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9.2 Black Holes Once Again

 

 

 

369

vector of the following type:

 

 

 

 

p1

 

p

 

 

p > 0, q < 0

or

p < 0, q > 0

BPS

p

 

 

0

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

q

 

 

 

 

q

 

 

 

 

 

 

1

3

p > 0, q > 0

or

p < 0, q < 0

non-BPS

 

 

=

0

 

 

q2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(9.2.104) and let us consider the solution of the attractor equations (9.2.40) with the above charge vector.

Non-BPS Case For p and q having the same sign it is easily verified that there is no solution of the equation Zz = 0 and hence no BPS attractor point. On the other hand there is a solution of the critical point equation (9.2.40) with both Zz = 0 and Z = 0. It corresponds to the following simple fixed value:

 

 

 

 

zfixed = −i

 

p

 

 

 

 

(9.2.105)

 

 

 

 

 

 

 

 

 

 

 

q

 

 

 

With such fixed value the i-invariant take the following values:

 

 

{i1, i2, i3, i4, i5} =

1

 

 

 

 

 

3

 

 

 

 

 

pq3

 

 

 

 

 

p

 

p

 

 

 

 

 

q2,

 

 

 

q2, 0,

 

, 3pq3

(9.2.106)

2

q

2

q

2

which satisfy the relations (9.2.48) characterizing a non-BPS attractor point of

type II. Furthermore the quartic invariant I (p, q) = −pq3 < 0 is negative in this

4

case and we expect that the horizon area will be proportional to −I4. This is indeed the case as we verified few lines above. Furthermore if we calculate the limiting value of the scalar field (9.2.98), (9.2.99) at τ → −∞ we precisely find the fixed value (9.2.105).

BPS Case If p and q have opposite signs there is just one solution of the equation Zz = 0 with Z = 0. Hence we a have a BPS attractor. The fixed point is:

p

 

zfixed = −i q

(9.2.107)

which perfectly fits the general formula (9.2.79). Moreover calculating the i- invariants at the fixed point we obtain:

{i1, i2, i3, i4, i5} = (2 pq3, 0, 0, 0, 0) (9.2.108)

which fulfills the relations (9.2.47) proper of the BPS attractors.

9.2.10 An Example of a Small Black Hole

Let us consider the exact solution of the σ -model variational equations encoded in the functions displayed below that depend on four parameters y, σ , ξ , κ:

370

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

9 Supergravity: An Anthology of Solutions

The Metric

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

y2 + 1 3 + 1

 

 

 

 

 

U (τ ) = −

 

 

log 2σ τ

(9.2.109)

 

 

 

 

2

The Complex Scalar Field

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

z(τ )

=

ξ(2(y2

+ 1)σ τ (y2 1)2 + 1) 4y(y4 1)κσ τ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2(y2 + 1)σ τ (y2 1)2 + 1

 

 

 

 

 

2(y

 

1)σ τ (y 1)

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

i

κ

 

 

2σ τ (y2 + 1)3

+ 1

 

 

 

 

 

 

 

 

 

(9.2.110)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

+

 

 

 

2

 

2

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The Electromagnetic Fields

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

y2((y2 1+ 2yξ )σ

 

 

 

 

 

Z1(τ )

=

6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

˙

 

 

 

κ3/2(2σ τ (y

2

+

1)3

+

1)2

 

(9.2.111)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8

 

 

 

 

 

 

 

 

 

 

 

 

Z2(τ )

 

 

 

 

 

 

 

2

y3σ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

+

 

 

 

 

 

 

 

 

 

˙

 

 

=

 

κ3/2(2σ τ (y2

1)3

1)2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The Charges Using the general formulae discussed in previous pages we find that the Taub-Nut charge is zero:

n = 0

while for the electromagnetic charges we get:

 

p1

 

 

 

2

2

 

2

 

3

2

 

 

 

 

 

 

 

 

2 3(y

κ3/2

 

 

p2

 

 

 

1)(ξy

 

+2κy+ξ ) σ

 

2(ξy +3/2

+

 

 

 

 

 

 

 

 

 

κ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2κy ξ ) σ

q

 

2

 

(y2

1)2(ξy2

 

2κy

 

 

3

 

ξ )σ

 

1

=

 

 

 

 

 

 

κ

3/2

 

q

2

 

 

 

 

 

 

 

 

 

1)3

σ

 

 

 

 

 

 

 

 

 

 

 

2(y2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

κ3/2

 

 

 

 

(9.2.112)

(9.2.113)

Structure of the Charges and Attractor Mechanism Observing the right hand side of (9.2.113), we realize that in this solution the electromagnetic charges satisfy the following two algebraic constraints:

q12 +

 

 

p1q2

= 0

 

 

 

 

(9.2.114)

3

 

 

 

 

3

 

 

2

 

= 0

 

 

 

 

(9.2.115)

 

 

 

 

 

 

 

p1

+ 3

 

3p2 q2

 

 

 

 

which can be solved for qΛ in terms of pΣ . Explicitly we have:

 

{q1, q2} =

 

2

 

 

 

p3

 

 

p

1

,

1

2

(9.2.116)

 

 

 

 

 

3p2

 

3

3p2

 

 

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