Sierpinski (<1K) The Dirac Particle

    "This is the kind of idea that can ruin a young scientists career. It appears to be too importanrt to keep quiet about. But if you try and explain it to most phsysicists, they are likely to dismiss you as some kind of crankpot."     --     David Hestenes, 1986

Introduction
    In this chapter we will establish the 5D Dirac-Hestenes equation Ñpyp   =   (m - qap*)yp     with 5D pseudovector potential ap* º ape12345-1 and yp a <0;3;4>-grade scaled idempotent Â4,1 multivector field for a particle of mass m and charge q.
    There are ten 3-blades in any extended basis for Â4,1 , five plussquare ones involving e4 and five minsusquare "spacial" ones. There are five 4-blades four of which are minussquare. Thus <0;3;4> pensities are formed from six plussquare blades (including the unit scalar) and nine minussquare blades.
    However, by replacing any 3-blade by i=i times its dual we can consider <0;3;4>-grade Â4,1 multivectors as residing in even Clifford algera C4,1 +.

     Â4,1 + contains ten plussquare even blades , or eleven if you include the unit scalar ; and ten minusquare even blades. This is suggestive of the the ten or eleven "dimensions" required by "String Theory". [  Symbol Check: ½ denotes the value half. denotes exponential conjugation. y denotes greek psi. ]

The 4D Dirac Equation
    To encompass a four-velocity in a particle's pensity we must move to a <0;3;4>-pensity, but we will approach this indirectly in a traditional QM manner via the Dirac equation. The traditional ket representor y for a ½-spin particle or Dirac spinor is a 4×4 complex matrix having zeroes everywhere but in the first column, effectively a 4-D complex 1-vector.

The Free Dirac Equation
    Dirac derives the Free Dirac equation applicable in the absence of an electrostatic field via a somewhat obsfucated factorisation of the Â1,3 Free Klein Gordan Equation . The Free Klein Gordan equation in a Â1,3 timespace is ( (/t)2 - Ñ2 + a2)y = 0 where Ñ2 denotes the usual Euclidean 3D Laplacian . We must regard this as Ñ2yp = -a2yp in Â1,3 and as Ñ2yp = a2yp in Â3,1.

    Ñp2 yp = -a2 yp factorises as (iÑp+a)(iÑp-a)yp  =  0       where real scalar a = 2pmch-1 = mch-1 = m in natural units .
    Subequation (iÑp - a)yp = 0 Û  (2p)-1he4(iÑp - a)yp = 0 Û  ( me4 - e14 me1 - e24 me2 + e34 me3 - e4mc)yp = 0 , where mei º hÐei = (2p)-1hi Ðei is the momentum operator traditionally denoted by pi. Dirac wrirtes this as ( me4 -a1 me1 -a2 me2 -a3 me3 -b)y= 0     [ PQM 67.7 ]     , seeking 4x4 complex matrices a1 ,a2 ,a3 ,b which anticommute and have ai2=1 , b2 = mc = m . Matrix representors of e01, e02, e03, and e0 (mc)½ clearly fulfill these requirements given an appropriate geometric i and h º h(2p)-1i = hi .

The Dirac Equation
    Dirac incorporates a 1-potential field ap by following the "classical rule" of replacing me0 with me0 - qc-1 V and mej by mej - qc-1 Aj for jÎ{1,2,3} where V and A are the scalar and 3D 1-vector electrodynamic potentials at p.  The relativistic analogue of this requires replacing md by md - qc-1 d-1¿ap for 1-vector "four-potential" ap.
    The Dirac equation for a particle of mass m and charge q in a electrostatic field is thus, in Dirac's original terminology (but retaining m for momentum)
    ( me0-qV -a1( me1-qA1) -a2( me2-qA2) -a3( me3-qA3) -b )yp = 0     [ PQM 67.11 ] ,

    A more common form of the C4×4 Dirac equation is obtained by setting g0 = (mc)-1b ; gj = -(mc)-1baj   for j=1,2,3 (which also anticommute); and exploiting mek º ih Ðxk = ih (/xk) where h=h(2p)-1 . It is
    åk=03 ( ih gk(/xk) - gkqc-1ak ) y = mcy
where the gk are four p-independant anticommuting 4x4 complex Dirac matrices satisfying g02=1 ; gj2=-1 for j=1,2,3 ; a0=V, aj=-Aj are the real scalar coordinates of a Â1,3 1-vector "four-potential"   ; c is the speed of light ; h is the Planck constant; and i=Ö-1 is an "imaginary" scalar that commutes with everything including the gj.

    Any solution to Ñpy = ±ay is a solution to Ñp2y = a2y and an e4-static origin-centred solution is given by yp = |^e4(p)|-1 b(± a|^e4(p)|) [  Where a º ea denotes exponentiation ] . The + solution is unbounded and so rejected, leaving Yukawa potential
    yp = |P|-1 b (-a|P|)   =   |P|-1 b (-a|P|) , where p = P + p4e4. A rapidly damped function of e4-spacial distance from 0.

Dirac Algebra

   
    There are various equivalent "suites" of Dirac matrices (aka. gamma matrices) in the QM literature. We adopt a coventional set here (see eg. Hannabuss) and construct them from the 2×2 Pauli matrices as
g0= æ 10 ö ; gj= æ 0 sj ö
è 0-1 ø è - sj0 ø
though RgjR-1 provides another set for any invertible 4×4 matrix R.     We list our gj explicitly for reference.
g0= æ 1000 ö    ; g1= æ 0001 ö    ; g2= æ 000-i ö ; g3= æ 0010 ö    
ç 0100 ÷ ç 0010 ÷ ç 00i0 ÷ ç 000-1 ÷
ç 00-10 ÷ ç 0-100 ÷ ç 0i00 ÷ ç -1000 ÷
è 000-1 ø è -1000 ø è -i000 ø è 0100 ø
We observe that they all have zero trace corresponding geometrically to having zero scalar part. Because we are here interested in Â3,1 spacetime rather than Â1,3 timespace we define matrices f j º igj j=0,1,2,3 . We will also change the "timelike" suffix from 0 to 4 via f4ºf0 , x4=x0.
    For distinct i,j,k Î {1,2,3} the Pauli construction gives
g_jk º gjgk= æ - sj sk0 ö =-eijki æ si0 ö     ;     gijk =eijki æ 0-1 ö     ;     g0123 =i æ 0-1 ö     ;     g5º ig0123 = æ 01 ö
è 0- sj sk ø è 0 si ø è 10 ø è -10 ø è 10 ø
    Note that the primary impotent (aka. primitive idempotent) 4×4 unit corner matrix having 1 as the top left element and zeroes elsewhere is given by
    u  º  ½(1+ig12)½(1+g0)   =   ½(1+g0)½(1+ig12)   =   ¼(1+g0 +ig12 +ig012)
    We can construct matrices having 1 in the second, third, and fourth elements of the first column (and zeroes elsewhere) as ig23u = -g23g12u = g13u  =  -e13u , g3u  =  e1245u, and -g1u  =  -e2345u respectively. Ones in only the second, third, and fourth entries of the top row are given by the matrix-transpose (Hermitian multivector conjugation) of these, specifically iug23 = ug13, ug3, and ug1 respectively.

Dirac Conjugation as Clifford Conjugation

    Dirac conjugation (aka. Dirac adjoint) of a 4×4 matrix y is traditionally defined by y»  º  g0 yT^ g0-1   =   g0 yT^ g0 corresponding to transpose, conjugation, and negation of the off-lead-diagonal blocks [ Lounesto 10.3 ] . For a ket left-column matrix Dirac spinor y, the leftmost g0 is superflous.
    Dirac conjugation preserves 1 and the gj, negates double and triple gj products, and preserves g0123 and so corresponds to Â1,3 reversion §.
    With regard to the f k, it preserves 1, negates the f k and double products, preseverving triple and quadruple products and so corresponds to Clifford conjugation §#. Of course, §#=§ when restricted to even multivectors.

    We can thus, albeit somewhat informally, regard a Dirac spinor ket matrix y as a "product" ypu where yp is an even Â3,1 multivector which we must "recast" into a 4×4 complex matrix via ej ® f j prior to matrix multiplication by 4×4 primitive idempotent u . The corresponding Dirac bra matrix y» is given by uyp§# . We cannot instead cast u into a Â3,1 multivector because we do not have a Â3,1 equivalent for i1.

The Dirac-Hestenes Equation
    We will adopt the natural Â3,1 analog of an Â1,3 approach due to Hestenes. First note that we can express any 4×4 complex matrix as a complex-weighted sum of matrix products of the gj .
[ Proof : Left multiplication by g0 negates the bottom two rows (ie. the bottom two 2×2 "subblocks") while right multiplication by g0 negates the left two columns (ie. the left two subblocks). This enables us to "zero" all but a given 2×2 "corner" of any 4×4 matrix via appropriate left and right multiplications by ½(1±g0) . We can thus construct a matrix having any sj or 1 in either off-lead-diagnonal block, or any i sj or i1 in either lead-diagnonal block, and zeroes in all other blocks. Since complex-weighted summations of the Pauli matrices can generate any 2×2 complex matrix, we can generate y as a complex-weighted sum of the gj  .]
    Because y has nonzeroes only in the leftmost column we can safely right-multiply by the unit corner matrix to obtain
    y =yu =y½(1+ig12)½(1+g0)
so it is logical to factorise y as y'u where 4×4 complex matrix y' may contain nonzero elements outside the leftmost column but is nonetheless expressible as a complex-weighted sum of gj products.

    The ½(1+g0) acts as a source or sink of g0's which commute across the ½(1+ig12) and so we can ensure that y' is a complex-weighted sum of only even gj products.
    Define f j º igj
    Since i(1+ig12) = -g12(1+ig12) , we can replace i weightings in y' by rightmultiplications by -g12=f12 ; thus y' can safely be considered a real-weighted sum of even gj products and hence also a real-weighted sum of even f j products. The Dirac equation thus becomes
     åk=03 [ih(2p)-1 gkÐxk(y'u) - gkqc-1aky'u] = mcy'u
    Adding a unit -i2 factor to the ak term and then converting just one of the i into the right multiplication of y' by -g12=f12 we obtain
     åk=03 [ih(2p)-1 gkÐxk(y')u + igkqc-1aky'f12u] = mcy'u   .     Replacing igk with f k and changing timelike suffix from 0 to 4 we obtain
     åk=14 [h(2p)-1 f kÐxk(y') + f kqc-1aky'f12]u = mcy'u .     Exploiting u=g0u=-if4u=-f4f12u  we have
     åk=14 [h(2p)-1 f kÐxk(y') + f kqc-1aky'f12]u = -mcy'f412u .
    Thus a sufficient (though perhaps not necessary, since u is noninvertible) condition for y to solve the Dirac equation is for real-weighted combination of even f j matrix products y' to solve åk=14 [h(2p)-1 f kÐxk(y') + f kqc-1aky'f12] = -mcy'f412    since the u will absorb the f412

    We can interpret 4×4 complex Hermitian matrices f j as an orthonormal frame of Â3,1 1-vectors. We further observe that as any frame f j will provide a set of Dirac matrices, we are free to interpret f j as ej where ek are the Â3,1 frame with regard to which our xk coordinates are constructed (p=x=åk=14 xkek) without regard to signatures.
    Taking c=1 in accordance with natural units (but retaining h) we can thus write the Dirac equation using the Â3,1 1-gradient operator Ñp as
    h(2p)-1 Ñpyp + qapype12 = -mype412     where yp is y' considered as a Â3,1 + multivector. Since e12=e12; and e4=-e4) we can rearrange as
    Ñpyp = 2ph-1 (-qapyp + mype4)e12
    [ Had we favoured Â1,3 we would have obtained  åk=03 [-h(2p)-1 gkÐxk(y')g12 - gkqc-1aky']u = mcy'g0u
    Þ Ñpyp = (2p)h-1 (qapyp + mypg0)g12 which is known as the Â1,3 Dirac-Hestenes equation. ]

    This result differs in sign of ap from Lounesto [13.4] who asserts Ñpyp = 2ph-1 (qapyp + mype4)e12 for the Â3,1 + Dirac equation.

    Assuming natural units, our final Â3,1+ idempotent-stripped Hestenes-Dirac equation is thus
    Ñpyp = (mype4 - qapyp)e12     with yp Î Â3,1 + . Particles having no mass or charge (such as, perhaps, neutrinos and photons)   thus solve Ñpyp = 0 even in the presence of an electromagnetic field.

    We will assume that our u factor contains "all the idempotency" of y so that Â3,1+ multivector yp is invertible and accordingly expressible canonically as yp = (rpebpe1234)½Rp where rp is a positive scalar , bp is a scalar , unit rotor Rp = (½l(ae1234)c) (ie. an even multivector with RpRp§=1) for unit plussquare bivector c, and e1234 is the Â3,1 unit pseudoscalar (preserved by both » and § and so not taking the QM i role) . There is also the particular case Rp=±(1+a) for null bivector a.
    Scalar bp is known as the Yvon-Takabayasi angle bp with bp=0 giving the Dirac particle having magnitide rp , and bp=p giving the Dirac antiparticle with yp»yp = uR§p§rpe1234»e1234Rpu =-rpu.

4D Dirac-Hestenes Solutions


Particle as spinning frame field
    We examine some geometric kinematic consequences of Ñpyp   =   mype124 - qapype12 when yp§ is grade preserving (bp=0 or p) and preserves orthogonality (a conseqeunce of Rp§Rp=1) . We then have
    Ñp yp§(e1)   =   2(myp§(e24) + qapyp§(e2))    ;     Ñp yp§(e2)   =   -2(myp§(e14) + qapyp§(e1))
    Ñp yp§(e3)   =   -2qap¿yp§(e123)   =   2q(apÙyp§(e4))*     provided i=e1234 commutes with yp§.
    Ñp yp§(e4)   =   -2(myp§(e12) + qap¿yp§(e124))   =   -2myp§(e12) + 2q(apÙyp§(e3))* .
[ Proof : Because we are in Â3,1 our kinemetic rule relates only to scalar parts yielding
    Ñp*ypbyp§   =   2((Ñpyp)byp§)   =   2((mype124-qapype12)byp§)<0>   =   -2q(apype12byp§)<0>   =   2qap¿yp§(b¿e12) .
    Thus Ñp ¿ yp§(e3) = Ñp ¿ yp§(e4) = 0 while Ñp ¿ yp§(e1) = 2qap¿yp(e2) and Ñp ¿ yp§(e2) = -2qap¿yp(e1)
       ÑpÙ(ypbyp§)   =   ((Ñpyp)byp§  - ypb(Ñpyp)§ )<2>
      =   m(ype124byp§  - ypbe124§yp§)<2> - q(apype12byp§  - ypbe12§yp§ap§ )<2>
      =   myp§(e124b+be124)<2> - q(apype12byp§  + ypbe12yp§ap )<2>   =   2myp§(b¿e124) - q(apype12byp§  + ypbe12yp§ap )<2>
    Ñp Ùyp§(e1) = 2(myp§(e24) + qapÙyp§(e2))    ;     Ñp Ùyp§(e2) = 2(-myp§(e14) - qapÙyp§(e1))
    Ñp Ùyp§(e3) = -2q(ap~yp§(e123))<2> = -2qap¿yp§(e123)    ;     Ñp Ùyp§(e4) = -2myp§(e12) -2qap¿yp§(e124)  .]

    Thus when ap=0, spacelike spin current wp=yp§(e3) satisfies Ñpwp = 0 suggesting that it is is more fundamental than timelike probability current jp º yp§(e4) which has Ñp jp = -2mSp where 2-blade Sp º yp§(e12) = -2mrp(bpe1234)Sp~ with unit 2-blade Sp~=R§p(e12) [  Ñpjp = -4ph-1mSp in unnatural units ]
    Ñp Sp   =   -(Ñp mrp)(mrp)-1 - 4mjp + 4qaprp so when ap=0 and rp=r we have Ñp2 jp = 8m2jp (ie. (25/2ph-1m)2 jp ) The "Laplacian eigenvalue" 25/2ph-1m is collosal, and we must consider jp as oscillating extremely rapidly but with a small amplitude about a "time-averaged" timelike four-momentum.
[ Proof : Ñp yp§(e12)   =   Ñp rp-1 yp§(e1) yp§(e2)   =   (Ñp rp-1)rp yp§(e12) + rp-1(Ñpyp§(e1))yp§(e2) - rp-1(Ñpyp§(e2))yp§(e1)
      =   -(Ñp rp) rp-1 yp§(e12) + 2rp-1( (myp§(e24) + qapyp§(e2))yp§(e2) - (-myp§(e14) - qapyp§(e1))yp§(e1))
      =   -(Ñp rp)rp-1 yp§(e12) + 2( -myp§(e4) + qapyp§(1) -(myp§(e4) - qapyp§(1)) )
      =   -(Ñp mrp)(mrp)-1 - 4myp§(e4) + 4qaprp  .]

    jp=yp§(e4) naturally decomposes as ½yp§(e4+e3) + ½yp§(e4-e3) as the sum of two independantly conserved opposite spin forward null currents.
    wp=yp§(e3) naturally decomposes as ½yp§(e4+e3) + ½yp§(-e4+e3) as the sum of two independantly conserved same spin oppositely charged (ie. forward and backward) null currents.

    Suppose ap=0 and we are given only a null field sp º yp§(e3+e4). We can recover mrp via Ñpsp = -2mrpSp~ for unit minussquare 2-blade Sp~.
    Ñp2sp = -(Ñp mrp)(mrp)-1 + 4myp§(e4) then provides mjp = mrpvp and hence mrpwp = mjp¿Sp~* where * denotes dual in e1234.
    But sp = rp(vp+wp) so we can isolate rp from m as rp = -vp¿sp = wp¿sp. It is then natural to regard probability rp as the "scaling" or "strength" of a nullvector and mass m as a constant frame-independant scalar measure of the curvature of the null trajectory.
    Similarly, if we are given jp we can reconstruct Sp and mrp and hence mrpwp from Ñpjp and derive rp as (-jp2)½ .
    If we are given wp = yp§(e3) however, Ñpwp=0 and we must resort to wp2 = rp2 and Ðwp~wp .

    If we are given nullvector field spº yp§(e3+e4) then
    Ñpsp   =   -2myp§(e14+e12)   =   2myp§((e3+e4)e1)   =   2mrp-1spyp§(e1)
    Ñp2sp     =   2m(Ñprp-1)spyp§(e1) + 4m2rp-2spyp§(e1)2 - 4m2rp-1yp§(e34)sp
      =   -2m(Ñprp)rp-2spyp§(e1) + 4m2sp - 4m2yp§(e34(e3+e4))   =   2m(Ñprp)rp-2spyp§(e1) + 8m2sp
      =   2mrp-2(Ñprp)¿(spÙyp§(e1)) + 8m2sp     since sp is orthogonal to yp§(e1) and Ñp2sp must be a 1-vector.

    Thus if rp=r , null 1-vector field sp=yp§(e3+e4) satisfies a Klein Gordan equation Ñp2 sp = (23/2m)2 sp .

Simple solutions
    Solutions to the Dirac equation tend to be oscillatory and multidimensional oscillations can be hard to undertand and secribe. We desrcibe a few here for completeness but the essential point is that they model the classical charged particle when "averaged" over nontiny time durations. The picture is one of wildly accelerating motions constrained into apparant smoothness. What appears to be a continuos timelike trajectory is actually the time-averaged value of and extremely tightly wound lightspeed trajectory, for example.

    We first observe that if yp solves Ñpyp = imyp for a fixed blade i commuting with yp, then so to does rpyp provided either  Ñprp = imrp or Ñprp = 0. The latter can always be atained by means of an e4-invarient spherical harmonic solution rp = YlmP)

    The simplest solution to Ñpyp = mype124 is the planewave yp=(m(p¿e4)e12) y0'     for some arbitary fixed y0' that commutes with e124. This solves Ñp2yp = -m2yp since e122 = -1 [  From Ñx2e(x*a)b = -a<1>2 e(p+x*a)b . ] and has constant magnitude over cotemporal plane p¿e4 = t. It can be viewed as representing a "particle" of exact "momentum" me4 and maximally arbitary (effectively nonexistant) "position" . This is more physically acceptable than solutions such as ((p¿e1)e24)y0' because e242=1 makes such such solutions hyperbolic unbounded with distance whereas ((p¿e4)e12) is trigononometric and bounded with time.

    For a more general solution consider first one factorising as yp=bpqpW2)   =   bpWqpe12)W§ for scalar field qp, even multivector field bp= (rp(bpe1234))½Rp with RpRp§ = 1 , and p-independant unit minussquare bivector W2 = W§(e12) for arbitary p-independant unit rotor W. Assume rp¹0 and define
    vp º R§p(e4)     ;     Sp º R§p(e12)    ;     Wp º R§p(W2) .
    We have  Ñpyp   =   Ñp(bp)(½qpW2) + ½Ñp(qp)ypW2 .
[ Proof : Decompose bp=bp ++bp - where bp + commutes with W2 while bp - anticommutes with W2 . Then
    Ñp(bpe½qpW2)   =   Ñp(bp)e½qpW2 + Ñp(e½qpW2)bp + + Ñp(eqpW2)bp -   =   Ñp(bp)e½qpW2 + Ñpqp)epqp)W2bp + - Ñpqp)epqp)W2bp -
    = Ñp(bp)e½qpW2 + Ñpqp)bp +epqp)W2 - Ñpqp)bp -eqpp)W2
      =   Ñp(bp)e½qpW2 + Ñpqp)bp +e½pqpW2bp + + Ñpqp)bp -qpW2)W2   =   Ñp(bp)(½qpW2) + Ñpqp)bpqpW2)W2  .]

    When bp=b is constant , or more generally when Ñpbp = 0,  we have Ñpyp   =   ½Ñp(qp)ypW2 and so right-multiplying the Dirac-Hestenes equation Ñpyp e21 = 2ph-1 (mype4 - qapyp) by yp§ gives Ñp(qp)ypW2e21yp§   =   4ph-1 (mrpvp - qaprp(bpe1234))     Þ     Ñp(qp) WpSp   =   4ph-1 (qap - mvp(-bpe1234)) .

    When W2=e12 this reduces to   (Ñpqp)   =   -4ph-1 (qap - mvp(-bpe1234)) which implies bp=0 or p and we have
½(Ñpqp) + 2ph-1 qap   =   ±  2ph-1mvp In natural units with 2ph-1=1 we have ½(Ñpqp) + qap   =   ± mvp . Squaring gives Hamilton-Jacobi equationÑpqp +qap)2 = m2 , while applying ÑpÙ gives   qÑpÙap = ± mÑpÙvp . Since vp¿(Ñpvp)= vp¿(ÑpÙvp) = v/t  where t is the natural parameterisation of the vp streamline we have m v/t = ± q vp¿(ÑpÙap) = -/+ q vp¿fp     for pure bivector field fp = -ÑpÙap ,  the standard trajectory equation for a spinless charged particle.

    Thus jp=rpvp looks like a conserved current of spinless charges following Maxwell's equations , but jp alone does not fully describe the flow

    If bp=0 and ap=0 and qp = wm~¿p so that Ñpqp = wm~ for p-independant nonzero w and unit timelike m~ we have   ½wm~   =   mvp Þ vp=m~, mw for a DeBroglie plane wave solution yp = rm~(m~+e4)~ ((m¿p)e12) for arbitary constant real rp=r .

    Now consider solutions factorising as yp = (½qpW1)bp . We can decompose bp=bp ++bp - via bp ± º ½(bp ± We12§(W§bp)) such that W§bp + commutes with e12 while W§bp - anticommutes with e12. If Ñpbp ± = 0 (a stronger condition than Ñpbp=0) then Ñpyp = ½(Ñpqp)W1yp and right-multiplying the Dirac-Hestenes equation by yp§ this time gives (Ñpqp)W1Sp   =   4ph-1 (qap - mvp(-bpe1234)) .
[ Proof : yp ± = Wqpe12)W§bp ± = bp ±(±½qpe12) Þ Ñpyp ±   =   (Ñpbp ±)(±½qpW1) + (Ñp½qp)W1yp ±
Þ Ñpyp = (Ñpbp +)(½qpW1) + (Ñpbp -)(-½qpW1) + (Ñp½qp)W1yp  .]

    Combining these results we have for the more general "singly phased" solution yp = (½qpW1)bpqpW2) that, provided
Ñpbp = Ñp We12§(W§bp)) = 0  where W1 = W§(e12) (certainly the case for constant bp=b) , we have
    Ñpyp = ½(Ñpqp)ypWp     and     (Ñpqp)WpSp   =   4ph-1 (qap - mvp(-bpe1234)) where Wp º W2 + R§p(W1) .

    If qp = wm~¿p so that Ñpqp = wm~ for p-independant nonzero w and unit timelike m~ we have
    wm~WpSp   =   4ph-1   (qap - mvp(-bpe1234) ) Þ wWpSp   =   4ph-1   (mvp(-bpe1234) - qm~ap)     where p-independant four-momentum m = mm~ is in general nonparallel with four-velocity vp .
    This grade decomposes into
wWp¿Sp = 4ph-1 (m¿vp cos(bp) + qm~¿ap) = ± 4ph-1 m¿vp for the free Dirac (anti)particle
wWp×Sp = 4ph-1 (mÙvp(-bpe1234) + qm~Ùap) =± 4ph-1 mÙvp for the free Dirac (anti)particle
wWpÙSp = -4ph-1 (m¿vp) sin(bp)e1234 =0 for the free Dirac (anti)particle

    Taking bp=0 so bp=rp½Rp with Rp<0>=1 , e12*Rp=0 , and Rp§Rp=1 and requiring jp=rpvp to solve conservation equation Ñp¿jp=0 provides
    ÐvpRp º (vp¿Ñp)RpÑ   =   -½(ÑpÙvpÑ)Rp .
[ Proof :  The condition Ñpbp=0 becomes (Ñprp½)Rp + rp½(ÑpRp) = 0 and since vp = Rpe4Rp§ Þ vprp½Rp = rp½Rpe4 we have
Ñp(vprp½Rp) = Ñp(rp½Rpe4)  = (Ñpbp)e4 = 0 . But Ñp(vprp½Rp)   =   (ÑpvpÑ)rp½Rp + 2(vp¿Ñp)(rp½Rp)Ñ - vpÑp(rp½Rp)Ñ and so  
    -½(ÑpvpÑ)rp½Rp   =   (vp¿Ñp)(rp½Rp)Ñ   =   ((vp¿Ñp)rp½Ñ)Rp + rp½(vp¿Ñp)RpÑ   =   ½rp((vp¿Ñp)rpÑ)Rp + rp½(vp¿Ñp)RpÑ .
    Ñp¿jp=0 provides (vp¿Ñp)rpÑ = -rp(Ñp¿vpÑ) hence -½(ÑpvpÑ)rp½Rp   =   -½rp(Ñp¿vpÑ)rpRp + rp½(vp¿Ñp)RpÑ
    Result follows when rp¹0.  .]

    We typically consuider the phase ½qp to vary with p much faster than rp½Rp and consider an orientated particle having timelike four-velocity vp = (Rpqpe12))§(e4) = Rp§(e4) and spacelike unit spin axis wp~ =  (Rpqpe12))§(e3) = Rp§(e3) which rotate in a comparatively leisurely manner in accordance with (vp¿Ñp)RpÑ   =   ± ½qm-1 fpRp while spacelike   (Rpqpe12))§(e1) = Rp§((qpe12)e1) and (Rpqpe12))§(e2) = Rp§((qpe12)e2) are rapidly rotating in the Rp§(e12) 2-plane.
    Spacelike wp = rpw