# New Equivalence Principles?

206 replies to this topic

### #205 Dubbelosix

Dubbelosix

Creating

• Members
• 3000 posts

Posted 08 August 2018 - 11:25 AM

https://entropytempe...Magnetic-Moment

The following writeup for the blog. Notice if you go to the earlier posts, you'll find discussions on the magnetic moment.

Edited by Dubbelosix, 08 August 2018 - 11:26 AM.

### #206 Dubbelosix

Dubbelosix

Creating

• Members
• 3000 posts

Posted 08 August 2018 - 03:11 PM

Shustaire,

Hello! I sent you a message just to give you the heads up.

### #207 Dubbelosix

Dubbelosix

Creating

• Members
• 3000 posts

Posted 08 August 2018 - 05:38 PM

One component of the power expression for the black hole that I derived was:

$\frac{Q^2a^2}{c^3}$

Notice the magnetic dipole (moment) equation I suggested for a black hole was

$\mathbf{m} = g\frac{QJ}{2m} = g \gamma J$

in which $\gamma$ is the ratio of charge to mass $(\frac{Q}{m})$ weighted by one-half.

The black hole expression has a squared charge to mass ratio in its alternative form:

$\frac{Q^2}{m^2c^3}(\frac{dp}{dt})^2$

Getting that squared value is just a nice application of notation:

$\gamma \mathbf{m} = g \frac{Q^2}{4m^2} \cdot J = g \gamma^2 J$

I was interested for a moment in a localization equation of the form

$\frac{Q^2}{4m^2} = \gamma^2 (\frac{J}{J_0})$

In which the commutation relations you can calculate the expected angular momentum through the relation

$[J,J_0] = \frac{J}{J_0}$

That could only be true from the last two equations if the square of the mass of black hole varies proportionally with the angular momentum of the form $m^2 \propto \frac{J}{J_0}$. While this might sound strange, it has been suggested in academia that the rotation of a black hole may vary since supermassive black holes are always observed to be spinning very close to the speed of light.

$\gamma \mathbf{m} = g \frac{Q^2}{4m^2} \cdot J = g \gamma^2 J$

$\frac{\gamma \mathbf{m}}{c^3}(\frac{dp}{dt})^2 = g \frac{Q^2}{4m^2c^3} \cdot J(\frac{dp}{dt})^2 = g \frac{\gamma^2 J}{c^3}(\frac{dp}{dt})^2$

This suggests we can form three equivalent near-Power expressions based on new takes of dynamics, this time properly generalized relativistically

$= \frac{\gamma \mathbf{m}}{c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

$= g \frac{Q^2}{4m^2c^3} \cdot J \frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

$= g \frac{\gamma^2 J}{c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

where $\gamma$ is the gyromagnetic ratio. And so a suggested power equation is

$P = \frac{\gamma \mathbf{m}}{J c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

A suggested power equation was

$P = \frac{\gamma \mathbf{m}}{J c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

In which the gyromagnetic moment is

$\mathbf{m} = \frac{\mathbf{B}_r \cdot V}{\mu_G}$

Plugging this in we get

$P = \frac{\gamma \mathbf{B}_r \cdot V}{J c^3 \mu_G}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

* Note the term $\mathbf{B}_r$ is not a magnetic field, but the residual flux density!

This is a power equation may become a power density equation (which is a rate of change in energy density) by simple rearrangement:

$\mathbf{P} = \frac{\gamma \mathbf{B}_r}{J c^3 \mu_G}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau} = \frac{Q \mathbf{B}_r \cdot}{2J m c^3 \mu_G}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Where we used a short notation $\mathbf{P} = \frac{P}{V}$.

The gravitational permittivity and permeability is defined as ~

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

$\frac{P}{\nu_G} = \frac{\gamma \mathbf{B}_r \cdot V}{J c^3 \mu_G \nu_G}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

$\frac{P}{\nu_G} = \frac{\gamma \mathbf{B}_r \cdot V}{J c}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

and so we find a simplification from all those constants:

$P = \nu_G(\frac{\gamma \mathbf{B}_r \cdot V}{J c})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

[1] - see notes

$P = \nu_G(\frac{\gamma \mathbf{B}_r \cdot V}{J c})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau} = \nu_G(\frac{Q \mathbf{B}_r \cdot V}{J mc})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

$\mathbf{B}_r$ has units of magnetic flux density (or)

$\mathbf{\phi} = \mathbf{B} \cdot dA$

where $\mathbf{B} = \frac{B}{V}$ as a magnetic field density term. This normalizes the volume term in the previous equation and features a different view on the power emission in terms of the magnetic dynamics:

$\nu_G(\frac{\gamma (B \cdot dA)}{J c})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Later i will investigate another equation I derived

$\gamma = \frac{\sqrt{G}m}{2m_p} = \frac{\mu_{G_{N}}}{\hbar}$

In which I will consider using the nuclear magneton for a black hole with photon nucleus as I did with my original investigations.

Let's recognize such a formula then say as:

$\gamma = \frac{\sqrt{G}m}{2m_p} = \frac{\mu_{G_{N}}}{\hbar}$

using the nuclear magneton for a black hole with photon nucleus.

Notes -

$P = \nu_G(\frac{\gamma \mathbf{B}_r \cdot V}{J c})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

can be written as

$P = \nu_G(\frac{\gamma \mathbf{B}_r \cdot V}{Er_g})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

and it is not hard to prove that this term $Gm^2$ can enter this since

$E = \frac{\hbar c}{r_g}$

and from Weyl invariance $\hbar c = Gm^2$ we use both equations to obtain

$Er_g = Gm^2$