Magnetic Moment of the Proton, Final, The Real Case #3

$\textbf{CASE 3: Rotating Sphere of Uniform Surface-Charge}$
$$\bar{M}={4\pi\over3}\sigma r_p^4\bar{\omega}$$
from http://www-personal.umich.edu/~pran/jackson/P505/F07_hw08a.pdf

$\sigma=$ surface charge density
$a=$ radius of sphere, using proton radius, $r_p$
$\bar{\omega}=$ angular velocity(speed)
$e=$ elementary charge
$\sigma={e\over A_{sphere}}$
$A_{sphere}=4\pi r_p^2$
$\bar{\omega}={\triangle\Theta\over\triangle t}={2\pi\over{2\pi r_p\over c}}={c\over r_p}$
$\bar{M}={4\pi\over3}{e\over4\pi r_p^2}r_p^4{c\over r_p}$
$\bar{M}={er_pc\over3}$
$m_pr_p={2h\over\pi c}$
$r_p={2h\over\pi cm_p}$
$\hbar={h\over2\pi}$
$\bar{M}={e2hc\over3\pi cm_p}={2eh\over3\pi m_p}={2e2\pi\hbar\over3\pi m_p}={8e\hbar\over3\times2m_p}$
$\mu_N={e\hbar\over2m_p}$
$\bar{M}={8\over3}\mu_N$
$\bar{M}=2.666\mu_N$

$\mu_p=2.666\mu_N$
QED.

Now, what is this quark model for spin all about?

My proton magnetic moment is 4.5% less than the measured.  Does this sound like the familiar proton radius puzzle error?

Using CODATA value for proton radius and $\bar{M}={er_pc\over3}$, we find:
$\bar{M}={er_pc\over3}=1.40119247\times10^{-26}\,m^2 A$
Google Calculator Link for CODATA calc of proton magnetic moment
(this is within 0.7% of CODATA value for proton magnetic moment)
(This, significantly, verifies the equation)


CODATA link for proton magnetic moment
($\mu_p=1.410 606 7873(97)\times10^{-26}\,J\,T^-1$)
(Google Calculated: using measured coeficient and fundamental constants
--google calculator link for proton magnetic moment alt calc --

However, the new accurate value for the proton's magnetic moment is very likely:
--google calculator link for proton magnetic moment NEW VALUE calc --
($1.34687565\times10^{-26}\,m^2\,A$ <~~ This is about 4.5% less than present CODATA value)

https://en.wikipedia.org/wiki/Proton_spin_crisis - solved

Addendum:

Should see a Larmor frequency of about 40.65MHz/Tesla (hydrogen - $H^{+1}$):
(close to 40.77MHz and this FCC Band: ITU # 8 - VFH Very high frequency – FM Radio, TV broadcasts, and aircraft communications (30-300 MHz))

Google calculator link to calculate Larmor frequency

(when calibrated against a proton radius of $0.8412fm$ instead of $0.8751fm$)

http://bio.groups.et.byu.net/LarmourFreqCal.phtml

https://physics.aps.org/synopsis-for/10.1103/PhysRevA.96.060501

The Surfer, OM-IV

©2017 Mark Eric Rohrbaugh & Lyz Starwalker © 2017

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Magnetic Moment of the Proton #3: RMS Current?

Using RMS for the current for case 1:

$$\bar{M}={\sqrt{2}\over2}4\mu_N$$

$$\bar{M}=2.8284271\mu_N$$

Which is closer.  Have to re-check current calculation.  

More later...
This is not justifiable in the derivation, as a DC circular loop current was assumed, however, it's fairly close to measured! More serious attempts to follow... 

Magnetic Moment of the Proton #2: Shape Factor (SF)

$$SF = {\bar{M}\over\mu_N}$$

For example, the shape factor, $SF_p$, of the proton is:

$$SF_p={2.79284735\mu_N\over\mu_N}$$

$$SF_p={2.79284735}$$

$$SF_{case\#1}=4.0$$

$$SF_{case\#2}=1.6$$
$$SF_{case\#4}=2.666$$

.

.

.

Could this be a repeat of the Silver Surfer's Crash and Burn???

http://phxmarker.blogspot.com/2015/03/surfers-down-but-not-out.html

to be continued...

The Surfer, OM-IV

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Magnetic Moment of the Proton #1

(this is an investigation, no conclusions yet)

The magnetic moment of the proton, $\mu_p$, is 

$$\mu_p=2.79284735\mu_N$$

$\mu_N={e\hbar\over2m_p}$

$\mu_p={2.79284735e\hbar\over2m_p}$ (*1)

$e=$ elementary charge

$\hbar = {h\over2\pi} =$ reduced Planck's Constant, $h=$ Planck's constant

$m_p=$ proton mass

from https://en.wikipedia.org/wiki/Proton_magnetic_moment

Wolfram: http://www.wolframalpha.com/input/?i=(proton+magnetic+dipole+moment)%2F((elementary+charge)*hbar%2F2m_p)

We can compare this to calculating the magnetic moment of various charge distributions of the proton and step through various configurations and stable patterns until the correct answer is found.

(this may have already been done before, however, I am going to do it again, step by step, to uncover if anything was missed that led to the formation of the quark model)

The proton is a vortex in the aether, thus, the flowing aether is the flowing charge.  The interesting thing will be the correct math for the charge vortex - a stable vortex in the yet to be defined superfluid aether.

Let's examine some simple cases first:  (with excess derivation steps for easy checking)

$\textbf{Case One: Ring of Current at Proton's Equator}$

Assume total charge of proton is at a point a distance $r_p$ from center, just like we did when deriving the proton radius from quantized angular momentum.  This is likely going to result in an answer that is EXCESSIVE magnetic moment, because the charge is actually distributed, not in a single point like this, however, we proceed:

(note: it can be shown that a linearly distributed charge "e" around the radius of the proton would results in the same magnitude circular loop current - TBD)

Circular loop of current of radius $r_p$ (proton radius):

$$\bar{M}=I\bar{A}$$

$\bar{M}=$ magnetic moment

$I=$ loop current, defined as elementary "e" moving at "c" at radius "$r_p$"

$I={e\over\triangle t}$

$\triangle t = {2\pi r_p \over c}$

$I={e\over{2\pi r_p\over c}}$

$I={ec\over2\pi r_p}$

$\bar{A}=$ area

$\bar{A}=\pi r_p^2$

$$\bar{M}=I\pi r_p^2$$

$$\bar{M}={ec\over2\pi r_p}\pi r_p^2$$

$$\bar{M}={ecr_p\over2}$$

$$r_p={2h\over\pi cm_p}$$

$$\bar{M} = {{ec{2h\over\pi cm_p}\over2}}$$

$$\bar{M} = {he\over\pi m_p}$$

$$\bar{M}= {2\pi\hbar e\over\pi m_p}$$

$$\bar{M}= {2e\hbar\over m_p}$$

re-writing to be in form of (*1):

$$\bar{M}= {4e\hbar\over2m_p}$$

$\mu_N={e\hbar\over2m_p}$

$$\bar{M}= 4\mu_N$$

We have a factor of 4 instead of  2.79284735, so this is too much magnetic moment.

$\textbf{Case 2: Rotating Uniformly Charged Solid Sphere of the Proton's Radius}$

Rotating uniformly charged solid sphere of Radius R, total charge Q, angular velocity $\omega$:

$$\bar{M}={1\over5}Q\omega R^2$$ 

from https://ocw.mit.edu/courses/physics/8-07-electromagnetism-ii-fall-2012/exams/MIT8_07F12_quizsol2.pdf

Using our proton radius solution, and assuming the speed of light, c, for the ve;ocity of our rotating proton at the point $r_p$ from the center of the proton:

$$\omega = {2\pi\over\triangle t}$$

$$\triangle t = {2\pi r_p\over c}$$

$$\omega={c\over r_p}$$

$$\bar{M}={1\over5}Q{c\over r_p}R^2$$ 

$$\bar{M}={1\over5}e{c\over r_p}r_p^2$$ 

$$\bar{M}={1\over5}ecr_p$$ 

$$r_p={2h\over\pi cm_p}$$

$$\bar{M}={1\over5}ec{2h\over\pi cm_p}$$ 

$$h=2\pi\hbar$$

$$\bar{M}={1\over5}e{4\hbar\over m_p}$$ 

$$\bar{M}={8\over5}{e\hbar\over2m_p}$$ 

$$\bar{M}=1.6{e\hbar\over2m_p}$$ 

$\mu_N={e\hbar\over2m_p}$

$$\bar{M}=1.6\mu_N$$ 

So, here in one evening, we have bracketed above and below the solution:

A too high estimate, 4.0, and a too low, 1.6, as compared to the measured coefficient of 2.79284735 for the magnetic moment of the proton.

These charge distributions that we have examined are not realistic charge distributions (stable vortex superfluid aether flow paths), so over the next series of posts, we will evaluate other more realistic charge flows/distributions and develop a Shape Factor (SF) to compare various possible stable patterns and solutions.  

If simple solutions do not fall out, we will move on to fluid dynamics and solutions of nonlinear systems Navier-Stokes equations.

The Surfer, OM-IV

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Proton Radius: Gooogle Calculator

#ProtonRadius

All of the fundamentals constants to calculate the proton radius are built into Gooogle Calculator.

Why does the mainstream insist there is a proton radius problem?

Gooogle Calculator Link for Proton Radius Calculation

"The 2010 result, which found a proton radius of 0.84087±0.00039 fm versus the CODATA value of 0.8751±0.0061 fm, formed the basis of what has been dubbed the proton-radius puzzle."

The Surfer, OM-IV

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Where It's All Going: Consciousness

The hottest topic in science and physics is consciousness.  The next big theory, or theory of everything must include consciousness.  The leading researches are focusing on consciousness, they have been for many decades.

https://www.amazon.com/Mind-Dynamics-Space-Elizabeth-Rauscher/dp/1892139448/

To see how the work of this blog relates to consciousness, see this paper:

The Unified Spacememory Network: from cosmogenesis to consciousness 

Haramein, N., Brown W., & Val Baker, A. K. F. (2016). The Unified Spacememory Network: from Cosmogenesis to Consciousness, Journal of NeuroQuantology

Download the PDF

Other supporting evidence for consciousness being key:

http://www.openminds.tv/lockheed-skunk-works-director-says-esp-is-the-key-to-interstellar-travel-video-1092/23042

Ben Rich: "We now have the technology to take ET home."

Lockheed Martin Skunk Works logo. (Credit: Lockheed Martin)

Connectedness  (@7:03)helps to explain consciousness and ESP.  There is no mystery to it in this framework.  

In the immense density of the superfluid-like vacuum aether, we are all already wormhole entangled, like I stated at beginning of blog.

The Surfer, OM-IV

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Proton Charge to Mass Ratio

$${e\over m_p}=\sqrt{\pi^2{r_p}^2c^3\alpha\epsilon_0\over{2h}}$$

$e=$ elementary electron charge

$m_p=$ proton mass

$r_p=$ proton radius

$c=$ speed of light

$\alpha=$ fine-structure constant

$\epsilon_0=$ permittivity of vacuum

$h=$ Planck's constant

$${e\over m_p}={\pi r_pcq\over{2h}}$$

$e=q=$ elementary charge

(pi*(0.841235640294664e-15)*c*(elementary charge)/2/h)=

Link for Google Calculation of charge to mass ratio of proton

elementary charge / proton mass =

95 788 332.2 s A / kg

(elementary charge)/m_p=

Proton Charge-to-Mass Ratioemp=9.57883358⋅107Ckg

First! (?)

CODAT value:
https://physics.nist.gov/cgi-bin/cuu/Value?esmp|search_for=proton

The Surfer, OM-IV

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The Proton: Superfluid Vortex Quantization

from: https://en.wikipedia.org/wiki/Quantum_vortex#Vortex-quantisation_in_a_superfluid

Vortex-quantization in a superfluid, circulation is quantized:

$$\oint_C\mathbf{v}\cdot d\mathbf{l}={2\pi\hbar\over m}n$$

$\mathbf{v}=$ velocity

$\mathbf{l}=$ path length, $d\mathbf{l}=rd\theta$, for dot product

$\hbar={h\over{2\pi}}$ = Reduced Planck's constant, and $h$, Planck's constant

$m=$ mass

$n=$ integer, quantized

for uniform circular motion, constant velocity:

$\oint_Cd\mathbf{l}={\int_0}^{2\pi}rd\theta=2\pi r$, circumference of circle

$$\oint_C\mathbf{v}\cdot d\mathbf{l}={\mathbf{v}2\pi r}$$

$${\mathbf{v}2\pi r}={2\pi\hbar\over m}n$$

$${m r}={2\pi\hbar\over {\mathbf{v}2\pi}}n$$

$${m r}={\hbar\over {\mathbf{v}}}n$$

$${m r}={nh\over {2\pi\mathbf{v}}}$$

For a proton that is a vibration in the superfluid vacuum aether, the phase velocity of circulation is the speed of light, $c$:

 $$\therefore{m r}={nh\over {2\pi c}}$$

the proton being the $n=4$ case:

 $$\therefore{m r}={2h\over {\pi c}}$$

which is the same equation derived for the proton using the quantized angular momentum approach and Haramein's team's geometrical information theory holofractographic approach.

This is implying that the vacuum is a superfluid.

Still need to derive why the proton mass is what it is... and why the proton is the $n=4$ case.

The Surfer, OM-IV

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XY = c , Mathematical Solutions

Math:
$XY=c$
Let $X=m(r)$
Let $Y=r(m)$
$c=constant$
m(r) is a function of r, $m={c\over r}$
r(m) is a function of m, $r={c\over m}$

What can mathematically be said about all the solutions to this type of equation? A product of two variables is a constant?

Time for math.

#Math

more later about this brief post,
- mr

Some related work here:
http://www.stumblingrobot.com/2016/02/22/find-the-orthogonal-trajectories-of-the-family-xy-c/

And, example #2 here:

https://proofwiki.org/wiki/Orthogonal_Trajectories/Rectangular_Hyperbolas

&

https://www.math24.net/orthogonal-trajectories-page-2/#example2

Quantized levels might look something like this, however, it must be reviewed:

The Surfer, OM-IV

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How Does the Mass Radius Product, MR, Become Specific Unique Mass and Radius Values?

(this post requires some rethinking)  7/29/17  -mr
It seems this still does not explain why the mass of the proton is what it is... back to drawing board!
$$mr={2h\over \pi c}$$

How does the mass become a specific value and the radius for a specific object or "particle", such as the proton?  In other words, why can't or why doesn't the mass come out twice as much and the radius half as much? Or, why can't the proton mass be 4% smaller, and the radius 4% bigger? It still works in the equation... (approximately.  Linearization errors...)

This is because the solution is a cymatics-like resonance in the vacuum, and the mass ratio, $\phi$, is actually an information-theory area ratio  divided by a geometric volumetric ratio: (Haramein's team's work):

$$\phi={\eta\over R}$$
$$\eta={Area_{objectSurface}\over A_{equitorialXsectionPSU}}$$
$$R={Volume_{object}\over Volume_{PSU}}$$
(the PSU is the Planck Sphereical Unit, a sphere of diameter Planck length, ${\ell}_{\ell}$)
for the proton, a factor of 2 is needed:

$$m_p={2{\eta\over R}m_{\ell}}$$
$m_p=$ Proton mass
$m_{\ell}=$ Planck mass

When the terms for eta over R, the mass ratio, are expanded, the results is:
$$m_pr_p={2h\over \pi c}$$
$r_p$ is Proton radius.
$h=$ Planck's constant
$c=$ Speed of Light
$\pi$ is good.

So what you have is a 2D area ratio term & a 3D volumetric ratio which resonates similar to cymatics-type oscillations in a substrate material, the substrate being the new superfluid like aether of the vacuum.

It's pretty simple math and this approach gives the same answer as the quantized angular momentum approach.  Thus, the higher-level 3D approach seems to be more inclusive, as it can be used to calculated the mass of black holes, protons, and electrons.  What's next?

The details of the math and derivation have been outlined previously:
http://fractalu.com/AdvancedGeometricPhysicsSolutions1.pdf

and more info here:
https://www.thenewenergyindustry.com/mass-ratio/

The Surfer, OM-IV

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A Scientific Revolution!

https://en.wikipedia.org/wiki/All_your_base_are_belong_to_us

The Scientific Revolution is ongoing:

https://plus.google.com/u/0/103725077496823758319/posts/6TfNvyqshZL

You won't hear Bill Nye the science denier or Neil deGrasse Tyson talk about the proton radius problem, the Achilles Heal (sic) of The Standard Model.  

The Surfer, OM-IV

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