PHZ3400 Sound

Vibrational modes of a one-dimensional chain

Harmonic approximation: inter-atomic forces as springs

• If we put energy into a crystal the atoms will begin to vibrate.
• We are able to put energy into the crystal in two ways: Mechanical and Thermal.
  • Mechanical energy are sound waves
  • Thermal energy can be used to measure resistivity.
• Periodicity helps to simplify the problem of lattice vibrations.
• One atom moves with the frequency $\omega _{0}={\sqrt {\frac {k}{m}}}$
  • If we have many atoms moving together than they effectively have a larger mass, therefore they will have a smaller frequency.
• Broken Symmetry
• Collective Phenomenon - behavior changes when in a large group

$H_{2}$ Example

Consider an $H_{2}$ molecule. Assume one is at rest while the other moves. $ma=-{\frac {dV(r)}{dt}}$

,where $V$ is potential energy and $r$ is radius. This can be rewritten as

$m{\frac {d^{2}r}{dt^{2}}}=-{\frac {dV(r)}{dt}}$

This equation cannot be solved via conventional methods, so we must somehow simplify it. Let us only worry about very small oscillations. This reduces our problem to a harmonic oscillator. Small oscillations can be described simply since it is parabolic at the minimum energy.

Now we expand $V(r)$ in Taylor Series (note $r_{0}$ is the radius with minimum V)

$V(r)=V(r_{0})+V'(r_{0})(r-r_{0})+{\frac {1}{2}}K(r-r_{0})^{2}$

,where $K={\frac {d^{2}V}{dt^{2}}}$.Notice that the second term is the derivative of $V$ at $r_{0}$, which is a minimum, therefore the derivative is zero and this term can be ignored. Now we have

$V(r)=V(r_{0})+{\frac {1}{2}}K(r-r_{0})^{2}$

Now let $u=(r-r_{0})$ and we have

$V(u)=V_{0}+{\frac {1}{2}}Ku^{2}$ and ${\frac {dV(u)}{du}}=Ku$

...

$\Rightarrow -\omega ^{2}u=-{\frac {k}{m}}u$

$\Rightarrow \omega ={\sqrt {\frac {k}{m}}}$

Two Atom Model

• Corresponds to putting $H_{2}$ in a potential well

• We can reduce this problem to particles on springs

• There is a spring with constant K holding the two atoms together. Also, we put them in a potential well such that we can imagine a spring connected each part individually to a wall, with spring constant Q

• Solve via eigenvalue analysis

• Two Modes
  • Both Particles move in concert
  • Both Particles move opposite to each other

• If one of these particles are displaced (ignore the effects of the second) it will act as a linear Harmonic Oscillator, so that it will act as a simple sine wave.
  • However, if you take into account the second particle, the problem will not be so simple since there will be a superposition of two since waves creating a beat pattern. If the displacement is described by one of the modes above then there will be a simple sine wave.

We can write the equations of motion for these two particles as

$m{\ddot {u}}_{1}=-Qu_{1}-Ku_{1}+Ku_{2}\!$

$m{\ddot {u}}_{2}=-Ku_{2}+Ku_{1}-Qu_{2}\!$

,where $K$ and $Q$ are the spring constants, $m$ is mass, and $u$ is displacement from the particles equilibrium position. This can be re-written as:

$m{\ddot {u}}_{1}=u_{1}(Q+K)+u_{2}(-K)\!$

$m{\ddot {u}}_{2}=u_{1}(-K)+u_{2}(Q+K)\!$

We can create a matrix representation for this system of equations by letting

${\overrightarrow {u}}={\begin{bmatrix}u_{1}\\u_{2}\end{bmatrix}}$

and choosing the interaction matrix, $\mathbb {M}$ to be

${\begin{bmatrix}Q+K&-K\\-K&Q+K\\\end{bmatrix}}$

Now we can rewrite our system of equations as

$m{\ddot {u}}=\mathbb {M} {\overrightarrow {u}}$

One dimensional mono-atomic chain

For this example, we imagine a chain of identical atoms, with mass $m$ connected by springs, with spring constant $K$. Since this is only in one dimension we assume the particles only move in a direction parallel to the chain.

There are two forces that contribute to the displacement of the $nth$ atom, $u_{n}$

1)The force from the spring on the left $K(u_{n}-u_{n-1})$

2)The force from the spring on the right $K(u_{n+1}-u_{n})$

Combining these we can write the total force as

$m{\ddot {u}}_{n}=-K[2u_{n}-u_{n-1}-u_{n+1}]\!$

...

$-m\omega ^{2}=-K[2u_{n}-u_{n-1}-u_{n+1}]\!$

$-m\omega ^{2}=-K[2-e^{iKa}-e^{-iKa}]\!$

$\omega (K)=2{\sqrt {\frac {K}{m}}}|sin(Ka)|\!$