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Re: Mobius Strip-Singularities last dance!
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Smile Re: Mobius Strip-Singularities last dance! - 01-16-2008, 08:18 PM

What steps are you on fellow reader of this post?Do you understand the rhymm of
mobius tune? When we look within for the answer,mobius quietly smiles.



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Re: Mobius Strip-Singularities last dance!
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Re: Mobius Strip-Singularities last dance! - 01-17-2008, 01:21 PM

Quote:
Originally Posted by mkirkpatrick
mobius quietly smiles.
Since Moebius smile is one-sided, it will stretch from here to eternity and back without frowning.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Re: Mobius Strip-Singularities last dance!
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Smile Re: Mobius Strip-Singularities last dance! - 01-17-2008, 02:44 PM

Quote:
Originally Posted by AntonioLao View Post
Since Moebius smile is one-sided, it will stretch from here to eternity and back without frowning.
Your right there my friend,had not thought of that angle!




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Re: Mobius Strip-Singularities last dance!
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Re: Mobius Strip-Singularities last dance! - 01-18-2008, 11:47 AM

Quote:
Originally Posted by mkirkpatrick
had not thought of that angle
Angular measurements generalized into imaginary phase factors in quantum mechanics remain unsolved even in quantum field theories.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: Mobius Strip-Singularities last dance! - 01-18-2008, 12:17 PM

Quote:
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Angular measurements generalized into imaginary phase factors in quantum mechanics remain unsolved even in quantum field theories.
Prehaps because there is no is no angular entry,and a rethink is needed!



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Re: Mobius Strip-Singularities last dance! - 01-18-2008, 12:35 PM

Quote:
Originally Posted by mkirkpatrick
there is no is no angular entry
Mathematicians use circular functions: sine, cosine, tangent, cotangent, secant, and cosecant replacing angular entries. All these can be found in trigonometry. The implied triangle inequality is obvious.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: Mobius Strip-Singularities last dance! - 01-18-2008, 12:37 PM

Quote:
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Mathematicians use circular functions: sine, cosine, tangent, cotangent, secant, and cosecant replacing angular entries. All these can be found in trigonometry. The implied triangle inequality is obvious.

That's just what I thought too!!!!!!!




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Re: Mobius Strip-Singularities last dance!
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Re: Mobius Strip-Singularities last dance! - 01-18-2008, 12:48 PM

Quote:
Originally Posted by mkirkpatrick
That's just what I thought too!!!!!!!
Your are a natural born mathematician just not practicing it as a trade.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: Mobius Strip-Singularities last dance! - 01-18-2008, 12:50 PM

Quote:
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Your are a natural born mathematician just not practicing it as a trade.
And you are a most generous Gentleman.


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Re: Mobius Strip-Singularities last dance! - 01-18-2008, 12:59 PM

Quote:
Originally Posted by mkirkpatrick
a most generous Gentleman
But with nothing to give away. Simply stating the obvious.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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