Shearing and geodesic axially symmetric perfect fluids that do not produce gravitational radiation

0.2 MB·English

by  L. Herrera

Shearing and geodesic axially symmetric perfect fluids that do not produce
gravitational radiation
L. Herrera∗
Escuela de F´ısica, Facultad de Ciencias, Universidad Central de Venezuela, Caracas 1050,
Venezuela and Instituto Universitario de F´ısica Fundamental y Matema´ticas,
Universidad de Salamanca, Salamanca 37007, Spain
A. Di Prisco†
Escuela de F´ısica, Facultad de Ciencias, Universidad Central de Venezuela, Caracas 1050, Venezuela
J. Ospino‡
Departamento de Matema´tica Aplicada and Instituto Universitario de F´ısica Fundamental y Matema´ticas,
Universidad de Salamanca, Salamanca 37007, Spain
5
1
0 J.Carot§
2 Departament de F´ısica, Universitat Illes Balears, E-07122 Palma de Mallorca, Spain
(Dated: January 12, 2015)
n
a Using a framework based on the 1+3 formalism we carry out a study on axially and reflection
J
symmetricperfectandgeodesicfluids,lookingforpossiblemodelsofsourcesradiatinggravitational

9 waves. Therefore, the fluid should be necessarily shearing, for otherwise the magnetic part of the
Weyltensor vanishes, leading toa vanishingof thesuper–Poynting vector. However,for thefamily
] of perfect, geodesic fluids considered here, it appears that all possible cases reduce to conformally
c
q flat, shear–free, vorticity–free, fluids, i.e Friedmann-Roberston-Walker. The super-Poynting vector
- vanishesandthereforenogravitationalradiationisexpectedtobeproduced. Thephysicalmeaning
r of theobtained result is discussed.
g
[
PACSnumbers: 04.40.-b,04.40.Nr,04.40.Dg
Keywords: RelativisticFluids,nonsphericalsources,interiorsolutions.
1
v
2
7 I. INTRODUCTION the Bondinews function[10](see [11]for adiscussionon
1 this point).
2 Inarecentpaper[1]the1+3formalism[2–5]hasbeen From the comments above, and as a further step
0 usedto developa generalframeworkforstudying axially towards the understanding of gravitationally radiating
.
1 symmetric dissipative fluids. Besides some results ex- sources, we shall consider in this work the simplest
0 hibited in [1], the above mentioned framework has been fluid distribution which we might believe to be compati-
5 applied to the shear-free case [6]. As the result of such ble with a non-vanishing super-Poyntingvector,namely:
1 a study, it follows that all geodesic and shear-free fluids, perfect fluid under the geodesic condition.
:
v are irrotational, and as consequence of this, also purely However, our investigation shows that all possible
i electric. Such a result holds for a general fluid (not nec- models, sourced by a perfect fluid (which of course in-
X
essarily perfect). cludes the pure dust configuration as a particular sub-
r
a Now,ifwedefineastateofintrinsicgravitationalradi- case),belonging to the family of the line element consid-
ation (at any givenpoint), to be one in which the super- eredhere,donotradiategravitationalwavesduringtheir
Poyntingvectordoesnotvanishforanyunittimelikevec- evolution.
tor [7–9], then since the vanishing of the magnetic part We shall discuss about this result and its relationship
of the Weyl tensor implies the vanishing of the super- withthe factthatthe processofradiation(including ab-
Poynting vector, it is clear that when looking for gravi- sorptionand/orSommerfeldtype conditions)isnotare-
tationally radiating sources (at least under the geodesic versible one.
condition)weshouldconsidershearingfluids. Itisworth We shall heavily rely on the general framework devel-
recalling that the tight link between the super-Poynting opedin[1],keepingthesamenotation,andjustreducing
vector and the existence of a state of radiation is firmly the general equations to the particular case considered
supported by the relationship between the former and here. These will be presented in an Appendix.
II. THE PERFECT, GEODESIC, FLUID
∗ [email protected]
† [email protected]
‡ [email protected] Weshallconsideraxiallyandreflectionsymmetricper-
§ [email protected] fect fluid distributions (not necessarily bounded). For
2
such a system, we assume that the line element may be
written in “Weyl spherical coordinates”,as:
Θ=Vα
;α
AB2 B˙ C˙
ds2 = A2dt2+B2 dr2+r2dθ2 +C2dφ2+2Gdθdt, (1) = r2A2B2+G2 r2 2B + C
− " !
where A,B,C,G are(cid:0)positive func(cid:1)tions of t, r andθ. We G2 B˙ A˙ G˙ C˙
+ + + . (5)
number the coordinates x0 =t,x1 =r,x2 =θ,x3 =φ. A2B2 B − A G C!#
The specific form of the line (1) deserves some com-
ments. Our original goal, here as well as in [1] and [6], The four acceleration
has been to describe the gravitational radiation process,
a =VβV =a K +a L , (6)
through the physical properties of its source. Such an α α;β I α II α
endeavour, should, eventually, lead to the obtention of
with vectors K and L having components:
a specific source of gravitational radiation. With this
purpose in mind, and in order to render the problem
√A2B2r2+G2
under consideration, analytically handable, we have im- K =(0,B,0,0); L =(0,0, ,0),
α α A
posed the highest degree of symmetry compatible with
(7)
the existence of gravitational radiation. For vacuum, it
and where the two scalar functions (a ,a ) are defined
I II
is represented by the Bondi metric [10]. Here, follow-
by (see eq.(17) in [1])
ing the framework developed in [1], we have restricted
the line element as muchas possible, alwaysallowing for A′
a = , (8)
the existence of gravitational radiation (at least for the I AB
most generalmatter distribution). As the result ofthese A G A˙ G˙ A
restrictions, we are able to manipulate the resulting ex- a = + + ,θ ,(9)
pressions,bypurelyanalyticalprocedures. However,this II √A2B2r2+G2 "A2 −A G! A #
isobtainedatthepriceofdealingwithaspace–time,that
whereas the vorticity vector is defined through a single
is not the more general one, compatible with axial sym-
scalar Ω, given by (see eq.(29) in [1])
metry (see the discussion below (50)).
The energymomentumtensorin the “canonical”form
reads: (AG′ 2GA′)
Ω= − , (10)
2AB√A2B2r2+G2
T =(µ+P)V V +Pg , (2)
αβ α β αβ
where primes denote derivatives with respect to r.
Weshallrestrictoursystemtothecaseofvanishingfour–
where as usual, µ,P and V denote the energy density,
β acceleration a =0. This condition implies that
α
theisotropicpressure,andthefourvelocity,respectively.
Asin[1]weareassumingthe fluidtobecomovinginthe
coordinates of (1). A′
a = =0 A=A˜(t,θ), (11)
The shear tensor is defined by two scalar functions I AB ⇒
σ ,σ , which in terms of the metric functions read (see
I II and
eqs.(20-25)in [1]):
G A˙ G˙ A
a =0 ( + )+ ,θ =0, (12)
II ⇒ A2 −A G A
3 B˙ C˙ or
2σ +σ = , (3)
I II A B − C!
2σII +σI = A2B2r32+G2 "AB2r2 BB˙ − CC˙ ! (cid:18)GA(cid:19). =−A,θ ⇒ GA =−Z A,θdt+G˜(r,θ). (13)
G2 A˙ G˙ C˙ Given that G(t,0,θ)=0, from (13) we find that
+ + , (4)
A −A G − C!# A =0 A=A˜(t) and G=G˜A˜. (14)
,θ
⇒
Inthiscase,reparametrizingthetimecoordinate,theline
where dots denote derivatives with respect to t.
element takes the form
Fortheotherkinematicalvariables(theexpansion,the
four acceleration and the vorticity) we have: ds2 = dt2+B2 dr2+r2dθ2 +2G˜(r,θ)dtdθ+C2dφ2,
−
the expansion (15)
(cid:0) (cid:1)
3
and the kinematical quantities become Thus,takingthetime derivativeofthe aboveequation
and combining with (A1), (A2) and (A4), we obtain
2B2r2+G˜2B˙ C˙
Θ= + , (16) σ σ 1
B2r2+G˜2 B C = I − II 3Ω2 Y (Θ+σ )(Θ+σ ) ,
EII−EI Θ+σ −EI − T − 3 I II
I (cid:20) (cid:21)
(26)
G˜′ and
Ω= , (17)
2B B2r2+G˜2 = σI −σII 3Ω2 Y 1(Θ+σ )(Θ+σ ) ,
EII−EI Θ+σ −EII − T − 3 I II
p II (cid:20) (cid:21)
(27)
B2r2+2G˜2B˙ C˙ where (5) has also been used.
σ = , (18)
I B2r2+G˜2 B − C In the above expressions, EI,II, are two of the three
scalar functions defining the electric part of the Weyl
tensor (see [1] for details) (the third scalar function is
3G˜2 B˙ denoted EKL ). Also, YT, is one of the structure scalars
σ σ = . (19) obtained from the orthogonal splitting of the Riemann
I − II B2r2+G˜2B
tensor which are defined in eqs.(38-50) in [1].The others
are denoted by Y ,X ,Z .
Now, from the regularityconditions, necessary to ensure I,II,KL T,I,II,KL I,II,III,IV
From the above equations and (21) it follows at once
elementary flatness in the vicinity of the axis of symme-
that as r 0
try, and in particular at the center (see [12], [13], [14]), ≈
we should require that as r ≈0 EI −EII = EI(n)(t,θ)−EI(nI)(t,θ) rn, (28)
Ω= Ω(n)(t,θ)rn, (20) nX≥4h i
nX≥1 and (22), unless Θ+σI =Θ+σII =0.
However, this last condition cannot be satisfied. In-
implying,becauseof(17)thatintheneighborhoodofthe deed the absence of singularities in H at r =0 requires
2
center from (A9), that σ r as r 0 which would produce
I,II
≈ ≈
Θ 0, implying in turn because of (A1)
G˜ = G˜(n)(θ)rn. (21) ≈
nX≥3 YT(r =0)=(µ+3P)r=0 =0. (29)
This last result in turn implies that as r approaches 0, Now, in order to satisfy (29) we have to assume, the
equation of state
σ σ = σ(n)(t,θ) σ(n)(t,θ) rn. (22)
I − II I − II (µ+3P)=0, (30)
nX≥4h i
at least in the neighborhood of the center. Excluding
Now, for the length of an orbit at t,θ constant, to be this possibility on physical grounds, we have to assume
2πr, close to the origin (elementary flatness), we may Θ+σ =0.
I,II
write, as r 0, Next,fro6 mthecombinationof(A10),(A12)and(A13)
→
we obtain
C rγ(t,θ), (23)
≈
implying G˜ H G˜′ B˙ C˙
2 +H =0,
C′ γ(t,θ), C rγ , (24) B2r2+G˜2 "2B B2r2+G˜2 1 B − C!#
,θ ,θ
≈ ≈ (31)
p p
where H ,H are the two scalar functions defining the
where γ(t,θ) is an arbitrary function of its arguments, 1 2
magnetic part of the Weyl tensor.
which as appears evident from the elementary flatness
From the above equation it follows that either G˜ =
condition, cannot vanish anywhere within the fluid dis-
0 Ω=0, or
tribution.
⇒
Finally, observe that a combination of (16)–(19) pro-
2σ +σ
duces H2Ω+H1 I II =0, (32)
3
(cid:20) (cid:21)
(σ σ )(Br2+G˜2)=G˜2(Θ+σ ). (25)
I II I where (17)–(19) have been used.
−
Since Ω goes to zero at the center, as r, we are left
We shall next make use of the full set of equations
with three possibilities to proceed further, namely:
deployed in [1], written for the specific case considered
here, they are given in the Appendix A. 1. The vorticity is assumed to vanish (G˜ =0).
4
2. G˜ = 0 and the term within the square bracket in whereas from (A2)–(A4), we obtain, respectively
6
(32) does not vanish at the center, meaning that
1 2
we may write σ˜˙ σ˜2+ Θσ˜+Y =0, (41)
I
− 3 3
σ = σ(n)rn, σ = σ(n)rn. (33)
I I II II
n=0 n=0
X X
Y =0 X =0, (42)
KL KL
⇒
3. G˜ = 0 and the term within the square bracket in
6 and
(32) does vanish at the center implying that
σI = σI(n)rn, σII = σI(nI)rn. (34) σ˜˙ 1σ˜2+ 2Θσ˜+Y =0. (43)
II
n≥1 n≥1 − 3 3
X X
Then, from (40)–(43) we get
Next,contracting(A.7)in[1]withKandLweobtain,
respectively 3C¨ Y
Y =Y =Y = + T, (44)
I II 2C 2
P′ =0 P =P(t,θ), (35) and using (42)–(45) and (47)–(50) in [1] it follows that
⇒
X =X =X, = = , Y = X = ,
I II I II
E E E − E
=X =Y =0. (45)
G˜P +P =0, (36) EKL KL KL
,t ,θ
Two constraint equations follow from (A6) and (A7)
which, due to the regularity conditions on the axis of
symmetry, implies that either G˜ = 0 and P = P(t), or C′
2Θ′ σ˜′ 3σ˜ =0, (46)
G˜ =0 and P =constant. − − C
6
We shallnow analyzethe three possible casesmentioned
before.
C
2Θ σ˜ 3σ˜ ,θ =0, (47)
,θ− ,θ − C
A. G˜ =0
whereas from (A8) and (A9) we get
From the well established link between radiation and σ˜ C σ˜ (σ˜C)
H = ,θ + ,θ = ,θ, (48)
vorticity (see [15] and references therein), it might be 1 −2Br C σ˜ −2BrC
(cid:18) (cid:19)
inferredthatno gravitationalradiationis expectedto be
emitted in this case.
In what follows we shall provide a formal proof of this σ˜ σ˜′ C′ (σ˜C)′
H = + = . (49)
result. 2 2B σ˜ C 2BC
(cid:18) (cid:19)
As mentioned before, in this case we have P = P(t);
then, from (16), (3) and (4) we get Fromthetwoequationsabovewefindthatifthemag-
netic part of the Weyl tensor vanishes, then
B˙ C˙ B˙ C˙
Θ=2 + , σ =σ =σ˜ = , (37)
B C I II B − C
σ˜C =ψ(t), (50)
implying:
where ψ(t) is an arbitrary integration function. The
3C˙ 3B˙ above equation implies, because of the regularity con-
Θ=2σ˜+ = σ˜. (38) dition C(t,0,θ)=0, that σ˜ =0.
C B −
This last result in turn implies (as mentioned before)
Next, equation (A1) is the Raychaudhury equation for that our line element (1) (or (15)) is not the more gen-
this case, which reads eralone, since inthe sphericallysymmetric limit, it does
not contain the Lemaitre-Tolman-Bondi metric. Also,
theSzekeresmetric[16–19](itsaxiallysymmetricversion)
1 2
Θ˙ + Θ2+ σ˜2+Y =0, (39) cannotbe recoveredfrom(15). Inotherwords,theshear
T
3 3
of the fluid (in our models) is sourced by the magnetic
or, using (38) part of the Weyl tensor. Therefore in the spherically
symmetric limit, we recover the shear–free case.
1 2 3C¨ Y Anotherimportantconclusionemergesfrom(48),(49).
σ˜˙ σ˜2+ Θσ˜+ + T =0, (40) Indeed, since B is regular at the origin (r 0), whereas
− 3 3 2C 2 ≈
5
C behaves as C r, then for the magnetic Weyl tensor Now, combining the two equations above with (58) we
≈
to be regular at the origin,we must demand there σ˜ r may write
≈
(at least). But then, (41) or (43) implies that r at
E ≈ C˙ C˙′
the origin. ,θ = C′ =γ(r,θ)C , (61)
Also, from (47), it follows that Θ,θ 0 (in the neigh- C,θ C′ ⇒ ,θ
≈
borhoodofthecenter),andthenfrom(38)itfollowsthat
implying
C =κ(t)C˜(r,θ), (62)
γ
,θ =l(θ), (51)
γ and, because of (57)
where l(θ) is an arbitrary function of its argument. H2 = rγ(r,θ)H1. (63)
−
Next, using (46) and (47) we may write
Then, using (61) in (58) we get
′
2(Θ+σ)′ = (3σ˜CC)′ ⇒2 BB˙ ! = (σ˜CC)′, (52) BB′ = BB,θγ(r,θ)+ǫ(r,θ), (64)
with
2(Θ+σ),θ = (3σ˜CC),θ ⇒2 BB˙ ! = (σ˜CC),θ. (53) ǫ= γ,θrr(1++γγr′2rγ32−) 1, (65)
,θ
implying
Feeding back the two equations aboveinto (48) and (49) B˙ T˙
we obtain B =T(t)B˜(r,θ) = =f(t), (66)
⇒ B T
′
B˙ B˙ where T(t) and B˜(r,θ) are arbitrary functions of their
BrH = , BH = . (54)
1 − B! 2 B! arguments.
,θ Using the above result in (54), it follows at once that
H = H = 0, which as mentioned before implies σ˜ =
On the other hand (A11) and (A16), produce, respec- 1 2
0. This last result in turn implies that the expansion
tively
scalar only depends on t and taking into account that
Y = 4π(µ + P), the Raychaudhury equation requires
2C B 2C′ (Br)′ T
H +H ,θ ,θ =r H′ +H , µ=µ(t).
2,θ 2 C − B 1 1 C − Br
(cid:18) (cid:19) (cid:20) (cid:18) (cid:19)(cid:21) Next, (A14) reads in this case
(55)
and 1 C′
′+ =0, (67)
3E C E
(H BrC2)′+(H BC2) =0, (56)
1 2 ,θ which due to the fact that 0 at the origin, implies
E ≈
=0.
whereasthecombinationoftheequationsabovewith(48) E
Thusourspacetimeisconformallyflat,shear–free,and
and (49) produces
due to the fact that the fluid is perfect, it is a FRW
H C′r+H C =0, (57) spacetime, in agreement with the result obtained in [6].
1 2 ,θ
For the sake of completeness we shall sketch another
proof of the above result in the Appendix B.
or
Now, fromthe fact that our system is conformallyflat
B˙ B˙ ′ it appears that it does not radiate gravitationally (ac-
C′+ C =0. (58) cordingtothecriteriumcommentedintheIntroduction).
− B! B! ,θ Indeed, the super-Poynting vector can be written
,θ
(ec.(55) in [1]) as
Next, using (37) in (52) and (53), it follows that
P =P K +P L , (68)
α I α II α
B˙ = B˙C,θ C˙,θ, (59) where according to (56) in [1], and (45), we have for the
B! BC − C two scalars defining the super-Poynting vector
,θ
P =2H Y; P = 2H Y. (69)
I 2 II 1
−
B˙ ′ B˙C′ C˙′ Thus, the vanishing of E, H1 and H2, as in our case,
= . (60) implies the vanishing of gravitationalradiation.
B! BC − C
6
B. G˜ 6=0 and the term within the square bracket in C. G˜ 6=0, and the term within the square bracket
(32) does not vanish at the center. in (32) does vanish at the center.
Inthis case,fromtheregularityofH atthecenterwe Then, it follows at once from (16) and (18), close to
2
may write, in the neighborhood of r 0, the center, that
≈
H2 = H2(n)rn, (70) B˙ C˙, Θ 3C˙ 3B˙ . (79)
n=0 B ≈ C ≈ C ≈ B
X
in which case, (32) implies
Next, from the lowest order of r in (A2) and (A4) it
appears that
H = H(n)rn. (71)
1 1
nX≥1 E1 = EI(n)rn, EII = EI(nI)rn, (80)
Replacing the two aboveexpressionsin a combinationof nX≥1 nX≥1
(A10)and(A12),andusing(21),(A3),(28),(22)wefind
and, (19), (A3), (26) and (27), as r 0, produce (22),
that the lowest order of r in H1 and H2 is raised as ≈
(28) and
H = H(n)rn, H = H(n)rn. (72)
1 1 2 2 = (n)rn. (81)
nX≥6 nX≥5 EKL EKL
n≥5
X
Next, from (A14) we may write (close to the center)
Excluding singularities of the scalars H ,H , at the
1 2
= (n)rn, = (n)rn, (73) origin, we may write:
EI EI EII EII
n≥1 n≥1
X X H = H(n)rn, H = H(n)rn, (82)
for otherwise there would be an inadmissible singularity 1 1 2 2
n=0 n=0
in the r-derivative of the energy density at the origin. X X
Feeding back the two expressions above in a combina- Then,lookingforthe lowestorderofr in(A10)–(A13)
tion of (A12) and(A13), it follows from the lowestorder and (A16) we obtain respectively:
in r, that in the neighborhood of the center
γ
H(0) = H(0) ,θ, (83)
(µ+P)σI O(r), (74) 1,θ − 1 γ
≈
which implies that,
2γ B
H(0)+H(0)( ,θ ,θ) H(0) =0, (84)
2,θ 2 γ − B − 1
(µ+P) O(r), (75)
≈
which of course is impossible unless we assume, close to
the center, the equation of state H(0) = H(0)(γ,θ B,θ), (85)
2 − 1 γ − B
(µ+P)=0. (76)
Excluding this possible situation from physical consider-
B
ations, we have to require that H(0)+H(0) ,θ H(0) =0, (86)
1,θ 1 B − 2
σ = σ(n)rn (77)
I I
n≥1
X
2γ B
which in turn implies 3H1(0)+H2(0,θ)+H2(0)( γ,θ + B,θ)=0. (87)
σ = σ(n)rn (78)
II II Next, from the lowest order in (A7) it follows that in
nX≥1 the neighborhood of the center
because of (22).
C˙ B˙
But of course this contradicts the main assumption of Θ 0 0, (88)
this case about the nonvanishing of the term within the ,θ ≈ → C! ≈ B! ≈
,θ ,θ
square bracket in (32), (Eq. (33)).
Thus we have to assume (34), implying that the term implying right there
within the square bracket in (32) vanishes at the center
as r. γ =f(t)α(θ), B =g(t)β(θ). (89)
7
Then integrating (83) we obtain
2γ B
H1(0) = αx((θt)), (90) 4H1(1)−H2(1,θ)−H2(1)( γ,θ + B,θ)=0. (101)
Then, proceeding exactly as we did before, using (89),
where x(t) is an integration function.
we are lead to
Anequationderivedfromthe combinationof(84) and
(87), can also be integrated to produce β β2
,θ =constant, (102)
α
y(t)β1/2(θ)
H(0) = , (91)
2 α2(θ) which cannot be satisfied because of the reflection sym-
metry, as argued before.
and a combination of (84) and (87) also produces Therefore we must put
2H(0)+H(0)B,θ =0. (92) H2 = H2(n)rn, (103)
1 2 B
n≥2
X
From the equations above it follows that
2x(t) = β,θ =constant. (93) H1 = H1(n)rn. (104)
− y(t) αβ1/2 nX≥2
At this point we have to stop the procedure followed
However, this last equation cannot be satisfied. In-
so far with equations (A10)–(A13) and (A16), since now
deed, because of the reflection symmetry, we have that
the lowest order in r in these equations, contains terms
β(0) = β(π), implying that β must have a change of
,θ not including H and H .
sign in the interval [0,π], whereas α and β are positive 1 2
Thus let us turn to equations (A6), (A8) (A9).
defined.
Thus we must put H(0) = H(0) = 0. Accordingly we From the lowest order in (A8) and (A9) we find, re-
1 2 spectively
have:
p(t)
H2 = H2(n)rn, (94) σI(1) =σI(1I) = α , (105)
n≥1
X
and
H = H(n)rn. (95) Ω(1) =q(t,θ)α, (106)
1 1
n≥1
X
with
Next, multiplying (A18) by 2 and subtracting from
2p(t)
(A14), we obtain at the lowest order of r q = , (107)
,θ α2
E1 = EI(n)rn, EII = EI(nI)rn. (96) where p(t) and q(t,θ) are arbitrary functions.
n≥2 n≥2 Now, by the same arguments based on the reflection
X X
symmetry exposed before, it is easily concluded that
Usingtheexpressionsabove,wearenowlookingforthe
p(t) = 0 implying σ(1) = σ(1) = 0. Using this result
lowest order of r in (A10)–(A13) and (A16), we obtain, I II
in the lowest order of r in (A6) we obtain Ω(1) = 0. We
respectively
can now feed these results back into (A8) and (A9), and
H(1) = H(1)γ,θ, (97) look for the lowest order in r. We obtain then that
1,θ − 1 γ
σ = σ(n)rn, σ = σ(n)rn. (108)
II II I I
n≥3 n≥3
X X
2γ B
2H(1) H(1) H(1)( ,θ ,θ)=0, (98)
1 − 2,θ − 2 γ − B Using this last result again in (A6), the lowest order in
r now implies Ω(2) = 0, which in turn, because of (A2)
and (A4) implies
γ B
2H2(1) =−H1(1)( γ,θ − B,θ), (99) E1 = EI(n)rn, EII = EI(nI)rn. (109)
n≥3 n≥3
X X
B Using the results above we can now return to (A10)–
H1(1,θ)+H1(1) B,θ −2H2(1) =0, (100) (A13)and(A16),sincenowthelowestorderinr,inthese
8
equations,onlycontainstermswithH andH . Doingso which of course reduces to (A5) if the four acceleration
1 2
we shall raise the lowest order in r of H and H , until vanishes.
1 2
the moment when the lowest order in these equations From(A5) it follows at once that if at any given time,
containstermswithoutH1andH2. Thenwecangoagain the vorticityvanishes, then it vanishes at any other time
throughthewholecycleabove. Now,itisasimplematter afterwards. Thus we should not expect gravitationalra-
to see that this procedure may be continued as many diation from a physically meaningful system, radiating
times as desired, to obtain that H(n) = H(n) = (n) = for a finite period of time (in a given time interval), for
1 2 EI
(n) = σn = σn = Ω(n) = 0 for any value of n 0, otherwise such a radiation will not be accompanied by
EII 1 II ≥ the presence of vorticity.
implying in turn that at the center, these quantities as
well as their r-derivatives of any order vanish. But, what happens for the perfect (non–dissipative,
Then, assuming that all relevant variables are of class non–geodesic) fluid?
Cω, i.e. that they equal their Taylor series expansion Inthislattercase,theconditionofthermalequilibrium
around the center, we can analytically continue the zero (absence of dissipative flux) reads (see eq. (57) in [1])
value atthe center to the whole configurationandthere-
fore, we obtain a conformally flat and shear–free space- a = hβΓ , (111)
µ − µ ,β
time (i.e. F.R.W.).
where Γ=lnT, and T denotes the temperature.
Feeding back (111) into (110) we get
III. CONCLUSIONS
1
Ω Vδ+ (2Θ+σ +σ +VµΓ )Ω=0. (112)
,δ I II ,µ
3
We have shown that all possible models compatible
with the line element (15) and a perfect fluid, are FRW,
Thus, even if the fluid is not geodesic, but is non–
and accordingly non–radiating (gravitationally). This
dissipative, the situation is the same as in the geodesic
clearly indicates that, both, the geodesic and the non–
case, i.e. the vanishing of vorticity at any given time
dissipativeconditions,arequiterestrictive,whenlooking
implies its vanishing for any time in the future.
for a source of gravitationalwaves.
This result is in full agreement with earlier works in-
Having arrived at this point, the relevant question is:
dicating that vorticity generation is sourced by entropy
does this result make sense from the physical point of
gradients [20]–[25]. At the same time we confirm, by
view?
invokingthe radiation–vorticitylink, the Bondi’s conjec-
To answer to such a question, let us first remember
ture about the absence of radiation for non–dissipative
that, already in the seminal Bondi’s paper on gravita-
systems.
tionalradiation(seesection6in[10]),itwasclearlystated
Finally, two comments are in order before concluding:
that,notonlyinthecaseofdust,butalsointheabsence
of dissipation in a perfect fluid, the system is not ex-
In a recent work [26], the role played by magnetic
pectedtoradiate(gravitationally)duetothereversibility
•
fields in the generation and survival of vorticity,
of the equation of state. The rationale supporting this
has been brought out. This strongly suggest that
conjecture is very clear: radiation is an irreversible pro-
the inclusion of magnetic fields in the discussionof
cess, this fact emerges at once if absorptionis taken into
gravitationally radiating sources, deserves further
accountand/orSommerfeldtypeconditions,whichelim-
attention.
inate inward traveling waves, are imposed. Therefore,
it is obvious that an entropy generator factor should be
presentinthedescriptionofthesource. Butsuchafactor Geodesic fluids not belonging to the class consid-
•
is absent in a perfect fluid, and more so in a collisionless ered here (Szekeres) have also been shown not to
dust. In other words, the irreversibility of the process of producegravitationalradiation[27]. Thisstrength-
emission of gravitational waves, must be reflected in the ens further the case of the non–radiative character
equationofstate throughanentropyincreasing(dissipa- of pure dust distributions.
tive) factor.
In order to delve deeper into this question, let us in-
voke here the tight relationship between radiation and
ACKNOWLEDGMENTS
vorticity mentioned before (see the beginning of Sec. II
A).
L.H. thanks Departament de F´ısica at the Universitat
Now, the equation (A5) in the general (non–geodesic)
de les Illes Balears, for financial support and hospital-
case (Eq. (B.5) in [1]),reads
ity. ADP acknowledges hospitality of the Departament
de F´ısica at the Universitat de les Illes Balears. J.O. ac-
1 knowledges financial support from the Spanish Ministry
Ω,δVδ+ (2Θ+σI +σII)Ω+K[µLν]aµ;ν =0, (110) of Science and Innovation (grant FIS2009-07238).
3
9
Appendix A: Summary of equations for the geodesic From B.3
case
1
Below we shall write the equations of the framework (σI σII)Ω+YKL =0. (A3)
3 −
developed in [1], for the geodesic and perfect fluid case.
Then, equations B.1– B.18 in [1], read, respectively From B.4
From B.1
1 2 1 2 2
Θ˙ + Θ2+ σ2+σ σ +σ2 +Y =2Ω2. (A1) σ˙ + σ2 + Θσ σ (σ +σ )+Y =Ω2. (A4)
3 9 I I II II T II 9 II 3 II−9 I I II II
(cid:0) (cid:1)
From B.2
From B.5
1 2 2
σ˙I + 9σI2+ 3ΘσI − 9σII(σI +σII)+YI =Ω2. (A2) Ω˙ + 1(2Θ+σI +σII)Ω=0. (A5)
3
From B.6
′
1 G˜C˙ C 1 2C′ B2r2+G˜2
Ω +G˜Ω˙ +Ω + ,θ + 2Θ′ σ′ σ +
− B2r2+G˜2 " ,θ C C !# 3B  − I − I C 2(cid:16) B2r2+G˜(cid:17)2 
 
p  (cid:16) ′ (cid:17)
 C′ B2r2+G˜2
σ =0.
− IIC − 2(cid:16) B2r2+G˜(cid:17)2 
 
 (cid:16) (cid:17) (A6)

From B.7
1 C′ 1
Ω′+Ω + (2Θ σ ) +G˜(2Θ σ ).
B C 3 B2r2+G˜2 − II ,θ − II
(cid:18) (cid:19)
n
+σ B,θ C,θ +G˜ B˙ C˙ pσ B,θ + 2C,θ +G˜ B˙ + 2C˙ =0.
I" B − C B − C!#− II" B C B C !#)
(A7)
From B.8
1 C′ G˜G˜′ 1
H = Ω′ Ω (2σ +σ )
1 −2B  − C − 2 B2r2+G˜2 − 6 B2r2+G˜2 I II ,θ
  n
+σ B,θ + C,θ  G˜ B˙ (cid:16) C˙ σ(cid:17)B,θ p2C,θ +G˜ 2B˙ 2C˙ .
I" B C − B − C!#− II" B − C B − C !#)
(A8)
From B.9
1 2C′ (Br)(Br)′ C′ 2(Br)(Br)′+3G˜G˜′
H =+ (σ +2σ )′+σ +σ +
2 6B  I II I(cid:20) C − B2r2+G˜2(cid:21) IIC 2 B2r2+G˜2 
 1 Ω Ω C,θ (cid:16)+G˜ B˙ + C(cid:17)˙ . (A9)
−2 B2r2+G˜2 ( ,θ− " C B C!#)
p
10
From B.10
1 4π
˙ + (µ+P)σ Ω + EI (3Θ+σ σ )+ EII (2σ +σ )
I I KL II I II I
3E 3 − E 9 − 9
1 C H C′ 2(Br)(Br)′+G˜G˜′
H +H ,θ 2 =0. (A10)
− B2r2+G˜2 (cid:18) 1,θ 1 C (cid:19)− B C − 2 B2r2+G˜2 
p  (cid:16) (cid:17) 
From B.11
Ω 1 2C′ 2(Br)(Br)′+G˜G˜′
2˙ + (2Θ σ σ )+ ( )+ H′ +H
EKL EKL − I − II 3 EI −EII B  1 1 C − 2 B2r2+G˜2 
 
1 H +H 2C,θ B,θ G˜(cid:16) B˙ C˙ (cid:17) =0. (A11)
− B2r2+G˜2 ( 2,θ 2" C − B − B − C!#)
p
From B.12
1 4π
˙ + (µ+P)σ +Ω + EII (3Θ+σ σ )+ EI (2σ +σ )
II II KL I II I II
3E 3 E 9 − 9
1 C′ H C B B˙ C˙
+ H′ +H + 1 ,θ ,θ G˜ =0.
B (cid:18) 2 2 C (cid:19) B2r2+G˜2 " C − B − B − C!#
p (A12)
From B.13
1 1 4π
( + ). ( + )Θ (µ+P)(σ +σ ) EI (2σ +σ ) EII (2σ +σ )
I II I II I II II I I II
− 3 E E − 3 E E − 3 − 9 − 9
1 B 1 (Br)(Br)′ +G˜G˜′
+ H +H ,θ H′ +H =0.
B2r2+G˜2 (cid:18) 1,θ 1 B (cid:19)− B ( 2 2" B2r2+G˜2 #)
p (A13)
From B.14
1 2C′ (B2r2+G˜2)′ C′ (B2r2+G˜2)′
′ + + +
3B (EI EI" C 2(B2r2+G˜2)# EII"C − 2(B2r2+G˜2)#)
1 2B C 2B˙ C˙
+ +G˜ ˙ + ,θ + ,θ +G˜ +
B2r2+G˜2 (EKL,θ EKL EKL" B C B C!#)
1 8π
p H (σ +2σ ) 3ΩH = µ′.
−3 2 I II − 1 3B
(A14)
From B.15
1 C′ (B2r2+G˜2)′ 1
′ + + + +G˜ ˙
B (EKL EKL"C B2r2+G˜2 #) 3 B2r2+G˜2 EII,θ EII
n
C B C˙ B˙ 2C p B 2C˙ B˙
+ ,θ ,θ +G˜ + ,θ + ,θ +G˜ +
EI" C − B C − B!# EII" C B C B!#)
1 8π
+ H (2σ +σ ) 3ΩH = (G˜µ˙ +µ ).
1 I II 2 ,θ
3 − 3 B2r2+G˜2
(A15)
p