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The time independent spherically symmetric solution of the Einstein equation revisited
Jean-Pierre Petit, Gilles d’Agostini and Sebastien Michea +
Manaty Research group
_________________________________________________________________________________________ Key words: Non contractible hypersurface. Throat sphere, space bridge. Sphericallysymmetricsolution.____________________________________________________________________________________________________
Abstract: Spherical symmetry does not immediately mean central symmetry. The timeindependent, spherically symmetric solution of the homogeneous Einstein equation isrevisitedwithcoordinateswhichkeepthesignature invariantandprevent timeandradialcoordinateinterchange.Theassociatedhypersurfaceisnotcontractibleandcorrespondstoa space bridge linking two Minkowski spacetimes though a throat sphere. As thedeterminant of the metric vanishes on that sphere one gets an orbifold structure. WhencrossingthatspheretheparticlesexperienceaPT,massandenergyinversions.____________________________________________________________________________________________________
Introductionandmainideaofthisarticle.In 1916 Karl Schwarzschild publishes [1] a solution of the vacuum Einstein equations(withoutsecondterm)correspondtotimetranslationinvarianceandsphericalsymmetry.Itis only in 1999 that an English translation of this article will be available [2] thanks toS.AntocietA.Loinger.Schwarzschilddecidestoexpressthis solutionbyusingafirstsetofrealvariables(a) t , x , y , z{ }∈!4
Thesolutionisthengivenintheform:(b) ds2 = Fdt2 − G (dx2 +dy2 +dz2 ) − H( xdx + ydy + zdz )2 Hethenintroducesanintermediaryvariable:(c) r = x2 + y2 + z2 Which,given(a)isessentiallypositive.He performs a new coordinates change that allows him to simply express the sphericalsymmetryhypothesis:(d) x = r sinθ cosϕ y = r sinθ sinϕ z = r cosθ
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Hethenobainstheform:(e) ds2 = Fdt2 − (G + H r2 ) dr2 − G r2 (dθ 2 +sin2θ dϕ 2 ) Inorderforthemetrictobelorentzianatinfinityitisnecessarythat:(f) r →∞ implies F→1, G →1 , H → 0 Heintroducesnextanewchangeofcoordinates:
(g) x1 =
r3
3, x2 = − cosθ , x3 =ϕ , x4 = t
Notethat(a)+(c)implythat x1 ≥ 0 Inthosenewcoordinatesthemetricbecomes:
(h) ds2 = f4 dx4
2 − f1 dx12 − f2
dx22
1− x22 − f3 dx3
2 ( 1− x22 )
where f1 , f2 = f3 , f4 are three functions of x1 ( hence of r ) that must satisfy theconditions:
-Forlarge x1 : f1 =
1r4 = (3x1)− 4/3 , f2 = f3 = r2 = (3x1) 2/3 , f4 = 1
-Thedeterminant f1 f2 f3 f4 = 1
-Themetricmustbeasolutionofthefieldequation.-Exceptfor x1 = 0 theffunctionsmustbecontinuous.
Hiscomputationleadshimto:
(i) f1 =
( 3x1 +α3 )− 4/3
1− α ( 3x1 +α3 )− 1/3 f2= f3 = ( 3x1 +α
3 ) 2/3 f4 = 1− 1− α ( 3x1 +α3 )− 1/3
α beinganintegrationconstant.Using(g)wecanrewritehissolutionas:
(j) ds2 = 1− α
( r3 +α 3 )1/3
⎡
⎣⎢
⎤
⎦⎥ dt2 − r4
( r3 +α 3 ) ( r3 +α 3 )1/3 − α⎡⎣ ⎤⎦dr2 − ( r3 +α 3 )2/3( dθ 2 + sin2θ dϕ 2 )
Thatwecanrewrite:
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(k)
ds2 = gt t dt2 − gr r dr2 − gθθ dθ 2 − gϕϕ dϕ 2
Thecoefficients
gt t , gr r et gθθ arefunctionsoftheonlyvariabler,thequantity gϕϕ beinga
functionofbothrandϕ .Thevariablervariesfrom0toinfinitywhilebeingstrictlypositivebydefinition(c).Letuslimitourselftothecase α > 0 andletconsiderapathcorrespondingtodt=dr=0.Ithasaperimeter:(l) p = 2π ( r3 +α 3 )1/3 Thisperimeterhasaminimalvalue p = 2πα .
Thehypersurfacesolutionisthereforenoncontractible.Thervariablecannotbeconsideredasa«radialdistance».Thishypersurfacedoesnothavea«centre»,theobjectcorrespondingtor=0hencetoàx=y=z=0isnotadotbutasphereofradiusα .WewillnowreplicateScharzschild'scalculationsbyreplacingthex,y,zandtvariablesbyGreekcharacters( ρ ishoweverkeptforalaterpurpose),sothatthereadernotbetemptedtoassimilatethemtodistances(x,yandz)ortime(t)andlosesightofwhattheyreallyare:realnumbers.
RevisitingSchwarzchildcomputation.LetusconsiderthezerosecondmemberEinsteinequation
Rµν = 0 intimeindependentand
sphericallysymmetricalconditions.Let ξ1 ,ξ2 ,ξ3 standforrectangularcoordinates,and ξo
as the time marker, with ( ξ0 , ξ1 , ξ2 , ξ3 )∈ R4 which stands real values for allcoordinates.Inadditionweassumethattherearenocrossedtermsinthelineelement,sothatthislastcanbewritten:(1)ds
2 = Fdξo2 −G (dξ12 + dξ22+ dξ32 )−H (ξ1dξ1 + ξ2dξ2 + ξ3dξ3 )2
whereF,G,Harefunctionsof ξ12 + ξ2
2 + ξ32 .
Atinfinitywemusthave(2) F and G →1 H → 0 Introducethefollowingcoordinatechange:
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(3) ζ = ξ12 + ξ2
2 + ξ32
(4) θ = arcos
ξ3
ζ⎛⎝⎜
⎞⎠⎟
(5)
ϕ = arccosξ1
ξ12 + ξ2
2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
whichgoeswith: ζ ∈ R + θ ∈ R ϕ ∈ R and:
ξ1 =ζ sinθ cosϕ ,ξ2 =ζ sinθ sinϕ ,ξ3 =ζ cosθ
thatwewillcall«pseudosphericalcoordinates».Itgives:(6) ds2 = Fdξo
2 − ( G + Hζ 2 ) dζ 2 − Gζ 2 ( dθ 2 + sin2θ dϕ 2 ) Introducethefollowingadditionalcoordinatechange::
(7)η1 =
ζ 3
3 , η2 = − cosθ , η3 =ϕ
Thenwehavethevolumeelement ζ2sinθ dζ dθdϕ = dξ 1 dξ 2 dξ 3 .Thenewvariablesare
thenpseudopolarcoordinatewith thedeterminant1.Theyhave theevidentadvantageofpolarcoordinatesforthetreatmentoftheproblem.
Inthenewpseudopolarcoordinates:
(8) ds2 = Fdξo
2 − ( Gζ 4 + H
ζ 2 )dη12 − Gζ 2 dη 2
2
1− η 22 + dη3
2 1− η 22( )⎡
⎣⎢⎢
⎤
⎦⎥⎥
forwhichwewrite:
(9)ds 2= f4dξo
2 − f1dη 12 − f2
dη 22
1−η 22 − f3 dη 3
2 (1−η 22)
Thenf1,f2=f3,f4arefunctionsofη1 whichhavetofullfillthefollowingconditions:
1–Forξ1 =∞ : f1 = 1
ζ 4 = (3η1)− 4
3 , f2 = f3 =ζ2= (3η1)
23 , f4 =1
2–Theequationofthedeterminant: f1 . f2 . f3 . f4 = 1
3–Thefieldequations.
4–Continuityofthef,exceptforη1 =0
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In order to formulate the field equations one must first form the components of thegravitational fieldcorrespondingtothe lineelement(9).Thishappensinthesimplestwaywhen one builds the differential equation of the geodesic line by direct execution of thevariation,andreadsoutthecomponentsofthese.Thedifferentialequationsofthegeodesiclineforthelineelement(9)immediatlyresultfromthevariationintheform:
(10)
0= f1
d 2η 1d s 2
+ 12∂ f4∂η1
dη4d s
⎛
⎝⎜⎞
⎠⎟
2
+ 12∂ f1∂η1
dη1d s
⎛
⎝⎜⎞
⎠⎟
2
− 12∂ f2∂η1
11−η22
dη2d s
⎛
⎝⎜⎞
⎠⎟
2
+ (1−η22)dη3d s
⎛
⎝⎜⎞
⎠⎟
2⎡
⎣
⎢⎢
⎤
⎦
⎥⎥
(11)
0= f2
1−η22d 2η2
d s 2+∂ f2∂η1
11−η12
dη1d s
dη2d s
+f2η2
(1−η12)2dη2d s
⎛
⎝⎜⎞
⎠⎟
2
+ f2η2dη3d s
⎛
⎝⎜⎞
⎠⎟
2
(12)
0= f2(1−η22)
d 2η3
d s 2+∂ f2∂η1
(1−η22)dη1d s
dη3d s
−2 f2η2dη2d s
dη3d s
(13)
0= f4
d 2η4
d s 2+∂ f4∂η1
dη1d s
dη4d s
Thecomparisonwith
(14)
d 2ηα
ds 2= − 12 Γµν
α
µ ,ν∑
dηµ
d sdην
d s
givesthecomponentsofthegravitationalfield:
(15a)Γ1 1
1 = − 12∂ f1∂η1
(15b)Γ2 2
1 = + 121f1
∂ f2∂η1
11−η22
(15c)Γ3 3
1 = + 121f1
∂ f2∂η1
(1−η22)
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(15d)Γ2 1
2 = − 121f2
∂ f2∂η1
(15e)Γ2 2
2 = −η2
1−η22
(15f)Γ3 32 = −η2 (1−η22)
(15g)Γ3 1
1 = − 121f2
∂ f2∂η1
(15h)Γ3 2
2 =η2
1−η22
(15i)Γ4 1
4 = − 121f4
∂ f4∂η1
The other ones are zero. Due to rotational symmetry it is sufficient to write the fieldequations only for the equator (η2 =0) , therefore, since they will be differentiated onlyonce, in the previous expressions it is possible to set everywhere since the begining
1−η22 =0 .Thenthecalculationofthefieldequationgives:
(16a)∂∂η1
1f1
∂ f1∂η1
⎛
⎝⎜⎞
⎠⎟= 12
1f1
∂ f1∂η1
⎛
⎝⎜⎞
⎠⎟
2
+ 1f2
∂ f2∂η1
⎛
⎝⎜⎞
⎠⎟
2
+ 1f4
∂ f4∂η1
⎛
⎝⎜⎞
⎠⎟
2
(16b)∂∂η1
1f1
∂ f2∂η1
⎛
⎝⎜⎞
⎠⎟=2+ 1
f1 f2
∂ f2∂η1
⎛
⎝⎜⎞
⎠⎟
2
(16c)∂∂η1
1f1
∂ f4∂η1
⎛
⎝⎜⎞
⎠⎟= 1f1 f4
∂ f4∂η1
⎛
⎝⎜⎞
⎠⎟
2
Besides these three equations the functions f1 , f2 , f3 must fullfill the equation of thedeterminant:
(17)f1 f2
2 f4 =1 i.e. 1f1
∂ f1∂η1
+ 2f2
∂ f2∂η1
+ 1f4
∂ f4∂η1
=0
Fornowweneglect(16b)anddeterminethethreefunctions f1 , f2 , f4 from(16a),(16c)and(13).Theequation(16c)canbetransposedintotheform:
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(18)∂∂η1
1f4
∂ f4∂η1
⎛
⎝⎜⎞
⎠⎟= 1f1 f4
∂ f1∂η1
∂ f4∂η1
Thiscanbeintegratedandgives
(19)1f4
∂ f4∂η1
= α f1 (α beinganintegrationconstant)
Theadditionof(12a)and(12c’)gives:
(20)∂∂η1
1f1
∂ f1∂η1
+ 1f4
∂ f4∂η1
⎛
⎝⎜⎞
⎠⎟= 1
f2
∂ f2∂η1
⎛
⎝⎜⎞
⎠⎟
2
+ 121f1
∂ f1∂η1
+ 1f4
∂ f4∂η1
⎛
⎝⎜⎞
⎠⎟
2
Bytaking(17)intoaccountweget:
(21)−2 ∂
∂η1
1f2
∂ f2∂η1
⎛
⎝⎜⎞
⎠⎟= 3 1
f2
∂ f2∂η1
⎛
⎝⎜⎞
⎠⎟
2
Byintegrating:
(22)
11f2
∂ f2∂η1
= 32η1 +
σ2 (σ integrationconstant)
or:
(23)1f2
∂ f2∂η1
= 13η1 +σ
Afterasecondintegration:
(24)f2 = λ (3η1 +σ )23 (λ integrationconstant)
Theconditionatinfinityrequiresλ =1.Then
(19)f2 = (3η1 +σ )23
Henceitresultsfurtherfrom(22)and(17)
(20)
∂ f4∂η1
= α f1 f4 =αf22 =
α
(3η1 +σ )43
Byintegrating,whiletakingaccounttheconditionatinfinity
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(25) f4 = 1−α (3η1 +σ )−13
Hence,from(17)
(26)
f1 =
(3η1 +σ )− 4
3
1−α (3η1 +σ )−13
As it canbe easily verified the equation (16b) is automatically fullfilledby the expressionthatwefoundforf1andf2
Thereforealltheconditionsaresatisfiedexcepttheconditionofcontinuity.
f1willbediscontinuouswhen1=α (3η1 +α3 )
−13 , 3η1 =α 3 −σ
Inorderthisdisconinuitycoincideswiththeorigin,itmustbe:
(27)σ =α 3
Therefore the condition of continuity relates in thisway the two integration constants σ andα .Wehave:
(28) 3η1 +α = ζ 3 +α
(29) f4 = 1− α
(ζ 3+α 3 )1/3
(30) f2 = (ζ 3+α 3 )2/3
(31)
dη 22
1− η22 = dθ 2
Finally:
(32)
ds2 = 1− α(ζ 3+α 3 )1/3
⎛⎝⎜
⎞⎠⎟
dξo2 − ζ 4
α 3 +ζ 3( ) α 3 +ζ 3( )1/3−α⎡
⎣⎢⎤⎦⎥
dζ 2 − α 3 +ζ 3( )2/3dθ 2 + sin2θ dφ 2( )
Whenζ → ∞ thelineelementtendstoLorentzform.
gt t tendstozerowhenζ tendstozero.
When ζ → 0 gζ ζ !
00.Anexpansionintoaseriesshowsthat
gζ ζ !
3ζα
→ 0
Consideralooplocatedintheplane θ = 0 ,with dξ0 = 0 .Itisnocontractile:forζ =0the
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perimeterofthelooptendsto 2πα .
Notice that ζ = ξ12 + ξ2
2 + ξ32 is definitely not a «radius». The point ζ = 0 does not
correspond to some «center of symmetry». The spherically symmetric requirement doesnotidentifyautomaticallytoacentralsymmetry,assuggestedbyDavidHilbert[5]1ζ isjustoneofthe«spacemarkers»,nothingelse.It’sanumber,notalength.Theonlylengthtobeconsideredisthequantitys.
Expressingthemetricinabettercoordinatesystem.
Letusconsiderthecoordinatesystemintroducedin[3].
Introducethenewspacemarker ρ throughthefollowingcoordinatechange:
(33) ζ = α 1+ Logcos hρ( )
(34) ρ = ± argch ( e1+ ζ
α
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
3
3 −1
)
ζ = 0 → ρ = 0
With ξo = c t thelineelementbecomes:
(35)
ds2 = Logcos hρ
1+ Logcos hρc2dt2 − 1+ Logcos hρ
Logcos hρα 2 tan h 2ρ dρ2 − α 2 (1+ Logcos hρ )2 ( dθ 2 + sin2θ dϕ 2 )
When ρ → ±∞ wegetthefollowingLorentzform:
(36) ds2 = c2dt2 − α 2 dρ2 − α 2 ρ2 ( dθ 2 + sin2θ dϕ 2 )
Wecan figure this geometricalobjet as a spacebridge linking twoMinkowski spacetimes,though a throat sphere whose perimeter is 2πα . We cannot think about its «radius»becausethatspherehasnocenter.
When ρ→ 0
1Wequote,page67ofthegermanedition:«DieGravitation
gµν istzentrischsymmetrischinBezugaufdenKooridinatenanfangspunkt.«
Englishtranslation:«Thegravitation gµν iscentrallysymmetricwithrespecttotheoriginof
coordinates.»
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(37) gt t =
Logcos hρ1+ Logcos hρ
c2 → 0
(38) gρ ρ = − 1+ Logcos hρ
Logcos hρα 2 tan h 2ρ → 0
0
(39) gθθ → α 2 gϕϕ → α 2 sin2θ
(40) gϕϕ =α 2 sin2θ
Wemayeasilyovercometheindetermination(33)throughanexpansionintoaseries,whichshowsthatwhen ρ → 0
(41) gρ ρ = − 1+ Logcos hρ
Logcos hρα 2 tan h 2ρ → − 2
Thedeterminantis:
(42) det gµν = − c2α 6 tan h2ρ sin2θ
Itvanisheson the throatsphere.Asaconsequence,on this lastwecannotdefinegaussiancoordinates,sothattheobjectisnolongeramanifoldbutanorbifold.Onthethroatspherethearrowoftimeandthespaceorientationcannotnedefined.Thiscanbeinterpretedasageometricstructurewherespaceandtimearereversedthroughthethroatsphere:whenaparticle crosses the throat sphere it experiences a PT-symmetry. According to Souriau’stheorem[4]thisT-inversiongoeswithamassinversion.
Inafuturepaperthephysicalinterpretationofsuchsolutionwillbeinvestigated.
References:
[1] K. Schwarzschild : Über das Gravitionalsfeld einer Kugel Aus incompressibler Flüssigkeit nach der Einsteinschen Theorie. Sitzung der phys. Math. Klasse v.23 märz 1916 [2] K. Schwarzschild : On the gravitational field of a sphere of incompressible fluid according to Einstein theory. Translation by S.Antoci and A.Loinger. arXiv :physics/9905030v1 [physics.hist-ph] 12 may 1999.
[3]J.P.Petit&G.D’Agostini:CancellationofthesingularityoftheSchwarzschildsolutionwithnaturalmassinversionprocess.Mod.Phys.Lett.Avol.30n°92015
[4]J.M.Souriau:Structuredessystèmesdynamiques.DunodEd.France,1970andStructureof Dynamical Systems. Boston, Birkhaüser Ed. 1997. For time inversion see page 190equation(14.67).
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[5] D.Hilbert : “DieGrundlagender Physij. (ZweiteMitteilung)” inNachchrichten vonderKöniglichenGesellschaftzuGöttingen.Math.-Phys.Klasse.1917,p.53-76.Presentedinthesessionof23december1916