I am trying to draw a spherical harmonics for my college project. The following formula I want to depict,
Y = cos(theta)
for that, I wrote this code
import numpy as np
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
def sph2cart(r, phi, tta):
''' r is from 0 to infinity '''
''' phi is from 0 to 2*pi '''
''' tta is from 0 to pi '''
x = r* np.sin(tta)* np.cos(phi)
y = r* np.sin(tta)* np.sin(phi)
z = r* np.cos(tta)
return x, y, z
# phi running from 0 to pi and tta from 0 to pi
phi = np.linspace(0, 2* np.pi, 25)
tta = np.linspace(0, np.pi, 25)
# meshgrid to generate points
phi, tta = np.meshgrid(phi, tta)
# THIS IS THE FUNCTION
Y = np.cos(tta)
# finally all things in cartesian co-ordinate system
# Note that "Y" is acting as "r"
x, y, z = sph2cart( Y, phi, tta)
# plotting :-
fig = plt.figure()
ax = fig.add_subplot( 111 , projection='3d')
ax.plot_surface(x, y, z, linewidth = 0.5, edgecolors = 'k')
And, get the sphere as a result. Which is not correct, because actual result is dumbbell like shape. See the second row of this image,
The picture in the Wikipedia article Spherical harmonics is obtained by using the absolute value of a spherical harmonic as the r coordinate, and then coloring the surface according to the sign of the harmonic. Here is an approximation.
x, y, z = sph2cart(np.abs(Y), phi, tta)
fig = plt.figure()
ax = fig.add_subplot( 111 , projection='3d')
from matplotlib import cm
ax.set_aspect('equal')
ax.plot_surface(x, y, z, linewidth = 0.5, facecolors = cm.jet(Y), edgecolors = 'k')
When you use Y itself as r, the two hemispheres (positive Y and negative Y) end up mapped onto the same half of the above surface.
The Y you are passing to the function needs to be an absolute value to make it r, else z = cos(theta)^2 is always positive. If r is to be the radius then this what you should be doing.
x, y, z = sph2cart(np.abs(Y), phi, tta)
来源：https://stackoverflow.com/questions/49158505/visualizing-spherical-harmonics-in-python