Introduction to Thermodynamics and Statistical Physics phần 2 ppt

... − ³ ∂ log Z gc ∂η ´ β  (∆X l ) 2 ®  (∆X l ) 2 ® = ∂ 2 log Z ∂ξ 2 l  (∆U) 2 ® = ∂ 2 log Z c ∂β 2  (∆U) 2 ® = ³ ∂ 2 log Z gc ∂β 2 ´ η  (∆N) 2 ® = ³ ∂ 2 log Z gc ∂η 2 ´ β σ σ = log Z + L P l=1 ξ l hX l i σ ... (1.69) Z c =exp µ βε 2 ¶ +exp µ − βε 2 ¶ =2cosh µ βε 2 ¶ , (1.75) thus using Eqs. (1.70) and (1.71) one finds hUi = − ε 2 tanh µ βε 2 ¶ , (1.76)...

Ngày tải lên: 08/08/2014, 15:21

... MMM ee ++ 2 p … 1 2 p q 1 1 p q M 1 1 +M e 2 1 p q M 2 1 +M e 21 MM e + … 2 2 1 p q M + 2 21 p q MM + 1 21 1 M qqqp +++= 21 1 12 MMM qqp ++ ++= 1 p 1 e 1 , ,, 21 M eee 1 1 p q 2 e 1 M e … 21 1 , , 1 MMM ee ++ 2 p … 1 2 p q 1 1 p q M 1 1 +M e 2 1 p q M 2 1 +M e 21 MM e + … 2 2 1 p q M ... of Statistical M echanics 127 4.1 ClassicalHamiltonian 127 4.1.1...

Ngày tải lên: 08/08/2014, 15:21

... 1 (2. 42) and employ the first order expansion log (1 + x)=x + O ¡ x 2 ¢ (2. 43) to obtain Eyal Buks Thermodynamics and Statistical Physics 53 2. 1. A Particle in a Box n Q = µ Mτ 2 ~ 2 ¶ 3 /2 . (2. 12) The ... by Z 1 = ∞ X n x =1 ∞ X n y =1 ∞ X n z =1 exp ³ − ε n x ,n y ,n z τ ´ = ∞ X n x =1 ∞ X n y =1 ∞ X n z =1 exp ¡ −α 2 ¡ n 2 x + n 2 y + n 2 z ¢¢ , (2. 6) wh...

Ngày tải lên: 08/08/2014, 15:21

... } π 4 /15 = 3π 4 5 Nτ ³ τ Θ ´ 3 , (3.80) Eyal Buks Thermodynamics and Statistical Physics 109 3 .2. Phonons in Solids mm ω 2 mm ω 2 mm ω 2 mm ω 2 m mm ω 2 mm ω 2 mm ω 2 mm ω 2 m Fig. 3 .2. 1D lattice. 3 .2 Phonons in Solids In ... given by σ = N µ log n Q n + 5 2 ¶ , (2. 304) where n Q = µ Mτ 2 ~ 2 ¶ 3 /2 , (2. 305) or as a function of τ and p σ = N Ã log ¡ M...

Ngày tải lên: 08/08/2014, 15:21

... = ε F Z 0 dεD(ε)= D (ε F ) ε 1 /2 F ε F Z 0 dεε 1 /2 = D (ε F ) ε 1 /2 F 2 3 ε 3 /2 F , thus ε F = ~ 2 2m µ 3π 2 N V ¶ 2/ 3 , (3.166) and therefore U = 3N 5 ~ 2 2m µ 3π 2 N V ¶ 2/ 3 . (3.167) Moreover, at ... Buks Thermodynamics and Statistical Physics 118 Chapter 3. Boso nic and Fermionic Systems √ ε 2 µ β α 2 ¶ 3 /2 dε = n 2 dn. Thus, b y introducing the d...

Ngày tải lên: 08/08/2014, 15:21

... (r) rdr =2 LA Z R 0 exp µ βMω 2 2 r 2 ¶ rdr = 2 LA βMω 2 · exp µ βMω 2 2 R 2 ¶ − 1 ¸ , (4.93) thus n (r)= NβMω 2 2πL h exp ³ βMω 2 2 R 2 ´ − 1 i exp µ β 2 Mω 2 r 2 ¶ . (4.94) 9. In the classical ... Similarly  v 2 ® = R ∞ 0 dvv 4 exp ³ − Mv 2 2τ ´ R ∞ 0 dvv 2 exp ¡ − Mv 2 2τ ¢ = 2 M R ∞ 0 dxx 3 /2 exp (−x) R ∞ 0 dxx 1 /2 exp (−x) = 2 M 3 2 , (4.99) thu...

Ngày tải lên: 08/08/2014, 15:21

... ˙x 2 2 ¢ 2 = m ¡ ˙x 2 + + ˙x 2 − ¢ 2 , (6.15) and the poten tial energy V is given by V (x 1 ,x 2 )= mω 2 x 2 1 2 + mω 2 x 2 2 2 + mΩ 2 (x 1 − x 2 ) 2 = mω 2 x 2 + 2 + m ¡ ω 2 +4Ω 2 ¢ x 2 − 2 . (6.16) The ... equipartition theorem yields mω 2  x 2 + ® 2 = m ¡ ω 2 +4Ω 2 ¢ x 2 − ® 2 = τ 2 , (6.17) thus D (x 1 + x 2 ) 2 E = 2...

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... 27 1.8. Solutions Set 1 Thus D (n −hni) 2 E = p 2 N (N − 1) + pN −p 2 N 2 = Npq , (1.164) and D (X − hXi) 2 E =4a 2 Npq . (1.165) 10. The total energy is given by E = kx 2 2 + m ˙x 2 2 = ka 2 2 , ... Z c ∂β = − 2Nεsinh (βε) 1+2coshβε , (1.190) b) and the va riance is D (U −hUi) 2 E = ∂ 2 log Z c ∂β 2 = − ∂ hUi ∂β =2Nε 2 cosh (βε) +2 [1 + 2 cosh (βε)] 2 . (...

Ngày tải lên: 08/08/2014, 15:21

... has log τ 2 τ 1 =log µ V 2 V 1 ¶ 1−γ . (2. 107) Thus τ 1 V γ−1 1 = τ 2 V γ−1 2 , (2. 108) or using Eq. (2 .55) p 1 V γ 1 = p 2 V γ 2 . (2. 109) Eyal Buks Thermodynamics and Statistical Physics 63 Chapter ... mean n umber of absorbed atoms is given by hN a i = N 0 +2 2 e 1+ + 2 e , (2. 147) where =1/ and = à M 2} 2 ả 3 /2 5 /2 p. (2. 148) 13. An ideal gas con...

Ngày tải lên: 08/08/2014, 15:21

... e −η V µ M 2 ~ 2 β ¶ 3 /2 , (2. 254) thus U = − µ ∂ log Z gc ∂β ¶ η = 3Nτ 2 . (2. 255) a) Using E q. (2. 2 52) D (∆E) 2 E = − ∂ ∂β 3N 2 = 3 2 N β 2 = 2U 2 3N . (2. 256) b) Using Eq. (2. 253) D (∆E) 3 E = ... τ l . (2. 264) 21 . Using Eq. (2. 264) A (τ h − τ l ) P = τ l τ h − τ l , (2. 265) thus τ 2 l − 2 l µ τ h + P 2A ¶ + τ 2 h =0, (2. 266) or τ l = τ h + P 2A ± s µ τ...

Ngày tải lên: 08/08/2014, 15:21