# Visualizing a vector field with Matplotlib

Matplotlib provides a function, streamplot, to create a plot of streamlines representing a vector field. The following program displays a representation of the electric field vector resulting from a multipole arrangement of charges. The multipole is selected as a power of 2 on the command line (1=dipole, 2=quadrupole, etc.)

It requires Matplotlib 1.5+ because of the choice of colormap (plt.cm.inferno): this can be replaced with another (for example plt.cm.hot) if using an older version of Matplotlib.

import sys
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Circle

def E(q, r0, x, y):
"""Return the electric field vector E=(Ex,Ey) due to charge q at r0."""
den = np.hypot(x-r0[0], y-r0[1])**3
return q * (x - r0[0]) / den, q * (y - r0[1]) / den

# Grid of x, y points
nx, ny = 64, 64
x = np.linspace(-2, 2, nx)
y = np.linspace(-2, 2, ny)
X, Y = np.meshgrid(x, y)

# Create a multipole with nq charges of alternating sign, equally spaced
# on the unit circle.
nq = 2**int(sys.argv[1])
charges = []
for i in range(nq):
q = i%2 * 2 - 1
charges.append((q, (np.cos(2*np.pi*i/nq), np.sin(2*np.pi*i/nq))))

# Electric field vector, E=(Ex, Ey), as separate components
Ex, Ey = np.zeros((ny, nx)), np.zeros((ny, nx))
for charge in charges:
ex, ey = E(*charge, x=X, y=Y)
Ex += ex
Ey += ey

fig = plt.figure()

# Plot the streamlines with an appropriate colormap and arrow style
color = 2 * np.log(np.hypot(Ex, Ey))
ax.streamplot(x, y, Ex, Ey, color=color, linewidth=1, cmap=plt.cm.inferno,
density=2, arrowstyle='->', arrowsize=1.5)

# Add filled circles for the charges themselves
charge_colors = {True: '#aa0000', False: '#0000aa'}
for q, pos in charges:

ax.set_xlabel('$x$')
ax.set_ylabel('$y$')
ax.set_xlim(-2,2)
ax.set_ylim(-2,2)
ax.set_aspect('equal')
plt.show()


The electric field of a dipole: $python efield.py 1 The electric field of an octopole: $ python efield.py 3

Current rating: 4.6

#### Stafford 7 years, 8 months ago

What clever program, Loved it. I had to change the 'inferno' to 'hot' as we were warned in the opening comments. Didn't
realise I was using an outdated version of matplotlib.

Current rating: 5

#### christian 7 years, 8 months ago

Thanks, Stafford – the issue of colormaps is one that exercises some people a great deal: for a good article about how awful the default colormap, jet, is, take a look here: https://jakevdp.github.io/blog/2014/10/16/how-bad-is-your-colormap/

Current rating: 4.3

#### Steven Armour 6 years, 11 months ago

Would you be opposed to me publishing a reproduction of this example that I made with the python yt lib (http://yt-project.org/) to the yt repository examples archive? (they need non-astro examples like nothing else) I have already made a prototype of what I am purposing for my E&M class last fall and would also like you to review the photo notebook as well as the final notebook to make sure there are no errors in trying to visualize the vector field volumetrically.

Sincerly
Steven Armour

Current rating: 5

#### christian 6 years, 11 months ago

Not at all – you're welcome to the example and any code here you find useful. I'd be happy to take a look at the Notebook you are preparing for your EM class.
Best wishes,
Christian

Current rating: 5

#### Youcef 6 years, 9 months ago

Hello,
i have some trouble running this program, it gives me this error if you can help me figure out what's the problem:

nq = 2**int(sys.argv[1])
IndexError: list index out of range

Current rating: 5

#### christian 6 years, 9 months ago

Hi Youcef,
The program expects the user to provide the order of the multipole on the command line (e.g. 1 for dipole, 2 for quadrupole etc.), so you should run the code with e.g.

python multipole.py 3

to plot a figure for an octopole.

Current rating: 5

#### Youcef 6 years, 9 months ago

Hello again Christian,

I did what you told me and it worked, so thanks a lot for the help and the quick response !

good day

Current rating: 5

#### Joaquín 5 years, 9 months ago

Hello Christian,

I am a beginner in python, what do you exactly would write in the code for a dipole ?

Current rating: 4

#### christian 5 years, 9 months ago

Dear Joaquín,
You need to save the code somewhere and open a command prompt (Terminal, whatever). If you navigate to the directory in which you've saved the code (on a Mac or *nix system, use the cd command; on windows I think it's chdir), you can just type:
python efield.py 1
to run the code for a dipole (2 for a quadrupole, 3 for a hexapole, etc), assuming you saved the code as efield.py and your python installation has the correct libraries installed. You might find it useful to install Anaconda (https://anaconda.org/anaconda/python) to manage your python installation.
I hope that helps,
Christian

Current rating: 5

#### Lina 6 years, 7 months ago

Is it possible to get the coordinates of the different streamlines ?
Also how can I interpolate between the streamlines ?

Current rating: 5

#### christian 6 years, 7 months ago

Hi Lina,
The call to streamplot returns a container object, stream_container, from which you can obtain the lines making up the plot as stream_container.lines, and get their coordinates from there.

If you want the lines to be closer together, you can set the density argument to streamplot when you call it, as described in the documentation: [http://matplotlib.org/api/pyplot_api.html#matplotlib.pyplot.streamplot]

Current rating: 5

#### Mark 6 years, 7 months ago

Perfect for creating custom field plots to add into my teaching notes.

Current rating: 5

#### christian 6 years, 7 months ago

I'm glad you find it useful – thanks for dropping by.

Current rating: 5

#### Keith 6 years, 7 months ago

In a couple of places hypot(x,y) should be used instead of sqrt(x**2+y**2).

Current rating: 5

#### christian 6 years, 7 months ago

That's a good idea – I've made the changes.
Thanks, Christian

Current rating: 5

#### Xinyuan Wei 6 years, 5 months ago

Hello,
I wonder In line 8: den = np.hypot(x-r0[0], y-r0[1])**1.5
if the exponent 1.5 should be replaced by 3?

Current rating: 4.5

#### christian 6 years, 5 months ago

Thanks for that catch, Xinyaun – you're absolutely right. I changed the previous code to use np.hypot without thinking it through. I've fixed it now.
Cheers,
Christian

Current rating: 5

#### mattes 5 years, 6 months ago

This is a nice code! But I have one question:
Not any steamline is equal. This is seen in your ocutpole (right top, red pole), where one steamline is straight. I want all lines look equally. Do you know how to do this?

Cheers,
mattes

Currently unrated

#### Zion_Samurai 4 years, 9 months ago

Hello, I have difficulty of plotting a magnetic dipole, could you please give me some clues?

Currently unrated

#### christian 4 years, 9 months ago

You could look at this follow-up post: https://scipython.com/blog/visualizing-the-earths-magnetic-field/

Current rating: 5

#### Valeri Stoyanov 3 years, 10 months ago

Hi, I'm trying to run this but it comes with an error for this command: nq = 2**int(sys.argv[1])

Error: Traceback (most recent call last):
File "C:\Users\valer\Documents\test\vectorfieldVisual.py", line 20, in <module>
nq = 2**int(sys.argv[1])
IndexError: list index out of range

Currently unrated

#### christian 3 years, 9 months ago

As stated in the description, you need to provide the multipole as a power of 2 on the command line. For example,

"python efield.py 1"

Currently unrated

#### Sergio 3 years, 9 months ago

Hi!, i am having the same problem, kinda new to this so not quite sure to what you mean whit the need to provide the multipole on the command line, if you can help it would be great.
Also taking an EM course so this come handy, tnks

Currently unrated

#### christian 3 years, 9 months ago

How are you executing the code? In an IDE or from an online service or from a terminal session?

Currently unrated

#### Maxim 3 years, 1 month ago

How can I do the same thing, but choose the sign of the charges on my own, for example, for two positive charges or three negative ones, and so on?

Currently unrated

#### christian 3 years, 1 month ago

The charge values and positions are set in a loop near the beginning of the code. Change:

charges.append((q, (np.cos(2*np.pi*i/nq), np.sin(2*np.pi*i/nq))))

to append tuples of the form (q, (x, y)) to the list charges instead of this line, which is for alternating charges on the unit circle.

charges = [(2, (0, 0)), (-1, (0, 1)), (-1, (0,-1))]

Current rating: 5

#### Maxim 3 years, 1 month ago

Thank you very much!

Currently unrated

#### Glenn 3 years ago

Good work! Can I ask what argument to add in order to change the position of the charges from the origin?

Currently unrated

#### christian 3 years ago

The charges are defined in the list of tuples called charges: each tuple gives the charge number and position: (q, (x, y)): you can set (x, y) as you like. In the code above, they are set to lie on the unit circle:

for i in range(nq):
q = i%2 * 2 - 1
charges.append((q, (np.cos(2*np.pi*i/nq), np.sin(2*np.pi*i/nq))))

Current rating: 5

#### Glenn 3 years ago

Can I adjust the placement of the charges while still maintaining the unit circle? The unit circle seems to change while I edit the tuples [] even though I haven't change the ((q, (np.cos(2*np.pi*i/nq), np.sin(2*np.pi*i/nq)))) part...

Thank you very much.

Currently unrated

#### Fernando Ramos 2 years, 11 months ago

Hi Christian!, I appreciate your code. It's useful!. But I have question about your definitions. Can you explain that?

First question: I dont understand the parameter "*charge"
in the for loop :

for charge in charges:
ex, ey = E(*charge, x=X, y=Y)
Ex += ex
Ey += ey

in the case that I eliminated " * " from " * charge" in the definition of the callback function , and then I run the scripts, It doesnt work . What is the meaning (or function )of "*charge" ?

Second question: What is the way to use r0?. I understand that define in this case (0,0), but if I need to create a (x,y) position, How can I do ? . In your code you only define in the "E function", but never more you use r0.

Thank you for sharing!

Currently unrated

#### christian 2 years, 11 months ago

The syntax *charge unpacks the tuple charge into the argument list of the function E:

E(*charge, x=X, y=Y)

is the same as

E(charge[0], charge[1], x=X, y=Y)

where the items inside charge are the charge value (charge[0]) and the charge position, r0 (charge[1]). In other words, r0 is set to the value determined by each charge in turn as the function is called.

I hope that helps!

Currently unrated

#### Martin Nieto 2 years, 1 month ago

Great code! I adapted it for my visualization of the magnetic field of a circular current loop.

Currently unrated

#### Frans 1 year, 11 months ago

Great code. Would like to visualize the vector field between a point and an object (preferably a 3D mesh). Could you give some tips?

Currently unrated

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