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.**

Started By
Dubbelosix
, Jul 10 2018 05:54 AM

206 replies to this topic

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.**

Posted 08 August 2018 - 03:11 PM

Shustaire,

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

Posted 08 August 2018 - 05:38 PM

One component of the power expression for the black hole that I derived was:[math]\frac{Q^2a^2}{c^3}[/math]Notice the magnetic dipole (moment) equation I suggested for a black hole was[math]\mathbf{m} = g\frac{QJ}{2m} = g \gamma J[/math]in which [math]\gamma[/math] is the ratio of charge to mass [math](\frac{Q}{m})[/math] weighted by one-half.The black hole expression has a squared charge to mass ratio in its alternative form:[math]\frac{Q^2}{m^2c^3}(\frac{dp}{dt})^2[/math]Getting that squared value is just a nice application of notation:[math]\gamma \mathbf{m} = g \frac{Q^2}{4m^2} \cdot J = g \gamma^2 J[/math]I was interested for a moment in a localization equation of the form[math]\frac{Q^2}{4m^2} = \gamma^2 (\frac{J}{J_0})[/math]In which the commutation relations you can calculate the expected angular momentum through the relation[math][J,J_0] = \frac{J}{J_0}[/math]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 [math]m^2 \propto \frac{J}{J_0}[/math]. 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.[math]\gamma \mathbf{m} = g \frac{Q^2}{4m^2} \cdot J = g \gamma^2 J[/math]

[math]\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[/math]

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

[math] = \frac{\gamma \mathbf{m}}{c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}[/math]

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

[math] = g \frac{\gamma^2 J}{c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}[/math]

where [math]\gamma[/math] is the gyromagnetic ratio. And so a suggested power equation is

[math]P = \frac{\gamma \mathbf{m}}{J c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}[/math]

A suggested power equation was

[math]P = \frac{\gamma \mathbf{m}}{J c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}[/math]

In which the gyromagnetic moment is

[math]\mathbf{m} = \frac{\mathbf{B}_r \cdot V}{\mu_G}[/math]

Plugging this in we get

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

* Note the term [math]\mathbf{B}_r[/math] 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:

[math]\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}[/math]

Where we used a short notation [math]\mathbf{P} = \frac{P}{V}[/math].

The gravitational permittivity and permeability is defined as ~

[math]\frac{1}{c^2} = \mu_G \nu_G[/math]

[math]\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}[/math]

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

and so we find a simplification from all those constants:

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

[1] - see notes

[math]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}[/math]

[math]\mathbf{B}_r[/math] has units of magnetic flux density (or)

[math]\mathbf{\phi} = \mathbf{B} \cdot dA[/math]

where [math]\mathbf{B} = \frac{B}{V}[/math] 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:

[math]\nu_G(\frac{\gamma (B \cdot dA)}{J c})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}[/math]

Later i will investigate another equation I derived

[math]\gamma = \frac{\sqrt{G}m}{2m_p} = \frac{\mu_{G_{N}}}{\hbar}[/math]

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:

[math]\gamma = \frac{\sqrt{G}m}{2m_p} = \frac{\mu_{G_{N}}}{\hbar}[/math]

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

Notes -

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

can be written as

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

and it is not hard to prove that this term [math]Gm^2[/math] can enter this since

[math]E = \frac{\hbar c}{r_g}[/math]

and from Weyl invariance [math]\hbar c = Gm^2[/math] we use both equations to obtain

[math]Er_g = Gm^2[/math]