsi 5 4 3 Given that TC 827aps7 15T 9 2 7 7 Tm mes 4 4 so the sound modes are unstable for T lt STCSi 46 a 2mT0 7 f r2 2 1 t2 r t Oexp 2R2 exp 1 2mT0 2m2Rg p ng 19 T7 tm2 2 fp7quott fp7 7vttfoezp7 ltgtexp7lt p gt By inspection7 the new temperature is l7iji T 7 T0 ng7 and is independent of 7 as it only depends on T1 Now7 7 2 2 1 t f foexdmhwtp mm P39ring Completing the squares7 7 T2 p 7 TrtR32 TTQtQ f fmpi miexp 2mTlttgt Mm This has the desired form With Ttr vltmgt Mg C75 fey 1 7 1 Tt2 R2t 7 R3 mR L nnn ggi RE REmRETot2 mTo WEETo tr R2 R Et21 m Note that the temperature now becomes 1 7 1 7 R2 1 1 T 7 T0 R3 TO7 BE T To i 0 f0 3 i P mvy p 27rh3d peXp 2R2 eXp 7 2mT 7 78 7 2 constTS2 exp 79 7 2 constexp i 80 C d This looks EXACTLY like the hydro solution It has the same form be cause the Boltzmann result has a locally thermalized distribution in mo mentum space Thus collisions do nothing because the distribution is already thermalizedi Thus7 the in nite collision rate and the zero collision rate are identical Note this would not be true if the initial gaussian was not spherically symmetric Oswdgrdspf 7 ln f7 32 3 W 4 7 Infoli Given that TR2 is a constant7 S is a constant 4 7 a fltPb7772 0671732me fp77 772 p phi 739 2 2 P5 739 C 77 eXp 27211113757 2 7 Tbi 72 f 07 T T 4000 K lt 0002 Ki 107 1 4 X 1010 3 d3 6mTp22mT f M ewelt1 7 e727 e mT27rmT32 47r2lT353 7r m 32 e mT V T 53 458 X 1041 1 W dgrd3p f1ilnf7 f 7 few232W 22quot p 7 p 7 mo T 103 d339rd3p l 7 ln f0 722132 p22mT e T22Rze p22mT 7r 4 8 WRQQET dgp e Te pz mT eHT c Lfic 197326Mev n R 5fmm 938 MeV At dNdtrl 551x105 finC 188 X1048 s 49 dedgpdt whim dNndgth WRQ 7 Where bmax is the maximum impact parameter for protons to reach the radius R using energy and angular momentum conservation7 bmaxp PM2mEpieQI37 272162 bmax R 17 i V 1021 Keeping in mind that the for p that makes argument of square root neg ative7 the absorption cross section is zero7 dedspdt 7 lt17 2m62gt e lt17 2m62gt dNndgp t 7 p2R p2R 410 a 300 X 138 X 10 23 7 7 77 D 7 AiTm73i0x10 911x107 0202111298515 R2 2Dt 82 R 2 84 mmi 83 b L 14E 2 7 if R 7 WW Tmvdrm7 39Udnft 2 mtcolly A tcoll 7 vtherm Tm7 vtherm 4TL 2 7 R 7 qE TL R 2473 Which is independent of electron mass or mean free path 22 411 a My t Aye yzW 8p 82p i Dwv A 2 39 3 2 PiT Ri D i v A 3D D Z PF Solving the second equation7 With Rt 0 R0 R2t R3 2Dt Solving the equations of motion for A7 dA 3D 1 7dt R32Dt A A0 exp 732 1n RR0 7 R0 3 AWgt b Adyye y22R2AR2 4AEamp N0 7 AoRgiRi R32Dt 412 a 19J 7 DTV2Pm V2762 267182 62 821V2 T ltm Uy m b 7 DT Dogy a n 7 D0 2 8739 7 Tofvnpm 3 7 L TBT 7 quot1n73973907 7 D0 2 7 as 7 TOWan sln7 Toi 23 quoti 84 85 86 87 c The solution can be a product of ld solutions7 p017 Mnm myv 739Pz77277397 Where each ld part is a solution of the ld diffusion equation7 1 D0 19 as 70 87712 7 Where the effective diffusion constant is DoTO and 8 plays the role of the time To correspond to a point source at To one can use the usual Gaussian solution that has zero Width at 739 7390 or equivalently s 07 M77127 W exp 7 Taking the product p pypz 1 l quot2 exp 77 i pquot 47rsD0739032 4D07390S CHAPTER 5 51 a 7T2 dN 6E 7T27 6 d67 V 47f Ne 727 3 27rh3 3W E P Pe V 3W2h37 VmPf d6 7T2h3 6E VT2 6amp3 LC LE 1 m Na V NadT pe2 3amp3 2m7r2T 2 Pf b mpf Cweleccmns VWTy 36T3 prhonons VMEQ Equatng the two7 T2 lmp njhgci 373 3654 2 3 V 3 Wm 6 4 3 3 V pf 7 4 90 7 Na7 108547 5 W 705 WD6W2Na7 67r2h3 7 5 how3 7 12 7T2Ef C V th 6 11X10 21J 0038 eV 44237 K 6W2N 13 W 4 gt d e 2301 pf Pm2hspe137 f 713 7 1i78 10 J 6f 2m X 7 1107 eV 7128 gtlt105 Ki 5 44273 T 7 7 7 54 K 12 2128 X 105 a tanh an7 60 sech2 an qJ606 an 52 a 1767man a 1e72 qju7 170 26 2 406a b If y ailT d9 1 71T 2 7 ET 0 e WNW lame 0 d29 2 1 71T 3 4 7 T70 lt7FFgte lT70721 1 6 mlx m dny 7 dTn T70 If one takes limit at some small but nonzero e eventually the terms in the Taylor expansion Will begin to grow and become relevanti a tanh qJU Jr MB 60 sech2 qJ60 M6B d0 6 sech2 dB W M 1 7 1 7 a2 qJ M 17 qJ 02M d0 M1 i 02 E T7TCa2TC 26 H 7 4Jltagt2 7 NMBltUgt7 a tanhqua 6MB 7 sech2 ltan MB qu7ggt 7 17 a2 W B 1 dH d0 g O39TCMB 1 dB 1 17 a2 2 NE 1 7 W1 02 UTC MB CV 55 a a sinthB MsinthMB 6 45 7 0B 7gt 0 07 0B 7gt 00 17 b 0T7gt0 037001 0T7gtoo wa 700 c First nd A wag FA65 an 1n7 A e osh MBI37 R E Usinh2 5MB 6743 39 7 74 JAe JMBSinhBMBMBs1nh Cosgf 2J6 82A 8A 2 6762 J 2 7 JA gt 43 MBy COShltI I 7 MB sinh Cos13 7 2Je 4 2 MB2 cosh2I3L E 1 NOV 9an 9an 7 7 2 2 BB Cvi 66 re 182A BABB 2 XTV ltTgt 016 Next7 calculate A X for each limit As B A 07 R A 672 A 7 6 67 X A JA72e JJJe jiei A a J 2X 7 JA W 4J2e 2 j s J2e e CV y 7 tanh2 J ASBHOO A A A A gagLB A gimme JA e jg e B J WWW ltJ MBVWWB CV J2M 2 J2 M 2 0 As gt0Tgtoo7 CV 5 W 7 m2 0 As gtooltgt0 R A 16m A gammy a X JE H uB e JMBe uB JMB63J3HB7 A a 2JJ MBMHWB 7 JQeWWHBMBeWB JMB2 B CV 6202 5 7 5202 WV 0 56 a aampd or bampc b 05 57 a Given that rst particle exists7 probability of n is probability of next n 7 1 bonds being unbroken tirne probability next bond is bustedi Pn 11 7 0 Note that 1710 1 Z P 1710 l1pp2p3l b it ZnP lipnpnil 17102 d p mm 17w 1710f CHAPTER 6 61 From Eqs 20 and 21 1 2 V0 ltp1p2vp1p2gt 47M 7 47rV0 7 39a From Eq 19 an ln Z0 7 BTTVe HHmN Where the trace covers all two particle states 92 M Imam dspldgpz elt 1 2gtTe2Wltp1pavlplp2 47W 1 mT 32 7 2 2 0 2m 3 ieT In Z0 59 p0 39a3 6 7 po 7 27rh3 d p6 lt27rh3gt 2 477103on 623 3a3 an07 Since an PQT one can inspect Eqs 21 22 and 23 to see that 47F 10573 B 7 7 2 Sag 7 47FV0570 A ii 2 3a3 62 a One need only consider connected diagrams With one particle in and one particle out ltprlpgt dgr e iP39reiP39r V Using Eq 19 for m 1 an ln Z0 7 5 2 V0 ln Z0 7 npoa m 17 b For higher m this gives 00 71 m an ln Z0 penal 1 VJ K c After using the fact that ln Z0 po e an po e l eivo n 7 dspeicwowT lt2whgt3 30 63 BC yield From Eq 33 sinka 6 07 6 71m d6 a 7dp7 hi dp eipzQMT dp 7r a232 2 7 2mT 7m dp 6 32 12 a2 27rmT NFL h 64 a From the BakerCampbeHHaussdorff lemma7 b C W a1ngt ltn11ngtltn1n2gt ei1n122enaquot0gt MW 1 TWO 6 gmlawa 1 gt 2 1 e W 2 7737quotme 110gt m 7 2 1 ne W ew 10gtn1ngt2 lt gt 3 3 a W QRoiw im ei1n122lt0 enn ngt ei1n12271n122MW 1 7 0 711 T 712 0 Wlt 1a WWW 1 gt nnlnmg 7 0 0 Wlt 1M gt nnlnmg xn11n217 7 2 6W ld Mm gtlt1ngt 7 11 1d11d equotquot zinninm W77r772177772 7r7777 m 1 1 7 2 gmmnj WW W Mn After substituting u E 7777 idmdelMWMn 1 6n1n2 du unle u 6 1 l2 65 a t r 7 am E e zhwa athaelhwa ath e zwa ataezwa a7 d r r f 76714 ataezwa at eizwa at 7 wa1a7aewa at 739 f 7 f we zhwa azae zhwa a7 d Edi zwat am aezwt eiiHotha aeiiHotheiwt b M 2 exp fah 0 W m mgwt Haw m emu exp fah m dt M ama emu Using the result from part a 271th Ho jtgtltaa w c From inspection of the rst line in b n ih dt Wm 77 ihdtei jt CHAPTER 7 71 A 7 92712 y7 7 c 7 A 2 22 H d9 2 7 dz2y 7a2 dz 7 0 Ay27a22 H dy 7 7 eddy dydx 5 7MINHMUM 37quot121 e 6 A 92 7 12 H dy mydz 5 dydz 5 0 g 7 92 7 a2 a 47312 7 z 7 r0 are is arbitrary itanywya 7 z 7 re p 7 p0 atanh a 7 10 72 a Let p probability of up7 irer7 p pp 5N 7 7plnp7 1701n17p W 7 P717P2P717 p712ltagt17p1 2ltagt 7W1nlt12lt0gtgt7 7gwlnlt172lt0gtgt b dFdltEagt7TS 0774Jltagt 1nlt12lt0gtgt771nlt172ltagtgt ltgt rncjm C ltagt lt62 qJltUgt 1 EQ qJltUgt 71 W 7 tanh 541W Which is the same was derived in the notes 33 d Minimize i 141 mviltgt2ifltwgti Umm Minimize Wiriti dadz d 7 va7vmin 12 dz 7 H2 da aimfumin12vafuminVZ pomjnda m e ForT07 7 1 2 7 1 110 7 iqu039 vmmiiiqi F 1 Z poxHqJ da 1702 71 Povm g 7i 1 fix 45W 2195 k k 2 7 A 2 1 dab F 7 dz H E A 1 H 1616 d1 2 Z Zed67k E I k E Z T510646 E I k W W sziJrikQ W2 By the equipartition theorem A k2 T lt z kgtlt H7gt 5 1 lt lt0gt ltzgtgt Z am qu ma 7 1 T iikx kZAHk2 1 m 00 71161 div QQEAH 27m 12 27 ie a x 2H0 7 4 Tr 63H fdgr 9Tp39r 0 lt gt W dltpgt 7 Tr 643130 0fd3T Mr 7 Tr emuMT 0 63 3T r d M 7 Tr 73 Tr eggt2 Tr d P 4 W ogtpltrgtgt Vltpltr 0gtgt2 dgr ltplt0gtpltrgtgt 7 W 75 a tanh an MB Since qJ 17 1 a m UBMB7 gltUBMB3 13 lt gt 13 a m lt3 A To For small Bl3 gtgt B7 7 Landau Field Theory Nothing is real and nothing to get hung about Strawberry Fields forever J Lennon P McCartney 71 What are elds Field theory represents a means by which the state of a system is expressed in terms of a quantity 1 which is a function of the position i The quantity 1 represents a measure of some quantity averaged over a small volume 6V One example of such a quantity is the average magnetization per unity volume 773f while another is the number density In each case when viewed microscopically the magnetization density and the number density are not smooth quantities but are carried by point like objects Thus eld theories are only applicable when looking at averages or correlations on scales larger than the microscopic scale of the matter As will be seen further ahead correlation length scales tend to go to in nity near a critical point a region where eld theories are especially valuable Finitetemperature eld theories are usually expressed in terms of the freeenergy density f F E d3z f gtV gt 1 f ltV gt2vlt gtgt The second statement that derivatives appear only quadratically is certainly an ansatz but as we will see below the assumption is motivated from physical arguments The choice of F for the free energy should not suggest that one is always interested in the Helmholtz free energy as for some applications it might represent the Gibbs free energy or the entropy If the parameter g5 changes slowly compared to the time required to come to thermal equilibrium constant temperature the use of the Helmholtz free energy is justi ed and if the pressure also equilibrates quickly compared to j the Gibbs free energy would be the appropriate choice The gradient term is warranted whenever there is an attractive short range interaction As an example one can look at the two dimensional X 7 Y model where spins on a lattice are denoted by an angle 0 with the following interaction between nearest neighbors v7 Von 7 coswi 7 0m 2 For small differences in the angle this can be approximated as V0 V 7 20 girl2 an girl2 3 Zahly V0 2 2 AL2 i 7 giarl l 0 7 0iy1l AI2 Zahly where AL is the lattice spacing Thus in this example it is easy to identify n with the microscopic quantity V0 Since the free energy is F E 7 TS terms that contribute to the energy density carry through to the free energy density A second example is that of a liquid or gas where 1 refers to the density In that case the attractive interaction is mainly the two particle potential 1277 v 1335 133 ltpfpfgtUfi 32 4 For a eld theory we are interested in a average density at a given point In terms of the averages the two body potential energy is v 13 dgz i f 0if viii 5 a4 7 ltpfpi gt CW new Here C73 is the correlation unity if uncorrelated for two particles to be separate by F In the low density limit particles interacting only two at a time the correlation is approximately Ci 32 z e WlT 6 Next one can make a gradient expansion for 6 7 7 a 7 a 1 7 a PW 3 PM 9 Mai0W 5 2m mm xj8iajpx 7 M Since V0 is an even function of F the odd terms in the expansion can be discarded leaving a quot9 7 a 7 a v dgz Meadow e pltzgtv2pltzgt lt8 1 r 730137 rzeTEUVTUU 9 where the approximation for the correlation Eq 6 has been used for the correlation function The factor of 13 in the expression for amp comes from assuming that 12F is rotationally symmetric so that the integral with r is one third the integral with r2 The term Vlocal is the contribution to the potential energy when density gradients are neglected Assuming that other contributions to the energy do not contribute to the V2 term the free energy becomes F ltigtv2 lt gtl lt10 W lt11 dgz MM 7 1335 120132 where the last step involved integrating by parts Since the de nition of density usually implies an average the bar over 6 can be dropped An obvious question concerns how one might justify dropping higher order gradients in the Taylor expansion of pi Certainly such terms should exist Calculating such higher order terms would involve nding expressions for coef cients like were found for a in Eq 9 above However rather than one extra power of r2 as in Eq 9 one would nd powers of T4 r6 If the potential vr falls off exponentially with a suf ciently short range these terms can be neglected and even if they fall off with a power law such as 17 6 for the Lennard Jones potential screening effects should give an exponential fall off and result in an exponential behavior at large T to keep the coef cients nite One common mistake students make when seeing a mean eld theory is to assume that the gradient term originated from kinetic terms in the microscopic equations of motion lnstead such terms originate from attractive short range interactions In fact if the interactions were repulsive amp would have the wrong sign The terms do however represent kinetic energy type terms in the free energy For instance if V is quadratic one can express the free energy as a sum of independent contributions from momentum modes where 6p N sinkz In that case the V2 term in the free energy behaves as k2 just as it would for free particles or sound modes 72 Calculating surface energies in eld theory As discussed in Chapter 3 two phases can co exist in equilibrium However there is a penalty associated with the boundary This is often referred to as a surface energy or a surface free energy To calculate the surface free energy we consider a large box where the left side m lt 0 is in a gaseous state and the right side m gt 0 is in a liquid state If A is the area of the y 7 2 plane the total Helmholtz free energy F E 7 TS is FA dz Mm gem 12 Our goal is to nd the density pro le that minimizes the free energy given the constraint that it has the density of the gas at z 700 and the density of the liquid at z 00 The Helmholtz free energy is the quantity one should minimize is a system is connected to a heat bath and the number of particles in the box is xed If the system can also exchange particles with a bath with chemical potential 0 one should include the free energy associated with the particles that have been added or subtracted into the bath In that case the number is no longer conserved and the quantity to be minimized is F 7 MN A dz We 7 Pop gltampgt21 13gt 7 A dz 7Pltpltzgtgt p 7 mp gem lt14gt The last step used the thermodynamic identity that the free energy density for a uniform gas is up 7 P Figure 1 illustrates a plot of 7 Pp T u 7 uop as a function of the density for xed temperature By picking the appropriate value of uo for the speci ed temperature T the two minima are at the same level and phase coexistence can occur In fact at the minima V7u0p 7P and the statement that both minima are equal is equivalent to saying that for the speci ed T uo was chosen so that two densities have the same pressure Figure 1 Here Pp T and up T are the pres sure and chemical potential of a gas at uniform density p and temperature T If the system is connected to a heat bath with chemical poten tial no the system will choose a density to mini mize 7 P u 7 uop lf uo equals the chemical potential for phase coexistence the two minima are degenerate and are located at the coexistence densities pgas and phq for the given T A bound ary between the two regions must traverse the region of less favorable free energies hence there is a surface free energy as described in Eq 19 In this expression the surface energy is found by I integrating over the densities between the two co pgas pliq existence values with the integrand being propor tional to the square root of the height illustrated p by the dashed lines PM MOP 390 O I The cost of the net free energy for a pro le pz that differs from the two coexistence densities AFA d P0 i P M mp g8p21 7 15 where P0 is the coexistence pressure for the temperature T The quantity V 7 uop P0 is represented by the height of the lines in the shaded region of Fig 1 After replacing dxlt8mp2 dp8mp the integration can be restated in terms of the densities P11 7 7 AFA qdp Wrgw 16 10 Pgas Next one chooses amp to minimize the free energy Whereas the rst term would be minimized by choosing a step function amp 00 the second term would be minimized by a smooth pro le Minimizing functional integral yields d P07PM7Mop H 78 0 17 aw l amp 2 mp amp Mgw 18 H Pliq AFA x2 dp P07Pu7u0p 19 pgamp5 Thus the coef cient a is responsible for the surface free energy If there were no gradient term in the functional for the free energy there would be no penalty and the density pro le between equilibrated phases would become a step function Looking back at the expression for a in the preVious section shows that surface energies will be larger for interactions with longer ranges EXAMPLE Let the function A I E P0 7 P u 7 p0p be approximated with the following form B 2 Ag 3 10 Pc2 7 0 2 39 Find the surface energy in terms of B 04 and re From Eq 19 the surface energy is P7pc7a 2 M Ma dp p 7 pg 7 a2 P7pc77or Dd VHB dm 0427m2 DL 2 3 304 VHB Note that B a and 042 would depend on the temperature Since two independent minima no longer exist for T gt To the coef cient 042 goes to zero at To which means that the surface energy also becomes zero 73 Correlations and the critical point The role of correlations eg CW7 V lt gtF gtT4gt7 20 is central to the study of critical phenomena Note that here we have assumed a complex eld which has two independent components 1 and in For this case lt gtV gtVgt lt r gtrf gt lt gt ltWMWgtgt 21 Thus if one were to solve for correlations of a one component real eld one would divide the expression derived here for complex elds by a factor of two We work with complex elds only because the Fourier transforms are somewhat less painful Near the critical point the characteristic length of such correlations approaches in nity To demonstrate this we consider a purely quadratic functional as the correlations can be found analytically The effect of higher order terms alters the behavior quantitatively eg different values for the critical exponents but does not alter the qualitative behavior shown here For quadratic potentials A W 3w lt22 the problem can be divided into individual modes 3 3 A 2 quot9 2 F d r f d r 3W lV gtl 23 1 2 2 52A3k gt 24 E where the Fourier and inverseFourier transforms are de ned by 7 1 3 13913 a 1 7 d r 5 Mr 25 Since the modes are independent and therefore uncorrelated the correlation in terms ofmomentum components is straight forward 2 gt 26gt T 515 5mlt in 6 where the equipartition theory was used for the last step One can now calculate the correlation in coordinate space 3 1 4 lt r gt gt V 6 T W Wlt E pgt 27 131 1 iETtm T v 5 A my k Now the sum over modes can be transformed to an integral lt ltFgt gtltWgtgt 213 d3k lt28 This integral can be performed by rst integrating over the angle between I and F7 i then performing a contour integration lt ltm gtltrogtgt macaw lt29 T 5776 5 rcA 4war Here 5 is referred to as the correlation length Near the critical point the curvature of the free energy wrt the order parameter 1 switches sign and as A goes to zero 5 a 00 The fact that this length becomes much larger than any driving microscopic length legitimizes the scaling arguments used in the discussion of critical phenomena in the next section In the case where 1 refers to the magnetization density m the correlation function also provides the susceptibility d a WWW 0gtgt lt30 d Tr e HOJrf damemmF 0 m Tr ei HoJrfda39mBmW 1 deT Tr e H0mFmF 0 1 T re HOmF 0 f 13 n e HOm T Tre Ho T Tr ei HOf dsr ltmf m0gt ltmF0gt2 31 22 gs XE where Eq 29 was inserted into the expression for the correlation in the last step Relations such as this are common in linear response theory and often come under the catch all phrase of the uctuation dissipation theorem Linear response theory is that the change of an observable 6A due to a small eld that couples to a second observable B Hint B6F This dependence can be expressed in terms of a generalized susceptibility x where 6A x6F de nes x The central result of linear response theory is that X can be expressed in terms of correlations of the type AOBr Although we will not have the time to cover linear response theory I recommend reading the rst two chapters of Forster 1 Equation 31 will be important in the next chapter as it will be used to connect the divergent behavior of the correlation function at the critical point to other quantities such as the susceptibilities and speci c heat The curvature used in the free energy A changes sign at the critical point One can parameterize this as AT at t T 7 TcTc 32 Thus the correlation length diverges as t lZ 5 Mg IltLat 12 33 2T X a The exponents describing the divergences here are known as critical exponents ln Landau theory all three dimensional systems would be have the same critical exponents 5 N W V 127 34 X N W777 v 1 The speci c heat is also divergent but to see that we must rst calculate the free energy Equation 23 gives the free energy for a particular set of amplitudes 11 12 To calculate the free energy of the entire system one must sum over all the possible values of 1 ftot iTln Z e F 1gt 2 39gtT 35 all con gurations 151 2quot39 1D Forster Hydrodynamic Fluctuations Broken Symmetry and Correlations Functions n Since F can be written as a sum over 1nd1v1dual modes k at F2111 ZaFkWT 36 I 1 As was shown in Chapter 3 the sum over states gtk can be written as an integral over the real and imaginary parts of 116 so that the partition function becomes eiFkT 2i d d 512AHk2l kl2 37 7139 N A nszl where the last expression ignores additive constants Finally the total free energy of the system will be at T Z lnA 2 38 E After inserting A at one can solve for the contribution to the speci c heat from the temperature dependence in A One must rst take a derivative wrt t to get an expression for the energy density then take a second derivative to nd the speci c heat The results is a2 3 1 1 7 7 N i 2 CVl 27031 k at may if 39 This last term represents another critical exponent CV N 7570 04 12 40 Another critical component concerns the rise of the mean magnetization for temperatures just below To In our approximation that we consider only small uctuations about the free energy minimum our choice of the mean magnetization comes from the mean eld approximation from the previous chapter ltmgt N W B 12 41 Finally for an external eld and T To mean eld theory gives ltmgt MEWS 6 3 42gt The divergence of the speci c heat is related to the divergence of the integrand above when t is set to zero Note that for 1 dimensions the integrand goes as dk kd 1k4 which will be divergent for d g 3 Thus in four dimensions or higher the speci c heat is well de ned at t a 0 whereas the behavior is more complicated for the more physical two and three dimensional cases Summarizing the critical exponents in Landau theory lt gt0 gtTgt Tz de T 5 17 V 127 Cu N t 0427d2 ltmgtt0 H15 63 45 ltmgtH0tlt0 25quot 512 X N 15 v 17 where d is the dimensionality We will see in the next two sections that Landau theory is not valid near To for d g 3 Nonetheless the features of Landau theory are qualitatively correct and most important the correct way of visualizing the behavior requires Landau theory for perspective 74 Validity of Landau Theory Near To The Ginzburg Criteria The exponents above were based on an assumption that one could ignore the quartic piece of the free energy density which would only be justi ed if uctuations of the order parameter were small compared to the mean value To see whether that is the case we consider the quartic case A B M 7W 7W 48 2 4 where A at is negative The free energy is then minimized for lt gtgt2 ABatB7 49 whereas the uctuations behaved as T lt gt0 gtTgt Tz de Tg 50 H The last expression was derived exactly for three dimensions previously see Eq 29 The integrals for the three dimensional integration can be generalized for an arbitrary number of dimensions 1 in the limit that r is large resulting in the expression above If one uses r E as a characteristic length scale for comparison the typical size for uctuations are T lt6 6 gttypica1 g 51 where dimensionless constants such as 27139d are ignored For mean eld theory to be valid one must justify ignoring the quartic piece in the free energy above or A B lt6 6 gttypical gtgt Zlt6 6 gtfypical7 52 T at gtgt B752 H Since 5 N Marl2 53 the criteria for validity becomes tHZ gtgt BchdZa tdZ 54 As the critical point is approached t a O and the criteria will be satis ed for d gt 4 However mean eld theory is unjusti ed in the critical region for three dimensions or fewer Despite the fact that the theory is unjusti ed Landau theory is still enormously useful First the theory is still justi ed away from the critical point depending on the parameters Secondly it is qualitatively correct in many aspects and represents a crucial perspective from which one can understand the more correct approaches based on scaling and renormalization which are also eld theories but differ in that they consider the effects of the 14 term 75 Critical Phenomena Scaling and Exponents Phase transitions can be classi ed as either rst or second order First order transitions are char acterized by co existing phases with the order parameter being different in the two phases In second order transitions the derivative or perhaps nth order derivative of the order parameter is discontinuous The term order parameter77 refers to any measure not even necessarily observable used to describe a system in a given state Examples are the magnetization density ferromagnetic transition number density liquid gas transition quark antiquark condensate chiral transition Higg s condensate electro weak transition or pair density superconductivity Quantities that involve comparing systems with different conditions such as the speci c heat which involves com paring the entropy at two different temperatures or the susceptibility which involves comparing the magnetization at two different external elds are not order parameters Transitions can belong to any of several universality classes a grouping determined by the symmetry associated with the phase transition For instance in the X 7 Y model the system breaks two dimensional rotational symmetry by choosing a speci c direction for aligning spins The three dimensional analog the Heisenberg model has a higher degree of symmetry hence belongs to a different universality class and different critical exponents which will be introduced further below At To Landau s eld theory gave the following form for the correlation function for magnetization density in the previous section 1 1W E ltmmr 0gt N 6 E wtA 55 7 where the free energy density was treated in the Gaussian approximation A f N 37712 gVm2 56 Here A is a function of temperature At To the sign of A will switch and the correlation length diverges as r T 7 To dAdT To 39 ln interacting theories the effects of interactions change the behavior of 5 near Tc quantitatively but not qualitatively Quantum aspects of the interactions lead to screening effects that can be understood in renormalization calculations but are beyond the scope of this class See Chapter 18 in Huang The resulting form for the correlation function near To is of the form m N TzidMe TE 5 N r 58 5 T 7 Tc 12 N 12 t 57 Here 1 is the number of dimensions In the three dimensional Landau calculation 1 3 and the two cn39tz39cal exponents are 710 112 59 However in the real world they vary from this amount depending on the universality class The other critical exponents Oz y 6 are all determined by 7 and 1 The relations can best be understood from dimensional analysis These arguments are based on two assumptions 1 The non analytic part of the free energy density scales proportional to N 153 From dimen sional grounds if E is the only relevant length scale and since the freeenergy density has dimensions of 1L3 it must behave as 5 to The magnetization and elds all behave as if they have effective dimensionalities such that changing t which causes a scale change of 5 will scale all quantities according to their effective dimensionality Following these arguments constrains all the other critical exponents as follows lt gtlt0gt gtltrgtgt WW6 5 r u 12 C7 N t7 042711 ltmgtt0 N H15 6d ltmgt HOtlt0 N 25quot BVdn722 X N 15 v Z2 i n The two assumptions above were nally justi ed with renormalization theory For this Ken Wilson was awarded the Nobel prize Renormalization theory is beyond what we will cover in this class but the interested student should read Chapter 18 of Huang 76 Symmetry Breaking and Universality Classes ln Landau theory the critical exponents were all determined by the dimensionality alone However in nature they are determined by the symmetry breaking involved in the problem For instance if there is a single real eld 1 whose free energy density is of the mexican ha 7 form A f 7 in 65 the system must choose whether to create a eld with lt gtgt gt 0 or lt 0 when A becomes negative For a complex eld with a free energy density ofthe same form the system must also choose the complex phase For a system where the spins are allowed to align in any direction the system must choose a direction 9 gt on which to align Each of these three examples represents a different symmetry and leads to different exponents The critical exponents are determined by the symmetry and problems of the same symmetry are referred to as being in the same universality class The breaking of the symmetry at low temperature is called spontaneous symmetry breaking as opposed to the breaking of the symmetry by adding an external eld which is referred to as explicit symmetry breaking For the broken re ection symmetry above the system chooses between two discrete minima whereas for the example of the complex eld or for the spin alignment the set of possible minima form an in nite continuous set These symmetries are called continuous symmetries and have the property that after being broken one can make incremental changes in the eld while maintaining a minimum in the free energy For instance consider the case of the complex eld Writing the free energy density in terms of the real and imaginary parts m and toy f go gt gt g i gt 2 aw lt66 where we consider the case where 04 lt 0 and minima appear for A is is gt3 7 67gt If one expands around the minima in the z direction one can rewrite the elds as Rewriting the free energy in terms of ow and My 1 82f 82f z a 76 27 f fmn2 aim 82 1 2 1 2 quot9 2 quot9 2 rm glam Ay6 gty 5mm ltV6 ygt Am 4Blal Ay O gm gm 69 Thus the y mode contributes to the free energy only through the term proportional to V6 gty2 or in momentum space the modes contribute to the free energy as k2 For relativistic eld theories modes for which the contribution vanishes as k a 0 are called massless modes as they have no rest energy These modes are also referred to as Goldstone bosons Goldstone derived the general idea that massless modes should be associated with breaking a continuous symmetry or Nambu Goldstone bosons Although the most famous example of such symmetry breaking is the electro weak transition in particle physics Higgs Goldstone and Nambu were concerned with the nuclear physics problem of chiral symmetry breaking and of course Landau was most concerned with problems in condensed matter physics It is no exaggeration to state that the most important problems addressed in the last 40 years in nuclear particle and condensed matter physics center around spontaneously broken symmetries Furthermore one of the most exciting astronomical measurements that of uctuations of the 3 degree background radiation concerns in ationary theory which might be associated with the breaking of GUT grand uni ed theories symmetries under which quarks and leptons appear in the same multiplet as do all gauge bosons 77 1 to Problems Consider the example for which the surface energy was calculated where A 2 AWEP07P km 3 pipc iazl Using Eq 17 solve for the density pro le pz between the two phases Consider the onedimensional lsing model with the total energy in the mean eld approxi mation being 1 E 7 E 7 J i 2g ltUgtUl Note the factor of 12 being added relative to the effective energy for one mode to account for double counting a Using the de nition of entropy 7 Zip lnp show that the entropy per spin is 7 1lta 1lta 170 170 SN 7 2 ln 2 2 ln 2 b The freeenergy F E 7 TS per spin is then FN 7qJltU2 T712ltUgtln 71297 T7172ltUgt1n 71 Show that minimizing the free energy wrt 0 gives QBqJlta ln w c Compare the expression above to that from the previous expression ltagt tanhww d If the density of spin sites per unit volume is p0 the free energy density is fltmTgt p0vltaTgtgltvagt2 VaT 7anzT1Ulnlt1TagtT lnlt1gagt Derive an expression for the surface free energy that is an integral over the density with limits from 7 Heel and aeq Write the integrand in terms of po rs T and Ra T E Va T 7 Vaeq T where aeq is the solution to the transcendental expression for the equilibrated value for a

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