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Photons: Why They Follow Geodesics


Willem F Esterhuyse

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Photons always propagate at locally, so do not experience proper acceleration anywhere, and hence they satisfy the geodesic equation - pretty much by definition. They cannot in fact do anything else, as they wouldn‘t be photons otherwise.

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I don't see any equations. Can you predict anything?

You state that c is the maximum speed - based on what?

You say "We have that the theory of defining photons my be tested by proving: there is a direction in which photons with the same orientation will not go." What direction is that?

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2 hours ago, Willem F Esterhuyse said:

That does not make total sense. What is the mechanism by which it does this?

Here is the manuscript attached. It is in .pdf format.

!

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Which bit of "present your discussion here" did you not understand?

 
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The program allows for uploading a file.

Anyway to answer the previous reply: 1) I can predict that photons see positive events of spacetime while anti-photons see negative events.

2) My model is still not solid on the "c" issue

3) The direction will depend on the experiment.

Anyway: here is the manuscript copied to here:

    Physics from Axioms.
                Mr. Willem F. Esterhuyse
            Email: talanum1@yahoo.com            
                Abstract:
  We introduce a definition of Time and Photons from four Axioms.  Basically you take a 4-dimensional manifold, transform them into two superimposed Riemann Spheres and isolate a circle (call this Pp) in one of the spheres. Then one specifies the circle to turn by a unit amount (the turn is an quantum rotation: turn from state A to state B without visiting the in between states) as measured along the circle, if the Pp encounters a space point. Space fluctuates and expands so this does not give a static circle Pp. The circle's infinity point stays at the north pole of the Riemann Sphere for any finite rotation since: infinity - constant = infinity. Using this one can define basic spacetime and from basic spacetime, Time can be defined if we require special particles to be in the particles of a clock.  We go on to define photons and antiphotons. The model predicts that there is a direction in which photons (from the same process and with the same orientation) are never emitted. We continue to define a pi-minus.
Keywords: time, photon, pi-minus.
Contents:
1. Defining Time.
2. Defining Photons and Anti-photons.
3. Defining a Pi-minus.
Bibliography.
1. Defining Time.
  Here are the four axioms we are going to use:
A1: Complex numbers exists. Call this C.
A2: x = x
A3: x + y = y + x
A4: A is a subset of B if B contains A and B - A not = the empty set.
  The following definitions are stated and will be used:
Definitions: "C x C" means "Comlex plane Cartesian product Complex plane".
             "RS <-> RS" means "Riemann sphere superimposed on Riemann sphere".
             "quantum rotation" means "a rotation from state A to state B without visiting the               states in between".
              By "event" I will mean: "point in spacetime".
  The format of the statements will be:
Index    Statement                                    Reason
  First we construct a Space. This space will be required in order to define a particle.
1    Construct S = C <-> C.                                A1, A2
1.1    S is 4 dimensional.                                1
1.2    Set the components of S = S1,2,3,4 in the following order: Re, Im, Re, Im.    1,     A2
  Cartesian product is a rotation of one C through 90 degrees and then superimposing the origins. This would cause the S3-axis to superimpose on the S1-axis if we turn the first C through 90 degrees. The reason that we could define this space is because of A1.
  We define a particle called Pp next.
2    S can transform into two Riemann Spheres.                    A1, 1
3    Construct two Riemann Spheres of S, call it RS <-> RS = Pp.            A1, 1
  We define a circle along the Imaginary axis of the second RS: S4.
4    Isolate a circle in the second RS namely S4 and call it PT.            A1, 3
4.1    I'm going to use physical terminology below.                Declaration
4.2    Construct "physical space" = SP = CxC/S4.                    A1, A2
  This gives physical space with Sp2 multiplied by i.
5    Let PT advance by one (rotate relative to S1,2,3 by one as measured along the circle) if encountering a space node and let the rotation be a quantum rotation. Call this "freq" = TS                                        A1, 4, 4.2, A2
  This rotation does not move infinity at the north pole of RS since: infinity - constant = infinity. This circle cannot have a charge of the particle Pp on it. Space fluctuates and stretches so this does not give a static PT.
7    Define "Change in freq" by TSf - TSi                        5
8    Let S1,2 be perpendicular to S3,4                            1
11    Construct {for all n = 1 to N: n(TSf - TSi)n} . Call this "Changes in freqs."        5,7
  Now we can define a basic time interval:
12    Define "basic time interval" = Delta tB = 1/[(1/N) \sum \limits_{n=1}^N n(TSf - TSi)n]                                            1-11, A3, A2
13    Construct MxTS, M element of Natural Numbers subset of C.            5, A4
  From these define "Basic time":
14    Define " Basic time" = tB = {1/[(1/M) (\sum \limits_{n=1}^M n#TSn)]}*Delta tB.    12, 5, 
A3
15    Couple tB to every point of SP and call the result "basic spacetime"= BST.    4.2, A2, A2
  Now we can make a similar construction in order to define Time:
15    Construct Si = C <-> C.                                A1
16    Construct RS <-> RS in Si, call it Pp.                            15.1, 2
17    Isolate a circle in Pp and call it PBT.                    A1, 16
18    Let PBT advance by one (rotate relative to Si1,2,3 by one measure along the curve of the circle) when encountering a BST event and let the rotation be a quantum rotation. Call this "freq2"= TBST.                                        7, A2
19    Construct KxTBST, K element of Natural Numbers, subspace of C.        18, A4
20    Define "Tim1" = t1 = 1/[(1/K)(\sum \limits_{n=1}^K n#TBSTn)].            A3, A2, 18
21    Pp is in every particle of the clock.                    Requirement
22    Tim1 advances like a clock, it depends on the Pp in the clock and on the route in BST.                                                18, 21
23    Tim1 = Time.                                    A2,  22
  In practice we only require that every particle of the clock has a circle with no charges on it that can serve as the particle clock.
  2. Defining Photons and Anti-photons.
  We go further to define photons. For this we need antiphotons as well. For this we need to define negative events of BST (the origin may then be constructed anywhere.)
23.1    Construct negative points of physical space as: SP- = (-C)x(-C)/Im {-C}        A1
23.2    Couple (-tB) to every point of SP-. Call the result BST-.                14, 23.1
23.3    Shift the origin of BST- in BST by an amount: min{ distance of two adjacent events of BST along any axis of BST}/2 and do the same for all four directions. Call the result CBST.    23.2
23.4    Define the events and negative events of CBST to have closest neighbours in a helix for any direction in CBST. This is not pictureable.                    23.3
24    Define a constant c = DSP/DtB                            4.2, A2
24.1    Let c be the maximum speed trough CBST i.e. the speed at which the particle sees no distance between succeeding events of CBST.                    4.2, 23.3
24.2    Construct S = C <-> C        .                        A1
25    From S, define a new RS <-> RS.                        24.2
29    Construct SAP = (-C) <-> (-C)                            A1
  This way the particle and antiparticle may look identical except for phase difference of 180 degrees (as if turned through 180 degrees).
30    Construct from SAP a RSAP <-> RSAP. Call it F1.                    29, 2
31    Let CBST  construct any vector in a RS <-> RS set =  F1, call it p. This is done by identifying four numbers in F1. Call such particle qFp1.                3, 18, 4.1
32    p is 4 dimensional                                31
33    Construct the same vector as in 31 x (-1) in F1. Call such particle qFp1.        31, 28
34    Identify a marker in F1's origin and at the origin in F1.                31, 33
35    Set Fp1 = qFp1 and leave out 2 distinguised events just below the unit circle crossing a curled up axis.     Call the two points A, B.                        24.2    
36     Set Fp1 = qFp1 and leave out 2 distinguised events just below the unit circle crossing a curled up axis.     Call the two points A, B.                        29
37    Let S1, S2 of Fp1 look like in Figure 1.1                        24.2, 35
 
Figure 1.1
  The little circles represent events of the circle that was left out. The figure shows an Fp1. The diamonds are positive events of CBST and the circles with dots in the centre are negative events of CBST, as the particle sees them. The little circles denotes passive events, this is accompliced by letting the Fp1 take four events of Fp1, now Fp1 would have four additions of events (see figure 1.2). The distance "d" is defined as a constant multiple of the interaction strength. The charges so generated (event exchanging) may be called: "passive mass" since it causes the photon to follow geodesics in spacetime. Passive mass reacts to curved spacetime but do not curve spacetime itself.
  In figure 1.1 CBST chose a momentum vector in the up direction, however it cannot go presicely in the up direction since this would require infinite momentum.
38    Let SAP1, SAP2 of Fp1look like in Figure 1.1, (just turned upside down and with events, negative events interchanged).                         29 -> 32.1    
39    Let the starting position (after one instance of time) of Fp1 and Fp1 be as drawn in figure 1.2 (only the curled up S1 and S2-direction shown).                29
 
Figure 1.2
  The figure shows a Fp1 and Fp1 with the Fp1 taking nodes from Fp1. We postuate that the Fp1 is made of negative nodes (S = (-C)x(-C)), so it carries the positive nodes (4 of them) from Fp1. It is easily seen that the two annihilate if becoming superimposed. They are defined to have momentum in opposite directions.
40    Let the two endpionts of c1 or A, c2 or B sense the closest two events of CBST in direction C and let them engage these events even if the whole Fp1 needs to turn or move linearly.                                        35
41    If four events were engaged: distinguish two new events and go to 40.    35
42    Let Fp1 move similarly to 40, just sensing nearest events of negative coordinates in the down direction.                                    35
43    Fp1 and Fp1 may be polarised: cicularly, transversely or longetudinally.    37
  43 is true since the point at infinity gives Fp1 an orientation in CBST.
45    Fp1 has spin 1.                                    44, 23.4
  This is true since Fp1 looks the same if turned through 360 degrees and because CBST is helical in any direction.
46    The events of CBST causes a force with nonzero component in the up direction. Define F = ma. With m = 0 we have infinite acceleration thus infinite speed. But infinite speed would saturate at c. Hence Fp1 goes upwards at the speed of light.                24.2, 37
47    That the movement of Fp1 causes Electro-Magnetic waves can be seen from the following figure. The F forces have a tiny reaction force in the up direction.    Figure 1.3 
47.1  To get a fuller wave we must have another Fp1 cooperating with this one such that "C" points in the up direction.                            Figure 1.3
47.2 To get a prependicular magnetic force we need to include events on the other circle as shown in Figure 1.1.                                Figure 1.1
47.3 The force F depends on the stiffness of spacetime.                Figure 1.3
 
Figure 1.3
48    Fp1 gets deflected if CBST is curved by gravity.                37
48.1    Let the other circle at C also have 4 events on it removed, so negative events remain. These events must be magnetic in nature.                    Figure 1.3
For this we need 4 types of events of CBST U {Magnetic field}.
49    Fp1 is a photon.                                43 -> 48
49.1    Fp1 is an antiphoton.                            43 -> 48
3. Defining a Pi-minus.
  Next we define a pi-minus:
50    Construct S = C <-> C                        A2, A1
51    Construct two Riemann Spheres from S, call it RS <-> RS  = G1    50, A2
52    Construct T = (-C) <-> (-C)                    A2, A1
53    Construct two Riemann Spheres from T, call it RS2 <-> RS2 = H1    A1, A2
54    Construct U = C <-> (-C)                        A2, A1
55    Construct two Riemann Spheres from U, call it RS3 <-> RS3 = g    54, A2
56    Construct a candidate for anti-ud. Call this I1. Let I1 be
    constructed from G1 <-> g <-> H1.                51, 53, 55
57    Let us label the circles in I1 as follows (left to right in 56):
    S1,2,3,4U1,2,3,4T1,2,3,4 in order Re, Im, Re, ...            56
58    Let the charges be added: Color charge: S1 and T1, Electric charge: S2 and T2, Mass: S4 and T4 in as balance with the left half, like in the following Figure:    57
 
Figure 1.4: I1.
  They are drawn like this but really the circles are all superimposed on each other so that one would see only two circles in three dimensions. The little stripes below the little circles and filled circles indicate they are active. Active events can influence events of spacetime external to the particle, passive eventss can only do that inside the particle. I1 must have 0 Weak Hypercharge.
59    Let the charges be ballanced by the antiparticle constructed as follows: right I1 is     constructed from copies of S, T, U.                51, 53, 55
60    Small circles are defined to be attracted to filled circles of the same charge type.                                        58, 59
61    A pi-minus has: electric charge = -1, mass = 139.570 MeV, decays into: electron and electron-antineutrino, interacts via: Strong, Weak, Electromagnetic, Gravity, has spin = 0 and parity = -1                            Pi-minus properties see: [1]
62    Define an I1 to decay to the particles in figure 1.5 and 1.6.Call the particle in figure 1.5 an I12 and the one in figure 1.6 an I13. We have that the strong force charge goes inactive in both particles, but they are still needed passively for keeping the particles together.
                                I1 decay definition.
 
Figure 1.5: I12.
  We have this decay to a left handed (I12) and right handed (I13) particle.
62.1    Define the particle's mas charge sphere to rotate twice for every revolution of the spin of I12.                                not bound together
 
Figure 1.6: I13.
   Mass charge devides in half. Space must give the I13 particle Right Handedness.
62.2    Define the sphere with mass charge to spin twice around for every total rotation of the particle.                                not bound together
63    I1 has charge -1.                        58
64    I1 has mass determinable with the Higgs field.    Define the mass charge
    by its ditance to sensed nodes and use the Higgs mechanism.    58
65    I1 decays to an electron and electron antineutrino.        62
66    I1 has Strong, Electromagnetic, Weak and Gravitational interactions        58
67    Spin 0 of I1 can be accomodated by defining the mass-charge to fill the entire Riemann sphere.
68    I1 has parity =  -1 since invering the axii puts infinity at the bottom.        58
  We must prove I12 is an electron before symmetry breaking:
  Decay from I1 to I12 can happen in two ways: rotate the I around the bottom point to produce left handed I12, or rotate around the topmost point (at infinity) to produce right handed I12.
70    I12 has weak, electromagnetic and gravitational interactions.            62
71    I12 has electric charge = -1.                            62
72    I12 is stable.                                    62
  This is since there is a gluon holding the particle together.
73    I12 has Weak Hypercharge = -1                            62
 74    I12 has spin 1/2                                     62.1
75    I12 is a left handed electron                            70 ->74
  We must prove I13 is an electron antineutrino:
76    I13 has spin- 1/2                                    62.2
77    I13 has charge = 0                                62
78    I13 has hypercharge = -1                                62
79    I13 is a right handed electron antineutrino                76 -> 78
80    I1 = left handed pi-minus (before symmetry breaking).        61,62 -> 68.1, 75, 79
81    I1 has Weak Hypercharge = 0, (Y = 2(Q - T3)) = 2(-1-(-1)) = 0         Figure 1.4
4.    Define protons.
  Protons can be defined from the above data for pi-minus using the charges for up and down quarks. Note that it is most natural to define the three quarks as superimposed on each other.
  It is now easy to define Hydrogen.
...    Define W and Z bosons
...    Define Gravitrons
 Comments:
  In trying to costruct photons by inserting a half circle on Pp one is led (because the half circle must come from a copy of space) to also contruct antiphotons and they are not made of anti-dimensions.
  After line 34 we have constructed a photon and an anti-photon and basic spacetime and time. We may postulate that EM comes from 3 dimensions of space x the 5'th dimension.
  We have that the theory of defining photons my be tested by proving: there is a direction in which photons with the same orientation will not go.
  We finally state that time defined by: "It is what a clock measure." has problems since a clock can be turned back or not tightly wound up i.e. clocks don't dictate time.
Bibliography:
[1] Kotz and Purcell. Chemistry and Chemical Reactivity. Saunders College Publishing, 1987
[2] www.sciforums.com. Username: NotEinstein.
[3] Nagashima Y, Elementary Particle Physics. Volume 1: Quantum Field Theory and Particles. Wiley-VCH Verlag GmbH & Co. KGaA. 2010.
[4] Hdjensofjfnen, Wikipedia, Internet: https://en.m.wikipedia.org/wiki/Pion. 2019.
[5] https://www.youtube.com/watch?v=Wmr4laNUeGc. Video name: Einstein's Relativity os WRONG Part 0.
 

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In order to prove your theory you should have been able to derive the geodesic equations and show the distinction between massive and massless particles in which geodesic equation they would follow.

I do not see any of this above. Nor do I see any applicable Langrangian equations

Edited by Mordred
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18 minutes ago, Willem F Esterhuyse said:

Anyway to answer the previous reply: 1) I can predict that photons see positive events of spacetime while anti-photons see negative events.

2) My model is still not solid on the "c" issue

3) The direction will depend on the experiment.

!

Moderator Note

These answers are completely inadequate.

Please show, exactly how (i.e. in mathematical detail) you predict that "that photons see positive events of spacetime while anti-photons see negative events." You might also want to explain what you mean by positive and negative "events".

Please show exactly how (i.e. in mathematical detail) "the direction will depend on the experiment."

If you are not able to provide proper answers to questions then the thread will be closed.

 
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Negative events are events lying on the negative axis, should some observer provide a zero point. The photons have events left out as shown in the picture. These left out points are located at 1-delta_1 - i*delta_2 and delta_1 - I*(1 - delta_2) as well as copies around the y-z plane. This is for the obvious coordinate system centered at the origin, with x pointing to the right. The left out events have by their nature the property of attracting events of spacetime, since left out events cancels out positive events. Then I think the stiffness of space will come into the formulas. The spacetime events engaged by the left out events create a force in the -x direction of size F = f (epsilon_0, d) - I must do dimensional analysis to determine this. I can state that negative events are equivalent to events left out of spacetime.

Negative events are similarly attracted by added positive events.

I need to draw pictures for the second question. It will take time.

I see now that the figures didn't copy, that is why I used the file attachment option.

Edited by Willem F Esterhuyse
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8 minutes ago, Willem F Esterhuyse said:

I see now that the figures didn't copy, that is why I used the file attachment option.

!

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You can attach the document for reference. You cannot just post a document with no explanation.

I see no mathematics in your answers. Just assertions. You need to do better.

 
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51 minutes ago, Willem F Esterhuyse said:

The program allows for uploading a file.

Anyway to answer the previous reply: 1) I can predict that photons see positive events of spacetime while anti-photons see negative events.

 

What do you mean by "positive" and "negative" events? "Negative events are events lying on the negative axis, should some observer provide a zero point." is not helpful

How do you tell the difference between a photon and an antiphoton, experimentally?

51 minutes ago, Willem F Esterhuyse said:

2) My model is still not solid on the "c" issue

That would be a problem, then, seeing as how this is a foundation of relativity.

 

51 minutes ago, Willem F Esterhuyse said:

3) The direction will depend on the experiment.

Evasiveness is not recommended here. If you aren't willing to present details and defend your idea, your stay will be quite short. Give some examples.

 

 

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34 minutes ago, swansont said:

What do you mean by "positive" and "negative" events?

I stated that negative events are events left out of the relevant spacetime. They have negative numbers associated with them. "Positive events" have positive numbers associated with them.

 

35 minutes ago, swansont said:

How do you tell the difference between a photon and an antiphoton, experimentally?

Inverse handedness.

36 minutes ago, swansont said:

Give some examples.

For example: an electron falling from a higher energy state at 12 o'clock in it's orbit would not emit a photon exactly in the "up" direction.

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1 hour ago, Willem F Esterhuyse said:

anti-photons

Layman question: In addition to the questions above, could you briefly state what current mainstream physics you reject? AFAIK there is no such thing as an anti-photon* in the standard model. Does that mean that the standard model does not apply when reading your paper? I've trouble to know what I should assume to be valid from current mainstream since you seem to introduce concepts that is not compatible with it.

 

*) There is no "anti-photon" with measurable differences from a photon

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22 minutes ago, Willem F Esterhuyse said:

I stated that negative events are events left out of the relevant spacetime. They have negative numbers associated with them. "Positive events" have positive numbers associated with them.

That has no meaning in the context of physics. Negative or positive numbers associated with them? That's ridiculous.

 

22 minutes ago, Willem F Esterhuyse said:

Inverse handedness.

That's not going to work. Photons and antiphotons are both spin 1 and can be either right-or left circularly polarized. I can change the polarization of a photon by passing it through the appropriate optics.

And what of linearly polarized photons?

 

22 minutes ago, Willem F Esterhuyse said:

For example: an electron falling from a higher energy state at 12 o'clock in it's orbit would not emit a photon exactly in the "up" direction.

First of all "12 o'clock" is an arbitrary direction, and more importantly, electrons do not have defined trajectories in atoms. There is no "orbit"

With " there is a direction in which photons with the same orientation will not go" you have not explained what you mean - photons with the same orientation as what?

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38 minutes ago, Willem F Esterhuyse said:

I stated that negative events are events left out of the relevant spacetime. They have negative numbers associated with them. "Positive events" have positive numbers associated with them.

!

Moderator Note

Last chance. Show, in mathematical detail, what "have numbers associated with them" means.

This sort of vague waffle is not good enough.

 
38 minutes ago, Willem F Esterhuyse said:

For example: an electron falling from a higher energy state at 12 o'clock in it's orbit would not emit a photon exactly in the "up" direction.

!

Moderator Note

Show in mathematical detail how this is derived.

 
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No, no, no …
Light travels in a 'straight' line.
Simply because the definition of 'straight' is given in terms of a light path.

By that definition, a geodesic has to always be straight.

( yes, I'm being facetious )

Edited by MigL
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21 hours ago, Mordred said:

In order to prove your theory you should have been able to derive the geodesic equations and show the distinction between massive and massless particles in which geodesic equation they would follow

I don't need to derive geodesic equations: I can simply state that the anti-photon engages the next two negative spacetime events thereby possibly changing its direction of motion.

19 hours ago, Strange said:

Last chance. Show, in mathematical detail, what "have numbers associated with them" means.

  We associate numbers to negative events simply by stating: CB("Name of Coordinate System", -ngx_1 - mgx_2 - lgx_3 - kgx_4), where x_i element of unit vectors on R^4, n, m, l, k element of N, g element of R.

19 hours ago, Strange said:

Show in mathematical detail how this is derived.

  From Figure 1.5 we see that an electron may emit a photon or an anti-photon (the two electric charge circles). To create a photon from an electron we need: an operator called a copier, a charge copier and a charge transport operator.


  We work in the reference frame of the Centre of Mass of the electron. The copier must copy the circle of S_2 and the Riemann Sphere of S_2 itself. Then the carge copier must make 2 copies of the charges on S_2, then the charge transporter must move two of the copies as follows: move the copied charge from (1 - idelta) to (-delta - i) and the one from (-1 - idelta) to (delta -i). Then the other copy of the right sided charge must move to (1 - i(delta + lambda)) and on the left to (- 1 - i(delta - lambda), with lambda chosen by the observer so that the line between the left and right charges is prependicular to the momentum direction. Since lambda >  0 we have the momentum not precisely in the "up" direction.

19 hours ago, swansont said:

With " there is a direction in which photons with the same orientation will not go" you have not explained what you mean - photons with the same orientation as what?

The same orientation as a reference photon.

2 hours ago, Markus Hanke said:

There is no such thing as anti-photons.

They are seen as phase shifted photons.

19 hours ago, swansont said:

Negative or positive numbers associated with them? That's ridiculous.

Space is made  of numbers.

20 hours ago, Ghideon said:

Layman question: In addition to the questions above, could you briefly state what current mainstream physics you reject? AFAIK there is no such thing as an anti-photon* in the standard model. Does that mean that the standard model does not apply when reading your paper? I've trouble to know what I should assume to be valid from current mainstream since you seem to introduce concepts that is not compatible with it.

I reject the strong force, I think we can do without it. See anti-photons as phase shifted photons going back in time. What is not compatible with the standard model?

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3 minutes ago, Willem F Esterhuyse said:

I don't need to derive geodesic equations: I can simply state that the anti-photon engages the next two negative spacetime events thereby possibly changing its direction of motion.

!

Moderator Note

You had your chance to bring some rigour and science to this discussion. Instead all you have is baseless assertions.

Clearly this "theory" is not ready for serious discussion. The thread is closed. Do not bring this subject up again. 

 
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