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NUCLEAR MAGNETIC RESONANCE

By Charles Rhodes, P.Eng., Ph.D.

This web page relies on parameters developed on the web page titled: PLANCK CONSTANT.

NUCLEAR MAGNETIC RESONANCE:
If an atomic charged particle is located within an externally applied magnetic field the net particle charge Q remains unchanged and the charge circulation velocity C remains unchanged, so the parameter that is free to change in response to the applied magnetic field is the charge hose length Lh.

A change in Lh produces a change in frequency Fh. The change in Fh corresponds to a change in particle potential energy.

The extent of the change in particle potential energy is dependent on the orientation of the spheromak axis with respect to the applied magnetic field. At one orientation extreme the applied magnetic field vectorially partially cancels the spheromak's poloidal magnetic field within the spheromak core. At the other orientation extreme the applied magnetic field vectorially adds to the spheromak's poloidal magnetic field within the spheromak core.

If RF energy of the appropriate frequency is applied the charged particles will tend to absorb photons and transfer from their lower energy state to their higher energy state. If the RF energy supply is removed the particles will tend to emit photons to transfer from their higher energy state to their lower energy state. The frequency of these photons is dependent on the energy difference between the two energy states that in turn is dependent on the strength of the extenally applied magnetic field.

There is also a density of states issue. When there is no external magnetic field the orientation of the spheromaks is completely random. As the strength of the externally applied magnetic field increases the probability of a particular spheromak occupying a lower energy state as compard to a higher energy state increases. However, there are a lot more available partially aligned states than fully aligned or fully non-aligned states. Hence the average change in energy is only about half of the energy difference between a full non-aligned spheromak and a fully aligned spheromak.

Consider charged particles such as free electrons or protons. With no external field the particle energy is approximately given by:
Ett = [Efs] (Ett / Efs)
= Upo (K Ro)^3 Pi^2 {1 - [(So -1)^2 / (So^2 + 1)]^2}
= (Bpo^2 / 2 Mu) (K Ro)^3 Pi^2 {1 - [(So -1)^2 / (So^2 + 1)]^2}

The change in spheromak energy due to a change in spheromakcore magnetic field with (K Ro) constant and hence Fh constant is:
d(Ett) = (2 Bpo / 2 Mu) (K Ro)^3 Pi^2 dBpo {1 - [(So -1)^2 / (So^2 + 1)]^2}

d(Ett) / (Ett) = [(2 Bpo / 2 Mu) (K Ro)^3 Pi^2 dBpo] / [(Bpo^2 / 2 Mu) (K Ro)^3 Pi^2]
= 2 dBpo / Bpo

If an external magnetic field Bx is applied the total particle energy equation becomes:
Ett = [(Bpo + Bx cos(theta))^2 / 2 Mu] (K Ro)^3 Pi^2 {1 - [(So -1)^2 / (So^2 + 1)]^2}
where (theta) is the angle between the external magnetic field direction and the particle's main axis of symmetry.

For Bx << Bpo the change in field energy occurring due to insertion of a particle into the external magnetic field is:
Delta(Ett) = [(2 Bpo Bx cos(Theta)) / 2 Mu](K Ro)^3 Pi^2 {1 - [(So -1)^2 / (So^2 + 1)]^2}
= [(2 Bpo Bx cos(Theta)) / 2 Mu] [Ett / (Bpo^2 / 2 Mu)]
= [(2 Bx cos(Theta))] [Ett / Bpo]

Consider N randomly oriented particles. These particles are evenly distributed over the surface of a sphere of radius R with the particle axis normal to the sphere surface. Then the surface density of particles is N / (4 Pi R^2).

The particles can be divided into two groups, those that are partially aligned to the applied magnetic field and those that are partially opposed to the applied magnetic field

For those particles which are partially aligned to the applied magnetic field the average influence of Bx on the particle internal field Bpo is given by:
Integral from theta = 0 to theta = (Pi / 2) of:
R d(theta) 2 Pi R cos(theta)(N / 4 Pi R^2) Bx sin(theta) / (N / 2)
= Integral from Theta = 0 to Theta = (Pi / 2) of:
sin(Theta) cos(Theta) Bx d(Theta)
= {[sin(Pi / 2)]^2 / 2 - [sin(0 / 2)]^2} Bx
Bx / 2

Thus when the particles transition from the aligned state to the anti-aligned state the average change in particle central magnetic field is:
(Bx / 2) - (- Bx / 2)
= Bx

The corresponding change in particle energy ie d(Ett).

The corresponding fractional change in system energy is:
d(Ett) / Ett = 2 dBpo Bpo / Bpo^2
= 2 Bx / Bpo

The corresponding change in (1 / Ro) is d(1 / Ro).

The corresponding change in frequency is dFh.
Thus:
dFh / Fh = d(Ett) / Ett or dFh = Fh d(Ett) / Ett
= d(Ett) / h
= (1 / h) [dEtt]

Thus we need an expression for the average change in particle energy dEtt produced by application of an external magnetic field Bx.

The particle characteristic frequency is:
Fh = C / Lh
or
dFh = - C dLh / Lh^2
= - Fh (dLh / Lh)
= - Fh (dRo / Ro)

E = (Bpo^2 / 2) (K Ro)^3 Pi^2 {1 - [(So -1)^2 / (So^2 + 1)]^2}
or
dE = Bpo dBpo (K Ro)^3 Pi^2 + 3 (Bpo^2 / 2) (K Ro)^2 Pi^2 dRo
or
h dFh = Bpo dBpo (K Ro)^3 Pi^2 + 3 (Bpo^2 / 2) (K Ro)^3 Pi^2 (K dRo / K Ro)
= Bpo dBpo (K Ro)^3 Pi^2 + 3 h Fh (dFh / - Fh)
or
4 h dFh = Bpo dBpo (K Ro)^3 Pi^2
or
dFh = Bpo dBpo (K Ro)^3 Pi^2 / 4 h
= (2 / Bpo) E dBpo / 4 h
= (E / 2 h)(dBpo / Bpo)
= (E / 2 h)(Bx / Bpo)

This equation allows determination of Bpo for protons in water which in turn allows determination of Ro and hence Rs for protons.
 

FOR A PROTON:
Experimental proton magnetic resonance data gives:
(dF / Bx) = (42.5781 X 10^6 Hz / T)

Hence:
dF / Bx = (Ett / 2 h)(1 / Bpo)
or
Bpo = (Ett / 2 h) / (dF / Bx)
= [(1.6726 10-27 kilograms X (2.997 X 10^8 m / s)^2 / [(2 X 6.626 X 10^-34 J-s) (42.5781 X 10^6 Hz / T)]
= 0.0266255 X 10^17 kg m^2 T / s^2 J s S^-1
= 0.0266255 X 10^17 T kg m^2 / s^2 (kg m^2 / s^2)
= 0.0266255 X 10^17 T

Thus Bpo for a proton is about 2.66255 X 10^15 T

Determination of Bpo for electrons is more difficult due to interfering magnetic fields caused by other electrons.

Recall that the total energy of an elementary particle is given by:
Ett = (Bpo^2 / 2 Mu) (K Ro)^3 Pi^2 {1 - [(So -1)^2 / (So^2 + 1)]^2}
or
(K Ro)^3 = [Ett 2 Mu] / {[Bpo^2 Pi^2] {1 - [(So -1)^2 / (So^2 + 1)]^2}}

This equation can be used to calculate the physical size of a proton.

For a proton:
(K Ro)^3 = [Ett 2 Mu] / {[Bpo^2 Pi^2]{1 - [(So -1)^2 / (So^2 + 1)]^2}}
 
= [(1.6726 10-27 kilograms X (2.997 X 10^8 m / s)^2 X 2 X (4 Pi X 10^-7 T m / A)]
/ {[(2.66255 X 10^15 T)^2 Pi^2][0.96]}
 
= [(1.6726 10-27 kilograms) X (2.997 X 10^8 m / s)^2 X (8 X 10^-7 T m / A)]
/ {[(2.66255 X 10^15 T)^2 Pi][0.96]}
 
= [16.95352 X 10^-48] kg m^2 T m / A s^2 T^2 / [0.96]
 

Hence:
(K Ro) = [2.56893 X 10^-16 m]
and
Rs = So (K Ro)
= So [2.56893 X 10^-16 m]
= 2.026 (2.56893 X 10^-16 m)
= 0.52 X 10^-15 m
= 0.52 fermi
 

This figure compares to an accepted "proton charge radius" in the range .84 fermi to 0.87 fermi.

This web page last updated September 26, 2016.

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