I am having a hard time understanding how to establish a connection between t and Z function in fresnel integrals. Here is the syntax for the Fresnel integrals where they basically give an array for t and it calculates the C(z) and S(z) coefficients.
In Here, they say basically argument Z is the same as t e.g. input parameter, and for the basic case where you wanna produce the simple graph in Wikipedia, it seems to work.
import matplotlib.pyplot as plt
import numpy as np
from scipy.special import fresnel
delta_nudot = 1.503e-13
t = np.linspace(0.01, 6.0, 501)
z = np.sqrt(np.pi * delta_nudot * t**2)/2
ss, cc = fresnel(t / np.sqrt(np.pi / 2))
scaled_ss = np.sqrt(np.pi / 2) * ss
scaled_cc = np.sqrt(np.pi / 2) * cc
ax.plot(z, scaled_ss, 'r', label=r'$S(z)$', linewidth=2)
ax.plot(z, scaled_cc, 'g--', label=r'$C(z)$', linewidth=2)
However, as it can be seen from this document, my Z is dependent on the delta \nudot and T which might vary. Therefore I am not sure, how can I come up with a way that my coefficients C(Z) and S(Z) change when I change for instance delta \nudot because as you can see t goes into the formula and not Z and when I try to add z into the formula in gives me a flat graph which is obviously wrong.
This is one of the graphs I would like to produce
delta_nudot = 1.503e-13
t = np.linspace(0.01, 5.0, 201)
z = np.sqrt(np.pi * delta_nudot * t**2)/2
ss, cc = fresnel(t / np.sqrt(np.pi / 2))
eq12 = (cc**2+ss**2)*(4/z**2)
scaled_ss = np.sqrt(np.pi / 2) * ss
scaled_cc = np.sqrt(np.pi / 2) * cc
eq12_scaled = (scaled_ss**2+scaled_cc**2)*(4/z**2)
eqlabel = r'$\frac{4}{Z}[C(z)^2 + S(z)^2]$'
ax.plot(z, eq12_scaled , 'b-.', label=eqlabel, linewidth=3)
But I think it is not producing the correct graphs because as I change my delta_nudot, only the z part in 4/z changes, and coefficients stays the same!
How do I make C(Z) and S(Z), depend on Z and not t in scipy integration of fresnel formulas?
