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Master
Join Date: Nov 2005 Posts: 620
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01-12-2006, 02:56 PM
| | You have come across a great and significant finding that may be one of those things that stays with you for some time. Some findings are that way.
As much as I want to know why too, I'm hoping this does not come under the category of behavior that is just accepted, never questioned. I'll be thinking about this one with you.
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Michelle | | | | Raider of the lost time
Join Date: Nov 2003 Posts: 6,036
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01-12-2006, 03:03 PM
| | no clue
Quote: |
Originally Posted by GUILLE Any clues? | No clue. But how do we find something that is already missing? If you find it you are welcome to keep it as long as it is still missing.
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Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
| | | | The Thinker
Join Date: Mar 2005 Posts: 3,278
48 | |
01-12-2006, 05:08 PM
| | The equation
from the well-knowned equation 
I started deriving, modifying, and mixing with others and end up with an interesting: the same equation must exist but with Energy instead of speed= 1/sqrt(EcB) where the 1 is the value for voltage, E is the electric field, c the speed of light in vacuum, B the magnetic field. Then I ahd several problems of consistency in my proof, where I have not yet closed up. When I do close the proof, I will post it as an article, it'll hopefully be at the end of the weekend after the one that is about to start. | | | | Raider of the lost time
Join Date: Nov 2003 Posts: 6,036
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01-13-2006, 02:45 PM
| | momentum of light
a simple energy derivation \(E=\frac{p}{\sqrt{\epsilon_0\mu_0}}\)
where p is the momentum of light.
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Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
| | | | The Thinker
Join Date: Mar 2005 Posts: 3,278
48 | |
01-14-2006, 04:50 AM
| Quote: |
Originally Posted by AntonioLao a simple energy derivation \(E=\frac{p}{\sqrt{\epsilon_0\mu_0}}\)
where p is the momentum of light. | The equation isn't clear, can you just right it like I wrote mine, the it'll be easier. | | | | Raider of the lost time
Join Date: Nov 2003 Posts: 6,036
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01-14-2006, 03:39 PM
| | wondering
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Originally Posted by GUILLE The equation isn't clear, can you just right it like I wrote mine, the it'll be easier. | I've been trying several version of latex but still couldn't make it work. Since my computer crashed over a month ago, no more latex can be emailed. It will take about $100 to remove the virus and still it's no guarantee that that's the problem. I had a Windows98 operating system. It's outdated. E=mc ² but the momentum of light is p and p=E/c such that E=pc and c=1/Öeomo therefore E=p/Öeomo
__________________
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
| | | | The Thinker
Join Date: Mar 2005 Posts: 3,278
48 | |
01-15-2006, 04:54 AM
| Quote: |
Originally Posted by AntonioLao I've been trying several version of latex but still couldn't make it work. Since my computer crashed over a month ago, no more latex can be emailed. It will take about $100 to remove the virus and still it's no guarantee that that's the problem. I had a Windows98 operating system. It's outdated. E=mc² but the momentum of light is p and p=E/c such that E=pc and c=1/Öeomo therefore E=p/Öeomo
| Interesting. My version just changes the momentum for voltage and the permitivity and permeavility by the magnetic and electric fields and adds a speed of light. From it, I derived E=mc^2, and even better, I found some interesting relationships... But I'm still in nowhere in defining it.
Now, if |f1|+|f2| doesn't equal |f1+f2| then also |f1xf2| doesn't equal |f1xf2|? Moreover, what would these two types of multiplications of the frequencies mean? | | | | The Thinker
Join Date: Mar 2005 Posts: 3,278
48 | |
01-15-2006, 08:06 AM
| | What could it mean?
If E=p/Öeomo and E=v/EcB we have to look back t at maxwell and even previous work. The equation VI/mo=J tells us that because E=(El+Fv)c^2=(El+VI)c^2 (this is because power is force times velocity and also voltage times current, and the El is the electric field which I write with the l to distinguish between E energy) then if there is no electric field, then the magnetic flux density J times mo is equal to the energy if and only if there is no speed (and because this is in vacuum, it's ok). But what about our gravity here? It really has no strength, but there is a speed for it: c^2. Moreover, q^2=EvxB so the energy that an electron carries times the energy of a magneton carries is equal to the speed of the photon that represents such electromagnetic wave.
I seem to walk blind, in the dark, but still I feel I'm getting somewhere... I hope somewhere only we know, so we may sleep in our valeys of equations undisturbed. | | | | Raider of the lost time
Join Date: Nov 2003 Posts: 6,036
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01-18-2006, 02:36 PM
| Quote: |
Originally Posted by GUILLE what would these two types of multiplications of the frequencies mean? | Sorry, I could not find any simple explanation for the product of two frequencies. What I'm working on is the ratio of spatial frequency over temporal frequency and this gives the speed of light.
__________________
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
| | | | Raider of the lost time
Join Date: Nov 2003 Posts: 6,036
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01-18-2006, 02:38 PM
| | photon to graviton
Quote: |
Originally Posted by GUILLE But what about our gravity here? | The revised development is that four spin-1 photons give a single spin-2 graviton.
__________________
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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