Jump to content


Photo
- - - - -

Is Magnetic Flux A Misnomer?


  • Please log in to reply
26 replies to this topic

#1 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 03 August 2019 - 08:35 AM

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

 

The Josephson constant is often named a ''magnetic flux'' but after much consideration into dimensional analysis, and through some basic arguments, it is not a magnetic flux at all but has to do with the flux of a charge:

 

[math]e^2 = Jc = Gm^2 = n \hbar c = \oint E \cdot dr[/math]

 

The question is, is it a dimensional coincidence that the ratio opening equation, has roots in charges? Secondly, I argue there cannot be a true magnetic flux, since the magnetic field is not quantized - it is very much like gravity in this sense, it is not mediated by a particle (ie. monopole) but is a phenomenon of frame-dependent motion.

 

Moreover, there is captivating reasons to think of this charge related to the index of refraction and the permittivity and permeability of free space:

 

[math]n = \sqrt{\frac{\mu_G \epsilon_G}{\mu_0 \epsilon_G}}[/math]

 

and

 

[math]\frac{1}{c} = \sqrt{ \mu_0 \epsilon_0}[/math]

 

The fact charge be related to a flux and be related to the speed of light in such a way that it may appease permittivity and permeability seems like a strong theoretical case to argue.


Edited by Dubbelosix, 03 August 2019 - 08:35 AM.


#2 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 03 August 2019 - 08:45 AM

To be clear, those relationships share the properties:

 

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

 

The Josephson constant is often named a ''magnetic flux'' but after much consideration into dimensional analysis, and through some basic arguments, it is not a magnetic flux at all but has to do with the flux of a charge:

 

[math]\frac{J}{2e^2} = \frac{1}{c}\frac{Gm^2}{2e^2} = \frac{n \hbar}{2 e^2}  = \frac{1}{c}\oint \frac{E}{2e^2} \cdot dr[/math]

 

And knowing that the formula presents an inverse of the speed of light we can also find the Doppler shift ~

 

[math]\frac{Jv}{2e^2} = \frac{v}{c}[/math]

 

Not only is [math]\frac{v}{c}[/math] a fine structure argument, it is also given in alternative notations such as the Dirac expression: [math]\beta \rightarrow \frac{v}{c}[/math]


Edited by Dubbelosix, 03 August 2019 - 08:49 AM.


#3 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 07 August 2019 - 04:35 AM

Ok so let's make some further statements: Using the relationships we obtained before, the square of the charge flux is:

 

[math]\phi^2 = \frac{n\hbar^2}{4 e^4} = \frac{1}{4c^2}\frac{G^2m^4}{e^4} = \frac{1}{c^2} = \epsilon_0 \mu_0[/math]

 

and I am being dragged away, will update the post instead of continuing in a new one later.  

 

Cont.

 

Since the index of refraction is

 

[math]n = \sqrt{\frac{\epsilon_G \mu_G}{\epsilon_0 \mu_0}}[/math]

 

(from previous investigations into the gravitational aether), the ratio is easily seen to be dimensionless and related to the index of refraction [theoretically] as:
 

[math]\frac{1}{n^2} = \frac{\phi^2}{\epsilon_G \mu_G} = \frac{1}{\epsilon_G \mu_G}\frac{n\hbar^2}{4 e^4} = \frac{1}{4c^2}\frac{1}{\epsilon_G \mu_G} \frac{G^2m^4}{e^4} = \frac{\epsilon_0 \mu_0}{\epsilon_G \mu_G}[/math]

 

What I found interesting was that a variation in the flux will led to different energy levels, very similar to the instructional equations I built for the transition of a black hole under discrete quantum processes:

 

[math]\frac{\Delta \phi}{\epsilon_G \mu_G} = \frac{\phi^2_1 - \phi^2_2}{\epsilon_G \mu_G} = \frac{1}{n^2_1} - \frac{1}{n^2_2} [/math]

 

This type of aether theory had the unusual properties discovered by various asian scientists that the speed of light is not technically a constant and that the permittivity and permeability, both for the electric and gravitational interpretations equally can vary: [math]\frac{\epsilon_0 \mu_0}{\epsilon_G \mu_G}[/math] - this allowed the new aether theory to allow light to escape black holes, similar to how it takes about 40,000 years for a photon to be radiated from the sun from a core, the radiation of a black hole insists the speed of light can only approach zero but never reach it. Using similar arguments from the transition formula for the black hole, the density of nodes (or fluctuations within a given volume) is calculated in the following way:

 

[math]\frac{n \mathbf{J} c}{\epsilon_G \mu_G}(\phi^2_1 - \phi^2_2) = \alpha\ \int\ (\frac{n\hbar c}{n^2_1} - \frac{n\hbar c}{n^2_2})\ k\ dk^3[/math]

 

with [math]\mathbf{J}[/math] as the spin density and [math]k[/math] being the wave number, or alternatively can be written with the difference of the flux charge density [math]\Phi[/math], and from gauge invariance, [math]n\hbar c = Gm^2[/math],

 

[math]\frac{n \hbar  c}{\epsilon_G \mu_G}(\Phi^2_1 - \Phi^2_2) = \beta\ \int\ (\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})\ k\ dk^3[/math]

 

Moreover, we created the transition formula for a black hole under discrete quantum processes from a set of laws which gave rise to the use of the Reynolds number - we know now from previous exploration of the drag formula that the Bejan number is related to the Reynolds number as:

 

[math]R_e = \frac{\rho \ell v}{\mu}[/math]

 

[math]\mu[/math] - dynamic viscocity with the drag related to the Bejan and Reynolds number as:

 

[math]f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{R^2_e}[/math]

 

with [math]B[/math] being the Bejan number and [math]R^2_e[/math] being the square of the Reynolds number. The transition formula for the black hole used the same quantum principles predicted from Reynolds own transition formula:

 

[math]\frac{1}{\Delta \lambda} = R_e(\frac{1}{n^2_1} - \frac{1}{n^2_2}) [/math]

 

Let's just cover those arguments for the transition formula for the black hole, as to derive the quantization of the black hole, in terms of discrete quantum mechanical processes, I defined the Rydberg constant in terms of the gravitational coupling constant we get:

 

[math]R_e = \frac{\alpha_G^2}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}[/math]

 

Even though the Rydberg constant was first applied to hydrogen atoms, it could be derived from fundamental concepts (according to Bohr). In which case we may hypothesize energy levels:

 

[math]\frac{1}{\Delta \lambda} = R_e(\frac{1}{n^2_1} - \frac{1}{n^2_2})[/math]

 

Plugging in the last expressions we get an energy equation:

 

[math]\Delta E_G = \frac{n\hbar c}{\Delta \lambda} = \frac{1}{4 \pi }\frac{m_0v^2}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{4 \pi \lambda_0}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2}) = \frac{p}{4 \pi \hbar}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})[/math]

 

The relativistic gamma appears from the definition of the deBroglie wavelength. The ultimate goal of creating this equation, was to explore an idea I can trace back to Lloyd Motz suggesting a stable black hole. Black holes, even the microscopic kind, cannot be mathematically ruled out from physics (but later I will show in a logical sense why the would be unstable) - and if the work of Hawking is to be taken seriously, including analysis provided by Motz, then a black hole system really is deduced from a discrete set of quantum processes - those discrete processes always lead to an increase in the entropy of a system like a black hole.

 

I noticed that the equation we derived:

 

[math]\frac{n \mathbf{J} c}{\epsilon_G \mu_G}(\phi^2_1 - \phi^2_2) = \alpha\ \int\ R^2_e(\frac{n\hbar c}{n^2_1} - \frac{n\hbar c}{n^2_2})\ k\  dk = \alpha\ \int\ B(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})\ k\  dk [/math]

 

is expressible by using the square of the Reynolds number, since it has units of inverse length and so by consequence is related to the Bejan number: From here, we can speculate that the wavenumbers carry the same information as the inverse of the front area:

 

[math]\frac{n \mathbf{J} c}{\epsilon_G \mu_G}(\phi^2_1 - \phi^2_2) = \alpha\ \int\ R^2_e(\frac{n\hbar c}{n^2_1} - \frac{n\hbar c}{n^2_2})\ \frac{1}{A_f} = \beta\ \int\ B(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})\ \frac{1}{A_f} [/math]

 

Dividing through by the back area term we obtain

 

[math]\frac{1}{\epsilon_G \mu_G}\frac{nJ c}{R}(\phi^2_1 - \phi^2_2) = \alpha\ \int\ R^2_e(\frac{n\hbar c}{n^2_1} - \frac{n\hbar c}{n^2_2})\ \frac{A_b}{A_f} = \beta\ \int\ B(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})\ \frac{A_b}{A_f} [/math]

 

We can see how we might obtain a drag term from the last two equalities by dividing through by the Reynolds number squared, 

 

[math] \alpha\ \int\ (\frac{n\hbar c}{n^2_1} - \frac{n\hbar c}{n^2_2})\ \frac{A_b}{A_f} = \beta\ \int\ \frac{A_b}{A_f}\frac{B}{R^2_e}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})\  [/math]

 

Since the Reynolds number is inverse, it has units of length squared as we obtain a simplification:

 

[math]\alpha\ \int\ (\frac{n\hbar c}{n^2_1} - \frac{n\hbar c}{n^2_2})\ \frac{A_b}{A_f} = \beta\ \int\ \frac{A_b}{A_f}\frac{B}{R^2_e}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})\  [/math]

 

By using the formula

 

[math]f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{R^2_e}[/math]

 

It can be written in the following way:

 

[math]\alpha\ \int\ (\frac{n\hbar c}{n^2_1} - \frac{n\hbar c}{n^2_2})\ \frac{A_b}{A_f} = \beta\ \int\ \frac{2F}{\rho v^2 A}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})  = \beta\ \int\ f\ (\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})  [/math]


Edited by Dubbelosix, 07 August 2019 - 07:53 AM.


#4 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 07 August 2019 - 07:53 AM

Just some curious formula's I have entertained from some dimensional analysis crunching and for educated reasons as well. 



#5 GAHD

GAHD

    Eldritch Horror

  • Administrators
  • 2649 posts

Posted 07 August 2019 - 06:00 PM

Gotta move this up to alt, you need to provide the HYPerlink biliOGRAPHY evidence parts of the hypography too. ;)

 


  • exchemist likes this

#6 exchemist

exchemist

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2722 posts

Posted 08 August 2019 - 03:10 AM

Gotta move this up to alt, you need to provide the HYPerlink biliOGRAPHY evidence parts of the hypography too. ;)

Given the algebra is riddled with errors, I’d even have said Strange Claims - but you’re the Mod. :)

#7 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 08 August 2019 - 04:45 AM

Gotta move this up to alt, you need to provide the HYPerlink biliOGRAPHY evidence parts of the hypography too. ;)

 

 

 

Meaning?

 

 

And no, the math is not riddled with errors, exchemist cannot demonstrate any errors, because there are none.



#8 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 08 August 2019 - 04:48 AM

I wrote a whole lot out. if GAHD. you mean references, then my blogs which have been moved into ''my spaces'' as a move in quora for all posters for only creative writers and contributors - have all the references... you'd be like asking me to go through years of work just to clarify something that hasn't been raised. I have said a few times, if something is not understood, then demonstrate a question, otherwise you'll attract people like exchemist to the site pretending they know what they are talking about, calling things out, without demonstrated evidence, not even in a math form which is more complexing since he is adamant that I cannot do math, yet he writes very little. 



#9 GAHD

GAHD

    Eldritch Horror

  • Administrators
  • 2649 posts

Posted 08 August 2019 - 05:34 AM

I wrote a whole lot out. if GAHD. you mean references, then my blogs which have been moved into ''my spaces'' as a move in quora for all posters for only creative writers and contributors - have all the references... you'd be like asking me to go through years of work just to clarify something that hasn't been raised. I have said a few times, if something is not understood, then demonstrate a question, otherwise you'll attract people like exchemist to the site pretending they know what they are talking about, calling things out, without demonstrated evidence, not even in a math form which is more complexing since he is adamant that I cannot do math, yet he writes very little. 

Read the Rules, In FULL. Remember them. It is not that hard to click on the link bottom right of page. Very specifically the second and eighth bold points in this situation.



#10 exchemist

exchemist

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2722 posts

Posted 08 August 2019 - 06:26 AM

Sigh.

To take a simple example, if J/2e^2=1/c, then it can only also equal 1/c . Gm^2/2e^2 if Gm^2/2e^2 = 1.

But Dubbelsox has previously stated e^2 = Gm^2. So that means that 1/2 = 1.

And so it will go on..and on...if anybody can be bothered to go through this rubbish.


This poster has found that by expressing himself in maths salad, rather than the more usual word salad, he can fly under the radar of some forums for a while before anybody can be bothered to point out it is gibberish.

Edited by exchemist, 08 August 2019 - 06:30 AM.

  • ralfcis likes this

#11 ralfcis

ralfcis

    Explaining

  • Members
  • PipPipPipPipPipPip
  • 885 posts

Posted 08 August 2019 - 07:22 AM

Hmm, I'm surprised XC knows any math at all. It'd be great if he challenged mine instead of dismissing it without any justification. I read the rules for the first time. It'd be nice if there was one against all the different types of motivations for deception I see on this forum. 


  • Dubbelosix likes this

#12 exchemist

exchemist

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2722 posts

Posted 08 August 2019 - 08:00 AM

Hmm, I'm surprised XC knows any math at all. It'd be great if he challenged mine instead of dismissing it without any justification. I read the rules for the first time. It'd be nice if there was one against all the different types of motivations for deception I see on this forum.

I got off your bus long before Sluggo. :)
  • Dubbelosix likes this

#13 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 08 August 2019 - 09:06 AM

Sigh.

To take a simple example, if J/2e^2=1/c, then it can only also equal 1/c . Gm^2/2e^2 if Gm^2/2e^2 = 1.

But Dubbelsox has previously stated e^2 = Gm^2. So that means that 1/2 = 1.

And so it will go on..and on...if anybody can be bothered to go through this rubbish.


This poster has found that by expressing himself in maths salad, rather than the more usual word salad, he can fly under the radar of some forums for a while before anybody can be bothered to point out it is gibberish.

 

Once again, you repeat your own misdoings in mathematics. You were actually correct in the first go concerning quantization of the drag coefficient....since then, you seem to be lost on the various examples of the fine structure constant..... let's see if you are capable of understanding why... a hint for now... both [math]\alpha[/math] and [math]\beta[/math] are fine structure constants.

 

 

It was shown from Bouyer that the exact value of the parameters of an electron revelled a direct relationship:

 

[math]energy\ of\ an\ electron\ = \alpha\ \frac{n \hbar c}{R_e}[/math]

 

 

Now explain if you are superior at math, what is [math]\frac{Gm^2}{e^2}[/math]....from gague invariance:

 

Let's see how you go???


Edited by Dubbelosix, 08 August 2019 - 09:26 AM.


#14 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 08 August 2019 - 09:08 AM

Read the Rules, In FULL. Remember them. It is not that hard to click on the link bottom right of page. Very specifically the second and eighth bold points in this situation.

 

 

Sorry what? I've been here for a while so it is likely I have forgotten the rules.... the main one is to out bad physics... not by derogatory comments, but if you are met with them, fire only beats fire. 



#15 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 08 August 2019 - 09:12 AM

Sigh.

To take a simple example, if J/2e^2=1/c, then it can only also equal 1/c . Gm^2/2e^2 if Gm^2/2e^2 = 1.

But Dubbelsox has previously stated e^2 = Gm^2. So that means that 1/2 = 1.

And so it will go on..and on...if anybody can be bothered to go through this rubbish.


This poster has found that by expressing himself in maths salad, rather than the more usual word salad, he can fly under the radar of some forums for a while before anybody can be bothered to point out it is gibberish.

 

 

If we kept c, then it would imply a Planck mass - however, since velocity is a variable subject, I do not understand where your confusion is borne from. 



#16 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 08 August 2019 - 09:16 AM

Just taking a small look at the history of [math]\frac{Gm^2}{e^2}[/math] we find a clear correspondence to the fine structure from Barrrow and Tipler:#

 

''

  • (4.5) in Barrow and Tipler (1986) tacitly defines α/αG as e2/Gmpme ≈ 1039. Even though they do not name the α/αG defined in this manner, it nevertheless plays a role in their broad-ranging discussion of astrophysicscosmologyquantum physics, and the anthropic principle;    '''

 

The inverse is what I have, it values at 

 

[math]\frac{\alpha_G}{\alpha}[/math] given through by [math]\frac{Gm_pm_e}{e^2} \approx 10^{-39}[/math]


Edited by Dubbelosix, 08 August 2019 - 09:25 AM.


#17 Dubbelosix

Dubbelosix

    Creating

  • Members
  • PipPipPipPipPipPipPip
  • 2975 posts

Posted 08 August 2019 - 09:17 AM

Exchemist... just a friendly warning, don't talk about things you have no idea about. Stick to you atomic quantization, the only thing you are good at.