I have a question in deriving Eqs. (3.6.15b) and (3.6.15c) in Polchinski's string theory vol I p. 105.


$$\Delta (\sigma,\sigma') = \frac{ \alpha'}{2} \ln d^2 (\sigma, \sigma') \tag{3.6.6}$$ where $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$.

It is said for the Weyl variation $\delta_W$,

$$\partial_a \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= \frac{ 1}{2} \alpha' \partial_a \delta \omega(\sigma) \tag{3.6.15a} $$ $$ \partial_a \partial_b' \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= \frac{ 1+ \gamma}{2} \alpha' \nabla_a \partial_b \delta \omega(\sigma) \tag{3.6.15b} $$ $$ \nabla_a \partial_b \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= - \frac{ \gamma}{2} \alpha' \nabla_a \partial_b \delta \omega(\sigma) \tag{3.6.15c} $$ Here $\gamma=-\frac{2}{3}$.

I know the reason for the factor of $1/2$ in the RHS of Eq. (3.6.15a) and I can derive it. But, putting $\nabla$ on LHS and RHS of Eq. (3.6.15a) will not lead to (3.6.15c). I guess it is because there is some higher-order terms in the geodesic distance where the treatment of (3.6.9)-(3.6.11) omitted. But I don't know how to get it. I tried $ds^2= g_{ab} dx^a dx^b= (\eta_{ab} + h_{ab})dx^a dx^b $ to separate the deviation of Minkowski metric, but I cannot get the factor $\gamma$

My question is, how to derive Eqs. (3.6.15b) and (3.6.15c) (especially there is a prime, $'$, namely $\partial'_b$ in the LHS of (3.6.15b) but not in the RHS)


Assuming the distance $d$ ($z$ is a formal notation) is: $$d(\sigma, \sigma') = \int_\sigma^{\sigma'} dz ~~e^{w(z)} \tag{1}$$

We have, developping at order 2, $w(z)$:

$$d(\sigma, \sigma') = e^{w(\sigma)} \int_\sigma^{\sigma'} dz ~ e^{(z-\sigma).\partial w(\sigma) + \frac{1}{2}(z-\sigma)^a(z-\sigma)^b \partial_a \partial_b w(\sigma)}\tag{2}$$

That is :

$$d(\sigma, \sigma') = e^{w(\sigma)} \int_\sigma^{\sigma'} dz ~ ( 1 + (z-\sigma).\partial w(\sigma) + \frac{1}{2}(z-\sigma)^a(z-\sigma)^b\partial_a \partial_bw(\sigma)+Q(w)) \tag{3}$$ where $Q(w)$ represents quadratic quantities in $\partial_a w \partial_b w$.

Because the metrics is diagonal, Christophel symbol are of king $\partial_x w$, so the difference between $\nabla_x \partial_y w$ and $\partial_x \partial_y w$ are precisely these quadratic quantities. We can only choose a inertial frame, so that $\partial_x w=0$, so we can neglect theses quantities $Q$, and working with standard derivatives (more details at the end of the answer).

After formal integration, we get :

$$d(\sigma, \sigma') = e^{w(\sigma)} |\sigma' - \sigma|(1 + \frac{1}{2}(\sigma'-\sigma).\partial w(\sigma) + \frac{1}{6}(\sigma'-\sigma)^a(\sigma'-\sigma)^b\partial_{ab}w(\sigma)+Q(w))\tag{4}$$

Reexpressing with an exponential, we have:

$$d(\sigma, \sigma') = e^{w(\sigma)} |\sigma' - \sigma| e^{\frac{1}{2}(\sigma'-\sigma).\partial w(\sigma) + \frac{1}{6}(\sigma'-\sigma)^a(\sigma'-\sigma)^b\partial_a \partial_bw(\sigma)+Q(w)} \tag{5}$$

So, now, we get, with $\Delta (\sigma, \sigma') = \frac{\alpha'}{2} \ln d^2(\sigma, \sigma')$:

$$\Delta (\sigma, \sigma') =\alpha'(w(\sigma) + \frac{1}{2}(\sigma'-\sigma).\partial w(\sigma) + \frac{1}{6}(\sigma'-\sigma)^a(\sigma'-\sigma)^b\partial_a \partial_bw(\sigma)+Q(w)+F(\sigma, \sigma'))\tag{6}$$ where $F(\sigma, \sigma')$ is a function of $\sigma, \sigma'$ which does not depend on $w$, so we don't care by looking at variations of $\Delta$ relatively to $w$. Choosing a inertial frame means that we don't care about the quadratic quantities $Q$ too.

We have an other problem, because here $\Delta (\sigma, \sigma')$ is not symmetric in $\sigma, \sigma'$, so we need to symmetrise it, so finally the relevant part of $\Delta$ is :

$$\Delta (\sigma, \sigma') =\alpha'( \frac{1}{2} [w(\sigma) + w(\sigma') ] + \frac{1}{4}(\sigma'-\sigma).[\partial w(\sigma) - \partial^{'} w(\sigma')] + \frac{1}{12}(\sigma'-\sigma)^a(\sigma'-\sigma)^b[\partial_a \partial_bw(\sigma)+\partial^{'}_a \partial_b^{'}w(\sigma')])\tag{6'}$$

So, taking variations, and derive relatively to $\sigma'_b$: $$\partial^{'}_b \delta_W \Delta (\sigma, \sigma') =\alpha'( \frac{1}{2}\partial^{'}_b \delta w(\sigma') + \frac{1}{4}[\partial_b w(\sigma) - \partial^{'}_b w(\sigma')] \\ - \frac{1}{4}(\sigma' - \sigma)^a \partial'_a \partial_b'w(\sigma')+ \frac{1}{6}(\sigma' - \sigma)^a[\partial_a \partial_bw(\sigma)+\partial^{'}_a \partial_b^{'}w(\sigma')] + O(\sigma - \sigma')^2\tag{7}$$

Then, we derive relatively to $\sigma_a$: $$\partial_a \partial^{'}_b \delta_W \Delta (\sigma, \sigma') =\alpha'( \frac{1}{4}[\partial_a \partial_b \delta w(\sigma)+ \partial^{'}_a \partial^{'}_b \delta w(\sigma')] - \frac{1}{6}[\partial_a \partial_b \delta w(\sigma)+ \partial^{'}_a \partial^{'}_b \delta w(\sigma')]+ O(\sigma - \sigma')\tag{8}$$

So, finally, when $\sigma' \rightarrow \sigma$, we have:

$$\partial_a \partial^{'}_b \delta_W \Delta (\sigma, \sigma')= \alpha' \frac{1}{6}\partial_a \partial_b\delta w(\sigma)\tag{9}$$

This is precisely the expression (3.6.15b), remembering that we calculate in a intertial frame, so $\nabla_x \partial_y w = \partial_x \partial_y w $

The same method, beginning from $(6')$ and applying two derivates $\partial_a \partial_b$ gives $3.16.15c$ (a $\frac{1}{3}$ term)


The quadratic quantities $Q$ could be written :

$$(\sigma'-\sigma)^a(\sigma'-\sigma)^b (\partial_a w)(\partial_b w)$$

When you derive 2 times this expression, you get :

$$ O(\sigma'-\sigma) + O(\partial_a w)$$ So, when $\sigma' \rightarrow \sigma$ and when we are in a inertial frame ($\partial_a w=0$), these quantities are not relevant.

  • $\begingroup$ Thanks a lot! I have two naive questions, (a) how do you get your equation (1)? is that just a formal notation? (since $\omega(\sigma)$ is arbitary(?)); (b) Is the symmetrisation of Eq. (6) ad hoc? $\endgroup$ – user26143 Aug 7 '13 at 19:05
  • $\begingroup$ $(1)$ : Equation $(1)$ is a better approximation of the zero-order version $d(\sigma, \sigma') = |\sigma' - \sigma| ~~e^{w(\sigma)} $, because the infinitesimal distance (which is in the integral) depends on the metrics at each point. (2) $\Delta$ must be symmetric, so if you begin with $(6)$ and symmetrize, you get automatically $(6')$ $\endgroup$ – Trimok Aug 7 '13 at 19:11
  • $\begingroup$ Since p 102 in Polchinski said $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$, there should be some formula to calculate it directly, how to see your Eq. (1) is for the geodesic distance? and I feel the symmetrization is added by hand, there might be better derivation for it (?) $\endgroup$ – user26143 Aug 7 '13 at 19:14
  • $\begingroup$ The geodesic distance is the integral of infinitesimal distance, that is why the $e^{w(z)}$ term is in the integral. I think you are right, there must be a symmetric version of this geodesic distance, but, in fact, it means simply symmetrize equation (2) or equation (3) or equation (4) or equation (5), so finally you wil make the symmetrization before, but this will not change $6'$ and the final result. $\endgroup$ – Trimok Aug 7 '13 at 19:18
  • $\begingroup$ Thanks a lot! Is the factor $e^{\omega(z)}$ follow the conformal gauge condition (3.3.4)? I am not sure where does it come from.... $\endgroup$ – user26143 Aug 7 '13 at 19:23

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