PHZ3400 Sound: Difference between revisions

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* We can reduce this problem to particles on springs
* We can reduce this problem to particles on springs
* Solve via eigenvalue analysis


* Two Modes
* Two Modes
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<math>m\ddot u_2 = ku_1 - 2ku_2\!</math>
<math>m\ddot u_2 = ku_1 - 2ku_2\!</math>


We can create a matrix representation by letting
Or
 
<math>m\ddot u_1 + 2ku_1 - ku_2 = 0\!</math>
 
<math>m\ddot u_2 - ku_1 + 2ku_2 = 0\!</math>
 
We can create a matrix representation for this system of equations by letting


<math>\overrightarrow u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}</math>
<math>\overrightarrow u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}</math>
So, we have
<math>k\begin{bmatrix} 2 & -1 \\ -1 & 2\end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end = 0{bmatrix}


==One dimensional mono-atomic chain==
==One dimensional mono-atomic chain==

Revision as of 23:16, 12 February 2009

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_2} Example

Consider an Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_2} molecule. Assume one is at rest while the other moves. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ma = -\frac{dV(r)}{dt}}

,where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} is potential energy and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} is radius. This can be rewritten as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\frac{d^2r}{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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(r)} in Taylor Series (note Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_0} is the radius with minimum V)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(r) = V(r_0) + V'(r_0)(r - r_0) + \frac{1}{2}K(r - r_0)^2}

,where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K = \frac{d^2V}{dt^2}} .Notice that the second term is the derivative of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_0} , which is a minimum, therefore the derivative is zero and this term can be ignored. Now we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(r) = V(r_0) + \frac{1}{2}K(r - r_0)^2}

Now let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = (r - r_0)} and we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(u) = V_0 + \frac{1}{2}Ku^2} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dV(u)}{du} = Ku}

...

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow -\omega^2u = -\frac{k}{m}u}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \omega = \sqrt{\frac{k}{m}}}

Two Atom Model

  • Corresponds to putting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_2} in a potential well
  • We can reduce this problem to particles on springs
  • 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\ddot u_1 = -ku_1 - k(u_1 - u_2)\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\ddot u_2 = -ku_2 - k(u_2 - u_1)\!}

,where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} is the spring constant, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} is mass, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} is displacement from the particles equilibrium position. We can rewrite these equations as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\ddot u_1 = -2ku_1 + ku_2\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\ddot u_2 = ku_1 - 2ku_2\!}

Or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\ddot u_1 + 2ku_1 - ku_2 = 0\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\ddot u_2 - ku_1 + 2ku_2 = 0\!}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overrightarrow u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}}

So, we have

<math>k\begin{bmatrix} 2 & -1 \\ -1 & 2\end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end = 0{bmatrix}

One dimensional mono-atomic chain

Sound waves - acoustic modes

One dimensional diatomic chain: optical modes