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Tuesday, October 31, 2017

Larmor Precession

Where does the 2.79 come from? The 42.5781MHz?
Papers like this are close, yet seems to skip some important things, like raw data:

The Surfer, OM-IV

Sunday, October 29, 2017

Magnetic Moment Basics

Atomic magnetic moment of Materials from Hamdard University Bangladesh

Some useful slides. <~~ data for various nucleons

Should see a Larmor frequency of about 40.7MHz/Tesla (hydrogen - H):
(when calibrated against a proton radius of $0.8412fm$ instead of $0.8751fm$)

The Surfer, OM-IV

Updating Fundamental Constants - The Connected Universe Style!

"Scientists update four key fundamental constants

New values of Planck constant, others bring world closer to revised measurement system..."

Updates to fundamental constants are carefully done, especially like the upcoming defining or redefining mass (more later on that in another post).  Many are interconnected, locked in fundamental relationships, so a change in one can not only imply, but require a change in another.  Recently, four $(4)$ were updated.

It will be interesting to see which fundamental constants need to be updated when the proton radius problem is finally "officially" solved:

$r_p=0.8412fm$ new value -4% present CODATA value
$r_p=0.8751fm$ present CODATA value

$\mu_p=2.6666\mu_N$ new value for proton magnetic moment, -4.5% present CODATA value
$\mu_p=2.7928\mu_N$ present CODATA value

Since magnetic moment, radius, charge, and mass are all interrelated, the proton's magnetic moment is locked to its radius.  The muonic measurements of the proton radius have been repeatedly verified since 2010, so this change in the proton's radius will ripple through physics WAY more than just an update to the proton's magnetic moment.

(See previous post for derivation of $\mu_p={8\over3}\mu_N$)

The Surfer, OM-IV

Thursday, October 26, 2017

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

$\textbf{CASE 3: Rotating Sphere of Uniform Surface-Charge}$
$$\bar{M}={4\pi\over3}\sigma r_p^4\bar{\omega}$$

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


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) - solved

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

The Surfer, OM-IV
©2017 Mark Eric Rohrbaugh & Lyz Starwalker © 2017

Wednesday, October 25, 2017

Magnetic Moment of the Proton #3: RMS Current?

Using RMS for the current for case 1:

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... 

Tuesday, October 24, 2017

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:

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

The Surfer, OM-IV

Magnetic Moment of the Proton #1

(this is an investigation, no conclusions yet)

The magnetic moment of the proton, $\mu_p$, is 
$\mu_p={2.79284735e\hbar\over2m_p}$ (*1)
$e=$ elementary charge
$\hbar = {h\over2\pi} =$ reduced Planck's Constant, $h=$ Planck's constant
$m_p=$ proton mass

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):
$\bar{M}=$ 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
$\bar{A}=\pi r_p^2$
$$\bar{M}=I\pi r_p^2$$
$$\bar{M}={ec\over2\pi r_p}\pi r_p^2$$
$$r_p={2h\over\pi cm_p}$$
$$\bar{M} = {{ec{2h\over\pi cm_p}\over2}}$$
$$\bar{M} = {he\over\pi m_p}$$
$$\bar{M}= {2\pi\hbar e\over\pi m_p}$$
$$\bar{M}= {2e\hbar\over m_p}$$
re-writing to be in form of (*1):
$$\bar{M}= {4e\hbar\over2m_p}$$
$$\bar{M}= 4\mu_N$$

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 $\omega$:
$$\bar{M}={1\over5}Q\omega R^2$$ 

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:
$$\omega = {2\pi\over\triangle t}$$
$$\triangle t = {2\pi r_p\over c}$$
$$\omega={c\over r_p}$$
$$\bar{M}={1\over5}Q{c\over r_p}R^2$$ 
$$\bar{M}={1\over5}e{c\over r_p}r_p^2$$ 
$$r_p={2h\over\pi cm_p}$$
$$\bar{M}={1\over5}ec{2h\over\pi cm_p}$$ 
$$\bar{M}={1\over5}e{4\hbar\over m_p}$$ 

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