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excited cold
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excited cold - 11-19-2007, 12:22 PM

As mentioned previously, the perfect cold fusion nuclear reaction is D + D ® He + energy. That is two deuterium atoms stripped off 2 electrons fused into a single Helium-4 nucleus and subsequently recaptured 2 electrons forming a neutral helium-4 atom plus excess energy. However, within the 40MK interior of the sun the process can be realized by 4 hydrogen nuclei plus 2 electrons.

Using space-time charges, hydrogen atoms can exist to two different quantum energy states: (12, 12) and (16, 16). On the other hand, helium atoms can exist in three quantum energy states: (38, 38), (46, 46), and (50, 50). To satisfy the abundance ratio of 3 to 1, these states can be combined into 6 distinct nuclear reactions: 1. (12, 12) ®(38, 38). 2. (12, 12) ® (46, 46). 3. (12, 12) ® (50, 50), 4. (16, 16) ®(38, 38). 5. (16, 16) ® (46, 46). 6. (16, 16) ® (50, 50). The space-time charge excess of the 1st is (2, 2), 2nd is (10, 10), 3rd is (14, 14). The 4th has a deficit of (10, 10), the 5th deficit of (2, 2). Finally, the 6th again has excess of (2, 2). The ratio of deficit to total reactions is 1/3; excess ratio is 2/3. Since the space-time charge configuration for a neutrino is (1, 1) and a photon is (4, 4), the best possible excited cold fusion reaction is the 3rd reaction with excess products of 3 photons and 2 neutrinos.

In this context, the meaning of the word ‘cold’ is absolute independence of temperature and density but not proximity. The concept of proximity is dependence only on the mean free path and the relaxation time between collisions that allow the grouping and formation for the 6 reactions among space-time charges. It remains to be seen whether a proximity theorem can be formulated awaiting empirical validation that the mean free paths are at most multiples of the Planck length and the relaxation times multiples of the Planck time. They are scales at which classical general relativity becomes ineffective and can be replaced by a quantum theory of gravity.


Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c˛
  
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Smile Re: excited cold - 11-19-2007, 12:24 PM

Lets hope for un-excitement then!



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Re: excited cold - 11-19-2007, 12:35 PM

Quote:
Originally Posted by mkirkpatrick
un-excitement then!
This is contrary to what a proximity theorem would suggest. The closer two people are the more exciting is the relationship. The farther apart, the more boring unless both people were using cellphones. Gone are the days of writing long narative letters and the time it takes to reach the intended parties.


Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c˛
  
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Smile Re: excited cold - 11-19-2007, 12:38 PM

Quote:
Originally Posted by AntonioLao View Post
This is contrary to what a proximity theorem would suggest. The closer two people are the more exciting is the relationship. The farther apart, the more boring unless both people were using cellphones. Gone are the days of writing long narative letters and the time it takes to reach the intended parties.
You are never alone with a cell phone,prehaps thats the reason I have not got one!




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Re: excited cold - 11-19-2007, 12:40 PM

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Originally Posted by mkirkpatrick
thats the reason I have not got one!
Do you mean you prefer being bored? The reason why I don't have one is because I can't afford it.


Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c˛
  
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Smile Re: excited cold - 11-19-2007, 12:45 PM

Quote:
Originally Posted by AntonioLao View Post
Do you mean you prefer being bored? The reason why I don't have one is because I can't afford it.
I could buy a dozen of the dam things if I wanted one,here in the UK,they almost give them away,Tesco sells them for about 20 dollars,I dont get bored,cannot ever remember that occuring.



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Re: excited cold - 11-19-2007, 12:49 PM

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cannot ever remember that occuring.
Seems to me you have reached the state of absolute excited cold as the highest state of perfect consciousness.


Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c˛
  
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Smile Re: excited cold - 11-19-2007, 01:36 PM

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Seems to me you have reached the state of absolute excited cold as the highest state of perfect consciousness.

And not even be aware of it?



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Re: excited cold - 11-20-2007, 11:55 AM

Quote:
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And not even be aware of it?
I finally finished Paul Davies 'Cosmic Jackpot' and I think he believes in a self-aware universe that understand itself by merging with the mind and absolute consciousness.


Time independence: [∂E(g)]˛=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c˛
  
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Smile Re: excited cold - 11-20-2007, 03:30 PM

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I finally finished Paul Davies 'Cosmic Jackpot' and I think he believes in a self-aware universe that understand itself by merging with the mind and absolute consciousness.

He seems to be on the right track then?



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