# User:Guy vandegrift/btvs/Effort 3

## General case

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:

{\displaystyle {\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 ${\displaystyle K_{S}=T/\Delta \ell }$. For atoms not touching the wall or the coupling spring connecting to the "atom":

${\displaystyle M_{S}{\ddot {y}}_{n}=-\omega ^{2}M_{S}\cdot y_{n}=K_{S}\cdot y_{n-1}-2K_{S}\cdot y_{n}+K_{S}\cdot y_{n+1}}$

The N-th mass is attached to the atom, and obeys:

${\displaystyle M_{S}{\ddot {y}}_{N}=-\omega ^{2}M_{S}\cdot y_{N}=K_{S}\cdot y_{N-1}-2K_{S}\cdot y_{N}+K_{S}\cdot y_{N+1}+K_{C}\cdot (y_{A}-y_{N})}$

and the atom obeys:

${\displaystyle M_{S}{\ddot {y}}_{n}=-\omega ^{2}M_{A}\cdot y_{N}=-K_{A}\cdot y_{A}+K_{C}\cdot (y_{N}-y_{A})}$

## Simple system

One atom. Two masses plus one massless object on the string. N=2 and 2N-1=3 masses on the string. There are 4 masses and we need 4 equations:

{\displaystyle {\begin{aligned}-\omega ^{2}M_{S}\cdot y_{1}&=-2K_{S}\cdot y_{1}\qquad \qquad +K_{S}\cdot y_{2}\\-\omega ^{2}\cdot 0\cdot y_{2}&=\;\;\;\,K_{S}\cdot y_{1}-(2K_{S}+K_{C})\cdot y_{2}+K_{S}y_{3}\quad +K_{C}\cdot y_{A}\\-\omega ^{2}M_{A}\cdot y_{A}&=\quad \qquad \qquad \qquad \qquad K_{C}\cdot y_{2}-\qquad \qquad (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.

## Establish two nearly degenerate modes

Use the symmetry of modes that are symmetric about the atom, ${\displaystyle y_{3}=y_{1}}$ to obtain a matrix equation:

a bit of algebra

{\displaystyle {\begin{aligned}-\omega ^{2}M_{S}\cdot y_{1}&=-2K_{S}\cdot y_{1}\qquad \qquad +K_{S}\cdot y_{2}\\-\omega ^{2}\cdot 0\cdot y_{2}&=\;\;\;\,K_{S}\cdot y_{1}-(2K_{S}+K_{C})\cdot y_{2}\quad +K_{C}\cdot y_{A}\\-\omega ^{2}M_{A}\cdot y_{A}&=\quad \qquad \qquad \qquad \qquad K_{C}\cdot y_{2}-\qquad (K_{A}+K_{C})\cdot y_{A}\\\end{aligned}}}

${\displaystyle {\begin{pmatrix}\omega ^{2}M_{S}-2K_{S}&K_{S}&0\\K_{S}&-2K_{S}-K_{C}&K_{C}\\0&K_{C}&\omega ^{2}M_{A}-K_{A}-K_{C}\end{pmatrix}}\cdot {\begin{pmatrix}y_{1}\\y_{2}\\y_{A}\end{pmatrix}}=0}$

Divide rows 1 and 2 by MS and row 3 by MA:

${\displaystyle {\begin{pmatrix}\omega ^{2}-2{\frac {K_{S}}{M_{S}}}&{\frac {K_{S}}{M_{S}}}&0\\{\frac {K_{S}}{M_{S}}}&-2{\frac {K_{S}}{M_{S}}}-{\frac {K_{C}}{M_{S}}}&{\frac {K_{C}}{M_{S}}}\\0&{\frac {K_{C}}{M_{A}}}&\omega ^{2}-{\frac {K_{A}}{M_{A}}}-{\frac {K_{C}}{M_{A}}}\end{pmatrix}}\cdot {\begin{pmatrix}y_{1}\\y_{2}\\y_{A}\end{pmatrix}}=0}$

${\displaystyle \omega _{S}^{2}=K_{S}/M_{S}}$,    ${\displaystyle \omega _{A}^{2}=K_{A}/M_{A}}$,    ${\displaystyle \gamma ^{2}=K_{C}/M_{A}}$,    ${\displaystyle \beta \gamma ^{2}=K_{C}/M_{S}}$,    where ${\displaystyle \beta =M_{A}/M_{S}>>1}$, to obtain:

The result is:

${\displaystyle {\begin{pmatrix}\omega ^{2}-2\omega _{S}^{2}&\omega _{S}^{2}&0\\\omega _{S}^{2}&-2\omega _{S}^{2}-\beta \gamma ^{2}&\beta \gamma ^{2}\\0&\gamma ^{2}&\omega ^{2}-\omega _{A}^{2}-\gamma ^{2}\end{pmatrix}}\cdot {\begin{pmatrix}y_{1}\\y_{2}\\y_{A}\end{pmatrix}}}$

where ${\displaystyle \omega _{S}^{2}=K_{S}/M_{S}}$,    ${\displaystyle \omega _{A}^{2}=K_{A}/M_{A}}$,    ${\displaystyle \gamma ^{2}=K_{C}/M_{A}<<\omega _{A}^{2}}$,    ${\displaystyle \beta \gamma ^{2}=K_{C}/M_{S}}$,    and ${\displaystyle \beta =M_{A}/M_{S}>>1}$

### Finding the normal modes

Let: ${\displaystyle x=\omega ^{2}}$, ${\displaystyle \sigma =\omega _{S}^{2}}$, ${\displaystyle \alpha =\omega _{A}^{2}}$, and ${\displaystyle \epsilon =\gamma ^{2}<<1}$

${\displaystyle {\begin{pmatrix}x-2\sigma &\sigma &0\\\sigma &-2\sigma -\beta \epsilon &\beta \epsilon \\0&\epsilon &x-\alpha -\epsilon \end{pmatrix}}\cdot {\begin{pmatrix}y_{1}\\y_{2}\\y_{A}\end{pmatrix}}=0}$

The determinant equals A-B-C, where:

${\displaystyle A=\left(x-2\sigma \right)\cdot \left(-2\sigma -\beta \epsilon \right)\cdot \left(x-\alpha -\epsilon \right)}$

${\displaystyle B=\sigma ^{2}x-\sigma ^{2}\alpha -\epsilon \sigma ^{2}}$ correct

${\displaystyle C=\epsilon ^{2}\beta x-2\epsilon ^{2}\beta \sigma }$

### Calculate term A in determnant

algebra

${\displaystyle A=\left(-2\sigma x-\epsilon \beta x+4\sigma ^{2}+2\epsilon \sigma \beta \right)\cdot \left(x-\alpha -\epsilon \right)}$

A=

${\displaystyle -2\sigma x^{2}-\epsilon \beta x^{2}+4\sigma ^{2}x+2\epsilon \beta x\sigma }$

${\displaystyle +2\sigma x+\epsilon \beta x-4\sigma ^{2}-2\epsilon \beta \sigma }$. ..........all times alpha

${\displaystyle +2\sigma x+\epsilon \beta x-4\sigma ^{2}-2\epsilon \beta \sigma }$......all times epsilon

A=

${\displaystyle -2\sigma x^{2}}$ ${\displaystyle -\epsilon \beta x^{2}}$ ${\displaystyle +4\sigma ^{2}x}$ ${\displaystyle +2\epsilon \beta x\sigma }$

${\displaystyle +2\sigma \alpha x}$ ${\displaystyle +\epsilon \beta \alpha x}$ ${\displaystyle -4\sigma ^{2}\alpha }$ ${\displaystyle -2\epsilon \beta \sigma \alpha }$

${\displaystyle +2\epsilon \sigma x}$ ${\displaystyle +\epsilon ^{2}\beta x}$ ${\displaystyle -4\epsilon \sigma ^{2}}$ ${\displaystyle -2\epsilon ^{2}\beta \sigma }$

Group according to power of x

A=

${\displaystyle -2\sigma x^{2}}$ ${\displaystyle -\epsilon \beta x^{2}}$

${\displaystyle +4\sigma ^{2}x}$ ${\displaystyle +2\epsilon \beta x\sigma }$ ${\displaystyle +2\sigma \alpha x}$ ${\displaystyle +\epsilon \beta \alpha x}$ ${\displaystyle +2\epsilon \sigma x}$ ${\displaystyle +\epsilon ^{2}\beta x}$

${\displaystyle -4\sigma ^{2}\alpha }$ ${\displaystyle -2\epsilon \beta \sigma \alpha }$ ${\displaystyle -4\epsilon \sigma ^{2}}$ ${\displaystyle -2\epsilon ^{2}\beta \sigma }$

Combine like terms and group with small terms last

A=

${\displaystyle -2\sigma x^{2}}$ ${\displaystyle -\epsilon \beta x^{2}}$

${\displaystyle +4\sigma ^{2}x}$ ${\displaystyle +2\sigma \alpha x}$ ${\displaystyle +2\epsilon \beta \sigma x}$ ${\displaystyle +\epsilon \beta \alpha x}$ ${\displaystyle +2\epsilon \sigma x}$ ${\displaystyle +\epsilon ^{2}\beta x}$

${\displaystyle -4\sigma ^{2}\alpha }$ ${\displaystyle -2\epsilon \beta \sigma \alpha }$ ${\displaystyle -4\epsilon \sigma ^{2}}$ ${\displaystyle -2\epsilon ^{2}\beta \sigma }$

Switch all signs and factor

${\displaystyle A=}$ ${\displaystyle \left(-2\sigma -\epsilon \beta \right)x^{2}}$

${\displaystyle +\left(4\sigma ^{2}+2\sigma \alpha +2\epsilon \beta \sigma +\epsilon \beta \alpha +2\epsilon \sigma +\epsilon ^{2}\beta \right)x}$

${\displaystyle +\left(-4\sigma ^{2}\alpha -2\epsilon \beta \sigma \alpha -4\epsilon \sigma ^{2}-2\epsilon ^{2}\beta \sigma \right)}$

### Normal mode frequencies

algebra

${\displaystyle -B=-\sigma ^{2}x+\sigma ^{2}\alpha +\epsilon \sigma ^{2}}$correct

${\displaystyle -C=-\epsilon ^{2}\beta x+2\epsilon ^{2}\beta \sigma }$

Place the B and C stuff at end, spaced with \qquad

${\displaystyle A-B-C=}$ ${\displaystyle \left(-2\sigma -\epsilon \beta \right)x^{2}}$

${\displaystyle +\left(4\sigma ^{2}+2\sigma \alpha +2\epsilon \beta \sigma +\epsilon \beta \alpha +2\epsilon \sigma +\epsilon ^{2}\beta \qquad -\sigma ^{2}-\epsilon ^{2}\beta \right)x}$

${\displaystyle +\left(-4\sigma ^{2}\alpha -2\epsilon \beta \sigma \alpha -4\epsilon \sigma ^{2}-2\epsilon ^{2}\beta \sigma \qquad +\sigma ^{2}\alpha +\epsilon \sigma ^{2}+2\epsilon ^{2}\beta \sigma \right)}$

All the new terms combine with terms already present

${\displaystyle A-B-C=}$ ${\displaystyle \left(-2\sigma -\epsilon \beta \right)x^{2}}$

${\displaystyle +\left(3\sigma ^{2}+2\sigma \alpha +2\epsilon \beta \sigma +\epsilon \beta \alpha +2\epsilon \sigma \right)x}$

${\displaystyle +\left(-3\sigma ^{2}\alpha -2\epsilon \beta \sigma \alpha -3\epsilon \sigma ^{2}\right)}$

Normal modes at ${\displaystyle ax^{2}+bx+c=0}$, where: ${\displaystyle a=\left(2\sigma +\epsilon \beta \right)}$

${\displaystyle b=-\left(3\sigma ^{2}+2\sigma \alpha +2\epsilon \beta \sigma +\epsilon \beta \alpha +2\epsilon \sigma \right)x}$

${\displaystyle c=\left(3\sigma ^{2}\alpha +2\epsilon \beta \sigma \alpha +3\epsilon \sigma ^{2}\right)}$

### Check with matlab

${\displaystyle 0=(2\sigma +\beta \epsilon )x^{2}-(2\alpha \sigma +3\sigma ^{2}+\alpha \beta \epsilon +2\beta \epsilon \sigma +2\epsilon \sigma )x+(3\alpha \sigma ^{2}+2\alpha \beta \epsilon \sigma )+3\epsilon \sigma ^{2}}$

## Degeneracy at zero coupling

If ${\displaystyle \epsilon =0}$, the modes are decoupled, and we have,

${\displaystyle a=2\sigma }$,     ${\displaystyle b=-\left(3\sigma ^{2}+2\sigma \alpha \right)}$,     ${\displaystyle c=3\sigma ^{2}\alpha }$

WLOG[1] we may set ${\displaystyle \sigma =1}$ and consider:

${\displaystyle {\tilde {a}}=2}$,     ${\displaystyle {\tilde {b}}=-\left(3\sigma +2\alpha \right)}$, and     ${\displaystyle {\tilde {c}}=3\sigma \alpha }$,

so that:  ${\displaystyle {\tilde {b}}^{2}-4{\tilde {a}}{\tilde {c}}=9\sigma ^{2}+12\alpha \sigma +4\sigma ^{2}-24\alpha \sigma =(3\sigma -2\alpha )^{2}}$

${\displaystyle x={\frac {3\sigma +2\alpha }{4}}\pm {\frac {3\sigma -2\alpha }{4}}={\frac {3\sigma +2\alpha }{4}}=\alpha ={\tfrac {3}{2}}\sigma }$ if ${\displaystyle 3\sigma =2\alpha }$.

### check with matlab

If σ=1, then we have: ${\displaystyle 0=2x^{2}-(2\alpha +3)x+3\alpha }$

## Simple system σ=2, α=3= x0

Under construction. I think x=x0+x1 where x0=3 and

x1=0

0 = 4*epsilon - 4*x_1 + beta*epsilon - beta*epsilon*x_1

${\displaystyle x_{1}={\frac {4\epsilon +\beta \epsilon }{4+\beta +\epsilon }}???}$

## next

1. Wouldn't it be great if the industry standard was to use a Wikpedia compatable markup language so that teachers can freely use jargon and let the student decide what words need to be defined in textbooks?