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 | 1 | 0 | ö ; | gj= | æ | 0 | sj | ö |
| è | 0 | -1 | ø | è | - sj | 0 | ø |
| g0= | æ | 1 | 0 | 0 | 0 | ö ; | g1= | æ | 0 | 0 | 0 | 1 | ö ; | g2= | æ | 0 | 0 | 0 | -i | ö ; | g3= | æ | 0 | 0 | 1 | 0 | ö |
| ç | 0 | 1 | 0 | 0 | ÷ | ç | 0 | 0 | 1 | 0 | ÷ | ç | 0 | 0 | i | 0 | ÷ | ç | 0 | 0 | 0 | -1 | ÷ | ||||
| ç | 0 | 0 | -1 | 0 | ÷ | ç | 0 | -1 | 0 | 0 | ÷ | ç | 0 | i | 0 | 0 | ÷ | ç | -1 | 0 | 0 | 0 | ÷ | ||||
| è | 0 | 0 | 0 | -1 | ø | è | -1 | 0 | 0 | 0 | ø | è | -i | 0 | 0 | 0 | ø | è | 0 | 1 | 0 | 0 | ø |
| g_jk º gjgk= | æ | - sj sk | 0 | ö | = | -eijki | æ | si | 0 | ö ; | gijk = | eijki | æ | 0 | -1 | ö ; | g0123 = | i | æ | 0 | -1 | ö ; | g5º ig0123 = | æ | 0 | 1 | ö | |
| è | 0 | - sj sk | ø | è | 0 | si | ø | è | 1 | 0 | ø | è | -1 | 0 | ø | è | 1 | 0 | ø |
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=bp(½qpW2)↑ = bpW(½qpe12)↑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(e-½qpW2)bp -
= Ñp(bp)e½qpW2 +
Ñp(½qp)e(½p+½qp)W2bp + - Ñp(½qp)e(½p-½qp)W2bp -
= Ñp(bp)e½qpW2 + Ñp(½qp)bp +e(½p+½qp)W2
- Ñp(½qp)bp -e(½qp-½p)W2
= Ñp(bp)e½qpW2 + Ñp(½qp)bp +e½p+½qpW2bp +
+ Ñp(½qp)bp -(½qpW2)↑W2
= Ñp(bp)(½qpW2)↑ + Ñp(½qp)bp(½qpW2)↑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 equation
(½Ñpqp +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~, m=½w 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 ± = W(½qpe12)↑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)↑bp(½qpW2)↑
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-½((vp¿Ñp)rpÑ)Rp + rp½(vp¿Ñp)RpÑ .
Ñp¿jp=0 provides
(vp¿Ñp)rpÑ = -rp(Ñp¿vpÑ) hence
-½(ÑpvpÑ)rp½Rp =
-½rp-½(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 = (Rp(½qpe12)↑)§(e4) = Rp§(e4)
and spacelike unit spin axis wp~ = (Rp(½qpe12)↑)§(e3) = Rp§(e3)
which rotate in a comparatively leisurely manner in accordance
with (vp¿Ñp)RpÑ = ± ½qm-1 fpRp
while spacelike
(Rp(½qpe12)↑)§(e1) = Rp§((qpe12)↑e1) and
(Rp(½qpe12)↑)§(e2) = Rp§((qpe12)↑e2) are rapidly rotating
in the Rp§(e12) 2-plane.
Spacelike wp = rpw