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)

[REMARK]

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.

This post imported from StackExchange Physics at 2014-03-11 07:56 (UCT), posted by SE-user Trimok