User:Tomdehoop/sandbox

The Comprehensive Action Principle

Compliance to the CAP of any correct analysis is a necessary Requirement to REALLY UNDERSTAND our REALITY with EASY "LINEAR" so-called "mathematical" tools!
First read a short description of the CAP by Paul A.M. Dirac: http://www.quantumuniverse.eu/Tom/GR_CHAPTER30.pdf

Compliance to the CAP is required in any mathematical analysis of our investigated Scientific Research!

The CAP is equivalent to including the gravitational action in any physical model.

During his later life Albert Einstein severely protested the not-understood Copenhagen interpretation of Quantum Mechanics because to him it turned "theoretical" physics into a game of dice. God doesn't play dice he once commented.

When re-writing QM in compliance with the CAP, the basic building-blocks called "Elementary Particles", together with all their required characteristics, can be derived and understood completely. In this way Einstein's frustrations concerning QM can be overcome.

The graviton is an invisible massless elementary particle. Besides the spin1 photon which represents the ElectroMagnetic field as a field build from uncountable many elementary force-particles called bosons, the graviton is the only massless elementary particle. Only the spin of the spin2 graviton is different from the spin of the spin1 photon.

However, the value of the spin of an elementary particle is very important for its mathematical characteristics.

To understand spin, please take the time to read: http://www.quantumuniverse.eu/Tom/intrinsic_spin.pdf

The spin of an elementary particle implies a carried symmetry of this particle: After rotating the wave-function of a spin s (multiplied by ħ, Dirac's constant) particle over an angle of

Δφ = 2π/s                                (1)

radians, its wave-function repeats itself. This is why the wave-functions of the spin1 photon and weak nuclear-forces can be analyzed on easy mathematical grounds. The spin½ elementary Leptons require a more thorough investigation of their characteristics: Anti-neutrino's only occur with right-handed chirality, while neutrino's are only observed with left-handed chirality. So only half of their possible eigen-states appear to be possible in the case of elementary spin½ particles.

In the case of the also elementary spin2 graviton a doubling of degrees of freedom is present as a result of symmetry (1). This explains on mathematical grounds why gravitation curves 4D-spacetime. All orthogonal axes of 4D-spacetime now cannot be straight 1D-lines anymore, but must be described mathematically with curved lines in mathematical described 2D-planes orhtogonal to the direction of the coordinate. And this spin2 curvature-effect should be included mathematically as a spin2 "dual" effect, i.e. in 2 mathematical orthogonal ways!

The description of the paths of the planets of our solar-system around the sun was for the first time described by https://en.wikipedia.org/wiki/Karl_Schwarzschild. He started with a description with point-planets and a point-sun and used polar-coordinates to yield the most easy to solve equations of motion using https://en.wikipedia.org/wiki/General_relativity. Later his description was improved with included sizes and rotations of the sun and the planets. I.e. this was a "single" mathematical 4D-spacetime analysis. This description is a macroscopic mathematical analysis. The second "dual" effect of gravitational curvature seems to be a mathematical "microscopical" effect: An up-to-this very day still not analyzed QM-effect:

Describe all possible elementary particles as harmonic oscillating waves in the 2D-plane orthogonal to the analyzed direction of motion. Any http://www.quantumuniverse.eu/Tom/What%20are%20so-called%20elementary%20particles.pdf can always be analyzed in a description with a non-zero speed of motion. And massless particles can't even be analyzed without speed other than the speed-of-light.

The QM CAP-complying harmonic oscillating extensiveness of all elementary particles can be described from the inertial-frame moving with origin at the average position of the described particle. Let's choose the direction of motion in the positive z-direction. From the inertial-frame we now have:

z = z' = 0                                (2)

The time measured from the origin of the chosen inertial-frame just is the eigen-time τ of the particle in any SR QM description. Therefore, use this easy imaginable time for the time-coordinate. Polar-coordinates yield the most easily to solve D(ifferential) E(quations).

We now have to solve DE describing ideal harmonic oscillating motion in the 2D-plane orthogonal to the z-axis specified by (2), i.e. we have to find solutions for:

(icτ, ρ, φ, z)                                (3)

And because this description of QM "very-small" space, the mathematical analysis can always neglect "macroscopic" curvature as an higher order effect. Even just outside the Schwarzschild-radius. This yields easy to solve mathematical, i.e. linear, SR analysis of the proven solvable DE

The Differential Equations of the Harmonic Oscillating point
Use symmetrical polar-coordinates (3), i.e. with xμ = xμ identical, this at-once shows why time and space are mathematical orthogonal.

The DE describe relativistic harmonic oscillation in the 2D-plane orthogonal to direction of motion of the described elementary particle. From the used inertial-frame the z-axis DE are independent and given by (2).

For the massive harmonic oscillator we have in SR:

Set I:

pμ = m0dxμ/dτ = (E, cp)                       (4)

Fμ = dpμ/dτ = m0dvμ/dτ                   (5)

pμFμ = 0                                                       (6)

Written with the SR notation with τ the propertime. Bold symbols are 3 dimensional space vectors and m0 is the rest mass of the particle. Not 1 second-order time derivative, but 2 consecutive first-order time derivatives.

Equation (6) gives the required harmonic demand for the massive particle.

Fμ is the 4-force acting on the particle.

In the following, let’s use Natural Units for simplicity.

Suppose the worldline of an inertial frame is given by:

yμ = y0μ +(τ - τ0)tμ			  (7)

Here tμ is a constant time-like 4-velocity and τ is the proper time. The y0μ is a constant 4-vector giving the position on the worldline at τ = τ0. And as time τ passes, the origin of the inertial frame moves in Minkowski space as given in (7). Then a spacelike 4-vector rμ, which represents the particle’s instantaneous center of energy with respect to the inertial frame defined by (7), is given by:

rμ = (xμ - y0μ) – (xν - y0ν)tνtμ                          (8)

The conservative force acting on the particle as measured by an observer following the worldline of the inertial frame is a space-like 4-vector fμ, i.e.:	fμtμ = 0.

The work done by the force is:

W = - fμpμ/(pνtν)						(9)

The force of (5) may now be split into 2 components. One along the 4-velocity tμ and another projected onto the spacelike plane orthogonal to tμ:

Fμ = (pνtνfμ - fνpν tμ)/m0                 (10)

The equations of motion (4), (5) and (6) may then be written as set (II):

Set II:

r’μ = pμ/(pνtν) - tμ 									  (10)

p’μ - (p’νtν)tμ = fμ									  (11)

p’μpμ = 0										  (12)

The accentuation mark over a letter denotes differentiation with respect to the proper time τ in the inertial frame defined by (7). These equations are also valid for massless particles, only the variable τ can’t be interpreted as proper time anymore. A massless particle is always observed as moving with the speed of light. The description of the extended character of this massless CAP-compliant particle from the inertial frame with origin moving with the particle on its average position may have the speed of light at its closest position related to the origin. At all other positions on its path the speed will always be less than the speed of light. Therefore, in general the massless extended wavelike particle is observed as moving with the speed of light, but in the description of this wave the mathematical point in general moves with speeds smaller than the speed of light.

Remark: The SR description of massless particles results in dτ → 0. In fact the variable τ now is a parameter which describes set II, but can’t be interpreted as the particle’s proper time anymore! The 4-force Fμ used in (5) is exchanged for the spacelike 4-force fμ in (11) to end up with harmonic oscillating motion. This projection of a 4-vector in 2 sets of 2, is exactly the same as the usual explanation of polarization vectors of the photon in QED (see [3], 5.1). Here the timelike component of the wave-vector is combined with the spacelike component in the direction of motion of the photon to describe the instantaneous Coulomb interaction in a covariant description. The force fμ in (11) actually is 2 dimensional, in the plane orthogonal to the particle’s worldline, as follows from (12). The timelike force-component is orthogonal to this plane, just like the observed direction of motion, so it doesn’t appear in set (II).

The spacelike force fμ depends on the distance of the center of energy of the point-particle with respect to the origin of the inertial frame that moves with the particle on its average position, i.e. its worldline described by (8): fμ = fμ(rν)	  (13)

The first force expression allowing ideal harmonic oscillation is given by:

F = -kρ                      (14)

This spacelike-force will always be directed towards the origin of the used inertial-frame.

The center of energy 4-vector rμ depends on the location of the inertial frame and on the location of the particle in Minkowski space. Therefore the assumption in (13) implies that the inertial frame, with the worldline given by (7) is not arbitrary but constitutes a source producing the force acting on the particle. This reasoning led Andreas Bette, to the model with more particles than 1 particle, where the other particle(s) are the source of the force. However, here I use this description for just 1 mathematical point-particle, and choose the force such that the particle’s motion is harmonic. In this way the extended character of all elementary particles is described using an exact mathematical point-description. The space-like force (11) is the source of the particle’s harmonic motion and this is also the source of the energy of a massless particle proportional to its frequency. In the case of the photon it just describes its EM-field. So, the photon is itself the non-reducible mathematical representation of all characteristics of the spin1 EM-field.

In the inertial frame of the source we have:

rμ = (0, r)  Λ   fμ(rν) = (0, f(r))								   (15)

Here both r and f(r) are in the 2D-plane orthogonal to the z-axis.

Set (III) now reads:

r’ = p/E    Λ     p’ = f(r)      Λ     p.p’ = 0	           (16)

We also have:	pμpμ = E2 – p.p = m02									  (17)

If the force f(r) is conservative we have:	f = -&#8711;U(r)							   (18)

Where U is the potential energy of the particle: U = ½kx.x.

From (16) up to (18) one obtains:	E’ = -(r’.&#8711;U(r))					(19)

This implies conservation of total energy:	H = E(p) + U(r)							  (20)

With a central-force one may write:	U(r) = U(|r|) = U(r)         (21)

From (16), (20) and (21) one then obtains (as in the non-relativistic case) that the orbital angular-momentum from the inertial-frame moving with the particle on its average position is a constant of a CAP-compliant explicit described elementary particle:

S = r x p = (H – U(r)) r x r’ = constant	        (22)

This constant is the well-known constant of Planck h (B7) multiplied by the particles (half-)integer spin s of elementary (fermions or) bosons.

We are searching the extended character of the bosonic massless particle here! This is performed describing the particle as an oscillating point-like particle ‘observed’ from its average position given by its worldline, where the inertial frame moves with the particle on its worldline with the speed of light. This harmonic movement is the source of the intrinsic angular momentum (spin) which is parallel with the particle’s worldline, i.e. it’s momentum as observed by an observer. I.e. the angular momentum (22), following from set (II), represents the spin (helicity) of the particle which is in the same direction as the moving average position of the particle given by its worldline. So we observe the massless particle from the inertial frame with origin at the average position of the particle given by its worldline. In this frame the point-particle oscillates harmonic, so the massless particle is not at rest with respect to the inertial frame! However the origin of the inertial frame is moving with the average position of the massless particle, or its position as seen as non-extended point-particle. In this way it is possible to choose a descriptive frame moving with the massless particle with origin at its average position. It took me quite some time to realize this is possible!

The removal of the derivatives of velocities with respect to proper time and the appearance of the constant of Planck, are in fact related to the 2D harmonic oscillator, where the average sweep of the oscillator has to be set equal to the Planck length. The reason is that choosing an inertial frame moving with the particle as it is observed as point-particle is never done in QM.

In local field theory (i.e. no curvature of space-time) a fermion is specified by its 4-momentum pμ and a boson by its wave 4-vector kμ, even if the boson has mass. All fermions must have mass. Even the neutrinos must have small masses because this is the only way to characterize the neutrinos of the 3 different families. And due to the fact that the neutrino masses are small and therefore the differences between the masses are even smaller it may be possible that there are neutrino flips between the neutrinos of the 3 families. Bosons of the strong- and weak- nuclear forces have masses. See appendix (E) for the proportionality between a particle’s energy and frequency. Only the force particles of the symmetrical spin 2 part (gravitational field) and the anti-symmetrical spin 1 part (EM-field) have no mass. So in fact only for the graviton and the photon set (II) should be used. All fermions can be dealt with using set (I). The equations of motion for the massive elementary particles will be analyzed later on, but in any case for all massive particles set (I) can be used as a starting point. The fact that all bosons are described using the wave-vector kμ and all fermions with the energy-momentum 4-vector pμ follows from the fact that bosons have to be described using closed B(oundary) C(onditions) and fermions have to be described using open BC. The extra, positive integer, degree of freedom in the case of open BC explains the appearance of more families of particles, which only differ in values of their rest-masses. The different statistic character of fermions and bosons is also related to their BC.

This idea has some similarities with Superstring theory, but as I see this universe now, there are just time and 3D space as orthogonal coordinates, and all elementary particles possess a 2D extendedness orthogonal to their worldline. And as a result of that required extended character the amount of uncertainty relations does not increase, but all well-known QM uncertainty relations are explained. The 2 dimensional freedom orthogonal to the average path, described by the worldline, of all particles can be seen as Albert Einstein’s assumed hidden variables to explain the statistical characteristics of QM. Where it should be understood that all possible Bose force-fields in Q(uantum)F(ield)T(heories) are fields build from uncountable amounts of identical (both elementary and compound "gluon") force-particles, which as a result of this must possess mathematical uncertainty in their actions.

The QM uncertainty relations require the harmonic oscillators to be of the order of the Planck length. This fact shall be shown to be valid in the following derivation.

In short:	The momentum p (16), i.e. set (III), is the point-particle’s momentum as observed in the frame moving with the extended particle on its average position. From this inertial frame which is not accessible by anyone, the oscillating motion results in an intrinsic angular momentum (helicity) in the direction of the particle’s worldline. I.e. this represents the particle’s spin and the usual ad-hoc added QM spin is described here explicitly. I.e. spin follows automatically looking at the particle in a way which complies to the "always required" CAP.

The velocity of the massless particle has the velocity of light, this is the speed of the point giving the center of energy, i.e. it's the actual position observed from the inertial frame moving with the particle and with origin on the observed position on the worldline. When the potential energy (21) is the highest, also curvature is the highest. Even though the speed always is the speed of light, curvature varies harmonic, just like the potential energy (21). I.e. when curvature is the smallest, the motion is more linear and this results in higher kinetic energy at smaller distances from the worldline. Microscopic, i.e. QM, curvature can be analyzed completely by logical analysis.

In case of a boson the measured momentum pμ is proportional with the wave-vector: pμ = h kμ, in which h is Planck’s constant and kμ is the wave-vector. The oscillating motion in the 2D-plane is orthogonal to the direction of motion. Therefore kμ follows from the solution of the 2D oscillating motion. If one observes a photon, one now has an extended character described as an harmonic oscillating point in the 2D-plane orthogonal to the observed direction of motion. The frequency of this motion, just is the frequency of any detected quantity of this photon. From this point of view the EM-field carried by the photon is a direct result of the required 2D extendedness of this elementary particle to comply to Einstein's CAP. And the only possible 2 allowed polarization modes are in the plane orthogonal to the direction of motion of the photon.

For the 2D-spacelike force of the 2D-oscillators we can re-write (18): f(x) = -&#8711;U(x). Where x = (x, y) ↔ (ρ, φ) now is a 2D-vector orthogonal to the worldline coordinate chosen as the z-axis. And U(x) is the potential energy of the oscillator which has to be central, as given in (21): U(x) = U(|x|) = U(ρ) = ½kρ2, with k a suitable chosen force-constant and with the used polar coordinates (ρ, φ, z = 0) described from the chosen inertial-frame.

The usage of the constants of Planck now is to give values to the constants of motion of the harmonic oscillating point particle. Here it should be realized that in the used description of (I) and (II) one only has to insert the velocity of light c in correct powers to end up with correct dimensionality.

I.e. the only added c’s appear in τ → cτ, p → cp and S → cS, with c = |c| the norm of the constant light-speed. The constant of Planck h = 2πħ does not appear anymore, but is needed as a proportionality constant giving values to the constant total energy H (22) and "intrinsic" angular momentum S (22).

The equations of motion set (III) can now be re-written by set (VI), which describes the oscillator in the easiest way:

Set VI:

dx/dτ = p/E							     (23)

dp/dτ = f(|x|)							(24)

f(|x|).p = 0			                               (25)

The used parameter τ is the time measured from inertial frame (7), moving with the harmonic oscillating point representing the center of energy of the observed particle. Only the space-like vectors are analyzed now. In this way the source of the uncertainty relations of QM is described, which is necessary in set (III) because this set describes the massless harmonical movement of the particle which is one of the "dual" sources of the QM uncertainty relations. So the SR 4 dimensional space-time equations of motion reduce to simple 2D-spacelike equations, which are 2 consecutive first-order time derivatives. This 2D extendedness follows from the CAP and even though the solution is simple, it’s the only way to explain QM.

The 4D-momentum-energy equation is:			E2 – p2 = m02 = constant					  (26)

For massless particles we therefore have:              E = |p| = p						   (27)

From (20) we have conservation of energy. Here one should realize that this is the so-called "intrinsic" energy of a massless particle. For any observer this particle always moves with the velocity of light. This is the measured speed of the particle along its average position give by its worldline. The harmonic oscillating motion is orthogonal to the particle’s worldline. Therefore this motion is Poincaré invariant. Only the particle’s detected frequency depends on the chosen inertial frame. Due to this invariant character the total energy in (20) has to be put equal to the detected kinetic energy: H = hf (B1) = ħω (B7). The frequency of the harmonic oscillating particle in the inertial frame moving with the average position of the point-like particle has to be equal to the detected frequency of any observed quantity of the particle. For example the harmonic oscillating motion of a photon is the source of it’s produced EM-field. Any elementary particle only has 1 frequency for all its investigated quantities, which result from its CAP enforced harmonic oscillating motion.

Differentiating the first equation of (16) using the second equation, (19) and (20) yields:

(H – U(ρ))x” + &#8711;U(ρ) – x’(x’&#8711;U(ρ)) = 0				  (28)

Here, x” is the second derivative with respect to the ‘proper time’ and x’ is the first derivative with respect to the ‘proper time’ of the oscillating massless particle as observed from the inertial frame moving with the particle on its worldline.

This is the non-linear equation of motion of the "intrinsic"-QM 2 dimensional harmonic oscillator. From (16), (20) and (21) we conclude that the angular momentum given in (22) is a constant. This angular momentum S just is the intrinsic angular momentum in QM, i.e. the helicity or spin in the direction of the particle’s motion. However, now mathematically written out explicitly to make the description comply to the CAP.

Let’s continue to follow the massless particle on its inertial worldline.

As seen from the inertial frame moving with the particle, which is chosen as the z-axis, S is directed in the z-axis. Now choose cylindrical coordinates (ρ, φ), also see (21) with ρ = |r|, on the spatial plane perpendicular to the z-axis.

The equations of motion now read with S = |S|:

d2ρ/dτ2- (dU/dρ)((dρ/dτ)2 – 1)/(H – U(ρ)) – S2/{( H – U(ρ))2ρ3} = 0           (29)

dρ/dτ = S/{(H – U(ρ))ρ2}   (30)

dz/dτ = z = 0                         (31)

Using (16) we have for the speed of the massless particle in NU: | x’| = 1    (32)

In cylindrical coordinates this yields (in NU):	(dρ/dτ)2 + ρ2(dφ/dτ)2 = 1        (33)

Inserting (33) in (30) yields:	 (dρ/dτ)2 = 1 – S2/{(H – U(ρ))2ρ2}  (34)

Integrating gives the time observed from inertial frame (7) of the harmonic oscillator:

τ(ρ) = ±ʃdρ[(1 – S2/{(H – U(ρ))2ρ2})]-½      (35)

Here the potential energy of the oscillator is: U(ρ) = ½kρ2                (36)

And the polar angle:	φ(ρ) = ±Sʃdρ[(H – U(ρ))ρ2(1 – S2/{(H – U(ρ))2ρ2})]-1	    (37)

Both (35) and (37), the equations of motion of the harmonic oscillator, are completely integrable.

In solving (35), use the following variables:

A = 2H/k       [m2]                         (81)

B = (3L/k)2  [m6]                (82)

Integral (35) can now be written as:

τ(x=ρ2) = ±1½ʃ{(A-x)/(9x3-18Ax2+9A2x-4B)½}dx                      (83)

Until further notice x is the squared polar distance.

Only the squared polar distance can be solved, because the polynomial of ρ is of higher degree than 4. This integral is solved using t(x15).mws with Maple 10, also see appendix (D). The solution contains a third-root of the following form:

(6B+2((3B)2-3B A3)-A3)⅓ = (2(3B+((3B)2-3B A3))-A3)⅓ = A(2(b+(b2-b))-1)⅓  f A, with: b  (3B/A3)   (84)

(1-f2) (x) (x) (x) This results in a solution: 	(x) = i —————————— A2{2(f2-f+1)EF- (3(f2+1)+i3(f2-1))EE}		  (85) 62 f2 (x(x-A)2-(⅔)2B)

With fraction f given in (84) and EF and EE incomplete elliptic integrals of the first (EF) and second (EE) kind. All arguments of all used elliptic integrals are identical.

z	    dt				          z   (1-2t2) I.e.	EF = F(z, ) = ———————    and 	EE = E(z, ) = —————dt				   (86) 0 (1-t2)(1-2t2)			        0     (1-t2) With arguments:

z = {½(1+i(6fx/A+f2-4f+1)/(3(1-f2)))} 								  (87)  = {2/(1+i3(1+f2)/(1-f2))}									  (88)

Giving the (x), (x) and (x) functions specifies (x) given in (85) completely:

i (x)(x) = {(i3+(6fx/A+f2-4f+1)/(f2-1))(i3(1+f2)/(1-f2)-(6fx/A+f2-4f+1)/(1-f2))}			  (89) (x) {(6fx/A-2f2-2-4f)/(-3(1+f2)+i3(1-f2))}							  (90) In short: {(3(f2-1)2+(6fx/A+f2-4f+1)2)(2(1+f)2-6fx/A)} (x) = A —————————————————————*{2(f2-f+1)F(z, )-(3(f2+1)+i3(f2-1))E(z, )}	  (91) (6f)2{(3(1+f2)+i3(f2-1))((⅔)2B/A3-x(1-x/A)2/A)}

This solution is only real if fraction f = 0, i.a.w. when the particle just is a point-particle moving on its SR worldline. I.e. the 2D extendedness of particles in the plane orthogonal to the direction of motion of the particles, given with the SR wordline results in a complex solution with all characteristics of the Hilbert-space used in QM. So, it explains why all solutions of QM have to use Hilbert-space, i.e. it answers problem 6 of the well-known Hilbert’s problems. The well known Golden Ratio (Phi = ½(5+1) or phi = ½(5-1) = 1/Phi) appears to show very beautiful symmetries when one chooses: f = Phi = ½(5+1)										  (92)

The definition f  (2(b+(b2-b))-1)⅓ now results in (b+(b2-b)) = Phi+1  b = Phi				  (93) From the definition in (84) we see that the real b > 1  f > 1 First I assumed f = phi, but further analysis showed f > 1, which resulted in (92). At this moment it is not yet proven that (92) is valid. But if every elementary particle is extended in the 2 dimensional plane orthogonal to the observed direction of motion, as follows from Einstein’s Comprehensive Action Principle. And the mathematical solution uses Phi, then it is a fundamental proportionality constant in the mathematical description of all elementary particles of our universe and as a result also shows up in many observed relations.

Using (92), the following result of (x=2) follows:

{(-(½3-½15)i-(3(5+1)x/A-1½5+½))((2½3+½15)i+(3(5+1)x/A-1½5+½))(3(5+1)x/A-1½5-3½)} (x) =A ———————————————————————————————————————————— * (185+54) {((⅔)2B/A3- x(1-x/A)2/A)½(15+35+i(3+15))}

*{4F(z, )-½(35+15+i3(5+1))E(z, )}		  (94)

If (92) is valid, both arguments of both incomplete elliptic integrals are such that the integrals exist:


 * z| < 1		|| = (½) < 1								   (95)

From (77) the extreme values of  follow, also see appendix G:

extreme(f,) = (A/3)(1/(fei½)+fei½), with:		 { 0, ⅔, 1⅓}					  (96) With: 	max(f,= 0) = (A/3)(1/f+f)									  (97)

And the real min must be extracted from the complex conjugated solutions of  {⅔, 1⅓}. The max and min are not assumed to depend on Phi, i.e. (92) is not used in this solution. Only when both max and min are given we have enough equations to be able to express force-constant k as function of physical constants like Planck’s constant. From (96) (also see (G6)) one is able to derive the minimum value of :

min = |½{extreme(f, ⅔)+extreme(f,1⅓)}| = ½|extreme(f, ⅔)+extreme(f,1⅓)| = ½max 			  (98)

First the solution of equation (80) will be given. This solution also is in the complex plane. In fact the SR solution of the extendedness of particles in this non-reducible way results in a solution in Hilbert space of QM. This time only a 3rd kind incomplete elliptic integral appears in the solution. But it now appears that all elliptic integrals are needed. The arguments have complex contributions, but the norms are such that solutions exist. The used Maple10 Worksheet Angle(rho)03.mws is given in (D.2). This solution has to be multiplied with (L/k): (2L/k)				 EP(z, , ) (x=2)= ————————— ’ (x) ’ (x) ’ (x) ———————, with:				  (99) {x(x-2H/k)2-(2L/k)2}	   	           (4f-1-f2-i3(1-f2))

{-6f x/A+2(f2+1)+4f} ’(x)= {6f x/A+f2+1-4f-i3(1-f2)}  ’(x)= —————————  ’(x)= {6f x/A+f2+1-4f+i3(1-f2)}	 (100) {-3(f2+1)+i3(1-f2)}

Here (’(x)’(x)) , but f > 1, so ’. And EP(z, , ) is the incomplete elliptic integral of the third kind. Its arguments are:

1				1	z = {½+i(4f-f2-1-6f x/A)/(23(f2-1))}   = ——————————   = ——————————	 (101) (½+i(4f-f2-1)/{23(f2-1)}	         {½+i½3(1+f2)/(1-f2)}

The arguments z and  are the same for all used elliptic integrals and only z depends on x, just as in the case of (x). Explicit: z		dt EP(z, , ) = EllipticPi(z, , ) = —————————— 						 (102) 0  (1-t2)(1-t2)(1-2t2) So, EP(z, =0, ) = F(z, ) given in (86). On taking f = Phi again:

(2L/k)					          	 {6(5+1)(x/A-1) -8} (x=2)= —————————{6(9+35)(x/A)2-6(7+5)(x/A)+16}—————————— * {x(x-2H/k)2-(2L/k)2}					{15+35+i(15+3)} 23		 1		    2	*EP(z ={½+½i(4-5-6x/A)/3}, = ——————,  = —————) * ————————		  (103) (3+i(4-5))	       (½-½i15)	      (35-1+i(15+3))

This result is such that it shows many characteristics of being real. This still has to be checked. Dimensional consistency only requires inclusion of c in   c and L  cL, which follows from p  cp. The constant of Planck does not appear anymore because we describe the extended character of massless bosons, which is the source of this QM-constant! Planck’s constant appears when giving values to the constant angular momentum L and detected energy H. From (72) and (73):  > 0  the particle can never be on its average position given by its worldline.

In the chosen inertial frame we now require () to be fixed to the Planck length: 2<> = lh = (hG/c3) = O(10-35) [m], see (B4). As will be showed later this is only possible using a proportionality constant. All solutions are integral solutions, so one always needs 2 values x1,2 and the results of (103) must be subtracted. The maximum polar distance (97) yields for the angle :  L		1							 1 (x=max2) = ———*——————————————{-6x/A+4+25}{(6x/A-4+5)2+3}————— * k(A3) (2+5){x/A(1-x/A)2-2(5+1)/27}				        {35-i3} {-6x/A+4+25}	         35+i3          35+i3 * EP(z = ———————, = ————,  = ————)					 (104) {35+i3}	          4+25	     35-i3

This expression shows that this angle is, as required, real. Argument z is the upper border of the elliptic integral and contains the complex conjugate factor 1/{35+i3} of the only complex argument in the multiplier to end up with  so the expression is real at this extreme of . Right now I assume the expression to be real for all possible polar distances, because otherwise my conclusions about how this universe can be described mathematically are wrong. Dimensional analysis shows that the argument of (x) is in radians.

Solutions (91) and (99) are only given to be able to determine force-constant k. This constant k must be chosen such that the average <()> is fixed to the Planck length: 2<()> = (,s)lh	 (105) Here (,s) is a constant which depends on the total intrinsic energy H = h and intrinsic angular momentum L = sh of the massless particle. It is easily extracted from max and min:

<> = ½(max + min)		(max(k) + min(k)) = (,s)lh					 (106)

From (97), (98) and (106) we have: {3H/(2k)}(1/f+f) = (,s)(Gh/c3). This yields:

3	           ((,s))2G	           3	        1   	            c3	— (2 +  — + f) = ————     k() = — (2 + — + f) ———— 					  (107) 2k	f	     c3		           2	        f	     G((,s))2 If f = Phi, one can use (1/f + f) = 5 This frequency dependency of the force-constant k = k() results in fundamental constant extendedness of all massless bosons proportional to the frequency. Only constant (, s) still has to be inserted. If in (107) the Golden Ratio (92) is used, we also have (93): b = ½(5+1), equal to the Golden Ratio. Fraction f(b) given in (84) can be inverted: 	b(f) = ¼(2+f3+1/f3) 						 (108)

Choosing f = Phi = ½(5+1) we have using the Fibonacci-series F(n):

f n = F(n-1)f+F(n-2), n > 1  f -n = (-1)n+1{F(n-1)f-F(n)},n > 0					 (109)

Equation (107) gives:

(sc)2 k			   2	            3 (3/2)3—— — = ¼(2+f3+1/f3)  k = —(2+f3+1/f3) —— h [kg rad2/s2]					 (110) 3   h			  33	          (sc)2

From (84) we have: 33 (s c h)2 	            22(5+1)h3 b = 3B/A3 = ————— k  k = ————— [kg rad2/s2]						 (111) 23 (h)3		 33 (s c)2

From (106), using (97) and (98), we have: 3 			G	 3c3		 1 —— (2+f+1/f)=((,s))2 —  k = ——(2+f+1/f) ———   [kg/s2]					  (112) 4 k			c3	 4G	          ((,s))2

Using (110) we get an expression for (,s):

34(2+f+1/f)   c5   s2	((,s))2 = —————  —  —	[1/rad2]								  (113) 22(2+f3+1/f3) hG 2

Using (109) the linearity constant (,s) between extendedness and Planck-length (105) is: s       32	  s	(, s) = (3/2)2Phi —   = —(5+1) —, with Planck-time: th = (Gh/c5)				  (114) th    23             th

And the force-constant k becomes: (⅔)3	k(,s) = Phi ——h [rad2kg/s2]									 (115) (sc)2 All proportionality constants depend on both the intrinsic energy H, given by frequency  and intrinsic angular momentum L specified by spin s. So equations (110) and (112) which follow from the independent demands (105) and (108) proof that taking fraction f = Phi is the only possible solution. The often observed value Phi = ½(5+1), i.e. the Golden Ratio, appears to be a characteristic of all elementary particles, which as interacting many particles collections describe all we can observe. This explains why this characteristic Golden Ratio is observed so often in our universe. Inserting (114) into (105) shows the following extendedness dependency of spin s and angular frequency :

2<> = (3/2)2Phi(sc)/ [m/rad]									 (116)