Generally Covariant Generalization of the Dirac equation that does not require gauges

Postulate 1 . z=zz (algebraic definition of 1)

 Dirac in 1928 made his equation(1) flat space(2). But space is not in general flat, there are forces. So over the past 100 years people have had to try to make up for that mistake by adding adhoc convoluted gauge force after gauge force until theoretical physics became a mass of confusion, a train wreck, a junk pile. So all they can do for ever and ever is to rearrange that junk pile with zero actual progress in fundamental theoretical physics being made,.. forever. By the way note that Newpde (3)

is NOT flat space (4) so it cures this problem(5).


(4)  Here  koo=1-rH/r =1/krr,  r=(2e2)(1040N) /(mc2). The N=..-1,0,1,.. applies to the Nth Fiegenbaum point self-similar fractal scale. Newpde from Occam’s razor optimized postulate1:                                                                                  
 z=zz is the algebraic definition of 1 and adding (small) constant C is trivial in  z=zz+C, δC=0 (1)
A  Substitute  z=1+δz into equation 1 and get  2D Dirac equations for e,v (sect.1.1).
B  Substitute left side z into right side zz repeatedly in equation 1 and get the 2D Mandelbrot set.
A,B together gives 4D Newpde.  So postulate 1->Newpde

Click on pink titles below for backup physics:

  Postulate 1 


 Part I     1U1 states

Part II    1U1U1 states 

Part III   Mixed State

 Rest       Miscellaneous


Start with an: Occam’s razor optimized (i.e., δC=0, ||C||=noise)
       POSTULATE OF 1                   So                                                                                   
        z=zz (1) is the algebraic definition of 1,o. So add constant C (i.e., z’=z’z’-C, δC=0) (2))    Solve eqs.1,2 for z’.       (Use this z’=1+δz to find δz=psi in the Newpde.) So just postulate1               

This is an Occam’s razor optimized (δC=0, z=zz not z=zzzzz) postulate of 1. You aren’t going to do any better than “optimize” because of real noiseC 

  Solve eq.1,2 for z’ with iteration and successive approximation.

Iteration Plug z’=1+δz into eq.2 get                                δz+δzδz=C      (3) The eq.3 quadratic eq. solution  for realC<<-¼ is   δz=dr+idt   (4)                                               imbedding real line C in the complex plane. Substitute z’ on right (eq.2) into left z’z’ repeatedly& get zN+1=zNzN+C. δC=0 requires us to reject the Cs for which δC=δ(zN+1-zNzN)=δ(∞-∞) NOT 0. Initial  z=zo=0 is then Mandelbrot set CM (fig2).  Eq1 solution is 1,soinitial  z=zo=0 gets the Mandelbrot set CM (fig1) out to some ||C|| distance from C=0.  C is found from δC=(dCM/d(||C||))d||C||=0 extremum giving the Fiegenbaum point ||CM|| =   ||-1.40115..|| global max given this  ||CM|| is largest in fig.1.  The intersection of real C<-¼ line with CM is at the (dense CM) Fiegenbaum pt ‘providing at least tiny real line C=CM displacements’ dr’dt’=|D|2<<1). Successive approximation z=1+dz=0 so δz=-1. Thus from eq.3 δ(δz+δzδz)=0= ddz(1)+2(ddz)(dz)=δδ(1)+2(δδz(-1+D)= (δδz(-1+2D)= ½δ(δzδz)+δδzD~½δ(dzdz)=0=  Therefore plug eq.4 in here: δ(dz+dzdz)~½δ(dzdz)=½δ[dr2-dt2+i(drdt+dtdr)]=0  (5) 

 real and Imaginary->Dirac equation->spin½ lepton                                         gi are the Dirac matrices here. The positive scalar drdt in eq.7, (fig.3, quadrant I) implies the eq.5 non infinite extremum imaginary drdt+dtdr=0= gidrgjdt+ gjdtgidr= (gigj+gjgi)drdt so Clifford algebra (gigj+gjgi)=0, i≠j. Also factor eq.5 Minkowski real  δ(dr2-dt2)=  δ[(dr+dt)(dr-dt)] =0=[[δ(dr+dt)](dr-dt)]+[(dr+dt)[δ(dr-dt)]]=0 get

(+-e)    dr+dt=ds, dr-dt=ds  =ds1     for    +ds-> I, IV quadrants       (7)   (light cone v) dr+dt=ds, dr=-dt,   for   +ds->   II quadrant             (8)       “        “             dr-dt=ds,  dr=dt,      “         “    III quadrant            (9)   (->vacuum)       dr=dt,       dr=-dt      so dt=0=dr (So eigenvalues of dt, dr=0 in eq.11) (10) 

We square eq.7  ds12=(dr+dt)(dr+dt) =[dr2+dt2] +(drdt+dtdr) =ds2+ds3=ds12. Since ds3 (is max or min) and ds12 (from eq.6) are invariant then so is Circle ds2=dr2+dt2 =ds12-ds3. Note this separate ds is a minimum at 45deg and so Circle=dz=dseiA=dsei(dA+Ao)= dsei((cosAdr+sinAdt)/(ds)+Ao),  Ao=45deg. We define k=dr/ds,  w=dt/ds, sinA=r, cosA=t. dsei45deg=ds’. So take the ordinary dr derivative (since flat space) of ‘Circle’  so , ). Cancel that ei45deg coefficient then multiply both sides of eq.11 by h and define dz=psi, p=hk. then implies (Hermitian) operator hobservables formalism(QM,  also appendix B3). Eqs 7-11 and eq.6 Clifford algebra imply 2D Dirac equations for e,v.     z=1 small C’ in eq.2 must Lorentz gamma boost Reldz to small scales Explicitly D=C’≈Reldz‘>> Rel(dz’dz’) in eq.6. Thus strongly nonuniform Reldz’=CM/m=CM/m1=C’=rH defining mass m1 and charge CM. So for Ao=45deg then perturbed uniform C eq.7 reads:      (dr-dz’)+(dt+dz’) =dr’+dt’=ds           (12)                                          since ds invariant. Define krr=(dr/dr’)2=(dr/(dr-(CM/m1)))2=1/(1-rH/r)=A1/(1-rH/r) +A2/(1-rH/r)2  r=dr,N=-1. The partial fractions AI term can be split off from RN and so krr=1/[1-((CM/m1)r))] (13)     So we have: ds2=krrdr’+koodt’+..(14).   From eq.6 dr’dt’=sqrtkrrdr’sqrtkoodt’=drdt so  krr=1/koo  (15) So given this 2D perturbation we get curved space 4D. For 4D our eq.3 Clifford algebra then implies (gxsqrtkxxdx+gysqrtkyydy+gzsqrtkzzdz+gtsqrtkttidt)2=kxxdx2+kyydy2+kzzdz2kttdt2=ds2. Multiplying the bracketed term by 1/ds & δz=psi so eq 11 implies 4D Newpde   gi(sqrtkii)dpsi/dxi=(w/c)psi  for e,v                                       (16) (covariant derivative still ordinary since psi (complex) scalar)Therefore postulate1>Newpde  


 koo=1-rH/r=1/krrrH=e2X1040N/m. N(=. -1,0,1.) is the CM fractal scale. So    postulate 1->Newpde is the solution to the above 3 term math problem.

For spherical symmetry curved space kookrr:    (gxsqrt(kxx)dx+gysqrt(kyy)dy+gzsqrt(kzz)dz+gtsqrt(ktt)idt)2= kxxdx2+kyydy2+kzzdz2kttdt2=ds2
In contrast kxx=kyy=kzz=ktt=1 is flat space, Minkowski,  in the ordinary 1928 Dirac equation. The difference between the ordinary Dirac equation and the Newpde is that we didn’t just drop that kmn. There actually are forces after all!

To make up for this mistake we have as result for a 100 years been adding gauge force after gauge force creating a adhoc convoluted mess, a train wreck, a junk pile which we are stuck in forever.


Figure 1 solves the puzzle of physics and real#math (just postulate 1) and figure 5 explains why. 

(Ockam’s razor motivated) Postulate 1 <->Physics   :

Equation 1 z=zz+C, δC=0 gets the New pde and physics (left side of fig.1)and yet the z=zz algebraic definition of 1 is also the small C limit of equation 1 (middle of fig.1). So eq.1 hints strongly that the (Ockam’s razor motivated) Postulate 1->Physics   is correct (right side of fig.1). fig.1
See sect.4 of Postulate 1 file above implies real#math. fig.2
Uniqueness: Need at least some measurable “observables” (Hermitian operator fig.3 formalism eq.6) and Cauchy sequence of rational numbers to derive real#(eigenvalues) math as well. So New pde is the ONLY result of postulate 1.


Mandelbrot Set Plug in the left side (of eq.1) z into the right side zz repeatedly and use  δC=0 and get the Mandelbrot set iteration formula.                                                                                               The Mandelbrot set Cis (and from the postulated  δCM=0)zN+1=zNzN+CM (since  δ(z’-zz)=δ(zN+1-zNzN)= δ(∞-∞)=0). (the z=zz) z=0=zo. Fiegenbaum point CM smallest real line Mandelbulb on next smaller (baseline) scale. Mandelbulb areas (drdt) for smallest Clifford algebra extremum drdt.


Boost  m1=tau+mu+me=3e from S=½ to L=1 at stable r=rH, 2P3/2 with m1/2=mreduced mass. See above sect.2 of above postulate 1 file for some more applications. See PartI, partII, PartIII for the rest.      

Comparison and contrast with mainstream concerns:

The origin of the many postulates of quantum mechanics is considered to be a mystery by one and all. 

Not really:  1) Quantum mechanics arises from a circle.(see below)

 The connection between general relativity and quantum mechanics is stated over and over again to be the greatest unsolved problem in physics.

Not really:  2) General Relativity(GR) comes out of quantum mechanics in 1 step.(see below)

A century has been devoted to the origin of high energy particle physics by thousands of people using billions of dollars worth of particle accelerators. It is supposed to be quarks, gluons, qcd SU(3) gauge forces, require many free parameters,..and so on and so forth.

Not really:  3) (high energy 3e composite) Particle physics is just a ortho and para state (see below)

The latest version “Standard electroweak Model” is now encumbered by pages of Lagrangian density terms, and so literally hundreds of postulates.

Not really: 4) Rotate e,v to get the Standard electroweak Model (it’s merely composite e,v). (see below)

Expansion of the above answers: 

(1)Plug z=1+δz into  z=zz+C, δC=0 (eq.1) and get δ(δz+δzδz)=0 which splits into  Im Clifford Algebra and Real Minkowski metric components which imply that circle ds2=dr2+dt2 so δz=dseiA. Write in angle ‘A’ radian measure dependence on r,t, take the r partial derivative and rearrange and you get a Hermitian operator and so “observables” formalism and then the Dirac equation of quantum mechanics (given that Clifford algebra). See sect.1 of above postulate 1 file.

(2)  Note square roots of koo=1-rH/r=1/krr in the new pde (so same kij math functionality as in the Schwarzschild metric gij but coming from QM instead.).  So squaring (actually iterating) the New pde on the next higher(cosmological)  fractal scale, where rH=2GM/c2, gets the Schwarzchild metric gij of GR, a one step derivation. General relativity comes out of quantum mechanics in one simple step!

(3) The two positrons in 2P3/2 on the r=rH shell are moving rapidly, the central electron is not moving so much at the center, and this New pde 2P3/2 state is stable because of h/e flux quantization(PartII) and r=rH motion. So a Clebsch Gordon two body analysis with Paschen Back effect excitation, in a straightforward way, leads to a ortho(s,c,b) and para(t) states with ground state(u/d), the proton. Use Frobenius series solutions (Ch.9) individually for each of the s,c,b,t associated Nucleonic doublets (eg.,P,N for u/d) to get the particle multiplets. See PartII also.

4) The 4 axis (New pde) e,v rotations (giving consecutively the 4 Bosons: Zo,W+,W-,photon) at r=rH on the r,t plane are implied by eq.7 (also see appendix A). We even get Maxwell’s and Proca’s equations in these iterative rotations through the New pde e,v solutions which means taking a first derivative at each 45deg. rotation.  

Conclusion: Ultimate Okcam’s razor postulate 1 implies math AND physics. We finally figured it out.