# Second Essay For The Gravitational Research Foundation

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### #1 Dubbelosix

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Posted 18 August 2018 - 03:27 AM

I have pretty much wrote up all the details of my next paper I will submit and if you would like to see it just follow:

https://blackholeradiation.quora.com/

It covers related topics, of gravimagnetic and spin coupling, the quantization of the black hole for a true analogue theory of the ground state hydrogen atom. It covers why, if black hole particles can exist, they would need to obey Larmor radiation in the form of Hawking emission and I will argue why there are no stable black holes in nature from the equations I derived. I still need to write out conclusions, and sort footnotes and marking the references or even just sorting the references. The main equations are:

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

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

2. A spin density obtained from the master equation:

$e (\nabla \times \mathbf{B}) = -\frac{\mathbf{J}}{r^3}$

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

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

4. Angular precession of a particle due to torsion is

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

5. Curl of the torsion field is:

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

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

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

7. It's also true as:

$\Omega \times v = \gamma (\mathbf{B} \times v) = \frac{G}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}$

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

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

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

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

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

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

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

11. Field strength is found as

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

12. With an equivalent formula:

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

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 ~

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

14. Sciama's theory can be implemented  on the field strength as a cross product:

$\mathbf{H} \times (\frac{\phi}{c^2}) = \frac{m}{r^2} = \frac{1}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3} \approx \mathbf{E}$

15. There is a scalar triple product::

$\nabla \cdot (\mathbf{H} \times (\frac{\phi}{c^2})) = (\frac{\phi}{c^2}) \cdot (\nabla \times \mathbf{H}) = \frac{m}{r^3} = \frac{1}{2c}\frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^4} \approx \nabla \cdot \mathbf{E} = 4 \pi \rho$

16. The pseudo-quantization of the field is:

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

Edited by Dubbelosix, 20 October 2018 - 12:36 PM.

### #2 VictorMedvil

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Posted 21 August 2018 - 07:48 AM

Good luck Dubbel I know that sometimes they get rejected and need to be revised.

### #3 Dubbelosix

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Posted 02 September 2018 - 10:54 AM

People may want to chase up the new developments in the paper. This time, we have constructed a fuller theory that unites the conducting sphere model and the discrete transition of a system.

https://blackholeradiation.quora.com/

### #4 VictorMedvil

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Posted 15 September 2018 - 09:22 PM

Did you fall into the Blackhole Dubbel?

### #5 Dubbelosix

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Posted 20 September 2018 - 08:00 AM

Not at all, I have been working very hard.

I am almost finished, I think now it contains enough math, I am searching now for knowledgeable scientists in the circle I speak to, to basically proof read it.

As meticulous as I think I have been, I still think errors are abound somewhere.

https://blackholeradiation.quora.com/

### #6 VictorMedvil

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Posted 20 September 2018 - 08:54 PM

Not at all, I have been working very hard.

I am almost finished, I think now it contains enough math, I am searching now for knowledgeable scientists in the circle I speak to, to basically proof read it.

As meticulous as I think I have been, I still think errors are abound somewhere.

https://blackholeradiation.quora.com/

Well, good I wasn't going to attempt a rescue as not even light can escape the black hole, I thought you may have fallen into.

Edited by VictorMedvil, 20 September 2018 - 09:04 PM.

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### #7 Super Polymath

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Posted 20 September 2018 - 09:19 PM

Well, good I wasn't going to attempt a rescue as not even light can escape the black hole,

Edited by Super Polymath, 20 September 2018 - 09:19 PM.

### #8 Super Polymath

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Posted 01 October 2018 - 10:19 AM

In case anyone was wondering what 006's J symbol means it's the magnetic flux analogue. Although he will deny it's true origin.

### #9 Dubbelosix

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Posted 01 October 2018 - 11:25 AM

$J$ is an angular momentum which when coupled to the torsion field is a Hamiltonian

$H = J \cdot \Omega$

.$e^2$ is just the charge squared or gravitational charge squared $Gm^2 = \hbar c$

Edited by Dubbelosix, 01 October 2018 - 11:25 AM.

### #10 Dubbelosix

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Posted 01 October 2018 - 11:26 AM

https://en.wikipedia...ic_flux_quantum

and related, more specifically written, the Von Klitzing constant related to the Josepson constant

https://en.wikipedia...tum_Hall_effect

### #11 Dubbelosix

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Posted 01 October 2018 - 11:29 AM

You can think of the angular momentum being related to the gravitational spin of objects, especially the celestial type. The rotation of an object in a gravimagnetic field gives rise to what we call the Coriolis force.

### #12 Dubbelosix

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Posted 03 October 2018 - 05:09 PM

Ok... so this following derivation has won me over, and it has come from applying Sciama's model into the understanding of the dimensions of the equation (so purely empiracle and theoretical).

The torsion I find as

$e\mathbf{B} = m\Omega = \frac{3Gm^2 \omega}{5c^2 r^3} xy$

$\Omega = (\frac{e}{m})\mathbf{B} = \frac{3Gm \omega}{5c^2 r^3} xy$

$\Omega = (\frac{e}{m})\mathbf{B} = \frac{3G\omega}{5c^2 r} \frac{m}{r^2} xy$

$\Omega = (\frac{e}{m})\mathbf{B} = \frac{3G\omega}{5c^2 r} \frac{m}{r^2} xy = \frac{3G\omega}{5c^2 r} \frac{\phi}{c^2} xy = -\frac{3G\omega}{5c^2 r} \frac{a}{G} xy$

This approach feels the most natural way to implement his gravielectric field $\frac{m}{r^2}$. We also recognize that

$\frac{G}{c^2 r} = \frac{1}{m}$

$\Omega = \frac{3\omega}{5m} \frac{\phi}{c^2} xy = \frac{3\omega r^2}{5m} \frac{m}{r^2}$

The charge to mass ratio is well-known to have a relationship as;

$\frac{e}{m} = \frac{\mathbf{E}}{\mathbf{B}^2r}= \frac{v}{\mathbf{B}r}$

and a direct relationship to velocity through the ratio of the electric and magnetic field

$v = \frac{\mathbf{E}}{\mathbf{B}}$

So the dimensional difference between the two fields is only a velocity term:

$v\mathbf{B} = \mathbf{E}$

Just some useful things to keep under the hat.

Edited by Dubbelosix, 03 October 2018 - 07:06 PM.

### #13 Super Polymath

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Posted 06 October 2018 - 12:10 PM

****ing beautiful equations. Every-time you use the term "Master Equation" I'm reminded of this speech:

### #14 Dubbelosix

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Posted 06 October 2018 - 01:35 PM

''master equation'' is just a generic term for a ''main equation'' on which you work with, in case you did not know.

### #15 Super Polymath

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Posted 06 October 2018 - 01:40 PM

''master equation'' is just a generic term for a ''main equation'' on which you work with, in case you did not know.

shhhuuuuuure, ''main equation"

### #16 Dubbelosix

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Posted 06 October 2018 - 01:41 PM

And yeah they are nice equations, with an undeniable torsion of the form $\frac{1}{mc^2}\frac{d^2U}{dt^2}$.

### #17 Super Polymath

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Posted 07 October 2018 - 07:14 PM

''master equation'' is just a generic term for a ''main equation'' on which you work with, in case you did not know.

What does A =?