# Permittivity And Permeability In Gravitoelectromagnetism

relativity quantum mechanics

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### #1 Dubbelosix

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Posted 21 March 2018 - 09:21 AM

There is a special relationship between permittivity and permeability in the Sciama model that I am not aware that Sciama himself was aware of.

There does exist in literature, a direct analogy of the gravitational permeability and permittivity. It is stated that $\epsilon_G = \frac{1}{4 \pi G}$ is the gravitational analogue of the vacuum permittivity $\epsilon$. Gravitational waves have been experimentally varified to move at the speed of light - this is related in analogy (Sivaram, Sabatta) with electromagnetism $c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$ and that a corresponding gravitational interpretation is $c = \frac{1}{\sqrt{\epsilon_G \mu_G}}$. This implies the gravitational permeability is $\mu_G = \frac{4 \pi G}{c^2}$. Sciama defined the gravielectric field as using cgs units for the gravitational constant:

$(\frac{\Phi + \phi}{c^2}) = -\frac{1}{G}$

*the units chosen for G ensures that the ratio of the scalar and the speed of light squared is not dimensionless, as often encountered in physics.

This equation may be understood when reformulated to represent an inverse gravitational permeability:

$\Phi = -\frac{c^2}{4 \pi G}$

That mean's Sciama's gravielectric scalar potential is itself the definition of the inverse permeability. A unique property of these special units is that the ratio and inverse ratio have the relationships:

$\frac{c^2}{\Phi} = \mu_G c^2 = -\frac{4 \pi G c^2}{c^2} = -4 \pi G = \frac{1}{\epsilon_G}$

$\frac{\Phi}{c^2} = \frac{1}{\mu_G c^2} = -\frac{c^2}{4 \pi G c^2} = -\frac{1}{4 \pi G} = \epsilon_G$

Both these quantities appear within the framework of a single gravitomagnetic coupling equation to angular momentum

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

Sciama never attributed the inverse G to the permeability but doing so makes Sciama's equation an approximate.

Edited by Dubbelosix, 28 March 2018 - 10:36 AM.

### #2 Dubbelosix

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Posted 21 March 2018 - 09:29 AM

### #3 exchemist

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Posted 21 March 2018 - 03:17 PM

In case any readers might be confused, the relationship between the speed of light, c, and the electrostatic permittivity, ε₀ and magnetic permeability, μ₀ of free space is:

c = 1/√(ε₀.μ₀),

not  c = √(ε₀.μ₀), which anyone with a reasonable grasp of physics will immediately realise would lead to a laughably slow speed for light!

Edited by exchemist, 21 March 2018 - 03:23 PM.

### #4 Dubbelosix

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Posted 21 March 2018 - 04:24 PM

In case any readers might be confused, the relationship between the speed of light, c, and the electrostatic permittivity, ε₀ and magnetic permeability, μ₀ of free space is:

c = 1/√(ε₀.μ₀),

not  c = √(ε₀.μ₀), which anyone with a reasonable grasp of physics will immediately realise would lead to a laughably slow speed for light!

Yes. Permittivity is

$\frac{1}{4 \pi G} = \epsilon_G$

Permeability is

$\frac{4 \pi G}{c^2} = \mu_G$

Those relationships must be combined as

$c^2 = \frac{1}{\epsilon_G \mu_G}$

Or, as exchemist has shown

$c = \frac{1}{\sqrt{\epsilon_G \mu_G}}$

Edited by Dubbelosix, 21 March 2018 - 04:30 PM.

### #5 Dubbelosix

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Posted 21 March 2018 - 04:29 PM

I have edited the main work to account for the discrepancy. It didn't effect the work, I did after all know what I meant

### #6 Dubbelosix

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Posted 21 March 2018 - 04:36 PM

Just to update, though we covered this equation, the previous posts here may suggest that was the only spin magnetic equation that took into respect those gravimagnetic and gravielectric constants. The curl of the gravielectric field is just as important and just as dynamically interesting

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

I take some time in my blogs to go over why these couplings could be important - for instance, one obvious way forward was in itself a unification interpretation (too strong to call it a true attempt so far at unification), but is a treatise with all previous work for full gravitomagnetic and gravielectric spin orbit couplings which I have never encountered before:

https://consciousnes...tein-Relativity

Edited by Dubbelosix, 28 March 2018 - 10:38 AM.

### #7 exchemist

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Posted 22 March 2018 - 02:45 AM

I have edited the main work to account for the discrepancy. It didn't effect the work, I did after all know what I meant

Affect.

### #8 Dubbelosix

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Posted 22 March 2018 - 08:13 PM

We can continue a little further now since we have identified the curl of the gravielectric force in post 6. That same curl can in principle be applied to the gravimagnetic field, yielding like terms:

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

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

The current density $\mathbf{j}$ has a relationship given using the Lorentz gauge, Maxwell's equations can be written in terms of the potential and the scalar potential

$\nabla^2 \phi - \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} = - \frac{\rho}{\epsilon_G}$

$\nabla^2 \mathbf{A} - \frac{1}{c^2} \frac{\partial \mathbf{A}}{\partial t^2} = - \mu_G j$

Really, this is no different to a four dimensional curl. From it you can identify the Coulomb gauge condition:

$\nabla^2 \mathbf{A}_x = -\mu_G j_x$

$\nabla^2 \mathbf{A}_y = -\mu_G j_y$

$\nabla^2 \mathbf{A}_z = -\mu_G j_z$

Edited by Dubbelosix, 23 March 2018 - 03:20 AM.

### #9 Dubbelosix

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Posted 23 March 2018 - 09:36 PM

A full write up of the entire theory as I have studied it can be found here: https://spinorbitcoupling.quora.com/

They give not only the spin orbit couplings, but a wide range of other investigations which included the gravitational charge - you can't avoid viewing mass as a charge in gravielectromagnetism.

### #10 Super Polymath

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Posted 24 March 2018 - 12:27 AM

The quantum eraser deletes the past, it deletes time, which is a part of space. Einstein's spacetime ether is permeatitive gravity, it's also non-vanishing in contrary to planck mass, this permativity you refer to.

So in my DID theory posted last night I said that if de sitter space has 2.6 dimensions, ADS will have 2.4, & motion/gravitoelectromagnetism will not exist because this bi-dimensional ether is stable, nothing is getting quantum mechanically erased.

However, at points with low thermodynamic conductivity in de sitter space where the dimensions are greater than 2.6 the ether has to stabilize via the quantum eraser. Where they were less than 2.4, the dimensions had to compensate via expansion.

I showed this by adding a special case bi-brane inequality operation that held up to every possible piece of factual LCDM data I could throw at it, even regarding what's known about expansion, electromagnetism, the nuclear forces, and quantum electrodynamics.

Edited by Super Polymath, 24 March 2018 - 12:29 AM.

### #11 Dubbelosix

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Posted 27 March 2018 - 09:06 AM

I like doing this in bits here because I know not everyone can follow the vocabulary or even the math. The math isn't too difficult a lot of the time as I tend to see beauty in simplicity.

I want to draw on relationships between the identity of the curl of the gravimagnetic field

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

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

and that which is already given through Maxwell's equations

$\nabla \times \mathbf{B} = \mu_0 \mathbf{j} + \epsilon_0\mu_0\frac{\partial \mathbf{E}}{\partial t}$

or

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

This formulation is of course explaining how motion in the magnetic field gives rise to a changing electric field. In fact, a changing electric field also gives rise to magnetism as they are relative. In our formulation, we took the gravielectric field's definition explicitely as it exists within the framework of spin-orbit coupling equations. What we found was part of Maxwell's identities. Certainly the current part with a coefficient of permeability is defined clearly in literature as equalling the Coulomb gauge $-\nabla^2\mathbf{A}$. Knowing Maxwell's relationships, we can see richer physics within the structure of our equation. Certainly a current term will be involved for any motion of a charged system. Our equations says something else - not only does a curl of the magnetic field give rise to a changing electric field $\frac{\partial \mathbf{E}}{\partial t}$ in Maxwell's theory but our equation says something about a curl of a gravimagnetic field being proportional to a curl in the gravielectric field $\nabla \times \mathbf{B} \propto \nabla \times \mathbf{E}$ through a proportionality constant of $\frac{\mathbf{J}}{mc^2r} = \frac{1}{c}$.

Edited by Dubbelosix, 27 March 2018 - 09:18 AM.

### #12 Dubbelosix

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Posted 27 March 2018 - 09:34 AM

Let's do some theorizing, let's assume this part plays the role of the current:

$-\frac{\mathbf{J}}{mc^2 r}\frac{\partial \mathbf{B}}{\partial t} = \mu_G \mathbf{j}$

We speculate this would be the component playing the role of the current since it has the coefficient associated to our approximated Sciama equation $\Phi = \frac{c^2}{4 \pi G} = \mu_G$Then we know from Maxwell's equations there are additional terms. We could speculate including those like

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

$= \nabla \times (-\frac{1}{mc^2 r}\frac{1}{4 \pi G}\frac{\partial v}{\partial t} \mathbf{J}) + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t}$

$= -\frac{\mathbf{J}}{mc^2 r}\frac{\partial \mathbf{B}}{\partial t} + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t} = \mu_G\mathbf{j} + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t} = \nabla(\nabla \cdot A) - \nabla^2 \mathbf{A} + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t} = - \nabla^2 \mathbf{A} + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t}$

I'm being very careful to track dimensions of the equation so far, just looking at it, nothing actually seems wrong, but there could be a factor of something missing somewhere which I missed I'll take a better look at a later date. Assuming nothing is wrong, then we surprisingly see that the current is formulated with derivatives associated to the curl of the Sciama gravielectric field from the spin-orbit coupling equations.

Edited by Dubbelosix, 27 March 2018 - 09:40 AM.

### #13 Dubbelosix

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Posted 27 March 2018 - 09:40 AM

fixed a sign problem in the current.

### #14 Dubbelosix

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Posted 27 March 2018 - 09:57 AM

Now, let's just take a look at the current relationships as the curl of the gravielectric field (is first of all, unusual because currents have in them definitions rooted from the curl of the magnetic field).

$\nabla \times \mathbf{B} = \frac{\mathbf{J}}{mc^2 r}( \nabla \times \mathbf{E}) + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t}$

And what we will do, is see if there is any physics, but before that we'll check to see what the dimensions are and whether first of all its right. All that when I come back which will be soon I need to leave for an hour.

Edited by Dubbelosix, 27 March 2018 - 10:00 AM.

### #15 Dubbelosix

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Posted 27 March 2018 - 10:38 AM

Right, I am back. While walking down the town I worked out the dimensions in my head, and they all seem to be in harmony:

$\frac{1}{ct} \mathbf{B} = \frac{1}{c}(\frac{1}{ct} \mathbf{E}) + \frac{1}{c}\frac{\mathbf{E}}{ct}$

They work absolutely fine under these units. So what we found about a relationship between the curl of the gravimagnetic field definition for the current only depends on the relationships between gravitational permeability and permittivity. In this sense, we have found a new definition of the current with a source term found in the curl of the gravielectric field. It may not be significant, but it could possibly be, since it was found from non-trivial spin-orbit analogy equations which following the rules we worked with, found a relationship of the curl of the gravimagnetic field being proportional to that of the gravielectric field - but this all rests on assumptions we made about associating the gravimagnetic of the form we gave to be equivalent to a current

$-\frac{\mathbf{J}}{mc^2 r}\frac{\partial \mathbf{B}}{\partial t} = \mu_G \mathbf{j}$

Edited by Dubbelosix, 27 March 2018 - 10:39 AM.

### #16 Dubbelosix

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Posted 27 March 2018 - 10:43 AM

However I should note it may not be entirely unusual to think of the current relationship since it has changes in the definition of the magnetic field. So. of course, its not unusual to think of a magnetic current as arising from a varying magnetic field.

### #17 Dubbelosix

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Posted 27 March 2018 - 12:33 PM

The constant of proportionality $\frac{1}{c}$ can be interpreted as $\mu_G$ in

$\nabla \times \mathbf{B} = -\frac{\mathbf{J}}{mc^2 r}\frac{\partial \mathbf{B}}{\partial t} + \epsilon_G\mu_G \frac{\partial \mathbf{E}}{\partial t}$

simply because the product $\epsilon_G \mu_G = \frac{1}{c^2}$ exists, so we may happily plug this in

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

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

$-\frac{\mathbf{J}}{mc^2 r} = -\frac{1}{c} = \mu_G$

The sign arises to satisfy the same reasons why the sign exists in Sciama's definition, except he never went this far with it and he never considered his gravielectric and gravimagnetic fields as actually approximates to a more fundamental relationship using the gravi-permeability and the gravi-permittivity.

Edited by Dubbelosix, 27 March 2018 - 12:43 PM.