Second Essay For The Gravitational Research Foundation

[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].

 

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]

[Dubbelosix]

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 October 14, 2018 by Dubbelosix

[Dubbelosix]

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 October 15, 2018 by Dubbelosix

[Dubbelosix]

Okay, my paper has now been updated for an alternative interpretation of the redshift, at least partly. It has been extended for a summary on my work on models towards a fundamental cosmological theory which involved making it part of the rotation group and so part of the full Poincare group of symmetries. It's also been updated to now include the bivector theory of gravity since it too is an example of a gravielectromagnetric theory under linearized gravity.

 

https://blackholeradiation.quora.com/The-Relativity-of-Black-Hole-Thermodynamics-and-Linearized-Gravity?utm_source=quora&utm_medium=referral

[exchemist]

Just glancing quickly down your paper, I have not studied it. But noted dark matter is not mentioned, and Dark energy gets one comment "Might it be there is no such discrepancy and dark energy is really off-shell zero point fluctuations? ". You also have a variable gravitational constant, and a "technically" variable speed of light. How do you justify these statements. 

 

"The answer may be surprisingly simple - zero point fluctuations do not generally live long enough to interact with real matter in the vacuum so the presence of this energy is completely shielded from our experimental prowess… Well almost shielded, since experimentation shows we cannot freeze a system to absolute zero, an indirect evidence of the fluctuations." This statement would over throw all of Verlindes ideas out the window.

Ha. It is also complete bunk to suggest that zero point energy prevents the attainment of absolute zero.

 

Zero point energy does not contribute to temperature in the first place, as by definition it can never be extracted from the body in which it is present. 

 

Experiment simply shows that one can only approach absolute zero asymptotically, as would be expected purely classically. It is not a quantum phenomenon at all. 

 

P.S. More on this here: https://en.wikipedia.org/wiki/Absolute_zero

Edited January 2, 2019 by exchemist

[Dubbelosix]

I have justified the change in the gravitational constant from experimentation: It's commonly known that all experiments conducted to measure the exact value of G has shown it to vary over time, as though it shuffles around a minimum, significantly at times. Following on from the statement that gravitational waves have now been detected, this means the speed of light must vary according to the permittivity and permeability.  A consequence of the gravitational aether involves key concepts that involve the two parameters (for instance, permittivity and permeability) and predicts that light can escape black holes, because the speed of light can only approach zero, but never reach it.

 

So it has some good application to solve the information paradox.

[Dubbelosix]

Ha. It is also complete bunk to suggest that zero point energy prevents the attainment of absolute zero.

 

Zero point energy does not contribute to temperature in the first place, as by definition it can never be extracted from the body in which it is present. 

 

Experiment simply shows that one can only approach absolute zero asymptotically, as would be expected purely classically. It is not a quantum phenomenon at all. 

 

 

Exchemist, you are a chemist yes? Let me fix you in on something, zero point fields are taught at Stanford, they are spoke about by only the top physicists of the world and the experimental consequence which has been well-tested, is that gases cannot be made to reach any such thing as a zero point, there is no such thing as a perfect vacuum.

 

It's nice learning yes?

[Dubbelosix]

At extremely cold temperature, the quantum jiggling of particles can be made to go very close to some kind of definition of a zero point field however... some of the coldest places in the universe can only be found in the lab! The minimum jiggling in the ground state, is due to a combination of the uncertainty principle and also a temperature contrubution of [math]\frac{1}{2}\hbar \omega[/math]. Planck first suggested it, later Einstein used it for a gas.

[exchemist]

The energy of the electrons in the ground state of any atom is zero point energy.

 

As is the residual energy in the ground vibrational state of any chemical bond.

 

There is nothing exotic about it.

 

The energy of such states does not contribute to temperature, as it is unextractable from the atom or molecule. As the Wiki article explicitly states, zero point energy remains at absolute zero.  

 

So the inability to attain absolute zero cannot and does not have anything to do with zero point energy.

Edited January 2, 2019 by exchemist

[OceanBreeze]

Do you have a reference for the gravitational constant G varying over time.

LIGO has demonstrated that both gravitational waves and light travel at the same speed ie c. 

 

 

 

 

 

Of course he has a reference.

 

It is his own work!

 

which of course, nobody else recognizes.

[Dubbelosix]

Do you have a reference for the gravitational constant G varying over time.

LIGO has demonstrated that both gravitational waves and light travel at the same speed ie c. 

 

 

 

Are you ignoring the causes of Dark Energy ?  or what might be the dark matter effect ?

 

The passage you are looking for is titled gravitational aether, and it has all the references you need. It is a real model, contrary to what the professor above says.

 

Yes, and no. I don't believe dark energy is what we understand it, the elasticity of spacetime depends on a varying metric tensor, which by definition must change as a universe expands (see Carrol for similar remarks). Dark energy in my own model, not particularly new, but the modified Friedmann equation is new: It takes into consideration a non-conservation which has implication to how we understand other equations, like the wheeler de witt equation. 

 

You'll find reference on the net to a varying cosmological G. It's a big thing now in physics because of the evidence supporting it.

[Dubbelosix]

The energy of the electrons in the ground state of any atom is zero point energy.

 

 

Chemists often say this, but it is fundamentally wrong.

[Dubbelosix]

It is certainly wrong within the context of quantum mechanics, as an approximation for chemistry, its often erroneously called the zero point state, but there is no such zero point state, there is only a ground state and it is NOT absolute.

 

But you don't know this, because you are a chemist.

[Dubbelosix]

Very well, the link is here: https://phys.org/news/2015-04-gravitational-constant-vary.html

 

I don't actually raise the issue of dark matter in my paper, but I do have a suspicion as I have written before, that there is a polarization of the gravitational field which is proportional to the mass of the supermassive black hole. The typical energy of a supermassive black hole like our own for a spiral galaxy, is the same energy required for the gravitational binding.

[exchemist]

It is certainly wrong within the context of quantum mechanics, as an approximation for chemistry, its often erroneously called the zero point state, but there is no such zero point state, there is only a ground state and it is NOT absolute.

 

But you don't know this, because you are a chemist.

Cornell university appears to agree with me, not you, about this: http://muchomas.lassp.cornell.edu/8.04/Lecs/lec_Heisenberg/node7.html

 

Any oscillator or system resembling one has a zero point energy. This includes vibrational states of molecules and also electronic ground states. 

 

Wikipaedia says the same thing: https://en.wikipedia.org/wiki/Ground_state

 

I quote the first line: "The ground state of a quantum-mechanical system is its lowest-energy state; the energy of the ground state is known as the zero-point energy of the system."

 

That goes, as I say, for any QM bound system in which the ground state is not at zero energy. Rotation in QM does have a stationary ground state so there is no zero point energy for rotation. In fact E(J) = (h/2π)².J(J+1), so when J =0, E=0.  But for vibration and for electronic states there is an energy in the ground state, and so there is a zero point energy. 

Edited January 2, 2019 by exchemist

[Dubbelosix]

I don't believe in standard dark matter, I never have. certainly, I do not think of dark matter as some variable parameter of particles. Dark matter is probably, as I said, something related to the gravitational influence of the supermassive black hole. The variable gravitational constant is a natural consequence of the gravitational aether and the notion that space is made from permittivity and permeability variables.

[Dubbelosix]

Cornell university appears to agree with me, not you, about this: http://muchomas.lassp.cornell.edu/8.04/Lecs/lec_Heisenberg/node7.html

 

Any oscillator or system resembling one has a zero point energy. This includes vibrational states of molecules and also electronic ground states. 

 

Wikipaedia says the same thing: https://en.wikipedia.org/wiki/Ground_state

 

I quote the first line: "The ground state of a quantum-mechanical system is its lowest-energy state; the energy of the ground state is known as the zero-point energy of the system."

 

That goes, as I say, for any QM bound system in which the ground state is not at zero energy. Rotation in QM does have a stationary ground state so there is no zero point energy for rotation. In fact E(J) = (h/2π)².J(J+1), so when J =0, E=0.  But for vibration and for electronic states there is an energy in the ground state, and so there is a zero point energy. 

 

Cornell university appears to agree with me, not you, about this: http://muchomas.lassp.cornell.edu/8.04/Lecs/lec_Heisenberg/node7.html

 

Any oscillator or system resembling one has a zero point energy. This includes vibrational states of molecules and also electronic ground states. 

 

Wikipaedia says the same thing: https://en.wikipedia.org/wiki/Ground_state

 

I quote the first line: "The ground state of a quantum-mechanical system is its lowest-energy state; the energy of the ground state is known as the zero-point energy of the system."

 

That goes, as I say, for any QM bound system in which the ground state is not at zero energy. Rotation in QM does have a stationary ground state so there is no zero point energy for rotation. In fact E(J) = (h/2π)².J(J+1), so when J =0, E=0.  But for vibration and for electronic states there is an energy in the ground state, and so there is a zero point energy. 

 

 

Some rouge scientists think that fluctuations do not exist, I can wholeheartedly assure you, that the majority of scientists in the field, not only understand zero point fields as a ''limit'' that can not be reached, but additionally that field theory strongly predicts fluctuations that exist for very short time periods. The fact is, if you want to recite papers, I have a whole wad of them which support the existence of zero point fluctuations. You have one.