Visualizing the Earth's dipolar magnetic field

(9 comments)

The magnetic field due to a magnetic dipole moment, $\boldsymbol{m}$ at a point $\boldsymbol{r}$ relative to it may be written $$ \boldsymbol{B}(\boldsymbol{r}) = \frac{\mu_0}{4\pi r^3}[3\boldsymbol{\hat{r}(\boldsymbol{\hat{r}} \cdot \boldsymbol{m}) - \boldsymbol{m}}], $$ where $\mu_0$ is the vacuum permeability. In geomagnetism, it is usual to write the radial and angular components of $\boldsymbol{B}$ as: $$ \begin{align*} B_r & = -2B_0\left(\frac{R_\mathrm{E}}{r}\right)^3\cos\theta, \\ B_\theta & = -B_0\left(\frac{R_\mathrm{E}}{r}\right)^3\sin\theta, \\ B_\phi &= 0, \end{align*} $$ where $\theta$ is polar (colatitude) angle (relative to the magnetic North pole), $\phi$ is the azimuthal angle (longitude), and $R_\mathrm{E}$ is the Earth's radius, about 6370 km. See below for a derivation of these formulae.

With this definition, $B_0$ denotes the magnitude of the mean value of the field at the magnetic equator on the Earth's surface, about $31.2\; \mathrm{\mu T}$. With these definitions, we can plot the dipole component of the Earth's magnetic field as a Matplotlib streamplot. We first construct a meshgrid of $(x,y)$ coordinates and convert them into polar coordinates with the relations: $$ \begin{align*} r &= \sqrt{x^2 + y^2},\\ \theta &= \mathrm{atan2}(y/x), \end{align*} $$ where the two-argument arctangent function, $\mathrm{atan2}$, is implemented in NumPy as arctan2. We can then use the formulae above to calculate $B_r$ and $B_\theta$ and convert back to Cartesian coordinates to plot the streamplot. The $\theta$ coordinate is offset by $\alpha = 9.6^\circ$ to account for the current tilt of the Earth's magnetic dipole with respect to its rotational axis.

Earth's dipolar magnetic field

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

# Mean magnitude of the Earth's magnetic field at the equator in T
B0 = 3.12e-5
# Radius of Earth, Mm (10^6 m: mega-metres!)
RE = 6.370
# Deviation of magnetic pole from axis
alpha = np.radians(9.6)

def B(r, theta):
    """Return the magnetic field vector at (r, theta)."""
    fac = B0 * (RE / r)**3
    return -2 * fac * np.cos(theta + alpha), -fac * np.sin(theta + alpha)

# Grid of x, y points on a Cartesian grid
nx, ny = 64, 64
XMAX, YMAX = 40, 40
x = np.linspace(-XMAX, XMAX, nx)
y = np.linspace(-YMAX, YMAX, ny)
X, Y = np.meshgrid(x, y)
r, theta = np.hypot(X, Y), np.arctan2(Y, X)

# Magnetic field vector, B = (Ex, Ey), as separate components
Br, Btheta = B(r, theta)
# Transform to Cartesian coordinates: NB make North point up, not to the right.
c, s = np.cos(np.pi/2 + theta), np.sin(np.pi/2 + theta)
Bx = -Btheta * s + Br * c
By = Btheta * c + Br * s

fig, ax = plt.subplots()

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

# Add a filled circle for the Earth; make sure it's on top of the streamlines.
ax.add_patch(Circle((0,0), RE, color='b', zorder=100))

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

Derivation

For now, consider the dipole to lie along the $y$-axis: $\boldsymbol{m} = m\boldsymbol{\hat{\jmath}}$ and define $\boldsymbol{B_0}$ to be the field perpendicular to $\boldsymbol{m}$ at a distance $R_\mathrm{E}$ from it (i.e. on the Earth's magnetic equator): $$ \boldsymbol{B_0} = -\frac{\mu_0m}{4\pi R_\mathrm{E}^3} \boldsymbol{\hat{\jmath}} = B_0\boldsymbol{\hat{\jmath}}. $$ Then, $$ \boldsymbol{B}(\boldsymbol{r}) = -B_0 \left(\frac{R_\mathrm{E}}{r}\right)^3 [3\cos\theta \boldsymbol{\hat{r}} - \boldsymbol{\hat{\jmath}}]. $$ The radial component of the magnetic field is therefore $$ B_r = \boldsymbol{B}\cdot\boldsymbol{\hat{r}} = -B_0\left(\frac{R_\mathrm{E}}{r}\right)^3[3\cos\theta - \cos\theta] = -2B_0\left(\frac{R_\mathrm{E}}{r}\right)^3\cos\theta, $$ Its angular component is perpendicular to this: $$ B_\theta = \boldsymbol{B}\cdot\boldsymbol{\hat{\theta}} = - B_0 \left(\frac{R_\mathrm{E}}{r}\right)^3 [-\boldsymbol{\hat{\jmath}} \cdot \boldsymbol{\hat{\theta}}] = -B_0\left(\frac{R_\mathrm{E}}{r}\right)^3\sin\theta. $$ Finally, $B_\phi = 0$: the field is symmetric about the axis of the dipole.

Further Reading

Current rating: 4.9

Comments

Comments are pre-moderated. Please be patient and your comment will appear soon.

Zion_Samurai 5 years, 2 months ago

As I see it, as you may allow. I think that the magnetic field of the Earth is reversed, and some of the beginning and tail end are not connected to the sphere. Is there any chance could you please improve that?

Link | Reply
Current rating: 5

christian 5 years, 2 months ago

Thank you for your comment: there's a great article about the north magnetic pole ("dip pole") vs the north geomagnetic pole ("dipole") here: https://www.scientificamerican.com/article/the-earth-has-more-than-one-north-pole/

The streamlines are plotted according to Matplotlib's own algorithm and where they converge some tend to get dropped. I think the overall impression is clear enough, though.

Link | Reply
Current rating: 5

pilis 5 years, 1 month ago

If I have an electric field that has E_r and E_theta components but E_phi = 0 (the potential has azimuthal symmetry), can it be plotted in the same way?

Why graph it in x and y but not in x and z for example? Why does theta become the azimuthall angle?

(Srry for my English)
Thx :D

Link | Reply
Current rating: 5

pilis 5 years, 1 month ago

Why did you change spherical coordinates to polar?

Link | Reply
Current rating: 5

christian 5 years, 1 month ago

I used polar coordinates because of the cylindrical symmetry: once you agree that the azimuthal component of the magnetic field (expressed in spherical polar coordinates) is constant, then it's really a two-dimensional problem with coordinates r and theta. It doesn't matter much how you choose to label the Cartesian plane that these coordinates are transformed onto. I chose (x, y) where, unfortunately, y coincided with the old x (hence North pointed to the right) and x with the old z (hence the azimuthal / polar confusion). I think I might choose differently if I did it again...

Link | Reply
Current rating: 5

Ambar 6 months, 1 week ago

Hello, I'm trying to model the deformation of the Earth's magnetic field due to the solar wind.
So I'm just starting to model the field using your code, and I wanted to know if you had any idea how to include in your code (I don't know maybe in the equation of Br and Btheta or in the transformation to Cartesian coordinates...) this distortion. I'm really trying, but it's a bit difficult.
And that's what I wanted to do with a simplified template like yours.
If maybe you have a idea or a recommendation, that would be really great !
Thank you very much!!!

Link | Reply
Currently unrated

christian 5 months, 1 week ago

I think in general this would be quite complex, though simple models for the magnetosphere have been proposed, e.g. https://www.physics.purdue.edu/~lyutikov/Liter/1979_jgr_4405.pdf

Link | Reply
Currently unrated

Pawel J 5 months, 1 week ago

Sorry for bothering, but I see what Zion_samurai meant in his comment 4 years ago. Indeed the field in this model is reversed in comparison to the Earth's one - the magnetic field lines go from south pole to the north pole, not the other way like in the picture above. I guess that this is just a "normal" dipole model, while the Earth is reversed (the northern magnetic pole is actually at the South Pole geographically, see e.g. Geomagnetic pole article in wikipedia)

Link | Reply
Currently unrated

christian 5 months, 1 week ago

What makes you both think that the north pole is "up"? 😛

Link | Reply
Current rating: 5

New Comment

required

required (not published)

optional

required