Calculating equation of motion using path integral

Question

Suppose my action integral is $S=\int d^4x(\nabla \times A)^2$ and $\delta S$ gives $\delta S =\int d^4x [2(\nabla \times A).(\nabla \times \delta A)]$ I would like to calculate the coefficient of $\delta A$ from this action integral . But I am stuck . How can I separate the $\delta A$ from the term like this ?

Answer 1

Let's do what Heidar says and write it with indices, and identify the Lagrangian. $$ L=\frac{1}{2}(\vec{\nabla}\times \vec{A})^2 = \frac{1}{2}\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\partial_l A_m $$ where, if you haven't heard of it yet, you pretend there is a summation symbol for each repeated index. Then since there are no bare $A_i$ sitting by themselves, only $\partial_i A_j$s the only part of Lagrange's equations that will contribute are $$ \partial_q \frac{\partial L}{\partial (\partial_q A_p)} $$ which we set equal to zero following the equations. Then $$ \frac{\partial L}{\partial (\partial_q A_p)}=\frac{1}{2}(\epsilon_{ijk}\delta_{jq}\delta_{kp}\epsilon_{ilm}\partial_l A_m+\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\delta_{lq}\delta_{mp}) $$ using $$ \frac{\partial (\partial_i A_j)}{\partial (\partial_q A_p)}=\delta_{iq}\delta_{jp}. $$ Then we have $$ \partial_q \frac{\partial L}{\partial (\partial_q A_p)}=\partial_q(\epsilon_{iqp}\epsilon_{ijk}\partial_i A_j) = \partial_q ((\delta_{qj}\delta_{pk}-\delta_{qk}\delta_{pj})\partial_{i} A_j)=\partial_q (\partial_q A_p - \partial_p A_q)=0 $$ where i used the contracted epsilon identity and changed the repeated indices as i needed them in order to combine terms. Hope this helps.

EDIT:

Well, I'll still try and help out, hopefully I don't make anything any worse. $$ \delta S = \int d^3 x \frac{1}{2}\epsilon_{ijk}\epsilon_{ilm}\delta(\partial_j A_k)\partial_l A_m+\int d^3 x\frac{1}{2}\epsilon_{ijk}\epsilon_{ilm}\partial_j A_k \delta(\partial_l A_m) $$ Now with the variations $\delta$ we can interchange the order of $\partial$ and $\delta$ $$ \delta(\partial_i A_j)=\partial_i (\delta A_k) $$ So with the two terms multiplied above we get $$ \partial_j( \delta A_k)\partial_l A_m=\partial_j (\delta A_k \partial_l A_m)-\delta A_k \partial_j \partial_l A_m $$ from the product rule. This helps isolate the variation of the field. Please (everyone) let me know if this is still confusing and/or wrong. Hope this helps.