A Solution to Symmetric Teleparallel Gravity(5)

2021-04-05 10:00

Teleparallel gravity models, in which the curvature and the nonmetricity of spacetime are both set zero, are widely studied in the literature. We work a different teleparallel theory, in which the curvature and the torsion of spacetime are both constrained

equationseasily,theinvariantdescriptionprovidesthecorrectunderstandingofthephysicalcontentsofasolution.

Sincemetricandconnectionareindependentquantitiesinnon-Riemannianspacetimes,wehavetopredictseparatelyappropriatecandidatesforthem.Thereforewe rstwritealineelementinordertodeterminethemetric.Wenaturallystartdealingwiththecaseofsphericalsymmetryforrealisticsimplicity,

g= F2dt2+G2dr2+r2dθ2+r2sin2θd 2

whereF=F(r)andG=G(r).Aconvenientchoiceforatetradreads

e0=Fdt,

1

1

Gr)e,1(31)(32)cotθe1=Gdr,e2=rdθ,e3=rsinθd .Inaddition,forthenon-RiemannianconnectionwechooseΛ12= Λ21=

FG1e,1e3,1Λ23= Λ32= G)e1,Λ11=

)e1,Λ22=others=0.G

Thesegaugecon gurations(32)and(33)satisfytheconstraintequationsRab(Λ)=0,0.OnecancertainlyperformalocallyLorentztransformationΛ33=

ea→Labeb,Λab→LacΛcdL 1db(33)Ta(Λ)=(34)+LacdL 1cb

whichyieldstheMinkowskigaugeΛab=0.Thismaymeanthatweproposeasetofconnectioncomponentsinaspecialframeandcoordinatewhichseemscontrarytothespiritofrelativitytheory.Howeverinphysicallynaturalsituationswecanchooseareferenceandcoordinatesystematourbestconvenience.

Wededucefromequations(32)-(33)

ω01=

Q00=FF′′rG

r1(1 e2,1ω13= 1(1 1re3r(1 1

(35)

whosecomponentsreadexplicitly

Zerothcomponent

Firstcomponent

Secondcomponentothers=0.FGGGWhenweput(35)into(30)weobtain,withτa=0 bc bfcbfcdq+2ωf∧q+qf∧q∧(ea∧eb∧ec)=0 2(G 1)′ r2G22F′r2G2rFG2

F′+(G 1)′ e1∧e2∧e3=0e0∧e2∧e3=0e0,q12= 1)e2,q13=1r 1)e3,(36)(37)(38) (F′G 1)′

+FGrGe0∧e1∧e2=0.(40)


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