steveupson

Additional Question About Surfaces in Higher Dimensions

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36 minutes ago, steveupson said:

Correct.

It is an identity in three dimensions.

It doesn't exist in two dimensions.

 

 

You made an assertion concerning three numeric variables.

You gave them Greek letters which makes writing about them more difficult, they could just as easily be called a, b and c or x1, x2 and x3.

 

Whatever, fixing the first two has a greater impact than you think on the third variable, you call alpha.

Fixing phi and lambda to the values I gave reduces alpha to numbers which make [math]\cos \alpha [/math] zero or the contents of the second brakcet cannot reduce to the square root of 1.

There is only one value of alpha that satisfies this (along with its cyclic values).

Dimensions have no bearing upon identities. Perhaps you should look up the meaning of the term in mathematics.

Your formula is an equation not an identity.

As such it has solutions for some phi, lambda and alpha, but not all.

 

 

Edited by studiot

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When the three Greek variables exist in three dimensions (the set of Greek variables) it is an identity.

 

When the number is reduced to two (when any of the three dimensions is zero) the identity does not exist.

 

 

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2 minutes ago, steveupson said:

When the three Greek variables exist in three dimensions (the set of Greek variables) it is an identity.

 

When the number is reduced to two (when any of the three dimensions is zero) the identity does not exist.

What on Earth does this mean? How does a variable "exist in three dimensions"?

But lets assume you mean that none of the variables can be equal to zero (so the equation is not an identity).

If we set all three to 1 then the result is approximately 5.798 which, it might surprise you to know, is not equal to 1.

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there are three angles represented by Greek letters

these three angles can either exist in two dimensions (co-planar) or not

when they are not in the same plane it is an identity

 

Edited by steveupson

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3 minutes ago, steveupson said:

there are three angles represented by Greek letters

these three angles can either exist in two dimensions (co-planar) or not

when they are not in the same plane it is an identity

How is this represented in the equation? In other words how do you (or, more importantly, the equation) know if these three scalar values are co-planar or not? 

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The answer to the previous question might be contained in the methods that were used in order to solve for the underlying function.  The model was solved for a family of functions with a fourth variable [latex]\upsilon[/latex].  In order to compose the function [latex]\alpha=f(\lambda)[/latex], this fourth variable was held constant. The process for coming up with the identity was this:

analysis of the geometric model gives two simultaneous equations:

[latex]\cot\alpha = \cos\upsilon\tan\frac{\phi}{2 }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\:(1)[/latex]

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

The function [latex]\alpha=f(\lambda)[/latex] is:

[latex]\alpha={\cot}^{-1 }(\cos\upsilon\tan{\sin}^{-1}(\frac{\sin\frac{\lambda}{ 2}}{ \sin\upsilon}))\quad\quad\quad\quad\:\:(3)[/latex]

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].

Equations (1) and (2) can be rewritten in the form of:

[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]

[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]

If there are mistakes, they are probably caused by my lack of skill with the algebra.  In any event, the original simultaneous equations may be what determines how the number of dimensions are decided.  I can't really tell.

Edited by steveupson

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How did that extra + sign get there?  

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

 

sorry people, I want to say that it wasn't my fault

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[math]{\sin ^2}\nu  + {\cos ^2}\nu  = 1[/math]


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

So is


[math]\sin (\alpha  + \beta ) = \sin \alpha \cos \beta  + \cos \alpha \sin \beta [/math]


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


[math]\alpha  = \frac{\pi }{4}\quad \varphi  = \frac{\pi }{2}\quad \lambda  = \frac{\pi }{2}[/math]


Then your expression 6 becomes


[math]{\left( {\sqrt {\frac{2}{4}} *\sqrt {\frac{2}{4}} } \right)^2} + {\left( {1*\sqrt {\frac{2}{4}} } \right)^2} = \frac{3}{4} \ne 1[/math]

 

 

Edited by studiot

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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

 

 

First, I want to thank you folks for taking time to look at the math.  I know that it doesn’t seem worth your while, but if we can talk more about the math then we can probably begin to make some real progress.

A projection from 3D to 2D is commonly described as taking a wire frame of the 3D object and placing it on a piece of paper that has been laid on a table.   Light passing through the wire frame will cast a shadow on the paper, and depending on where the light source is oriented in relation to the object there can be different results, or different shadows cast.  When the light source is directly above the object and the rays of light are parallel, this projection is often called an oblique projection.

If we take the example that was given in your post and project it onto a 2D plane then the projection contains the identical values as the 3D object.  This is because the values that were chosen are for an instance where the small circle is at maximum (has become a great circle) and this causes all of the angles in the model to collapse into a 2D construction.  In other words, the expression is expressing a relationship that exists in 2D and the expression is not an identity in 2D.

For some further illumination, it is possible for the function to go beyond a great circle, but when this occurs it actually becomes a small circle on the back side of the sphere.  This issue can probably dealt with mathematically with the use of imaginary numbers.

The use of imaginary numbers to describe this part of the function was first discussed publicly back in July of last year I only mention this in order to show anyone who thinks I’m making this all up in order to engage in some tit-for-tat rhetorical argument that this is not the case.  The math is the math.  It’s what it is.  And it isn’t my fault that it is what it is, either.

Our research has progressed a great deal since then, but we are coming to a big slow down in our efforts because we need help with the math.  Understand that our research is way ahead of where the math currently is.  We need to get the math caught up with where we’re at.

Edited by steveupson

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Why is it so important to you that your expression be an identity?

 

It may well be an expression that correctly models some phenomenon.

But an identity relationship does not depend upon the use you put the expression to.

Either the expression always equal one, whatever you use it for and it is then an identity.

Or the expression doesn't always equal one, in which case it is not an identity.

 

I am not sure what you are trying to achieve with this projective stuff.
Can you elaborate?

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The identity is important because it defines which particular aspects of Euclidean 3-space are currently being ignored in our mathematical models of spacetime.

The stuff that the equation shows to be an identity in 3D is the stuff that is not currently being captured in our relativistic transformations.  The stuff that the equation includes that is not an identity is the only stuff that is being included in our spacetime models.

In other words, our spacetime models only represent the mathematical relationships that occur in the instances of geometric descriptions that are like the one that you gave.  Our spacetime models do not represent the rest of the plethora of additional geometric relationships that are present in Euclidean 3-space. 

We appear to be ignoring what Euclidean 3-space actually is when we go about trying to bend it.  We should apply this new information to the model before we try to come up with relativistic transformations or they will never be very accurate.

One more time; how can we mathematically represent the effects of gravity on spacetime when we are not using an accurate model of space?

The projection is 2D vs 3D representation of the values.  I don't understand your question.

 

edited to add:

We have three points a,b,c in space.  There are three angles that can be formed using these points as vertices, abc, bca, cab with values x,y,z.  If these thee angles in space can be parallel projected onto a plane where x=x’, y=y’, z=z’ then we have a case where the values (identical values) exist in 2D and 3D.  In this case there is not an identity.  In every other case there is.

I don’t know the word for this.  I’m quite sure that there is no word for this. The lack of proper nomenclature is another issue, in addition to the math.

second edit:

tridenty?

Edited by steveupson

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second edit:

tridenty?

 

third edit>>>

I need to correct the first edit. What we have in 3-space are the three points a,b,c and the origin of the Cartesian coordinate system.  These points also make angles with the origin.  When all of these angles can be projected into 2D without change, then the identity does not exist.  This is what distinguishes 2D geometric functions or rules (Pythagoras, pi, trig) from the 3D functions or rules.  All of the 2D functions and rules are applicable to 3D, plus and entire new set of new functions and rules.

The reason we never knew about them is probably because we always just reapply the 2D rules over and over again thinking that this is sufficient.  It isn't enough; there's a lot more to 3-space.

Edited by steveupson

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7 hours ago, steveupson said:

The identity is important because it defines which particular aspects of Euclidean 3-space are currently being ignored in our mathematical models of spacetime.

Perhaps the fact that it isn't an identity shows that it doesn't reveal anything that is being ignored in Euclidean geometry.

You seem convinced that your idea has some value, will allows us to do things in Euclidean geometry that were not possible before. But you have provided no evidence or other rationale to support this. Until you do that, I doubt anyone is going to accept that assertion.

7 hours ago, steveupson said:

In other words, our spacetime models only represent the mathematical relationships that occur in the instances of geometric descriptions that are like the one that you gave. 

Most space-time models are non-Euclidean.

You have an even bigger challenge if you want to show that your equation will contribute to the field of differential geometry (because that is a massively complex area of mathematics).

Good luck. 

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On 12/1/2017 at 2:47 PM, Strange said:

How is this represented in the equation? In other words how do you (or, more importantly, the equation) know if these three scalar values are co-planar or not? 

These scalar values are directions.  They have no length component.  Therefore, direction is a quantity.

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2 minutes ago, steveupson said:

These scalar values are directions.  They have no length component.  Therefore, direction is a quantity.

So. It still seems that the "identity" only holds for a subset of values of those. (So it isn't an identity.) How do you define which values the equation is valid for?

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30 minutes ago, Strange said:

Perhaps the fact that it isn't an identity shows that it doesn't reveal anything that is being ignored in Euclidean geometry.

You seem convinced that your idea has some value, will allows us to do things in Euclidean geometry that were not possible before. But you have provided no evidence or other rationale to support this. Until you do that, I doubt anyone is going to accept that assertion.

Most space-time models are non-Euclidean.

You have an even bigger challenge if you want to show that your equation will contribute to the field of differential geometry (because that is a massively complex area of mathematics).

Good luck. 

Are there any spacetime models that are not non-Euclidean?

 

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13 minutes ago, steveupson said:

Are there any spacetime models that are not non-Euclidean?

Not that work on cosmological scales.

Do you know that your concerns about the shortcomings of Euclidean geometry are not addressed by the more general approach of differential geometry? 

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31 minutes ago, Strange said:

So. It still seems that the "identity" only holds for a subset of values of those. (So it isn't an identity.) How do you define which values the equation is valid for?

I don't know of any simple way to go about this explanation except with the mathematical truths.  Every time I try to put context to the math I am called a crackpot.  Let me try one more time.

In 2D direction is always a vector or ratio quantity.  In 3D direction is always a scalar or magnitude quantity.  Sometimes the two quantities for expressing direction are the same, but most of the time they aren't.  In curved spacetime  this mathematical reality causes issues with how the equations commute .  It appears that there is only one instance in any reference frame where the the two are the same, or where  time and space commute (where the distance and direction from event A to event B is the same as the distance and direction from B to A.)  The tridentity identifies the difference between these two sets of relationships or orientations, the ones that can only exist in 3-space and the the ones that can exist in both 2-space and in 3-space.

A mathematical definition of a turn in 3-space is a different mathematical thing than the definition of a turn in a plane.  The reason that they are different is because of the relationships that are identified by the tridentity.

11 minutes ago, Strange said:

Not that work on cosmological scales.

Do you know that your concerns about the shortcomings of Euclidean geometry are not addressed by the more general approach of differential geometry? 

So, was that a yes or a no?  I'd be extremely interested if there are any spacetime models that are not non-Euclidean.

Edited by steveupson

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1 minute ago, steveupson said:

In 3D direction is always a scalar or magnitude quantity.

That makes no sense. 

2 minutes ago, steveupson said:

So, was that a yes or a no?

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.

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There may be no way for the math to make sense to you.  It will probably make more sense when it's more fully developed and discussed.  Many members at many fora have made discussion of the math almost impossible, which is a shame.

Edited by steveupson

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Well when you say meaningless things like “In 3D direction is always a scalar or magnitude quantity” ...

Scalar means there is no direction associated with a value. Direction is represented in the same way in 2d and 3D. 

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Of course direction is represented the same way in 2D and 3D.  That's the whole issue.  That's why it's important to look at the mathematics that underlie that treatment.

It isn't me who says nonsensical things, it's the numbers, the math, the geometry.  A turn in three dimensions is mathematically different than a turn in two dimensions.  That's not me saying it.  That's what the math proves.  That's how you can have an identity that occurs in three dimensions that doesn't occur in two dimensions.  I'm not making any of this up.  It's the way the universe is built.  And it makes absolutely no difference at all until you try and model spacetime because space with direction as a scalar value reacts differently with time than  space with direction as a vector.  You must know that scalars and vectors are mathematically different.

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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

 

 

Edited by studiot

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