Electrically Charged Black-holes for the Heterotic String Compactified on a $(10-D)$-Torus

17 Pages·0.17 MB·English

by  Pablo M. Llatas

ELECTRICALLY CHARGED BLACK-HOLES FOR THE HETEROTIC STRING
7
9
9 COMPACTIFIED ON A (10 D)-TORUS
1 −

n
a
J

1
2

2
v
8
5 Pablo M. Llatas1
0
5
Department of Physics
0
6 University of California
9 Santa Barbara, CA 93106-9530
/
h
t
-
p
e We show that the most general stationary electrically charged black-hole solutions of the
h
heterotic string compactified on a (10 D)-torus (where D > 3) can be obtained by using
:
v −
the solution generating transformations of Sen acting on the Myers and Perry metric. The
i
X
conserved charges labeling these black-hole solutions are the mass, the angular momentum
r
in all allowed commuting planes, and 36 2D electric charges. General properties of these
a
−
black-holes are also studied.
May, 1996
1 [email protected]
1. Introduction
Black-hole solutions emerging in string theory have been extensively studied in the
recentliterature. Oneofthereasonsforsuchattentionisthesuggestionthattheelementary
massive string states (with Planck masses) could be identified with black-holes ([1] and
references therein). Thisidentificationisnotnecessarilymadeinthesamestringtheorybut
usuallyinvolvesdualpictures: aclassicalblack-holesolutionofastringtheory(“soliton”)is
identified with a quantum (bound or elementary) state on a dual theory. Identifications of
this type have been recently employed to provide a statistical derivation of the Bekenstein-
Hawking entropy by identifying the black-holes solutions with bound configurations of
D-branes ([2,3,4,5,6,7]).
In the present work, we generate the most general rotating electrically charged black-
hole solution of the heterotic string compactified on a (10 D)-torus (conforming the “no
−
hair” theorems). We follow the work of [8], where the particular case D = 4 was studied.
Related works (using the same rotating technique to generate new solutions from a given
one) can be found in [9,10,11,12], where cases of different dimensions, charges and number
of rotation planes where studied. In the present work, we generate a family of solutions
depending on a mass, D 1 angular momenta ( denotes the integer part) and 36 2D
2− −
h i h i
electrical charges. 16 electrical charges come from the U(1)16 of the heterotic string in
ten dimensions (on a general point on the moduli space of compactifications) and the
remaining 2 (10 D) electric charges come from the compactified dimensions (any time
· −
we compactify a spatial dimension there appears generically two new U(1) gauge fields:
the Kaluza-Klein U(1) field comming from the metric G and the “winding” U(1) field
µν
originating from the antisymmetric tensor field B ). These 1 + D 1 + 36 2D are
µν 2− −
the largest number of parameters labeling these stationary black-hol(cid:2)es con(cid:3)forming the “no
hair” theorems. In section 2 we introduce the action for the heterotic string compactified
on a (10 D)-torus. The solutions we are about to introduce satisfy the equations of
−
motion derived from this action. In section 3 we describe the generalization to any D of
the solution generating technique of [8] which generates new solutions to the equations of
motion from a given one. We will show that the generating boosts depend on 36 2D
−
free parameters, in such a way that if we rotate the Myers and Perry metric [13] (which
already depends on 1+ D 1 parameters corresponding to the mass andangular momenta
2−
h i
associated to rotations in all commuting planes) we generate the most general solution we
want to construct. In section 4 we carry out the rotation and obtain expressions for the
family of new solutions. Finally, in section 5 we study general features of these general
black-hole solutions, like the mass, electric charges, angular momentums, ergosphere and
horizons.
2. The Heterotic String Compactified on a (10 D)-Torus.
−
The low energy effective theory of the ten-dimensional heterotic string corresponds to
N = 1 supergravity coupled to U(1)16 super Yang-Mills (on a generic point of the space
of compactification). The bosonic content of this theory is given by the metric G , the
AB
dilaton Φ, the antisymmetric tensor B and the U(1) gauge bosons AI (where A,B :
AB A
1
0,1,...,9, and I : 1,2,..,16). When we compactify on a (10 D)-torus, the metric G
AB
−
induces the D-dimensional metric G , 10 D U(1) Kaluza-Klein gauge bosons G and
µν µi
−
(11 D)(10 D)/2 scalars G ( here µ,ν : 0,1,..D 1, and i,j : 1,2,..10 D). The gauge
ij
− − − −
U(1)bosonsAI originate16U(1)gaugebosonsAI and16 (10 D) scalarsAI. Finally, the
A µ · − i
antisymmetric tensor field B induces the two form B , 10 D U(1) “winding” gauge
AB µν
−
bosons B and (10 D)(9 D)/2 scalars B . Then, the total number of gauge bosons
µi ij
− −
of the compactified theory is 36 2D. These gauge bosons can be arranged in a column
−
(a)
matrix of vectors A (where a : 1,..,36 2D). The total number of scalars (excluding
µ
−
the dilaton) is 260 36D+D2. These scalars can be arranged in a (36 2D) (36 2D)
− − × −
matrix M = MT valued on the group G given by:
O(10 D,26 D)
M G, G = − − (2.1)
⊂ O(26 D) O(10 D)
− × −
This matrix fulfills MLMT = L where L is the matrix given by:
I 0
L = − 26−D (2.2)
(cid:18) 0 I10 D(cid:19)
−
(I is the n n unit matrix). One can easily check that the dimension of G is precisely
n
×
260 36D+D2, in such a way that the scalars of the theory (excluding the dilaton) fit in
−
(a)
G. In terms of Φ, M, A , B and L, the action for the heterotic string compactified on
µ µν
a (10 D)-torus takes the form [14]:
−
1
S = C dDx√ Ge Φ R +Gµν∂ Φ∂ Φ+ GµνTr(∂ ML∂ ML)
− G µ ν µ ν
Z − 8
(cid:2)
1 Gµµ′Gνν′Gρρ′HµνρHµ′ν′ρ′ (2.3)
− 12
Gµµ′Gνν′F(a)(LML) F(b)
− µν ab µ′ν′
(cid:3)
where
F(a) = ∂ A(a) ∂ A(a) (2.4)
µν µ ν ν µ
−
and
H = ∂ B +2A(a)L F(b) +cyclic. (2.5)
µνρ µ νρ µ ab νρ
It is straightforward to check that this action is invariant under global rotations Ω
leaving the matrix L invariant (ΩLΩT = L):
M ΩMΩT, A(a) Ω A(b) (2.6)
µ ab µ
→ →
Under these rotations, G , B and Φ remain invariant.
µν µν
3. The Solution-Generating Technique, “Bar” Fields.
Here we describe in some detail the solution-generating technique by following the
work of Sen [8] where the case D = 4 was studied (see also [12,9,10,11]). This technique
2
can be used to, given one time-independent solution of the equations of motion of the
action (2.3), generate a new family of time-independent solutions. If we restrict ourselves
to time-independent backgrounds we find two new U(1) gauge fields in the theory: B
0m
and G0m. The first U(1) gauge field is related to the time-independent gauge invariance
G00
δB = ∂ λ ∂ λ = ∂ λ . The second gauge field is related to invariance under
m0 m 0 0 m m 0
−
local time-independent translation of the time coordinate. One can add these two new
(a)
U(1) gauge fields to our 1 (36 2D) column matrix of vectors A and define the
µ
× −
1 (38 2D) matrix of vectors A¯(a¯) (a¯ : 1,2,..38 2D) given by:
µ
× − −
G
A¯(a) A(a) 0nA(a) 1 a 36 2D
n ≡ n − G 0 ≤ ≤ −
00
1G
A¯(37 2D) 0n (3.1)
n − ≡ 2 G
00
1
A¯(38 2D) B +A(a)L A¯(b)
n − ≡ 2 0n 0 ab n
(a)
(note that A is a scalar under time-independent general coordinate transformations).
0
Also, the (36 2D) (36 2D) G-valued matrix M is promoted to a (38 2D)
(38 2D) matrix M−¯. The×elem−ents of the M¯ matrix are given by (M¯ = M¯T): − ×
−
(a) (b)
A A
M¯ M +4 0 0 1 a,b 36 2D
ab ab
≡ G ≤ ≤ −
00
(a)
A
M¯ 2 0 1 a 36 2D
a,37 2D
− ≡ − G00 ≤ ≤ −
1
M¯
37 2D,37 2D
− − ≡ G00
(3.2)
(b) (c)
(A L A )
M¯ 2(ML) A(b) +4 0 bc 0 A(a) 1 a 36 2D
a,38−2D ≡ ab 0 G00 0 ≤ ≤ −
(b) (c)
(A L A )
M¯ 2 0 bc 0
37 2D,38 2D
− − ≡ − G00
(A(b)L A(c))2
M¯ G +4A(b)(LML) A(c) +4 0 bc 0
38−2D,38−2D ≡ 00 0 bc 0 G00
Also, one defines a “bar” metric by:
G G
G¯ G 0n 0m (3.3)
nm nm
≡ − G
00
and a “bar” antisymmetric tensor:
1 G 1
B¯ B + 0n(A(a)L A(b) B ) (n m) (3.4)
nm ≡ 2 nm G m ab 0 − 2 0m − ↔
00
Finally, the “bar” dilaton is defined through:
3
1
Φ¯ = Φ ln( G ) (3.5)
00
− 2 −
For time-independent field configurations, the action (2.3) can be written in terms of
the “bar” fields as:
1
S = C dt dD 1x G¯e Φ¯ R +G¯mn∂ Φ¯∂ Φ¯ + G¯mnTr(∂ M¯L∂ M¯L)
− − G¯ m n m n
Z Z − 8
p (cid:2)
1 G¯mm′G¯nn′G¯ll′H¯mnlH¯m′n′l′ (3.6)
− 12
G¯mm′G¯nn′F¯(a¯)(L¯M¯L¯) F¯(¯b)
− mn a¯¯b m′n′
(cid:3)
where now:
F¯(a¯) = ∂ A¯(a¯) ∂ A¯(a¯) (3.7)
mn m n n m
−
H¯ = ∂ B¯ +2A¯(a¯)L¯ F¯(¯b) +cyclic. (3.8)
mnl m nl m a¯¯b nl
(1 a¯,¯b 38 2D) and L¯ is given by:
≤ ≤ −
L 0 0
L¯ =  0 0 1 (3.9)
0 1 0
 
(L is the one in (2.2)). Again, one straightforwardly verifies that this action is invariant
under the O(11 D,27 D) rotations leaving the matrix L¯ invariant (Ω¯L¯Ω¯T = L¯):
− −
M¯ M¯ = Ω¯M¯Ω¯T
′
→ (3.10)
A¯(a¯) A¯(a¯) = Ω¯ A¯(¯b)
n → ′n a¯¯b n
Once more, Φ¯, G¯ and B¯ remain invariant under this O(11 D,27 D) rotation.
nm nm
− −
The spirit of the solution generating technique is the following. The rotation (3.10)
in the space of time-independent backgrounds mixes the “genuine” initial (36 2D) U(1)
−
(a)
gauge fields A with the components G and G of the metric tensor and the com-
µ 00 0n
ponents B of the antisymmetric tensor (as can be seen in (3.1)). Then, through this
0n
mixing one is able to generate new non-trivial solutions to the equations of motion of (2.3)
(non-trivial means that we do not produce equivalent solutions). Note that due to the
fact that we do not have magnetic charges (monopoles) the gauge fields are well defined
globally and so, the new metrics obtained after rotation (with whom the original gauge
fields mix) are also well defined.
Let usnow provethatthenon-trivialgeneratingrotationsleavinginvarianttheasymp-
totic behaviour of the solution are labeled by 36 2D free parameters. As we noted above,
not all the O(11 D,27 D) rotations are allo−wed. Ω¯ must satisfy Ω¯L¯Ω¯T = L¯. Let us
− −
study this condition (we follow the lines of [8]). The first observation is that the U matrix
given by:
4
I 0 0
36 2D
U =  0− 1 1  (3.11)
√2 √2
0 1 1
−
 √2 √2 
diagonalizes L¯:
I 0 0 0
26 D
− −
0 I 0 0
L¯ UL¯UT =  10 D  (3.12)
d ≡ 0 0− 1 0
 0 0 0 1
 
−
Then, we can write the condition Ω¯L¯Ω¯T = L¯ as (UTU = I ):
38 2D
−
Ω¯ L¯ Ω¯T = L¯ (3.13)
d d d d
where we have defined Ω¯ UΩ¯UT. In an asymptotically free space-time (where, at
d
≡
infinity, G η and A(a) 0), M¯ has the asymptotic form:
µν µν µ
→ →
I 0 0
36 2D
M¯  0− 1 0  (3.14)
→ −
0 0 1
 − 
(where we take M = I ). From (3.12) and (3.14) we conclude [8] that Ω¯ (and then,
36 D d
Ω¯ itself) belongs to a sub−group O(1,26 D) O(10 D,1) of O(11 D,27 D). The
− × − − −
O(1,26 D) rotates the first 26 D elements and the element 37 2D of a column vector
− − −
between themselves whereas the O(10 D,1) rotates the elements 27 D,..,36 2D and
− − −
38 2D of a column vector between themselves. However, not all the elements of the
−
subgroup O(1,26 D) O(10 D,1) generate new solutions. We have to quotient by
− × −
the global symmetry transformations Ω in (2.6) of the action (2.3) that do not generate
new solutions. Then, we have to quotient O(1,26 D) O(10 D,1) by the O(26 D)
− × − −
rotations mixing the first 26 D elements of a column vector between themselves and
−
the O(10 D) rotations mixing the elements 27 D,..,36 2D between themselves.
− − −
Summarizing, the rotation group in the 38 2D internal space generating inequivalent
solutions of the equations of motion associate−d to the action (2.3) is given by G¯:
O(1,26 D) O(10 D,1)
¯
G = − × − (3.15)
O(26 D) O(10 D)
− × −
¯
One can easily verify that the dimension of G is precisely 36 2D.
−
The strategy to generate the most general electrically charged black-hole solutions in
the heterotic string compactified on a (10 D)-torus is now clear. First, we take as the
−
generating solution the one given by:
G = GMP
µν µν
M = I (3.16)
36 2D
−
Φ = B = A(a) = 0
µν µ
5
where GMP is the metric of Myers and Perry (see next section), which from now on we
µν
will refer as the MP metric. This metric is designed for dimensions D > 3 and we will
assume from the rest of the paper that this is the case. The MP metric already depends on
1+ D 1 parameters: the mass and the angular momentums associated to the maximun
−
2
allow(cid:2)ed n(cid:3)umber of rotating planes. M has to be an element of G and I36 2D is the
−
simplest choice. (3.16) is trivially a solution of the equations of motion of the action (2.3)
(essentially, the constant fields satisfy automatically the equations of motion leaving as the
only non-trivial equations the Einstein equations in vacuum, to which GMP is a solution).
µν
The next step is to use the solution (3.16) to define the “bar fields” (3.1) and (3.2).
In particular A¯(a¯) is given by (substitute (3.16) in the definitions (3.1) and (3.2)):
n
1G
A¯(a¯) = 0nδ (3.17)
n 2 G00 a¯,37−2D
and M¯ is given by:
I 0 0
36 2D
M¯  −0 (G ) 1 0  (3.18)
00 −
→
0 0 G
00
 
One then performs the Ω¯ rotation (depending on 36 2D free parameters) and gets A¯′n(a¯)
¯ −
and M . Finally one goes backwards in the relations (3.1) and (3.2) and recovers the
a¯′¯b
(a)
new solutions G , M , Φ , A and B . The metric G so obtained depends on
′µν ′ ′ ′µ ′µν ′µν
1+ D 1 +36 2D parameters, and then is the most general electrically charged black-
−
2 −
hole(cid:2)solut(cid:3)ion allowed conforming the “no hair” theorems.
¯
In the next section we explicitly carry out the G rotation and find the new solutions.
4. The General Solutions.
Let us first suppose that we have performed the rotations Ω¯ in (3.10) and obtained
the vector A¯′n(a¯) and the matrix M¯a¯′¯b. We can use the relations defining the “bar fields” in
(3.1) and (3.2) to obtain the new solutions in terms of these A¯′n(a¯) and M¯a¯′¯b. For instance,
in (3.2) we see that G is just given by:
′00
1
G = (4.1)
′00 M¯
3′7 2D,37 2D
− −
(a)
As another example, we easily obtain A from (3.2):
′0
¯
A′(0a) = −G2′00M¯a′,37 2D = −2M¯Ma′,37−2D 1 ≤ a ≤ 36−2D (4.2)
− 3′7 2D,37 2D
− −
In a similar way one can compute the expressions for all the fields of the new solutions
in terms of the elements of the rotated vector A¯′n(a¯) and the rotated matrix M¯a¯′¯b. Let us
collect the result. The new metric is given by:
6
¯(37 2D)
1 An′ −
G = , G = 2
′00 M¯ ′0n M¯
3′7 2D,37 2D 3′7 2D,37 2D
− − − − (4.3)
¯(37 2D) ¯(37 2D)
G0nG0m An′ − Am′ −
G = G +4
′nm nm − G M¯
00 3′7 2D,37 2D
− −
The dilaton is given by:
1
¯
Φ = Φ ln(G M ) (4.4)
′ − 2 00 3′7−2D,37−2D
The expressions for the U(1) gauge fields are:
(a) 1 M¯a′,37 2D
A′0 = −2M¯ − 1 ≤ a ≤ 36−2D
3′7 2D,37 2D
− − (4.5)
¯
M
A′(na) = A¯′n(a) − M¯ a′,37−2D A¯n′(37−2D)
3′7 2D,37 2D
− −
The scalar fields of the new solution are given by:
M¯ M¯
M′ab = M¯a′b − aM′¯,37−2D b′,37−2D 1 ≤ a,b ≤ 36−2D (4.6)
3′7 2D,37 2D
− −
And finally, the new antisymmetric tensor is given by:
M¯
B′0n = 2A¯n′(38−2D) + M¯ a′,37−2D LabA¯n′(b)
3′7 2D,37 2D
− −
1 G G
B = B + 0nA(a)L A(b) +B 0m
′nm 2 nm G m ab 0 0n2G
00 00
1
+ A¯(37 2D)A¯(a)L M¯ +2A¯(37 2D)A¯(38 2D) (n m)
M¯3′7 2D,37 2D n′ − ′m ab b′,37−2D n′ − m′ − − ↔
− −
(4.7)
(note that this last expression simplifies in our case, because we are going to rotate the
(a)
solution (3.16) where Φ = A = B = 0). What remains now is to compute the rotated
µ µν
vector A¯′n(a¯) and matrix M¯a¯′¯b and substitute the result in the expressions (4.3)-(4.6) to
obtain the final solution. In (3.15) we saw that the generating boosts are elements of the
group
O(1,26 D) O(10 D,1)
¯
Ω − × − (4.8)
∈ O(26 D)O(10 D)
− −
We also saw that Ω¯ = UΩ¯UT is such that the O(1,26 D) rotates the first 26 D
d
− −
elements and the element 37 2D between themselves whereas the O(10 D,1) rotates
− −
the elements 27 D,..,36 2D and 38 2D between themselves. The O(26 D) and
− − − −
O(10 D) that we are quotienting out are the rotations between the first 26 D elements
− −
7
and the rotations between the elements 27 D,..,36 2D respectively. The simplest way
to parametrize this Ω¯ is the following [8]:− −
d
Ω¯ = UΩUT = R(p¯,q¯) B(α,β) (4.9)
d
·
where B(α,β) performs a boost of angle α between the elements 26 D and 37 2D and
− −
a boost of angle β between the elements 36 2D and 38 2D:
− −
B(α,β) = δ +(coshα 1)(δ δ +δ δ )
a¯,¯b a¯,¯b − a¯,26−D ¯b,26−D a¯,37−2D ¯b,37−2D
+sinhα(δ δ +δ δ )
a¯,26 D ¯b,37 2D a¯,37 2D ¯b,26 D
− − − − (4.10)
+(coshβ 1)(δ δ +δ δ )
− a¯,36−2D ¯b,36−2D a¯,38−2D ¯b,38−2D
+sinhβ(δ δ +δ δ )
a¯,36 2D ¯b,38 2D a¯,38 2D ¯b,36 2D
− − − −
¯
andR(p¯,q¯)isgivenbyR(p¯,q¯) = R (p¯) R (q¯) I whereR (k)isaN-dimensional
26 D 10 D 2 N
− ⊕ − ⊕
rotation matrix rotating the N-dimensional column unit vector (0,0,..,0,1) to the N-
dimensional unit vector (k ,k ,..,k ):
1 2 N
k 2
¯ i
[R (k)] =δ 1 (1 δ )+k δ
N ij ij iN N iN
− 1+k −
(cid:2)(cid:0) N(cid:1) (cid:3) (4.11)
k k
i j
+(1 δ ) (δ +δ 1) +k δ k δ
ij iN jN i jN j iN
− − 1+k −
N
(cid:2) (cid:3)
As a consistency check of this parametrization of the generating boosts, note that the total
number of parameters of the matrix Ω¯ are 25 D coming from the unit vector p¯, 9 D
d
− −
coming from the unit vector q¯, and the two boost angles α and β, making a total of 36 2D
−
parameters (precisely the dimension of the group (4.8)).
Now, we are in position to perform the boost of the vector A¯(a¯) and the matrix M¯ .
n a¯¯b
First, we boost A¯′n(a¯). We need to compute:
A¯(a¯) = Ω¯ A¯(¯b) = (UTR(p¯,q¯)B(α,β)U) A¯(¯b) (4.12)
′µ a¯¯b µ a¯¯b µ
From our expressions (3.11), (4.10), (4.11), and (3.17) one gets the following result:
sinhαG
A¯(a) = 0npa 1 a 26 D
′n 2√2 G ≤ ≤ −
00
sinhβ G
A¯(a+26 D) = 0nqa 1 a 10 D
′n − 2√2 G ≤ ≤ −
00 (4.13)
1 G
A¯(37 2D) = (coshα+coshβ) 0n
n′ − 4 G
00
1 G
A¯(38 2D) = (coshα coshβ) 0n
n′ − 4 − G
00
Now, we compute M¯ . We need to calculate, in this case:
′
M¯ = Ω¯M¯Ω¯T = UTR(p¯,q¯)B(α,β)UM¯UTBT(α,β)RT(p¯,q¯)U (4.14)
′
8
The result is:
M¯ = δ +(1+G+)sinh2αpapb 1 a,b 26 D
a′b ab
≤ ≤ −
M¯ = sinhαsinhβG paqb 1 a 26 D, 1 b 10 D
a′,b+26−D − ≤ ≤ − ≤ ≤ −
M¯ = δ +(1+G+)sinh2βqaqb 1 a,b 10 D
a′+26 D,b+26 D ab ≤ ≤ −
− −
a+
M¯ = 1 pa 1 a 26 D
a′,37−2D G00 ≤ ≤ −
a+
M¯ = 2 qa 1 a 10 D
a′+26 D,37 2D G ≤ ≤ −
− − 00 (4.15)
∆
M¯ =
3′7 2D,37 2D G
− − 00
a
M¯ = −1 pa 1 a 26 D
a′,38 2D G ≤ ≤ −
− 00
a
M¯ = −2 qa 1 a 10 D
a′+26 D,38 2D G ≤ ≤ −
− − 00
δ
M¯ = M¯ =
3′7 2D,38 2D 3′8 2D,38 2D G
− − − − 00
where we have defined:
1 1
G ( G )
± 00
≡ 2 G ±
00
sinhα
a coshα coshβ +2coshαG +(coshα coshβ)G 2
±1 ≡ 2√2 ± 00 ∓ 00
(cid:2) (cid:3)
sinhβ
a coshα coshβ 2coshβG (coshα coshβ)G 2 (4.16)
±2 ≡ 2√2 ± ± 00 − ∓ 00
(cid:2) (cid:3)
1 G G2
∆ (coshα+coshβ)2 + 00(sinh2α+sinh2β)+ 00(coshα coshβ)2
≡ 4 2 4 −
1 1
δ (cosh2α cosh2β)(1+G 2)+ (sinh2α sinh2β)G
00 00
≡ 4 − 2 −
Now, we areready tosubstitute these resultsin(4.1)-(4.7)to obtainthefinal solution.
Here we collect the result. The boosted metric reads:
G 1 G
00 0n
G = , G = (coshα+coshβ)
′00 ′0n
∆ 2 ∆
(4.17)
(coshα+coshβ)2 G G
0n 0m
G = G + 1
′nm nm
4∆ − G
h i 00
The boosted dilaton is:
1
Φ = ln∆ (4.18)
′
−2
9