# 1.5.1.2. 2017-04 Group Meeting¶

## 1.5.1.2.1. 2017-04-13¶

1. Show the new dispersion relations as well as the $$\omega(n)$$ plots.

2. Crossing change the number of solutions for different n.

3. Ask about how to identify the plus and minus modes before the calculation in the 2013 paper: Duan, H. (2013). Flavor oscillation modes in dense neutrino media. Physical Review D - Particles, Fields, Gravitation and Cosmology, 88(12), 1–7.

## 1.5.1.2.2. 2017-04-19¶

1. The unstable points

### 1.5.1.2.2.1. Some Discussion on Scaling¶

Code

The corresponding Mathematica code is:

codebase.git/mma/linear-stability/growth-in-fourier-modes.nb

This means that the time evolution of slope is quite complicated. But we should compare the slope at the same time for different parameters. For each time snapshot we should be able to guess the function of parameters that describes the slope.

As a very preliminary result, scale as I defined, changes how fast the perturbation is propagated into higher moments. So does the off diagonal element.

meh

The eigenvalues of this pseudo Hamiltonian

The key is actually the eigenvectors. If we assume each of the modes are plane waves, the eigenvalues are the ones that determines how fast the grow is while the eigenvectors tells us which element will grow. As I increase the scale, the eigenvectors seems to be more clean, i.e., fewer but hierarchial excited elements.

Decreasing the off diagonal elements have similar effect on eigenvectors which explains the similar structure of fourier modes for different off diagonal elements. However, they should have different growth rate as a function of time.

More explicitly, we plot out the logplot of eigenvector elements and we will see the exponential behavior.

Just for reference, the positive values of eigenvectors also have this exp behavior.

Do all the eigenvectors have the same slope?

Qualitatively we can see why the exponantial behavior in Fourier modes through the eigenvectors.

To guess an expression for it, I need to compare different scales.

As reference, we plot mat[dim,1,1,1].

Here is a plot that shows the slope of $$\text{Log@Abs@Eigenvalues}\sim \text{Eigenvalue Element ID}$$.

Ratio of Scale and Off diagonal?

Is the ratio of scale and off diagonal element responsible for the slope?

Or the plot with scale/off-diag = 1 to 10.

## 1.5.1.2.3. 2017-04-22¶

This is an update for the Fourier mode growth.

I recalculated the eigenvectors using a Hamiltonian without random numbers so that the Hamiltonian matrix is defined as

$\begin{split}H_{ij} =\begin{cases} od, & i= j\pm 1\\ scale * i, & i=j \end{cases}.\end{split}$

What I found is that if $$1/ratio = od/scale$$ is small, the slope is related to the ration through

$slope = k \ln (ratio) + b,$

or in other words,

$ratio = e^{( slope - b)/k }.$

In the calculations I did, $$k\sim 1$$.

The dimension seems to change the interception.

Change the dimension seems to only slightly change the slope increasing rate.

Enough for these eigenvectors. To make a real connection with the time evolution or the long time limit of the time evolution, we need to find how the system evolves.

The idea is that all Fourier modes are oscillations. In the beginning of oscillations, it is simply a linear growth in time since Taylor expansion of $$e^{-i E t}$$ is $$E t$$.

Given an initial condition $$\psi(0)$$ with only nonzero in the middle element, we can expand it using the eigenvectors $$\psi_i (0)$$, i.e.,

$\psi(0) = \sum_i A_i \psi_i (0).$

What I find is that the coefficients $$A_i$$ is actually a exponential shape.

If I start from all 1’s as initial condition, the evolution won’t be of this wedge shape and it will oscillation.

The corresponding decomposition of intial condition is shown in :numfig:fourier-mode-growth-initial-all-1-decompose.

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