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They show the forces tangential to the surface of the proton, going around one way near the "surface" and the other way in the middle. However, the hairy ball t

They show the forces tangential to the surface of the proton, going around one way near the "surface" and the other way in the middle. However, the hairy ball theorem says there must be something like "poles" in this case. https://en.wikipedia.org/wiki/Hairy_ball_theorem https://en.wikipedia.org/wiki/Hairy_ball_theorem I'm wondering if their proton map covers that, and if the "axis" corresponds to anything familiar.

jeffwass 3y ago
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.”
- arbitrandomuser 3y ago
  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 .
  - perihelions 3y ago
    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.
  - itishappy 3y ago
    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.
- itishappy 3y ago
  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.)
  - bmacho 3y ago
    > (You can build a globally consistent map on a torus.) No, you can not, there is no global map on the torus.
    - itishappy 3y ago
      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.svg https://en.wikipedia.org/wiki/Torus#/media/File:Torus_cycles...
- bmacho 3y ago
  What example of what common problem? That's a true statement, and its proof is right above it, in the wiki page?
- Aerroon 3y ago
  Is this a consequence of tan(90) being undetermined? (approaches +infinity from one side and -infinity from the other)
- a1369209993 3y ago
  > 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".
  - Sharlin 3y ago
    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.
owendlamb 3y ago
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.
antognini 3y ago
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.
neovialogistics 3y ago
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.
- vecter 3y ago
  Why does the proton not meet this criteria?
  - neovialogistics 3y ago
    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.
pyinstallwoes 3y ago
The Toroidal Physical Model
gus_massa 3y ago
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.
- deleted 3y ago
  [deleted]

jeffwass 3y ago

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.”

arbitrandomuser 3y ago
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 .
- perihelions 3y ago
  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.
- itishappy 3y ago
  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.
itishappy 3y ago
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.)
- bmacho 3y ago
  > (You can build a globally consistent map on a torus.) No, you can not, there is no global map on the torus.
  - itishappy 3y ago
    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.svg https://en.wikipedia.org/wiki/Torus#/media/File:Torus_cycles...
bmacho 3y ago
What example of what common problem? That's a true statement, and its proof is right above it, in the wiki page?
Aerroon 3y ago
Is this a consequence of tan(90) being undetermined? (approaches +infinity from one side and -infinity from the other)
a1369209993 3y ago
> 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".
- Sharlin 3y ago
  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.

owendlamb 3y ago

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.

antognini 3y ago

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.

neovialogistics 3y ago

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.

vecter 3y ago
Why does the proton not meet this criteria?
- neovialogistics 3y ago
  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.

pyinstallwoes 3y ago

The Toroidal Physical Model

gus_massa 3y ago

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.

deleted 3y ago
[deleted]