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The photon and the constant of Planck mathematically



The photon and the constant of Planck mathematically

When you are inter­ested in physics you must read “Unbe­liev­able”!

A pho­ton, with enough energy, can trans­form, split up into a positron and an elec­tron. The assump­tion that a pho­ton can be described as two charges in oscil­la­tion seems accept­able. The Coulomb-​force between the two charges is:

If we try to cal­cu­late the energy of the oscil­la­tion with F we have to solve a few prob­lems first. Unknown are the size of the charges Qf, the dis­tance Rf at the begin­ning of the oscil­la­tion and at what dis­tance F becomes inac­tive (so the sin­gu­lar­ity is avoided).

We assume the max­i­mum dis­tance between the two charges Rf is ¼ λ. The force is then:

This is the total force between both charges. When +Qf moves R to -Qf, then -Qf also moves R to +Qf. The force while both par­ti­cles moved R will be:

When we inte­grate F to a dis­tance R=0 a sin­gu­lar­ity arises. Both the force and energy become infi­nite. This prob­lem was not solv­able before. Now we have an indi­ca­tion for the dis­tance to which F is active: the radius of the neu­tron Rn or the radius of the pro­ton Rp.

The diam­e­ter, the dis­tance between +Qf and -Qf when F=0, when both charges pen­e­trate the point-​volume, is twice the radius of the point-​volume and there­fore approx. twice the cal­cu­lated radius (2Rn) of the neu­tron or pro­ton. The energy of the oscil­la­tion will be:

The last for­mula demon­strates that the arbi­trary dis­tance Rf=1/4λ , the max­i­mum dis­tance between the two oppo­site charges, is of no influ­ence for the energy Wψ of the oscil­la­tion because 8Rn«λ. The oscil­la­tion energy will be:

The only unknown fac­tor is Qf. We assume that Qf=Qe/n where n is an arbi­trary number:

The charge Qf is replaced by Qe/​n. The unknown fac­tor is now n. The most straight­for­ward assump­tion is that , because the radius of the elec­tron with charge -Qe is Re and the wave­length λ has to be pro­por­tional with . Sub­sti­tu­tion gives:

Like elec­tron, pro­ton and neu­tron the pho­ton has two degrees of free­dom to store energy: oscil­la­tion and move­ment. So the total energy of the pho­ton Wf will be equal to 2.

while the energy of the pho­ton is:

Wf = hν

with H the derived con­stant and h the con­stant of Planck. The devi­a­tion between h and H is 6%, so the rela­tion is not exact, but fair enough when one con­sid­ers the radius of the neu­tron is play­ing a cru­cial part. The coin­ci­dence that the con­stant of Planck will derived within a 6% accu­racy, when the ether is no more than a wild fan­tasy, is very unlikely. There are many pos­si­ble causes to explain the dif­fer­ence, like:

  • The math­e­mat­i­cal con­stant π and the nat­ural log­a­rithm e have to arise some­where. Maybe they arise in the point-​volume?
  • The elec­tric field of the with speed c mov­ing charges of the pho­ton is forced in a two dimen­sional cir­cle. Is the three dimen­sional equa­tion for the elec­tro­sta­tic attrac­tion still com­pletely valid?
  • Can the point-​volume be trans­formed in some way when it becomes a neutron?

The pic­ture the ether showed us so far is that the point-​volume is not rigid.

We assume that the con­stant of Planck (h) deter­mines the radius of the point-​volume rather than the cal­cu­lated, and prob­a­bly deformed, radius of the pro­ton or neutron.

When we use the above derived for­mula and replace H with the exper­i­men­tal derived value for Planck’s con­stant h we can cal­cu­late Rplanck; the radius of the point-​volume accord­ing to the exper­i­men­tal value of h:

The cal­cu­lated radius of the point-​volume; the Planck-​radius (Rplanck), con­sid­er­ing that

and , becomes:

Next chap­ter: QM and the Ether: The Deriva­tion of Planck’s Constant

When you are interested in physics you must read “Unbelievable“!

A photon, with enough energy, can transform, split up into a positron and an electron. The assumption that a photon can be described as two charges in oscillation seems acceptable. The Coulomb-force between the two charges is:

If we try to calculate the energy of the oscillation with F we have to solve a few problems first. Unknown are the size of the charges Qf, the distance Rf at the beginning of the oscillation and at what distance F becomes inactive (so the singularity is avoided).

We assume the maximum distance between the two charges Rf is ¼ λ. The force is then:

This is the total force between both charges. When +Qf moves R to –Qf, then –Qf also moves R to +Qf. The force while both particles moved R will be:

When we integrate F to a distance R=0 a singularity arises. Both the force and energy become infinite. This problem was not solvable before. Now we have an indication for the distance to which F is active: the radius of the neutron Rn or the radius of the proton Rp.

The diameter, the distance between +Qf and –Qf when F=0, when both charges penetrate the point-volume, is twice the radius of the point-volume and therefore approx. twice the calculated radius (2Rn) of the neutron or proton. The energy of the oscillation   will be:

The last formula demonstrates that the arbitrary distance Rf=1/4λ , the maximum distance between the two opposite charges,  is of no influence for the energy Wψ of the oscillation because 8Rn<<λ. The oscillation energy will be:

The only unknown factor is Qf. We assume that Qf=Qe/n where n is an arbitrary number:

The charge Qf is replaced by Qe/n. The unknown factor is now n. The most straightforward assumption is that , because the radius of the electron with charge –Qe is Re and the wavelength λ has to be proportional with . Substitution gives:

Like electron, proton and neutron the photon has two degrees of freedom to store energy: oscillation and movement. So the total energy of the photon Wf will be equal to 2Wψ.

while the energy of the photon is:

Wf  = hν

with H the derived constant and h the constant of Planck. The deviation between h and H is 6%, so the relation is not exact, but fair enough when one considers the radius of the neutron is playing a crucial part. The coincidence that the constant of Planck will derived within a 6% accuracy, when the ether is no more than a wild fantasy, is very unlikely. There are many possible causes to explain the difference, like:

  • The mathematical constant π and the natural logarithm e have to arise somewhere. Maybe they arise in the point-volume?
  • The electric field of the with speed c moving charges of the photon is forced in a two dimensional circle. Is the three dimensional equation for the electrostatic attraction still completely valid?
  • Can the point-volume be transformed in some way when it becomes a neutron?

The picture the ether showed us so far is that the point-volume is not rigid.

We assume that the constant of Planck (h) determines the radius of the point-volume rather than the calculated, and probably deformed, radius of the proton or neutron.

When we use the above derived formula and replace H with the experimental derived value for Planck’s constant h we can calculate Rplanck; the radius of the point-volume according to the experimental value of h:

The calculated radius of the point-volume; the Planck-radius (Rplanck), considering that

and , becomes:

Next chapter: QM and the Ether: The Derivation of Planck’s Constant

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