The charged dust solution of Ruban -- matching to Reissner--Nordström and shell crossings

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by  Andrzej Krasiński

Noname manuscript No.
(will be inserted by the editor)
The charged dust solution of Ruban –
matching to Reissner–Nordstr¨om and shell crossings
Andrzej Krasin´ski · Gabriel Giono
thedateofreceiptandacceptance shouldbeinsertedlater
2 Abstract Themaximallyextended Reissner–Nordstro¨m(RN)manifoldwith e2 <m2 begs forattachinga
1 materialsourcetoitthatwouldpreservetheinfinitechainofasymptoticallyflatregionsandevolvethrough
0 the wormholebetween the RN singularities.So far, the attemptswerediscouraging.Here we try one more
2
possible source – a solution found by Ruban in 1972 that is a charged generalisationof an inhomogeneous

n Kantowski–Sachs-typedustsolution.ItcanbematchedtotheRNsolution,andthematchingsurfacemust
a stayallthetimebetweenthetwoRNeventhorizons.However,shellcrossingsdonotallowevenhalfacycle
J
of oscillationbetween the maximaland the minimalsize.

2

]
c
q 1 Motivation
-
r
Avoidingthe Big Bang singularityin cosmologicalmodels had been a recurring idea in the literature.The
g
[ hopes for constructing a model without singularity were largely dashed by the singularity theorems of

Hawking and Penrose (see Ref. [1] for a review). They were temporarily revived by the finding of Vickers
3
[2] that in a charged dust model generalising that of Lemaˆıtre [3] and Tolman [4] (LT) the Big Bang
v
can be prevented by the charge distribution,1 provided that the absolute value of the charge density is, in
7
9 geometricalunits,smaller thanthemassdensity.Thisfindingcreatedanotherexpectation,thatafiniteball
8 of charged dust matched to the Reissner [6] – Nordstr¨om [7] (RN) solution would be able to collapse and
4 bouncethroughthewormholeofthemaximallyextendedRNmanifold,therebygivingphysicalmeaningto
. the infinite chain of black holes and asymptoticallyflat regions.2 However, this renewed hope was dashed
7
0 againin1991byOri[12]whoprovedthat,underexactlythesameconditionsthatpreventtheBigBang,shell
1 crossings will inevitably appear and destroy the dust ball before it enters the wormhole. Then, Krasin´ski
1 and Bolejko [13,14] found a gap in Ori’s assumptions and tried to improve upon his result. Namely, Ori
v: assumed that the ratio of the absolute value of the charge density to the mass density (call this ratio α)
i is smallerthan 1 everywhere, includingthe centre of symmetry.Krasin´ski and Bolejkoconsidered the case
X
when α 1 at the centre, whilebeing <1 everywhereelse. The situationturned out to be better, but not
→
r definitively. With initial conditions carefully tuned, the charged dust ball could go through the wormhole
a
justonce,beingdestroyedbyshellcrossingssoonafterreachingthemaximalsizeinthenextasymptotically
flat region. Moreover, a direction-dependent singularity necessarily appeared at the centre of the ball at
the instant of minimalsize. This is not a satisfactorysituation,but the best result achieved so far.
AndrzejKrasin´ski
N.CopernicusAstronomicalCentre,PolishAcademyofSciences,Bartycka18,00716Warszawa, Poland,
email:[email protected]
·GabrielGiono
D´epartementdePhysique,Universit´eClaudeBernardLyon1,14rueEnricoFermi69622Villeurbanne,France,
email:[email protected]
1 This charged generalisation of the LT model was first found as a solution of the Einstein – Maxwell equations by
MarkovandFrolovin1970[5].
2 TheextensionoftheRNsolutioncomposedoftwocoordinatepatcheswasfirstcalculatedbyGravesandBrillin1960
[8].Theinfinite mosaicof conformal diagramsshown inFig.1firstappeared inthe paper byCarter [9]in1966, and was
describedindetailinareviewarticlebyCarter[10].SeeRef.[11]foranother pedagogical introduction.
In an attempt to find a better model, we now tried to match the Ruban [15] charged dust solution
to the RN metric and see what results. The charged Ruban solution is a generalisationto nonzero charge
of the nonstatic dust solution investigated earlier by Ruban [16,17], which in turn is an inhomogeneous
generalisation of the Kantowski – Sachs model [18].3 The matching of these two solutions is very easily
achieved, with an interesting result. The Ruban solution represents a pulsating charged dust ball whose
outer surface must remain foreverbetween the twoevent horizons of the RN solution.It touches the inner
eventhorizonatitsminimalsizeandtheouterhorizonatitsmaximalsize.However,asimpleinvestigation
showsthat shellcrossingsareinevitableinsidetheball and do not alloweven a half cycleof an oscillation,
appearingbetweenthemaximumandminimumsizebothintheexpansionphaseandinthecollapsephase.
Thus, the field is still open for finding a physical nonstatic source for the RN solution that would
proceed through the wormhole. The next thing to do is to find a charged perfect fluid solution of the
Einstein – Maxwell equations in which pressure gradients would prevent the formation of shell crossings.
This, however,promises to be an extremely difficult task.
Inthispaper,wefirstpresenttheRNsolutionincoordinatesadaptedtothematching,thenthecharged
Rubansolutionasfoundbytheauthor,thenweprovethatthematchingcanbedone,andfinallyweprove
that shell crossings in the Ruban solution are inevitable. We also investigate the limit of zero charge of
the Ruban – RN configuration. Then the dust source is matched to the Schwarzschild solution across a
hypersurfacethatremains insidethe Schwarzschildhorizon,only touchingit frominside at themomentof
maximalexpansion.Shellcrossingscanthenbeavoidedbyanappropriatechoiceofthearbitraryfunctions
in the Ruban model. However, this model has a finite time of existence, being born in a Big Bang that
matches to the past Kruskal [21] – Szekeres [22] singularity and crushing into a Big Crunch that matches
to the future KS singularity.
Weusethesignature(+ ).Wedonotassume“unitsinwhichG=c=1”;whenevertheseconstants
−−−
are absent, this means they were absorbed into other symbols. For example, our time coordinate will be
t =c [the physical time], our M will denote (G/c2) [the mass], etc. The labelling of the coordinates is
× ×
(x0,x1,x2,x3)=(t,r,ϑ,ϕ).
2 2
2 The Reissner–Nordstr¨om solution with e <m and its maximal extension
The Reissner [6] – Nordstr¨om[7] solutionin its standard form is
ds2 =Fdt2 (1/F)dr2 r2 dϑ2+sin2ϑdϕ2 (2.1)
− −
(cid:16) (cid:17)
where
F(r) d=ef 1 2m + e2 + 1Λr2 (2.2)
− r r2 3
(weincludeΛforawhile).Forreference,Fig.1(adaptedfromRef.[11])showstheCarter–Penrosediagram
ofthemaximalanalyticextensionoftheunderlyingspacetimeforthecaseΛ=0ande2<m2,towhichwe
will limit our attention in the main part of this paper. The ragged lines in the diagram are the curvature
singularitiesat r=0, the lines marked “inf” are the past and future null infinities, r+ are the outer event
horizons at
r=r+ =m+ m2 e2, (2.3)
−
and r− are the inner event horizons at p
r=r− =m m2 e2. (2.4)
− −
These two values of r are zeros of the function F whenpΛ = 0. Regions I and III are asymptotically flat,
regions II and IV are contained between the two horizons. The thin curved lines are those on which r
is constant; they are timelike in the asymptotically flat regions where r > r+ and inside the inner event
horizon where r<r−; between the horizons where r− <r<r+ they are spacelike.
In the following we will be interested in the region between the horizons, where F < 0, so r in (2.1)
becomes the time coordinate and t becomes a spacelike coordinate. It is easy to verify that the lines on
which (t,ϑ,ϕ) are all constant are geodesics.
For later convenience, we rename the coordinates in (2.1) as follows
(t,r)=(ρ,R), (2.5)
3 TheneutraldustsolutioninvestigatedbyRubanwasfirstfoundasasolutionoftheEinsteinequationsbyDattin1938
[19],butinstantlydismissedasbeingof“littlephysicalsignificance”.
2
rr--
IIII
iinnff
IIIIII rr++ II
iinnff
IIVV
Fig. 1 The conformal diagram of the maximally extended Reissner–Nordstr¨om spacetime with Λ = 0 and e2 <m2. It
can either beinterpreted as an infinite chainof copies of the segment withinthe rectangle, or one can identifythe upper
sideoftherectanglewiththelowerside,obtainingamanifoldwithclosedtimelikeandnulllines.Moreexplanationinthe
text.
and then transform the R-coordinatein the region F <0 to τ(R) defined by
dR 2 2m e2 1
= F = 1+ ΛR2. (2.6)
dτ − − R − R2 − 3
(cid:18) (cid:19)
After this, the metric in this region becomes
ds2 =dτ2 [ F(R(τ))]dρ2 R2(τ) dϑ2+sin2ϑdϕ2 , (2.7)
− − −
(cid:16) (cid:17)
with R(τ) being defined by (2.6).
When Λ=0, eq. (2.6) can be integratedto give the followingexplicit function τ(R):
dR R m
τ τ0 =µ = µ R2+2mR e2+µmarcsin − , (2.8)
− Z 1+ 2m e2 − p− − (cid:18)√m2−e2(cid:19)
− R − R2
r
3
where µ= 1 and τ0 is an arbitraryconstant.
±
For some purposes it is more convenient to introduce the parameter η by (R m)/√m2 e2 =sin(η
− − −
π/2) cosη, and then
≡−
R=m m2 e2cosη, τ τ0 =µ mη m2 e2sinη , (2.9)
− − − − −
p (cid:16) p (cid:17)
where τ0 = τ0 µmπ/2. This shows that as τ increasees, R is oscillating between the r− and r+ given by
−
(2.3) and (2.4).
e
3 The charged Ruban solution [15]
To derive this solution, we assume comoving coordinates, spherical symmetry and a charged dust source.
For the metric we assume a less general form than spherical symmetry would allow,namely
ds2 =eC(t,r)dt2 eA(t,r)dr2 R2(t) dϑ2+sin2(ϑ)dϕ2 , (3.1)
− −
h i
where C(t,r), A(t,r) and R(t) are functions to be found from the Einstein – Maxwell equations. The
limitation of generality is in R(t); in the most general case it would depend on r as well and the field
equations would lead to a charged generalisationof the LT model.
DetailsofthecalculationaregiveninRef.[11].ThefieldequationsshowthatR,tC,r=0.WithR,t=0,
we obtain an (electro-) vacuum solution, which is presented in Appendix A – it is a coordinate transform
ofthewell-knownRobinsonsolution[20].ThusC,r=0,whichmeansthatthedustis movingongeodesics.
A transformationof t can then be used to achieve C =0. The other equations imply
2M Q2 1
R,t2 =−1+ R − R2 − 3ΛR2, (3.2)
whereM is a constantof integration,and Q is theelectricchargewithinthesphereof coordinateradiusr.
Since all other quantities in this equation are independent of r, it follows that Q must be constant. This
implies that the charge density is zero everywhere. Thus, the dust particles must be neutral, they only
movein an exteriorelectric field.
For the other metric function we obtain the followingsolution:
dt
eA/2 =R,t(cid:20)X(r)Z RR,t2 +Y(r)(cid:21), (3.3)
where X(r) and Y(r) are arbitraryfunctions, and the expression for matter density in energy units is
2X
κǫ= . (3.4)
R2eA/2
Equations (3.1) – (3.4) define the charged Ruban solution, first semi-published in 1972 [15], and then
mentioned in a later paper [23]. Note that (3.2) is identical to (2.6), so the function R(t) defined by (3.2)
will be the same as R(τ) defined by (2.6). Note also the following:
1.WithX =0,eq.(3.4)showsthattheRubansolutionbecomeselectro-vacuum;infactitthenbecomes
the Reissner – Nordstr¨om solutionexpressed in the (τ,ρ) coordinates defined by (2.5) – (2.6).
2. When Y =BX, where B is a constant (B =0 being allowed),the r-dependence in (3.4) cancels out
′
and by defining a new r by r = X(r)dr we make also the metric independent of r. The spacetime then
becomes spatiallyhomogeneouswiththeKantowski– Sachssymmetry;infactitis then thegeneralisation
R
of the Kantowski– Sachs solution to nonzero chargeand cosmologicalconstant.
3. The geometry of a 3-space t= const in (3.1) is that of a 3-dimensional cylinder whose sections r =
const are spheres, all of the same radius, and the coordinate r measures the position along the generator.
Thespace is inhomogeneousalongther-direction,and theelectricfield has its only componentalso in the
r-direction.The radius R of the cylinder evolves with time according to Eq. (3.2).
4.Thematterdensityinthissolution,givenby(3.4),depends onr andiseverywherepositiveifX >0.
Thus,theamountofrestmasscontainedinsideaspherer=r0 =constdoesdependonthevalueofr0,and
is an increasingfunctionof r.Nevertheless,as seen from (3.2),theactivegravitationalmass M thatdrives
the evolution is constant. Ruban [17] interpreted this property as follows: the gravitationalmass defect of
any matter added exactly cancels its contributionto the activemass.
4
4 Matching the Ruban and RN solutions
To match two spacetime regions to each other we have to prove that on the hypersurface H forming the
borderbetweenthemboth4-dimensionalmetricsinducethesame3-dimensionalmetricandthesamesecond
fundamentalform. The formulaewe use here are derived in Ref. [11].
We will show that the RN metric in the coordinates of (2.6) – (2.7) and the Ruban metric defined by
(3.1) – (3.3) withC =0 can be matchedalonganyH of constantρ=r. On each such H the3-metricswill
be identical when τ and t are identified and
m=M, e=Q, (4.1)
since then the R(τ)of (2.7) and the R(t)of (3.1) willobey the sameequation,(2.6) and (3.2) respectively.
The coincidence of the second fundamentalforms requires that at H [11]
1 dhij 1 dhij
= , (4.2)
dr dρ
N (cid:12)Ruban N (cid:12)RN
(cid:12) (cid:12)
wherei,j =0,2,3and isther-componentof(cid:12)(cid:12)theunitnormalv(cid:12)(cid:12)ectortoH,thus grr 2 =1oneachside
N | |N
of H. But the relevantcomponents of the Ruban metricdo not depend on r, and the relevant components
oftheRNmetricdonotdependonρ,so(4.2)isfulfilledinatrivialway:itreducesto0=0.Note,however,
that the matching is possible only in that region of the RN manifold where F <0 in (2.1) – (2.2). In the
subcase Λ=0, this is the region between the two event horizons.
Equation (2.9) applies also in the Ruban region, with m = M and e = Q. It is independent of r, so
each constant-r shell evolves by the same law. This means that all shells, including the outer surface of
the Ruban region, oscillate between R = r− and R = r+, where r− and r+ are given by (2.3) – (2.4). If
the solution could be continued to infinite values of t, the conformal diagram of the complete manifold
would look as in Fig. 2. However, shell crossings make the completion of even half a cycle of oscillations
impossible,see next section.
5 Shell crossings in the charged Ruban solution are inevitable
For the rest of this articlewe assume Λ=0. In this case the Ruban metric is
ds2 =dt2 eA(t,r)dr2 R2(t) dϑ2+sin2ϑdϕ2 , (5.1)
− −
(cid:16) (cid:17)
where the solution for R(t) can be written as (cf. eq. (2.9)):
R=M M2 Q2cosη, t t0 =µ Mη M2 Q2sinη , (5.2)
− − − − −
p (cid:16) p (cid:17)
whereµ=+1 intheexpansionphaseandµ= 1in thecollapsephase.WithΛ=0,thefunctioneA/2 can
−
be calculated explicitly.For this purpose, we find from (3.2)
2M Q2
R,t=µ −1+ R − R2. (5.3)
r
Note that this solution exists only when Q2 < M2.4 Now we use this in the integral in (3.3) rewritten as
follows
dt dR
. (5.4)
Z RR,t2 ≡Z RR,t3
Then we obtain
1 M/R 2M Q2 R M
eA/2 =X(r) 2+Q2 − 1+ arcsin −
" M2−Q2 −r− R − R2 M2−Q2!#
2M Q2 p
+µY(r) 1+ . (5.5)
− R − R2
r
4 With Q2 =M2, Eq. (5.3)has the static solution R=M =|Q|.As remarkedabove (3.2), this leads to the Robinson
solution[20];seeAppendixA.
5
Fig. 2 The maximally extended Reissner–Nordstr¨om spacetime with a hypothetical charged matter source matched to
it.Theextentofthesourceinthisdiagramisunknown,evenifitisthechargedRubansolution–becauseofthearbitrary
functionsitcontains.Thisiswhytheleftedgeisschematicallymarkedwithadashedline.Thetwolongwavycurvesshow
possiblematchingsattwodifferentvaluesoftheradialcoordinate;thiscoordinateistherof(5.1)andtheρof(2.7).Note:
these are the curves on which r and ρ are constant. It is R that pulsates between the r− and r+ of (2.3) and (2.4), and
itchanges along thesecurves.ThelinesofconstantRarethehorizontalhyperbola-likearches.Ifthematter sourceisthe
chargedRubansolution,thenshellcrossingsappear withineachhalf-cycleofoscillation.Thepositionsoftwoofthemare
schematicallymarkedwiththinhorizontal straightlines.
A shell crossing is where eA/2 = 0, because at such a location two matter shells whose r-coordinates
differ by dr are at a zero distance, as seen from (5.1), and stick together.From (3.4) one sees that at such
a location the mass density becomes infinite, so this is a curvature singularity. The question we seek to
answerisnow:canthefunctionsX(r)andY(r)in(5.5)bechoseninsuchawaythateA/2 =0everywhere.
6
Unfortunately, the answer given by the formulae above is a definitive “no”, i.e. shell crossings are
inevitable.To see this, let us first recall that R changes between the values
R− =M M2 Q2, and R+ =M+ M2 Q2 (5.6)
− − −
p p
6
(this is a copy of (2.3) – (2.4) rewritten in the notation for the Ruban model). At both these values the
expression under the square root in (5.5) is zero, and in the whole range 0 < R− < R < R+ the function
eA/2 given by (5.5) is continuous. Now, at R=R− we have
eA/2 = X(r) R− , (5.7)
R=R− − M2 Q2
(cid:12) −
(cid:12)
while at R=R+ we have (cid:12) p
eA/2 =X(r) R+ . (5.8)
R=R+ M2 Q2
(cid:12) −
(cid:12)
Thus, whatever the sign of X(r), the sign(cid:12)s of eA/2 at tphe ends of the range [R−,R+] are opposite, i.e.
eA/2 = 0 for some value of R within this range. This is a shell crossing. Note that it appears every time
when R traversesthis range, which means that theshell crossingswillnot allowR to go througheven half
a cycle of oscillation.
The geometrical nature of this shell crossing is different than in the LT and Szekeres models [24,25],
wheretheyhavebeeninvestigatedquitethoroughly[26,27,28].In theLTandSzekeresmodels,thespheres
thatcollideareone withintheother.At a shell crossing,the smallersphere catches up withthelargerone
duringexpansion(orviceversaduringcollapse).IntheRubanmodel,thespheresaresurfacesofconstantr
inthe3-dimensionalcylindert=constant,andtheymoveupanddownalongthegeneratorsasthecylinder
expandsorcollapses.It turnsouttheywillcollidebeforethecylindermanagestoproceedallthewayfrom
the minimal radius to maximal, or from maximal to minimal. One more property of such a shell crossing
is noteworthy: its locus does not depend on r, which means that all the spheres with different values of r
collideat the same moment.
KantowskiandSachs [18] notedthepossibleoccurrenceofthis singularityin thesubcaseQ=Y(r)=0
of the Ruban model, where they said “Thesingularitiesfor the closed models areof twokinds. (...)In one
kind of a singularity the cylinder squashes to a disk, in the other it contracts to a line.” What we called
shell crossing here is the “squashing to a disk”.
The shell crossings could possibly be prevented by pressure with a nonzero gradient in the r-direction,
but such exact solutions are not yet known. Then, assuming that the pressure and its gradient would be
zeroatthesurfaceofthechargedfluidball,itssurfacewouldfollowatimelikegeodesicintheRNspacetime,
and the conformaldiagram could really look likeFig. 2.
6 Absence of shell crossings in the Ruban solution with zero charge
When Q = 0, the charged Ruban solution goes over into the Datt – Ruban neutral dust solution [19,17,
11]. Then from (2.9), adapted to the notationfor the Ruban solution,we obtain
R=M(1 cosη), t t0 =µM(η sinη). (6.1)
− − −
This shows that now R starts from 0 at t = t0, then increases to 2M at t = t0+µMπ, and decreases to
0 again at t = t0+2µMπ. At R = 0 the model has a Big Bang/Crunch type singularity. Equation (5.5)
simplifies to5
eA/2 = 1+ 2M X(r) 2 arcsin R 1 +µY(r) d=ef 1+ 2M (t,r). (6.2)
r− R   −1+ 2RM − (cid:18)M − (cid:19)  r− R F
q 
Now the spacetime is singular at R = 0, so the model exists only for the finite time T = 2πM between
the Big Bang and Big Crunch, but, unlikein the charged case, the functions X(r) and Y(r) can be chosen
so that within this interval there are no shell crossings. We show below that, with an appropriate choice,
>0 during the whole evolution.
F
Namely, with R=0 at the start of the expansion phase, where µ=+1, we have
π
=X +Y. (6.3)
F|R=0 2
5 Equation (6.2) is equivalent to (19.101) in Ref. [11] under the renaming Y = µ(Y −π/2), in virtue of the identity
arcsin(R/M−1)+π/2≡2arcsin R/(2M). e
p
7
Thus, X >0 and Y >0 will guaranteethat >0. We also have
F|R=0,µ=+1
d X
F = , (6.4)
dR R 2M 1 3/2
R −
which will be positive for all 0 R < 2M if X >(cid:0)0, becom(cid:1)ing + as R 2M. Since also + as
≤ ∞ → F → ∞
R 2M, the switch from expansion to collapse, i.e. from µ = +1 to µ = 1, will keep positive at the
→ − F
beginning of collapse. So we only have to guarantee that > 0 at the end of the collapse phase, when
F
R 0 again.This will be the case when
→
π
F|R=0,µ=−1 =X2 −Y >0. (6.5)
Consequently,both (6.3) and (6.5) will be positiveduring the wholecycle when
π
0<Y <X (6.6)
2
at all values of r. The fact that the shell crossings become absent when Q = 0 shows that the limiting
transitionQ 0 is discontinuous;just as it was in the RN Schwarzschildtransition.
→ →
TheneutralRuban solutioncan be matchedto the Schwarzschildsolution,as shown by Ruban himself
[17]. The matchinghypersurfacestays withinthe Schwarzschildradius except at R=2M, when it touches
the horizon from inside. However, this configurationhas a finite time of existence, just as the Kruskal [21]
– Szekeres [22] manifold between its two singularities.
7 Summary
We investigated the charged Ruban solution [15] as a possible matter source for the maximally extended
Reissner – Nordstr¨om solution with e2 <m2. The matching of these two solutions is easily achieved. The
hypersurface that forms the boundary between the Ruban and RN regions stays all the time between the
twoRN event horizons,touchingthe inner oneat its minimalradius and the outerone at maximalradius.
However, shell crossings make the completion of even half a cycle of such an oscillation impossible; they
appear between each [r−,r+] and each [r+,r−] pair.
Shell crossings can be avoided in the limit of zero charge, when the arbitrary functions in the Ruban
solution obey a simple inequality. Then the Ruban solution can be matched to the Schwarzschild solution
inside the event horizon, but the model has a finite time of existence.
TheproblemoffindingamattersourcetothemaximallyextendedRNsolutionthusstillremainsopen,
with one more possibilitybeing now elliminated.
2 2
A The limiting case Q = M ,Λ=0
As stated inthe footnote below (5.3), this leads to the static solution R=M =|Q|.This limitisnot asubcase of (3.3),
and the Einstein –Maxwell equations have to besolved anew. With R,t=0 they lead to the electro-vacuum solution, in
which
eA/2=α(r)cos(t/M)+β(r)sin(t/M), (A.1)
whereα(r)andβ(r)arearbitraryfunctions,andthemetricis
ds2=dt2−eAdr2−M2 dϑ2+sin2ϑdϕ2 (A.2)
(cid:0) (cid:1)
(werecallthat M =|Q|,whereQistheelectriccharge). Thisshouldbecomparedtothe Nariaisolution[29]intheform
foundbyKrasin´skiandPleban´ski[30]:
ds2=[a(t)cos(ρ/ℓ)+b(t)sin(ρ/ℓ)]2dt2−dρ2−ℓ2 dϑ2+sin2ϑdϕ2 , (A.3)
(cid:0) (cid:1)
where ℓ is an arbitrary nonzero constant. This is a vacuum solution with the cosmological constant Λ = 1/ℓ2. Equation
(A.3)isrelatedto(A.1)–(A.2)bytheinterchange oftandr,andhas thesamegeometrical structure–itisaCartesian
productoftwosurfacesofconstant curvature.
Infact,(A.1)–(A.2)isacoordinatetransformoftheRobinsonsolution[20].Namely,byarathercomplicatedsequence
of coordinate transformations, strictly analogous to those used in Ref. [30] for the Nariai solution, (A.1) – (A.2) may be
transformedto
ds2= ℓ2 dt2−dr′2−r′2 dϑ2+sin2ϑdϕ2 , (A.4)
r′2 h (cid:0) (cid:1)i
whichisidenticalwithoneofRobinson’sformulae,withRobinson’sλ=1/ℓ.
Acknowledgement:TheresearchofAKwassupportedbythePolishMinistryofEducationandSciencegrantnoN
N202104 838. Whilethe paper was being prepared,GG was asummerinternat theN. Copernicus Astronomical Center
inWarsaw.
8
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