# Permittivity And Permeability In Gravitoelectromagnetism

relativity quantum mechanics

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### #18 Dubbelosix

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Posted 27 March 2018 - 01:05 PM

So, now let's compare Maxwell's theory

$\nabla \times \mathbf{B} = \mu_0 \mathbf{j} + \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}$

with our results:

$\nabla \times \mathbf{B} = -\mu_G \frac{\partial \mathbf{B}}{\partial t} + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t}$

This implies that $\frac{\partial \mathbf{B}}{\partial t} \equiv \mathbf{j}$... so is this true (see Biot-Savart Law)? Well yes as one such factor of the inverse speed of light which must account for the permeability coefficient. Because

$\nabla \times \mathbf{B} = \frac{1}{ct} \mathbf{B}$

under dimenisonal analysis, yet we also have

$\nabla \times \nabla \times \mathbf{A} = \frac{1}{ct} \mathbf{B} = \frac{1}{(ct)^2} \mathbf{A}$

And we know that

$\frac{1}{(ct)^2} \mathbf{A} = \nabla \times \nabla \times \mathbf{A} = -\nabla^2 \mathbf{A} = \mu_G \mathbf{j}$

under the Coulomb gauge.

Edited by Dubbelosix, 28 March 2018 - 10:32 AM.

### #19 Dubbelosix

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Posted 28 March 2018 - 09:47 AM

So yes after many scribbles and association to like terms within my equations and Maxwell's we do learn something new.

Maxwells theory tells you that a curl in the magnetic field is proportional the rate of the change of the electric field. Likewise, in relativity, the curl of a magnetic field gives rise to the proportionality to a change in the electric field. And a change in the magnetic field is again the curl of the magnetic... it tells you the two are inescapable for non-static systems.

When we chose a specific spin-orbit magnetic equation, we found a new relationship. Not only do we find in Maxwells theory dependence between the two curls of the electric and magnetic forces to the rate of changes in said fields, we also find from my formalism that a curl in the gravimagnetic field is in fact proportional the gravielectric field as an approximated Sciama equation, but here, for proportionality issues it is simply:

$\nabla \times \mathbf{B} \propto \nabla \times \mathbf{E}$

I haven't seen this definition even with new formalisms of Maxwell's equation. It strongly suggests for cyclotron motion, that the system is subject to both curls in both fields and must be both non-vanishing. In regards to this proportionality if no one has declared it before me, appears to be a law in itself of the fundamental relationships for spin-orbit equations.

Edited by Dubbelosix, 28 March 2018 - 10:45 AM.

### #20 Dubbelosix

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Posted 28 March 2018 - 10:26 AM

Perhaps I haven't seen this proportionality before because it applies only to spin-orbit coupling dynamics.

### #21 Dubbelosix

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Posted 28 March 2018 - 10:56 AM

So, since I have now convinced myself with the relationship

$\nabla \times \mathbf{B} \propto \nabla \times \mathbf{E}$

as being some fundamental relationship for spin-orbit equations, I suppose I need to now warrant an investigation into what role the $\nabla \times \mathbf{E}$ plays when spin orbit equations are taken into consideration. Right now, I can think of a few examples in which non-zero electric forces may be experienced in a non-linear motion. Also, the relationship we just gave, really shouldn't be a facet of the theory we should be surprised with considering the general interdependence of both the simplified equations:

$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$

$\nabla \times \mathbf{B} = -\frac{\partial \mathbf{E}}{\partial t}$

Edited by Dubbelosix, 28 March 2018 - 10:56 AM.

### #22 Dubbelosix

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Posted 03 April 2018 - 02:33 PM

these curls should not be the basis of my investigations... it has stole a lot of time from me since I made this last post, I cannot understand yet why the spin orbit coupling allows the curl of an electric field... but my first guess is that it has to do with the gravimagnetic curl relationship giving rise to those gravielectric interactions. I have more later I will write up regarding other investigations in this area.

### #23 Dubbelosix

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Posted 03 April 2018 - 03:51 PM

First let's understand the dynamics with even more depth.

The gravimagnetic field relationship with the electric field in which a charge moves through is

$\mathbf{B} = -\frac{v \times \mathbf{E}}{c^2}$

(which is not) some speculative function, but actually part of the first principles of the objects required to explain spin orbit coupling, so this must be its gravielectric analogue.

$v$ is the velocity of an electron. It's a non-relativistic limit though its shows that the gravielectric field are combined as a radial expression of the form:

$\mathbf{E} = |\frac{E}{r}| \mathbf{r}$

substition of the momentum $p = mv$ followed by changing the order of cross products will give

$\mathbf{B} = \frac{r \times p}{mc^2} |\frac{E}{r}|$

It is said from here in literature that one can express the electric field as the gradient of an electrostatic potential, an object no less we have investigated before vigorously:

$\mathbf{E} = -\nabla V$

where the usual central potential arises from circular gauges

$\frac{\partial V}{\partial r} = \frac{1}{e} \frac{\partial U(r)}{\partial r} \mathbf{J}$

It is important to note this has featured in previous equations which explored gravimagnetic couplings in the spin orbit regime.

### #24 Dubbelosix

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Posted 03 April 2018 - 03:55 PM

However... and here's the rub... with it comes a consequence of assumption...

I quote wiki very quickly:

''Here we make the central field approximation, that is, that the electrostatic potential is spherically symmetric, so is only a function of radius.''

Which from my investigations is not a general assumption, (meaning it applies to all systems). The spherical configuration for instance, of a universe and even a bubble are both radius and temperature dependant.

Edited by Dubbelosix, 03 April 2018 - 04:01 PM.

### #25 Dubbelosix

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Posted 03 April 2018 - 03:59 PM

The equivalence of that gravimagnetic field with the primary gravimagnetic equation may be implied as:

$- \frac{v \times \mathbf{E}}{c^2} = \frac{1}{mc^2 r}(\frac{\Phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} \approx \frac{1}{mc^2 r}(\frac{1}{c^2} \frac{c^2}{4 \pi G})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{mc^2 r}\frac{1}{4 \pi G}\frac{\partial v}{\partial t} \mathbf{J}$

The coupling of $v \times \mathbf{B}$ as the usual gravimagnetic coupling is also proportional to the gravielectric coupling of velocity

Edited by Dubbelosix, 03 April 2018 - 04:14 PM.

### #26 Dubbelosix

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Posted 03 April 2018 - 04:03 PM

sorry had to edit there, I meant a coupling to electric field, all fixed

### #27 Dubbelosix

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Posted 03 April 2018 - 04:26 PM

And this remains as an obvious assumption of dynamics

$\nabla \times \mathbf{B} = \nabla \times (- \frac{v \times \mathbf{E}}{c^2}) = \nabla \times (\frac{1}{mc^2 r}(\frac{\Phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J}) \approx \nabla \times (\frac{1}{mc^2 r}(\frac{1}{c^2} \frac{c^2}{4 \pi G})\frac{\partial v}{\partial t} \mathbf{J})$

### #28 Dubbelosix

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Posted 03 April 2018 - 05:25 PM

The equivalence of that gravimagnetic field with the primary gravimagnetic equation may be implied as:

$- \frac{v \times \mathbf{E}}{c^2} = \frac{1}{mc^2 r}(\frac{\Phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} \approx \frac{1}{mc^2 r}(\frac{1}{c^2} \frac{c^2}{4 \pi G})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{mc^2 r}\frac{1}{4 \pi G}\frac{\partial v}{\partial t} \mathbf{J}$

The coupling of $v \times \mathbf{B}$ as the usual gravimagnetic coupling is also proportional to the gravielectric coupling of velocity

Now... it's not unknown for me to search for simplifications... do you see a simplification here based on the dimensions of this equation? Well, one ''common factor'' we can pull out of both sides of the equations is the squared speed of light.

$- v \times \mathbf{E} = \frac{1}{mr}(\frac{\Phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} \approx \frac{1}{mc^2 r}(\frac{c^2}{4 \pi G})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{mr}\frac{1}{4 \pi G}\frac{\partial v}{\partial t} \mathbf{J}$

Notice from early investigations we found magnetisms analogue to this from gravitational coupling to magnetism in the Sciama and later the Motz model as:

$v \times \mathbf{B} = - \frac{\omega \times v}{\sqrt{G}}$

This is of course a very close component of the Lorentz force.

### #29 Dubbelosix

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Posted 11 April 2018 - 06:49 AM

Either a discrepancy has been made with one of the equations, or an equation I have noted has set a constant to 1. The reason why I think this goes as following:

I take us to recall two equations of interest for me this time around:

$\frac{\partial V}{\partial r} = \frac{1}{e} \frac{\partial U(r)}{\partial r} \mathbf{J}$

$\mathbf{B} = \frac{1}{mc^2 r}(\frac{\Phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} \approx \frac{1}{mc^2 r}(\frac{1}{c^2} \frac{c^2}{4 \pi G})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{mc^2 r}\frac{1}{4 \pi G}\frac{\partial v}{\partial t} \mathbf{J}$

Since both

$\mathbf{E} \equiv \frac{\partial V}{\partial r} = \frac{1}{e} \frac{\partial U(r)}{\partial r} \mathbf{J}$

and the Sciama gravielectric field is

$\mathbf{E} = (\frac{\Phi}{c^2})\frac{\partial v}{\partial t}$

and (actually may still hold to be approximate if the gravitational permittivity and permeability relationships hold, which they seem to do). Plug our new relatonship in

$\mathbf{B} = \frac{1}{mc^2 r}\frac{1}{e} \frac{\partial U(r)}{\partial r} \mathbf{J} \cdot \mathbf{S} = \frac{1}{mc^2 r}\frac{\partial V}{\partial r} \mathbf{J}$

Crucially what it shows more important though, is that the spin orbit coupling has derivatives which rely on the former equation to be an approximate to a more fundamental underlying theory involving the gravitational permeability and permittivity. The fact our terms could match the usual Maxwell equation only seems to varify that the Sciama definition of the field was related to those constants from the first place a certain fact it seems, he seemed to have missed about his theory.

$\approx \frac{1}{mc^2 r}(\frac{1}{c^2} \frac{c^2}{4 \pi G})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{mc^2 r}\frac{1}{4 \pi G}\frac{\partial v}{\partial t} \mathbf{J}$

Notice, the former results in the gravimagnetic spin orbit weak coupling! Now, something strange has popped out of the equation, which I think has been a result of Sciama's application of cgs units to the Newton constant $G$. Not sure yet, will need to investigate it, because the normal gravielectric field is

$\mathbf{E} = \frac{\partial V}{\partial r} = \frac{1}{e}\frac{\partial U}{\partial r}$

Which means the first equation is at odds with the dimensions...

$\frac{\partial V}{\partial r} = \frac{1}{e} \frac{\partial U(r)}{\partial r} \mathbf{J}$

I wonder how this happened, unless the previous equation has set a Planck constant equal to 1. I'll find the answer to this soon.

Edited by Dubbelosix, 11 April 2018 - 06:50 AM.

### #30 Dubbelosix

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Posted 25 April 2018 - 02:11 PM

So yeah, I have been away from this for a while, but I have been drawing up new ideas. The gravielectric and gravimagnetic couplings, do appear to be linked dynamically to Sciama's definition of the ratio of the gravitational potential and the speed of light squared through a dynamic factor of a gravitational interpretation of the permittivity and permeability. He didn't seem to realize this because you can't really ''do away'' with the factor of $4 \pi$ that comes from using those definitions and the fact that they seem to play those direct roles would actually make the Sciama theory an approximate.

It has been argued by Susskind, that gravity could play a significant role at the fundamental level. I have also shown in previous investigations (and the main topic for my essay to the GRF), Penroses idea's concerning whether gravity was able to collapse a wave function. I argued in my essay a thought experiment in which a particle shielded from all other fundamental fields other than gravity, could a wave function collapse upon its own weight, interpreted by a center of gravity which fluctuated around its absolute square value?

In the thought experiment, a standing wave is actually more compact with respects to its own interference closer to the center of where the wave diverges from. Since this implies that gravity is not in equilibrium, the tugging of the superpositioned states (if considered in a real physical sense) would be directed towards the center. However the center is in fact not defined either due to the wave function which is why the wave function will fluctuate around a square value of its wave function. You can consider, at any moment the square of the value of its wave function exists, is theoretically its path through space. Certainly, in the pilot wave theory, it is clear this is totally consistent. It's not so much that the particle is smeared through space as a wave - there is an extra dynamical feature of the vacuum which appears to be essentially the foundation for it all. There is a ''response'' of the vacuum due to the motion of a particle - remember, all things are in motion in relativity, this is even true in quantum mechanics due to the uncertainty principle, so even a standing wave is not truly standing... at least no more than Einstein's notion of a ''rest system.''

This response of a moving particle generating waves around it, which further dictates the motion of the particle itself, is actually well understood to have a complete analogy in nature. (Chase my blogs on that stuff). This idea that the particle is always there, moving through space, generating the waves which consequently dictates its evolution, totally explains why every time we try and locate a particle, it is seen as a paricle. Copenhagen interpretation, doesn't care about such things. It states, what cannot be seen may as well be ignored.

Edited by Dubbelosix, 27 April 2018 - 02:48 PM.