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String Theory and M-Theory

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202 T-duality <strong>and</strong> D-branes<br />

X µ<br />

R = xµ − ˜x µ<br />

+<br />

2<br />

1<br />

2 l2 s p µ (τ − σ) + i<br />

2 ls<br />

1<br />

m<br />

m=0<br />

αµ me −im(τ−σ) .<br />

The mode expansions for the fields with Neumann <strong>and</strong> Dirichlet boundary<br />

conditions are<br />

X i = X i L + X i R <strong>and</strong> X I = X I L − X I R ,<br />

respectively. The two ends of the string have<br />

X I (0, τ) = X I (π, τ) = ˜x I ,<br />

which specifies the position of the D-brane. In uncompactified space-time<br />

there can be no winding modes, so p I = 0.<br />

The energy–momentum tensor<br />

T++ = ∂+X i ∂+Xi + ∂+X I ∂+XI = ∂+X µ<br />

L ∂+XLµ<br />

has the same mode expansion as in Chapter 2, independent of p, <strong>and</strong> thus<br />

the Virasoro generators, the zero-point energy, <strong>and</strong> the mass formula, are<br />

the same as before<br />

M 2 = 2(N − 1)/l 2 s .<br />

The main difference is that this is now the mass of a state in the (p + 1)dimensional<br />

world volume of the Dp-brane, whereas in Chapter 2 only the<br />

space-time-filling p = 25 case was considered. The mass squared of the<br />

open-string ground state therefore is<br />

M 2 = −2/l 2 s = −1/α ′ .<br />

EXERCISE 6.2<br />

Consider a relativistic point particle with mass m <strong>and</strong> electric charge e<br />

moving in an electromagnetic potential Aµ(x). The action describing this<br />

particle is<br />

S =<br />

<br />

− m − ˙ X µ ˙ Xµ − e ˙ X µ <br />

Aµ dτ.<br />

Suppose that one direction is compactified on a circle of radius R. Show<br />

that a constant vector potential along this direction, given by<br />

A = − θ<br />

2πR ,<br />

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