From: hackbard Date: Wed, 4 Jun 2008 08:03:32 +0000 (+0200) Subject: nearly finished X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=cd77e3b6e7e35332d6fedd7093cc9bdb86f53277;p=lectures%2Flatex.git nearly finished --- diff --git a/solid_state_physics/tutorial/2_03s.tex b/solid_state_physics/tutorial/2_03s.tex index fb0e83b..f1b880d 100644 --- a/solid_state_physics/tutorial/2_03s.tex +++ b/solid_state_physics/tutorial/2_03s.tex @@ -171,41 +171,77 @@ w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}. \right)\nonumber\\ &=&-\frac{1}{V}\frac{\partial}{\partial \beta} ln \prod_{{\bf k}s} \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)} - {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber + {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\ + &=&-\frac{1}{V}\frac{\partial}{\partial \beta} \sum_{{\bf k}s} ln + \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)} + {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\ + &=&-\frac{1}{V}\sum_{{\bf k}s} + \frac{1-\exp(-\beta\hbar\omega_s({\bf k}))} + {\exp(-\beta\hbar\omega_s({\bf k})/2)}\nonumber\\ + &&\times + \frac{(1-e^{-\beta\hbar\omega_s({\bf k})}) + e^{-\beta\hbar\omega_s({\bf k})/2}(-\hbar\omega_s({\bf k})/2)+ + e^{-\beta\hbar\omega_s({\bf k})/2} + e^{-\beta\hbar\omega_s({\bf k})}(-\hbar\omega_s({\bf k}))} + {(1-e^{-\beta\hbar\omega_s({\bf k})})^2}\nonumber\\ + &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k}) + \frac{e^{-\beta\hbar\omega_s({\bf k})}- + \frac{1}{2}(1-e^{-\beta\hbar\omega_s({\bf k})})} + {1-e^{-\beta\hbar\omega_s({\bf k})}}\nonumber\\ + &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k}) + \frac{\frac{1}{2}e^{-\beta\hbar\omega_s({\bf k})}-\frac{1}{2}} + {1-e^{-\beta\hbar\omega_s({\bf k})}} + =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2} + \frac{1+e^{\beta\hbar\omega_s({\bf k})}} + {e^{\beta\hbar\omega_s({\bf k})}-1}\nonumber\\ + &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2} + \frac{1+e^{\beta\hbar\omega_s({\bf k})}} + {e^{\beta\hbar\omega_s({\bf k})}-1} + =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2} + \frac{2+e^{\beta\hbar\omega_s({\bf k})}-1} + {e^{\beta\hbar\omega_s({\bf k})}-1}\nonumber\\ + &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k}) + (\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1} + +\frac{e^{\beta\hbar\omega_s({\bf k})}-1} + {2(e^{\beta\hbar\omega_s({\bf k})}-1)}) + =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k}) + (\underbrace{\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1}}_{n_s({\bf k})} + +\frac{1}{2})\nonumber \end{eqnarray} + $n_s({\bf k})$: Mean excitation number of the normal mode ${\bf k}s$ at + temperature $T$. - \item Evaluate the expression of the energy density. - {\bf Hint:} - The energy levels of a harmonic crystal of N ions - can be regarded as 3N independent oscillators, - whose frequencies are those of the 3N classical normal modes. - The contribution to the total energy of a particular normal mode - with angular frequency $\omega_s({\bf k})$ - ($s$: branch, ${\bf k}$: wave vector) is given by - $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$ with the - excitation number $n_{{\bf k}s}$ being restricted to integers greater - or equal zero. - The total energy is given by the sum over the energies of the individual - normal modes. - Use the totals formula of the geometric series to expcitly calculate - the sum of the exponential functions. - \item Separate the above result into a term vanishing as $T$ goes to zero and - a second term giving the energy of the zero-point vibrations of the - normal modes. - \item Write down an expression for the specific heat. - Consider a large crystal and thus replace the sum over the discrete - wave vectors with an integral. - \item Debye replaced all branches of the vibrational spectrum with three - branches, each of them obeying the dispersion relation - $w=ck$. - Additionally the integral is cut-off at a radius $k_{\text{D}}$ - to have a total amount of N allowed wave vectors. - Determine $k_{\text{D}}$. - Evaluate the simplified integral and introduce the - Debye frequency $\omega_{\text{D}}=k_{\text{D}}c$ - and the Debye temperature $\Theta_{\text{D}}$ which is given by - $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$. - Write down the resulting expression for the specific heat. + \item \[ + w=w_{\text{eq}}+ + \frac{1}{V}\sum_{{\bf k}s}\frac{1}{2}\hbar\omega_s({\bf k})+ + \frac{1}{V}\sum_{{\bf k}s} + \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1} + \] + \item \[ + c_{\text{V}}=\frac{1}{V}\sum_{{\bf k}s}\frac{\partial}{\partial T} + \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1} + \] + Large crystal: ($\lim_{V\to\infty}\frac{1}{V}\sum_{{\bf k}}F({\bf k}) + =\int\frac{d{\bf k}}{(2\pi)^3}F({\bf k})$) + \[ + \Rightarrow + c_{\text{V}}=\frac{\partial}{\partial T} + \sum_s\int\frac{d{\bf k}}{(2\pi)^3} + \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1} + \] + \item \begin{itemize} + \item Debye dispersion relation: $w=ck$ + \item Volume of $k$-space per wave vector:\\ + $(2\pi)^3/V \Rightarrow (2\pi)^3N/V=4\pi k_{\text{D}}^3/3 + \Rightarrow n=\frac{N}{V}=\frac{k_{\text{D}}^3}{6\pi^2}$ + \item Debye frequency: $\omega_{\text{D}}=k_{\text{D}}c$ + \item Debye temperature: + $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$ + \end{itemize} + Integral: + \[ + c_{\text{V}}=\ldots + \] \end{enumerate} \end{document}