04.06.2013 Views

Gravity and Strings

Gravity and Strings

Gravity and Strings

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

580 String black holes in four <strong>and</strong> five dimensions<br />

Now, to add a wave, we use the procedure studied on page 327 <strong>and</strong> boost the above<br />

solution in the direction y 1 , obtaining [208, 550]<br />

d ˆs 2 s = H − 1 2<br />

D1 H − 1 <br />

2 −1<br />

D5 HW Wdt2 − HW[dy1 + αW(H −1<br />

W − 1)dt]2<br />

− H 1 2<br />

D1 H 1 2<br />

D5 [W −1 dr 2 + r 2 d 2 (3) ]<br />

− H 1 2<br />

D1 H − 1 2<br />

Ĉ (2) ty 1 = αD1e − ˆφ0<br />

D5 [(dy2 ) 2 + (dy3 ) 2 + (dy4 ) 2 + (dy5 ) 2 ],<br />

−1<br />

HD1 − 1 , Ĉ (6) ty1 ···y5 = αD5e− ˆφ0 −1<br />

HD5 − 1 ,<br />

e−2 ˆφ −2 ˆφ0 = e HD5/HD1, Hi = 1 + r 2 i<br />

,<br />

r 2<br />

ω<br />

W = 1 +<br />

r<br />

ω = r 2 i (1 − α2 i ), i = D1, D5, W.<br />

2 ,<br />

(20.15)<br />

In the extreme limit ω = 0werecover a D1 D5 W that could have been constructed<br />

using the harmonic-superposition rule.<br />

Let us now dimensionally reduce this solution in the direction y 1 .Inthis reduction a<br />

modulus field that measures the size of that direction arises, 3<br />

k1<br />

k10<br />

=|ˆg y 1 y 1| 1 2 = H 1 2<br />

W<br />

H 1 4<br />

D1 H 1 4<br />

D5<br />

, (20.16)<br />

<strong>and</strong> k10 = R1/ℓs. Inthe reduction on T 4 we obtain another modulus field associated with<br />

its volume, 4<br />

k V 4 =|ˆg y 2 y 2 ˆg y 3 y 3 ˆg y 4 y 4 ˆg y 5 y 5| 1 2 = HD1/HD5. (20.17)<br />

The d = 5 dilaton is given by<br />

e −2φ = e −2 ˆφ k1 = e −2φ0<br />

H 1 2<br />

W<br />

H 1 4<br />

D1 H 1 4<br />

D5<br />

, e −2φ0 = e −2 ˆφ0 k1,0. (20.18)<br />

3 Before reducing, we can rescale y 1 so that it takes values in [0, 2πℓs]. We will do the same systematically<br />

in common worldvolume directions, but not in relative transverse directions, since, in order to apply<br />

Eqs. (11.124), the coordinates have to take values in [0, 2π R]. The value at infinity of the corresponding<br />

modulus is 1.<br />

4 All we are doing here is a st<strong>and</strong>ard toroidal compactification of the kind we have performed in Section 16.5<br />

<strong>and</strong> studied in general in Section 11.4. In general we would obtain a bunch of moduli fields coming from the<br />

internal metric. For this <strong>and</strong> the solutions that will follow, the internal metric is proportional to the identity<br />

<strong>and</strong> there is only one non-trivial modulus: its determinant. Its square root is kv. Wehave not performed the<br />

toroidal reduction of RR fields, but it is clear that they give rise to a series of form potentials of equal <strong>and</strong><br />

lower ranks. Since the potentials in this <strong>and</strong> the other solutions that we are going to study have components<br />

only in compact directions (plus time) only the time component of the vector fields that originate from the<br />

reduction will be non-trivial <strong>and</strong> have the obvious value.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!