Analogue of the Theorem of Birkhoff in a Metric Nonsymmetric Theory
of the Gravitational-Electromagnetic Field
S. Ragusa1 and D. Bosquetti2
1. Instituto de F´ısica de S˜ao Carlos, Universidade de S˜ao Paulo, C.P. 369, 13560-970 S˜ao Carlos, SP, Brazil 2. Faculdade de Tecnologia de Mococa, Mococa, SP, Brazil
Received on 6 January, 2007
We prove that a spherically symmetric exterior solution of the field equations of a metric nonsymmetric theory of gravitation coupled with the electromagnetic field is necessarily static.
Keywords: Birkhoff; Nonsymmetric; Gravitation; Electromagnetism
I. INTRODUCTION
Recently [1] we have shown that the solution of the field equations of a metric nonsymmetric theory of gravitation [2-I,II] outside a spherically symmetric time-dependent matter distribution is necessarily static. This is the analogue of the theorem of Birkhoff of general relativity (GR). Here we ex-tend the result to the situation in which electromagnetic field is also present, that is, when the time-dependent source is charged . This is the situation for instance for a radially pul-sating charged sphere. The proof of the result, which we will call the static theorem, will run along the same lines of GR as presented, for instance, in [3].
The form of the field equations when the electromagnetic is included into the theory was studied in [4] and the solution for a point charged mass was obtained. In what follows we will be making use of the field equations of that paper.
The paper is organized as follows. In Sec. II we present the vacuum field equations of the theory and in Sec. III, we prove the static theorem. In Sec. IV we draw our conclusions.
II. THE VACUUM GRAVITO-ELECTROMAGNETIC FIELD EQUATIONS
When the electromagnetic field is present, the vacuum grav-itational field equations of the theory become [4] (κ=8πG)
Uαβ+Λg(αβ)= κ
4πE(αβ), (2.1)
g(αβ),γ+g(ασ)Γβ(σγ)+g(βσ)Γα(σγ)−g(αβ)Γσ(σγ)=0, (2.2)
g[αβ],β=0 , (2.3)
Λg[αβ,γ]= κ
4πE[αβ,γ] (2.4) and
∂ν{√−g f[g(µα)g(νβ)+ (2Z−1)g[µα]g[νβ]
+(1−Z)g[αβ]g[µν]]Fαβ}=0 . (2.5)
We use the notationa(αβ)= (aαβ+aβα)/2 anda[αβ]= (aαβ− aβα)/2 for the symmetric and antisymmetric parts ofaαβand the notationa[αβ,γ]=a[αβ],γ+a[γα],β+a[βγ],α for the curl of
a[αβ]. In the first equation
Uαβ=Γσ(αβ),σ−Γσ(σα),β+Γ(ραβ)Γσ(ρσ)−Γρ(ασ)Γσ(ρβ), (2.6) symmetric because the second term is (see (2.14) below) and containing only the symmetric part of the connection, is the analogue of the Ricci tensor. Λis the cosmological constant and
Eαβ = 1 4
µ
f gαβ−2 ∂f ∂gαβ
¶
gλσgρµ(ZFλρFσµ +(1−Z)[FσρFλµ+FλσFρµ])
−f gµν¡
ZFαµFβν+ (1−Z)[FανFβµ+FµνFαβ]¢
. (2.7)
Here and in (2.5)Z is an arbitrary parameter. However, the point charge solution turned out to be independent ofZ.Fµν=
Aν,µ−Aµ,νis the electromagnetic field strength tensor and
f = √
−g
√ −gs
, (2.8)
whereg=det.(gαβ) andgs=det.(g(αβ)).Eαβis a traceless ten-sor,gαβEαβ=0, because we have the relation [4]
gαβ ∂f
∂gαβ =0 . (2.9)
The next two equations involve the symmetric and antisym-metric parts ofgαβ=√−ggαβwheregαβis the inverse ofgαβ as defined by
gαβgαγ=gβαgγα=δβγ . (2.10) The second field equation, (2.2), can be solved for the sym-metric part of the connection giving [2-I],
Γσ(αβ)=1 2g
(σλ)¡
sαλ,β+sλβ,α−sαβ,λ¢
+³g(σλ)sαβ−δασδλβ−δλαδσβ
´
Here
C=1 4ln
s
g (2.12)
andsαβ,symmetric and with determinants,is the inverse of g(αβ)as defined by
sαβg(αγ)=δγβ. (2.13) In deriving (2.11) from (2.2) we come across the relation
Γσ(σα)= (ln√−g
−s),α, (2.14)
which can be re-obtained from that equation. One then sees that the second term on the right of (2.6) is in fact symmetric. Equation (2.5) is the inhomogeneous Maxwell equation in the presence of the metric nonsymmetric field.
Inside the sources of the field the right-hand side of equa-tions (2.1), (2.3) (2.4) and (2.5) would contain, respectively, the source terms 8πGT¯(αβ), 4π√−gSα, 8πGT¯[αβ,γ]/Λ and −4π√−gJµwith
¯
Tαβ=Tαβ−1
2gαβT . (2.15)
Tαβis the stress tensor andT=gµνTµν,Sαis the fermion num-ber current density andJµis the electromagnetic current den-sity.
III. SPHERICALLY SYMMETRIC VACUUM SOLUTION: THE STATIC THEOREM
We will then consider the time-dependent spherically symmetric gravitational-electromagnetic field satisfying the empty-space field equations (2.1)-(2.5). The metric is the same one considered in [1], with the componentsg00,g11and
g[01]depending onrandt,
g00=γ(r,t) =eν(r,t), g11=−α(r,t) =−eµ(r,t),
g22=−r2, g33=−r2sin2θ,
g01=−g10=−ω(r,t) (3.1) and all other components equal to zero. The non-zero compo-nents of the inverse matrix are then
g00= α
αγ−ω2 , g 11=
−αγγ
−ω2 ,
g22=−1
r2 , g
33=− 1
r2sin2θ,
g01=−g10= ω
αγ−ω2. (3.2)
We need alsosαβ,whose non-zero components are
s00 =
αγ−ω2
α , s11=− αγ−ω2
γ ,
s22 = −r2, s33=−r2sin2θ. (3.3) The determinants have values
g=−(αγ−ω2)r4sin2θ (3.4) and
s=−(αγ−ω 2)2
αγ r
4sin2θ. (3.5)
With these relations equation (2.12) gives
C=1 4ln
αγ−ω2
αγ . (3.6)
As the sources are spherically symmetric only the radial component of the electric field is present, that is,
F01=E(r.t) (3.7)
and all other components ofFαβ vanish. Then the non-zero components of the electromagnetic stress tensor are given by [4]
E00=
E2 2α(1+
ω2 αγ)(1−
ω2 αγ)−
1/2, (3.8a)
E11=− α
γE00, (3.8b)
E22=
E2r2 2αγ(1−
ω2 αγ)
−1/2
, (3.8c)
E33=E22sin2θ (3.8d) and
E01=−E10= ωE2
αγ (1− ω2 αγ)−
1/2. (3.8e)
Let us see the form acquired by the field equations. From the onset we see, from the last relation in (3.1) and (3.8e), that (2.4) is identically satisfied because it involves only the componentsg[01]andE[01]. We consider now (2.3). Due to the last relation in (3.2) there are only two non-trivial components of that equation, that is, forα=0 andα=1.They are
g[01],r=0 ;g[10],0=0 . (3.9)
From (3.4) and the last relation in Eq. (3.2) we have
g[01]= ωr 2
p
From the equations in (3.9) we see that the quantity
ωr2
p
αγ−ω2=F (3.11)
is a constant, with respect to both space and time. F is the conserved fermion charge number. From here it follows that the combination
αγ
αγ−ω2=1+
F2
r4 (3.12)
is time-independent. From this result it follows that the quan-tityCin (3.6) depends only onr.Therefore onlyC,r survives in equation (2.11). We then have
C,λ=δλr
F2
r(r4+F2). (3.13) Using these two equations we obtain [4] the following values for the non-zero connection components(a′=a,r),
Γ0(01)= γ ′
2γ+ F2 r(r4+F2), Γ1 00= γ′ 2α+ γ α F2 r(r4+F2), Γ111=
α′ 2α+
F2 r(r4+F2), Γ1
22=−
r α ,
Γ1 33=−
rsin2θ
α ,
Γ2(12)=Γ3(13)= r 3
r4+F2, Γ233=−sinθcosθ, Γ3
(23)=cotθ,
and(a.=a,t) Γ0
00= 1 2 .
ν, Γ0
11= 1 2e
µ−νµ., Γ1
(01)=
1 2 .
µ . (3.14) These three last relations come from
Γ000= 1 2
α αγ−ω2
∂ ∂t
γ(αγ−ω2)
γα , Γ0 11= 1 2 α αγ−ω2
∂ ∂t
α(αγ−ω2) αγ
and
Γ1(01)=1 2
γ αγ−ω2
∂ ∂t
α(αγ−ω2)
αγ , (3.15)
when use is made of (3.12).
Let us see now what are the relations that they will bring about to the field equation (2.1). Putting for short
B=r4+F2 (3.16)
we obtain the following non-trivial four relations, corre-sponding first toα=β=0,
à γ′ 2α+ γ α F2 rB !′ + à γ′ 2α+ γ α F2 rB ! à α′ 2α− γ′ 2γ+
2r3 B
!
+γΛ−1 2 .. µ+1
4 . νµ. −1
4 . µ2= κ
4πE00, (3.17a) then toα=β=1,
Ã
γ′ 2γ+
F2+2r4 rB !′ − Ã γ′ 2γ+ F2 rB ! Ã α′ 2α− γ′ 2γ !
−2r 3 B Ã α′ 2α+ F2 rB− r3 B !
+αΛ−eµ−ν(1 2 .. µ−1
4 . νµ. +1
4 . µ2) = κ
4πE11, (3.17b) next toα=β=2,
−³αr´
′
−αr
Ã
γ′ 2γ+
α′ 2α+
2F2 rB
!
+1−Λr2
= κ
4πE22 (3.17c)
and finally toα=0 andβ=1
r3 F2+r4
.
µ=0 , (3.17d)
because according to equation (3.8e)E(01)=0.All these
re-lations reduce to those of GR [3] whenF=0. From the equation above we see that
.
µ=0 . (3.18)
Therefore, the additional last terms on the left-hand side of both equations (3.17a) and (3.17b) drop out. Asµ is time-independent it follows from the second relation in (3.1) that this is so also forα, that is, it depends only onr,α=α(r). Multiplying (3.17a) byα/γand adding the result to (3.17b) we see, on account of (3.8b), that the right-hand side will vanish, leading then to the same relation as before [4],
γ′ 2γ+
α′ 2α =
F2
Integration gives
γα=k(t)
µ
1+F 2
r4
¶−1/2
. (3.20)
Therefore,
γ(r,t) =k(t)¯γ(r). (3.21) where
¯ γ(r) =1
α
µ
1+F 2
r4
¶−1/2
. (3.22)
The fact thatγ(r,t)can depend on time only through a time-dependent factor could be guessed before integration because (3.19) tell us thatγ′/γis time-independent and, consequently, γ has to have the form written in (3.21). From this we see that only the time part of the line elementds2will contain a time-dependent coefficient,γdt2=k(t)¯γ(r)dt2.However, this time-dependent factor can be made equal to unity by a change of the time coordinate. In fact it is sufficient to use the new time coordinate ¯tgiven by
dt¯=pk(t)dt (3.23) to obtain the time-independent metric in [4],
ds2=γ(¯ r)dt¯2−α(r)dr2−r2dθ2−r2sinθdφ2. (3.24) Starting from here and following the same steps of the begin-ning of this section, we obtain Eq. (3.11) with ¯γ replacing γ,
ωr2
p
α¯γ−ω2=F. (3.25)
This tell us thatωis also time-independent and given by the same expression as before in [4],
ω(r) = rF (r4+F2)34
. (3.26)
This follows from (3.25), which gives
ω2= α¯γF2
r4+F2 (3.27)
and Eq. (3.22).
We consider now equation (2.5)for the electric field in the redefined metric with ¯γreplacingγ. Asgs=−α¯γr4sin2θ equa-tion (2.8) gives, recalling (3.4),
f =
s
1−ω 2
α¯γ . (3.28)
Only two non-trivial relations follows from (2.5), those for µ=0 andµ=1. They are
∂ ∂r(
E2r2 √
α¯γ) =0 (3.29)
and
∂ ∂t(
Er2
√α¯γ) =0 . (3.30)
¿From this last relation we see that∂E/∂t=0,meaning that Eis time-independent.
Therefore, all quantities, the metric coefficients and the electric field, are independent of time. This proves the sta-tic theorem: the exterior spherically symmetric gravitational-electromagnetic field is necessarily static.
We go now to equation (3.17c) and solve forα(r).Using in (3.8c) the results in (3.19) and the relation
(1−ω 2
α¯γ)
−1/2= (1+F2
r4) 1/2
, (3.31)
that follows from (3.27), we have
E22=
Q2 2r2(1+
F2 r4)
1/2
. (3.32)
Then taking this into (3.17c) together with (3.16), we obtain, neglecting the small cosmological constant,
−³r α
´′
−r
α 3F2
r(F2+r4)+1=G
Q2 r4(1+
r4 F2)
1/2. (3.33)
Choosing the constant of integration equal to−2MG(Mis the mass of the charged particle) in such a way that the Reissner-Nordstr¨om (RN) result appears whenF=0, we obtain, as in [4],
1 α(r)=
µ
1+F 2
r4
¶
−
µ
1+F 2
r4
¶3/4
(2MG
r +
GQ2 r h(r)),
(3.34) where
h(r) =
Z dr
r(F2+r4)1/4 , (3.35) which, as it should, goes to −r−1 when F vanishes. Then from (3.22), we obtain
¯ γ(r)=
µ
1+F 2
r4
¶1/2
−
µ
1+F 2
r4
¶1/4
(2M r +
GQ2 r h(r)).
(3.36) Equation (3.35) can be put in the closed form:
h(r) = 1 2|F|1/2[tan
−1
µ
1+ r 4
F2
¶1/4
+1 2ln
³
1+r4 F2
´1/4
−1
³
1+r4
F2
´1/4
+1
−π
2], (3.37)
Now the electric field. From (3.29) it then follows that the quantity
E(r)r2
√
α¯γ =Q (3.38)
is a constant, the charge of the system. Using (3.22) we have
E(r) =Q r2
µ
1+F 2
r4
¶−1/4
. (3.39)
This reproduces in fact the point charge RN resultQ/r2forE whenF=0,that is wheng[αβ]=0, the situation in which we recover the Einstein-Maxwell field of GR. At large distances, r>> |F |12,E(r)goes into the RN Coulomb field but for
small values ofr, it behaves asr−1. So all the results of the time-independent regime are recuperated.
[1] S. Ragusa and D. Bosquetti, Braz. J. Phys.36, 1223 (2006). [2] S. Ragusa, Phys. Rev. D56, 864 (1997).∆in this paper is now
namedΓ.Also the factor 2/3 in equation (6.7) is dropped here; Gen. Relat. Gravit.31, 275 (1999 ). The first term inside the first brackets of equation (23) should beγ´/2γ.The exponent ofrin the expression in the middle of the second line below equation (33) should be 2 instead of 3.These papers will be referred to
as I and II, respectively. Equations (1,I-n) and (1,II-n) stand for equations in the corresponding references.
[3] See for instance, A. PapapetrouLectures on General Relativity (Dordrecht, Holland, 1974).
