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origin of hyperbolic boost
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origin of hyperbolic boost - 02-26-2008, 12:01 PM

The simplified Lorentz transformations are expressed using hyperbolic functions. These are necessary for describing boosts and rotations near the vacuum speed of light. However, the one real compelling reasons is really to breakaway the complex domain of imaginary linear independence in order to add up all the Hamiltonian energy contributions of all space-time points of the gravitational field, which is not necessarily quantized.
On the other hand, both quantum mechanics and any of the quantum field theories, for example, quantum electrodynamics (QED) or quantum chromodynamics (QCD), work extremely well when expressed in circular functions of ordinary trigonometry using Euler’s formula describing the imaginary phase factors of wavefunctions. In this case, the approach to real is by conjugate boost. However, it unavoidably makes the square absolute wavefunctions into proper fractions of probabilistic superposition requiring normalization. Whose absolute certainty can only be derived by using Feynman path integral and his space-time diagrams of particle interactions and taking the Lagrangians of energy differences (subtracting instead of adding as in a Hamiltonian mechanism).
Hyperbolic and conjugate boosts seem to work very well for all spin half fermions for finding solutions to Dirac equation using imaginary quantum vectors called spinors. However, extension to imaginary space-time vectors called twistors could not dissolve the inconvenient of the cosmological constant of general relativity field equations, unless, of course, by placing postulates for the existence of dark matter and energy analogous to the existence of antimatter in quantum mechanics. Nevertheless, the crucial critical indifference is that antimatter can be detected while dark matter and energy cannot.


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: origin of hyperbolic boost
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Smile Re: origin of hyperbolic boost - 02-26-2008, 12:14 PM

I have always been puzzled about the origin of the hyperbolic boost,thanks for enlightening me?

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Re: origin of hyperbolic boost - 02-26-2008, 12:26 PM

Quote:
Originally Posted by mkirkpatrick
I have always been puzzled about
That is good for you but I'm still puzzled by the reluctance of theorists to letgo of its uselessness.


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: origin of hyperbolic boost
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Smile Re: origin of hyperbolic boost - 02-26-2008, 12:47 PM

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Originally Posted by AntonioLao View Post
That is good for you but I'm still puzzled by the reluctance of theorists to letgo of its uselessness.

Sometimes Antonio where there is a theory a grant will follow??



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Re: origin of hyperbolic boost - 02-26-2008, 01:08 PM

Quote:
Originally Posted by mkirkpatrick
where there is a theory a grant will follow
Or we could also say that for every granter there is a grifter.


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: origin of hyperbolic boost - 02-26-2008, 02:06 PM

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Or we could also say that for every granter there is a grifter.

You could indeed say that! I need a boost shall I call hyperbolic?


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Re: origin of hyperbolic boost
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Re: origin of hyperbolic boost - 02-27-2008, 11:42 AM

Quote:
Originally Posted by mkirkpatrick
I need a boost shall I call hyperbolic
That would be an obvious admission of extended hyperbole. By the way, the boost as defined in physics is just a parallel displacement in an inertial system at constant uniform velocity in an arbitrary direction.


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: origin of hyperbolic boost - 02-27-2008, 12:04 PM

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That would be an obvious admission of extended hyperbole. By the way, the boost as defined in physics is just a parallel displacement in an inertial system at constant uniform velocity in an arbitrary direction.

Thanks I understand it better now.


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Re: origin of hyperbolic boost - 02-27-2008, 12:13 PM

Quote:
Originally Posted by mkirkpatrick
I understand it better now
The next thing for me is to understand how a boost can change into absolute acceleration.


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: origin of hyperbolic boost - 02-27-2008, 12:19 PM

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The next thing for me is to understand how a boost can change into absolute acceleration.

Thats the sort of understanding that speeds up progress?


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