Table Of ContentA perturbative no-hair of form fields for higher dimensional static black holes
Tetsuya Shiromizu and Seiju Ohashi
Department of Physics, Kyoto University, Kyoto 606-8502, Japan
Kentaro Tanabe
Yukawa Institute for Theoretical Physics, Kyoto University, Kyoto 606-8502, Japan
In this paper we examine the static perturbation of p-form field strengths around higher dimen-
sional Schwarzschild spacetimes. As a result, we can see that the static perturbations do not exist
when p ≥ 3. This result supports the no-hair properties of p-form fields. However, this does not
excludethepresence of the black objects having non-sphericaltopology.
PACSnumbers:
1
1
0 I. INTRODUCTION deformed black holes with spherical topology do not ex-
2 ist. But, this does not exclude the presence of the black
hole solution with non-spherical topology. As the dis-
n Motivatedbytherecentprogressofsuperstringtheory,
cussion in the appendix, if the solution exists, it seems
a higher dimensional black holes has been actively stud-
J to have both of the electric and magnetic hair of p-form
3 ied so far [1]. Different from the four dimensional cases, field strengths simultaneously.
the conventional uniqueness theorem does not holds in
1 The rest of this paper is organized as follows. In
stationary higher dimensional black holes. Indeed, there
Sec. II, we describe the model, boundary conditions
areseveraldifferentblackhole/ringspacetimeswithsame
] andhypersphericalharmonicfunctions(harmonicsforthe
c massandangularmomentum[2,3]. SeeRef. [4]foranew
q brevity). In Sec. III, we analyse the Maxwell fields from
approach“blackfolds”. But, if one considers static (elec-
- the pedagogical point of view. Then, in Sec. IV, we
r tro)vacuum cases, the uniqueness theorem holds [5, 6]
g will discuss the static perturbation of general form field
(See also Refs. [7, 8]) and then the spacetimes is the
[ and show that there are no regular solutions. In Sec. V,
Schwarzschild-Tangherlini solution [9] (The higher di-
we also have a little consideration of no-hair in asymp-
3 mensionalReissner-Nordstro¨msolutioninelectrovacuum
totically (anti)deSitter spacetimes. Finally we will sum-
v cases). However, there are open questions even in static
marise our work and discuss future issues in Sec. VI. In
1 cases. Ifoneputsothermatterfields,itbecomesdifficult
2 the appendix, we try to show the no-hair theoremin the
toshowtheuniquenessingeneral(SeealsoRefs. [10,11]).
1 cases with both of electric and magnetic parts of p-form
For example, one might be interested in the higher form
1 field strength. But, we fail to do it.
. fields(say, p-form field strengths). According to the re-
1
centwork[12], one canshowthe no-hairtheoremfor the
0
cases with (n + 1)/2 ≤ p ≤ (n − 1) in n-dimensional
1 II. FIELD EQUATIONS, BOUNDARY
asymptotically flat spacetimes. Note that the Maxwell
1 CONDITIONS AND HARMONICS
: field(p = 2) is out of the condition on p and consistent
v
with the presence of the Maxwell hair of charged black
i A. Model
X holes. However,there is a mystery aboutthe presenceof
thehairsfor2<p<(n+1)/2. Weshouldalsonotethat
r
a the cases with p ≥ 3 cannot have the conserved charge We consider the system described by the Lagrangian
associatedwithH . Therefore,weintuitivelyguessthat
(p) 1
the monopole component of H does not exist. In sta- L=R− H2 , (1)
(p) p! (p)
tionarycases,thereisthe exactsolutionwithdipolehair
[13]. where R is the n-dimensional Ricci scalar and H is
(p)
In this paper, using the perturbation analysis, we will the p-form field strength. H(p) has the (p−1)-form field
consider the possibility of the back hole spacetime with potential as
non-trivial p-form field strength hair. Since the back-
groundspacetimes are vacuum one, the p-formfield per- H(p) =dB(p−1). (2)
turbations are decoupled with the metric perturbations.
The field equations are
So this set-up makes the analysis much easier than the
cases of the perturbation analysis of “charged” black 1 p−1
holes. Theanalysiswillshowusthatthestaticperturba- Rµν = p! pHµρ1···ρp−1Hνρ1···ρp−1 − n−2gµνH(2p) (3)
tions of p-form field strength around the Schwarzschild- (cid:16) (cid:17)
Tangherlini spacetime does not exist [23]. See the study and
on stationarymetric perturbationfor the Schwarzschild-
Tangherlinispacetimes[15]. Ourresultsuggeststhatthe ∇µHµν1ν2···νp−1 =0, (4)
2
where∇ is the covariantderivativewithrespectto g . Thus, from Eqs. (7) and (8), the regularity implies
µ µν
As in Ref. [12], we can include the dilation field too.
However, the effect from the dilaton does not affect our kij|V=0 =Dρ|V=0 =0 (9)
result. Then, for simplicity, we will not include the dila-
ton fields in this study. H0i1···ip−1H0i1···ip−1|V=0 =O(V2). (10)
and
B. Boundary conditions
Hi1···ipHi1···ip|V=0 =O(1). (11)
Let us consider the boundary conditions. In general,
In this paper we focus on the static perturbation
the metric of static spacetimes can be written as
aroundvacuumandsphericalsymmetricsolutions. That
ds2 =g dxµdxν =−V2(xi)dt2+g (xk)dxidxj, (5) is,thebackgroundp-formfielddoesnotexist. Theback-
µν ij
ground metric is given by
where xi are spatial coordinate and t is time coordinate.
Since we mainly focus onasymptoticallyflatspacetimes, ds20 =−f(r)dt2+f(r)−1dr2+r2dΩ2n−2, (12)
wesupposethatthe asymptoticboundaryconditionsare
where f(r) = 1−(r /r)n−3 and dΩ2 =: σ dxAdxB
given by 0 n−2 AB
is the metric of the (n−2)-dimensional unit sphere. In
m
V =1− +O(1/rn−2) this specific form, the static perturbation should satisfy
rn−3
g = 1+ 2 m δ +O(1/rn−2), (6) H0rA1···Ap−2|V=0 =O(1) (13)
ij (cid:16) n−3rn−3(cid:17) ij H0A1···Ap−1|V=0 =O(V)=O( f) (14)
wdihreecrtelym. FisrotmhethAeDaMsymmpatsost.icWfleatwneilslsn,Hot ussehtohueldadbeocvaeys HrA1···Ap−1|V=0 =O(V−1)=Op(1/ f) (15)
at the infinity. Although we mainly disc(up)ss the asymp- HA1···Ap|V=0 =O(1). p (16)
totically flat cases,we will address the no-hairof H in
(p)
asymptotically (anti)deSitter spacetimes shortly.
C. Hyperspherical harmonic functions
The boundary condition on the event horizon V = 0
comes from the regularity. To see this, we compute the
Since the background spacetimes has spherical sym-
Kretschmann invariant
metry, we can decompose all quantities in terms of hy-
R Rµνρσ = 4R Ri0j0+R Rijkl perspherical harmonics defined on the sphere Sn−2 [16–
µνρσ i0j0 ijkl
18]. In general, there are three type of harmonics, that
4
= V2DiDjVDiDjV +RijklRijkl is, scalar, vector and tensor types. The scalar harmonic
function Y follows
4 1 1 2
= V2 ρ2kijkij + ρ4(ni∂iρ)2+ ρ4(Dρ)2 D2Y =−ℓ(ℓ+n−3)Y. (17)
" #
+RijklRijkl The vector harmonic function VA satisfies
4 1 1 DAV =0 (18)
= k kij + (k−ρD2V)2 A
V2"ρ2 ij ρ2 D2VA =−[ℓ(ℓ+n−3)−1]VA. (19)
+ 2 (Dρ)2 +R Rijkl, (7) Since quantities which we will consider are often asym-
ρ4 ijkl metrictensor,weconsiderthetotallyanti-symmetricten-
#
sor harmonic function only
where we used R = VD D V in the second line and
i0j0 i j
Di is the covariant derivative with respect to gij. In the DA1TA1···Aq =0 (20)
third line kij = hkiDknj with ni = ρDiV. For the last D2TA1···Aq =−[ℓ(ℓ+n−3)−q]TA1···Aq. (21)
line, we may be going to use the Einstein equation
Notethatthestaticperturbationofmetricandp-form
R00 = VD2V fields are decoupled each others. This is due to the non-
= (nn−−2)p(p−−11)!H0i1···ip−1H0i1···ip−1 kpnreoswentcheaotftthheepbaocsskigbrloeusntdatfiicelpdeorftupr-fboartmionfieoldfst.hSeinmceetwrice
p−1 are ℓ = 0,1 modes. So if the mass is fixed, ℓ = 0 static
+(n−2)p!V2Hi1···ipHi1···ip. (8) modesvanishes. Theℓ=1modescanbeabsorbedtothe
redefinition of the coordinate, that is, it corresponds to
D isthecovariantderivativewithrespecttotheinduced the choice of the “center” of the coordinate. Therefore,
i
metric h =g −n n . wewillnotconsiderthestaticperturbationofthemetric.
ij ij i j
3
III. MAXWELL FIELDS B. Solutions
As a pedagogical exercise, we will first consider Under the gauge condition of Eq. (25), the Maxwell
the Maxwell fields. As already known, the unique- equation becomes
ness theorem of charged black hole (the higher dimen-
n−2 1
sionalReissner-Nordstro¨msolution)holdsinthissystem. ∂2A + ∂ A + D2A =0 (32)
Therefore, the static monopole perturbation is only per- r t r r t r2f t
mitted. We will confirm this fact in this section. See
and
Refs. [19, 20] for the perturbation analysis of Reissner-
Nordstro¨m spacetimes. n−4 f′ D2−(n−3)
∂2A + + ∂ A + A =0. (33)
TheeachcomponentsoftheMaxwellequationbecome r A r f r A r2f A
(cid:16) (cid:17)
∂ F + n−2F + 1 DAF =0 (22) Here we expand At,AA in terms of harmonics as
r rt r rt r2f At
A =G(r)Y, A =H(r)V . (34)
t A A
DAF =0 (23) Let us first solve the equation for At. Introducing the
Ar
new variable x defined by
and r n−3
0
x:= , (35)
n−4 1 r
∂rFrA+ r FrA+ r2DBFBA =0. (24) (cid:16) (cid:17)
the solution can be written in the analytic form of
G(r)=Br−(n+ℓ−3)F(α,β,γ;x)+CrℓF(α′,β′,γ′;x),
A. Gauge conditions (36)
where F(α,β,γ;x) is the hypergeometric function, and
We employ the following gauge
ℓ
α= (37)
A =DAA =0. (25) n−3
r A
n+ℓ−3
β = (38)
This can been achievedby following standardargument. n−3
There is the gauge freedom of Aµ → A˜µ = Aµ +∂µχ 2(n+ℓ−3)
γ = =α+β+1 (39)
Then if we choose χ as
n−3
and
χ=− drA (r,xA)+η(xA) (26)
r
Z ′ ℓ
α =− (40)
we can set n−3
n+ℓ−3
′
β =− (41)
A =0. (27) n−3
r
2ℓ
′ ′ ′
γ =− =α +β +1. (42)
Eq. (23) implies n−3
D2A −∂ (DAA )=0, (28) Fromtheasymptoticflatness,wemustsetC =0andthe
r r A
solution becomes
and then
A =Br−(n+ℓ−3)F(α,β,γ;x)Y. (43)
t
∂ (DAA )=0 (29)
r A Now we compute the field strength
that is, DAAA does not depends on the coordinate of F = −(n+ℓ−3)r−(n+ℓ−2)F(α,β,γ;x)Y
r. Using the remaining gauge freedom of A → A˜ = r0
A A
x d
AA+∂Aη(xB) satisfying −(n−3)r−(n+ℓ−3) F(α,β,γ;x)Y
r dx
D2η =−DAA , (30) = −(n+ℓ−3)r−(n+ℓ−2)F(α,β,γ;x)Y
A
αβ x
−(n−3) r−(n+ℓ−3) ×
we can set
γ r
d
DAA =0. (31) F(α+1,β+1,γ+1;x)Y (44)
A
dx
4
Let us examine the behavior on the horizon. Since DAHArA1···Ap−2 =0. (57)
Γ(β+1)Γ(γ−α−β−1)
F(α+1,β+1,γ+1;1)= (45) and
Γ(γ−α)Γ(γ−β)
n−2p f′
andΓ(γ−α−β−1)=Γ(0), itdivergesexceptforℓ6=0. ∂rHrA1···Ap−1 + r + f HrA1···Ap−1
This means that the second term in the right-hand side
(cid:16) (cid:17)
coafsEe,q.α(4v5a)ndisihveesrgaesn.dTthheencatsheeofseℓc=on0distesrpmecidails.apInpetahriss +r21fDBHBA1···Ap−1 =0. (58)
and the solution will be regular everywhere outside of
theblackholes. Thus,themonopolecomponent(ℓ=0)is
onlypermitted. Ofcourse,thisisthecaseoftheReissner- A. Gauge conditions
Nordstro¨m solution.
Next we solve the equation for AA and then we have Using the gauge freedom of Bµ1···µp−1 → B˜µ1···µp−1 =
the analytic solution as Bµ1···µp−1 + ∂[µ1Cµ2···µp−1], we can show that one can
H(r)=Br−(ℓ+n−4)F(α,β,γ;x)+Crℓ+1F(α′,β′,γ′;x), choose the following gauge condition
where (46) DABtAA1···Ap−3 =0 (59)
ℓ+n−4 DABrAA1···Ap−3 =0 (60)
α= n−3 (47) DBBBA1···Ap−2 =0. (61)
ℓ+n−2
β = (48) With Eq. (59), the field equation shows
n−3
ℓ+n−3
γ =2 =α+β (49) BtrA1···Ap−3 =0. (62)
n−3
Thedetailcanbeenseenbythefollowingargument. The
and
gauge transformation gives us
ℓ+1
′
α =− (50)
n−3 DAB˜tAA1···Ap−3
ℓ−1
β′ =−n−3 (51) =DABtAA1···Ap−3
′ 2ℓ −[D2−(p−3)(n−p+1)]CtA1···Ap−3, (63)
γ =− . (52)
n−3
where we already imposed DA1CtA1···Ap−3 =0. Then we
Fromtheasymptoticflatness,wemustsetC =0andthe take CtA1···Ap−3 satisfying
solution becomes
AA =Br−(ℓ+n−4)F(α,β,γ;x)VA. (53) [D2−(p−3)(n−p+1)]CtA1···Ap−3 =DAB˜tAA1···Ap−3
(64)
Since γ =α+β, on the event horizon,
Γ(γ)Γ(0) NotethatthereexiststhesolutionsforCtA1···Ap−3. Then
F(α,β,γ;1)= (54) this implies
Γ(γ−α)Γ(γ−β)
diverges. Therefore there is no regular solution. DABtAA1···Ap−3 =0 (65)
As a conclusion, the regular solution is ℓ = 0 mode
only ofA which correspondsto the Reissner-Nordstro¨m In this case, Eq. (55) becomes
t
solution. This is well-known fact.
[D2−(p−3)(n−p−1)]BtrA1···Ap−3 =0. (66)
IV. HIGHER FORM FIELDS In terms of harmonics, BtrA1···Ap−3 will be expanded as
In this section we examine the static perturbation of BtrA1···Ap−3 =J(r)TA1···Ap−3. (67)
H fields with p≥3. The field equations are
(p) Then
DAHAtrA1A2···Ap−3 =0 (55) (ℓ+p−3)(ℓ+n−p)J(r)=0. (68)
n−2(p−1) Except for the special case with ℓ = 0,p = 3, it is easy
∂rHrtA1A2···Ap−2 + r HrtA1A2···Ap−2 to see that
1
+r2fDBHBtA1···Ap−2 =0 (56) J(r)=0 (69)
5
holds. We can also show J(r) = 0 even for ℓ = 0,p = 3 Here we expandBtA1···Ap−2 interms ofthe harmonicsas
case by a distinct argument. In fact, we can use the
remaining gauge freedom of B˜tr =Btr−∂rCt(r). Then, BtA1···Ap−2 =K(r)TA1···Ap−2. (81)
taking
Then the above equation becomes
r
n−2(p−1)
Ct(r)= drBtr(r), (70) ∂2K+ ∂ K
r r r
Z
we can set (ℓ+p−2)(ℓ+n−p−1)
− K =0. (82)
r2f
B (r)=0. (71)
tr
The solution has been found in the analytic form of
Therefore, without loss of generality, we can conclude
that K(r) = Br−(ℓ+n−p−1)F(α,β,γ;x)
+Crℓ+p−2F(α′,β′,γ′;x), (83)
BtrA1···Ap−3 =0 (72)
where
holds.
ℓ+p−2
Next we will ask if we can take the gauge condition of
α= (84)
Eq. (60). To see this, we first look at n−3
ℓ+n−p−1
DAB˜rAA1···Ap−3 β = n−3 (85)
=DABrAA1···Ap−3 +∂r(DACAA1···Ap−3) γ =2ℓ+n−3 =α+β+1 (86)
−[D2−(p−3)(n−p+1)]CrA1···Ap−3, (73) n−3
and
where we imposed DACrAA1···Ap−3 =0. ℓ+p−2
Using Cµ1···µp−2 satisfying α′ =− (87)
n−3
∂r(DACAA1···Ap−3)=0 (74) β′ =−ℓ+n−p−1 (88)
[D2−(p−3)(n−p+1)]CrA1···Ap−3 n−3
ℓ
=DABrAA1···Ap−3, (75) γ′ =−2n−3. (89)
we can set
From the asymptotic flatness, the solution will be
DABrAA1···Ap−3 =0. (76) BtA1···Ap−2 =r−(ℓ+n−p−1)F(α,β,γ;x)TA1···Ap−2. (90)
Finally we consider the following transformation Now we can compute the field strength HrtA1···Ap−2
DAB˜AA1···Ap−2 HrtA1···Ap−2
=DABAA1···Ap−2 =−(ℓ+n−p−1)r−(ℓ+n−p)F(α,β,γ;x)TA1···Ap−2
+[D2−(n−p)(p−2)]CA1···Ap−2, (77) −(n−3)αβxr−(ℓ+n−p−1)×
γ r
CwAhe1·r·e·Apw−e2 ismatpisofsyeidngDACAA1···Ap−3 = 0. Then, taking F(α+1,β+1,γ+1;x)TA1···Ap−2. (91)
(92)
[D2−(n−p)(p−2)]CA1···Ap−2 =−DABAA1···Ap−2,(78) Since
we can adopt the gauge of
Γ(γ+1)Γ(γ−α−β−1)
F(α+1,β+1,γ+1;1)= ,
DABAA1···Ap−2 =0. (79) Γ(γ−α)Γ(γ−β)
(93)
B. Solutions and Γ(γ−α−β−1)=Γ(0)=−∞, the second term in
the right-hand side of Eq. (91) diverges at the horizon.
Thus, there are no regular solutions.
Now, under the current gauge conditions, Eq. (56)
In the current gauge, Eq. (58) becomes
becomes
∂r2BtA1···Ap−2 + n−2(rp−1)∂rBtA1···Ap−2 ∂r2BA1···Ap−1 + n−r2p + ff′ ∂rBA1···Ap−1
(cid:16) (cid:17)
D2−(n−p)(p−2) D2−(n−p−1)(p−1)
+ r2f BtA1···Ap−2 =0. (80) + r2f BA1···Ap−1 =0. (94)
6
Let us expand BA1···Ap−1 in terms of harmonics as A. deSitter cases
BA1···Ap−1 =L(r)TA1···Ap−1. (95) First we consider the cases with positive cosmologi-
Then Eq. (94) becomes cal constant. In this case, the background spacetime is
n−2p f′ higherdimensionalSchwarzschild-deSitterspacetimeand
∂2L+ + ∂ L f(r) in the metric becomes
r r f r
(ℓ+p(cid:16)−1)(ℓ+n−(cid:17)p−2) f(r)=1−(r /r)n−3−r2/a2. (108)
0
− L=0. (96)
r2f
Underacertaincaseswithparameterr anda,thereare
0
The solution is given by two horizons, black hole and cosmologicalhorizons at r
h
L(r) = Br−(ℓ+n−p−2)F(α,β,γ;x) and rc. Note that rc >rh.
From Eq. (82), we have the following relation
+Crℓ+p−1F(α′,β′,γ′;x), (97)
rc
where drrn−2(p−1) (∂ K)2
r
ℓ+n−p−2 Zrh "
α= (98)
n−3 (ℓ+p−2)(ℓ+n−p−1)
ℓ+n+p−4 + K2
β = (99) r2f #
n−3
n+ℓ−3 = rn−2(p−1)K∂ K rc (109)
γ =2 n−3 =α+β (100) r rh
h i
Since the presence of the cosmologicalconstantdoes not
and
disturb the behavior the horizons, the same regularity
ℓ+p−1
α′ =− (101) conditionsareimposedonthebothofthehorizons. Thus
n−3 wecanseethattheboundarytermintheright-handside
′ ℓ−p+1 vanishes and then
β =− (102)
n−3
K =0. (110)
ℓ
′ ′ ′
γ =−2 =α +β . (103)
n−3
In the same way, from Eq. (96), we have
The asymptotic flatness implies C =0 and then we see
rc
BA1···Ap−1 =Br−(ℓ+n−p−2)F(α,β,γ;x)TA1···Ap−1.(104) drrn−2p f(∂rL)2
Zrh "
Since
(ℓ+p−1)(ℓ+n−p−2)
Γ(γ)Γ(0) + L2
F(α,β,γ;1)= , (105) r2
Γ(γ−α)Γ(γ−β) #
we see the singular behaviors of the field strength as = rn−2pfL∂rL rc. (111)
rh
h i
HrA1···Ap−1 =O(1/(r−r0)) (106) Since the boundary term vanishes, we can see that
and
L=0 (112)
HrA1···Ap−1 =O(1/(r−r0)). (107)
holds. Therefore,therearenoregularstaticperturbation
Asaconclusionwecanshowthatblackholescannothave in the region of r ≤r≤r .
h c
the hair of the p-form field strengths.
B. anti-deSitter cases
V. ASYMPTOTICALLY (ANTI)DESITTER
SPACETIMES
Nextletusconsiderasymptoticallyanti-deSittercases.
In this case, f(r) = 1−(r /r)n−3 +r2/a2. Then, near
0
So far we concentrated on asymptotically flat space-
the infinity, K follows the equation approximately
times and could have the analytic solution for the equa-
tion of static perturbation. On the other hand, this is n−2(p−1)
∂2K+ ∂ K ≃0. (113)
notthecaseonceoneturnsonthecosmologicalconstant. r r r
Withoutsolvingtheequations,however,wecanaskifthe
Then the solution is approximately given by
solutionexists. Notethattheequationsforthestaticper-
turbations does not changedexcept for the expressionof A
f(r) in the metric of the background spacetimes. K ≃ rn−2p+1. (114)
7
In the current case, Eq. (82) gives us Appendix A: no-dipole-hair theorem revisit
∞
drrn−2(p−1) (∂ K)2
r
Zrh " In this appendix, we revisit the no-hair theorem of p-
form fields strengths in static asymptotically flat space-
(ℓ+p−2)(ℓ+n−p−1)
+ K2 times [12]. In the theorem, one supposes the presence
r2f
# of the electric part only. Then, if p ≥ (n + 1)/2, we
= rn−2(p−1)K∂ K ∞. (115) canshow that the p-formhair does not exist. From this,
r if one supposes the presence of the magnetic part of p-
rh
h i formfieldstrengthsonly,weexpectthatsimilartheorem
Sincetheboundarytermneartheinfinityisroughlyesti-
matedas dr1/rn−2p+2,onehastoimpose(n+1)/2>p holds. In fact, its dual version is the electric part of
(n−p)-form field strengths. Therefore, we would guess
inordertomakeitfinite. Thus,ifweimpose(n+1)/2>
R that, if (n + 1)/2 ≥ (n −p), the magnetic parts of p-
p,theboundarytermvanishesandthenwecanconclude
form field strengths does not exist. The above condition
K =0. (116) is rearrangedas p≥(n−1)/2. The above consideration
indicates the breakdown of the proof of the no-hair of
Similar result will be obtain for L, that is, L=0.
p-form field strengths if both parts exist. On the other
As a consequence, we can see that there no static per-
hand, the argument in the main text indicates the no-
turbationofp-formfield strengthinasymptoticallyanti-
hair of p-forms except for p=2. Or it may suggests the
deSitter spacetimes too.
presence of the solutions which cannot be explained by
the perturbation on the Schwarzschild spacetime.
VI. SUMMARY AND DISCUSSIONS
Let us examine if the no-hair theorem holds in the
details. Here we includes the dilation to the system de-
In this paper we studied the static perturbation of
scribed by the Lagrangian
p-form field strengths for the Schwarzschild-Tangherlini
spacetime and then we could show that the black holes
cannothavep-formhairexceptfortheMaxwellcases(p=
2,ℓ=0). Thisworkisinitiatedbyremainingissueinthe 1 1
no-hair theorem [12] of p-form fields in higher dimen- L=R− 2(∇φ)2− p!H(2p), (A1)
sional black hole spacetimes. That is, therein, there is
a limitation of p as p ≥ (n +1)/2 in the proof of the
no-hair theorem. Therefore, it was natural to ask if the
no-hair properties with p<(n+1)/2 is. Our currentre- where φ is the dilation field. The Einstein equation is
sult supports no-hair properties of p-form field strength
with p≥3 regardless of such limitation.
Ouranalysisisbasedontheperturbationandthenthe
ttohpinoklosgoyfoafnobtlahcekr htoopleosloigsylimlikiteebdlatockbreinsgp,htehree.reSaoriefsotnilel Rµν = 12∇µφ∇νφ+ p1!e−αφ pHµα1···αp−1Hνα1···αp−1
possibilitytohaveasolution. Accordingtotheappendix,
however, the solution may have the both electric and −p−1g H2 . (A2)
magnetic hairs if it exists. They will be addressed in n−2 µν (p)!
near future study.
Acknowledgments Since we will not the equations for the p-form fields and
dilation,wedonotwritedowntheseequations. Different
TS is partially supported by Grant-Aid for from Ref. [12], we will not assume that the p-form fields
Scientific Research from Ministry of Educa- have the electric components only. The metric of static
tion, Science, Sports and Culture of Japan spacetimes is written as Eq. (5). From the Einstein
(Nos.21244033,21111006,20540258and19GS0219). SO equation, then, we can see
thanks Professor Takashi Nakamura for his continuous
encouragement. KT is supported by JSPS Grant-Aid
for Scientific Research (No. 21-2105). SO and KT are
R = VD2V
supported by the Grant-in-Aid for the Global COE 00
PUrnoivgerarsmali“tTyhaenNdexEtmGeerngeernactei”onfroofmPhythsiecs,MSipnuisntryfromof = (nn−−2)p(p−−11)!e−αφH0i1···ip−1H0i1···ip−1
E(MduEcXaTtio)no,f JCauplatunr.e, Sports, Science and Technology +(np−−21)p!V2e−αφHi1···ipHi1···ip (A3)
8
and where
Rij = (n−1)Rij − V1DiDjV Ω± = 1±V n−23 =:ω±n−23. (A7)
2
1
(cid:16) (cid:17)
= D φD φ
i j
2 Then
+(p−12)!e−αφ Hi0k1···kp−2Hj0k1···kp−2 Ω2(n−1)R˜
±
(cid:16)
1
−n−2gijH0k1···kp−1H0k1···kp−1 =(n−1)R∓ 2(n−2)ω±−1D2V
n−3
+(p−11)!e−αφ Hik1···kp−1Hjk1·(cid:17)··kp−1 = 1(Dφ)2+ 1 e−αφ λ±H i1···ip−1H0i1···ip−1
(cid:16) 2 (p−1)! V2 ω± 0
p−1
−p(n−2)gjkHk1···kp−1(cid:17) (A4) + p1!e−αφωµ±±Hi1···ipHi1···ip, (A8)
hold. Moreover, we can compute the Ricci scalar of g
ij
where
as
(n−1)R = 12(Dφ)2+ (p−11)!eV−α2φH0i1···ip−1H0i1···ip−1 λ± = 1∓ 3n2−n−4p3−1 (A9)
1
+p!e−αφHi1···ipHi1···ip. (A5) and
woTrkhse. oWutelifinrestocfonthsiedeprrtohoef cwoinllforbmealastrafonlslfoowrms aitfioint µ± = 1± n2−n4−p3+1. (A10)
of t =constant hypersurfaces so that the Ricci scalar is If (n−1)R˜± ≥ 0, we can proceed the proof. However, we
non-negativeand the ADM mass vanishes. Then we will cannotdothat. The sufficientconditionfor(n−1)R˜± ≥0
apply the positive mass theorem [21, 22] and then show are λ± ≥0 and µ± ≥0. Each conditions become
that the conformally transformed spacetime is flat and
the p-form hair does not exist. We know that the vac- n+1 n−1
p≥ and p≤ , (A11)
uum black hole spacetimes with conformally flat static 2 2
slices must be spherical symmetry. Thus, the resulted
spacetimes is the Schwarzschild spacetime. respectively. The both conditions together do not hold
Letus lookatthe details. For the proofofno-hair,we manifestly. Therefore, we can say nothing about the no-
willconsiderthetwoconformaltransformationsgivenby hair for the cases having both of electric and magnetic
p-form fields. This results may suggest the presence of
g˜± =Ω2g , (A6) the p-form hairy static black object solutions.
ij ± ij
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