but you already think you know it all.

# Second Essay For The Gravitational Research Foundation

### #35

Posted 08 October 2018 - 02:23 PM

### #36

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

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) L^{2}(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

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) L

^{2}(Radius of Universe) ?

or

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

The dimensions are wrong:

[math]mvr \ne mR^2[/math]

### #39

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

Posted 09 October 2018 - 08:07 PM

### #41

Posted 10 October 2018 - 02:34 PM

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

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

Identifying

[math]\Omega^2 = \frac{1}{mc^2}\frac{\partial^2 U}{\partial t^2} [/math]

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

### #43

Posted 10 October 2018 - 04:31 PM

ÐÏà¡±á

### #44

Posted 10 October 2018 - 05:20 PM

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

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

Identifying

[math]\Omega^2 = \frac{1}{mc^2}\frac{\partial^2 U}{\partial t^2} [/math]

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

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

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

[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]

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

### #45

Posted 11 October 2018 - 09:11 PM

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

[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]

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 [math]\mathbf{B} \times v[/math]. This coupling from the last equation yields the gravitational field strength:

[math]\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}[/math]

and

[math]\gamma (\mathbf{B} \times v) = \mathbf{H}[/math]

where [math]\gamma[/math] is once again the gyromagnetic ratio.

terms cancel

[math]\frac{\mathbf{J}}{e^2}\ c = \alpha_G \approx 1[/math]

With the field strength defined as:

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

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:

[math]|\mathbf{E}| = \frac{\alpha_G}{e} \frac{\partial U}{\partial r} = \alpha_G \frac{\partial V}{\partial r}[/math]

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

### #46

Posted 11 October 2018 - 09:39 PM

### #47

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 [math]\mathbf{B} \times v[/math]. This coupling from the last equation yields the gravitational field strength:

[math]\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}[/math]

and

[math]\gamma (\mathbf{B} \times v) = \mathbf{H}[/math]

where [math]\gamma[/math] is once again the gyromagnetic ratio.

terms cancel

[math]\frac{\mathbf{J}}{e^2}\ c = \alpha_G \approx 1[/math]

With the field strength defined as:

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

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:

[math]|\mathbf{E}| = \frac{\alpha_G}{e} \frac{\partial U}{\partial r} = \alpha_G \frac{\partial V}{\partial r}[/math]

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

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

### #48

Posted 11 October 2018 - 10:28 PM

### #49

Posted 11 October 2018 - 11:01 PM

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

**reference vs. reference**

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

### #50

Posted 11 October 2018 - 11:26 PM

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

### #51

Posted 12 October 2018 - 12:09 AM

**EDITED**

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