[math-fun] 3D version of central angle theorem
Is there a version of the central angle theorem that applies to spheres and solid angles? The central angle theorem (in one formulation) tells us that if C is a circle containing the point p and C' is a circle centered at p, then projection from C to C' through p halves all angles. if S is a sphere containing the point p and S' is a sphere centered at p, does projection from S to S' through p halve all solid angles? By the way, I spent about three minutes fruitlessly searching the internet for an answer, and I figure that that's enough, since one of you has probably thought about this before and can answer my question in under three minutes. But if some of you think that I (and other posters) should do more googling before we post our questions, please let me know. Jim Propp
On Dec 30, 2015, at 8:56 PM, James Propp <jamespropp@gmail.com> wrote:
The central angle theorem (in one formulation) tells us that if C is a circle containing the point p and C' is a circle centered at p, then projection from C to C' through p halves all angles.
For some reason I'm having trouble imagining this projection properly. Is this "central angle theorem" the same as the theorem that says: An angle inscribed in a circle subtends a central angle that's twice its size ? —Dan
Dan, Yes, your central angle theorem is the same as my central angle theorem; I was reformulating it in a non-standard (and, I now see, confusing) way that I thought would be likely to generalize. Maybe I should just ask the question of whether there's ANY formulation of the central angle theorem that extends to solid angles... Jim On Thursday, December 31, 2015, Dan Asimov <asimov@msri.org> wrote:
On Dec 30, 2015, at 8:56 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
The central angle theorem (in one formulation) tells us that if C is a circle containing the point p and C' is a circle centered at p, then projection from C to C' through p halves all angles.
For some reason I'm having trouble imagining this projection properly.
Is this "central angle theorem" the same as the theorem that says:
An angle inscribed in a circle subtends a central angle that's twice its size
?
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Just now trying a right circular cone, 45-degrees apart from its axis, whose vertex lies on a unit sphere. If the axis of the cone is normal to the sphere, it subtends a hemisphere. But at the other extreme, if the axis of the cone is tangent to the sphere, the cone subtends less than a full hemisphere. So there can't be a simple relationship. —Dan P.S. Everytime I'm reminded of the inscribed angle theorem, I am struck by how counter-intuitive it is. Not so much the factor of 2 per se, but the fact that no matter how a given inscribed angle is situated, it always subtends the same amount of central angle.
On Dec 31, 2015, at 5:45 AM, James Propp <jamespropp@gmail.com> wrote:
Yes, your central angle theorem is the same as my central angle theorem; I was reformulating it in a non-standard (and, I now see, confusing) way that I thought would be likely to generalize.
Maybe I should just ask the question of whether there's ANY formulation of the central angle theorem that extends to solid angles...
On Thursday, December 31, 2015, Dan Asimov <asimov@msri.org> wrote:
On Dec 30, 2015, at 8:56 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
The central angle theorem (in one formulation) tells us that if C is a circle containing the point p and C' is a circle centered at p, then projection from C to C' through p halves all angles.
For some reason I'm having trouble imagining this projection properly.
Is this "central angle theorem" the same as the theorem that says:
An angle inscribed in a circle subtends a central angle that's twice its size
?
In two dimensions, another way of thinking about the central angle theorem is: "Q: Given points A and B, what is the locus of all points C such that the size of angle ACB is a given theta? A: The locus is two arcs of the two circles with centers O and O', where angles AOB and AO'B are size 2*theta, and O and O' are on the perpendicular bisector of AB." This question can certainly be generalized to three dimensions. Probably the most natural way is to ask: "Given points A,B,C, what is the locus of all points D such that the solid angle at D subtended by ABC has measure theta?" The answer is evidently not of the form "All points on some piece of some sphere", but it is certainly *some* surface... --Michael On Thu, Dec 31, 2015 at 1:26 PM, Dan Asimov <asimov@msri.org> wrote:
Just now trying a right circular cone, 45-degrees apart from its axis, whose vertex lies on a unit sphere.
If the axis of the cone is normal to the sphere, it subtends a hemisphere.
But at the other extreme, if the axis of the cone is tangent to the sphere, the cone subtends less than a full hemisphere.
So there can't be a simple relationship.
—Dan
P.S. Everytime I'm reminded of the inscribed angle theorem, I am struck by how counter-intuitive it is. Not so much the factor of 2 per se, but the fact that no matter how a given inscribed angle is situated, it always subtends the same amount of central angle.
On Dec 31, 2015, at 5:45 AM, James Propp <jamespropp@gmail.com> wrote:
Yes, your central angle theorem is the same as my central angle theorem; I was reformulating it in a non-standard (and, I now see, confusing) way that I thought would be likely to generalize.
Maybe I should just ask the question of whether there's ANY formulation of the central angle theorem that extends to solid angles...
On Thursday, December 31, 2015, Dan Asimov <asimov@msri.org> wrote:
On Dec 30, 2015, at 8:56 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
The central angle theorem (in one formulation) tells us that if C is a circle containing the point p and C' is a circle centered at p, then projection from C to C' through p halves all angles.
For some reason I'm having trouble imagining this projection properly.
Is this "central angle theorem" the same as the theorem that says:
An angle inscribed in a circle subtends a central angle that's twice its size
?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.
Instead of the three points A,B,C, consider a closed loop C. What is the family of surfaces such that for all points P on a surface, the solid angle C subtends at P is constant? If C is planar and encloses an infinitesimal area dA, then the solid angle is (d Omega) = (dA cos(theta))/r^2, where r is the distance between C and P, and theta is the inclination angle of dA as seen from P. The family of surfaces is thus cos(theta)/r^2 = constant, or r = constant * sqrt(cos(theta)). Note that, in contrast, r= constant * cos(theta) is a sphere. Now suppose dA is filled with an electric dipole layer. The potential of a single charge is 1/r, and the potential of a dipole is the gradient of this along the dipole axis. But this is just cos(theta)/r^2, and the strength of the dipole is proportional to dA, so the potential at a point P is proportional to the solid angle subtended by dA as seen from P. For a finite curve bounding an area filled with a constant dipole layer, just integrate over the area, and the potential is again proportional to the total solid angle. The electric field is the gradient of the potential. Alternatively, consider the infinitesimal curve to be carrying a current. This is a magnetic dipole, and the magnetic field has the same spatial dependence as in the case of the electric dipole. And the magnetic field is the gradient of a potential (called the scalar magnetic potential) which again is proportional to the solid angle (d Omega). A finite curve C can be filled by little current loops, each carrying the same current. The internal currents cancel, leaving only the current flowing through C. The magnetic field is the gradient of a potential proportional to the solid angle subtended by C, without regard as to the choice of surface that C bounds. The electric potential is single valued, but discontinuous across the dipole layer. In the magnetic case, there is no distinguished surface, and the potential is multivalued along a path that links the curve. The surfaces of constant solid angle are equipotential surfaces, and so are orthogonal to the field lines. Up close to the curve, the curve looks like a straight line, and the magnetic field consists of circles centered on C. The orthogonal surfaces are planes through C. Thus, in general, the surfaces of constant solid angle pass through C. -- Gene From: Michael Kleber <michael.kleber@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, December 31, 2015 7:14 PM Subject: Re: [math-fun] 3D version of central angle theorem In two dimensions, another way of thinking about the central angle theorem is: "Q: Given points A and B, what is the locus of all points C such that the size of angle ACB is a given theta? A: The locus is two arcs of the two circles with centers O and O', where angles AOB and AO'B are size 2*theta, and O and O' are on the perpendicular bisector of AB." This question can certainly be generalized to three dimensions. Probably the most natural way is to ask: "Given points A,B,C, what is the locus of all points D such that the solid angle at D subtended by ABC has measure theta?" The answer is evidently not of the form "All points on some piece of some sphere", but it is certainly *some* surface... --Michael
On 31/12/2015 04:56, James Propp wrote:
Is there a version of the central angle theorem that applies to spheres and solid angles?
I don't think so. Draw a picture like this. You have the unit sphere, centre O. You have an area element dA somewhere on it, say at P. You have a point Q somewhere else on the sphere. Let alpha be angle OPQ (= angle OQP). The solid angle subtended at O is (by definition) dA. The solid angle subtended at Q is smaller because of two factors: - The distance is 2 cos alpha instead of 1. - The area element's normal isn't along PQ but at an angle alpha to it. This means that the ratio of subtended angles is 1/(2 cos alpha)^2 from the first factor, times cos alpha from the second factor. The cosine factors don't cancel the way they do in two dimensions; you get 1 / 4 cos alpha. -- g
Intuitive picture: You're inside a big sphere on the equator. You run the center of your conical flashlight beam along the equator. By the 2D thm, the spot has constant width. But the height varies, being circular only at the antipode. --rwg On 2015-12-31 06:34, Gareth McCaughan wrote:
On 31/12/2015 04:56, James Propp wrote:
Is there a version of the central angle theorem that applies to spheres and solid angles?
I don't think so.
Draw a picture like this. You have the unit sphere, centre O. You have an area element dA somewhere on it, say at P. You have a point Q somewhere else on the sphere. Let alpha be angle OPQ (= angle OQP).
The solid angle subtended at O is (by definition) dA. The solid angle subtended at Q is smaller because of two factors:
- The distance is 2 cos alpha instead of 1. - The area element's normal isn't along PQ but at an angle alpha to it.
This means that the ratio of subtended angles is 1/(2 cos alpha)^2 from the first factor, times cos alpha from the second factor. The cosine factors don't cancel the way they do in two dimensions; you get 1 / 4 cos alpha.
That's very helpful. Thanks! Jim On Fri, Jan 1, 2016 at 11:31 AM, rwg <rwg@sdf.org> wrote:
Intuitive picture: You're inside a big sphere on the equator. You run the center of your conical flashlight beam along the equator. By the 2D thm, the spot has constant width. But the height varies, being circular only at the antipode. --rwg
On 2015-12-31 06:34, Gareth McCaughan wrote:
On 31/12/2015 04:56, James Propp wrote:
Is there a version of the central angle theorem that applies to spheres
and solid angles?
I don't think so.
Draw a picture like this. You have the unit sphere, centre O. You have an area element dA somewhere on it, say at P. You have a point Q somewhere else on the sphere. Let alpha be angle OPQ (= angle OQP).
The solid angle subtended at O is (by definition) dA. The solid angle subtended at Q is smaller because of two factors:
- The distance is 2 cos alpha instead of 1. - The area element's normal isn't along PQ but at an angle alpha to it.
This means that the ratio of subtended angles is 1/(2 cos alpha)^2 from the first factor, times cos alpha from the second factor. The cosine factors don't cancel the way they do in two dimensions; you get 1 / 4 cos alpha.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Isn't there also a "circular" height at the other extreme — where one edge of the conical beam is tangent to the sphere, at least for the case of the cone (one nappe) being pi/4 from its axis. —Dan
On Jan 1, 2016, at 8:31 AM, rwg <rwg@sdf.org> wrote:
Intuitive picture: You're inside a big sphere on the equator. You run the center of your conical flashlight beam along the equator. By the 2D thm, the spot has constant width. But the height varies, being circular only at the antipode. --rwg
On 2015-12-31 06:34, Gareth McCaughan wrote:
On 31/12/2015 04:56, James Propp wrote:
Is there a version of the central angle theorem that applies to spheres and solid angles?
I don't think so. Draw a picture like this. You have the unit sphere, centre O. You have an area element dA somewhere on it, say at P. You have a point Q somewhere else on the sphere. Let alpha be angle OPQ (= angle OQP). The solid angle subtended at O is (by definition) dA. The solid angle subtended at Q is smaller because of two factors: - The distance is 2 cos alpha instead of 1. - The area element's normal isn't along PQ but at an angle alpha to it. This means that the ratio of subtended angles is 1/(2 cos alpha)^2 from the first factor, times cos alpha from the second factor. The cosine factors don't cancel the way they do in two dimensions; you get 1 / 4 cos alpha.
On 2016-01-01 12:44, Dan Asimov wrote:
Isn't there also a "circular" height at the other extreme -- where one edge of the conical beam is tangent to the sphere, at least for the case of the cone (one nappe) being pi/4 from its axis.
--Dan I think not. (Therefore I aren't) If the curve lies on the cone, it would develop a corner to fit the apex as it approached, no? gosper.org/sphere+cone.png [1] --rwg
On Jan 1, 2016, at 8:31 AM, rwg <rwg@sdf.org> wrote:
Intuitive picture: You're inside a big sphere on the equator. You run the center of your conical flashlight beam along the equator. By the 2D thm, the spot has constant width. But the height varies, being circular only at the antipode. --rwg
On 2015-12-31 06:34, Gareth McCaughan wrote:
On 31/12/2015 04:56, James Propp wrote:
Is there a version of the central angle theorem that applies to spheres and solid angles?
I don't think so. Draw a picture like this. You have the unit sphere, centre O. You have an area element dA somewhere on it, say at P. You have a point Q somewhere else on the sphere. Let alpha be angle OPQ (= angle OQP). The solid angle subtended at O is (by definition) dA. The solid angle subtended at Q is smaller because of two factors: - The distance is 2 cos alpha instead of 1. - The area element's normal isn't along PQ but at an angle alpha to it. This means that the ratio of subtended angles is 1/(2 cos alpha)^2 from the first factor, times cos alpha from the second factor. The cosine factors don't cancel the way they do in two dimensions; you get 1 / 4 cos alpha.
Links: ------ [1] http://ma.sdf.org/gosper.org/sphere+cone.png
The solid nappe of that cone intersects the sphere in a region that includes one semicircle, in fact a semicircle one of whose endpoints is exactly at the cusp ("corner") you mention. If the cone has its apex at (0, 0, -1) and is tangent to the plane z = -1, then this semicircle will have its endpoints at (0, 0, -1) and (0, 0, 1), with that cusp at (0, 0, -1). —Dan
On Jan 2, 2016, at 1:56 AM, rwg <rwg@sdf.org> wrote:
On 2016-01-01 12:44, Dan Asimov wrote:
Isn't there also a "circular" height at the other extreme -- where one edge of the conical beam is tangent to the sphere, at least for the case of the cone (one nappe) being pi/4 from its axis.
I think not. (Therefore I aren't) If the curve lies on the cone, it would develop a corner to fit the apex as it approached, no? gosper.org/sphere+cone.png [1] --rwg
On Jan 1, 2016, at 8:31 AM, rwg <rwg@sdf.org> wrote:
Intuitive picture: You're inside a big sphere on the equator. You run the center of your conical flashlight beam along the equator. By the 2D thm, the spot has constant width. But the height varies, being circular only at the antipode. --rwg
On 2015-12-31 06:34, Gareth McCaughan wrote:
On 31/12/2015 04:56, James Propp wrote:
Is there a version of the central angle theorem that applies to spheres and solid angles?
I don't think so. Draw a picture like this. You have the unit sphere, centre O. You have an area element dA somewhere on it, say at P. You have a point Q somewhere else on the sphere. Let alpha be angle OPQ (= angle OQP). The solid angle subtended at O is (by definition) dA. The solid angle subtended at Q is smaller because of two factors: - The distance is 2 cos alpha instead of 1. - The area element's normal isn't along PQ but at an angle alpha to it. This means that the ratio of subtended angles is 1/(2 cos alpha)^2 from the first factor, times cos alpha from the second factor. The cosine factors don't cancel the way they do in two dimensions; you get 1 / 4 cos alpha.
Links: ------ [1] http://ma.sdf.org/gosper.org/sphere+cone.png _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (7)
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rwg