The image halfway through the article says the forces are "twisting shear forces," which "twist one way [or] the other"—only two ways to twist!

Maybe by "twisting" the author means that the field is one of torques rather than of linear forces. I guess you can make a continuous field of torques tangent to the surface of a sphere (as long as you're speaking of the "wheel" of the torque, not its pseudovector axis, being tangent to the sphere).

In addition, you can only speak of two "ways" any particular torque in such a field can go: clockwise or counterclockwise, as viewed from, say, a point inside the sphere. That would explain the one-way-or-the-other language.

It's not applicable. The theorem applies to the boundary of 1+2n dimensional balls - surface of an ordinary sphere, bulk of a 5-ball, 6-surface of a 7-ball, etc.

Why does the proton not meet this criteria?

I see where my post above was unclear. For the theorem to apply to a 1+2n dimensional object, the vector field on the 2n dimensional surface of the object must be restricted to the surface - it must be tangent to the object everywhere.

The proton is fully 3-dimensional AFAICT so the vector field on the surface (if it has a surface, I'm not a physicist) can have non-tangent components, pointing inwards or outwards.

Below quote from your wiki link. I’m not a graphics guy, but would appreciate if someone with experience in computer graphics could please give an example of this common problem.

“A common problem in computer graphics is to generate a non-zero vector in R3 that is orthogonal to a given non-zero vector. There is no single continuous function that can do this for all non-zero vector inputs.”

Humans have an intuitive understanding of "up", so there's usually a single obvious way to orient a camera when taking photos. However, what happens when you point a camera straight up or down? There's no longer an obvious choice, any direction you choose is reasonable (orthogonal)!

Another way to think about it is assigning cardinal directions to the Earth. Which way is north from the north pole? There's no possible way to create a map that has defined directions at every point.

The pictures in the Wikipedia article give a great intuitive understanding, particularly if you can figure out why a sphere and torus behave differently. (You can build a globally consistent map on a torus.)

> (You can build a globally consistent map on a torus.)

No, you can not, there is no global map on the torus.

My language may be imprecise.

What I mean is that you can assign a direction to each point of the torus, and have it be consistent with it's neighbors (free of discontinuities) throughout the entire surface. This is in contrast to a sphere, which will always have tufts (poles) at at least one point.

Note that this only applies within the surface itself, not to it's embedding in 3d space (the donut shape we're all familiar with). If north points up on the outside edge, it'll point down to us on the inside edge, but an ant on the surface would experience no contradictions.

https://en.wikipedia.org/wiki/Torus#/media/File:Torus_cycles...

Is this a consequence of tan(90) being undetermined? (approaches +infinity from one side and -infinity from the other)

i might be misunderstanding , but it seems easy if you want a vector orthogonal to A , generate a random vector B non co-linear to A and take AxB (cross product). AxB is orthogonal to A .

Right: and that's not a single, continuous function that works for all inputs—specifically, it fails on the input B itself. BxB = 0.

Any solution will have a discontinuity in its output vector angles. I don't know how this problem is applied in computer graphics, but you probably want to avoid rendering objects in the vicinity of a discontinuity: you'd get some kind of flickering artifact when you cross it, with small ɛ-displacements being amplified into something much larger.

This does not work on all non-zero vectors, hence your "non co-linear" comment. If the vectors point in the same direction, the cross product is zero, and you have an extra degree of freedom when choosing your "up" vector.

What example of what common problem? That's a true statement, and its proof is right above it, in the wiki page?

> There is no single continuous function that can do this

Nitpick: that should be "no single continuous deterministic function"; it's (relatively) very easy to sample uniformly randomly from the unit circle orthogonal to a given non-zero vector, but that won't give, for example, approximately the same result on two consecutive video frames, such that you could usefully orient the camera with that direction "up".

How is that continuous? Sure, there’s a trivial continuous mapping from a vector to the set of vectors orthogonal to it, but that’s hardly relevant to the problem st hand. Anyway, functions are deterministic by definition unless explicitly specified otherwise in some context.

If I understand their diagram correctly I would guess that it is somewhat nuanced. Those shear forces are probably related to the internal angular momentum of the proton. But in quantum mechanics you cannot precisely measure the axis of a particle's angular momentum. You can only measure the total magnitude and the component along one axis. Because of this there wouldn't be any regions you can point to that are "poles" where there is no angular momentum.

The Toroidal Physical Model

It's not very clear and this is not my area, but I think it's relevant that the proton has spin != 0. [1]

<guess> I think that the graphic assumes that the spin on the proton is pointing up (perpendicular to the sheet of paper) and the forces that are drawn are parallel to the "equator". In the "north pole"and "south pole" there are no forces.</guess>

[1] The spin is 1/2, but I guess the exact value is not important for this, only that it's not null.

The article says it has been discovered that it is lower than 1/2 but above 0. No idea what this means.