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PLANCK CONSTANT

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

RADIATION AND MATTER:
Atomic charged particles are spheromaks. Spheromaks can absorb or emit radiation photons. During radiation absorption and emission total energy is conserved. Every packet of potential energy, including both charged particles and photons, has a characteristic frequency. During photon emission the emitting particle's characteristic frequency decreases and the number of photons present increases. During photon absorption the absorbing particle's characteristic frequency increases and the number of photons present decreases.

The energy versus frequency behavior of charged particles and photons is determined by the Planck constant, which lies at the heart of quantum mechanics. Analysis of the Planck constant provides insight into the mechanism by which nature stores energy in rest mass and the reasons for quantum mechanical behavior.

In our local universe there is an overall tendency for high energy photons to gradually change into a larger number of lower energy photons. This tendancy determines the direction of evolution of the universe.
 

PLANCK CONSTANT:
The Planck Constant relates an amount of potential energy to its characteristic frequency.

The Planck Constant is a fundamental parameter in quantum mechanics. The origin of the Planck Constant lies in the manner in which energy is stored in an atomic particle spheromak. The Planck Constant is a frequently reoccurring composite of other physical and mathematical constants. This web page shows the origin of the Planck Constant.

A spheromak's field structure allows individual quantized charges to act as stable stores of potential energy. The behavior of these energy stores is governed by electrodynamics. This web page shows the mathematical relationship between spheromaks and quantum mechanics.

It is shown herein that the total potential energy Ett of a charged atomic particle at steady state is given by:
Ett = h Fh
where Fh is the characteristic frequency of the particle and h is a composite constant known as the Planck constant.

When a charged particle gains or loses potential energy by absorption or emission of a photon the particle transitions from state "a" with particle potential energy Etta and natural frequency Fha to state "b" with particle potential energy Ettb and natural frequency Fhb. The change in particle potential energy is:
(Ettb - Etta) = h (Fhb - Fha)

When a charged particle absorbs a photon with energy Ep it also absorbs that photon's linear momentum. From Einstein's fameous special relativistic relationship:
E^2 = P^2 C^2 + Mo^2 C^4
the momentum Pp of a photon with no rest mass but with energy Ep is:
Pp = Ep / C
where:
C = speed of light.
 

PHOTON ABSORPTION:
If a charged particle at rest with initial potential energy Etta absorbs a photon with energy Ep to conserve momentum the charged particle with combined total energy (Etta + Ep) also acquires the photon momentum Pp. Hence: (Etta + Ep)^2 = Pp^2 C^2 + Ettb^2
or
(Etta + Ep)^2 = Ep^2 + Ettb^2
where Ettb is the charged particle potential energy after absorption of the photon.

Hence:
(Etta + Ep)^2 = Ep^2 + Ettb^2
or
Etta^2 + 2 Etta Ep = Ettb^2
or
Ettb = [Etta^2 (1 + 2 Ep / Etta)]^0.5
= Etta (1 + 2 Ep / Etta)^0.5

Hence:
(Ettb - Etta) = Etta (1 + 2 Ep / Etta)^0.5 - Etta
= Etta [(1 + (2 Ep / Etta))^0.5 - 1]
~ Etta [1 + (Ep / Etta) - [(2 Ep / Etta)^2 / 8] - 1]
= Ep - (Ep^2 / 2 Etta)
= Ep [1 - (Ep / 2 Etta)]

Hence for photon absorption:
Ep = (Ettb - Etta) / [1 - (Ep / 2 Etta)]
 

PHOTON EMISSION:
If a charged particle at rest with initial potential energy Etta emits a photon with energy Ep to conserve momentum the charged particle with the new total energy (Etta - Ep) acquires the photon momentum Pp. Hence: (Etta - Ep)^2 = Pp^2 C^2 + Ettb^2
or
(Etta - Ep)^2 = Ep^2 + Ettb^2
where Ettb is the charged particle potential energy after emission of the photon.

Hence:
(Etta - Ep)^2 = Ep^2 + Ettb^2
or
Etta^2 - 2 Etta Ep = Ettb^2
or
Ettb = [Etta^2 (1 - 2 Ep / Etta)]^0.5
= Etta (1 - 2 Ep / Etta)^0.5

Hence:
(Etta - Ettb) = Etta - Etta (1 - 2 Ep / Etta)^0.5
= Etta [1 - (1 - 2 Ep / Etta)^0.5]
~ Etta [ 1 - (1 - (Ep / Etta) - (2 Ep / Etta)^2 / 8)]
= Etta [ (Ep / Etta) + (2 Ep / Etta)^2 / 8)]
= Ep + (Ep^2 / 2 Etta)
= Ep [1 + (Ep / 2 Etta)]

Hence for photon emission:
Ep = (Etta - Ettb) / [1 + (Ep / 2 Etta)]
 

EXPERIMENTAL MEASUREMENT OF h:
Ep = h Fp
or
h = Ep / Fp

If the experimental methodology involves measurement of the frequency of emitted photons the formula that should be used for determining h is:
h = (Etta - Ettb) / {Fp [1 + (Ep / 2 Etta)]}
Note that on emission of a photon the change in charged particle potential energy (Etta -Ettb) is slightly greater than the photon energy Ep.

If the experimental methodology involves measurement of the frequency of absorbed photons the formula that should be used for determining h is:
h = (Ettb - Etta) / {Fp [1 - (Ep / 2 Etta)]}

Note that on absorption of a photon the change in charged particle potential energy (Ettb -Etta) is slightly less than the photon energy Ep.
 

EXPERIMENTAL ERROR:
Most high resolution experimental measurements of h rely on precise spectroscopic measurement of the frequency of photons emitted by excited electrons. In such experiments personnel usually incorrectly implicitly assume that the term:
[1 + (Ep / 2 Etta)] = 1

However, at resolutions in measurement of h with 5 or more significant figures that assumtion is wrong and the claimed experimentally measured values of h are consistently larger than the precise theoretically calculated value.

Using spheromak theory we can precisely calculate:
h = (Ettb - Etta) / (Fhb - Fha)
= dEtt / dFh

In high resolution measurements of h it is also necessary to account for the charged particle recoil kinetic energy on absorption or emission of a photon.

A secondary object of this web page is to investigate measurement of h and spheromak core magnetic field strength Bpo via use of an externally applied magnetic field.
 

EXPERIMENTAL VALUE FOR PLANCK CONSTANT:
The published experimentally measured value for the Planck Constant h is:
h = 6.62606957 10-34 m^2 kg / s.

The accuracy of measurement of a proton charge has been limited by the experimental error in determination of h. Hence it is highly desirable to find a precise expression for h.
 

ORIGIN OF PLANCK CONSTANT:
The parameter h is a function of:
Mu = permiability of free space;
C = speed of light in a vacuum;
Q = proton charge;
Pi = (circumference / diameter) of a circle
= 3.141592653589793
Pi^2 = 9.869604401

To understand the relationship of spheromak parameters to the Planck constant it is necessary to derive a closed form expression for the total field energy of a spheromak and hence an elementary atomic charged particle.
 

PARAMETER DEFINITIONS:
Define for a spheromak in free space:
Rc = minimum radius of inner spheromak wall;
Rs = maximum radius of outer spheromak wall;
(K Ro) = (Rs Rc)^0.5 = radius of atomic particle where the potential energy well is deepest;
So = [Rs / K Ro] = [K Ro / Rc] = spheromak shape parameter;
So^2 = (Rs / Rc);
Hs = distance of spheromak wall from the equatoral plane;
Hf = maximum value of |Hs|
2 Hf = spheromak overall length;
Lh = charge hose length;
Np = number of poloidal charge motion path turns contained in Lh;
Nt = number of toroidal charge motion path turns contained in Lh;
Nr = Np / Nt;
Rf = wall radius at H = Hf and H = -Hf;
Lp = Pi (Rs + Rc);
Lt = Pi (Rs - Rc);
Bpo = poloidal magnetic field strength at the center of the spheromak;
Upo = (Bpo^2 / 2 Mu) = maximum field energy density at the center of the spheromak;
Upo = (Uo / K^4) = maximum field energy density at the center of the spheromak;
 

As shown on the web page titled ELECTROMAGNETIC SPHEROMAK the peak magnetic field strength Bpo at the center of a spheromak can be expressed as:
Bpo = [(Mu C Qa) / (4 Pi K^2 Ro^2)]

As shown on the web page titled ELECTROMAGNETIC SPHEROMAK the peak magnetic field strength Bpo at the center of a spheromak can also be expressed as:
Bpo = I [(Mu Qs C) / (2 Pi^2 Rc^2)] {Nr / {[Nr (So^2 + 1)]^2 + [So^2 - 1]^2}^0.5}
= I (Mu Qs C) / (2 Pi^2 Ro^2)(Ro / Rc)^2 {Nr / {[Nr (So^2 + 1)]^2 + [So^2 - 1]^2}^0.5}
= I (Mu Qs C / (2 Pi^2 K^2 Ro^2) So^2 {Nr / {[Nr (So^2 + 1)]^2 + [So^2 - 1]^2}^0.5}
where:
I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 Z - So^2 + Z]^1.5)

where:
So^2 = (Rs / Rc)
and
Nr = (Np / Nt)
where:
Np = integer number of poloidal turns
and
Nt = integer number of toroidal turns.

In order to determine the spheromak operating point for each value of So^2 find the corresponding value of Nr^2 using the common boundary condition formula:
Nr^2 = {(8 / Pi^2) - [(So^2 - 1) / (So^2 + 1)]^2} / {1 - (16 / [Pi (So^2 - 1)]^2)}
which formula is derived on the web page titled: ELECTROMAGNETIC SPHEROMAK and then do a numerical integration to determine I.
 

RELATIONSHIPS BETWEEN Efs, Ett and Fh:
The maximum possible spheromak energy Efs. As shown on the web page titled: SPHEROMAK ENERGY the maximum possible energy Efs that can be trapped by a spheromak is given by:
Efs = (Bpo^2 / 2 Mu) (K Ro)^3 Pi^2

As shown on the web page titled: SPHEROMAK ENERGY the actual energy trapped by a spheromak at steady state is given by:
Ett = Efs {1 - [(So -1)^2 / (So^2 + 1)]^2}
where:
So ~ 2.026

Thus:
(Ett / Efs) = {1 - [(So -1)^2 / (So^2 + 1)]^2}

The characteristic frequency Fh of an atomic particle spheromak is given by:
Fh = (C / Lh)
= C / {[2 Pi Np (Rs + Rc) / 2]^2 + [2 Pi Nt (Rs - Rc) / 2]^2}^0.5
= C / Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5
= C / Pi Nt {[Nr (Rs + Rc)]^2 + [(Rs - Rc)]^2}^0.5
= C / Pi Nt Rc {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5
= C Ro / Pi Nt Rc Ro {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5
= [C / (Pi Nt K Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Hence:
[1 / K Ro] = Fh Pi Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [C So]

Thus:
Efs = (Bpo^2 / 2 Mu) (K Ro)^3 Pi^2
= (Bpo^2 / 2 Mu) (K Ro)^4 Pi^2 [1 / K Ro]
= (Bpo^2 / 2 Mu) (K Ro)^4 Pi^2 Fh Pi Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [C So]

As shown on the web page ELECTROMAGNETIC SPHEROMAK from far field spheromak energy density matching:
Bpo^2 = [Mu^2 C^2][Qa / (4 Pi K^2 Ro^2)]^2

Thus:
Efs = (Bpo^2 / 2 Mu) (K Ro)^4 Pi^2 Fh Pi Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [C So]
= ([Mu^2 C^2][Qa / (4 Pi K^2 Ro^2)]^2 / 2 Mu) (K Ro)^4 Pi^2 Fh Pi Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [C So]
= Fh ([Mu C Qa^2] / 32) Pi Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]
= Fh ([Mu C Qa^2] / 4 Pi) [Pi^2 / 8] Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]

 

Ett = (Ett / Efs) Efs
= (Ett / Efs) Fh ([Mu C Qa^2] / 4 Pi) [Pi^2 / 8] Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]
= h Fh
where the Planck constant h is given by:
h = ([Ett / Efs] [Mu C Qa^2] / 4 Pi) [Pi^2 / 8] Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]
= ({1 - [(So -1)^2 / (So^2 + 1)]^2} [(Mu C Qa^2) / (4 Pi)] [Pi^2 / 8] Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]

In this formula So at steady state adopts the value that minimizes Ett. As shown by the following graph of the So dependent term of Ett vs So:
So ~ 2.026

In the expression for the Planck constant:
So ~ 2.026
Pi = 3.141592653589793
and
Pi^2 = 9.869604401

As shown on the web page titled: SPHEROMAK ENERGY at the spheromak minimum energy operating point the approximate values are:
{1 - [(So -1)^2 / (So^2 + 1)]^2} {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So] ~ [2.2882]
So ~ 2.026
So^2 ~ 4.104676
Nr^2 ~ 0.5297583716
Nr ~ 0.7278450189
Ett = [Mu C Qa^2 / 4 Pi] [Pi^2 / 8] Fh Nt [2.2882]
Ett = h Fh

The published experimentally measured value of h is:
h = 6.62606957 10-34 m^2 kg / s
= 6.62606957 10-34 J-s

Qa = 1.60217662 X 10^-19 coulombs
C = 2.99792458 X 10^8 m / s
Mu = 4 Pi X 10^-7 T^2 m^3 / J
[Pi^2 / 8] = 1.23370055

Hence:
Nt = h / {[Mu C Qa^2 / 4 Pi] [Pi^2 / 8] [2.2882]
= 6.62606957 10-34 J-s / {10^-7 X 2.99792458 X 10^8 m / s X (1.60217662 X 10^-19 coulombs)^2 x 1.23370055 X 2.2882}
= 6.62606957 10-34 J-s / {(7.695582224 X 10^-37) X 1.23370055 X 2.2882}
= 305.0076759 turns

Then:
Np = Nr Nt
= 0.0.7278450189 (305.0076759)
= 221.9983176 turns

Since Np and Nt must be integers with no common factors the only possible exact solution for Np, Nt is:
Nt = 305
Np = 222

which gives the precise value of Nr as:
Nr = (Np / Nt)
= 222 / 305
= 0.7278688525
 

EXACT VALUES:
Since Np and Nt are integers we know that the exact value of Nr is given by:
Nr = (Np / Nt)
= 222 / 305
= 0.7278688525

Hence the exact value of Nr^2 is given by:
Nr^2 = (0.7278688525)^2
= 0.5297930664

Since we now have an accurate value for Nr^2 we can accurately determine So^2 using the common boundary condition. With an accurate value of So^2 we can calculate an accurate value of h in terms of Mu, C and Q.
 

Recall that from the common boundary condition:
Nr^2 = ([8 - {[Pi]^2 [(So^2 - 1) / (So^2 + 1)]^2}] / {[Pi]^2 - [16 / (So^2 - 1)^2]})

Using the precise value of Nr^2 find the corresponding precise value of So^2. In order to do so use the approximate value of So^2 in combination with the slope of the Nr^2 versus So^2 curve at the approximate solution to find the exact solution.

Let X = So^2 - 1
Then:
Nr^2 = {8 - [Pi]^2 [X / (X + 2)]^2} / {[Pi]^2 - [16 / (X^2)]}

Recall that:
Pi^2 = 9.869604401

Try So = 2.0260000
Then:
X = So^2 - 1
= 3.104676
and
X^2 = 9.639013065
giving a trial value of Nr^2 as:
Trial Nr^2 = {8 - [Pi^2] [X / (X + 2)]^2} / {[Pi]^2 - [16 / (X^2)]}
= {8 - [9.869604401] [3.104676 / (5.104676)]^2} / {[9.869604401] - [16 / (9.639013065)]}
= 4.34913352 / {8.209683422}
= 0.5297565444

Recall that the target value of Nr^2 is:
(222/ 305)^2 = 0.5297930664

Thus for a trial value of So = 2.02600000 the target value of Nr^2 exceeds the trial value of Nr^2 by: 0.5297930664 - 0.5297565444 = 3.652198 X 10^-5

We need to make the trial value of Nr^2 slightly larger which implies making the trial value of So slightly smaller.

 

Try So = 2.02590000
Then:
X = So^2 - 1
= 3.10427081
and
X^2 = 9.636497262
giving a trial value of Nr^2 as:
Trial Nr^2 = {8 - [Pi^2] [X / (X + 2)]^2} / {[Pi]^2 - [16 / (X^2)]}
= {8 - [9.869604401] [3.10427081 / (5.10427081)]^2} / {[9.869604401] - [16 / ( 9.636497262)]}
= 4.349506903 / 8.209250066
= 0.5298299928

Recall that the target value of Nr^2 is:
(222 / 305)^2 = 0.5297930664

Thus for a trial value of So = 2.02590000 the trial value of Nr^2 exceeds the target value of Nr^2 by: 0.5298299928 - 0.5297930664 = 3.692635 X 10^-5
 

FINDING EXACT VALUES:
Thus:
d(Trial value of Nr^2) / dSo = (0.5298299928 - 0.5297565444) / (2.02590000 - 2.0260000)
= 7.34484 X 10^-5 / (- 10^-4)
= - 0.734484

Hence: [So|target] = [So|trial] + [dSo / dNr^2] {[Nr^2|target] - [Nr^2|trial]}
= 2.02600000 + [- 1 / 0.734484] [0.5297930664 - 0.5297565444]
= 2.02600000 - [4.972470469 X 10^-5]
= 2.02600000 - .00004972470469
= 2.025950275

The corresponding value of X is:
X = So^2 - 1
= 3.104474517

X^2 = 9.637762026

We can check this calculation by recalculating Nr^2.
Nr^2 = {8 - [Pi^2] [X / (X + 2)]^2} / {[Pi^2] - [16 / (X^2)]}
= {8 - [9.869604401] [3.104474517 / (5.104474517)]^2} / {[9.869604401] - [16 / (9.637762026)]}
= 4.349319182 / 8.209467955
= 0.529793064

Hence:
Nr = 0.7278688508 as compared to target value of:
(222 / 305) = 0.7278688525

Thus the calculated precise So value gives a precise Nr value that is accurate to eight significant figures.

Thus at the spheromak operating point So is:
So = 2.025950275
and
So^2 = 4.104474517

Recall that the theoretical value of the Planck constant h is given by:
h = [(Mu C Qa^2) / (4 Pi)] [Pi^2 / 8] Nt
{1 - [(So -1)^2 / (So^2 + 1)]^2}
{[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]

Evaluation of the terms of h gives:
[(Mu C Qa^2) / (4 Pi)] [Pi^2 / 8] Nt
= {(7.695582224 X 10^-37) X 1.23370055 X 305 J-s
= 2.895683427 X 10^-34 J-s

 

{1 - [(So -1)^2 / (So^2 + 1)]^2}
= {1 - [(1.025950275)^2 / (5.104474517)]^2}
= {1 - 0.0425209706}
= 0.9574790294

{[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]
= {[0.5297930664 (5.104474517)^2] + [(3.104474517)]^2}^0.5 / [2.025950275]
= {[13.80410806] + [9.637762027]}^0.5 / [2.025950275]
= 2.389831854

Hence the theoretical value of h is:
h = [2.895683427 X 10^-34 J-s] [0.9574790294] [2.389831854]
= 6.625943023 X 10^-34 J-s

By comparison the published experimental value for h is:
h = 6.62606957 10-34 m^2 kg / s.

These two values for h are in agreement to an accuracy of 2 parts in 100,000. The discrepency is due to conservation of linear momentum on the emission of a photon by an excited charged particle. It is necessary to examine exactly how h is experimentally measured to understand this matter. The fractional discrepency is:
[(6.62606957 - 6.625943023) / 6.62606957] = 1.9098 X 10^-5

When a charged particle reduces its potential energy by photon emission part of the drop in particle potential energy will become recoil kinetic energy. Hence there is a slight difference between the decrease in charged particle potential energy and the energy carried away by the photon. That difference is the charged particle recoil kinetic energy.

Consider an electron with rest mass Me. The rest potential energy of the electron is:
Etta = Me C^2.

The ionization energy of a gas is typically of the order of:
Ep = 20 eV.

For electrons: Etta = Me C^2 = 9.1 X 10^-31 kg X (3 X 10^8 m / s)^2 X 1 eV / 1.602 X 10^-19 J
= 51.12 X 10^4 eV

Hence if h is measured via photon emission from an ionized gas and if the term:
[1 +(Ep / 2 Etta)]
is assumed to be unity we can expect an error in the experimentally determined value of h of about:
(Ep / 2 Er) = 20 eV / [2 (51.12 X 10^4 eV)]
= 2 / (102.24 X 10^3)
~ 1.95 X 10^-5

Thus the small discrepency between the theoretical and experimental values of h is primarily caused by failure of experimentalists to properly take into account recoil kinetic energy when an electron emits a photon. When an electron emits a photon about 2 / 100,000 of the change in potential energy becomes electron linear kinetic energy rather than photon energy. If this error is uncorrected the experimentally observed value of h will be slightly larger than the theoretical value calculated herein. Note that the h value calculated herein is actually the change in charged particle potential energy with respect to a change in charged particle natural frequency. The energy carried away by the photon will be slightly less than the decrease in charged particle potential energy due to the small increase in charged particle kinetic energy on emission of a photon. This increase in charged particle kinetic energy must occur to satisfy the law of conservation of linear momentum.
 

CONCLUSION:
The spheromak model of a charged particle provides a means of accurately calculating the Planck constant h in terms of Pi, Mu, Q and C. In highly accurate experimental measurements of h it is necessary to take into account the charged particle recoil kinetic energy. When recoil kinetic energy is properly taken into account there is agreement between the theoretical and experimental determinations of the Planck constant h to at least 7 significant figures.

Note that for a spheromak at steady state conditions the Planck constant is independent of the charged particle nominal radius (K Ro) and hence is also independent of the charged particle total energy Ett.
 

This web page last updated September 30, 2016.

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