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() ax = fig.add_subplot(111) # 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.add_artist(Circle(pos, 0.05, color=charge_colors[q>0])) 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`

## Comments

## Stafford 1 year, 10 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

Link | Replyrealise I was using an outdated version of matplotlib.

## christian 1 year, 10 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/

Link | Reply## Steven Armour 1 year, 1 month 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.

Link | ReplySincerly

Steven Armour

## christian 1 year, 1 month 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.

Link | ReplyBest wishes,

Christian

## Youcef 11 months, 1 week ago

Hello,

Link | Replyi 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

Thanks for your time

## christian 11 months, 1 week ago

Hi Youcef,

Link | ReplyThe 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.

## Youcef 11 months, 1 week ago

Hello again Christian,

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

good day

## Lina 9 months, 3 weeks ago

Is it possible to get the coordinates of the different streamlines ?

Link | ReplyAlso how can I interpolate between the streamlines ?

## christian 9 months, 3 weeks ago

Hi Lina,

Link | ReplyThe 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]

## Mark 9 months, 1 week ago

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

Link | Reply## christian 9 months, 1 week ago

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

Link | Reply## Keith 9 months, 1 week ago

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

Link | Reply## christian 9 months, 1 week ago

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

Link | ReplyThanks, Christian

## Xinyuan Wei 7 months, 1 week ago

Hello,

Link | ReplyI 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?

## christian 7 months, 1 week 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.

Link | ReplyCheers,

Christian

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