Second Essay For The Gravitational Research Foundation

[Dubbelosix]

Since I have fixed that problematic equation, there is still truth behind the velocity coupling to gravimagnetic field yielding the 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]

 

It is also true that

 

[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]

[Dubbelosix]

 

[math]\mathbf{D}[/math] before had been defined as no different to the potential [math]\mathbf{A}[/math] by me, by looking at it as an analogue of the full equation for the gravielectric field. That was only a preliminary observation, let's try and find the true definition behind it. The field is defined as

 

[math]\mathbf{D} = \epsilon_G \Gamma[/math]

 

[math]\epsilon_G = \frac{1}{4 \pi G} \Gamma[/math]

 

In this case, from a trusted reference, [math]\mathbf{D}[/math] is known as the gravitational displacement - I won't go as far to say my previous speculations had been wrong, but I will say that the definition of [math]\mathbf{D}[/math] may differ from source to source. 

 

Refreshing our minds in the relationships:

 

[math]\Omega = \mu_G \mathbf{H}[/math]

 

[math]\Omega = \frac{\Gamma}{c}[/math]

 

There is also the relationship (within standard gravielectromagnetism):

 

[math]\sqrt{\mu_G}\ \mathbf{H} = \sqrt{\epsilon_G}\ \Gamma[/math]

 

This equation determines wave impedance which clearly forms:

 

[math]\frac{\Gamma}{\mathbf{H}} = \sqrt{\frac{\mu_G}{\epsilon_G}} = c\ \mu_G[/math]

Edited October 12, 2018 by Dubbelosix

[Dubbelosix]

The gravimagnetic potential is well-known to be assembled as:

 

[math]\mathbf{A} = \frac{G(\mathbf{J} \times r)}{r^3c}[/math]

 

and we have shown that the acceleration is also satisfied by:

 

[math]\vec{a} = c\ \Omega = \nabla \times \frac{Gm}{r} = \nabla \times \mathbf{A}[/math]

 

[math]\nabla \times \mathbf{A} = \frac{G(\mathbf{J} \times r)}{r^3c^2t}[/math]

 

from it we can form:

 

[math]\frac{\phi}{c^2}\frac{\partial v}{\partial t} = \frac{m}{r^2} = \frac{(\nabla \times \mathbf{A})}{G} = \frac{(\mathbf{J} \times r)}{r^3c^2t}[/math]

 

Which as you can see, satisfies Sciama's definition of the field. This simplification could come in handy when deriving more equations. One immediate application can be seen from our master equation, solved as:

 

[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}[/math]

 

In which ~

 

[math]e\mathbf{B} = \frac{1}{m} \frac{(\nabla \times \mathbf{A})}{G} \mathbf{J} = \frac{\mathbf{J}}{mc^2} \frac{(\mathbf{S} \times r)}{r^3t} [/math]

 

(the last expression is a type of spin-orbit coupling)

 

... instead of deriving, for a moment I will answer a question that may have crossed your mind at least once... why gravielectromagnetism? Well, I believe gravielectromagnetism is actually the route to unification of the forces. In my work, I explain that gravielectromagnetism, is in fact, a linearized theory of gravitation, meaning that only linear terms are significant... and while at first hand you may not understand why this may be, the strange thing is that the linearization still makes quite good approximations of the physics at hand. In other words, while we have been conditioned to think of gravity in terms of geometrization of the metric, the linear approximation just makes as good a case as any and is in fact exploiting the symmetries between gravitation and electromagnetism, in such a way I ask near the end of my essay in conclusions:

 

''At least ponder that perhaps, gravity and electromagnetism is already unified, instead of thinking the other way around, which implies unification happened at some earlier time, or can only be understood as a ‘’phase’’ in the early universe. Perhaps the fact that any force can live side by side, must in a sense of symmetry, unify in some way already.''

 

I say perhaps they are unified together, because in some way, we are already seeing this through the symmetries and success of linear approximation. 

Edited October 12, 2018 by Dubbelosix

[Dubbelosix]

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

 

It's hard not to see similarity to a term we have seen in the last expression of the first equation ~

 

[math]\mathbf{A} = \frac{G(\mathbf{J} \times r)}{r^3c}[/math]

 

At the equitorial plane, the first equation reduces to (to make clear),

 

[math]\Omega = \frac{G \mathbf{J}}{2c^2r^3}[/math]

 

In this simpler demonstration, you can pull out like terms and simplify (with [math]r[/math] becoming the operator curl of the potential)::

 

[math]\Omega = \frac{1}{2c} (\nabla \times \mathbf{A})[/math]

Edited October 12, 2018 by Dubbelosix

[Dubbelosix]

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

 

It's hard not to see similarity to a term we have seen in the last expression of the first equation ~

 

[math]\mathbf{A} = \frac{G(\mathbf{J} \times r)}{r^3c}[/math]

 

At the equitorial plane, the first equation reduces to (to make clear),

 

[math]\Omega = \frac{G \mathbf{J}}{2c^2r^3}[/math]

 

In this simpler demonstration, you can pull out like terms and simplify (with [math]r[/math] becoming the operator curl of the potential)::

 

[math]\Omega = \frac{1}{2c} (\nabla \times \mathbf{A})[/math]

 

A conclusion of a previous derivation not long ago suggested a velocity coupling to torsion 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]

 

Again we identify like terms:

 

[math]\mathbf{A} = \frac{G(\mathbf{J} \times r)}{r^3c}[/math]

 

and again at the equitorial plane, the first equation reduces to

 

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

 

Again, the curl of the gravimagnetic potential removes one such term of the cross product:

 

[math]\nabla \times \mathbf{A} = \frac{G\mathbf{J}}{r^3c}[/math]

 

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

 

we see that the two formula present the same physics as acceleration terms.

Edited October 12, 2018 by Dubbelosix

[Super Polymath]

Okay so, using the time/length dilation/contraction of the planck force upon an omnipresent photo-electric esque quantum foam spacetime aether (in which a tensioactive deleterious fabric of infinitesimal parallel & everted 3D brane-worlds, within the super-strings of the quantum foam) in order to create 84 elementary particles - w/28 +, 28 neutral, & 28 - each denser than the last by factors of 9. Then we can start the cosmos at a few googol years ago within the "Big Anti-Photon" at t=0 and continuously compress super lp by super lp, frame by frame, fill the neutrally charged, - charged, & + charged photons in the "Big Anti-photon" accordingly at t=0 and continue frame by frame, lp by lp, tp by tp, until we get to the Big Expandding Positron that is the current universe at t=

 

 

 

This Big Positron will be a 3D graph of that measures the positions and simultaneously the momentum of every subatomic particle exactly over a volume of 27 orders of magnitude greater than the current observable universe in cubic meters. A simulation that can predict the future. 

Edited October 13, 2018 by Super Polymath

[Super Polymath]

 006 can check that I know the radius is going to be >10^36 ly

 

t should = 1 + (3x10^-89) compressions of the Big Photon. 1.3861775e+66 or super tp per big bounce, not sure how many secs that is off the top of my head, it's over a googol years several thousand Big Bounces, each taking less time to complete than the former as is the nature of E=MC^2 energy being the current universe mass being a Big Crunch or "Blue Epoch" neutral charge being a CMB.

Edited October 13, 2018 by Super Polymath

[Dubbelosix]

Radius of what?

 

Most of what you have said amounts to word salad. I keep asking you to learn physics, instead of pretending to. But you don't seem to want to put any effort in, but instead ''play professor'' games. 

[Dubbelosix]

 

The definitely correct equation for torsion is:

 

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

 

The probably correct field strength is:

 

[math]\mathbf{H}  =\frac{c^2}{4 \pi G}[\Gamma \times \Omega] = \mu_g^{-1}(\Gamma \times \Omega)[/math]

 

Plug in directly:

 

[math]\mathbf{H}  = \frac{1}{8 \pi}[\Gamma \times \frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}] = \mu_g^{-1}\frac{G}{2c^2}(\Gamma \times \frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3})[/math]

 

EDITED

 

 

Just to note, I have edited this post because the previous approach was unnecessarily complicated. Notice we have the nice solution for the field strength:

 

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

Edited October 13, 2018 by Dubbelosix

[Dubbelosix]

 

Under the various analysis I have done, it appears that 

 

[math]\mathbf{H} = \frac{G\mathbf{J}}{cr^3}[/math] 

 

would have the field strength as an acceleration since

 

[math]\mathbf{H} = c\ \Omega = \frac{G\mathbf{J}}{cr^3}[/math] 

 

[math]\Omega = \frac{G\mathbf{J}}{c^2r^3}[/math]

 

(if) it truly is an acceleration, then it has the same dimensons as

 

[math]\Gamma = c\ \Omega[/math]

 

(the gravitational field). If still true, it would also satisfy a spin density:

 

[math] \frac{c^2}{4 \pi G} \Omega = \frac{\mathbf{J}}{r^3}[/math] 

 

From the master equation it was possible to form:

 

[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]

 

and can also be written in the 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]

 

and if the field strength is an acceleration, then we find:

 

[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]

 

The task is now to discern how true these statements are, which involves more analysis. 

[Dubbelosix]

 

The definitely correct equation for torsion is:

 

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

 

The probably correct field strength is:

 

[math]\mathbf{H}  =\frac{c^2}{4 \pi G}[\Gamma \times \Omega] = \mu_g^{-1}(\Gamma \times \Omega)[/math]

 

Plug in directly:

 

[math]\mathbf{H}  = \frac{1}{8 \pi}[\Gamma \times \frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}] = \mu_g^{-1}\frac{G}{2c^2}(\Gamma \times \frac{\mathbf{J} - 3(\mathbf{J} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3})[/math]

 

EDITED

 

 

The issue I have concerns the 'energy flux density' described here:

 

https://en.wikiversity.org/wiki/Gravitational_torsion_field

 

But after some reading, it appears that this is a case of units as in the cgs system of units, the field strength truly is an acceleration term ~ which may mean we cannot just plug my equation into

 

[math]\mathbf{H}  =\frac{c^2}{4 \pi G}[\Gamma \times \Omega] = \mu_g^{-1}(\Gamma \times \Omega)[/math]

 

since this seems to be deemed as an energy density flux. The saving grace is that it appears then if we treat our equations under cgs units, we have the right results for post 20. 

Edited October 13, 2018 by Dubbelosix

[Super Polymath]

Radius of what?

 

Most of what you have said amounts to word salad. I keep asking you to learn physics, instead of pretending to. But you don't seem to want to put any effort in, but instead ''play professor'' games. 

The "Big Positron" that our current observable universe inhabits from beyond one end of the cosmic event horizon to beyond the other. In this cosmic iteration you have constituent particles like the normal positron yet upon the next scalar cosmic iteration these "subatomic particles" existing as density fluctuations of the photo-electron aether (the spacetime foam) are fractals that recur on larger scales, i.e. the "Big Electron"

 

I've reinvented physics to nullify QM entirely and achieve graphical unification of gravity & the S/W nuclear electromagnetic by applying scale relativity in a continuous spacetime foam to the gravitational photo-electron aether analogous to fluctuations as time/length dilation/contraction in the planck force arising from the cancellation in surface area of subplanck or microscopic dimensionally everted spherical braneworlds existing in parallel universes (rank e12+ in the Lie algebra). 

[Super Polymath]

Okay so, using the time/length dilation/contraction of the planck force upon an omnipresent photo-electric esque quantum foam spacetime aether (in which a tensioactive deleterious fabric of infinitesimal parallel & everted 3D brane-worlds, within the super-strings of the quantum foam) in order to create 84 elementary particles - w/28 +, 28 neutral, & 28 - each denser than the last by factors of 9. Then we can start the cosmos at a few googol years ago within the "Big Anti-Photon" at t=0 and continuously compress super lp by super lp, frame by frame, fill the neutrally charged, - charged, & + charged photons in the "Big Anti-photon" accordingly at t=0 and continue frame by frame, lp by lp, tp by tp, until we get to the Big Expandding Positron that is the current universe at t=

 

 

 

This Big Positron will be a 3D graph of that measures the positions and simultaneously the momentum of every subatomic particle exactly over a volume of 27 orders of magnitude greater than the current observable universe in cubic meters. A simulation that can predict the future. 

fixed

[Dubbelosix]

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].

[Super Polymath]

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].

That's super polymath to you, sir. You should be thanking me, after all because without my insights into the existential madness Victor & yourself would be mathematically lost right now. 

[Dubbelosix]

That's super polymath to you, sir. You should be thanking me, after all because without my insights into the existential madness Victor & yourself would be mathematically lost right now. 

 

You're too delusional for me to even take seriously. My efforts have zero [anything] to do with your ramblings. 

[Super Polymath]

You're too delusional for me to even take seriously. My efforts have zero [anything] to do with your ramblings. 

shuuuure. dun worry, our secret's safe with meh!