2_debye_model_solutions.md 3.68 KB
 Radoica Draškić committed Feb 09, 2020 1 2 3 4 5 6 ``````# Solutions for lecture 2 exercises ### Exercise 1: Debye model: concepts. 1, 2, 3. `````` T. van der Sar committed Feb 13, 2020 7 ``````See the lecture notes. `````` Radoica Draškić committed Feb 09, 2020 8 9 10 11 12 13 14 15 16 `````` 4. \$\$ g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega} = \frac{1}{v}\frac{dN}{dk}. \$\$ We assume that in \$d\$ dimensions there are \$d\$ polarizations. `````` Anton Akhmerov committed Apr 09, 2020 17 ``````For 1D we have that \$N = \frac{L}{2\pi}\int_{-k}^{k} dk\$, hence \$g(\omega) = \frac{L}{\pi v}\$. `````` Radoica Draškić committed Feb 09, 2020 18 19 20 21 22 23 24 25 26 `````` For 2D we have that \$N = 2\left(\frac{L}{2\pi}\right)^2\int d^2k = 2\left(\frac{L}{2\pi}\right)^2\int 2\pi kdk\$, hence \$g(\omega) = \frac{L^2\omega}{\pi v^2}\$. For 3D we have that \$N = 3\left(\frac{L}{2\pi}\right)^3\int d^3k = 3\left(\frac{L}{2\pi}\right)^3\int 4\pi kdk\$, hence \$g(\omega) = \frac{3L^3\omega^2}{2\pi^2v^3}\$. ### Exercise 2: Debye model in 2D. 1. `````` T. van der Sar committed Feb 13, 2020 27 ``````See the lecture notes. `````` Radoica Draškić committed Feb 09, 2020 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 `````` 2. \$\$ E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant}. \$\$ 3. High temperature implies \$\beta \rightarrow 0\$, hence \$E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + T \text{ independent constant}\$, and then \$C = \frac{k_BL^2\omega^2_D}{2\pi v^2} = 2Nk_B\$. We've used the value for \$\omega_D\$ calculated from \$2N = \int_{0}^{\omega_D}g(\omega)d\omega\$. 4. In the low temperature limit we have that \$\beta \rightarrow \infty\$, hence \$E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant} = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + T \text{ independent constant}\$. Finally \$C = \frac{6\zeta(3)k^3_BL^2}{\pi v^2\hbar^2}T^2\$. We used the fact that \$\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx = 2\zeta(3)\$ where \$\zeta\$ is the Riemann zeta function. ### Exercise 3: Different phonon modes. 1. \$\$ g(\omega) = \sum_{\text{polarizations}}\frac{dN}{dk}\frac{dk}{d\omega} = \left(\frac{L}{2\pi}\right)^3\sum_{\text{polarizations}}4\pi k^2\frac{dk}{d\omega} = \frac{L^3}{2\pi^2}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\omega^2 \$\$ \$\$ E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^3}{2\pi^2\hbar^3\beta^4}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx + T \text{ independent constant}. \$\$ 2. Note that we can get \$\omega_D\$ from \$3N = \int_{0}^{\omega_D}g(\omega)\$ so everything cancels as usual and we are left with the Dulong-Petit law \$C = 3Nk_B\$. 3. In the low temperature limit we have that \$C \sim \frac{2\pi^2k_B^4L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3\$. We used that \$\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}\$. ### Exercise 4: Anisotropic sound velocities. \$\$ E = 3\left(\frac{L}{2\pi}\right)^3\int d^3k\hbar\omega(\mathbf{k})\left(n_B(\beta\hbar\omega(\mathbf{k})) + \frac{1}{2}\right) = 3\left(\frac{L}{2\pi}\right)^3\frac{1}{v_xv_yv_z}\int d^3\kappa\frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1} + T \text{ independent part}, \$\$ where we used the substitutions \$\kappa_x = k_xv_x,\kappa_y = k_yv_y, \kappa_z = k_zv_z\$. Finally \$\$ E = \frac{3\hbar L^3}{2\pi^2}\frac{1}{v_xv_yv_z}\int_0^{\kappa_D} d\kappa\frac{\kappa^3}{e^{\beta\hbar\kappa} - 1} + T \text{ independent part} = \frac{3L^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1} + T \text{ independent part}, \$\$ `````` Radoica Draškić committed Feb 11, 2020 76 ``hence \$C = \frac{\partial E}{\partial T} = \frac{6k_B^4L^3T^3}{\pi^2\hbar^3}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1}\$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor \$1/v_xv_yv_z\$ instead of \$1/v^3\$.``