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How to do the first OPE in Polchinski?

+ 1 like - 0 dislike
219 views

I can't get the first OPE in Polchinski's String Theory book. It is $$:\partial X^\mu(z)\partial X_\mu(z)::\partial'X^\nu(z')\partial'X_\nu(z'):=:\partial X^\mu(z)\partial X_\mu(z)\partial'X^\nu(z')\partial'X_\nu(z'):$$ $$-4\cdot \frac{\alpha'}{2}(\partial\partial'\ln|z-z'|^2):\partial X^\mu(z)\partial' X_\mu(z'):+2\cdot\eta^\mu_{\;\mu}\left(-\frac{\alpha'}{2}\partial\partial'\ln|z-z'|^2\right)^2$$ $$\sim\frac{D\alpha'^2}{2(z-z')^4}-\frac{2\alpha'}{(z-z')^2}:\partial' X^\mu(z')\partial' X_\mu(z'):-\frac{2\alpha'}{z-z'}:\partial'^2 X^\mu(z')\partial'X_\mu(z'):$$ I can see how the first equality comes about from Eq. (2.2.9) and Polchinski's hint. However, I don't know how to get part of the asymptotic. Thanks to Prahar I got the first term. I definitely don't know how to get the other terms. I'm not really sure how to make use of his hint to Taylor expand. So if you want to just give a hint, that's fine, just please don't tell me to Taylor expand ;)

Any help would be greatly appreciated.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
asked Feb 1, 2015 in Theoretical Physics by 0celo7 (50 points) [ no revision ]
retagged Feb 1, 2015 by dimension10

Related: this question and links therein.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Qmechanic

Note that $\partial_z \partial_w \log |z-w|^2 = \frac{1}{(z-w)^2}$ since $\log|z-w|^2 = \log(z-w) + \log({\bar z} - {\bar w})$.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@Prahar: I see, thank you. That clears that up. That also gives me the second term once I figure out how $:\partial X^\mu(z)\partial' X_\mu(z'):\longrightarrow :\partial' X^\mu(z')\partial' X_\mu(z'):$

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
The latter happens because we only want to keep singular terms. So, if I Taylor series the first around $z=z'$ and keep terms only singular as $z \to z'$ you'd get that.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@0celo7 - I read your question only after I commented. But ya, Taylor expansion is basically the answer.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@Prahar: Cool! How about $:\partial X^\mu(z)\partial X_\mu(z)\partial'X^\nu(z')\partial'X_\nu(z'):$?

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
@0celo7 - I can't tell you everything. The only thing you really have to do is to Taylor expand all functions around $z = z'$ and keep only the singular terms. Once you do that, you will correctly reproduce Polchinski's formula.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@Prahar: How do I know which terms are singular? If I Taylor (not Laurent) expand, how can any terms be singular? After all, aren't they all polynomials?

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
@0celo7 - Because each term is multiplied by factors of the form $\frac{1}{(z-z')^n}$ which for $n > 0$ are singular.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@Prahar: I don't see how I can get singular terms by Taylor expanding. Do I have to do something to $:\partial X^\mu(z)\partial X_\mu(z)\partial'X^\nu(z')\partial'X_\nu(z'):$ before Taylor expanding?

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
@0celo7 - If I give you $\frac{1}{(z-w)^2} f(z) f(w)$ and ask you what the singular terms are as $z \to w$, you would taylor expand $f(z)$ around $w$ and then keep just the singular terms in the WHOLE expression. I can't help you anymore than that. I've basically answered your question.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@Prahar: I'm in high school. I'm afraid I need a little more help than that! I don't have an expression of the form $\frac{1}{(z-w)^2}f(z)f(w)$, or if I do, then I don't see it anywhere.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
$f(z) = \partial X^\mu \partial X_\mu$

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
OK. Modify your question and show me all the work you have done and exactly where you are stuck.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@Prahar: I figured it out while typing up my work. The term $-4\cdot \frac{\alpha'}{2}(\partial\partial'\ln|z-z'|^2):\partial X^\mu(z)\partial' X_\mu(z'):$ actually contributes two terms to the asymptotic. Thank you for all your help. Post one of your hints below if you want the check.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
@Prahar: Sorry for pulling the high school card. I just get frustrated when P.S.E. people won't answer questions because they think I am a lazy college student or whatever. When I ask a question, then I'm really stuck. Its not like I can ask anyone I know.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
@0celo7 - For future reference, no one is going to help you unless you actually show what you have done and ask specific questions about where you are stuck. You did not do that at all in this question. For HW help, you are supposed to elaborate extensively on EVERYTHING that you have tried.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar
@Prahar: I see what you mean in this question. If I had gone through the Taylor expansion fully in the OP, I probably could have figured it out right then.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user 0celo7
@0celo7 - Exactly! That is why we require that it be written up. More often than not, this clears it up for the OP.

This post imported from StackExchange Physics at 2015-02-01 13:24 (UTC), posted by SE-user Prahar

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