# Plotting Magnetic Field Lines in 2D? Here’s what you need to know.

One of the major difficulties of understanding plasmas is just getting a handle on what the magnetic field is doing. If you’re lucky enough to be running simulations in 2D, you’ll have the value of the magnetic field at every point on your grid, but it still can be hard to understand the geometry of it all. That’s where field lines come in.

There’s this trick that is passed around by word of mouth amongst theoretical plasma physicists, but I can’t find it online or in print anywhere. Hopefully the solid use of SEO in my blog title will help it become more widely known (Don’t @ me, Michael Barbaro). The trick makes calculating field lines in 2D very fast, but as you’ll see, it only works for magnetic or other divergence-free 2D vector fields.

In 2D, the field lines in the \(xy\) plane can be calculated as lines of
equipotential of \(A_z\). The trick is: **you can calculate \(A_z\) at any
point in the grid by integrating \(-B_y\) along \(x\), and \(B_x\) along
\(y\).**

If you want to just believe me and skip the vector calculus, scroll to the bottom for a simple python snippet showing how to use this trick efficiently. Otherwise, let’s take a little detour to prove we’ve made a valid construction of \(A_z\).

### Simple calculation of the of the magnetic vector potential, \(A_z\).

Recall Maxwell’s EQ’s. \[ \mathbf{E} = \nabla V - \frac{\partial \mathbf{A}}{\partial t} \] \[ \mathbf{B} = \nabla \times \mathbf{A} \]

We are free to choose any \(A\) that satisfies the above equation, due to gauge freedom. Calculate curl of \(A\) \[ \nabla \times \mathbf{A} = \hat{i}\left[\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right] -\hat{j}\left[\frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z}\right] +\hat{k}\left[\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right] \] Now because the simulation is \(2D\), we know that \( \frac{\partial}{\partial z} \) is zero for all quantities. Furthermore, we only need to know \(A_z\), so we just need to construct any continuous and infinitely differentiable \(A_z\) that is a valid solution to the following set of partial differential equations. \[ B_x = \frac{\partial A_z}{\partial y} \] \[ B_y = -\frac{\partial A_z}{\partial x} \] One choice of \(A_z\) that satisfies the above equations is the following: \[ A_z(x,y) = - \int_0^{x}{B_y(x’,0)\ dx’} + \int_0^{y}{B_x(x,y’)\ dy’}.\]

Hopefully you can convince yourself that \(\frac{\partial A_z}{\partial y} = B_x \), So we’ll just show the algebra for \[B_y = -\frac{\partial A_z}{\partial x} = B_y(x,0) - \int_0^y{ \frac{\partial B_x}{\partial x}\ dy’}. \]

Because \(\nabla \cdot B = 0\) and fields are 2D; \(\frac{\partial B_x}{\partial x} = -\frac{\partial B_y}{\partial y}.\) Substituting this in… \[-\frac{\partial A_z}{\partial x} = B_y(x,0) + \int_0^y{ \frac{\partial B_y}{\partial y}\ dy’} = B_y(x,0) + B_y(x, y) - B_y(x,0) \] or \[B_y = -\frac{\partial A_z}{\partial x}.\]

Which is precisely what we needed to show, so we have proven our choice of \(A_z\) is valid. However, you explicitly use the fact the field is solenoidal to calculate the field lines, so this trick will not work for electric fields. You’ll need to do the full Helmholtz decomposition or trace them out as streamlines for that.

Let’s say you have two 2D numpy arrays, `bx`

and `by`

, corresponding
to the \(x\) and \(y\) components of the field respectively. With some
Numpy magic and disregarding units, \(A_z\) can be calculated and field lines
plotted with:

```
import numpy as np
import matplotlib.pyplot as plt
bydx = -np.cumsum(by[0, :])
bxdy = np.cumsum(bx[:, :], axis=0)
Az = bydx[np.newaxis, :] + bxdy
plt.contour(Az)
```

Isn’t it beautiful to program with multidimensional arrays? Thank you, Kenneth E. Iverson