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- 12-23-2016, 11:21 AM #21
## Re: Cross product vs Tensor product

Pythagorean triples are not exceptions

That means that with the exception of Pythagorean triples, the coefficients each vector are a ring without division, which is why fractions consisting only of individual number lines as coefficients of vectors are independent.

E.g. For fractions such as z = y/x, x and y are generally independent (which it is why z is the slope of a straight line in two dimensions), but not fractions such as z = f(x,y)/g(x,y). So they are also independent for , which is why they can form the bases of linear systems consisting of polynomials with constant coefficients.

- 12-23-2016, 11:29 AM #22

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## Re: Cross product vs Tensor product

Last edited by BuleriaChk; 12-23-2016 at 11:40 AM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 12-23-2016, 11:33 AM #23

- 12-23-2016, 12:54 PM #24

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- 12-23-2016, 01:14 PM #25

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## Re: Cross product vs Tensor product

So "superior" is anything you want to thnk about instead of anything I want to think about..

I think not... (that is intellectual bullshit)...

(It is true that I like some people a whole lot better if I don't know them very well.)

(and you are hardly a shining light either socially or intellectually.)

---------------------------------

That said, the concept of z = x/y where x and y are function variables as coefficients of a tensor is the subject of Einstein's application of differential geometry to linear algebera in the form of tensor analysis, where z = x/t where x and t are the "space-time" variables and and c is an invariant for STR. This means that , where

( where for x = r in radial coordinates.

For STR, and for x=ct (inertial frames).

-------------------------------

Note that the area of the unit circle can be thought of as the cross product beteween the quotient of the circumference and the circumference, since

,

which represents the gradient to the unit circle in the plane , and so a third dimension.

**Hold on, more editing coming; this is starting to get interesting. Back in a while.**-------------------------------

For GTR, the question arises:

Suppose

However, for mass in STR, so

Suppose where the components of the fraction are now variables... ?

This brings up some interesting questions, which are too long for this thread..... (But ultimately are related to my proof of Fermat's theorem in terms of the relativistic unit circle).Last edited by BuleriaChk; 12-23-2016 at 07:13 PM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 12-23-2016, 04:55 PM #26
## Re: Cross product vs Tensor product

What they meant was, you haven't engaged those subjects yet, on your intellectual journey

(It is true that I like some people a whole lot better if I don't know them very well.)

(and you are hardly a shining light either socially or intellectually.)

---------------------------------

That said, the concept of z = x/y where x and y are function variables as coefficients of a tensor is the subject of Einstein's application of differential geometry to linear algebera in the form of tensor analysis, where z = x/t where x and t are the "space-time" variables and and c is an invariant for STR. This means that , where

( where for x = r in radial coordinates.

For STR, and for x=ct (inertial frames).

Note that the area A of a circle can be thought of as the cross product beteween the diameter and the circumference, since

where at the point where the diametter intersects the circumference (so that and are connected ( at an "origin", i.e., not affine)).

For GTR, the question arises:

Suppose

However, for mass in STR, so

Suppose where the components of the fraction are now variables... ?

This brings up some interesting questions, which are too long for this thread..... (But ultimately are related to my proof of Fermat's theorem in terms of the relativistic unit circle).

- 12-23-2016, 05:43 PM #27

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## Re: Cross product vs Tensor product

It comes from D=2r, but the equation is wrong (tex dyslexia) I'll fix it shortly...

Thank you .

This is the correct version, I think (fixed/edited in previous post #25)

but hold on, more editing coming... (this is getting interesting)

(The vector is unrelated to the scalar r, which was my point; it could just as easily have been designated

(Phrase deleted for now, will be back)Last edited by BuleriaChk; 12-23-2016 at 07:04 PM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 12-23-2016, 06:54 PM #28
## Re: Cross product vs Tensor product

The 2 is still extraneous. With it, that expression doesn't equal what is before it, nor does it equal what comes after. Just get rid of the "2"

but hold on, more editing coming... (this is getting interesting)

(The vector is unrelated to the scalar r, which was my point; it could just as easily have been designated

Since it is unrelated (i.e., independent), it is orthogonal to the area A, and thus its magnitude represents the scalar gradient at that point (on the circumference, but since it is a vectors, can be translated to the center of the circle - the concept of "parallel transport").

- 12-23-2016, 07:01 PM #29

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## Re: Cross product vs Tensor product

Last edited by BuleriaChk; 12-23-2016 at 07:05 PM.

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 12-23-2016, 07:06 PM #30

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