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Second Essay For The Gravitational Research Foundation


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#69 Dubbelosix

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Posted 14 October 2018 - 03:44 AM

Repeating what you say does not mean I am going to read or listen to anything you say. All you are doing is spamming polymath and it's not cool.

 

Back to the work. 

 

The gravimagnetic force is directed, just like a Lorentz force which is perpendicular to both the velocity and the strength of the gravitomagnetic field ~

 

[math]\mathbf{F} = \frac{m}{c}(v \times \mathbf{H}) = \frac{Gm}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

This is right, working the units out in the last expression gives [math]\frac{Gm^2}{r^2}[/math].

 

Under these units, the set of equivalences hold:

 

[math]\mathbf{H} = \Gamma = \nabla \times \mathbf{A} = \Omega \times v = \gamma (\mathbf{B} \times v)[/math]



#70 Dubbelosix

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Posted 14 October 2018 - 08:24 AM

Under these units, the set of equivalences hold:

 

[math]\mathbf{H} = \Gamma = \nabla \times \mathbf{A} = \Omega \times v = \gamma (\mathbf{B} \times v)[/math]

 

Loosely-speaking more terms are equivalent to the curl of the gravimagnetic potential, as there is one more such term of a gradient of the gravielectric field, 

 

[math]\nabla \times \mathbf{A} \approx \frac{1}{c}\frac{\partial \phi}{\partial t} \approx \mathbf{E}[/math]


Edited by Dubbelosix, 14 October 2018 - 08:24 AM.


#71 Dubbelosix

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Posted 15 October 2018 - 04:28 AM

Ok let's sum the most important equations up now. I really am coming to an end with the essay. Here are the key equations for the gravielectromagnetic part of the essay. 

 

 

1. The gravimagnetic field for rotating systems is obtained from the master equation:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{me}\frac{a}{G} \mathbf{J} = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}[/math]

 

2. A spin density obtained from the master equation:

 

[math]e (\nabla \times \mathbf{B}) = -\frac{\mathbf{J}}{r^3} [/math]

 

3. The Von Klitzing factor appears invariant through many of the equations I looked at:

 

[math]e \mathbf{B} = \frac{ \mathbf{J}}{e^2 } \frac{\partial U}{\partial r}= \frac{1}{m}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{m}\frac{a}{G} \mathbf{J} = -\frac{1}{m}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m} \frac{m}{r^2} \mathbf{J}[/math]

 

4. Angular precession of a particle due to torsion is

 

[math]\omega = -\frac{\Omega}{2} = \frac{e \mathbf{B}}{2m} = \frac{\mathbf{J}}{2e^2} \frac{1}{m}\frac{\partial U}{\partial r} = \frac{1}{2m^2}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{Gm^2}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{2m^2}\frac{a}{G} \mathbf{J} = -\frac{1}{2m^2}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{2m^2} \frac{m}{r^2} \mathbf{J}[/math]

 

5. Curl of the torsion field is:

 

[math]\nabla \times \Omega = \gamma \frac{\partial \mathbf{B}}{\partial r} \frac{e}{2m} \frac{\partial \mathbf{B}}{\partial r} = \frac{\mathbf{J}}{2e^2} \frac{1}{mc^2}\frac{\partial^2 U}{\partial t^2} [/math]

 

6. Velocity coupling to gravimagnetic field is shown with coupling constants (gravitational fine structure):

 

[math]\mathbf{B} \times v = \frac{\alpha_G}{e} \frac{\partial U}{\partial r} = \alpha_G \frac{\partial \mathbf{V}}{\partial r} = \frac{m}{ e}\frac{G}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

7. It's also true as:

 

[math]\Omega \times v = \gamma (\mathbf{B} \times v) = \frac{\alpha_G}{m} \frac{\partial U}{\partial r} = \frac{G}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

8. A Hamiltonian spin-orbit coupling equation is presented as:

 

[math]H = \frac{1}{2}\Omega \cdot \mathbf{L} = \frac{e \mathbf{B} \hbar}{2m} = \frac{\mathbf{J} \cdot \mathbf{S}}{2e^2} \frac{1}{m}\frac{\partial U}{\partial r} = \frac{1}{2m^2}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} \cdot \mathbf{S} [/math]

 

[math] = -\frac{1}{Gm^2}\frac{\partial v}{\partial t} \mathbf{J} \cdot \mathbf{S} = -\frac{1}{2m^2}\frac{a}{G} \mathbf{J} \cdot \mathbf{S} = -\frac{1}{2m^2}\frac{\omega^2 r}{G} \mathbf{J} \cdot \mathbf{S} = -\frac{1}{2m^2} \frac{m}{r^2} \mathbf{J} \cdot \mathbf{S}[/math]

 

9. The traditional definition for the torsion field finds one such term from the master equation:

 

[math]\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J}= \frac{m}{ e} \frac{1}{mc^2} \frac{\partial U}{\partial t} = \frac{m}{ e}\frac{G}{2c^2}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

10. Related to the previous gravimagnetic field, an equivalent form:

 

[math]\gamma \mathbf{B} = \frac{e\mathbf{B}}{2m} = \frac{1}{m^2c^2} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J}= \frac{1}{mc^2} \frac{\partial U}{\partial t} = \frac{G}{2c^2}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

11. Field strength is found as

 

[math]\mathbf{H} = \Omega \times v = \frac{G}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

12. With an equivalent formula:

 

[math]\mathbf{H} = \gamma (\mathbf{B} \times v) = \frac{e(\mathbf{B} \times v)}{2m} = \frac{G}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

13. The gravimagnetic force is directed, just like a Lorentz force which is perpendicular to both the velocity and the strength of the gravitomagnetic field ~

 

[math]\mathbf{F} = \frac{m}{c}(v \times \mathbf{H}) = \frac{Gm}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

14. Sciama's theory can be implemented  on the field strength

 

[math]\mathbf{H} \cdot (\frac{\phi}{c^2}) = \frac{m}{r^2} = \frac{1}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}[/math]

 

15. The pseudo-quantization of the field is:

 

[math]n \hbar = e\oint_S\ \mathbf{B} \cdot dS = \frac{\mathbf{J}}{e^2} \int \int_S\ \frac{\partial U}{\partial r} \cdot dS[/math]


Edited by Dubbelosix, 15 October 2018 - 03:57 PM.


#72 Super Polymath

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Posted 15 October 2018 - 11:46 AM

You MAD HAt ter



#73 Super Polymath

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Posted 15 October 2018 - 11:47 AM

*** - - - HAt



#74 Super Polymath

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Posted 15 October 2018 - 11:47 AM

tAH neutrino

 

get to tAH choppa

 

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t=time