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On page 209 of Polchinski's string theory book he writes down the expectation value of a product of vertex operators on the torus; equation $(7.2.4)$. The derivation is analogous to an earlier calculation on the sphere, equation $(6.2.17)$, and I'm perfectly happy with the result except for the factor of $2\pi/\partial_\nu \vartheta_1(\nu\vert\tau)$.

Can anyone give me an insight into how this term appears? Thanks.

EDIT: Following Lubos's answer.

The expectation value we wish to calculate is

\begin{align} \Bigg< \prod_{i=1}^n :e^{ik_i \cdot X(z_i,\overline z_i)}:\Bigg>_{T^2} &= iC^X_{T_2}(\tau) (2\pi)^d \delta^d(\sum_i k_i) \\& \exp \Big(-\sum_{i<j} k_i \cdot k_j \, G'(w_i,w_j) - \frac{1}{2}\sum_i k_i^2 G_r'(w_i,w_i) \Big) \end{align}

The second line follows just as in eq. $(6.2.17)$, and the Green functions are

$$ G'(w,w') = -\frac{\alpha'}{2} \ln \Bigg\vert \vartheta_1\Big(\frac{w-w'}{2\pi}\Big\vert \tau\Big) \Bigg\vert^2 + \alpha' \frac{[Im(w-w')]^2}{4\pi\tau_ 2}$$

\begin{align} G'_r(w,w)&=G'(w,w)+\alpha'\omega(w)+\frac{\alpha'}{2}\ln \vert w-w'\vert^2 \\&= -\frac{\alpha'}{2}\ln\Bigg\vert \frac{\partial_\nu\vartheta_1(0|\tau)}{2\pi} \Bigg\vert^2 +\alpha'\omega(w) \end{align}

Where we have used

$$ \left. \vartheta_1 \left( \frac{w-w'}{2\pi} | \tau \right)\right|_{w\to w'} \to \partial_\nu\vartheta_1(0|\tau)\cdot \left(\frac{w-w'}{2\pi} \right) $$

as explained by Lubos. Substituting these into the original equation and taking the curvature to infinity $\omega\to 0$, we find

\begin{align} \Bigg< \prod_{i=1}^n :e^{ik_i \cdot X(z_i,\overline z_i)}:\Bigg>_{T^2} &= iC^X_{T_2}(\tau) (2\pi)^d \delta^d(\sum_i k_i) \\& \times\prod_{i<j} \Bigg\vert \frac{2\pi}{\partial_\nu \vartheta_1(0\vert\tau)}\vartheta\Big(\frac{w_{ij}}{2\pi}\Big\vert\tau\Big)\exp\Big[-\frac{(Im w_{ij})^2}{4\pi\tau_2}\Big] \Bigg\vert^{\alpha' k_i \cdot k_j} \end{align}

As in equation $(7.2.4)$.

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    $\begingroup$ It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open Polchinski's book to understand the question. $\endgroup$ – Qmechanic May 14 '15 at 21:09
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    $\begingroup$ I'm confused. Have you gone and edited the answer from a post below into the question? Why? That makes this especially unclear. Are you trying to ask another question? $\endgroup$ – Jim May 15 '15 at 15:14
  • $\begingroup$ No, sorry. I was trying to make the question self contained as Qmechanic asked, I thought since the question had already been answered I may as well add the addition steps to the solution :P sorry for the confusion. $\endgroup$ – Haz May 15 '15 at 15:21
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This extra factor arises from the analogy of the conformal factor $\alpha'\omega$ term in (6.2.16). The required $\omega$ is $$\omega = \ln \left ( \frac{2\pi}{\partial_\nu\vartheta_1}\right) $$ and substituting it to the exponential we get $$ \exp\left( -\frac{\alpha'}{2}\sum_ik_i^2 \cdot \ln \frac{2\pi}{\partial_\nu\vartheta_1} \right) = \left(\frac{2\pi}{\partial_\nu\vartheta_1} \right)^{-\alpha' \sum_i k_i^2/2} = \left(\frac{2\pi}{\partial_\nu\vartheta_1} \right)^{+\alpha' \sum_{i\lt j} k_i k_j} $$ which gives exactly the factor whose origin you wanted to trace. As Joe says, this factor you asked about comes from "normalized self-contractions", which refers to the $\sum_i$ in the exponent, but because $\sum_i k_i = 0$ (and its inner-product square is zero, too), we may convert this $\sum_i$ to $\sum_{i\lt j}$ above.

The aforementioned required $\omega$ is determined as follows. For the self-contractions, the following factor we use for the contractions should be substituted with $w-w'=0$ $$ \vartheta_1 \left( \frac{w-w'}{2\pi} | \tau \right) $$ but it vanishes at that point, so the value has to be computed by l'Hospital rule – or the first term from the Taylor expansion as a function of $w-w'$, if you wish: $$ \left. \vartheta_1 \left( \frac{w-w'}{2\pi} | \tau \right)\right|_{w\to w'} \to \partial_\nu\vartheta_1(0|\tau)\cdot \left(\frac{w-w'}{2\pi} \right) $$ This expression must coincide with the $\omega$-dependent "self-contraction" exponential from (6.2.17) which fixes the value of $\omega$.

Effectively, if you got a result omitting the factor mentioned in the question, it is analogous to saying that $f(0)\to x$ if $f(0)=0$ but the right leading approximation is $f(0)\to f'(0) x$ for such functions.

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  • $\begingroup$ Thank you. Btw, the sign of the exponent of the last them should be +. $\endgroup$ – Haz May 14 '15 at 11:55
  • $\begingroup$ Absolutely! Do you understand my - perhaps too concise - answer? It's been years since I was calculating it, so I slightly blindly parroted myself. ;-) $\endgroup$ – Luboš Motl May 14 '15 at 11:56
  • $\begingroup$ Yes. Is there any particular reason to choose this value of $\omega$? $\endgroup$ – Haz May 14 '15 at 12:00
  • $\begingroup$ Yes, of course, there's a way to determine this only good $\omega$.I am adding it to the answer. $\endgroup$ – Luboš Motl May 15 '15 at 8:25

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