User:Guy vandegrift/btvs/Effort 4

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Transverse modes[edit]

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:

where . For atoms not touching the wall or the coupling spring connecting to the "atom":

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

and the atom obeys:

rearranging for convenience later on

This is all we need.

Consider only symmetric modes[edit]

In this simple case Since we are only considering modes that are symmetrical with respect to the atom,

We don't need the third equation for modes symmetric about the atom.

writing in matrix form

Divide the first two rows by MS and the third by MA:

Define the fundamental frequencies and beta[edit]

,    ,    ,    ,    where , to obtain:

Matrix equation for normal modes[edit]

Let: , , , and

Finding the determinant of a 3x3 matrix[edit]

click to view a great deal of albebra

This equals A-B-C, where:

Distribute the factor in front for each line

Now we group each line by factors of x

Now factor, for each of the three lines:

Regroup so that each line corresponds to a different factor of x:

Regroup and put the ε terms last since they are small

Now recall that the determinant is A-B-C, where

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.

Solving the cubic equation for the uncoupled case[edit]

The exact uncoupled equations (no approximations made) are:

If the coupling (epsilon) equals zero:

From physical grounds we know one answer is x=α. It can be shown that:

Two ways to be degenerate (uncoupled case only)[edit]

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


The other two roots for zero coupling are solutions to:


The high beta approximation[edit]

Simply by making very large.


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

Now we see what happens if we go to the original matrix and make the same simplification (should get same answer).


Remove small elements


Approximate solutions to cubic equations[edit]

Failed efforts[edit]

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