User:Guy vandegrift/btvs/Effort 4

Transverse modes Longitudinal modes in a lattice of springs and masses near the ground state form a quantized phonon field that interacts with the atom via a weak linear coupling spring at the center.

Place springs at ℓ = 0 ±Δℓ, ±2Δℓ, ... ±L, so that the length of the string is 2L. Often we shall count from the left, with the center at (N+1)Δℓ, so that there are 2N-1 little mass that make up the string, plus one "atomic" mass. Counting from the left:

{\begin{aligned}M_{S}{\ddot {y}}_{1}=-\omega ^{2}M_{S}\cdot y_{1}&=-(T/\delta \ell )\cdot y_{1}-(T/\delta \ell )\cdot (y_{1}-y_{2})\\&=-K_{S}\cdot y_{1}-K_{S}\cdot (y_{1}-y_{2})\\&=-2K_{S}\cdot y_{1}+K_{S}\cdot y_{2}\end{aligned}} where $K_{S}=T/\delta \ell$ . For atoms not touching the wall or the coupling spring connecting to the "atom":

$M_{S}{\ddot {y}}_{n}=-\omega ^{2}M_{S}\cdot y_{n}=K_{S}\cdot y_{n-1}-2K_{S}\cdot y_{n}+K_{S}\cdot y_{n+1}$ The N-th mass is attached to the atom, and obeys:

$M_{S}{\ddot {y}}_{N}=-\omega ^{2}M_{S}\cdot y_{N}=K_{S}\cdot y_{N-1}-2K_{S}\cdot y_{N}+K_{S}\cdot y_{N+1}+K_{C}\cdot (Y_{A}-y_{N})$ and the atom obeys:

$M_{S}{\ddot {y}}_{n}=-\omega ^{2}M_{A}\cdot y_{N}=-K_{A}\cdot Y_{A}+K_{C}\cdot (y_{N}-Y_{A})$ rearranging for convenience later on

{\begin{aligned}-\omega ^{2}M_{S}\cdot y_{1}&=-2K_{S}\cdot y_{1}+K_{S}\cdot y_{2}\\-\omega ^{2}M_{S}\cdot y_{2}&=K_{S}\cdot y_{1}-2K_{S}\cdot y_{2}+K_{S}\cdot y_{3}+K_{C}\cdot (Y_{A}-y_{2})\\-\omega ^{2}M_{A}\cdot Y_{A}&=-K_{A}\cdot Y_{A}+K_{C}\cdot (y_{2}-Y_{A})\end{aligned}} This is all we need.

Consider only symmetric modes

In this simple case Since we are only considering modes that are symmetrical with respect to the atom, $y_{3}=y_{1}$ {\begin{aligned}-\omega ^{2}M_{S}\cdot y_{1}&=-2K_{S}\cdot y_{1}\qquad \qquad +K_{S}\cdot y_{2}\\-\omega ^{2}M_{S}\cdot y_{2}&=\;\;2K_{S}\cdot y_{1}-(2K_{S}+K_{C})\cdot y_{2}\quad \qquad +K_{C}\cdot Y_{A}\\-\omega ^{2}M_{A}\cdot Y_{A}&=\quad \qquad \qquad \qquad \qquad K_{C}\cdot y_{2}-\;(K_{A}+K_{C})\cdot Y_{A}\\&...\\-\omega ^{2}M_{S}\cdot y_{3}&=-2K_{S}\cdot y_{3}\qquad \qquad +K_{S}\cdot y_{2}\\\end{aligned}} We don't need the third equation for modes symmetric about the atom.

writing in matrix form

$\left[{\begin{array}{c|c}\omega ^{2}M_{S}-2K_{S}&K_{S}&0\\\hline 2K_{S}&\omega ^{2}M_{S}-2K_{S}-K_{C}&K_{C}\\\hline 0&K_{C}&\omega ^{2}M_{A}-K_{A}-K_{C}\end{array}}\right]\cdot \left[{\begin{array}{c|c}y_{1}\\\hline y_{2}\\\hline y_{A}\end{array}}\right]=0$ Divide the first two rows by MS and the third by MA:

$\left[{\begin{array}{c|c}\omega ^{2}-2K_{S}/M_{S}&K_{S}/M_{S}&0\\\hline 2K_{S}/M_{S}&\omega ^{2}-2K_{S}/M_{S}-K_{C}/M_{S}&K_{C}/M_{S}\\\hline 0&K_{C}/M_{A}&\omega ^{2}-K_{A}/M_{A}-K_{C}/M_{A}\end{array}}\right]\cdot \left[{\begin{array}{c|c}y_{1}\\\hline y_{2}\\\hline y_{A}\end{array}}\right]=0$ Define the fundamental frequencies and beta

$\omega _{S}^{2}=K_{S}/M_{S}$ ,    $\omega _{A}^{2}=K_{A}/M_{A}$ ,    $\gamma ^{2}=K_{C}/M_{A}$ ,    $\beta \gamma ^{2}=K_{C}/M_{S}$ ,    where $\beta =M_{A}/M_{S}$ , to obtain:

$\left[{\begin{array}{c|c}\omega ^{2}-2\omega _{S}^{2}&\omega _{S}^{2}&0\\\hline 2\omega _{S}^{2}&\omega ^{2}-2\omega _{S}^{2}-\beta \gamma ^{2}&\beta \gamma ^{2}\\\hline 0&\gamma ^{2}&\omega ^{2}-\omega _{A}^{2}-\gamma ^{2}\end{array}}\right]\cdot \left[{\begin{array}{c|c}y_{1}\\\hline y_{2}\\\hline y_{A}\end{array}}\right]=0$ Matrix equation for normal modes

Let: $x=\omega ^{2}$ , $\sigma =\omega _{S}^{2}$ , $\alpha =\omega _{A}^{2}$ , and $\epsilon =\gamma ^{2}<<1$ $\left[{\begin{array}{c|c}x-2\sigma &\sigma &0\\\hline 2\sigma &x-2\sigma -\beta \epsilon &\beta \epsilon \\\hline 0&\epsilon &x-\alpha -\epsilon \end{array}}\right]\cdot \left[{\begin{array}{c|c}y_{1}\\\hline y_{2}\\\hline y_{A}\end{array}}\right]=0$ Finding the determinant of a 3x3 matrix

click to view a great deal of albebra

$\det \left[{\begin{array}{c|c}x-2\sigma &\sigma &0\\\hline 2\sigma &x-2\sigma -\beta \epsilon &\beta \epsilon \\\hline 0&\epsilon &x-\alpha -\epsilon \end{array}}\right]=A-B-C$ This equals A-B-C, where:

$A=\left(x-2\sigma \right)\cdot \left(x-2\sigma -\beta \epsilon \right)\cdot \left(x-\alpha -\epsilon \right)$ $B=2\sigma ^{2}x-2\sigma ^{2}\alpha -2\epsilon \sigma ^{2}$ $C=\epsilon ^{2}\beta x-2\epsilon ^{2}\beta \sigma$ $A=\left(x^{2}-2\sigma x-\epsilon \beta x-2\sigma x+4\sigma ^{2}+2\epsilon \sigma \beta \right)\cdot \left(x-\alpha -\epsilon \right)$ $A=$ $x\left(x^{2}-2\sigma x-\epsilon \beta x-2\sigma x+4\sigma ^{2}+2\epsilon \sigma \beta \right)$ $-\alpha \left(x^{2}-2\sigma x-\epsilon \beta x-2\sigma x+4\sigma ^{2}+2\epsilon \sigma \beta \right)$ $-\epsilon \left(x^{2}-2\sigma x-\epsilon \beta x-2\sigma x+4\sigma ^{2}+2\epsilon \sigma \beta \right)$ Distribute the factor in front for each line

$A=$ $x^{3}-2\sigma x^{2}-\epsilon \beta x^{2}-2\sigma x^{2}+4\sigma ^{2}x+2\epsilon \sigma \beta x$ $-\alpha x^{2}+2\alpha \sigma x+\alpha \epsilon \beta x+2\alpha \sigma x-4\alpha \sigma ^{2}-2\epsilon \alpha \sigma \beta$ $-\epsilon x^{2}+2\epsilon \sigma x+\epsilon ^{2}\beta x+2\epsilon \sigma x-4\epsilon \sigma ^{2}-2\epsilon ^{2}\sigma \beta$ Now we group each line by factors of x

$A=$ $x^{3}+\left(-2\sigma x^{2}-\epsilon \beta x^{2}-2\sigma x^{2}\right)+\left(4\sigma ^{2}x+2\epsilon \sigma \beta x\right)$ $-\alpha x^{2}+2\alpha \sigma x+\alpha \epsilon \beta x+2\alpha \sigma x-4\alpha \sigma ^{2}-2\epsilon \alpha \sigma \beta$ $+\left(-\epsilon x^{2}\right)+\left(2\epsilon \sigma x+\epsilon ^{2}\beta x+2\epsilon \sigma x\right)+\left(-4\epsilon \sigma ^{2}-2\epsilon ^{2}\sigma \beta \right)$ Now factor, for each of the three lines:

$A=$ $x^{3}+\left(-2\sigma -\epsilon \beta -2\sigma \right)x^{2}+\left(4\sigma ^{2}+2\epsilon \sigma \beta \right)x$ $-\alpha x^{2}+\left(2\alpha \sigma +\alpha \epsilon \beta +2\alpha \sigma \right)x+\left(-4\alpha \sigma ^{2}-2\epsilon \alpha \sigma \beta \right)$ $-\epsilon x^{2}+\left(2\epsilon \sigma +\epsilon ^{2}\beta +2\epsilon \sigma \right)x+\left(-4\epsilon \sigma ^{2}-2\epsilon ^{2}\sigma \beta \right)$ Regroup so that each line corresponds to a different factor of x:

$A=x^{3}$ $+\left(-2\sigma -\epsilon \beta -2\sigma \right)x^{2}-\alpha x^{2}-\epsilon x^{2}$ $+\left(4\sigma ^{2}+2\epsilon \sigma \beta \right)x+\left(2\alpha \sigma +\alpha \epsilon \beta +2\alpha \sigma \right)x+\left(2\epsilon \sigma +\epsilon ^{2}\beta +2\epsilon \sigma \right)x$ $+\left(-4\alpha \sigma ^{2}-2\epsilon \alpha \sigma \beta \right)+\left(-4\epsilon \sigma ^{2}-2\epsilon ^{2}\sigma \beta \right)$ $A=x^{3}$ $+\left(-2\sigma -\epsilon \beta -2\sigma -\alpha -\epsilon \right)x^{2}$ $+\left(4\sigma ^{2}+2\epsilon \sigma \beta +2\alpha \sigma +\alpha \epsilon \beta +2\alpha \sigma +2\epsilon \sigma +\epsilon ^{2}\beta +2\epsilon \sigma \right)x$ $+\left(-4\alpha \sigma ^{2}-2\epsilon \alpha \sigma \beta -4\epsilon \sigma ^{2}-2\epsilon ^{2}\sigma \beta \right)$ Regroup and put the ε terms last since they are small

$A=x^{3}$ $+\left(-4\sigma -\alpha -\epsilon \beta -\epsilon \right)x^{2}$ $+\left(4\sigma ^{2}+4\alpha \sigma +\epsilon \alpha \beta +2\epsilon \sigma \beta +4\epsilon \sigma +\epsilon ^{2}\beta \right)x$ $+\left(-4\alpha \sigma ^{2}-2\epsilon \alpha \sigma \beta -4\epsilon \sigma ^{2}-2\epsilon ^{2}\sigma \beta \right)$ Now recall that the determinant is A-B-C, where

$-B=-2\sigma ^{2}x+2\sigma ^{2}\alpha +2\epsilon \sigma ^{2}$ $-C=-\epsilon ^{2}\beta x+2\epsilon ^{2}\beta \sigma$ When we subtract B and C from A. This modifies the x term and the constant term.

x1 term: sigma term in B reduces the four to a 2 in front, while the epsilon squared term in C cancels a term near the end.
x0 term in B: the alpha term and the epsilon term change both 4s to twos in the constant term below:
x0 term in C: the epsilon term squared term cancels the last term.

$A-B-C=x^{3}$ $+\left(-4\sigma -\alpha -\epsilon \beta -\epsilon \right)x^{2}$ $+\left(2\sigma ^{2}+4\alpha \sigma +\epsilon \alpha \beta +2\epsilon \sigma \beta +4\epsilon \sigma \right)x$ $+\left(-2\alpha \sigma ^{2}-2\epsilon \alpha \sigma \beta -2\epsilon \sigma ^{2}\right)$ $0=x^{3}$ $-\left(4\sigma +\alpha +\epsilon \beta +\epsilon \right)x^{2}$ $+\left(2\sigma ^{2}+4\alpha \sigma +\epsilon \alpha \beta +2\epsilon \sigma \beta +4\epsilon \sigma \right)x$ $-\left(2\alpha \sigma ^{2}+2\epsilon \alpha \sigma \beta +2\epsilon \sigma ^{2}\right)$ Solving the cubic equation for the uncoupled case

The exact uncoupled equations (no approximations made) are:

$0=x^{3}$ $-\left(4\sigma +\alpha +\epsilon \beta +\epsilon \right)x^{2}$ $+\left(2\sigma ^{2}+4\alpha \sigma +\epsilon \alpha \beta +2\epsilon \sigma \beta +4\epsilon \sigma \right)x$ $-\left(2\alpha \sigma ^{2}+2\epsilon \alpha \sigma \beta +2\epsilon \sigma ^{2}\right)$ If the coupling (epsilon) equals zero:

$0=x^{3}-\left(4\sigma +\alpha \right)x^{2}+\left(2\sigma ^{2}+4\alpha \sigma \right)x-2\alpha \sigma ^{2}$ From physical grounds we know one answer is x=α. It can be shown that:

$x^{3}-\left(4\sigma +\alpha \right)x^{2}+\left(2\sigma ^{2}+4\alpha \sigma \right)x-2\alpha \sigma ^{2}=(x-\alpha )\left(x^{2}-4\sigma x+2\sigma ^{2}\right)$ Two ways to be degenerate (uncoupled case only)

If there is no coupling, the atom's frequency will match one of the two even string mode frequencies if:

proof

The other two roots for zero coupling are solutions to: $x^{2}-4\sigma x+2\sigma ^{2}=0$ $x=\left(2\pm {\sqrt {2}}\right)\sigma$ check

$\left(2\sigma -{\sqrt {2}}\sigma -x\right)\left(2\sigma +{\sqrt {2}}\sigma -x\right)=4\sigma ^{2}-2\sigma ^{2}-4\sigma x+x^{2}=2\sigma ^{2}-4\sigma x+x^{2}$ $\alpha =\left(2\pm {\sqrt {2}}\right)\sigma$ The high beta approximation

Simply by making $\beta =M_{A}/M_{S}$ very large.

$0=x^{3}$ $-\left(4\sigma +\alpha +\epsilon \beta \right)x^{2}$ $+\left(2\sigma ^{2}+4\alpha \sigma +\epsilon \alpha \beta +2\epsilon \sigma \beta \right)x$ $-\left(2\alpha \sigma ^{2}+2\epsilon \alpha \sigma \beta \right)$ proof

Above was obtained by dropping terms with epsilon but not beta.

$0=x^{3}$ $-\left(4\sigma +\alpha +\epsilon \beta +\epsilon \right)x^{2}$ $+\left(2\sigma ^{2}+4\alpha \sigma +\epsilon \alpha \beta +2\epsilon \sigma \beta +4\epsilon \sigma \right)x$ $-\left(2\alpha \sigma ^{2}+2\epsilon \alpha \sigma \beta +2\epsilon \sigma ^{2}\right)$ Now we see what happens if we go to the original matrix and make the same simplification (should get same answer).

Exact

$\left[{\begin{array}{c|c}x-2\sigma &\sigma &0\\\hline 2\sigma &x-2\sigma -\beta \epsilon &\beta \epsilon \\\hline 0&\epsilon &x-\alpha -\epsilon \end{array}}\right]\cdot \left[{\begin{array}{c|c}y_{1}\\\hline y_{2}\\\hline y_{A}\end{array}}\right]=0$ Remove small elements

$\left[{\begin{array}{c|c}x-2\sigma &\sigma &0\\\hline 2\sigma &x-2\sigma -\beta \epsilon &\approx 0\\\hline 0&\approx 0&x-\alpha -0\end{array}}\right]\cdot \left[{\begin{array}{c|c}y_{1}\\\hline y_{2}\\\hline y_{A}\end{array}}\right]=0$ Appendix

Failed efforts

Special:Permalink/1600364 is a version that used deltas instead of epsilons for the coupling coefficient. Subpages