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.


1 Answer 1


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.

There is another subtlety with this calculation - the on-shell graviton is pure gauge and strictly speaking this amplitude is zero. But with appropriate limiting procedure you can extract the tension as the proportionality constant involved in a slightly off-shell amplitude (which is slightly ill-defined). This is another reason why the annulus calculation is cleaner.

  • $\begingroup$ Good, +1, but could you please add a comment about the counting of the ghost number of this disk diagram? Of course, the answer could be uninteresting by your saying that you may insert any PCO operators anywhere to cancel it but one should still know why, what the rules are. $\endgroup$ Apr 1, 2011 at 20:56
  • $\begingroup$ @Luboš: No time at the moment, if you want to write a more complete answer, I'll delete mine. $\endgroup$
    – user566
    Apr 1, 2011 at 21:03
  • $\begingroup$ Thanks for the answer, I think I understand why the CKG is finite. But wouldn't there still be a c\tilde{c} needed to fix the graviton vertex operator, which has a vanishing expectation value? @Lubos, is this what you're referring to? $\endgroup$
    – user2888
    Apr 1, 2011 at 21:11
  • $\begingroup$ No, @Moshe, thanks for your kind offer but I would have to study it again because I don't have the full answer on the top of my head now. It was interesting enough for me to read another person's answer carefully, but not interesting enough for me to review all the ghost numbers of various diagrams and what to do with them haha. $\endgroup$ Apr 2, 2011 at 12:56
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 2, 2011 at 13:02

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