Assignment 1. Tyler Shendruk January 31, 2010

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1 Assignment 1 yler Shendruk January 31, Problem 9 Superconductor phase transition. 1.1 Problem 9a he heat capacity of a metal going to it s super conducting state at low enough temperatures has a heat capacity { C s V α 3 superconducting phase C( ) C s V [ β 3 + γ ] (1) normal phase. In general, we know that the heat capacity is C x dq d () x and when no work is being enacted on the system we have dq ds. (3) By putting these two together, we find the entropy by integrating ds x C x d which for us is really: x ds B,µ C d B,µ () i.e. we find the entropy in the case when we had a fixed field which Kadar says is zero. he integration results in V α 3 S s ( ) S s (0) d 0 V α 3 3 V β 3 + γ S n ( ) S n (0) d 0 [ ] β V γ 1

2 By the hird Law (lim 0 S 0), we write these as 1. Problem 9b S s ( ) V α 3 3 (5) S n ( ) V [ ] β γ When there latent heat is zero (L 0), we are dealing with a second order transition. Since there is no latent heat there is no change in entropy. herefore, at a critical temperature, c, the phase change occurs and the entropies are equal (6) S n ( c ) S s ( c ). (7) hat is enough to determine what the critical temperature is S n ( c ) S s ( c ) V α 3 3 c V [ β 3 3 c + γ c 1 3 [α β] c γ c 3γ α β 3γ c ± α β. We reject the negative option as unphysical and are left with the phase transition temperature 3γ c α β. (8) 1.3 Problem 9c So we ve shown that when Cooper pairs are weakly coupled together the entropy does decrease due to this increased order. his of course has an effect on the energy. We assume the volume of the metal isn t changing considerably and we still have no magnetic field so ] de ds + J i dx i ds. (9)

3 By Eq. 1.1 we know ds C x d. herefore and de Cd de n Cdt E n ( ) E n (0) V [ β 3 + γ ] dt [ β V + γ ] but the question tells us E s ( ) E s (0) V α E( 0) { E n (0) E 0 E s (0) E 0 V. (10) herefore the energies are 1. Problem 9d E n V he Gibbs free energy is defined [ β + γ ] + E 0 (11) [ ] α E s V + E 0 (1) G E S J i x i. (13) We still say that we are in the zero field limit. here is no chemical potential difference between phases and so the Gibbs free energy at c must be equal for both phases: V [ β c + γ c ] + E 0 c V G n G s E n c S n E s + c S s [ ] β 3 c 3 + γ c V [ α c β c + γ c β 3 c γ c α c α 3 c ] + E 0 c V α 3 3 c γ c + c (α β) c 3 γ c c (α β). 1 (α β) 3

4 We also know the critical temperature from Eq. 8 which we substitute in: γ 3γ α β α β ( ) 3γ 1 α β 1.5 Problem 9e γ γ 3 α β 3 α β γ 3 α β. (1) We repeat the exact same Gibbs free energy balance but add a magnetic term (only to the superconduction phase) G n G s (15) E n S n 0 E s S s + V 8π B (16) So we know E n, S n, E s, S s, c and from previous questions. Sub them in (keep the temperature as but remember when you put in that it depends on c not, just a warning). hen solve for B. A comment for those of you who are concerned about these sorts of things: he entropy remain unchanged (Eq. 7) as long as the heat capacity has no field dependence. he heat capacity probably does have a field dependence. But we re assuming it s neglible. I suppose this would always be true at small enough fields and so the field must be small enough that the heat capacity s (and therefore the entropy s) dependance on it is neglegible. Furthermore, we are using the energy from Eq. 9 which must assume the field doesn t vary. It doesn t have to be zero but it must be constant. Ok, back to the solution. he Gibbs free energies equal each other at the critical field - there is a different critical field for each temperature V [ β + γ ] [ β + E 0 V 3 + γ E n S n E s S s + V 8π B c ] [ ] α V β + γ β 3 γ α B c 8π α 3 + B c 8π + (α β) 1 γ + E 0 V α 3 + V 8π B c 3 γ + (α β) α β 1 γ ( ) ( α β 6γ 1 α β + ( ) [ α β 3γ ] 1 α β ( ) α β [ ] c 1 ) 9γ (α β)

5 And we re done. he order of operation inside the () doesn t matter. By setting B 0 π(α β)/3 c πγ, the critical temperature is ( ) ] B c B 0 [1 c (17) Problem 10 Photon gas. Dont worry about what it is just consider the PV diagram..1 Problem 10a he work is just the area of the cycle.. Problem 10b he heat exchange is.3 Problem 10c Q E W W dp dv. (18) de + P dv [ E V dv + E ] d +P dv V E V dv + 0 +P dv Q We can express the efficiency, η, in two equivalent ways: [ ] E V + P dv (19) 1.. η W Q η 1 1 d (0) (1) 5

6 By equating the two we can find η d W Q dp dv [ E V + P ] dv dp E V + P Rearranging we find dp ( E V + P ) d (). Problem 10d We ve been told and so we then know P A (3) dp A 3 d. () By substituting these two into Eq., we can arrive at an expression for the energy. ( ) E dp V + P d ( A 3 d ) ( ) E V + A d A E V + A E V 3A de 3A dv de 3A dv Where we remembered that we are keeping constant and use that E( 0, V ) 0 to find E(, V ) 3AV. (5) 6

7 .5 Problem 10 e Now it s adiabatic so đq 0 but it s not necessarily along an idotherm. So we have đq 0 de + P dv E d + E V V dv + P dv 3AV d + 3AV V V dv + P dv 1AV 3 d + 3A dv + P dv ( ) ( ) dp P 1AV 3 A 3 + 3A dv + P dv A 3V dp + P dv. From here we can rearrange and integrate. Let K and K the integration constants 3 Problem 11 Kadar dp P dv 3 V ln P 3 ln V + K ln P ln V /3 + K P K V.3 P V /3 constant. (6) An irreversible process between two substances at different temperatures. 3.1 Problem 11a Show that the change of entropy must be positive. We realize that the total change of entropy is the sum of the change of entropy in both substances: S S 1 + S but the process of heat exchange is irreversible for both substances so that S i i đq i i. herefore, we must have the condition S 1 đq đq. (7) 7

8 he heat that is recieved by one substance is lost by the other so we use the notation đq đq 1 đq. So integrating over the temperature, the inequality reads đq 1 S + đq 1 đq + đq 1 ( + ) 1 đq đq (8) 1 Keeping in mind our notation (Eq. 3.1), let s consider two cases: 1 > : Heat flows from substance 1 to substance. herefore } đq > 0 đq > 0 but đq 1 < 0 1 > 0 therefore ( 1 ) đq > 0 therefore S > 0 > 1 : Heat flows from substance to substance 1. herefore } đq < 0 đq < 0 and đq 1 > 0 1 < 0 therefore ( 1 ) đq > 0 therefore still have S > 0 So for all cases, we have S 0 (9) 3. Problem 11b I was only able to demonstrate that the Gibbs free energy is an extrema point at equallibrium, not necessarily a minimum. We expect that it will be a minimum and never a maximum so this isn t really a big deal. Problem Sethna Blackhole has mass M (and so energy E Mc ) and a radius of R G M c. he hole emits radiation at a temperature of where we have defined K c 3 /8πGk B. c 3 8πGMk B K M (30) 8

9 .1 Problem a o calculate the specific hwe use that eat (which maybe negative) we use the definition of a heat capacity as already stated in Eq. and the fact that no work is being done so So then the heat capacity is đq de đw de C đq d E Mc ( ) Kc Kc OR C c5 1 8πGk B (31) C 8πGk B M (3) c 9

10 . Problem b o calculated the entropy of the black hole, we that S(M 0) 0 and that S E 1 ds de 1 [ ( )] 1 Kc d Kc 3 d ds Kc d 3 S(M) S(M 0) Kc S(M) 0 Kc (K/M) S(M) c M K c M c3 8πGk B πgk B M c πgk B c A k Bc G. c R G So the entropy is directly related to the area - the rest are just universal constants. 10

8.333: Statistical Mechanics I Problem Set # 1 Due: 9/18/13 Thermodynamics. J = ax bt + ctx,

8.333: Statistical Mechanics I Problem Set # 1 Due: 9/18/13 Thermodynamics 1. Filament: For an elastic filament it is found that, at a finite range in temperature, a displacement x requires a force J =