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34 {\LARGE {\bf Materials Physics II}\\}
39 {\Large\bf Tutorial 3 - proposed solutions}
44 \section{Specific heat in the classical theory of the harmonic crystal -\\
45 The law of Dulong and Petit}
50 w&=&-\frac{1}{V}\frac{\partial}{\partial \beta}
51 ln \int d\Gamma \exp(-\beta H)
52 =-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
53 \frac{\partial}{\partial \beta} \int d\Gamma \exp(-\beta H)\nonumber\\
54 &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
55 \int d\Gamma \frac{\partial}{\partial \beta} \exp(-\beta H)\nonumber\\
56 &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
57 \int d\Gamma \exp(-\beta H) (-H) \qquad \textrm{ q.e.d.} \nonumber
59 \item Potential energy:
61 U=\frac{1}{2}\sum_{{\bf RR'}}\Phi({\bf r}({\bf R})-{\bf r}({\bf R'}))
62 =\frac{1}{2}\sum_{{\bf RR'}}
63 \Phi({\bf R}-{\bf R'}+{\bf u}({\bf R})-{\bf u}({\bf R'}))
66 $U_{\text{eq}}=\frac{1}{2}\sum_{{\bf R R'}} \Phi({\bf R}-{\bf R'})$:
69 \frac{1}{2}\sum_{{\bf RR'}}({\bf u}({\bf R})-{\bf u}({\bf R'}))
70 \nabla\Phi({\bf R}-{\bf R'})+
71 \frac{1}{4}\sum_{{\bf RR'}}
72 [({\bf u}({\bf R})-{\bf u}({\bf R'})) \nabla]^2
73 \Phi({\bf R}-{\bf R'}) + \mathcal{O}(u^3)
76 The coefficient of ${\bf u}({\bf R})$ is
77 $\sum_{\bf R'}\nabla\Phi({\bf R}-{\bf R'})$
78 which is minus the force excerted on atom ${\bf R}$
79 by all other atoms in equlibrium positions.
80 There is no net force on any atom in equlibrium.
81 The linear term is zero.\\\\
85 a\nabla \sum_u a_u \frac{\partial\Phi}{\partial r_u}=
86 \sum_v \frac{\partial \sum_u a_u
87 \frac{\partial\Phi}{\partial r_u}}{\partial r_v} a_v=
88 \sum_{uv}\frac{\partial}{\partial r_v} a_u
89 \frac{\partial \Phi}{\partial r_u} a_v=
90 \sum_{uv}a_u \frac{\partial^2\Phi}{\partial r_u \partial r_v} a_v$\\
92 U_{\text{harm}}=\frac{1}{4}\sum_{\stackrel{{\bf R R'}}{\mu,v=x,y,z}}
93 [u_{\mu}({\bf R})-u_{\mu}({\bf R'})]\Phi_{\mu v}({\bf R}-{\bf R'})
94 [u_v({\bf R})-u_v({\bf R'})],
95 \quad \Phi_{\mu v}({\bf r})=
96 \frac{\partial^2 \Phi({\bf r})}{\partial r_{\mu}\partial r_v}.
98 \item Change of variables:
100 {\bf u}({\bf R})=\beta^{-1/2}\bar{{\bf u}}({\bf R}), \qquad
101 {\bf P}({\bf R})=\beta^{-1/2}\bar{{\bf P}}({\bf R})
105 d{\bf u}({\bf R})=\beta^{-3/2}d\bar{{\bf u}}({\bf R}), \qquad
106 d{\bf P}({\bf R})=\beta^{-3/2}d\bar{{\bf P}}({\bf R}), \qquad
108 Kinetic energy contribution:
110 H_{\text{kin}}=\frac{{\bf P}({\bf R})^2}{2M}
114 \int d\Gamma \exp(-\beta H)=
115 \int d\Gamma \exp\left[-\beta\left(\sum \frac{{\bf P}({\bf R})^2}{2M}+
116 U_{\text{eq}} + U_{\text{harm}}\right)\right]
121 \section{Specific heat in the quantum theory of the harmonic crystal -\\
124 As found in exercise 1, the specific heat of a classical harmonic crystal
125 is not depending on temeprature.
126 However, as temperature drops below room temperature
127 the specific heat of all solids is decreasing as $T^3$ in insulators
128 and $AT+BT^3$ in metals.
129 This can be explained in a quantum theory of the specific heat of
130 a harmonic crystal, in which the energy density $w$ is given by
132 w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}.
135 \item Show that the energy density can be rewritten to read:
137 w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \sum_i \exp(-\beta E_i).
139 \item Evaluate the expression of the energy density.
141 The energy levels of a harmonic crystal of N ions
142 can be regarded as 3N independent oscillators,
143 whose frequencies are those of the 3N classical normal modes.
144 The contribution to the total energy of a particular normal mode
145 with angular frequency $\omega_s({\bf k})$
146 ($s$: branch, ${\bf k}$: wave vector) is given by
147 $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$ with the
148 excitation number $n_{{\bf k}s}$ being restricted to integers greater
150 The total energy is given by the sum over the energies of the individual
152 Use the totals formula of the geometric series to expcitly calculate
153 the sum of the exponential functions.
154 \item Separate the above result into a term vanishing as $T$ goes to zero and
155 a second term giving the energy of the zero-point vibrations of the
157 \item Write down an expression for the specific heat.
158 Consider a large crystal and thus replace the sum over the discrete
159 wave vectors with an integral.
160 \item Debye replaced all branches of the vibrational spectrum with three
161 branches, each of them obeying the dispersion relation
163 Additionally the integral is cut-off at a radius $k_{\text{D}}$
164 to have a total amount of N allowed wave vectors.
165 Determine $k_{\text{D}}$.
166 Evaluate the simplified integral and introduce the
167 Debye frequency $\omega_{\text{D}}=k_{\text{D}}c$
168 and the Debye temperature $\Theta_{\text{D}}$ which is given by
169 $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$.
170 Write down the resulting expression for the specific heat.