Rotating Black Holes in Metric-Affine Gravity

0.38 MB·English

by  Peter Baekler

February 7, 2008 3:49 WSPC/Guidelines Magaxi13
International JournalofModernPhysicsD
(cid:13)c WorldScientificPublishingCompany
6
0 ROTATING BLACK HOLES IN METRIC-AFFINE GRAVITY
0
2

n PETERBAEKLER∗
a
Fachbereich Medien, Fachhochschule Du¨sseldorf
J
Universityof Applied Sciences
6
Josef-Gockeln-Str. 9
1
40474 Du¨sseldorf, Germany

1
FRIEDRICHW.HEHL†
v
3 Institute for Theoretical Physics
6 Universityof Cologne
0 50923 K¨oln, Germany
1 and
0 Department of Physics and Astronomy
6 Universityof Missouri-Columbia
0 Columbia, MO 65211, USA
/
c
q
Withintheframeworkofmetric-affinegravity(MAG,metricandanindependentlinear
-
r connection constitute spacetime), we find, for a specific gravitational Lagrangian and
g by using prolongation techniques, a stationary axially symmetric exact solution of the
: vacuumfieldequations.ThisblackholesolutionembodiesaKerr-deSittermetricandthe
v
post-Riemannian structures of torsion and nonmetricity. The solution is characterized
i
X by mass, angular momentum, and shear charge, the latter of which is a measure for
violatingLorentzinvariance.file Baekler/Magaxi13.tex
r
a
Keywords:Metric-affinegravity,prolongation,exactsolutions,Kerr-deSittermetric,tor-
sion,nonmetricity
PACS:03.50.Kk;04.20.Jb;04.50.+h;11.15.-q
1. Introduction
In the spirit of a gauge-theoretical approach to gravity, a metric-affine gauge field
theory of gravitation (“metric-affine gravity” MAG) has been proposeda based on
the metric g and the affine group A(4,R), i.e., the semi-direct product of the four
∗[email protected],http://www.et.fh-duesseldorf.de/home/baekler/index.html
†[email protected],http://www.thp.uni-koeln.de/gravitation/
aReviewshavebeenprovidedinRef.25andRef.21.ThegroupofDereli,Tucker,andWang17,58,59
appliedsuchtheoriestothedarkmatterproblem,interalia,Minkevichetal.33,37,34,35,36,Puet-
zfeldetal.48,49,50,51,andBabourova&Frolov3,2,4 mainlytocosmologicalsolutions.Newexact
solutions werefound, amongst others, by Vassiliev& King31,60,61,43.The determination of the
energy and of the other conserved quantities of exact solutions of MAG has been developed in
particular by Nester and his group14,63,39,64,65, see also Baekler et al.9. Comparison with ob-
servationshavebeenpioneeredbyPreuss46 andSolanki55,seealsoPuetzfeld, loc.cit..
1
February 7, 2008 3:49 WSPC/Guidelines Magaxi13
2 PeterBaekler and Friedrich W. Hehl
dimensionaltranslationalgroupR4 andthe generallinear groupGL(4,R).Besides
the usual “weak” Newton-Einstein type gravity, additional “strong gravity” pieces
will arise that are supposed to be mediated by additional geometrical degrees of
freedom related to the independent linear connection 1-form Γ β =Γ βdxi. Here
α iα
α,β,... = 0,1,2,3 denote frame (or anholonomic) indices and i,j,... = 0,1,2,3
coordinate (or holonomic) indices. Alternatively, the strong gravity pieces can also
be expressed in terms of the nonmetricty 1-form Qαβ = Q αβdxi and the torsion
i
2-formTα = 1T αdxi dxj.The propagatingmodes relatedtothe new degreesof
2 ij ∧
freedom manifest themselves in post-Riemannian pieces of the curvature R β.
α
2. Geometrical structures of a metric-affine spacetime
Webrieflysummarizethebasicnotionsofmetric-affinegeometry.Letusstartfrom
an-dimensionaldifferentiablemanifoldM .AteachpointP M wecanconstruct
n n
∈
the n-dimensional tangent vector space T (M ) with vector basis (or frame) e .
P n α
In the space T∗(M ), dual to T (M ), we introduce a local one-form basis (or
P n P n
coframe) ϑα such that
e ϑβ =δβ, (1)
α⌋ α
where symbolizes the interior product. Generally, the coframe is not integrable,
⌋
i.e., we have
1
Cα :=dϑα = C αϑµ ϑν =0, (2)
µν
2 ∧ 6
where the 1-form Cα is a measure of the anholonomity.
Weassumethatthemanifoldisendowedwithametric g.Wedecomposeitwith
respect to the coframe ϑα and find
g =g ϑα ϑβ. (3)
αβ
⊗
Furthermore, we assume that the manifold M carries additionally a metric-inde-
n
pendent linear connection Γ β. Accordingly, nonmetricity and torsion emerge as
α
geometrical field strengths, to be defined as
Q := Dg (4)
αβ αβ
−
and
Tα :=dϑα+Γ α ϑµ =Dϑα, (5)
µ
∧
respectively, together with the curvature 2-form
R β :=dΓ β Γ µ Γ β. (6)
α α α µ
− ∧
Here D is the exterior covariantderivative with respect to the connection Γ β.
α
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Rotating black holes inmetric-affine gravity 3
The geometrical field strengths give rise to integrability conditions, namely to
Bianchi and Ricci identities. We have, withb Z :=R ,
αβ (αβ)
DDg = DQ = 2R µg = 2Z , (7)
αβ αβ (α β)µ αβ
− − −
DDϑα =DTα =R α ϑµ, (8)
µ
∧
DR β =0; (9)
α
DDTα =(DR α) ϑµ+R α Tµ =R α Tµ, (10)
µ µ µ
∧ ∧ ∧
DDQ = 2R µ Q =:S , (11)
αβ (α β)µ αβ
− ∧
DS =DDDQ = 4R µ Z . (12)
αβ αβ (α β)µ
− ∧
Wecan,intermsofthemetricfieldofMAG,alwaysconstructaRiemannian(or
Levi-Civita) connection. Therefore, for the purpose of a comparison with general
relativity, e.g., it is useful to decompose the connection Γ := Γ µg into a
αβ α βµ
Riemannian piece Γ and a tensorial post-Riemannian piece N ,
αβ αβ
Γ =Γ +N . (13)
e αβ αβ αβ
The distortion1-formN allows us to recovernonmetricity andtorsionaccording
αβ e
to
Q =2N , Tα =N α ϑβ. (14)
αβ (αβ) β
∧
Explicitly, we have
1 1
N = e T + (e e T )ϑγ +(e Q )ϑγ + Q , (15)
αβ − [α⌋ β] 2 α⌋ β⌋ γ [α⌋ β]γ 2 αβ
25
see Ref. , Eq.(3.10.7).
In a metric-affine spacetime we can separate the curvature 2-form R into a
αβ
tracefree symmetric part (“shear”) Z , a trace part (“dila[ta]tion”) Z, and an
6 αβ
antisymmetric part (“rotation”) W according to
αβ
1
R =R +R = Z + Zg +W , (16)
αβ (αβ) [αβ] 6 αβ 4 αβ αβ
with the definitions
1
Z :=R , Z :=Z Zg , Z :=Z α, W :=R . (17)
αβ (αβ) 6 αβ αβ − 4 αβ α αβ [αβ]
The symmetric part Z represents the post-Riemann-Cartan part of the curva-
αβ
ture,thatis,it vanishestogetherwith Q ,whereasW includes the Riemannian
αβ αβ
contributions, inter alia.
Wenowspecializeto4-dimensionalspacetimewithLorentzsignature( +++).
−
Quitegenerally,nonmetricityQ ,torsionTα,andcurvatureR canthenbesplit
αβ αβ
into smaller pieces, they can be decomposed irreducibly under the Lorentz group,
bParenthesessurroundingindices(αβ):=(αβ+βα)/2denotesymmetrizationandbrackets[αβ]:=
(αβ−βα)/2antisymmetrization.
February 7, 2008 3:49 WSPC/Guidelines Magaxi13
4 PeterBaekler and Friedrich W. Hehl
see Appendix A. In the following table, we will give, for n=4, an overview of the
number of independent components of these quantities:
Table 1. Number of components of the irreducible pieces.
Q (1)Q (2)Q (3)Q (4)Q - - Q
αβ αβ αβ αβ αβ αβ
16 16 4 4 - - Σ=40
Tα (1)Tα (2)Tα (3)Tα - - - Tα
16 4 4 - - - Σ=24
R (1)W (2)W (3)W (4)W (5)W (6)W W
[αβ] αβ αβ αβ αβ αβ αβ αβ
10 9 1 9 6 1 Σ=36
R (1)Z (2)Z (3)Z (4)Z (5)Z - Z
(αβ) αβ αβ αβ αβ αβ αβ
30 9 6 6 9 - Σ=60
The exterior products of the coframe ϑα are denoted by ϑαβ := ϑα ϑβ, etc..
∧
Since a metric is prescribed, we can define a Hodge star operator ⋆ which maps,
in 4 dimensions, p-forms into (4 p)-forms. Then, we can introduce the eta-basis
−
η := ⋆1, ηα := ⋆ϑα, ηαβ := ⋆ϑαβ, etc..
3. Lagrangian and field equations
We consider in the first order Lagrangian formalism the geometrical variables
g ,ϑα,Γ β tobeminimallycoupledtomatterfields,collectivelycalledΨ,such
αβ α
{ }
that the total Lagrangian,i.e., the geometrical part plus the matter part, reads
L =V(g ,ϑα,Q ,Tα,R β)+L (g ,ϑα,Ψ,DΨ). (18)
tot αβ αβ α matter αβ
By using the excitations as place holders,
∂V ∂V ∂V
Mαβ = 2 , H = , Hα = , (19)
− ∂Q α −∂Tα β −∂R β
αβ α
25
the field equations of metric-affine gravity can be given in a very concise form:
DMαβ mαβ =σαβ (δ/δg ), (20)
αβ
−
DH E =Σ (δ/δϑα), (21)
α α α
−
DHα Eα =∆α (δ/δΓ β), (22)
β β β α
−
δL
=0 (matter). (23)
δΨ
On the right-hand-sides of each of the three gauge field equations (20) to (22),
there act the material currents as sources, on the left-hand-side there are typical
Yang-Mills like terms governing the gauge fields, their first derivatives, and the
February 7, 2008 3:49 WSPC/Guidelines Magaxi13
Rotating black holes inmetric-affine gravity 5
correspondinggauge field currents.The gauge currents turn out to be the metrical
(Hilbert) energy-momentum of the gauge fields
∂V
mαβ:=2 =ϑ(α Eβ)+Q(β Mα)γ T(α Hβ) R (α Hγ|β)+R(β|γ Hα) ,(24)
γ γ γ
∂g ∧ ∧ − ∧ − ∧ ∧
αβ
the canonical (Noether) energy-momentum of the gauge fields
∂V 1
E := =e V +(e Tβ) H +(e R γ) Hβ + (e Q )Mβγ, (25)
α ∂ϑα α⌋ α⌋ ∧ β α⌋ β ∧ γ 2 α⌋ βγ
and the hypermomentum of the gauge fields
∂V
Eα := = ϑα H g Mαγ, (26)
β ∂Γ β − ∧ β − βγ
α
respectively.
Themost general parityconservingquadratic Lagrangian,whichisexpressedin
termsofthe4+3+6+5irreduciblepiecesofQ ,Tα,W β,andZ β,respectively,
αβ α α
reads
1
V = a Rαβ η 2λ η
MAG 0 αβ 0
2κ − ∧ −
3 4
(cid:2)
+Tα ⋆ a (I)T +Q ⋆ b (I)Qαβ
I α αβ I
∧ ! ∧ !
I=1 I=1
X X
4
+2 c (I)Q ϑα ⋆Tβ +b (3)Q ϑα ⋆ (4)Qβγ ϑ (27)
I αβ 5 αγ β
!∧ ∧ ∧ ∧ ∧ #
XI=2 (cid:16) (cid:17) (cid:16) (cid:17)
6 5
1
Rαβ ⋆ w (I)W + z (I)Z
I αβ I αβ
−2ρ ∧
I=1 I=1
X X
9
+w ϑ (e (5)Wγ )+z ϑ (e (2)Zγ )+ z ϑ (e (I−4)Zγ ) .
7 α γ β 6 γ α β I α γ β
∧ ⌋ ∧ ⌋ ∧ ⌋ !
I=7
X
see Refs.18,41,24,27 and the literature quoted there. Here κ is the dimensionful
“weak” gravitational constant, λ the “bare” cosmological constant, and ρ the di-
0
mensionless “strong” gravity coupling constant. The constants a ,...a , b ,...b ,
0 3 1 5
c ,c ,c , w ,...w , z ,...z are dimensionless and are expected to be of order
2 3 4 1 7 1 9
unity. The constant a can only have the values 1 or 0 depending on whether a
0
Hilbert-Einstein term is present or not.
We ordered the Lagrangian(27) in the following way: In the first line, we have
the linear pieces, a Hilbert-Einstein type term and the cosmological term. Some
algebra yields Rαβ η = (6)Wαβ η , that is, only the curvature scalar is left
αβ αβ
∧ ∧
over, as expected. In the second line, we have the pure Yang-Mills type terms for
torsion and nonmetricity. If we expand them, we find a (1)Tα ⋆(1)T +...+
1 α
∼ ∧
b (1)Qαβ ⋆(1)Q +... . For a Yang-Mills field strength F we have always the
1 αβ
∧
Lagrangian F ⋆F.In ourcase,forTα andQαβ,the fieldstrengtharereducible
∼ ∧
and we can put open weighting factors in front of each square piece. Nevertheless,
February 7, 2008 3:49 WSPC/Guidelines Magaxi13
6 PeterBaekler and Friedrich W. Hehl
the second line is the obvious analog of a Yang-Mills Lagrangian for Tα and Qαβ.
Inthethirdline,wehave“interactions”betweenQ andTαandbetweendifferent
αβ
irreducible pieces ofQ . Inthe fourth line,we havethe pure Yang-Mills terms for
αβ
the rotational and the strain curvature w (1)Wαβ ⋆(1)W +...+z (1)Zαβ
1 αβ 1
∼ ∧ ∧
⋆(1)Z +...,and,eventually,inthelastline,“exotic”interactionsbetweendifferent
αβ
irreducible pieces of the curvature enter that we will drop subsequently. In other
words, we restrict ourselves to
w =z =z =z =z =0. (28)
7 6 7 8 9
Taking into consideration (28), the various excitations Mαβ,H ,Hα are
α β
{ }
found to be
Mαβ = 2⋆ 4 b (I)Qαβ
I
−κ !
I=1
X
2 1
c ϑ(α ⋆(1)Tβ)+c ϑ(α ⋆(2)Tβ)+ (c c )⋆Tgαβ
2 3 3 4
−κ ∧ ∧ 4 −
(cid:20) (cid:21)
b 1
5 ϑ(α ⋆(Q ϑβ)) gαβ⋆(3Q+Λ) , (29)
−κ ∧ ∧ − 4
(cid:20) (cid:21)
1⋆ 3 4
H = a (I)T + c (K)Q ϑβ , (30)
α −κ I α K αβ ∧ !
I=1 K=2
X X
6 5
a
Hα = 0ηα + w ⋆(I)Wα + z ⋆(K)Zα . (31)
β β I β K β
2κ
I=1 K=1
X X
The last equation can be slightly rewritten as
5
a w 1
Hα = 0 6 W ηα + w ⋆(n)Wα +z ⋆(n)Zα , (32)
β β n β n β
2κ − 12ρ ρ
(cid:18) (cid:19) nX=1(cid:16) (cid:17)
where (6)Wαβ = Wϑαβ/12 corresponds to the curvature scalar W.
−
4. Master equation for solving the field equations algebraically
Generally, it is a very delicate task to solve the nonlinear partial differential equa-
tions(20)to(22)forthesetofvariables g ,ϑα,Tα,Q .Evenforhighsymme-
αβ αβ
{ }
tries,therewillbeveryfewchancestofindexactsolutions.Therefore,wedeveloped
an algebraic method for solving the field equations.
The main observation is that we can construct an algebraic relation between
torsion and nonmetricity. This is a result of Prolongation Theory that has been
22
applied very successfully in the context of Einstein’s field equation by M.Gu¨rses
11
and Bilge et al. , amongst others. Application of this method to the Poincar´e
gauge field theory, i.e., to MAG with vanishing nonmetricity, Q = 0, leads to
αβ
the construction of stationary axisymmetric solutions with dynamic torsion, see
February 7, 2008 3:49 WSPC/Guidelines Magaxi13
Rotating black holes inmetric-affine gravity 7
Baekler et al.7,32,8,5. This method has been developedfurther systematically and,
10
inref. ,weusedaquitegeneralansatzforsolvingalsothefieldequationsofMAG.
Ithasbeenshown6,10 that the followinglinear relationship between nonmetric-
ity and torsion canbe exploitedfor solvingthe fieldequationsof MAGstraightfor-
wardly:
4
Tα = ξ˜ (A)Qα ϑµ+(3)Tα. (33)
A µ
∧
A=2
X
The parameters ξ˜ have to be determined by the field equations.
A
Todemonstratetheconsequencesofsuchanansatz,wewillconsiderasimplified
version of (33) in the form of
Tα =ξ Qα ϑµ+ξ Q ϑα+(3)Tα. (34)
0 µ 1
∧ ∧
We name this equation master equation. The constants ξ and ξ will be picked
0 1
later in the context of solving the field equations.
Alternatively, we can write it as
Tα =ξ Qα ϑβ +(ξ +ξ )Q ϑα+(3)Tα. (35)
0 β 0 1
6 ∧ ∧
The Weyl covector Q and the traceless nonmetricity Qα are defined by
β
6
1
Q:= Qα , Qα :=Qα Qδα. (36)
4 α 6 β β − β
Ifwemakeuseofthe2-formPαof(A.2)andthe1-formΛof(A.1),eq.(35)translates
intoc
ξ
Tα =ξ Pα 0Λ ϑα+(ξ +ξ )Q ϑα+(3)Tα. (37)
0 0 1
− 3 ∧ ∧
We compute the trace of this equation by contracting it with the frame e . Since
α
e Pα =0 and e (3)Tα =0, we find
α α
⌋ ⌋
T =ξ Λ 3(ξ +ξ )Q, (38)
0 0 1
−
with the 1-form T :=e Tα.
α
⌋
Empirically,aspecialcaseofrelation(38)hasbeenusedinMAGforconstruct-
ing exact solutions in the form of the triplet ansatz42,62,41,19
Q/k =Λ/k =T/k , (39)
0 1 2
with some constants k ,k ,k . We refere here to the triplet of 1-forms Q,Λ,T.
0 1 2
42
Spherically symmetric solutions were found as well as stationary axially sym-
metric ones.62,41,19 A deeper understanding of why this ansatz (38) could work
successfully in those approaches has been elaborated systematically by Baekler et
cWecouldslightlygeneralizethisexpressionforthetorsionto
Tα=ξ˜0Pα+ξ˜1Λ∧ϑα+ξ˜2Q∧ϑα+(3)Tα,
withsuitableconstants ξ˜0,ξ˜1,andξ˜2.
February 7, 2008 3:49 WSPC/Guidelines Magaxi13
8 PeterBaekler and Friedrich W. Hehl
10 27
al. , see also Heinicke et al. , demonstrating that the key is to look for further
integrability conditions. Especially the first Bianchi identity (8) turns out to be
helpful in answering this question.
We turn now to the connection and thus to the distortion 1-form N . We
αβ
eliminate the torsion T from (15) by means of our master equation (35). After
α
some algebra we find
1 1 1 1
N = Q 2 ξ Q ϑγ 2 ξ +ξ Q ϑ e (3)T .(40)
αβ 2 αβ − 0− 2 6 [αβ]γ − 0 1− 2 [α β]− 2 [α⌋ β]
(cid:18) (cid:19) (cid:18) (cid:19)
Note that e Q = Q and e Q = Q . Moreover, by means of (35), we can
α βγ αβγ α α
⌋ ⌋
alsoexpressthe firsttwoirreduciblepiecesofthetorsion(A.10)and(A.9)interms
of the nonmetricity,
1
(1)Tα =ξ (Qα ϑµ+ Λ ϑα)=ξ Pα, (41)
0 µ 0
6 ∧ 3 ∧
1
(2)Tα = [ξ Λ 3(ξ +ξ )Q] ϑα, (42)
0 0 1
−3 − ∧
with the 2-form Pα of (A.2). Note that both, (1)Tα and Pα, have 16 independent
components. Eq.(42) is equivalent to (38).
Further insight into the structure of the metric-affine field equations can be
gained if we take care of the master equation (34) in the excitations (29) and (30).
Let us first turn to the simpler expression (30). With our master equation, we
derived (1)Tα and (2)Tα in (41) and (42), respectively. We substitute these two
pieces, together with the irreducible decompositions (A.4), (A.5), and (A.6), into
(30). We find
κH =⋆ a (1)T +a (2)T +a (3)T +c (2)Q ϑβ +c (3)Q ϑβ
α 1 α 2 α 3 α 2 αβ 3 αβ
− ∧ ∧
(cid:16)+c (4)Q ϑβ
4 αβ
∧
=⋆ a ξ P a2 [ξ Λ(cid:17) 3(ξ +ξ )Q] ϑ +a (3)T 2c2 e P ϑβ
1 0 α− 3 0 − 0 1 ∧ α 3 α− 3 (α⌋ β) ∧
(cid:26)
+4c3 ϑ e Λ 1g Λ ϑβ +c Q ϑ . (cid:0) (cid:1) (43)
9 (α β)⌋ − 4 αβ ∧ 4 ∧ α
(cid:18) (cid:19) (cid:27)
Nowweorderthe right-handsideintermsofP ,Λ,andQ.After somealgebra,we
α
have
1
κH =⋆ (a ξ +c )P (a ξ +c )Λ ϑ +[a (ξ +ξ )+c ]Q ϑ
α 1 0 2 α 2 0 3 α 2 0 1 4 α
− − 3 ∧ ∧
(cid:26)
+a (3)Tα . (44)
3
o
The evaluation of the excitation (29) is a bit more complicated. In expanded
form, eq.(29) reads
κMαβ =⋆ b (1)Qαβ +b (2)Qαβ +b (3)Qαβ +b (4)Qαβ
1 2 3 4
− 2
(cid:16) 1 (cid:17)
+c ϑ(α ⋆(1)Tβ)+c ϑ(α ⋆(2)Tβ)+ (c c )⋆Tgαβ
2 3 3 4
∧ ∧ 4 −
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Rotating black holes inmetric-affine gravity 9
b 1
+ 5 ϑ(α ⋆(Q ϑβ)) gαβ⋆(3Q+Λ) . (45)
2 ∧ ∧ − 4
(cid:20) (cid:21)
Now we substitute into this equation the irreducible pieces (2)Q ,(3)Q ,(4)Q
αβ αβ αβ
from(A.6),(A.5),(A.4),respectively,(1)Tα,(2)Tα from(41),(42),andT from(38):
κ 2b 4b 1
Mαβ =b ⋆(1)Qαβ 2⋆ e(α Pβ) + 3 ⋆ ϑ(αeβ) Λ gαβ⋆Λ
1
− 2 − 3 ⌋ 9 ⌋ − 4
+b ⋆Qgαβ +c ξ ϑ(cid:16)(α ⋆Pβ(cid:17)) c3ϑ((cid:20)α (cid:16)⋆ [ξ Λ (cid:17)3(ξ +ξ )Q(cid:21)] ϑβ)
4 2 0 0 0 1
∧ − 3 ∧ − ∧
+c3−c4 [ξ ⋆Λ 3(ξ +ξ )⋆Q]gαβ n o
0 0 1
4 −
b 1
+ 5 ϑ(α (eβ) ⋆Q) gαβ(3⋆Q+⋆Λ) . (46)
2 ∧ ⌋ − 4
(cid:20) (cid:21)
With some algebrad we can order Mαβ in a similar way as we did with H :
α
κ 1
Mαβ =b ⋆(1)Qαβ + (2b +3c ξ )ϑ(α ⋆Pβ)
1 2 2 0
−2 3 ∧
1 1
+ (4b +3c ξ )Λ(αηβ) (8b +18c ξ +6c ξ +9b )gαβΛ ηµ (47)
3 3 0 3 4 0 3 0 5 µ
9 − 72
1 1
[2c (ξ +ξ )+b ]Q(αηβ)+ [8b +2(c +3c )(ξ +ξ )+b ]gαβQ ηµ.
3 0 1 5 4 3 4 0 1 5 µ
−2 8
This completes our simplifications of H and Mαβ.
α
Incidentally, the distortion (40) exhibits the special role of the choice ξ =1/2,
0
ξ =0 or Q=0, and (3)Tα =0. In either case, the connection reduces to
1
1 1
Γ =Γ + Q , with ξ = , ξ =0 or Q=0, (3)Tα =0. (48)
αβ αβ αβ 0 1
2 2
Metric-affine spacetimes with such a simple connection have already been studied
e
before.e We will come back to such a connection later in the discussion of our new
exact solution in Sec.9.
Eventually,we canalsosubstitute the master equation(34) andthe choice(28)
intothe Lagrangian(27).Again,likewiththe excitationsH andMαβ,weexpress
α
the Lagrangianin terms of Pα, Λ, and Q. We find
1
V = a Rαβ η 2λ η+b ⋆(1)Qαβ (1)Q
0 αβ 0 1 αβ
2κ − ∧ − ∧
n
dNote that for any 1-form Φ = Φγϑγ we have ⋆ ϑα eβ⌋Φ = Φβ⋆ϑα = Φβηα. A bit more
complicatedisthecomputationofϑα∧⋆ Φ∧ϑβ .IfΩisanother1-form,wehavethegeneralrule
(cid:2) (cid:0) (cid:1)(cid:3)
⋆Φ∧Ω=⋆Ω∧Φ.Moreover,wehavetherulesfortheHodgestarforanyform⋆(Φ∧ϑα)=eα⌋⋆Φ
(cid:0) (cid:1)
and(infourdimensions)⋆(eα⌋Φ)=−⋆Φ∧ϑα.Consequently,
ϑα∧⋆ Φ∧ϑβ = ϑα∧ eβ⌋⋆Φ =−eβ⌋(ϑα∧⋆φ)+gαβ⋆Φ=eβ⌋(⋆ϑα∧φ)+gαβ⋆Φ
(cid:16) (cid:17) = ηαβ∧(cid:16)Φ−ηα(cid:17)Φβ+gαβ⋆Φ.
Uponsymmetrization,wefindϑ(α∧⋆ Φ∧ϑβ) =−η(αΦβ)+gαβ⋆Φ.
eThesespacetimesemergeinthefollowingcontext:WedefinethePalatini3-formPαβ :=−δ(ηµν∧
Rµν/2)/δΓαβ and find Pαβ = −Pβα(cid:0)= Dηα(cid:1)β/2 = −Q∧ηαβ +Tγ ∧ηαβγ/2. If we require
e[α⌋Pγβ]=0,then23 Γαβ =Γαβ+Qαβ/2.
e
February 7, 2008 3:49 WSPC/Guidelines Magaxi13
10 PeterBaeklerand Friedrich W. Hehl
1
+ (a ξ +c )ξ + (2b +3c ξ ) Pα ⋆P
1 0 2 0 2 2 0 α
3 ∧
(cid:20) (cid:21)
1 4
+ (a ξ +c )ξ +c ξ + b ΛµΛ η
2 0 3 0 3 0 3 µ
3 3
(cid:20) (cid:21)
+[3(a ξ +c )ξ +3c ξ +4b ]QµQ η
2 0 4 0 4 0 4 µ
4
2(a (ξ +ξ )+c )ξ +2(a ξ +c )(ξ +ξ ) a ξ (ξ +ξ )+b
2 0 1 4 0 2 0 3 0 1 2 0 0 1 5
− − 3
(cid:20) (cid:21)
QµΛµη +VR2. (49)
× }
We put here a =0.
3
5. Finding solutions by nullifying the excitations
WecanfindexactsolutionsofthefieldequationsofMAGstraightforwardlyinavery
simple manner.We willask for non-trivialfieldconfigurationswiththe propertyof
vanishing field excitations, i.e., we will require
H =0, Mαβ =0, Hα =0. (50)
α β
And indeed, because of the inhomogeneity of the excitations in terms of the field
strengths, it will be possible to generate solutions with non trivial curvature.
Table 2. The case Q=0.
Excitation constraints
a ξ +c =0
1 0 2
H =0 a ξ +c =0
α 2 0 3
a =0 or (3)Tα =0
3
b =0
1
Mαβ =0 2b +3c ξ =0
2 2 0
4b +3c ξ =0
3 3 0
b +2c ξ =0
5 4 0
w =w =w =w =w =0
1 2 3 4 5
Hα =0 z =z =z =z =z =0
β 1 2 3 4 5
6ρa κw W =0
0 6
−
If we substitute this into the sourcefree field equations (21) and (22), only the
following truncated equation is left over:
E = e V =0. (51)
α α
⌋
Since ϑ E = 0, this equation has only 10 independent components. The field
[α β]
∧
equation(20)isredundantbecause(21)and(22)arefulfilled.Accordingly,wehave
just to solve the algebraic relation (51).