Gauging gravity with SO(1,3) for spin-1/2 particles

0.2 MB·English

by  Arpan Saha

Gauging gravity with SO(1,3) for spin-1/2 particles
Arpan Saha†
2
1 Banibrata Mukhopadhyay‡
0
2 January 24, 2012

n
a
J
Abstract

3
2 We demonstrate, by analogy with electromagnetism, that the geometric content in the theory of
gravity is an indirect consequence of the fact that the gauge group in question is the Lorentz group

] SO(1,3). We hence construct field equations for gravity and a spin-1/2 particle in a gravitational field
c
based on gauge considerations. Furthermore, we derive the weak field and Schr¨odinger limits of the
q
- Diracequationoftheparticleinthegravitationalfield,especiallyinFerminormalcoordinatesandonthe
r equatorialplaneoftheKerrgeometry,followingwhichweidentifythetermstowhichtheelectromagnetic
g
potentials A~ and Φ are analogous.
[

1
v
1 Introduction
9
4
6 One of the principal attractions of Einstein’s theory of gravity is the fact that it is geometric in nature
4 – the gravitational field, instead of being regarded as a force field acting upon particles, is seen as being
.
1 manifested as the intrinsic curvature of the spacetime semi-Riemannian manifold. As geometryforms one of
0 thefundamentalarenaswhereinwehistoricallylearnthowtoexerciselogic,itwashopedthatsuchgeometric
2 insight would carry over to theories describing the other three forces of Nature – electromagnetic, weak and
1
strong. This unfortunately turned out not to be the case. As a result, many physicists [1] started to regard
:
v treatmentsofgravitythat aregeometricinnature fromthe outsetwith acertainlevelofsuspicion. Building
Xi the theoryfromthe principle ofequivalence provesto be a better strategyasit is afar morereliableconnect
amongst the four forces than geometry.
r
a
In the above spirit, this paper takes up the semi-classical description of a spin-1/2 particle in a gravita-
tionalfieldandshowsthatmuchoftheinformationaboutthedynamicsofparticlesandfieldscanbeobtained
fromgaugeconsiderationsalone. Bythiswemeanthatifwerequirethatthedynamicsonspacetimeremains
unchanged when the connection associated with the principal bundle corresponding to the gauge group in
questionundergoesagaugetransformation,wecaneffectivelyreconstructthegoverningdynamicalequations
or at least impose constraints strong enough on their form, so that straightforwardarguments may be used
to eliminate the candidates which are clearly not physical. The group associated with the internal space of
†DepartmentofPhysics,IndianInstituteofTechnologyBombay,Mumbai;arpan [email protected]
‡DepartmentofPhysics,IndianInstituteofScience,Bangalore;[email protected]
1
symmetry in the case of electromagnetism is U(1), while that in the case of gravity is the Lorentz group
SO(1,3). Thisiswhatimpartsgeometriccontenttothetheoryofgravityi.e.thefactthatinformationabout
the gravitationalfield is encoded in the metric.
Thus, our twofold claim is as follows.
1. The fact that the metric carries information about the gravitationalfield is an indirect consequence of
the fact that the gauge group is SO(1,3) and certain other conditions of physicality, as we shall see.
2. The field equations for gravity may be constructed through gauge considerations without any a priori
geometric assumptions.
Followingthedemonstrationoftheabove,weshalltakeuptheDiracequationforaparticleinagravitational
field that we will obtain as a result of our analysis, and investigate its Klein-Gordon form and the limits
thereof in the weak field and Schr¨odinger regime (when the velocity of the particle is much lower than the
speed of light). We finally conclude by looking at the magnetic analogue of the gravitational potential in
Fermi normal coordinates and on the equatorial plane of the Kerr geometry.
2 Electromagnetism and U(1)
In this section, we will recall how the dynamical equations for an electromagnetic field and a spin-1/2 par-
ticle in it may be developed, given that the gauge group in question is U(1). The motive is to illustrate the
key steps in our derivation (which will carry over to the case for gravity) in a setting that we understand
relativelywellandwhichisnotgeometricinnature. Thiswillenableus tobetterhighlightthe analogiesand
differences between electromagnetism and gravity.
We enumerate the aforementioned steps as follows.
1. The Lagrangiandensity for the particleandthe field is writtenas a sumoftwo parts– the Lagrangian
density for the particle and the Lagrangian density for the field. While the former may be obtained
by simply replacing the derivative in the no-field Lagrangian density by a ‘gauge covariant derivative’
whereinaconnectiontermisincludedsothatwemaycompareobjectslyinginfibersovertwodifferent
points, the latter is not known a priori.
2. The elements of the gauge group are allowed to act on the wavefunction spinors (and other objects
dwelling in the internal space of symmetry). The particle Lagrangian density must remain invariant
under such transformations; this gives us the transformation of the connection.
3. The field Lagrangian density is stipulated to be a gauge-invariant function of the connection. Scalars
constructed out of the ‘gauge curvature’ (i.e. commutators of the gauge covariant derivatives) fit this
requirement rather well.
4. The resulting action is varied around the stationary ‘points’. This yields the equations governing the
behaviour of the particle and the field.
First the Lagrangiandensity is split as
L
= + (1)
1/2 EM
L L L
2
where istheLagrangiandensityofthefield,whoseformistobedetermined,and istheLagrangian
EM 1/2
L L
density of the particle given by [2]
i
= ψ¯γµ(ψ iqA ψ) (ψ¯ +iqψ¯A )γµψ mψ¯ψ (2)
L1/2 2 ,µ− µ − ,µ µ −
(cid:2) (cid:3)
where ψ is the wavefunction spinor, ψ¯ is its adjoint spinor, m and q are the mass and charge of the parti-
cle respectively, γµ are the metric-dependent Dirac matrices and A is the electromagnetic connection field
µ
i.e. the 4-vector potential.
Now, the elements of the gauge group U(1) are represented by eiqχ. Therefore, under the action of such
an element, the following transformations occur: ψ ψ′, ψ¯ ψ¯′ and A A′ , where
→ → µ → µ
ψ′ =eiqχψ (3a)
ψ¯′ =ψ¯e−iqχ (3b)
Moreover,the Dirac matrices γµ act as mapping operators from physical spacetime (i.e. the tangent bundle
T of the semi-Riemannian manifold that is spacetime) to the internal space of symmetry associated
M M
with each point, and hence must transform as well when the internal space of symmetry is acted upon by
the group. Therefore, we have
γ′µ =eiqχγµe−iqχ (3c)
Of course,here since the Dirac matrices commute with the elements of U(1), this does not change anything.
But this shall not be the case with gravity, as we shall see in the next section.
The result of the above transformations is
i
′ = ψ¯′γ′µ(ψ′ iqA′ψ′) (ψ¯′ +iqψ¯′A′)γ′µψ′ mψ¯′ψ′
L1/2 2 ,µ− µ − ,µ µ −
= i (cid:2)ψ¯e−iqχeiqχγµe−iqχ((eiqχψ) iqA′eiqχψ) (cid:3)
2 ,µ− µ
(cid:2) ((ψ¯e−iqχ) +iqψ¯e−iqχA′ )eiqχγµe−iqχeiqχψ mψ¯e−iqχeiqχψ
− ,µ µ −
i
= ψ¯γµ(iqχ ψ+ψ iqA′ψ) (ψ¯ iqψ¯χ +(cid:3)iqψ¯A′ )γµψ mψ¯ψ (4)
2 ,µ ,µ− µ − ,µ− ,µ µ −
(cid:2) (cid:3)
′
Requiring the Lagrangian density to be invariant under such transformations (i.e. = ) yields
L1/2 L1/2 L1/2
the following transformation law for A
µ
′
A =A +χ (5)
µ µ ,µ
The gaugecurvature F =A A effectively takescare ofthe χ term, andwe stipulate that the field
µν ν,µ µ,ν ,µ
−
LagrangiandensityisascalarconstructedoutofF . Wealsostipulate,outofempiricalconsiderations,that
µν
the constructed scalar minimises the number of contractions taking place while not identically vanishing (as
otherwise,thatwouldleadtolossofallphysicalinformation). Inthepresentcase,thisisquitestraightforward
and given by
=KgµλgνσF F (6)
EM µν λσ
L
where gµλ denotes the metric tensor and K is a constant of proportionality.
3
The action, hence, is
S = √ g d4x
L −
Z
= + √ g d4x
1/2 EM
L L −
Z
(cid:0) i (cid:1)
= ψ¯γµ(ψ iqA ψ) (ψ¯ +iqψ¯A )γµψ
,µ µ ,µ µ
2 − −
Z (cid:18)
(cid:2) mψ¯ψ+KgµλgνσF F √ g(cid:3)d4x (7)
µν λσ
− −
Here, g = detgµν and the factor √ g is included to account for the fa(cid:1)ct that the infinitesimal volume d4x
−
does not transform like a scalar under general coordinate transformations.
Here, the independent variables are taken to be ψ, ψ¯ and A . The Euler-Lagrange equations are then
µ
obtained as
∂( √ g) ∂( √ g)
∂ L − L − =0 (8a)
µ
∂ψ − ∂ψ
(cid:18) ,µ (cid:19)
∂( √ g) ∂( √ g)
∂ L − L − =0 (8b)
µ ∂ψ¯ − ∂ψ¯
(cid:18) ,µ (cid:19)
∂( √ g) ∂( √ g)
∂ L − L − =0 (8c)
µ
∂A − ∂A
(cid:18) ν,µ (cid:19) ν
We may use the fact that (∂ √ g)= 1√ ggλνg to factor out √ g and obtain
µ − 2 − λν,µ −
1 ∂ ∂ ∂
gλρg L +∂ L L =0 (9a)
λρ,µ µ
2 ∂ψ ∂ψ − ∂ψ
(cid:18) ,µ(cid:19) (cid:18) ,µ(cid:19)
1 ∂ ∂ ∂
gλρg L +∂ L L =0 (9b)
2 λρ,µ ∂ψ¯ µ ∂ψ¯ − ∂ψ¯
(cid:18) ,µ(cid:19) (cid:18) ,µ(cid:19)
1 ∂ ∂ ∂
gλρg L +∂ L L =0 (9c)
λρ,µ µ
2 ∂A ∂A − ∂A
(cid:18) ν,µ(cid:19) (cid:18) ν,µ(cid:19) ν
4
We now compute the derivatives as
∂ i
L = iqψ¯γµA (ψ¯ +iqψ¯A )γµ mψ¯
µ ,µ µ
∂ψ 2 − − −
∂ i (cid:2) (cid:3)
L = [γµ(ψ iqA ψ) iqA γµψ] mψ
∂ψ¯ 2 ,µ− µ − µ −
∂
L =qψ¯γνψ
∂A
ν
∂ i
L = ψ¯γµ
∂ψ 2
,µ
∂ i
L = γµψ
∂ψ¯ −2
,µ
∂
L =Kgαλgβρ δνδµ δνδµ F +F δνδµ δνδµ
A β α− α β λρ αβ ρ λ − λ ρ
ν,µ
=4KFµν h(cid:16) (cid:17) (cid:0) (cid:1)i
The Euler-Lagrangeequations (9) thus boil down to the following three equations
i
i(ψ¯ +iqψ¯A )γµ+mψ¯= gλρg ψ¯γµ (10a)
,µ µ λρ,µ
−4
i
iγµ(ψ iqA ψ) mψ = gλρg γµψ (10b)
,µ µ λρ,µ
− − −4
q 1
Fµν ψ¯γνψ = gλρg Fµν (10c)
,µ− 4K −2 λρ,µ
The three equations we have obtained are in fact the adjoint Dirac equation, the Dirac equation and the
source-dependent Maxwell equation (the source-free one becomes tautologous) respectively. The terms on
the right side of the equality represent corrections accounting for (possibly) noncartesian coordinates. All
thisisconsistentwithwhatwealreadyknowaboutthedynamicsofaspin-1/2particleinanelectromagnetic
field.
3 The spinor connection
Not surprisingly, various complications arise when it comes to gravity. A chief reason for this is the fact
that the gauge group for gravity SO(1,3) (dealt with by means of the spin-1/2 spinor representation) is
nonabelian unlike U(1), and that the matrix representatives of its elements do not commute with the Dirac
matrices. However, the essential features of the derivation in the last section undergo no change. We once
again split the Lagrangiandensity
= + (11)
1/2 G
L L L
with being the field Lagrangian density whose form is to be determined and being the particle
G 1/2
L L
Lagrangiandensity given by
i
= ψ¯γµ(ψ +Γ ψ) (ψ¯ +ψ¯Γ¯ )γµψ mψ¯ψ (12)
L1/2 2 ,µ µ − ,µ µ −
(cid:2) (cid:3)
5
Here, Γ is the gravitational connection analogous to the electromagnetic potential A . As it would be
µ µ
expected of a connection form associated with a principal G-bundle, it takes values in the Lie algebra g
associatedwithG. Inthe caseofgravity,Gisthe LorentzgroupSO(1,3),andits Liealgebraso(1,3)carries
a matrix representation wherein the space of matrices is spanned by σ = i [γ ,γ ], when the matrices
AB 2 A B
γ being the metric-independent Dirac matrices (the indices are denoted with the capital Latin alphabet to
A
denote that they are internal).
Next, we perform the following gauge transformations: ψ ψ′, ψ¯ ψ¯′, γµ γ′µ, Γ Γ′ and
Γ¯ Γ¯′ where → → → µ → µ
µ → µ
′
ψ =Lψ (13a)
ψ¯′ =ψ¯L−1 (13b)
γ′µ =LγµL−1 (13c)
The last two transformations are to be determined.
The matrix L is a spin-1/2 representative of a Lorentz transformation, given by exp iǫABσ , where
AB
ǫAB is the generator of the Lorentz transformation in its rank 2 tensor representation.
(cid:0) (cid:1)
The result of the above transformations leads to
i
′ = ψ¯′γ′µ(ψ′ +Γ′ ψ′) (ψ¯′ +ψ¯′Γ¯′ )γ′µψ′ mψ¯′ψ′
L1/2 2 ,µ µ − ,µ µ −
= i (cid:2)ψ¯L−1LγµL−1((Lψ) +Γ′Lψ) (cid:3)
2 ,µ µ
(cid:2) ((ψ¯L−1) +ψ¯L−1Γ¯′)LγµL−1Lψ mψ¯L−1Lψ
− ,µ µ −
i
= ψ¯γµ(L−1L ψ+ψ +L−1Γ′ Lψ) (cid:3)
2 ,µ ,µ µ
(cid:2) (ψ¯ +ψ¯L−1L+ψ¯L−1Γ¯′L)γµψ mψ¯ψ (14)
− ,µ ,µ µ −
′
Since, the Lagrangian density L1/2 is required to be invariant (i.e(cid:3). L1/2 = L1/2), we have the following
transformation laws:
Γ′ =LΓ L−1 L L−1 (15a)
µ µ − ,µ
Γ¯′ =LΓ¯ L−1 LL−1 (15b)
µ µ − ,µ
Furthermore, as we have L L−1 +LL−1 = LL−1 = 1 = 0, we note that Γ¯ transforms exactly like
,µ ,µ ,µ ,µ µ
Γ . Since the spinor connection is defined by its transformation law alone (any information not deducible
− µ (cid:0) (cid:1)
from that is a question of representation), we set
Γ¯ = Γ (16)
µ µ
−
Therefore, all we require is
Γ′ =LΓ L−1+LL−1 (17)
µ µ ,µ
WehavealreadyremarkedthatthespinorconnectionisanelementoftheLiealgebraspannedbyσ . Thus,
AB
Γ may be represented as a linear combination of σ
µ AB
i
Γ = ωABσ (18)
µ 4 µ AB
6
wherethefactori/4hasbeenintroducedtomaintainnotationalconsistencywithKaku[3]andthecoefficients
ωAB are referredto as the Fock-Ivanenko coefficients (which may be taken to be antisymmetric in indices A
µ
and B as they are contracted with A and B in σ which is antisymmetric in A and B).
AB
AsitisonlytheFock-Ivanenkocoefficientswhichundergochangesundergaugetransformationsandwhich
exhibit coordinate dependence, rather than the matrices σ , it is clear that all gravitationaldynamics will
AB
beencodedinthem. Thus,ournextstepwillbetoconstructanexpressionintermsofthesecoefficientsthat
is invariant under gauge transformations.
4 The gauge curvature
MotivatedbyouranalysisinSection2,webeginbyexaminingthe gaugecurvaturewhichinthepresentcase
is given by
Ω =Γ Γ +[Γ ,Γ ] (19)
µν ν,µ µ,ν µ ν
−
To see how it transforms, we replace all the quantities above with their primed counterparts and use the
transformation law (17) given by
′ ′ ′ ′ ′
Ω =Γ Γ +[Γ ,Γ ]
µν ν,µ− µ,ν µ ν
= LΓ L−1 + LL−1 LΓ L−1 LL−1
ν ,µ ,ν ,µ− µ ,ν − ,µ ,ν
(cid:0) + [LΓ(cid:1)L−1+(cid:0) LL−1(cid:1),LΓ (cid:0)L−1+LL(cid:1)−1] (cid:0) (cid:1)
µ ,µ ν ,ν
=LΓ L−1 LΓ L−1+L[Γ ,Γ ]L−1
ν,µ µ,ν µ ν
−
=LΩ L−1 (20)
µν
The gauge curvature may be written in terms of the Fock-Ivanenko coefficients.
i i i i
Ω = ωABσ ωABσ + ωIJσ , ωKLσ
µν 4 ν,µ AB − 4 µ,ν AB 4 µ IJ 4 ν KL
(cid:20) (cid:21)
i i
= ωABσ ωABσ + ωIJωKL[σ ,σ ]
4 ν,µ AB − µ,ν AB 4 µ ν IJ KL
(cid:20) (cid:21)
i i
= ωABσ ωABσ + ωIJωKL ωIJωKL [σ ,σ ]
4 ν,µ AB − µ,ν AB 8 µ ν − ν µ IJ KL
(cid:20) (cid:21)
i 1 (cid:0) (cid:1)
= ωAB ωAB ωIJωKL ωIJωKL δPAδQBη σ (21)
4 ν,µ − µ,ν − 4 µ ν − ν µ IJ KL PQ AB
(cid:20) (cid:21)
(cid:0) (cid:1)
where η is the usual Minkowski metric and δPA is the generalised Kronecker delta given by
PQ IJ
δPA =δPδA δPδA
IJ I J − J I
The proof of the fact
[σ ,σ ]= 2iδPAδQBη σ
IJ KL − IJ KL PQ AB
used above is given in the Appendix.
7
If we let
1 1
RAB = ωAB ωAB ωIJωKL ωIJωKL δPAδQBη (22)
µν 4 ν,µ − µ,ν − 4 µ ν − ν µ IJ KL PQ
(cid:20) (cid:21)
(cid:0) (cid:1)
the gauge curvature may be succinctly written as
Ω =iRABσ (23)
µν µν AB
It may be easily verified that RAB is antisymmetric in A and B.
µν
Now, under a gauge transformation, elements in the Lie algebra associated with the gauge group get
′
mapped to other elements in the said Lie algebra. In particular, Ω gets mapped to Ω . The basis
µν µν
matrices σ , A < B, prior to the gauge transformation, however, continue to serve as a basis for the
AB
transformedelements. It is therefore a matter ofchoice that we willpreservethe Dirac commutatorsσ as
AB
they are and let only the coefficients transform. Thus, we have
Ω′ =iR′ABσ (24)
µν µν AB
Substituting the above into (20) we obtain
R′ABσ =LRUVσ L−1
µν AB µν UV
i
= RUV Lγ L−1,Lγ L−1
2 µν U V
i (cid:2) (cid:3)
= RUV ΛA γ ,ΛB γ
2 µν U A V B
=ΛA ΛB(cid:2)RUVσ (cid:3) (25)
U V µν AB
where ΛA is the rank 2 tensor representativeof the Lorentz transformationdenoted by L and the third line
U
follows from the second by the virtue of the form-invariance of the Dirac equation in flat spacetime. If we
shuffle the above onto one side we obtain
R′AB ΛA ΛB RUV σ =0
µν − U V µν AB
We may use the fact that all the matr(cid:0)ices σ with A<B(cid:1)are linearly independent to assert. Hence
AB
R′AB =ΛA ΛB RUV (26)
µν U V µν
This is exactly how a contravariantLorentz tensor of rank 2 would transform.
ThequantityRAB istheexactanalogueoftheelectromagneticfieldtensorF –itis,inadditiontobeing
µν µν
antisymmetric in the indices A and B as we had observedearlier, also antisymmetric in µ and ν. Motivated
by our observations in Section 2, we might consider η η gµλgνρRABRMN as a probable choice for a
AM BN µν λρ
gauge-invariant diffeomorphism-invariant scalar. But again, empirical considerations (more precisely, the
fact that in the weak field Newtonian limit, the 1/r2 decay of the gravitational‘force’ holds across all scales
[1]) dictate that we reduce the number of contractions as far as possible while ensuring that our scalar does
not identically vanish. In this case, owing to the presence of two internal indices A and B in addition to the
spacetimeindicesµandν, wecandosoatthe expenseofintroducinganewindependentfieldoflinearmaps
eµ that takes basis vectors in the internal space of symmetry associated with a point to coordinate basis
A
8
vectors dwelling in the tangent space at the point. By construction, such maps would behave as Lorentz
1-formsundergaugetransformationsandasvectorsundergeneralcoordinatetransformations. Therefore,we
take eµeνRAB as the required gauge-invariantdiffeomorphism-invariant scalar and let the field Lagrangian
A B µν
density for gravity be
=KeµeνRAB (27)
LG A B µν
where K is a constant of proportionality.
While we ought to be wary about introducing new independent fields leading to new degrees of freedom,
we have not explicitly put in geometric information by hand – no assumptions have been made regarding
the possible dependence of eµ on the metric. Hence, as long as we treat these fields as just ‘scaffolds’of sort
A
that carry no physical meaning a priori, our construction is a legitimate one.
5 The Palatini action
Theactionassociatedwiththe Lagrangiandensity ofspinorsinpresenceofgravitationalinteractionwehave
arrived at is thus
S = √ g d4x
L −
Z
= + √ g d4x
1/2 G
L L −
Z
(cid:0) i (cid:1)
= ψ¯γµ(ψ +Γ ψ) (ψ¯ ψ¯Γ )γµψ
,µ µ ,µ µ
2 − −
Z (cid:18)
(cid:2) mψ¯ψ+KeµeνRAB √ g(cid:3)d4x (28)
− A B µν −
(cid:1)
This is known as the Palatini action [4] (strictly speaking, it is known as the Palatini action once the maps
eµ have been identified with the vierbein fields; that is an issue that will be addressed later). On carrying
A
out a variation about its stationary points with respect to ψ, ψ¯, ωAB and eν, we obtain the Euler-Lagrange
ν A
equations
∂( √ g) ∂( √ g)
∂ L − L − =0 (29a)
µ
∂ψ − ∂ψ
(cid:18) ,µ (cid:19)
∂( √ g) ∂( √ g)
∂ L − L − =0 (29b)
µ ∂ψ¯ − ∂ψ¯
(cid:18) ,µ (cid:19)
∂( √ g) ∂( √ g)
∂ L − L − =0 (29c)
µ ∂ωAB − ∂ωAB
(cid:18) ν,µ (cid:19) ν
∂( √ g) ∂( √ g)
∂ L − L − =0 (29d)
µ ∂eνA,µ !− ∂eνA
9
Again using the fact that (∂ √ g)= 1√ ggλνg , we may factor out √ g to obtain
µ − 2 − λν,µ −
1 ∂ ∂ ∂
gλρg L +∂ L L =0 (30a)
λρ,µ µ
2 ∂ψ ∂ψ − ∂ψ
(cid:18) ,µ(cid:19) (cid:18) ,µ(cid:19)
1 ∂ ∂ ∂
gλρg L +∂ L L =0 (30b)
2 λρ,µ ∂ψ¯ µ ∂ψ¯ − ∂ψ¯
(cid:18) ,µ(cid:19) (cid:18) ,µ(cid:19)
1 ∂ ∂ ∂
gλρg L +∂ L L =0 (30c)
2 λρ,µ ∂ωAB µ ∂ωAB − ∂ωAB
(cid:18) ν,µ (cid:19) (cid:18) ν,µ (cid:19) ν
1 ∂ ∂ ∂
gλρg L +∂ L L =0 (30d)
2 λρ,µ ∂eνA,µ! µ ∂eνA,µ!− ∂eνA
We compute the derivatives
∂ i
L = ψ¯γµΓ (ψ¯ ψ¯Γ )γµ mψ¯
µ ,µ µ
∂ψ 2 − − −
∂ i (cid:2) (cid:3)
L = [γµ(ψ +Γ ψ)+Γ γµψ] mψ
∂ψ¯ 2 ,µ µ µ −
∂ 1 K
L = ψ¯ γν,σ ψ eξ eζ δIJ δνωKL ωIJδKLδν
∂ωAB −8 { AB} − 16 U V AB ξ ζ − ξ AB ζ
ν
δIJ δνωKL+ωIJδKLδν(cid:0)δPUδQVη
− AB ζ ξ ζ AB ξ IJ KL PQ
∂
L =2KeξRAU (cid:1)
∂eν U νξ
A
∂ i
L = ψ¯γµ
∂ψ 2
,µ
∂ i
L = γµψ
∂ψ¯ −2
,µ
∂ K
L = eξ eζ δµδν δµδν δUV
ωAB 4 U V ξ ζ − ζ ξ AB
ν,µ
(cid:16) (cid:17)
K
= (eµeν eνeµ)
2 A B− A B
∂
L =0
∂eν
A,µ
10