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Dispersion relation
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jhillairet authored Jul 16, 2017
2 parents 3d8dbf9 + 3762edb commit e2302e6
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56 changes: 55 additions & 1 deletion Dispersion_Relation.tex
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\chapter{Dispersion Relation}
\section{Vacuum dispersion relation}
\section{Vacuum Dispersion Relation}
In a non-dispersive isotropic homogeneous non-lossy dielectric medium, such as vacuum (or air), the wavevector $\mathbf{k}=k\mathbf{\hat{k}}$ direction is given by:
\begin{equation}
\mathbf{\hat{k}}\cdot\mathbf{E}=0
Expand All @@ -13,6 +13,60 @@ \section{Vacuum dispersion relation}

The condition for that system of 3 equations and 3 unknowns $(E_{x},E_{y},E_{z})$ to have a non-trivial solution, consists in solving the \emph{dispersion relation} between the wavevector $\mathbf{k}$ and the angular frequency $\omega$, i.e. $\mathbf{k}(\omega)$. In this medium, this relation is simple (TODO demo): $k=\sqrt{\mu\varepsilon}\omega=\frac{c}{n}\omega$\parencite[(7.4)]{Jackson1998}\footnote{While in a lossy medium $k^{2}=\mu\varepsilon\omega^{2}-j\omega\mu\sigma$ \parencite[sec. 8.2]{Bladel2007}.}.

\subsection{Cold Plasma Dispersion Relation}
Starting from Maxwell equations expressed in the $k-\omega$ domain (\ref{eq:k-omegaMaxwellEquations}) and assuming that the dispersion relations are of the form (\ref{eq:k-omegaDispersionRelation}), with the supplementary assumption that the medium is non magnetic (i.e. $\boldsymbol{\mu}=\mu_0$), one has thus:
\begin{subequations}
\begin{align}
\mathbf{k} \times \boldsymbol{\mathcal{E}} (\mathbf{k}, \omega)
=&
\omega \mu_0 \boldsymbol{\mathcal{H}}(\mathbf{k}, \omega)
\\
\mathbf{k} \times \boldsymbol{\mathcal{H}} (\mathbf{k}, \omega)
=&
\left(
-\omega \boldsymbol{\varepsilon} (\mathbf{k}, \omega)
+
j\boldsymbol{\sigma} (\mathbf{k}, \omega)
\right) \cdot
\boldsymbol{\mathcal{E}}(\mathbf{k}, \omega)
\\
\mathbf{k} \cdot \boldsymbol{\varepsilon}(\mathbf{k}, \omega) \cdot \boldsymbol{\mathcal{E}} (\mathbf{k}, \omega)
=& jq_v(\mathbf{k}, \omega)
\\
\mathbf{k} \cdot \boldsymbol{\mathcal{H}} (\mathbf{k}, \omega)
=& 0
\end{align}
\end{subequations}
combining the first two equations leads to the propagation equation:
\begin{equation}
\mathbf{k} \times \mathbf{k} \times \boldsymbol{\mathcal{E}} (\mathbf{k}, \omega)
+
\left(
\omega^2 \mu_0 \boldsymbol{\varepsilon} (\mathbf{k}, \omega)
-
j\omega\mu_0\boldsymbol{\sigma} (\mathbf{k}, \omega)
\right) \cdot
\boldsymbol{\mathcal{E}}(\mathbf{k}, \omega)
=
\boldsymbol{0}
\end{equation}
Let's define the equivalent relative dielectric tensor $\mathbf{K}$ such as:
\begin{equation}
\mathbf{K}(\mathbf{k},\omega)
=
\mathbf{I}
-
\frac{j}{\omega\varepsilon_0}\boldsymbol{\sigma} (\mathbf{k}, \omega)
\end{equation}
then,
\begin{equation}
\mathbf{k} \times \mathbf{k} \times \boldsymbol{\mathcal{E}} (\mathbf{k}, \omega)
+ \frac{\omega^2}{c^2}
\mathbf{K}(\mathbf{k}, \omega) \cdot
\boldsymbol{\mathcal{E}}(\mathbf{k}, \omega)
=
\boldsymbol{0}
\end{equation}

% Dans un plasma froid magnétisé, la situation est radicalement différente,
% car les courants de polarisation générés par les mouvements électroniques
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69 changes: 69 additions & 0 deletions Figures/Fusion Cross Section Plots/Fusion Cross Section.py
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# -*- coding: utf-8 -*-
"""
Created on Thu Jan 8 15:27:56 2015
@author: JH218595
"""

from pylab import *

"""
Plot the Reaction rates in cm^3 s^-1 as a function of
E, the energy in keV of the incident particle
[the first ion of the reaction label]
Data taken from NRL Formulary 2013.
"""
E,DD,DT,DH=loadtxt('reaction_rates_vs_energy_incident_particle.txt',
skiprows=1, unpack=True)

figure(num=1)
clf()
loglog(E, DD, E, DT, E, DH, lw=2)
grid()
xlabel('Temperature [keV]', fontsize=16)
ylabel('Reaction Rates $<\sigma v>$ [$cm^3.s^{-1}$]', fontsize=16)
xticks(fontsize=14)
yticks(fontsize=14)
ylim([1e-26, 2e-15])
legend(('D-D', 'D-T', 'D-He$^3$'), loc='best', fontsize=18)

"""
Plot the total cross section in m^2 for various species vs incident energy in keV
"""

def cross_section(E, A):
"""
The total cross section in barns (1 barns=1e-24 cm^2) as a function of E,
the energy in keV of the incident particle.
Formula from NRL Formulary 2013.
"""
sigma_T = (A[4]+((A[3]-A[2]*E)**2+1)**(-1) * A[1])/(E*(exp(A[0]/sqrt(E))-1))
return(sigma_T)

A_DD_a = [46.097, 372, 4.36e-4, 1.220, 0]
A_DD_b = [47.88, 482, 3.08e-4, 1.177, 0]
A_DT = [45.95, 50200, 1.368e-2, 1.076, 409]
A_DHe3 = [89.27, 25900, 3.98e-3, 1.297, 647]
A_TT = [38.39, 448, 1.02e-3, 2.09, 0]
A_THe3 = [123.1, 11250, 0, 0, 0]

E = logspace(0, 3, 501)
barns2SI = 1e-24 * 1e-4 # in m^2

sigma_DD = barns2SI*(cross_section(E, A_DD_a) + cross_section(E, A_DD_b))
sigma_DT = barns2SI*cross_section(E, A_DT)
sigma_DHe3 = barns2SI*cross_section(E, A_DHe3)

figure(num=2)
clf()
loglog(E, sigma_DD, E, sigma_DT, E, sigma_DHe3, lw=2)
grid()
xlabel('Deuteron Energy [keV]', fontsize=16)
ylabel('Cross-section $\sigma$ [$m^2$]', fontsize=16)
legend(('D-D', 'D-T', 'D-He$^3$'), loc='best', fontsize=18)
ylim([1e-32, 2e-27])
xticks(fontsize=14)
yticks(fontsize=14)
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Temperature [keV] D�D D-T D-He3
1.0 1.5e-22 5.5e-21 1.0E-26
2.0 5.4e-21 2.6e-19 1.4e-23
5.0 1.8e-19 1.3e-17 6.7e-21
10.0 1.2e-18 1.1e-16 2.3e-19
20.0 5.2e-18 4.2e-16 3.8e-18
50.0 2.1e-17 8.7e-16 5.4e-17
100.0 4.5e-17 8.5e-16 1.6e-16
200.0 8.8e-17 6.3e-16 2.4e-16
500.0 1.8e-16 3.7e-16 2.3e-16
1000.0 2.2e-16 2.7e-16 1.8e-16
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