From: hackbard Date: Wed, 21 May 2008 07:37:00 +0000 (+0200) Subject: final version X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=commitdiff_plain;h=2fea80ba4b623fdfb421e2c727b671d742873155 final version --- diff --git a/solid_state_physics/tutorial/2_03.tex b/solid_state_physics/tutorial/2_03.tex index b4a3571..c33fe43 100644 --- a/solid_state_physics/tutorial/2_03.tex +++ b/solid_state_physics/tutorial/2_03.tex @@ -48,13 +48,15 @@ Thus, the specific heat at constant volume $V$ is given by \[ c_V = \frac{\partial w}{\partial T} \] -in which $w$ is the energy density of the system. +in which $w$ is the internal energy density of the system. +In the following the contribution to the specific heat due to the +degrees of freedom of the lattice ions is calculated. \section{Specific heat in the classical theory of the harmonic crystal -\\ The law of Dulong and Petit} In the classical theory of the harmonic crystal equilibrium properties -can no longer be evaluated by simply assuming that each ion sits quitly at +can no longer be evaluated by simply assuming that each ion sits quietly at its Bravais lattice site {\bf R}. From now on expectation values have to be claculated by integrating over all possible ionic configurations weighted by @@ -73,7 +75,7 @@ of the ions whose equlibrium sites are ${\bf R}$. \begin{enumerate} \item Show that the energy density can be rewritten to read: \[ - u=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \int d\Gamma \exp(-\beta H). + w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \int d\Gamma \exp(-\beta H). \] \item Show that the potential contribution to the energy in the harmonic approximation is given by @@ -111,13 +113,56 @@ $\Phi_{\mu v}({\bf r})= Which parts of the integral do not contribute to $w$ and why? \end{enumerate} - \section{Specific heat in the quantum theory of the harmonic crystal -\\ The Debye model} +As found in exercise 1, the specific heat of a classical harmonic crystal +is not depending on temeprature. +However, as temperature drops below room temperature +the specific heat of all solids is decreasing as $T^3$ in insulators +and $AT+BT^3$ in metals. +This can be explained in a quantum theory of the specific heat of +a harmonic crystal, in which the energy density $w$ is given by +\[ +w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}. +\] \begin{enumerate} - \item - \item + \item Show that the energy density can be rewritten to read: + \[ + w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \sum_i \exp(-\beta E_i). + \] + \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 + $\Theta_{\text{D}}=\hbar\omega_{\text{D}}$. + Write down the resulting expression for the specific heat. \end{enumerate} \end{document}