# Second Essay For The Gravitational Research Foundation

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### #35 Super Polymath

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Posted 08 October 2018 - 02:23 PM

but you already think you know it all.

### #36 Dubbelosix

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Posted 08 October 2018 - 02:31 PM

Oh ok, touch my weak side because you know were I live... I read your post the other day... You love sex with women and why should you be ashamed of it? Well I agree, but why limit yourself to one gender ... That's the way I feel you are like with my equations, I feel like you having some depraved sexual liaison with my equations, with interpretations that I strictly would never allow. I know that was a strange analogy but sometimes you have speak a common language so we get each other.

### #37 VictorMedvil

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Posted 08 October 2018 - 04:40 PM

So, dubbel have you solved for the rotation speed of the universe yet, what is the value of J? Just as there is a value of expansion there should be one for rotation.

J (Universe Rotation) = m(Mass of Universe) L2(Radius of Universe) ?

or

J(Universe Rotation) = D(average density of universe)/r(Radius Universe) ?

Edited by VictorMedvil, 08 October 2018 - 04:55 PM.

### #38 Dubbelosix

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Posted 09 October 2018 - 04:34 PM

So, dubbel have you solved for the rotation speed of the universe yet, what is the value of J? Just as there is a value of expansion there should be one for rotation.

J (Universe Rotation) = m(Mass of Universe) L2(Radius of Universe) ?

or

J(Universe Rotation) = D(average density of universe)/r(Radius Universe) ?

The dimensions are wrong:

$mvr \ne mR^2$

### #39 Super Polymath

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Posted 09 October 2018 - 08:05 PM

From this point everyday is Halloween, we actually begin graphing where t=0 at the "Big Anti-Photon" until now which is t=

Hail Satan

### #40 Super Polymath

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Posted 09 October 2018 - 08:07 PM

### #41 Dubbelosix

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Posted 10 October 2018 - 02:34 PM

$e (\nabla \times \mathbf{B}) = \frac{1}{mc^2 } \frac{1}{r} \frac{\partial^2 U}{\partial r^2} \mathbf{J}$

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

Identifying

$\Omega^2 = \frac{1}{mc^2}\frac{\partial^2 U}{\partial t^2}$

and the conventional expanded definition under ordinary gravitomagnetism, for the torsion field it is:

$\Omega = \frac{G}{2c^2}\frac{\mathbf{L} - 3(\mathbf{L} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}$

then plugging in

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

### #42 Super Polymath

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Posted 10 October 2018 - 04:28 PM

### #43 Super Polymath

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Posted 10 October 2018 - 04:31 PM

ÐÏà¡±á

### #44 Dubbelosix

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Posted 10 October 2018 - 05:20 PM

$e (\nabla \times \mathbf{B}) = \frac{1}{mc^2 } \frac{1}{r} \frac{\partial^2 U}{\partial r^2} \mathbf{J}$

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

Identifying

$\Omega^2 = \frac{1}{mc^2}\frac{\partial^2 U}{\partial t^2}$

and the conventional expanded definition under ordinary gravitomagnetism, for the torsion field it is:

$\Omega = \frac{G}{2c^2}\frac{\mathbf{L} - 3(\mathbf{L} \cdot \frac{\mathbf{r}}{r}) \frac{\mathbf{r}}{r}}{r^3}$

then plugging in

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

A simpler version exists within the first term of the master equation as you can simply write:

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

Edited by Dubbelosix, 10 October 2018 - 09:44 PM.

### #45 Dubbelosix

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Posted 11 October 2018 - 09:11 PM

A simpler version exists within the first term of the master equation as you can simply write:

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

Another solution which continues this work nicely, an opportunity to express the coupling of the gravimagnetic field to the particle velocity, which was spoke about in the work of Sciama and later Motz who defined $\mathbf{B} \times v$. This coupling from the last equation yields the gravitational field strength:

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

and

$\gamma (\mathbf{B} \times v) = \mathbf{H}$

where $\gamma$ is once again the gyromagnetic ratio.

terms cancel

$\frac{\mathbf{J}}{e^2}\ c = \alpha_G \approx 1$

With the field strength defined as:

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

It's also important to point out that this curl (after simplifying terms) has yielded an equivalent term to the gravielectric field, defined through a voltage:

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

Edited by Dubbelosix, 11 October 2018 - 10:21 PM.

### #46 Dubbelosix

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Posted 11 October 2018 - 09:39 PM

It is said that the gravitational field (which needs to be distinguished from the field strength is:

$\Gamma = \nabla \psi + \frac{1}{c}\frac{\partial \mathbf{D}}{\partial t}$

but this notation doesn't sit well for me, nor is this last equation any different in structure to the potentials that define the gravielectric field:

$\mathbf{E} = \nabla \phi + \frac{\partial \mathbf{A}}{\partial r}$

From here it is said, using original notation that the torsion only depends on the vector potential $\mathbf{D}$, in which case we replace it with the gravimagnetic potential:

$\Omega = \nabla \times \mathbf{A}$

Knowing this approach, using a refined notation that cannot be mistaken for other things, we see easily that, with first identifying ~

$\mathbf{A} = \frac{Gm}{r}$

then we see an additional term of the speed of light fixes the units properly:

$c\ \Omega = \frac{Gm}{r^2} = acceleration$

So what the first two equations really describe, appears to be pretty much symmetrical even though notation found from other sources is at odds with the second equation, but as far as consistency, they appear totally the same, both satisfying a scalar and vector potential.

### #47 Dubbelosix

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Posted 11 October 2018 - 10:16 PM

Another solution which continues this work nicely, an opportunity to express the coupling of the gravimagnetic field to the particle velocity, which was spoke about in the work of Sciama and later Motz who defined $\mathbf{B} \times v$. This coupling from the last equation yields the gravitational field strength:

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

and

$\gamma (\mathbf{B} \times v) = \mathbf{H}$

where $\gamma$ is once again the gyromagnetic ratio.

terms cancel

$\frac{\mathbf{J}}{e^2}\ c = \alpha_G \approx 1$

With the field strength defined as:

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

It's also important to point out that this curl (after simplifying terms) has yielded an equivalent term to the gravielectric field, defined through a voltage:

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

A useful reference for the field strength equation: https://en.wikiversi...l_torsion_field

Edited by Dubbelosix, 11 October 2018 - 10:22 PM.

### #48 VictorMedvil

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Posted 11 October 2018 - 10:28 PM

### #49 Dubbelosix

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Posted 11 October 2018 - 11:01 PM

I've been concerned with definitions in the last post. First identify:

$eB = m\omega$

$\gamma B = \Omega$

$\Omega \times v = \mathbf{H}$

for this to be true, the field strength must have dimensions of acceleration, same as we have defined $\Gamma$.

That's just my analysis, yet elsewhere it has been defined with torsion:

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

Which is characteristically different.

and momentum density:

$\mathbf{p} = \frac{1}{c^2} \mathbf{H} =\frac{1}{4 \pi G}[\Gamma \times \Omega]$

Now, can I see why the last equation might arise with conjunction with the first?

In the link we see clearly it defined the field strength and torsion as differing by a factor of c,

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

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

From the momentum density equation you surely do have:

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

In which we can obtain:

$\mu_g \mathbf{H} = \Gamma \times \Omega$

If this is true, we have uncovered a problem with the dimensions found in the equation from the link:

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

They simply do not match up.

reference vs. reference

Edited by Dubbelosix, 11 October 2018 - 11:04 PM.

### #50 Dubbelosix

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Posted 11 October 2018 - 11:26 PM

I know for sure that

$\Omega = c^{-1}\Gamma$

and we have identified

$\frac{\mathbf{H}}{c^2} = \Gamma \times \Omega = \frac{acceleration}{time}$

therefore to match the units of the first equation we would have

$\mathbf{H} = \Omega \times (\Omega \times v) = \dot{\Gamma}$

So long as we are right with what we have done, this last equation (should have the same) units as the field strength $\mathbf{H}$. Now I must make sense of this with

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

and

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

According to these last two equations the field strength and torsion field differ by one factor of velocity - looking back at this equation:

$\mathbf{H} = \Omega \times (\Omega \times v) = \dot{\Gamma}$

we learn that from this last equation, the difference is giving by a factor of time,

$\mathbf{H}\ t = \Omega \times v = \Gamma$

so I stated before that

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

$= \mathbf{H}$

So something does at first glance appear wrong with the definition of the field strength:

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

But these are the kinds of things I take enjoyment figuring out.

Edited by Dubbelosix, 11 October 2018 - 11:29 PM.

### #51 Dubbelosix

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Posted 12 October 2018 - 12:09 AM

The definitely correct equation for torsion is:

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

The probably correct field strength is:

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

Plug in directly:

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

EDITED

Edited by Dubbelosix, 13 October 2018 - 05:20 AM.