# Graviton Emission from D-Branes

I'm working through Polchinski's book on string theory, and I ran into something that I don't think I understand. I'm hoping that someone who knows this stuff can help me out.

Before calculating the Dp-brane tension in Chapter 8, Polchinski says that we could have obtained the same result by calculating the amplitude for graviton emission from the D-brane (instead of calculating closed string exchange between two D-branes). It seems like we would do this by placing a graviton vertex operator on the disk with Dirichlet boundary conditions on (25-p) coordinates and Neumann on the rest. But doesn't the amplitude with only one vertex operator vanish, since there aren't enough ghost insertions to get a nonzero result? I must be misunderstanding something, because it seems like if we only fix the real and imaginary parts of the position of the vertex operator, then we have to divide by the volume of the rest of the CKG of the disk, which is infinity. Any ideas or hints? Thanks.

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After you fix the closed string vertex operator, the remaining group of isometries is one dimensional and its volume is finite. For example if the vertex operator is fixed at the center of the disk, the remaining isometries are rotations. The volume of that group in some units is $2\pi$. Figuring out the precise constants involved is a bit messy though.
Dear user2888, yes, exactly, I am talking about the required minimal number of $c$ and $\tilde c$ operators inserted somewhere on the disk which is needed for the amplitude to be nonzero. Note that the Veneziano amplitude vanishes for less than 3-4 open strings. I don't know what's the orthodox treatment right now but the $c,\tilde c$ simply have to be saturated by hand so that one gets a nonzero result because the right result is nonzero, and the residual conformal symmetry is not a problem because $U(1)$ that preserves the disk and the central point's vertex operator is compact. –  Luboš Motl Apr 2 '11 at 13:02