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Seeing this statement in this answer

There are at least two spots on Earth where the magnetic field of the planet is not horizontal, but is vertical.

has got me thinking. I don't mean to quote it out of context, I'm using it as a spring board for a different situation.

For a realistic spherical harmonic model of the Earth's field (e.g. World Magnetic Model of degree and order 12), is it a mathematical necessity that there be two points on an spherical approximation of the Earth's surface where the direction of the field is normal to the sphere?

What about if an ellipsoidal surface was chosen to better reflect the shape of the Earth's surface? Would there still be at least two points that were normal?

note: I'm not using the word "vertical" in the expanded statement of the question as it would then require a second model of the Geopotential, and then different people would choose differently if a term reflecting a pseudo potential reflecting rotation should be included or not in the definition of "vertical".


below: "Geodynamo Between Reversals", from here to illustrate that the source of the Earth's surface field is something that should not be thought of as a simple dipole.

enter image description here

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  • $\begingroup$ One could also ask: Must there necessarily be one point where the $B$-field vector is upwards directed (angle with outer-pointing surface normal is zero), and one point where the $B$-field vector points down (angle with outer surface normal is 180 degrees or $\pi$)? $\endgroup$ – Jeppe Stig Nielsen Jul 20 '17 at 14:29
  • $\begingroup$ Another question we could ask is Are there necessarily always at most two points where the Earth's magnetic field is vertical? As an example, Wikipedia's North Magnetic Pole states: "The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, ..." Why on Earth(!) is there only one? $\endgroup$ – Jeppe Stig Nielsen Jul 20 '17 at 15:02
  • $\begingroup$ @JeppeStigNielsen you can consider posting your questions as proper stackexchange questions, rather than as comments under my question. That way you can receive some feedback on your thoughts. $\endgroup$ – uhoh Jul 20 '17 at 15:45
  • $\begingroup$ I did that. By the way, your question is somewhat related to Must every magnetic configuration have a north and south pole?. $\endgroup$ – Jeppe Stig Nielsen Jul 21 '17 at 7:58
  • $\begingroup$ If you assume that the field is due to magnetic dipoles located deeper than (say) 1 metre below the surface then the situation mentioned by @ACuriousMind cannot occur. The "two zeros" question is then a well defined mathematical question - you could ask it on math.SE . $\endgroup$ – Keith McClary Jul 21 '17 at 14:01
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Your question is related to the hairy ball theorem, which states that every continuous tangent vector field defined on a sphere must have a zero. Let's consider the tangent part of the magnetic field, $\mathbf{B}_T$: according to the previous theorem, there is a point on Earth where $\mathbf{B}_T = 0$, or in other words there is a point on Earth where the magnetic field is normal to the surface of Earth.

We just obtained a lone point where the geomagnetic field is normal, but why should there be a second point where this property is satisfied? Here, the symmetry of Earth comes into play: the geomagnetic field is roughly symmetrical around the plane of the equator (slightly turned to match the inclination of the magnetic poles), and here the magnetic field is almost tangent so the first zero can't be in this plane. Thus, the symmetrical point of our first point is another point where the geomagnetic field is normal to Earth.

If we changed the model of Earth's surface, the result would remain the same: indeed, the hairy ball theorem can be demonstrated for shapes similar from a topological point of view to a sphere.


Notice that in the second paragraph, I used a physical argument instead of a purely mathematical one: indeed, the result is not true in the general case. One can find a continuous vector field that is vertical on a sphere in only one point, see this question on Math.SE for example. Thus, generally speaking, the geomagnetic field could be vertical in only one point.

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    $\begingroup$ The statement that there are two zeroes of the magnetic field does not follow from the hairy ball theorem, and in fact is in general false if the only restriction is that we're considering a field on a sphere. Put an infinitesimal dipole on the sphere - the resulting field has only one zero, namely at the location of the dipole, cf. this graphic on Wikipedia. If the statement in the question is true, then it has to have to do with special properties of the earth's magnetic field, and not just the hairy ball theorem. $\endgroup$ – ACuriousMind Jul 20 '17 at 9:14
  • $\begingroup$ You're totally right ! And, as you said, I used a special property of the geomagnetic field which is its symmetry. $\endgroup$ – Spirine Jul 20 '17 at 9:52
  • $\begingroup$ @Spirine sorry, I just want to double check that the real answer is the first paragraph, and that the physical discussion in the second paragraph reverts to a different, simplified problem where there is a symmetry that does not actually exist in the question? The distinction between the two is my question's motivation! You know even a dipole field would have to be offset from the center of the Earth to obtain the best fit to the field . See this answer and South Atlantic Anomaly. $\endgroup$ – uhoh Jul 20 '17 at 11:03
  • $\begingroup$ Mathematically I think you can prove this using the Intermediate Value Theorem if there's a point where the direction is pointing "down" and one where it's "up", and it's continuous and differentiable, then it must pass through horizontal. $\endgroup$ – davidsheldon Jul 20 '17 at 14:22
  • $\begingroup$ @davidsheldon The intermediate value theorem doesn't work that way for vector-valued functions. $\endgroup$ – ACuriousMind Jul 20 '17 at 21:20

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