View source for Problem:Classical lattice vibrations in one dimension

This problem is a simple illustration of classical lattice vibrations, as calculated in the section [[Scattering from lattice vibrations#Lattice vibrations, classical treatment|Lattice vibrations, classical treatment]]<!--~\ref{sect:phonon_classic}--> on the [[Scattering from lattice vibrations]] page. The problem owes much to Squires 3.1<ref name="squires">G.L. Squires, ''Thermal Neutron Scattering'' (Cambridge University Press, 1978)</ref><!--\cite{squires}-->.

We imagine a one-dimensional chain of atoms of lattice constant \(d\), connected by springs with force constant \(K\). We assume all atoms are equal, ''i.e.'' they have identical mass, \(M\). The displacement of the atoms from equilibrium is denoted by \(u_j(t)\) and can move only along the chain direction (longitudinal vibrations). This system is illustrated in the [[Scattering from lattice vibrations#label-fig:springs|spring figure]] on the [[Scattering from lattice vibrations]] page<!--Fig.~\ref{fig:springs}-->.

=====Question 1=====
Prove that \(M \ddot{u}_j(t) = K (u_{j-1}(t) + u_{j+1}(t) - 2 u_j(t)) \) is the correct equation of motion for atom \(j\).

{{hidden begin|toggle=right|title=Hint|titlestyle=background:#ccccff}}
Consider the sum of forces exerted on atom \(j\) and remember that the harmonic oscillator force on the atom with mass \(M\) is in the opposite direction of the dislocation.
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{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}

For the classical harmonic oscillator the force on mass \(M\) is in the opposite direction of the dislocation \(u\), i.e. \(F=-ku\). For atom \(j\) at time \(t\) the classical force is
<equation id="eq:eqmot">\(  \begin{align}
F_j=M\ddot{u}_j(t)=K(u_{j-1}(t)-u_j(t))+K(u_{j+1}(t)-u_j(t))=K(u_{j-1}(t)+u_{j+1}(t)-2u_j(t))
\end{align}  \)</equation>
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=====Question 2=====
We assume a vibrational motion of the plane wave type: \(u_j(t) = A_q \exp(i (q j a - \omega_q t)) \), where \(A_q\) is an arbitrary amplitude. Prove that this trial function is a solution of the equation of motion with a vibration frequency of \(\omega_q^2 = 4K/M \sin^2(q d/2) \).

{{hidden begin|toggle=right|title=Hint|titlestyle=background:#ccccff}}
Differentiate the trial function twice and insert the result into the left-hand side of the equation of motion. Then insert the trial function in the right-hand side of the equation and rearrange to find  \(\omega_q\).
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{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
Suppose \(u_j(t)=A_q e^{i(djq-\omega_q t)}\) solves \(M\ddot{u}_j(t)=K(u_{j-1}(t)+u_{j+1}(t)-2u_j(t))\).
We first work on the lefthand side of <xr id="eq:eqmot"></xr>. The first order differential with respect to \(t\) is then 

\( 
\dot{u}(t)=-i\omega_q A_q e^{i(djq-\omega_q t)}
\)

and from the second order differential we find

\( 
M\ddot{u}(t)=-M\omega_q^2 A_q e^{i(djq-\omega_q t)}
\)

The righthand side of <xr id="eq:eqmot"></xr> can be rewritten

\(  \begin{align}
K(u_{j-1}(t)+u_{j+1}(t)-2u_j(t))
&= K A_q e^{-i\omega_q t}\left( e^{idjq}e^{-idq}+e^{idjq}e^{idq}-2e^{idjq}\right)\\
&= K A_q e^{-i\omega_q t}e^{idjq}\left( e^{-idq}+e^{idq}-2\right)\\
&= K A_q e^{i(djq-\omega_q t)}\left( 2\cos{dq}-2\right)\\
&= -2K A_q e^{i(djq-\omega_q t)}\left( 1-\cos{dq}\right)\\
&= -4K A_q e^{i(djq-i\omega_q t)}\sin^2\left(\dfrac{dq}{2}\right)\\
\end{align}  \)

Hence \(u_j(t)=A_q e^{i(djq-\omega_q t)}\) solves <xr id="eq:eqmot"></xr> if

<equation id="eq:freq">\( 
\omega_q^2= \dfrac{4K}{M}\sin^2(\dfrac{qd}{2})
\)</equation>
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=====Question 3=====
Argue that the potential energy can be written as \(V = K/2 \sum_j (\Re(u_{j+1}(t))-\Re(u_j(t)))^2\).

{{hidden begin|toggle=right|title=Hint|titlestyle=background:#ccccff}}
At a given time the displacement of atom \(j\) can be written as the real part of the \(u\)'s, i.e. the displacement of \(j\) is \(\Re(u_{j+1})+\Re(u_{j-1})-2\Re(u_{j})\)
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{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}

The potential energy in a classical spring is \(V=\frac{1}{2}ku^2\).
At a given time the displacement of atom \(j\) can be written as the real part of the \(u\)'s, i.e. the displacement of \(j\) is \(\Re(u_{j+1})+\Re(u_{j-1})-2\Re(u_{j})\). When we sum over all atoms some of the terms cancel out and we arrive at

\( 
V=\dfrac{K}{2}\displaystyle\sum_j\left(\Re(u_{j+1})-\Re(u_{j}) \right)^2
\)
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=====Question 4=====
Prove that the time average of the potential energy and the kinetic energy, \(T = M/2 \sum_j \Re(\dot{u}_j(t))^2\) are equal for any \(q\).

{{hidden begin|toggle=right|title=Hint|titlestyle=background:#ccccff}}
The time average of the kinetic energy is given by \( \langle T\rangle_t = \dfrac{1}{t_p}\displaystyle\int_0^{t_p}\text{d}t \ T \) where \(t_p\) is a full period of the oscillation.
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{{hidden begin|toggle=right|title=Solution|titlestyle=background:#ccccff}}
The kinetic energy of the \(N\) atoms a specific \(q\) is

\(  \begin{align}
T &= \dfrac{M}{2}\displaystyle\sum_j\left( \Re(\dot{u}_j)\right)^2\\
&= \dfrac{M}{2}\displaystyle\sum_j \Re\left( -i\omega_q A_q(\cos{(djq-\omega_qt)}+i\sin{(djq-\omega_qt)})\right)^2\\
&= \dfrac{M}{2}\displaystyle\sum_j \left( -\omega_q A_q(\sin{(djq-\omega_q t)})\right)^2\\
&= \dfrac{M}{2}N \left( -\omega_q A_q(\sin{(djq-\omega_q t)})\right)^2\\
\end{align}  \)

The time average of the kinetic energy is

\( 
\langle T\rangle_t = \dfrac{1}{t_p}\displaystyle\int_0^{t_p}\text{d}t \ T
=\dfrac{M}{2}N(\omega_q A_q)^2\dfrac{\omega_q}{2\pi}\displaystyle\int_0^{\frac{2\pi}{\omega_q}}\text{d}t \ \sin^2(djq-\omega_qt)
=\dfrac{M}{2}N(\omega_q A_q)^2\dfrac{\omega_q}{2\pi}\displaystyle\int_0^{\frac{2\pi}{\omega_q}}\text{d}t \ \sin^2(\omega_qt)
\)

where \(t_p\) is a full period of the oscillation and the zero-point shift (\(djq\)) in the argument of \(\sin^2\) doesn't change it's integrated value over a full period. The equation is valid for all \(q\), hence we suppress the \(q\) index. Substituting \(z=\omega_qt \Rightarrow \text{d}t=\frac{\text{d}z}{\omega}\) we have

\(  \begin{align}
\langle T\rangle_t 
&= \dfrac{M}{2}N(\omega A)^2\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi}\text{d}z \sin^2(z)\\
&= \dfrac{M}{2}N(\omega A)^2\dfrac{\pi}{2\pi}\\
&= \dfrac{M}{4}N(\omega A)^2
\end{align}  \)

The potential energy can be rewritten (with \(q\) index supressed)

\(  \begin{align}
V&=\dfrac{K}{2}A^2\displaystyle\sum_j\left[\Re(u_{j+1})-\Re(u_{j})\right]^2\\
&=\dfrac{K}{2}A^2\displaystyle\sum_j\left[\Re(u_{j+1}-u_{j})\right]^2\\
&=\dfrac{K}{2}A^2\displaystyle\sum_j\left[\Re(e^{i(qjd-\omega t)}(e^{iqd}-1))\right]^2\\
&=\dfrac{K}{2}A^2\displaystyle\sum_j\left[\cos(qjd-\omega t)(\cos{qd}-1)-\sin(qjd-\omega t)\sin{qd}\right]^2\\
&=\dfrac{K}{2}A^2\displaystyle\sum_j\left[ \cos^2(qjd-\omega t)(\cos{qd}-1)^2+\sin^2(qjd-\omega t)\sin^2{qd}-2\cos(qjd-\omega t)(\cos{qd}-1)\sin(qjd-\omega t)\sin{qd} \right]
\end{align}  \)

The time average of the potential energy is (using the subtitution \(z=\omega_qt \Rightarrow \text{d}t=\frac{\text{d}z}{\omega}\))

\(  \begin{align}
\langle V\rangle_t 
&=\dfrac{K}{2}A^2\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi}\text{d}z \displaystyle\sum_j \bigg[\underbrace{\cos^2(qjd-z)}_{1}\left(1+\cos^2(qd)-2\cos(qd)\right)+\underbrace{\sin^2(qjd-z)}_{1}\sin^2{qd} -2\underbrace{\cos(qjd-z)\sin(qjd-z)}_{=0}(\cos(qjd)-1)\sin{qd}\bigg]\\
&=\dfrac{K}{2}A^2\displaystyle\sum_j\dfrac{1}{2}\left(2-2\cos{qd}\right)\\
&=\dfrac{K}{2}A^2N(1-\cos{qd})\\
&=\dfrac{K}{2}A^2N\left(\sin^2\left(\dfrac{qd}{2}\right)\right)\\
&=\dfrac{M}{4}N(\omega A)^2
\end{align}  \)

where the frequency found in <xr id="eq:freq"></xr> has been inserted in the last line. It is now seen that

\( 
\langle V\rangle_t=\langle T\rangle_t
\)

for all \(q\).
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