Problem 1. Solution: = show that for a constant number of particles: c and V. a) Using the definitions of P
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1 rol. Using t dfinitions of nd nd t first lw of trodynis nd t driv t gnrl rltion: wr nd r t sifi t itis t onstnt rssur nd volu rstivly nd nd r t intrnl nrgy nd volu of ol. first lw rlts d dq d t onstnt volu: Q t onstnt rssur: d d d d d d d dq Q or d d d dividing trougout y d t onstnt rssur: tus ro t fundntl diffrntil rltion for d nd t diffrntil rltion for loltz fr nrgy d d d sow tt for onstnt nur of rtils:
2 fundntl diffrntil rltion d d d king rtil drivtiv wit rst to volu wit onst: rrnging ro d d d v nd Diffrntiting wit rst to t otr vrils: nd On rrivs t Mxwll rltion wit wi Us t ov rsults to find for Idl Gs oining rsults fro nd on gts v rssur qution of stt for on ol of idl gs is tus
3 d Us t ov rsults to find n dr Wls gs n / / ow tt t onstnt rssur in t liit t rsult for of n dr Wls gs is t s s for of n idl gs. O /
4 rol. initil stt of onotoi idl gs is dsrid y nd t trtur rssur nd volu rstivly. gs is tkn ovr t t qusisttilly s sown in t skt. volus r rltd s.5. ind: ow u work dos t gs do on t t nd wt is t ng in its intrnl nrgy? xtrnl work: W ng in xtrnl nrgy: n n n n ow u t is sord in going fro? ro t first lw: 5 d dq d Q W Driv t xrssion for t ntroy ng for n ritrry ross. d d d wit n v nd n d n d n d n d n d n v ln n ln n ln n [ ] ln n ln sin onst for diti rvrsil ross. d If is n diti ross find t finl gs rssur nd t ntroy ng fro t gnrl xrssion otind in rt. or diti xnsion of idl gs onst wit 5 /
5 5 n If t ross is rvrsil on xts. [ ] ln ln ln n n n
6 rol. onsidr on-dinsionl in onsisting on >> sgnts s illustrtd in t skt. Lt t lngt of sgnt wn t long dinsion is rlll to t in nd zro wn t sgnt is vrtil i.. long dinsion is rndiulr to t in dirtion. sgnt s just two stts orizontl nd vrtil nd of ts stts is not dgnrt. distn twn t in nds is fixd. or givn lngt Ll <l< of in wt is t totl nur of irostts ssil y t syst nd wt is t ntroy of t syst s funtion of l? or t in lngt to l tr will l/ orizontls sgnt so tt totl nur of ossil irostts is: Ω! tn!! k ln Ω k ln l! l!! Writ down t rorit trodyni idntity for t syst quivlnt to t first lw nd dsri qulittivly ow on ould otin n xrssion for tnsion for nssry to intin t lngt l ssuing t joints turn frly fro t rsult otind in rt rorit trodyni idntity is d d dl It follows tt nd l l l Using t riroity rltion on gts l l lultd fro t rsult for in rt. wi ould l Otin t rltionsi twn tt tnsion for intining t distn l nd t trtur using t nonil nsl dsrition. nrgy ontriution for orizontl link is tus vrg link lngt is 6
7 l k k k for links: k l k k k k l l ln k l l d Undr wi onditions dos your nswr ld to ook s lw nd wt is orrsonding xrssion for t sring onstnt? t ig or low k k k k k k k l k k k k k k l wit sring onstnt rol. syst of two nrgy lvls nd is oultd y rtils t trtur. ssu tt > so tt - is ositiv. Driv n xrssion for t vrg nrgy r rtil s funtion of trtur. Dtrin t liiting vior nd vlu for vrg nrgy r rtil in t liits of nd. or so Wit liiting vlu li 7
8 8 or so Wit liiting vlu li Driv n xrssion for sifi t of t syst. or ol: k d out sifi t in t liits of nd. or : k or : k rol 5. onsidr n idl gs of sin-½ frions is onfind to n r in dinsions. onsidr t ground stt for su syst. Driv t xrssion for t nur of singl rtil irostts in t ontu intrvl d for t givn syst. In D n n Γ L n n k L k k Γ kdk kdk L dk dk k d dk k g Γ k dk d d d g or rions tr r two ossil sin stts for givn nrgy tus t dnsity of stts douls: d d g
9 9 ind t ri ontu nd il otntil of t syst. il otntil t is lld ri nrgy nd it is rltd to ri ontu s follows ε µ. totl nur of ltrons is d g n In t ground stt µ d d g µ µ ind t vrg nrgy r rtil for t syst. µ ε d d d g d g d ow ssu t gs syst is ld in t unifor gnti fild wi ks n dditionl singl rtil nrgy ontriution ±µ dnding on t sin orinttion. ind t ri ont for sin-u nd sin-down frions. singl rtil nrgy in t gnti fild would ± µ ri ontu is xiu ssil ontu so for t sin-u frions: µ tus µ in-down:
10 µ tus µ lult t vrg gntiztion r r for t syst. vrg nur of sin-down rtils: µ µ µ d µ lugin µ µ iilrly d µ µ vrg gntiztion r r µ µ µ
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