Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,847 answers , 20,601 comments
1,470 users with positive rep
501 active unimported users
More ...

Polchinski Equation (7.2.4)

+ 1 like - 0 dislike
187 views

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 $(7.2.4)$.

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Haz
asked May 14, 2015 in Theoretical Physics by Haz (5 points) [ no revision ]
retagged May 25, 2015
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.

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Qmechanic
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?

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Jim the Enchanter
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.

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Haz

1 Answer

+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Luboš Motl
answered May 14, 2015 by Luboš Motl (10,248 points) [ no revision ]
Thank you. Btw, the sign of the exponent of the last them should be +.

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Haz
Absolutely! Do you understand my - perhaps too concise - answer? It's been years since I was calculating it, so I slightly blindly parroted myself. ;-)

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Luboš Motl
Yes. Is there any particular reason to choose this value of $\omega$?

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Haz
Yes, of course, there's a way to determine this only good $\omega$.I am adding it to the answer.

This post imported from StackExchange Physics at 2015-05-25 09:00 (UTC), posted by SE-user Luboš Motl

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOver$\varnothing$low
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...