Goodbye Matlab, Hello Python : Emag by example

Here is an excerpt of some emag problems I worked on some time ago.

The point here is not to follow the math, but to get an idea of how to do matlab and mathematica things in Python.

I used python + sympy + numpy + matplotlib + jupyter to create a CAS mathematica like notebook.

To install and repeat, simply make sure that you have python3 and pip3.

pip3 install jupyterlab numpy sympy matplotlib

What you see below is my exported jupyterlab notebook session as markdown with latex+mathjax enabled.


from ipywidgets import *
import numpy as np
import matplotlib.pyplot as plt
from sympy import *
from mpl_toolkits.mplot3d import Axes3D
#enable inline to print
%matplotlib inline
#enable notebook for interactice features such as sliders
#matplotlib notebook
#Some physics stuff
from sympy.physics.vector import ReferenceFrame
from sympy.physics.vector import curl

#Arbitrary precision
from mpmath import mp, degrees
mp.dps = 15; mp.pretty = True

Some Constants

fE_0 =  Float(8.85*10**(-12))
mu_0 =  Float(1.256*10**(-6))
#Region 1 which is air
fE_1 = E_0
#Region 2 - pexiglass
fE_2 = 3.45*E_0

\displaystyle 8.85 \cdot 10^{-12}


\displaystyle 3.05325 \cdot 10^{-11}

Problem 1

We can get \eta_1 and \eta_2 using \eta = \sqrt{\frac {\mu} {\epsilon}}

#Get eta_1 and eta_2
eta_1 = sqrt(mu_0/fE_1)
eta_2 = sqrt(mu_0/fE_2)

Brewster's angle of incidence implies \eta_2 \cos(\theta^t) = \eta_1 \cos(\theta_B)

We derive:

\begin{align} \cos \theta^t = \sqrt{1 - \frac {\epsilon_1}{\epsilon_1 + \epsilon_2}} \end{align}
\begin{align} \cos \theta^i = \sqrt{\frac{\epsilon_1}{\epsilon_2} \cos^2 \theta^t} \end{align}

And we can solve for \cos \theta^i and \cos \theta^t below.

cos_theta_t = sqrt(1 - (fE_1)/(fE_1 + fE_2))
cos_theta_i = sqrt((fE_1/fE_2)*(cos_theta_t**2))

We need the following reflection coefficients both for the parallel and perpendicular cases to compute the power fractions.

\begin{align} \Gamma_{||} = \frac {\eta_2 \cos \theta^t- \eta_1 \cos \theta^i} {\eta_2 \cos \theta^t + \eta_1 \cos \theta^i} \end{align}
\begin{align} \tau_{||} = \frac {2 \eta_2 \cos \theta^i} {\eta_2 \cos \theta^t + \eta_1 \cos \theta^i} \end{align}
\begin{align} \Gamma_\perp = \frac {\eta_2 \cos \theta^i - \eta_1 \cos \theta^t} {\eta_2 \cos \theta^i - \eta_1 \cos \theta^t} \end{align}
\begin{align} \tau_\perp = \frac {2 \eta_2 \cos \theta^i} {\eta_2 \cos \theta^i + \eta_1 \cos \theta^t} \end{align}
ref_perp = (eta_2*cos_theta_i - eta_1*cos_theta_t)/(eta_2*cos_theta_i + eta_1*cos_theta_t)
ref_par = (eta_2*cos_theta_t - eta_1*cos_theta_i)/(eta_2*cos_theta_t + eta_1*cos_theta_i)
trans_par = (2*eta_2*cos_theta_i)/(eta_2*cos_theta_t + eta_1*cos_theta_i)
trans_perp = (2*eta_2*cos_theta_i)/(eta_2*cos_theta_i+eta_1*cos_theta_t)

Transmitted Power

Now we can solve for the transmitted power using \frac {1} {2} ( \tau_\perp^2 + \tau_{||}^2 )

(1/2)*(trans_par**2) + (1/2)*(trans_perp**2)

\displaystyle 0.245924885051478

Reflected Power

And a similar process for reflected power \frac {1} {2} ( \Gamma_\perp^2 + \Gamma_{||}^2 )

(1/2)*(ref_par**2) + (1/2)*(ref_perp**2)

\displaystyle 0.151559146572402


There is no reflected component. The transmitted wave is right hand circular.

Problem 2

To start off, we find values \epsilon_2 and \epsilon_1

\begin{align} \eta_1 = \sqrt{\frac {\mu_0}{70 \epsilon_0}} \end{align}
\begin{align} \eta_2 = \sqrt{\frac {\mu_0}{\epsilon_0}} \end{align}
#Some constants
feta_2 = Float(sqrt(mu_0/fE_0))
feta_1 = Float(sqrt(mu_0/(fE_0*70)))

part a.

Between the critical angles, the transmitted wave is able to pass the air-water boundary into the air.

\begin{align} \theta_c = sin^{-1}(\sqrt{\frac{\epsilon_2}{\epsilon_1}}) \end{align}

\displaystyle 6.86456633770632

We must limit \theta_i to [-\theta_c, \theta_c]

In degrees this is [-6.9, 6.9]

part b.

A parallel wave at the brewster angle provides the maximum power transmitted into the air.

The brewster angle is easily found with:

\begin{align} \theta_B = \sin^{-1} \sqrt{\frac{\epsilon_2}{\epsilon_1 + \epsilon_2}} \end{align}

The brewster angle in degrees is:


\displaystyle 6.81582191825822

Some Graphs

You can see what is happening with the reflection and transmission coefficients for both the parallel and perpendicular wave cases.

epsilon_1, epsilon_2, theta_i = var('epsilon_1 epsilon_2 theta_i')
Fi_to_t = asin((sqrt(epsilon_1)*sin(theta_i))/sqrt(epsilon_2))

\displaystyle \operatorname{asin}{\left(\frac{\sqrt{\epsilon_{1}} \sin{\left(\theta_{i} \right)}}{\sqrt{\epsilon_{2}}} \right)}

Reflection Perpendicular and Parallel

eta_1, eta_2, theta_t = var('eta_1 eta_2 theta_t')
F_ref_par = (eta_2*cos(theta_t) - eta_1*cos(theta_i))/(eta_2*cos(theta_t) + eta_1*cos(theta_i))
F_ref_par = F_ref_par.subs(theta_t,Fi_to_t)
F_ref_perp = (eta_2*cos(theta_i) - eta_1*cos(theta_t))/(eta_2*cos(theta_i) + eta_1*cos(theta_t))
F_ref_perp = F_ref_perp.subs(theta_t,Fi_to_t)
p = plot(
    abs(F_ref_perp.subs({eta_1:feta_1, eta_2:feta_2, epsilon_2:fE_0, epsilon_1:(70*fE_0)})),
    abs(F_ref_par.subs({eta_1:feta_1, eta_2:feta_2, epsilon_2:fE_0, epsilon_1:(70*fE_0)})),
p[0].line_color = 'green'
p[0].label = '$|\\Gamma \\perp|$'
p[1].label = '$|\\Gamma_{||}|$'
p.title = '$|\\Gamma|$ vs. $\\theta^i$'
p.xlabel = '$\\theta^i$'
p.ylabel = '$\\Gamma$'


Tranmission Perpendicular and Parallel

F_trans_perp = (2*eta_2*cos(theta_i))/(eta_2*cos(theta_i) + eta_1*cos(theta_t))
F_trans_perp = F_trans_perp.subs(theta_t,Fi_to_t)
#F_trans_par = F_trans_par.subs(theta_t,Fi_to_t)
F_trans_par = (2*eta_2*cos(theta_i))/(eta_2*cos(theta_t) + eta_1*cos(theta_i))
F_trans_par = F_trans_par.subs(theta_t,Fi_to_t)
#F_trans_par = F_trans_par.subs(theta_t,Fi_to_t)
p = plot(
    abs(F_trans_perp.subs({eta_1:feta_1, eta_2:feta_2, epsilon_2:fE_0, epsilon_1:(70*fE_0)})),
    abs(F_trans_par.subs({eta_1:feta_1, eta_2:feta_2, epsilon_2:fE_0, epsilon_1:(70*fE_0)})),
p[0].line_color = 'green'
p[0].label = '|$\\tau_\\perp|$'
p[1].label = '|$\\tau_{||}|$'
p.title = '$\\tau$ vs. $\\theta^i$'
p.xlabel = '$\\theta^i$'
p.ylabel = '$\\tau$'