# Calculating equation of motion from action

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?

• It might be useful to use index notation $(\nabla\times A)_i = \epsilon_{ijk}\partial_j A_k$ and integration by parts. – Heidar Sep 1 '12 at 14:10

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.
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.
• Sorry, your $=0$ at the very end is extremely confusing because it seems you are claiming that the previous expression is identically zero. It's surely not. Obviously, in general, the contribution is proportional to $\nabla\times B$. It would be pretty bad if if the (spatial) Maxwell term gave zero to the Maxwell equations. ;-) You shouldn't write it's zero because there are other terms in the equations from other terms in the action such as $j\cdot A$ and $(\partial_t A)^2$. – Luboš Motl Sep 1 '12 at 15:57