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The Phonon Heat Capacity for 1D, 2D and 3D System Masaatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: September 29, 2018)

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The Phonon Heat Capacity for 1D, 2D and 3D System Masaatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: September 29, 2018) The phonon heat capacity for 1D, 2D and 3D system Masaatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: September 29, 2018) When I taught the Phys.411-511 (Statistical Thermodynamics, 2016) using the textbook of Huang, Introduction to Statistical Mechanics, I realized that the definition of the density of states for the statistical mechanics is different from that for the solid state physics even if we solve the same problem. In solid state physics, the velocity of the transverse mode is in general different from that of the longitudinal mode, while these velocities are the same in the statistical mechanics. In the 2D phonon, we use the approximation 2 1 1 2 2 2 for the solid state physics v vT vL and v vT vL for the statistical mechanics. 1. 2D phonon Show that the Debye model of a 2-dimensional crystal predicts that the low temperature heat capacity is proportional to T 2 . Solve the problem by answering the following questions. (a) Find the expression of the density of states. (b) Find the expression (without any approximation) for the internal energy. (c) Find the expression of the heat capacity with integral (you do not have to calculate the integral). (d) Find the expression of the entropy. ((Solution-1)) (a) Density of states There is one allowed states per (2 /L)2 in 2D k-space. In other words, there are 1 2 2 2( /) L , states per unit area of 2D k space, for each polarization and for each branch The density of states is defined by 2 dkxdk y 2kdk L kdk D()d 2 2 2 2 2( /) L 2( /) L 2 , using the linear dispersion relation, vk , L2 D() 2 2v , which is proportional to . ky dk k kx 2pêL (a) 2 2 Fig. Density of states for the 2D k-space. There is one state per area of the reciprocal L lattice plane. (b) Internal energy We calculate the heat capacity of 2D systems in the Debye approximation. The thermal energy is given by D L2 1 U dD()ℏ n() d ℏ , 2 ℏ 0 2 v e 1 for each polarization type (1 TA, 1 LA) The phonon velocity for the longitudinal mode and the transverse mode is given by vL and vT , respectively. Then the total energy is L2 1 1 D 1 U ( ) dℏ 2 2 2 ℏ 2 vL vT 0 e 1 We assume that . 2 1 1 2 2 2 v vL vT and vL vT , leading to v vL vT . Then we get ℏL2 D 2 U 2 d 2 ℏ 2v 0 e 1 ℏL2 k 3T 3 / T x2 2 B dx 2 ℏ3 x 2v 0 e 1 ℏ ℏ where x , kB D We note that D D L2 L2 2 N dD() d D , 2 2 0 0 2 v 4v or N k 2 2 4v2 B 2 , D L2 ℏ2 Thus we have 2 U T / T x2 4k T dx B x N 0 e 1 In the low temperature limit, we have x2 dx 2 )3( = 2.40411, )3( .1 20205 x 0 e 1 2 T U 4NkBT )3( In the high temperature limit 2 U T / T 4k T dxx 2k T , B B N 0 since x2 x . ex 1 (c) The heat capacity The heat capacity is 2 2 U T T C 24 )3( NkB .28 8492NkB . T The entropy is calculated as C S dT T T T 24 )3( Nk dT B 2 0 2 T 12 )3( NkB 2 T 14.4246NkB (( Solution-2)) This method is correct only if the phonon velocity is independent of the polarization ( vt = vl = v). The density of states is defined by dk dk 2kdk L2kdk D ()d 2 x y 2 2 , 2 2( )/2 L2 2( )/2 L2 2 using the linear dispersion relation, vk , L2 D () , 2 v2 which is proportional to . Note that the factor 2 of the density of states comes from two types of polarization (1 TA and 1 LA). Then we have D L2 L2 2 2N D ()d d D 2 2 2 0 0 v 2v or N k 2 2 4v2 B 2 D L2 ℏ2 or k 22 N B L2 4v2ℏ 2 The total energy is D L2 1 U dD ()ℏ n() d ℏ 2 2 ℏ 0 v e 1 or ℏL2 D 2 U d 2 ℏ v 0 e 1 ℏL2 k 3T 3 / T x2 B dx 2 ℏ3 x v 0 e 1 The ratio U / N can be evaluated as U 4v2ℏ 2 ℏL2 k 3T 3 / T x2 B dx 2 2 2 2 ℏ3 x N kB L v 0 e 1 2 T / T x2 4k T dx B x 0 e 1 In the low temperature limit, we have x2 dx 2 )3( = 2.40411, )3( .1 20205 x 0 e 1 2 T U 4NkBT )3( . In the high temperature limit 2 U T / T 4k T dxx 2k T B B N 0 The heat capacity is 2 2 U T T C 24 )3( NkB 28.8492NkB T The entropy is calculated as C S dT T T T 24 (3)Nk dT B 2 0 2 T 12 (3)NkB 2 T 14.4246NkB 2. Exact expression of the heat capacity for the 2D system We start with the internal energy given by ℏL2 D 2 U d 2 ℏ v 0 e 1 The heat capacity is dU d U 1 ℏ2L 2 D 3eℏ C d 2 2 ℏ 2 dT dT kBT v 0 (e 1) or 4 1 ℏ2L 2 k T / T x3 ex C B dx k T 2 v2 ℏ (ex 1)2 B 0 L2 / T x3 ex k 3T 2 dx ℏ 2 2 B x 2 v 0 (e 1) Then the ratio C 4v2ℏ 2 L2 / T x3 ex k 3T 2 dx 2 2 2 ℏ2 2 B x 2 N kB L v 0 (e 1) 2 T / T x3 ex 4k dx B x 2 0 (e 1) or 2 T / T x3 ex C 4Nk dx B x 2 0 (e )1 When N N A , 2 C T / T x3 ex 4dx x 2 R 0 (e )1 This is a universal function of only a ratio T / . We make a plot of this function by using the Mathematica. 1.0 C 2R 0.8 0.6 0.4 0.2 T 0.5 1.0 1.5 2.0 Fig. Plot of C /(2R) as a function of T / for the 2D phonon system. The red line denotes the exact expression, while the blue line denotes the low temperature limit. C /(2R) tends to the unity in the high temperature limit. 3. 3D phonon Show that the Debye model of a 3-dimensional crystal predicts that the low temperature heat capacity is proportional to T 3 . Solve the problem by answering the following questions. (a) Find the expression of the density of states. (b) Find the expression (without any approximation) for the internal energy. (c) Find the expression of the heat capacity with integral (you do not have to calculate the integral). (d) Find the expression of the entropy. ((Solution-1)) (a) Density of states There is one allowed states per (2 /L)3 in 3D k-space. In other words, there are 1 , 2( )/3 L3 states per unit area of 3D k space, for each polarization and for each branch The density of states is defined by dk dk dk 4k 2dk L3 k 2dk D()d x y z , 2( )/3 L3 2( )/3 L3 2 2 using the linear dispersion relation, vk , L3 2 D() , 2 2v 3 (b) Internal energy We calculate the heat capacity of 2D systems in the Debye approximation. The thermal energy is given by UdD () ℏ n () D L3 2 1 dℏ , 2 3 ℏ 0 2 v e 1 ℏL3 1D 1 d 3 2 3 ℏ 2 v0 e 1 for each polarization type (2 TA, 1 LA) The phonon velocity for the longitudinal mode and the transverse mode is given by vL and vT , respectively. Then the total energy combining three branches is ℏL3 1 2D 1 U( ) d 3 2 3 3 ℏ 2 vL v T 0 e 1 We assume that . 3 1 2 3 3 3 33 v vL v T v If vL vT , we have v vL vT . Under this condition, we get ℏL3 D 3 U 3 d 32 ℏ 2 v 0 e 1 ℏL3 k 4T 4 / T x3 3 B dx 32 ℏ4 x 2 v 0 e 1 ℏ ℏ where x , kB D We note that D D L 23 L3 3 N dD() d D , for each branch, 32 32 0 0 2 v 6 v or 1/3 N k 3 ℏv6 2 N 3 6 v32 B 3 , D 3 ℏ3 L kB V or ℏ2 1/3 ℏ v6 N v 2 1/3 6 n (Debye temperature) kB V k B N with the number density, n . Then we have V 3 U T / T x3 9k T dx B x N 0 e 1 In the low temperature limit, we have x3 4 dx = 6.49394, x 0 e 1 15 and 3 3 4 T U NkBT 5 In the high temperature limit U 3NkBT since x3 x2 .
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