# Thread: The Relativistic Unit Circle

1. ## The Relativistic Unit Circle

--------------------------------------------------------------
The Relativistic Unit Circle on my Website. Updates to this file will be generally tracked in the document

RCU Update, also on my Website

The Relativistic Unit Circle is at the foundation of the Theory of Relativity, where c is declared a constant. However, for c a variable, the relation of the "time dilation" equation can be extended to the realm of real numbers, without any physical interpretation.

Since I reference this concept in my proofs of Fermat's Theorem, I have written a short pdf discussing its relevance independent of any particular application. There is much more to say (especially vis a vis Quantum Mechanics and Relativity), and I will be adding to it as I continue on my own journey, but for a start, it is located at:

(I am continuing to add to and update it, so refresh your browser occasionally).

(Understanding its significance is at the root of Quantum Field Theory, which rests on the Theory of Relativity and is the foundation of the Standard Model in Physics).

In other threads (and documents on my Website) I have shown its relation to Fermat's theorem and now will be returning to the interpretation of the Pauli/Dirac matrices given its foundation.

2. ## Re: The Relativistic Unit Circle

why do you keep crying about that stupid list no one gives a damn about? It is pathetic.

3. ## Re: The Relativistic Unit Circle

The the Relativistic circle, and the characterization of integer generation by Lorentz rotation is at the foundation of sampling theory, signal processing and analysis and analysis, (For the Lorentz Boost, imagine that the invariant initial condition ct is replaced by Planck's constant, and then ask yourself what happens to and h as goes to infinity.

(The fact that if where ct is the initial condition illustrates the problem of normaliztion in QFT ( ), and shows that the General Field Theory cannot conserve energy (god must be involved in a Lorenatz boost)

for the case n = 2 is the reason "Black Holes have no hair" [/tex] and why virtual particles are just that - figments of the mathematical imaginiation.

(We observe the universe throgh a peephole that has a relativistic area of of at the temperature of our parking lot or however much we can cool our optical devices relative to it.)

Consider a physical system modeled by an integer n. Then multiplication by the Green's function () yekds the unit impulse ; e.g. linear system characterized by 3n as the system multiplied by the impulse response This concept is then extended to the relativistic unit circle for Quantum Field Theory. (That is, with wave equations and all that... , both with Pauli/Dirac equations, with the Dirac equation extended to the relativistic interpretatio of the Schroedinger equation. The problem of normalization arises because in a Lorentz boost.

(Some people (including a few that pretend to be algebraicists) only live at the business end of a geodesic, and so have no peripheral vision... Medically, it is called tunnel vision But if they ever try to get a job in mathematics, they will soon discover it is the headlight of an oncoming train. Unfortunately they haven't yet received the news of the invention of the wheel....

IMO, the case for n = 2 is then extended to the concept of "up/down" quarks, the concept of "spin" for strange particles (mirror) and Red, Blue, and Green quarks in three dimensions. (All energies must be determined in three dimensions (a sphere) in therms of the energy level in the parking lot; i.e. Planck's constant, the zero point energy in the Standard Model, modeling local gravity as a relativistic quantum effect.

In the Standard Model, the question is whether one wants to include the graviational energy in the QFT energy or not.

When we finally characterize the Higgs boson, we will be able to calculate the radius of the LHC from first principles... (The problem is that a linear characterization can never be normalized because of the Lorentz boost, and if there is no Lorentz boost, the experimental result is unobservable (the RUC, or a "black hole").

TRUST me - I'm NOT from the government.....

4. ## Re: The Relativistic Unit Circle

I maintain the proof of Fermat's Theorem is trivial;

From the Binomial Expansion for n > 2, where rem(a,b,n) is everything not equal to or on the r.h.s., and is a positive integer (since there are no divisions or subtractions in the expansion). Therefore rem(a,b,n) > 0.

For Fermat's Theorem to be false, rem(a,b,n) must be equal to 0. (That is, )

Therefore, (Fermat's Theorem is true)

QED

I'm still waiting for an intelligent counter-argument. I think the idea that Fermat's Theorem hasn't been proven many times over is an urban legend, or a hoax promulgated by math depts. to get unwary citizens involved in mathematics to get money from them.

Either that, or it is a Jedi mind trick....

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

That said,

If the diagram of is NOT a right triangle (i.e., a Pythagorean right triangle for positive integers) then (by the Binomial Expension).

(from an idiot controversy on another thread)

In my paper rem(a,b,2) = 2ab corresponds to the Lorentz boost] () in one dimension) where , is the initial condition.

www.flamencochuck.com/files/Misc/RCUCircle.pdf

and where , where is the equation of the relativistic unit circle, valid for all real circles, integers or otherwise) and in particular, for the Lorentz rotation in the positive quadrant.

(i.e., two orthogonal unit vectors in the positive quadrant of the RUC, () with and

For the Lorentz boost,

where

so that

which can only be true if for (the radius of the relativistic unit circle as the initial condition).

(i.e., the relation is only true if the diagram is the relativistic unit circle) which is consistent with the interpretation from the Binomial Theorem where rem(a,b,n) = 2ab = 0 for b= 0) for the Pythagorean circle in two dimensions).

The Binomial expansion then follows for triangles that are not Pythagorean triangles for a,b,c integers and n=2. (i.e. , (a,b,c) not a right triangle

Proof of Fermat's theorem is then easily extended to n>2, since rem(a,b,n) is always a positive integer greater than zero, so cannot be an integer.

(expand , where rem(a,b,n) is everything not equal to or ).
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5. ## Re: The Relativistic Unit Circle

Originally Posted by BuleriaChk
I maintain the proof of Fermat's Theorem is trivial;

From the Binomial Expansion for n > 2, where rem(a,b,n) is everything not equal to or on the r.h.s., and is a positive integer (since there are no divisions or subtractions in the expansion). Therefore rem(a,b,n) > 0.

For Fermat's Theorem to be false, rem(a,b,n) must be equal to 0. (That is, )

Therefore, (Fermat's Theorem is true)

QED

I'm still waiting for an intelligent counter-argument.

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

That said,

If the diagram of is NOT a right triangle (i.e., a Pythagorean right triangle for positive integers) then (by the Binomial Expension).

(from an idiot controversy on another thread)
Idiot, singular

For any such triangle,

Less than, not equal
In my paper rem(a,b,2) = 2ab corresponds to the Lorentz boost] () in one dimension) where , is the initial condition.

www.flamencochuck.com/files/Misc/RCUCircle.pdf

and where , where is the equation of the relativistic unit circle, valid for all real circles, integers or otherwise) and in particular, for the Lorentz rotation in the positive quadrant.

(i.e., two orthogonal unit vectors in the positive quadrant of the RUC, () with and

For the Lorentz boost,

where

so that

which can only be true if for (the radius of the relativistic unit circle as the initial condition).

(i.e., the relation is only true if the diagram is the relativistic unit circle) which is consistent with the interpretation from the Binomial Theorem where rem(a,b,n) = 2ab = 0 for b= 0) for the Pythagorean circle in two dimensions).

The Binomial expansion then follows for triangles that are not Pythagorean triangles for a,b,c integers and n=2. (i.e. , (a,b,c) not a right triangle

Proof of Fermat's theorem is then easily extended to n>2, since rem(a,b,n) is always a positive integer greater than zero, so cannot be an integer.

(expand , where rem(a,b,n) is everything not equal to or ).
-------------------------

6. ## Re: The Relativistic Unit Circle

Originally Posted by grapes
Idiot, singular

For any such triangle,

Less than, not equal

That is the proof. For Fermat's theorem to be false, for some set of positive integers (a,b,c) and n > 2.

Fermat's Theorem is true.
-------------------------

(Not only that, your equation isn't even a triangle. A triangle would be )

Which I have also said many times over, except for that one post, which was a mistake that didn't change the substance of anything I said. And as a moderator, you locked me out so I couldn't correct it.

is the equation of a circle plus the area of a rectangle.

Again,

QED

Village idiot. And a myopic, vindictive shit to boot.

7. ## Re: The Relativistic Unit Circle

Originally Posted by BuleriaChk
That is the proof. For Fermat's theorem to be false, for some set of positive integers (a,b,c) and n > 2.

But (as you finally figured out)

Fermat's Theorem is true.

Village idiot.
I know you might be stressed, anxious, confused or disoriented, but read my post carefully:

Originally Posted by grapes
Idiot, singular

For any such triangle,

Less than, not equal
That's why your proof is nonsense.

8. ## Re: The Relativistic Unit Circle

I spy parenthesis.

9. ## Re: The Relativistic Unit Circle

Originally Posted by grapes
I know you might be stressed, anxious, confused or disoriented, but read my post carefully:

That's why your proof is nonsense.
Your equation isn't a triangle. A triangle would be )

The triangle inequality is irrelevant.

By the Binomial Theorem,

That is,
Are you questioning the Binomial Theorem?

Village idiot.

10. ## Re: The Relativistic Unit Circle

Originally Posted by BuleriaChk
Your equation isn't a triangle. A triangle would be )
Which means c<a+b

That's true for all triangles
The triangle inequality is irrelevant.

Village idiot.
Except it totally invalidates your entire proof

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