Jump to content
Science Forums

New Equivalence Principles?


Recommended Posts

 

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]

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...