How to do the first OPE in Polchinski?

Question

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

Comments

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

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@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

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@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

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@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

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@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

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@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

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@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