Introduction to Thermodynamics and Statistical Physics phần 7 pptx

2.9. Solutions Set 2 Using pV = Nτ yields η =1+ τ d − τ c τ b − τ a =1+ p 2 (V d − V c ) p 1 (V b − V a ) . (2.295) Along the isentropic process pV γ is constant, where γ = C p /C v ,thus η =1+ p 2 p 1 ³ p 1 p 2 ´ 1 γ (V a − V b ) V b − V a =1− µ p 2 p 1 ¶ γ−1 γ . (2.296) 28. No heat is exchanged in the isentropic processes, thus the efficiency is given b y η =1+ Q l Q h =1+ Q c→d Q a→b =1+ c V (τ d − τ c ) c V (τ b − τ a ) . (2.297) Since τV γ−1 remains unchanged in an isentropic process, where γ = c p c V , (2.298) one finds that τ b V γ−1 1 = τ c V γ−1 2 , (2.299) τ d V γ−1 2 = τ a V γ−1 1 , (2.300) or τ c τ b = τ d τ a = µ V 2 V 1 ¶ 1−γ , (2.301) thus η =1− µ V 2 V 1 ¶ 1−γ . (2.302) 29. Let V A1 = Nτ A /p (V B1 = Nτ B /p) be the initial volume of vessel A (B) and let V A2 (V B2 )bethefinal volume of vessel A (B). In terms of the final temperature of both vessels, wh ich is denote d as τ f ,onehas V A2 = V B2 = Nτ f p . (2.303) Eyal Buks Thermodynamics and Statistical Physics 95 Chapter 2. Ideal Gas The entropy of an ideal gas of density n = N/V , w hich contains N particles, is given by σ = N µ log n Q n + 5 2 ¶ , (2.304) where n Q = µ Mτ 2π~ 2 ¶ 3/2 , (2.305) or as a function of τ and p σ = N Ã log ¡ M 2π~ 2 ¢ 3/2 τ 5/2 p + 5 2 ! . (2.306) Thus the c hange in entrop y is giv en by ∆σ = σ final − σ initial = 5N 2 log τ 2 f τ A τ B . (2.307) In general, for an isobaric process the following holds Q = W + ∆U = p (V 2 − V 1 )+c V (τ 2 − τ 1 ) , (2.308) where Q is the heat that w as added to the gas, W theworkdonebythe gas and ∆U the change in internal energy of the gas. Using the equation of state pV = Nτ this can be written as Q =(N + c V )(τ 2 − τ 1 ) . (2.309) Since no heat is exchanged with the environmen t during this process the following holds Q A + Q B =0, where Q A =(N + c V )(τ f − τ A ) , (2.310) Q B =(N + c V )(τ f − τ B ) , (2.311) thus τ f = τ A + τ B 2 , (2.312) and therefore ∆σ = 5N 2 log (τ A + τ B ) 2 4τ A τ B . (2.313) Eyal Buks Thermodynamics and Statistical Physics 96 3. Bosonic and Fermionic Systems In the first part of this chapter we study two Bosonic systems, namely photons and phonons. A photon is the quanta of electromagnetic wa ves whereas a phonon is the quanta of acoustic waves. In the second part we study two Fermionic systems, namely electrons in metals and electrons and holes in semiconductors. 3.1 Electromagnetic Radiation In this section we study an electroma gnetic cavity in thermal equilibrium. 3.1.1 Electromagnetic Ca vit y Consider an empt y volume surrounded by conductive walls having infinite conductivity. T he Maxw ell’s equations in SI units are given by ∇ × H =  0 ∂E ∂t , (3.1) ∇ × E = −µ 0 ∂H ∂t , (3.2) ∇ · E =0, (3.3) and ∇ · H =0, (3.4) where  0 =8.85 × 10 −12 Fm −1 and µ 0 =1.26 × 10 −6 NA −2 are the permit- tivity and permeability respectively of free space, and the following holds  0 µ 0 = 1 c 2 , (3.5) where c =2.99 × 10 8 ms −1 is the speed of light in vacuum. In the Coulomb gauge, w here the v ector potent ial A is chosen such that ∇ · A =0, (3.6) Chapter 3. Bosonic and Fermionic Systems the scalar potential φ vanishes in the absence of sources (charge and current), and consequently both fields E and H canbeexpressedintermsofA only as E = − ∂A ∂t , (3.7) and µ 0 H =∇× A . (3.8) The gauge condition (3.6) and Eqs. (3.7) and (3.8) guarantee that Maxwell’s equations (3.2), (3.3), and (3.4) are satisfied ∇ × E = − ∂ (∇× A) ∂t = −µ 0 ∂H ∂t , (3.9) ∇ · E = − ∂ (∇· A) ∂t =0, (3.10) ∇ · H = 1 µ 0 ∇· (∇ × A)=0, (3.11) where in the last equation the general vector identity ∇· (∇ × A)=0has been employed. Subs tituting Eqs. ( 3 .7) and (3.8) in to the only remaining nontrivial equation, namely into Eq. (3.1), leads to ∇ × (∇ × A)=− 1 c 2 ∂ 2 A ∂t 2 . (3.12) Using the vector id entity ∇ × (∇ × A)=∇(∇ · A) −∇ 2 A , (3.13) and the gauge condition (3.6) one finds that ∇ 2 A = 1 c 2 ∂ 2 A ∂t 2 . (3.14) Consider a solution in the form A = q (t) u (r) , (3.15) where q (t) is independent on position r and u (r) is independent on time t. The gauge condition (3.6) leads to ∇ · u =0. (3.16) From Eq. (3.14) one finds that q∇ 2 u = 1 c 2 u d 2 q dt 2 . (3.17) Eyal Buks Thermodynamics and Statistical Physics 98 3.1. Electromagnetic Radiation Multiplying by an arbitrary unit vector ˆn leads to ¡ ∇ 2 u ¢ · ˆn u · ˆn = 1 c 2 q d 2 q dt 2 . (3.18) The left hand side of Eq. (3.18) is a function of r only while the right hand side is a function of t only. T herefore, both should equal a constan t, which is denoted as −κ 2 ,thus ∇ 2 u+κ 2 u =0, (3.19) and d 2 q dt 2 +ω 2 κ q =0, (3.20) where ω κ = cκ . (3.21) Equation (3.19) should be solved with the boundary conditions of a per- fectly conductive surface. Namely, on the surface S enclosing the cavit y we have H · ˆs =0andE × ˆs =0,whereˆs is a unit vector normal to the surface. To satisfy the boundary condition for E we require that u be normal to the surface, namely, u = ˆs (u · ˆs)onS. T his condition guarantees also that the boundary condition for H is satisfied. To see this we calculate the integral of the normal comp onent of H over some arbitrary portion S 0 of S. Using Eq. (3.8) and Stoke ’s’ theorem one finds that Z S 0 (H · ˆs) dS = q µ 0 Z S 0 [(∇ × u) · ˆs] dS = q µ 0 I C u·dl , (3.22) where the close curve C encloses the surface S 0 .Thus,sinceu is norma l to the surface one finds that the integral along the close curve C vanishes, and therefore Z S 0 (H · ˆs) dS =0. (3.23) Since S 0 is arbitrary we conclude that H · ˆs =0 on S. Each solut ion of Eq. (3.19) that satisfies the boundary conditions is called an eigen mode. As can be seen from Eq. (3.20), the dynamics of a mode amplitude q is the same as the dynamics of an harmonic oscillator having angular frequency ω κ = cκ. Eyal Buks Thermodynamics and Statistical Physics 99 Chapter 3. Bosonic and Fermionic Systems 3.1.2 Partition Function What is the partition function of a mode having eigen angular frequency ω κ ? We have seen that the mode amplitude has the dynamics of an harmonic oscillator having angular frequency ω κ . Thus, the quantum eigenenergies of the mode are ε s = s~ω κ , (3.24) where s =0, 1, 2, is an integer 1 . When the mode is in the eigenstate having energy ε s themodeissaidtooccupys pho tons. T he canonical partition function of the m ode is found using Eq. (1.69) Z κ = ∞ X s=0 exp (−sβ~ω κ ) = 1 1 − exp (−β~ω κ ) . (3.25) Note the similarity between this result and the orbital partition function ζ of Bosons given by Eq. (2.36). The a verage energy is found using hε κ i = − ∂ log Z κ ∂β = ~ω κ e β~ω κ − 1 . (3.26) The partition function of the entire system is given by Z = Y κ Z κ , (3.27) and the a verage total energy b y U = − ∂ log Z ∂β = X κ hε κ i . (3.28) 3.1.3 Cube Cavity For simplicity, consider the ca se of a cavity shaped as a cube of volume V = L 3 . We seek solutions of Eq. (3.19) satisfying the boundary condition 1 In Eq. (3.24) above the ground state energy was taken to be zero. Note that by taking instead ε s =(s +1/2) ~ω κ ,oneobtainsZ κ =1/2sinh(β}ω κ /2) and hε n i =(~ω κ /2) coth (β}ω κ /2). In some cases the offset energy term ~ω κ /2is very important (e.g., the Casimir force), howev er, in what follows we disregard it. Eyal Buks Thermodynamics and Statistical Physics 100 3.1. Electromagnetic Radiation that the tangential component of u vanishes on the walls. Consider a solution having the form u x = r 8 V a x cos (k x x)sin(k y y)sin(k z z) , (3.29) u y = r 8 V a y sin (k x x)cos(k y y)sin(k z z) , (3.30) u z = r 8 V a z sin (k x x)sin(k y y)cos(k z z) . (3.31) While the boundary c ondition on the walls x =0,y =0,andz =0is guaranteed to be satisfied, the boundary condition on the walls x = L, y = L, and z = L yields k x = n x π L , (3.32) k y = n y π L , (3.33) k z = n z π L , (3.34) where n x , n y and n z are i ntegers. This so lution clearly satisfies Eq. (3.19) where the eigen value κ is given by κ = q k 2 x + k 2 y + k 2 z . (3.35) Alternatively, using t he notation n =(n x ,n y ,n z ) , (3.36) one has κ = π L n, (3.37) where n = q n 2 x + n 2 y + n 2 z . (3.38) Using Eq. (3.21) one finds that the angular frequency of a mode characterized by the vector of integers n is given by ω n = πc L n. (3.39) In addition to Eq. (3.19) and the boundary condition, eac h solution has to satisfy also the transversality condition ∇·u = 0 (3.16), which in the present case reads n · a =0, (3.40) Eyal Buks Thermodynamics and Statistical Physics 101 Chapter 3. Bosonic and Fermionic Systems where a =(a x ,a y ,a z ) . (3.41) Thus, for each set of integers {n x ,n y ,n z } there are two orthogonal modes (polarizations), unless n x =0orn y =0orn z = 0. In t he latter case, only a single solution exists. 3.1.4 Average Energy The average energy U of the system is found using Eqs. (3.26), (3.28) and (3.39) U = X n ~ω n e β~ω n − 1 =2τ ∞ X n x =0 ∞ X n y =0 ∞ X n z =0 αn e αn − 1 , (3.42) where the dimensionless parameter α is given by α = β~πc L . (3.43) The following relation can be employed to estimate the dimensionless param- eter α α = 2.4 × 10 −3 L cm τ 300 K . (3.44) In the limit where α ¿ 1(3.45) the sum can be approximated b y the integral U ' 2τ 4π 8 ∞ Z 0 dnn 2 αn e αn − 1 . (3.46) Emplo ying the in tegration variable transformation [see E q. (3.39)] n = L πc ω, (3.47) allows expressing the energy per unit volume U/V as U V = ∞ Z 0 dωu ω , (3.48) Eyal Buks Thermodynamics and Statistical Physics 102 3.1. Electromagnetic Radiation where u ω = ~ c 3 π 2 ω 3 e β~ω − 1 . (3.49) This result is know as Plank’s radiation law. The factor u ω represents th e spectral distribution of the radiation. The peak in u ω is obtained at β~ω 0 = 2.82. I n terms of the wavelength λ 0 =2πc/ω 0 one has λ 0 µm =5.1 µ T 1000 K ¶ −1 . (3.50) 0.2 0.4 0.6 0.8 1 1.2 1.4 0246810 x The function x 3 / (e x − 1). The total energy is found by integrating Eq. (3.48) and by employing the variable transformation x = β~ω U V = τ 4 c 3 π 2 ~ 3 ∞ Z 0 x 3 dx e x − 1 | {z } π 4 15 = π 2 τ 4 15~ 3 c 3 . (3.51) 3.1.5 Stefan-Boltzmann Radiation La w Consider a small hole having area dA drilled into the conductive wall of an electromagnetic (EM) cavity. What is the rate of energy radiation emitted from the hole? We employ below a kinetic approach to answer this question. Considerradiationemittedinatimeintervaldt in the direction of the unit vector ˆu.Letθ be th e angle between ˆu and the normal to the surface of the hole. Photons emitted during that time interval dt in the direction ˆu came from the region in the cavity that is indicated in Fig. 3.1, which has volume Eyal Buks Thermodynamics and Statistical Physics 103 Chapter 3. Bosonic and Fermionic Systems θ dAcosθ c d t u ˆ θ dAcosθ c d t u ˆ Fig. 3.1. Radiation emitted through a small hole in the cavity wall. V θ =dA cos θ × cdt. (3.52) The average energy in that region can be found using Eq. (3.51). Integrating over all possible directions yields the total rate of energy radiation emitted fromtheholeperunitarea J = 1 dAdt 1 4π π/2 Z 0 dθ sin θ 2π Z 0 dϕ U V V θ = π 2 τ 4 15~ 3 c 2 1 4π π/2 Z 0 dθ sinθ cos θ 2π Z 0 dϕ | {z } 1/4 = π 2 τ 4 60~ 3 c 2 (3.53) In terms of the historical definition of temperature T = τ/k B [see Eq. (1.92)] one has J = σ B T 4 , (3.54) where σ B ,whichisgivenby σ B = π 2 k 4 B 60~ 3 c 2 =5.67 × 10 −8 W m 2 K 4 , (3.55) is the Stefan-Boltzmann constant. Eyal Buks Thermodynamics and Statistical Physics 104 [...]... Thermodynamics and Statistical Physics (3.68) 1 07 Chapter 3 Bosonic and Fermionic Systems Similarly to the EM case, the average total energy is given by X }ω n U = exp (β}ω n ) − 1 n 3π = 2 n ZD dn n2 0 (3.69) }ω n exp (β}ωn ) − 1 (3 .70 ) To proceed with the calculation the dispersion relation ω n (kn ) is needed Here we assume for simplicity that dispersion can be disregarded to a good approximation, and consequently... kn = 2πn , aN (3. 57) and n is integer ranging from −N/2 to N/2 0.8 0.6 0.4 0.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 x 0.6 0.8 1 The function |sin (πx/2)| Eyal Buks Thermodynamics and Statistical Physics 105 Chapter 3 Bosonic and Fermionic Systems What is the partition function of an eigen-mode having eigen angular frequency ωn ? The mode amplitude has the dynamics of an harmonic oscillator having angular... β}vπ 6N β}vπnD π = xD = L L Alternatively, in terms of the Debye temperature, which is defined as Θ = }v µ 6π2 N V ¶1/3 , (3 .75 ) one has Eyal Buks Thermodynamics and Statistical Physics 108 3.2 Phonons in Solids xD = Θ , τ (3 .76 ) and x ³ τ ´3 Z D x3 U = 9N τ dx Θ exp x − 1 (3 .77 ) 0 As an example Θ/kB = 88 K for Pb, while Θ/kB = 1860 K for diamond Below we calculate the heat capacity CV = ∂U/∂τ in two... dx ' 9Nτ Θ exp x − 1 0 | {z } π4 /15 = ³ τ ´3 3π4 Nτ 5 Θ Eyal Buks , Thermodynamics and Statistical Physics (3.80) 109 Chapter 3 Bosonic and Fermionic Systems and therefore CV = 12π4 ³ τ ´3 ∂U = N ∂τ 5 Θ (3.81) Note that Eq (3.80) together with Eq (3 .75 ) yield 3 π2τ 4 U = V 2 15}3 v3 (3.82) Note the similarity between this result and Eq (3.51) for the EM case 3.3 Fermi Gas In this section we study... Due to distinctions 1 and 2, the sum over all modes is substituted by an integral according to ∞ ∞ ∞ X X X nx =0 ny =0 nz 3 → 4π 8 =0 n ZD dn n2 , (3.66) 0 where the factor of 3 replaces the factor of 2 we had in the EM case Moreover, the upper limit is nD instead of infinity, where nD is determined from the requirement 3 4π 8 n ZD dn n2 = 3N , (3. 67) 0 thus nD = µ Eyal Buks 6N π ¶1/3 Thermodynamics and. .. relation can be assumed to be linear ω n = vkn , (3 .71 ) where v is the sound velocity The wave vector kn is related to n = q n2 + n2 + n2 by x y z kn = πn , L (3 .72 ) where L = V 1/3 and V is the volume In this approximation one finds using the variable transformation x= β}vπn , L (3 .73 ) that 3π U = 2 n ZD dn n2 0 = 3V τ 4 2}3 v3 π2 exp x ZD dx 0 ³ }vπn L ´ β}vπn L −1 x3 exp x − 1 (3 .74 ) where ¡ ¢1/3 β}vπ... particle due to internal degrees of freedom [see Eq (2 .71 )] As is required by the Pauli exclusion principle, no more than one Fermion can occupy a given internal eigenstate and a given orbital 3.3.2 Partition Function of the Gas The grandcanonical partition function of the gas is given by Y ζn Zgc = (3.86) n Eyal Buks Thermodynamics and Statistical Physics 110 3.3 Fermi Gas As we have seen in chapter... x ³ τ ´3 Z D x3 dx U = 9Nτ Θ exp x − 1 0 x ³ τ ´3 Z D dx x2 ' 9Nτ Θ 0 ³ τ ´3 x3 D = 9Nτ Θ 3 = 3Nτ , (3 .78 ) and therefore CV = ∂U = 3N ∂τ (3 .79 ) Note that in this limit the average energy of each mode is τ and consequently U = 3Nτ This result demonstrates the equal partition theorem of classical statistical mechanics that will be discussed in the next chapter Low Temperature Limit In the low temperature... solids We start with a onedimensional example, and then generalize some of the results for the case of a 3D lattice 3.2.1 One Dimensional Example Consider the 1D lattice shown in Fig 3.2 below, which contains N ’atoms’ having mass m each that are attached to each other by springs having spring constant mω 2 The lattice spacing is a The atoms are allowed to move in one dimension along the array axis... freedom, its grandcanonical Fermionic partition function is given by [see Eq (2.33)] ζ n = 1 + λ exp (−βεn ) , (3.83) where λ = exp (βµ) = e−η , (3.84) is the fugacity and β = 1/τ Taking into account internal degrees of freedom the grandcanonical Fermionic partition function becomes Y ζn = (1 + λ exp (−βεn ) exp (−βEl )) , (3.85) l where {El } are the eigenenergies of a particle due to internal degrees . }v µ 6π 2 N V ¶ 1/3 , (3 .75 ) one has Eyal Buks Thermodynamics and Statistical Physics 108 3.2. Phonons in Solids x D = Θ τ , (3 .76 ) and U =9Nτ ³ τ Θ ´ 3 x D Z 0 dx x 3 exp x − 1 . (3 .77 ) As an example. (3. 67) thus n D = µ 6N π ¶ 1/3 . (3.68) Eyal Buks Thermodynamics and Statistical Physics 1 07 Chapter 3. Bosonic and Fermionic Systems Similarly to th e EM case, the average total energy is given by U = X n }ω n exp. Buks Thermodynamics and Statistical Physics 109 Chapter 3. Bosonic and Fermionic Systems and therefore C V = ∂U ∂τ = 12π 4 5 N ³ τ Θ ´ 3 . (3.81) Note that Eq. (3.80) together w ith Eq. (3 .75 )