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Gravity and Strings

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346 The Kaluza–Klein black hole<br />

It is given by the action<br />

11.5.3 Example 3: an SL(2, R)/SO(2) σ-model<br />

<br />

ˆS =<br />

d ˆd<br />

<br />

<br />

ˆx |ˆg|<br />

ˆR + 1<br />

∂ ˆτ∂ˆτ ∗<br />

2 (Im(τ)) 2<br />

<br />

. (11.245)<br />

This action is invariant under global SL(2, R) fractional-linear transformations of ˆτ =â +<br />

ie −ˆϕ <strong>and</strong>, in particular, under constant shifts ˆτ →ˆτ + b, which act only on the real part.<br />

Furthermore, we have argued that, in many cases, we should consider as equivalent two<br />

values of τ related by SL(2, Z) transformations, which in particular means that a lives in a<br />

circle of unit length. In this case, the σ -model metric is regular for finite values of ˆϕ.<br />

The general recipe of GDR tells us to use the Ansatz<br />

ˆτ(ˆx) = τ(x) + N<br />

z,<br />

2πℓ<br />

(11.246)<br />

<strong>and</strong> we obtain the action<br />

<br />

S = d d x <br />

|g| k R − 1<br />

4k2 F 2 + 1<br />

2 (∂ϕ)2 + 1<br />

2 (Da)2 − 1<br />

2 N<br />

k 2 2πℓ<br />

−2 e −2ϕ<br />

<br />

,<br />

(11.247)<br />

where<br />

Dµa = ∂µa + N<br />

2πℓ Aµ, (11.248)<br />

<strong>and</strong> there is invariance under the massive U(1) gauge transformations<br />

δ Aµ = ∂µ, δa = N<br />

. (11.249)<br />

2πℓ<br />

Global SL(2, R) invariance is now clearly broken <strong>and</strong> the KK U(1) gauge invariance is<br />

also broken by the st<strong>and</strong>ard Stückelberg mechanism.<br />

Let us look for a vacuum solution that will have Aµ = 0 <strong>and</strong> a = a0, aconstant. The<br />

action for the remaining fields, in the Einstein frame, is<br />

<br />

S =<br />

d d x |g|<br />

<br />

R + 2(∂χ) 2 + 1<br />

2 (∂ϕ)2 − 1<br />

2 N<br />

e 2 2πℓ<br />

−2<br />

<br />

2 d−1<br />

d−2 χ−2ϕ<br />

<br />

. (11.250)<br />

The potential for the remaining scalars has no lower bound, but we can still look for<br />

(d − 2)-brane solutions. First, we diagonalize the potential by redefining the scalars:<br />

χ ′ <br />

d − 1 2 ϕ<br />

= χ + <br />

3d − 5 2(3d − 5) 2<br />

d − 2<br />

,<br />

ϕ ′<br />

2 =−<br />

<br />

2<br />

d − 1 ϕ<br />

χ +<br />

2(3d − 5) 3d − 5 2<br />

d − 2<br />

,<br />

(11.251)

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