Question 3.TC.1: Vibrations of a linear crystal lattice A very large number N......
Vibrations of a linear crystal lattice
A very large number N of movable identical point particles (N » 1) , each with mass m, are set in a straight chain with N + 1 identical massless springs, each with stiffness S, linking them to each other and the ends attached to two additional immovable particles. See Fig. 3 – 1. This chain will serve as a model of the vibration modes of a one – dimensional crystal. When the chain is set in motion, the longitudinal vibrations of the chain can be looked upon as a superposition of simple oscillations (called modes) each with its own characteristic mode frequency
(a) Write down the equation of motion of the nth particle.
(b) To attempt to solve the equation of motion of part (a) use the trial solution
where x_{n}(\omega) is the displacement of the nth particle from equilibrium, \omega the angular frequency of the vibration mode and A, k and a are constants; k and \omega are the wave numbers and mode frequencies respectively. For each k, there will be a corresponding frequency \omega . Find the dependence of \omega on k, the allowed values of k, and the maximum value of \omega . The chain’s vibration is thus a superposition of all these vibration modes. Useful formulas:
\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)\,\cos\alpha x=-\,\alpha\sin\alpha x\,,\;\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)\sin\alpha x\,=\,\alpha\mathrm{cos}\,\alpha x\,,\;\alpha=\, constant;
\sin(A+B)=\sin A\cos B+\cos A\sin B;
\cos\left(A+B\right)=\cos A\mathrm{cos}\,B-\sin A\mathrm{sin}\,B.
According to Planck the energy of a photon with a frequency of \omega is \hbar \omega, where \hbar is the Planck constant divided by 2π. Einstein made a leap from this by assuming that a given crystal vibration mode with frequency (\omega) also has this energy. Note that a vibration mode is not a particle, but a simple oscillation configuration of the entire chain. This vibration mode is analogous to the photon and is called a phonon. We will follow up the consequences of this idea in the rest of the problem. Suppose a crystal is made up of a very large (~10^{23} ) number of particles in a straight chain.
(c) For a given allowed \omega (or k) there may be no phonons; or there may be one; or two; or any number of phonons. Hence it makes sense to try to calculate the average energy ‹E( \omega) ›of a particular mode with a frequency (\omega). So ‹E ( \omega› is the average energy of a phonon with frequency \omega. Let P_p(\omega) ( \omega) represent the probability that there are p phonons with this frequency \omega. Then the required average is
\langle{ E}(\omega)\,\rangle\,=\,\frac{\displaystyle{\sum_{\rho=0}^{\infty}\rho}\hbar\omega~{ P}_{\mathrm{\scriptsize~p}}(\omega)}{\displaystyle{}\sum_{\rho=0}^{\infty}{ P}_{\mathrm{\scriptsize~p}}(\omega)}.
Although the phonons are discrete, the fact that there are so many of them (and the P_{p} becomes tiny for large P ) allows us to extend the sum to P = ∞ , with negligible error. Now the probability P_{p} is given by Boltzmann’s formula
P_{\mathrm{{p}}}(\omega)\ \propto\exp\biggl({\frac{- P\hbar\omega}{k_{\mathrm{{B}}}T}}\biggr),where k_{B} is Boltzmann’s constant and T is the absolute temperature of the crystal, assumed constant. The constant of proportionality does not depend on P. Calculate the average energy for phonons of frequency \omega. Possibly useful formula:
\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{f(x)}\,=\frac{\mathrm{d}f}{\mathrm{d}x}\mathrm{e}^{f(x)}.(d) We would like next to compute the total energy E_{T} of the crystal. In part (c) we found the average energy <E(w) > for the vibration mode \omega. To find E_{T} we must multiply <E(w) > by the number of modes of the crystal with frequency \omega and then sum up all these for the entire range from \omega = 0 to \omega_{max}. Take an interval Δk in the range of wave numbers. For very large N and for Δk much larger than the spacing between successive (allowed) k values, how many modes can be found in the interval Δk?
(e) To make use of the results of (a) and (b), approximate
\Delta k\ by\frac{\mathrm{d}k}{\mathrm{d}\omega}\mathrm{d}\omegaand replace any sum by an integral over \omega . (It is more convenient to use the variable \omega in place of k at this point. ) State the total number of modes of the crystal in this approximation. Also derive an expression E_{T} but do not evaluate it. The following integral may be useful:
\int_{0}^{1}{\frac{\mathrm{d}x}{\sqrt{1-x^{2}}}}={\frac{\pi}{2}}.(f) The molar heat capacity C_{V} of a crystal at constant volume is experimentally accessible:C_{V}\,=\,{\frac{\mathrm{d}E_{\mathrm{T}}}{\mathrm{d}T}} (T = absolute temperature). For the crystal under discussion determine the dependence of C_{V} on T for very large and very low temperatures (i. e. is it constant, linear or power dependent for an interval of the temperature?). Sketch a qualitative graph of C_{V}versus T, indicating the trends predicted for very low and very high T.
(b) Let x_{n}\;=\;A\;\;\sin\;n\,ka\;\;\cos(\omega\,t\,+\,\alpha)\,, which has a harmonic time dependence.
By analogy with the spring, the acceleration is \bar{x}_{n}=-\omega^{2}x_{n}.
Substitute into (a) :
-m A{\omega}^{2}\,\sin n k a\,=\,A\mathrm{S}[{\mathrm{sin}}(n+1)\,k a-2\,\sin n k a+\sin(n-1)\,k a]=-4\mathrm{S}A~\sin n k a~\sin^{2}\frac{1}{2}k a.
Hence {\boldsymbol{\omega}}^{2}={\frac{4 {S}}{m}}\mathrm{sin}^{2}{\frac{1}{2}}k{{a}}.
To determine the allowed values of k, use the boundary condition
\sin(N+1)k a\,=\,\sin\,k L\,=\,0.The allowed wave numbers are given by
k L\,=\,\pi,\;2\pi,\;3\pi,\;.\,.\,.\,,\;N\pi\,\,(N\;\mathrm{in~all})\,,and their corresponding frequencies can be computed from \omega=\omega_{0}\sin\frac{1}{2}k a\,,
in which \omega_{\mathrm{max}}=\omega_{0}=2\Bigl(\frac{ S}{m}\Bigr)^{\frac{1}{2}} is the maximum allowed frequency.
(\mathrm{c})\ \langle E(\omega)\,\rangle={\frac{\sum_{\rho=0}^{\infty}{ P} \hbar\omega P_{\mathrm{p}}(\omega)}{\sum_{\rho=0}^{\infty}\,P_{\mathrm{p}}(\omega)}}.
First method: (\mathrm{c})\ \langle E(\omega)\,\rangle={\frac{\sum_{n=0}^{\infty}{ P} \hbar\omega e^{\frac{n\hbar\omega}{k_{B}T}}}{\sum_{n=0}^{\infty}\,e^{\frac{\rho \hbar\omega}{k_{B}T}}}}.
=k_{B}T^{2}\ {\frac{\partial}{\partial T}}\ln\sum_{n=0}^{\infty}\mathrm{e}^{\frac{\rho\hbar\omega}{K_{B}T}}.The sum is a geometric series and is \left(1 -\mathrm{e}^{\frac{\hbar\omega}{k_{B}T}}\right)^{-1}\!\!.
We find \langle{ E}(\omega)\,\rangle\,=\,\frac{{\hbar}\omega}{\mathrm{e}^{\frac{\hbar\omega}{k_{B}T}}-1}.
Alternatively: denominator is a geometric series = (\,1-\mathrm{e}^{-\frac{\mathrm{\hbar\omega}}{k_{B}T}})^{-1}.
Numerator is k_{B}T^{2}\ {\frac{\mathrm{d}}{\mathrm{d}T}} (denominator) = \mathrm{e}^{-{\frac{\mathrm{\hbar\omega}}{k_{B}T}}}(1\;\mathrm{-\;e}^{-{\frac{\mathrm{\hbar\omega}}{k_{B}T}}})^{-2}
and result follows.
A non-calculus method:
Let D=1+\mathrm{e}^{-x}{+}\mathrm{e}^{-2x}{+}\mathrm{e}^{-3x}{+}{+}\mathrm{.}\ .\ . where x={\frac{\hbar\omega}{k_{\mathrm{B}}T}} This is a geometric
series and equals D={\frac{1}{1-\mathrm{e}^{-x}}}. Let N=\mathrm{e}^{-x}\!+\!2\mathrm{e}^{-2x}\!+\!3\mathrm{e}^{-3x}\!+\!\cdot\!\cdot\!\cdot The resuIt we want is \frac{N}{D} Observe
(D-1)\,\mathrm{e}^{-x}=\,\mathrm{e}^{-2x}+\mathrm{e}^{-3x}+\mathrm{e}^{-4x}+\mathrm{e}^{-5x}+\ldots.
(D-1)\,\mathrm{e}^{-2x}=\mathrm{e}^{-3x}+\mathrm{e}^{-4x}+\mathrm{e}^{-5x}+\ldots.
Hence { N}=(D-1)D\:\mathrm{or}\:\frac{{ N}}{D}=D-1=\frac{\mathrm{e}^{-x}}{1-\mathrm{e}^{-x}}=\frac{1}{\mathrm{e}^{x}-1}.
(d) From part (b), the allowed k values are{\frac{\pi}{L}}\,,\;{\frac{2\pi}{L}},\;\ \cdot\ \cdot\ \,\cdot\;,\;{\frac{N\pi}{L}}.
Hence the spacing between allowed k values is {\frac{\pi}{L}}\,, so there are {\frac{L}{\pi}}\Delta k
allowed modes in the wave-number interval Δk (assuming Δk»{\frac{\pi}{L}}\,,)
(e) Since the allowed k are {\frac{\pi}{L}}\,,\;\;\cdot\ \,\cdot\,\cdot\,,\;{\frac{N\pi}{L}},there are N modes.
Follow the problem:
\frac{\mathrm{d}\omega}{\mathrm{d}k}=\frac{1}{2}a\omega_{0}\,\cos\,\ \frac{1}{2}k a\,\,\mathrm{from~parts}\,\,({\bf a})\,\,\,\mathrm{and~(~b)}={\frac{1}{2}}a{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}\,,\;\omega_{\mathrm{max}}=\omega_{0}.
This second form is more convenient for integration.
The number of modes dn in the interval dω is
={\frac{L}{\pi}}\left({\frac{1}{2}}a\omega_{0}\cos{\frac{1}{2}}k a\right)^{-1}\mathrm{d}\omega
=\frac{L}{\pi}\ \frac{2}{a}\ \frac{1}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}\,\mathrm{d}\omega
=\frac{2(N+1)}{\pi}\,\frac{1}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}\,\mathrm{d}\omega.
Total number of modes =\int\!\mathrm{d}n=\int_{0}^{\omega_{max}}{\frac{2(N+1)}{\pi}}\,{\frac{\mathrm{d}\omega}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}}=N+1\approx N for large N.
Total crystal energy from (c) and dn of part (e) is given by
E_{T}=\frac{2N}{\pi}\int_{0}^{\omega_{max}}\frac{\hbar\omega}{\mathrm{e}^{\frac{\hbar\omega}{k_{B}T}}-1}\,\frac{\mathrm{d}\omega}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}.
(f) Observe first from the last formula that ET increases monotonically with temperature since (\mathrm{e}^{\frac{\hbar\omega}{k T}}-1)^{-1}.is increasing with T.
When T → 0, the term – 1 in the last result may be neglected in the denominator so E_{T}\approx_{T\rightarrow0}\,{\frac{2N}{\pi}}\int\hbar\omega\,\mathrm{e}^{{\frac{\hbar\omega}{k_{B}T}}}{\frac{1}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}}\mathrm{d}\omega
=\frac{2N}{\hbar\pi\omega_{\mathrm{max}}}(k_{B}T)^{2}{\int_{0}^{\infty}}\frac{x\mathrm{e}^{-x}}{\sqrt{1-\left(\frac{k_{B}T x}{\hbar\omega_{\mathrm{max}}}\right)^{2}}}{\mathrm{d}x},
which is quadratic in T (denominator in integral is effectively unity) hence C_{V} is linear in T near absolute zero.
Alternatively, if the summation is retained, we have
E_T=\frac{2 N}{\pi} \sum_\omega \frac{\hbar \omega}{ e ^{\frac{\omega}{k_B T}}-1} \frac{\Delta \omega}{\sqrt{\omega_{\max }^2-\omega^2}} .When T → 0 ,
E_{T}\approx\frac{2N}{\pi}\sum_{\omega}\hbar\omega\mathrm{e}^{\frac{\hbar\omega}{k_{B}T}}\,\frac{\Delta\omega}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}} ={\frac{2N}{\pi}}\,{\frac{(k_{B}T)^{2}}{\hbar\omega}}\,\sum_{y}\,\mathrm{e}^{-y}y\,\Delta y.
When T → ∞ , use \operatorname{e}^{x}\approx1+x in the denominator,
E_{T}\approx_{T\rightarrow\infty}\frac{2N}{\pi}\int_{0}^{\omega_{max}}\frac{\hbar\omega}{\frac{\hbar\omega}{k_{B}T}}\frac{1}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}\mathrm{d}\omega=\frac{2N}{\pi}k_{B}T\,\frac{\pi}{2},
which is linear; hence C_{V} →Nk_{B} = R, the universal gas constant. This is the Dulong – Petit rule.
Alternatively, if the summation IS retained, write denominator
as \mathrm{e}^{\frac{\hbar\omega}{k_{B}T}}-1\approx\frac{\hbar\omega}{k_{B}T} and when T→∞ ,
E_{T}\approx\frac{2N}{\pi}k_{B}T\sum_{\omega}\frac{\Delta\omega}{\sqrt{\omega_{\mathrm{max}}^{2}-\omega^{2}}}\,,
which is linear in T, so C_{V} is constant.
Sketch of C_{V} versus T:
