steveupson

Additional Question About Surfaces in Higher Dimensions

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3 hours ago, Strange said:

...There are Euclidean models of space-time. They only work locally, where the curvature can be assumed to be zero. They do not work on cosmological scales (where the curvature cannot be ignored).

I would suggest you take some time to study differential geometry. As I gather you are a genius it should only take two or three years of full time study to get to grips with the basics.

Calculus has no impact on this issue.  It's strictly plane and spherical trigonometry. Believe me, if calculus were required then there would be calculus in the equations.

The quantity of direction is something that you yourself have graphed on different occasions, although you remain unaware of the significance of those graphs.  The area under each 2D curve for different upsilon angles is a quantity that expresses the amount of direction that angle possesses in 3D space.  It is impossible to understand these words until the math is properly understood. 

For a sine curve (which occurs with [latex]\upsilon=\frac{\pi}{2}[/latex]) the quantity is [latex]\frac{\sqrt 2}{2}[/latex].  When the curves are "normalized" in order to eliminate the length differences associated with the radius of the small circle compared to the radius of the sphere (similar to the way standard vectors are normalized in order to generate direction vectors), then the area under each unique curve is properly scaled to represent the unique 2D quantity that represents that portion of a 3D turn.

This particular 2D (graphical) representation of the 3D direction (or angle or orientation) is very different from what we are used to in that it represents the mathematical quantity or magnitude that is associated with each change of direction in normal space.  It is mathematically a quantity or a scalar, and it can also be thought of that way in order to better understand the relationships that exist. 

1 hour ago, studiot said:

rIsn't this veering off topic?

 

A surface is a two dimensional object, insofar as it may or may not be embedded in a space of higher dimension.

You have referred to projective geometry.

One of the aspects of PG is that it is indifferent to distance.

That is why a diagram of Desargues Theorem works.

http://mathworld.wolfram.com/DesarguesTheorem.html

 

 

Yes, surfaces are two-dimensional objects.  The methods that we generally use in order to specify and manipulate the mathematical representation of direction in these two dimensional objects are called plane trigonometry.  

There is another type of trigometry that isn't plane geometry, called spherical trigonomety and it is used in order to examine triangles on a sphere or ball, or in other words it can be used to establish geometric relationship in either two or three dimensions. We're concerned with the interplay between these.  Normally, the sphere and ball have the same trigonometric identities.  They are actually very different mathematically or geometrically.  The ball has a center which makes it a three-dimensional object.  Therefore there are three-dimensional identities associated with it that cannot be associated with a sphere.

On 12/2/2017 at 5:05 AM, studiot said:


sin2ν+cos2ν=1


Is a trigonometric identity since there are no values of nu that do not satisfy the equality.

So is


sin(α+β)=sinαcosβ+cosαsinβ


again since there are no (pairs of) values of alpha and beta that fail to satisfy the equality.

 

But there are still values of (triples of) alpha, phi and lambda that fail to satisfy your equation 6.

 

So it is still not an identity.

For instance put


α=π4φ=π2λ=π2


Then your expression 6 becomes


(2424)2+(124)2=341

 

 

Returning to this issue for a minute, maybe someone who knows more about this stuff than I do can explain something to me.

We have

[latex]\sin{\upsilon}= \frac{\sin\frac{\lambda}{2}}{\cos\frac{\phi}{2}}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\:\:\:\:(4)[/latex]

and

[latex]\cos{\upsilon}= \frac{\cot\alpha}{\cot\frac{\phi}{2}}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\:\:\:\:(5)[/latex]

 

and since,

[latex]\sin^2 \upsilon+\cos^2\upsilon=1[/latex]

then we should get a new spherical trigonometry identity:

[latex](\cos\frac{\phi}{2}\,\sin\frac{\lambda}{2})^2 +(\cot\frac{\phi}{2}\,\cos\alpha)^2=1\quad\quad\:\:\:(6)[/latex]

 

Why isn't this all true?  Did we break trigonometry or did we break algebra?  What's up with this? Anyone?

Edited by steveupson

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3 hours ago, steveupson said:

A turn in three dimensions is mathematically different than a turn in two dimensions.

I wonder if this is a key to your difficulty.

I have hastily dashed of some sketches as I had to sneak onto the scanner. Sorry for the lack of quality.

Each of the three shows an angle as described between two straight lines in different orientations relative to ayz coordinate axes.

Each sketch shows what happens as you move the pair of lines and their included angle relative to each axis, but wihtout changing the angle itself or the relationship between the pair of lines.

I wondered if this had any bearing on what you are trying to describe.

anglerots.thumb.jpg.d43ffff965545def133277ec51b8f29c.jpg

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3 hours ago, studiot said:

rIsn't this veering off topic?

 

A surface is a two dimensional object, insofar as it may or may not be embedded in a space of higher dimension.

You have referred to projective geometry.

One of the aspects of PG is that it is indifferent to distance.

That is why a diagram of Desargues Theorem works.

http://mathworld.wolfram.com/DesarguesTheorem.html

 

 

In your link ,is the larger triangle a projection of the smaller triangle onto a plane containing the larger triangle?

 

And RST lies in the same plane?(as also does ABR)

 

You can "stack up" triangles in all sorts of orientations and any combination of angles between the two triangles  (without constraints)?

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The issue does not concern any single angle because a single angle occurs in 2D and we are talking about what happens in 3D.

The relationship is about two non-coplanar angles in 3D.

We solve the relationship by using a third angle that relates to both of the two angles independently.

A fourth angle is the primary angle of concern.  It is the angle that determines which member of the family of functions will be represented by the first two angles.

This is a 3D construction and there is no prior art in this genre of identities.

1 hour ago, geordief said:

In your link ,is the larger triangle a projection of the smaller triangle onto a plane containing the larger triangle?

 

And RST lies in the same plane?(as also does ABR)

 

You can "stack up" triangles in all sorts of orientations and any combination of angles between the two triangles  (without constraints)?

The only constraint will the plane in which the additional triangle lies.  The plane that it is in will be determined by the length of the sides (or the angle of the vertices as these two things correlate.)

I have been trying to figure out why the equations work the way they do.  It might be because when we collapse 3D space down to a single point it yields a different mathematical result than when we do this with 2D space, as was mentioned above in the discussion on spacetime that isn't non-Euclidean.

in the 3D animation of the 3D function, the North pole of the sphere SN is at the top of the sphere and the equator is where the tan plane intersects the sphere.  There is a point P that moves along a small circle where three more planes intersect.  There is an angle that we're calling [latex]\lambda[/latex] that is formed by SN, the sphere center SO, and point P.  There is another angle that we're calling [latex]\alpha[/latex] that is formed by the intersection of the blue and green planes. The green plane remains a longitude while the blue plane is tangent to P and contains SO.

In order to establish the relationship between [latex]\alpha[/latex] and [latex]\lambda[/latex] we must construct a third angle (not shown) that is formed by SN, the center of the small circle CO and P.  This angle [latex]\phi[/latex] is actually the arc length along the small circle from SN to P.

The fourth angle [latex]\upsilon[/latex] is formed by SN, SO, and CO. This angle is what determines the size of the small circle relative to the sphere.  (It is generally a small circle, but it can be made large enough to be a great circle, at which point the model collapses from 3D to 2D. 

NSTF_(1).gif

Edited by steveupson

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2 hours ago, steveupson said:

The issue does not concern any single angle because a single angle occurs in 2D and we are talking about what happens in 3D.

The relationship is about two non-coplanar angles in 3D.

We solve the relationship by using a third angle that relates to both of the two angles independently.

A fourth angle is the primary angle of concern.  It is the angle that determines which member of the family of functions will be represented by the first two angles.

This is a 3D construction and there is no prior art in this genre of identities.

 

 

Well I tried, even though I'm a bit sorry I bothered since you clearly didn't read my post properly.
 

What do you think the black loop curve is in your diagram?

Do you think that it could be used for spherical trigonometry?

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3 hours ago, studiot said:

 

 

Well I tried, even though I'm a bit sorry I bothered since you clearly didn't read my post properly.
 

What do you think the black loop curve is in your diagram?

Do you think that it could be used for spherical trigonometry?

Of course I read your post.  You are talking about an angle. It has no bearing at all on the math that I have presented. I think that all of us know about what you are showing in your post.  It's available in any book on rotations or basic kinematics.  Trust me that I know what you said.  You are the person who won't listen to me.

What I have shown you is not in any books.

I read every word you say.  We agree on almost everything.  Try and focus on the math that we disagree about. 

 

on edit>>> sorry, I forgot to answer your question.  The black loop is a small circle on the imaginary spherical surface that allows us to do spherical trigonometry on a 3D object (for instance a ball, or any other object, including empty space.)  The size of the small circle is related to the size of the sphere through the value or magnitude of [latex]\upsilon[/latex].  It is shown in the animation in order to help illustrate how the intersection of the three moving planes can be tracked along the spherical surface.  The slope of the tangent to this circle is [latex]\alpha[/latex].  This is the first of two angles, oriented in space, that relate to one another geometrically.  The other angle,  [latex]\lambda[/latex] is the angle PSOSN.  

Its used to solve this particular geometry by another method that does involve spherical trigonometry.  The circle is used in order to create an isosceles spherical triangle that is solved in order to find [latex]\alpha[/latex] and [latex]\lambda[/latex] in relation to [latex]\phi[/latex], which is the arc length from SN to P along the black loop curve, which is otherwise called the small circle.

Edited by steveupson

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3 minutes ago, steveupson said:

Of course I read your post.  You are talking about an angle. It has no bearing at all on the math that I have presented. I think that all of us know about what you are showing in your post.  It's available in any book on rotations or basic kinematics.  Trust me that I know what you said.  You are the person who won't listen to me.

 

Even with a much simpler shorter post you didn't read it properly.

Like you my post mentioned four angles. Further it didn't say you didn't read my post it said you didn't read it properly.

 

3 minutes ago, steveupson said:

 

What I have shown you is not in any books.

 

I didn't say it was. So what?

 

3 minutes ago, steveupson said:

 

I read every word you say.  We agree on almost everything.  Try and focus on the math that we disagree about. 

 

I also asked a question about (the mathematics of) your diagram but instead of answering you accuse me of not reading your post.

 

Considering I am not accusing you of being a crank but actually seeking to find out if you genuinely have a useful viewpoint to offer and trying to help you with it, despite your mathematical failings, this just extreme arrogance and disregard for others.

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I apologize.  I edited my previous post once I realized that I had not furnished an answer to your question.

I don't see anything new in your post at all.  Once more, what I'm showing you is new. And I'm sort of hurt that you think it is arrogance.  It's total and complete frustration.  I keep answering the same questions over and over again, using different approaches, and everyone seems to just ignore what I say and assume that I don't know what I am talking about.

What does the "identity" mean to you at this point?  Does it mean anything to you?

 

The Coordinate System

In the model as illustrated in this particular case, small circle C has a circumference that is 45º of latitude.   We will be using the conventional terminology where a circle on the surface of a sphere that is made by the intersection of the sphere with a plane passing through the sphere center is called a great circle.  All other circles, where the intersecting plane does not pass through the sphere center SO, are called small circles.

If we view the model such that the equator of sphere S is the intersection of the surface with the tan-colored plane, and where the main axis of the sphere is along the intersection of the green-colored and magenta-colored planes, then circle C can be said to be tilted 45º from the natural orientation of a circle of latitude.  In this case, circle C intersects the north pole SN and is tangent to the equator.

The center CO of circle C, together with point SO and point SN, form angle ∠COSOSN.  There is a main axis passing through SN and forming one ray of this angle which we'll call the cardinal axis, and there's another axis that passes through CO which we'll call the ordinal axis.

The relationship between these two axes is that they are 45º to one another in the instance shown, but a similar model can be constructed using a small circle that is other than a 45º latitude.   As an example, there could be a small circle with a circumference of 10º that is tilted to an angle of 10º from the horizontal, and this circle will also intersect the north pole.  In this case the angle between the cardinal and ordinal axes will be 10º.   This angle between the cardinal and ordinal axes is the upsilon angle, [latex]\upsilon[/latex].

A "casual observer" would think that the initial position of the colored rectangles in the animation represents the planes defined by the coordinate axes (the xy-, xz-, and yz-planes).  However, that’s not the correct interpretation.   The point where the rectangles meet (Fig. 2) is the center SO of unit sphere S, having position (0,0,sz), and is not, in general, the origin of the coordinate system (0,0,0).

Circle C is, in general, a small circle, and is the intersection of the xy-plane (which isn't a plane indicated by any of the animated rectangular planes) and sphere S.  Its orientation can be regarded as a circle of latitude of a specific size (in the example, 45º that has been tilted by the same angle (45º) such that it intersects north pole SN of sphere S.

Circle C can be described as the base of cone N (not shown), having sphere center SO as the apex, having the ordinal axis as the cone axis, and having the cardinal axis as a generatrix. Line COSO lies along the ordinal axis and line SNSO lies along the cardinal axis. In the model as illustrated, the angle between the axis of cone N and the axis of sphere S is 45º, giving an aperture of 90º (maximum angle between two generatrix lines) but this angle is variable.  The axis of cone N is the z-axis.  The circle center CO is the origin of the coordinate system (0,0,0).

North pole SN is facing toward the top in the animation, and lies along the x-axis, which is oriented such that the positive x-axis passes through SN.  The radius of circle C is r. Since sphere S is a unit sphere, 0 < r < 1

The y-axis is oriented such that the positive y-axis is toward the viewer in the illustration.

Figure 2.  Coordinate System

 

Figure 2.jpg

The Planes

The two planes that remain stationary in the animation each contain SO and a line tangent to circle C.  For tan-colored plane [latex]\mathbb{T}[/latex] this tangent is at (-r,0,0) and for magenta-colored plane [latex]\mathbb{M}[/latex] it is at SN which is also (r,0,0). Because these two planes contain the center of the sphere, which need not be in the xy-plane, these stationary planes are not parallel to, or perpendicular to, the xy-plane except for when r is at either limit r = 0, or r = 1.

The point on C where 3 moving planes intersect is P. Blue-colored plane [latex]\mathbb{B}[/latex] contains SO and a line tangent to the circle at point P. The animation starts with P at (r,0,0) and then moves P along a 180º arc of circle C.

Green-colored plane [latex]\mathbb{G}[/latex] contains the point P and pivots about an axis as P moves, but this axis is not one of the coordinate axes. It is an axis defined by a line containing the line segment from SO to SN, which is the main or cardinal axis of sphere S.

Yellow-colored plane [latex]\mathbb{Y}[/latex] (or light green) contains points SO and P, and is perpendicular to [latex]\mathbb{G}[/latex].

The Angles

Plane [latex]\mathbb{G}[/latex] can be considered the longitude plane because the great circle made by its intersection with sphere S is always a line of longitude. Plane [latex]\mathbb{B}[/latex] can be considered the tangent plane since it always contains a line tangent to circle C at point P. The dihedral angle between planes [latex]\mathbb{G}[/latex] and [latex]\mathbb{B}[/latex] is angle [latex]\alpha[/latex], the angle of interest.

Plane [latex]\mathbb{Y}[/latex] can be considered the elevation plane and it stays perpendicular to plane [latex]\mathbb{G}[/latex] and passes through sphere center SO and point P. The smallest angle between the cardinal axis of the sphere and plane [latex]\mathbb{G}[/latex] is angle [latex]\lambda[/latex] , or ∠SNSOP.

Point P can be defined as an arc length along circle C equal to angle

 

The Problem

The objective is to define a family of functions which are based on different values of [latex]\upsilon[/latex], and which express α as a function of [latex]\lambda[/latex] . The approach will be to find [latex]\alpha[/latex] and [latex]\lambda[/latex], each as a function of [latex]\phi[/latex].  One of the methods used here will be to solve a spherical isosceles triangle using spherical trigonometry.

At the top right corner of the animation there is a set of numbers representing [latex]\lambda[/latex] that change between 0° and 90°, while below them, in the rectangular box, there is another set of numbers representing [latex]\alpha[/latex] that also change between 0° and 90°. The mathematical model used for the animation defines a specific member of a family of functions. The z-coordinate for the center of sphere S defines which member of the family we are analyzing. The model shows [latex]\alpha[/latex] as a function of [latex]\lambda[/latex] with angle [latex]\upsilon[/latex] equal to 45°, or z = [latex]\frac{1}{\sqrt{2}}[/latex].


Also, because we are using a unit sphere,  r = sin [latex]\upsilon[/latex] = [latex]\sqrt{1-{s_z}^2}[/latex]

(Unit sphere center SO can move up or down along the z-axis and the north pole will remain the north pole and circle C will still pass though the pole SN because angle [latex]\upsilon[/latex] changes accordingly in order to keep this true.)

It should be mentioned that, although a sphere is used for construction of the model that is presented here, there actually is no sphere involved in the function itself. In other words, there is no two-dimensional surface involving spherical excess (parallel transport) or anything like that. The sphere is simply used as an aid in visualizing how the object is constructed and spherical trigonometry used in solving for some of the unknowns which can be found by analysis of the sphere’s gross occupation of Euclidean 3-space.

Lambda as a Function of Phi

Point SC is the intersection of the z-axis with sphere S such that ∠SNSOCO = [latex]\upsilon[/latex] = ∠SNSOSC. There is a spherical triangle ∆abc that, when solved, will express the relationship between [latex]\alpha[/latex], [latex]\phi[/latex], [latex]\lambda[/latex], and [latex]\upsilon[/latex] such that:

Side a is great circle arc of length [latex]\lambda[/latex]
Side b is great circle arc of length [latex]\upsilon[/latex], or ∠SCSOSN
Side c is great circle arc of length [latex]\upsilon[/latex], or ∠SCSOP


[latex]\alpha[/latex] = ∠[latex]\mathbb{BG}[/latex]
[latex]\phi[/latex] = ∠SNCOP = dihedral ∠SN-SOCO-P
[latex]\lambda[/latex] = ∠PSOSN
[latex]\upsilon[/latex] = ∠SNSOCO

Since both point P and point SN are on the unit sphere with center SO, the triangle that these 3 points form is an isosceles triangle with a pair of sides, PSO and SNSO, each having length = 1.  The angle between the two equal sides is [latex]\lambda[/latex].  If we take [latex]\lambda[/latex] as given, we can find the length h of the base of this isosceles triangle, which is the distance between SN and P.  Then, using length h we can find [latex]\phi[/latex] by regarding the solution as a 2-dimensional problem that is set in the xy-plane where circle C lies.

For the first step, h is the chord of the great circle arc [latex]\lambda[/latex] of sphere S, or h = [latex]{2}\sin\frac{\lambda}{2}[/latex].  For the second step, with a circle of radius r and center (0,0), h will be the chord length of the arc segment from (r,0) to P, or h = [latex]{2}{r}\sin\frac{\phi}{2}[/latex].  These equations give a parametric representation of [latex]\lambda[/latex] as a function of [latex]\phi[/latex].

Figure 3. Isosceles Spherical Triangle ∆abc

Figure 3.jpg

Alpha as a Function of Phi

Referring to Fig. 4, since b and c both have endpoints on circle C, Δabc is an isosceles triangle with dihedral angle [latex]\phi[/latex] between the two equal sides. If we construct a great circle arc from point SC to midpoint D of side a, we will bisect Δabc into two congruent right spherical triangles.


The bisected dihedral angle at SC[latex]=\frac{\phi}{2}[/latex], and side c = b = [latex]\upsilon[/latex]. We can solve for the remaining dihedral angle α at P, or at SN, since these are equal. Since side c is perpendicular (Fig. 4) to the great circle arc made by plane B, this dihedral angle at P will be α.

[latex]\cot\alpha=\cos\upsilon\:\tan\frac{\phi}{2}[/latex]

 

Then, solving for side a (which was bisected) we get:

[latex]\sin\frac{\lambda}{2}=\sin\frac{\phi}{2}\:\sin\upsilon[/latex]

 

Also, because we are using a unit sphere,

[latex]{r}=\sin\upsilon=\sqrt{{1}-{s_z}^2}[/latex]   (Unit sphere center SO can move up or down along the z-axis and the north pole will remain the north pole and circle C will still pass though the pole SN because [latex]\upsilon[/latex] angle changes accordingly in order to keep this true.

Figure 4. Dihedral Angle α

Figure 4.jpg

Variables

C  =  circle made by intersection of the xy-plane and unit sphere S

CO  =  (0,0,0)  =  origin =  center of circle C

r =  [latex]\sqrt{1-{s_z}^2}[/latex]  =  radius of circle C

(r,0,0)  =  intersection of positive x-axis with unit sphere S  =  SN (north pole)
(-r,0,0)  =  intersection of negative x-axis with unit sphere S
(0,r,0)  =  intersection of positive y-axis with unit sphere S
(0,-r,0)  =  intersection of negative y-axis with unit sphere S

z  =  length from origin to center of unit sphere

S  =  unit sphere

S =  (0,0, sz )  =  center of unit sphere S, where -1 < sz < 1

SN  =  (r,0,0)  =  north pole of unit sphere S

SC  =  (0,0,1-z)  =  projection of line segment between SO and CO to the surface of sphere S

N  =  cone having circle C as its base and sphere center SO as its apex

P =  (r cos[latex]\phi[/latex], r sin[latex]\phi[/latex],0)  =  tangent point on circle C where planes [latex]\mathbb{B}[/latex], [latex]\mathbb{G}[/latex], and [latex]\mathbb{Y}[/latex] intersect

h  =  arc length of circle C from P to SN
h  =  [latex]{2r}\sin\frac{\phi}{2}[/latex]   (from analyzing a chord in circle C)
h  = [latex]{2}\sin\frac{\lambda}{2}[/latex]   (from analyzing isosceles triangle ∆SNSOP, which has two sides  of length 1)

[latex]\mathbb{T}[/latex] =  plane containing point SO and also the equator of sphere S

[latex]\mathbb{M}[/latex] =  plane containing point SO and also the tangent (in the xy-plane) to circle C at point SN

[latex]\mathbb{B}[/latex] =  plane containing point SO and also the tangent (in the xy-plane) to circle C at point P

[latex]\mathbb{G}[/latex] =  plane containing point SO and point P and point SN

[latex]\mathbb{Y}[/latex] =  plane containing point SO and point P and plane [latex]\mathbb{Y}[/latex] is perpendicular to plane [latex]\mathbb{G}[/latex]

a  =  great circle arc λ
b  =  great circle arc υ
c  =  great circle arc ∠SCSOP

D =  midpoint of side a

[latex]\alpha[/latex] =  ∠[latex]\mathbb{BG}[/latex]

[latex]\phi[/latex] =  arc length in degrees from (r,0,0) to point P   =   ∠SNCOP  =  dihedral ∠SN-SOCO-P

[latex]\lambda[/latex] =  ∠PSOSN

[latex]\upsilon[/latex]  =  angle between the axis of cone N and the cardinal axis of sphere S  =   ∠SNSOCO.

Edited by steveupson

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Thank you for this, I will print it out to look at tomorrow.

On first sight it appear more rigorous than this comment in your previous diagram/post

 

6 hours ago, steveupson said:

in the 3D animation of the 3D function, the North pole of the sphere SN is at the top of the sphere and the equator is where the tan plane intersects the sphere.

 

There are no tangent planes in your diagram.

A sphere has tangent planes, a circle has tangent lines.

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3 minutes ago, studiot said:

Thank you for this, I will print it out to look at tomorrow.

On first sight it appear more rigorous than this comment in your previous diagram/post

 

 

There are no tangent planes in your diagram.

A sphere has tangent planes, a circle has tangent lines.

Right. Thanks.  I do tend to remove my genius spectacles from time to time.

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No, I hadn't seen it, but it isn't important to this problem.  In this problem the tangent plane is tangent to cone N at point P.  I think that it would be useful to apply some other method such as vector analysis or quaternions or something besides spherical trigonometry to validate the solution obtained through spherical trigonometry.  If there are discrepancies between methods then knowing what these discrepancies are would be useful.

We have already uncovered a major anomaly concerning the trigonometric identity.  We should probably work toward a better understanding of why the identity exists in 3D and not in 2D, and what the implications are concerning the spacetime question.   

Don't misunderstand me.  This information deserves great skepticism.  That's a good thing.  

 

Since everyone probably doesn't have a spherical trigonometry textbook:

textbook isoscoles spherical triangle (2).JPG

Edited by steveupson
edited to add solution for isosceles spherical triangle

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12 hours ago, steveupson said:

Returning to this issue for a minute, maybe someone who knows more about this stuff than I do can explain something to me.

We have

sinυ=sinλ2cosϕ2(4)

and

cosυ=cotαcotϕ2(5)

 

and since,

sin2υ+cos2υ=1

then we should get a new spherical trigonometry identity:

(cosϕ2sinλ2)2+(cotϕ2cosα)2=1(6)

 

Why isn't this all true?  Did we break trigonometry or did we break algebra?  What's up with this? Anyone?

I guess this is because the three variables [math]\upsilon[/math], [math]\lambda[/math], [math]\phi[/math]are not independent.

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Why should that matter?  We still have proper expressions for [latex]sin\upsilon[/latex] and [latex]cos\upsilon[/latex].  Why doesn't the identity hold true?  How does the identity know not to be in effect whenever [latex]\upsilon,\lambda,\phi[/latex] are in the same plane?

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Actually, thinking about it more, I think the trouble is that they are independent (to some extent).

I am not sure, but I suspect this is where the error is. I'm trying to think how to prove it.

The identity [math]\sin^2 \upsilon+\cos^2\upsilon=1[/math] is (obviously) only true if you have the same [math]\upsilon[/math] in both places.

You have:
[math]\sin{\upsilon}= \frac{\sin\frac{\lambda}{2}}{\cos\frac{\phi}{2}}[/math]
And:
[math]\cos{\upsilon'}= \frac{\cot\alpha}{\cot\frac{\phi}{2}}[/math]

Do we know that [math]\upsilon = \upsilon'[/math] (always)

In other words, do we know that:

[math]\arcsin{\frac{\sin\frac{\lambda}{2}}{\cos\frac{\phi}{2}}} = \arccos{\frac{\cot\alpha}{\cot\frac{\phi}{2}}}[/math]

Unless [math]\upsilon[/math] is uniquely determined by the values of [math]\lambda[/math] and [math]\alpha[/math]. And I'm not sure that is true. Can the three variables can change independently? Is the identity only true for the angles prescribed by your animation?

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10 hours ago, steveupson said:

Why should that matter?  We still have proper expressions for sin\upsilon and cos\upsilon .  Why doesn't the identity hold true?  How does the identity know not to be in effect whenever \upsilon,\lambda,\phi are in the same plane?

The analysis commits the classic error of division by zero.

Both the cosine and the cotangent are zero for angles equal to  [math]\left( {2n - 1} \right)\pi \quad :n = 1,2,3...[/math]

leading to division by zero in both the defining equations.

 

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The divide-by-zero condition only exists because the model of the object cannot be completely represented in only two dimensions.  Whenever the model is completely parameterized it must have three dimensions in order to describe it, and the divide by zero error does not occur.  This does explain why the tridentity is only an identity in three dimensions and cannot be represented in only two.

I see how the math works, but is that the reason why the function cannot be applied to surfaces?  Even surfaces of many dimensions?  I think this may be the case.  This would be the reason why tensors have not been able to accurately describe fields that alter 3-dimensional space.  They are unable to express this particular aspect of it.

 

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56 minutes ago, steveupson said:

.

I see how the math works,

 

Good to hear that.

So what happens if you shrink your black circle to a point so that each zero radius circle describes a single position on the sphere?

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On 12/2/2017 at 12:21 AM, steveupson said:

...It is a smooth function that approaches a sine curve when [latex]\upsilon\to0[/latex], and that approaches a hyperbola when [latex]\upsilon\to\frac{\pi}{2}[/latex]...

As for what happens when a geometric object becomes a point, I think it fails to exist. 

According to Euclid: "A point is that which has no part."

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2 minutes ago, steveupson said:

As for what happens when a geometric object becomes a point, I think it fails to exist. 

According to Euclid: "A point is that which has no part."

That doesn't mean it doesn't exist. Just that it is not further divisible. After all, mathematical points exist (as much as anything mathematical exists) and that is what Euclid was referring to - if they didn't exist it would have undermined much of the rest of his derivations. And electrons (for example) are point particles but they appear to exist.

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47 minutes ago, Strange said:

That doesn't mean it doesn't exist. Just that it is not further divisible. After all, mathematical points exist (as much as anything mathematical exists) and that is what Euclid was referring to - if they didn't exist it would have undermined much of the rest of his derivations. And electrons (for example) are point particles but they appear to exist.

You're not arguing that a point is a circle are you?  Because that was the question I was answering.

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4 minutes ago, steveupson said:

You're not arguing that a point is a circle are you?

No. Just pointing out that becoming a point is not the same as ceasing to exist.

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This a verbal description of what you get when you set r to zero.  The circle doesn't exist (or does it still exist as is being hinted at, because if it still exists then this answer will have to be rethought.)

When the circle does't exist then there is no arc length [latex]\phi[/latex] because this arc length is part of a nonexistent circle.

 [latex]\lambda, \upsilon, \alpha[/latex] will also fail to exist because they have either had one or two endpoints merged into the single point at the North poles SN.  

Also, since z is the distance of the center of the unit sphere from the center of the coordinate system, all we seem to have remaining is this unit sphere, without anything else to relate to anything else.  I am assuming that at this time we are left with [latex]0=f(0)[/latex].

One more time, the relationship either exists in 3D or it doesn't exist at all.  

If [latex]\upsilon[/latex] is one infinitesimal from zero the function is one infinitesimal from the sin function and when [latex]\upsilon[/latex] is one infinitesimal from [latex]\frac{\pi}{2}[/latex] then the function is one infinitesimal from the hyperbolic function.

I don't know if there are any more ways to express what is going on here.

 

on edit>>>>

probably should be:

[latex]\varnothing=F(\varnothing)[/latex]

Edited by steveupson

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55 minutes ago, steveupson said:

This a verbal description of what you get when you set r to zero.  The circle doesn't exist (or does it still exist as is being hinted at, because if it still exists then this answer will have to be rethought.)

When the circle does't exist then there is no arc length ϕ because this arc length is part of a nonexistent circle.

 λ,υ,α will also fail to exist because they have either had one or two endpoints merged into the single point at the North poles SN.  

Also, since z is the distance of the center of the unit sphere from the center of the coordinate system, all we seem to have remaining is this unit sphere, without anything else to relate to anything else.  I am assuming that at this time we are left with 0=f(0) .

One more time, the relationship either exists in 3D or it doesn't exist at all.  

If υ is one infinitesimal from zero the function is one infinitesimal from the sin function and when υ is one infinitesimal from π2 then the function is one infinitesimal from the hyperbolic function.

I don't know if there are any more ways to express what is going on here.

 

on edit>>>>

probably should be:

=F()

 

I'm glad you are thinking about the maths because you asked about surfaces and one of the difference between a sphere and a donut (or teacup or torus) is that you can shrink any circle (or any other curve that forms a single closed loop) to a point on the surface of the sphere but you can't on a torus.

 

This article should be accessible.

http://www.maths.ed.ac.uk/~aar/surgery/zeeman.pdfhttp://www.maths.ed.ac.uk/~aar/surgery/zeeman.pdf

Edited by studiot

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The concept that is being examined here in this thread is the one where it was stated that “we can establish a one-to-one mapping between any Rm and any Rn.”

The model for [latex]\alpha=f(\lambda)[/latex] is the example of [latex]R^3[/latex] that cannot be mapped to [latex]R^2[/latex], or at least that's how it appears.

The point [latex]S_O[/latex] at the origin, and unit sphere S, are all that exists in a form that can be mapped, and these are given, not derived.   There is nothing in that accessible paper that will help with this particular mapping.

Please, if it isn't too much to ask, can you tell me what you think I am saying?  We are not talking to one another.  What is it that you hear me saying?  What is it that the math is showing you?

As is the case with the circle becoming a point, the entire map vanishes in [latex]R^2[/latex].

Edited by steveupson

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