I'm currently working at the start of Alexei Tsvelik's book Quantum Field Theory in Condensed Matter Physics. I'm kinda stumped on a few essential steps.
Starting with the action:
$$S = \int dt \int d^3 x \mathcal{L}_{\mathrm{EM}} = \int dt \int d^3 x \left( -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + j_\mu A^{\mu}\right).$$
In the absence of any sources $j_\mu = 0$ and $c=1$, which after pulling in the negative sign implies
$$S = \frac{1}{2}\int dt \int d^3 x \,\,\left(E^2 – B^2\right)$$
The next step is to take $\vec{E} = -\vec{\nabla} \phi + \partial_t \vec{A}$ and $\vec{B} = \vec{\nabla} \times \vec{A}$. Letting $\phi = 0$ and directly inserting these into the above yields:
$$S = \frac{1}{2} \int dt \int d^3 x \,\left( (\partial_t \vec{A})^2 – (\vec{\nabla} \times \vec{A})^2 \right) $$
Now considering $\vec{A} = \vec{A}_0 + \delta \vec{A}$ as some minimal change in our vector potential, with $\vec{A}_0$ gives a minimum for the action and putting this into the above expression we have:
$$\delta S = \int dt \int d^3 x \left(\partial_0 \vec{A}_0 \partial_0 \delta \vec{A} + \left(\vec{\nabla} \times \vec{A}_0\right)\cdot\left(\vec{\nabla} \times \delta \vec{A} \right) \right)+ \mathcal{O}(A^2)\,\,\,\,\,\,\, (1)$$
where I have the term
$$\mathcal{O}(A^2) = \int dt \int d^3 x \left( (\partial_0 \vec{A}_0)^2 + (\partial_0 \delta \vec{A})^2 – (\vec{\nabla}\times\vec{A}_0)^2-(\vec{\nabla}\times \delta\vec{A})^2\right)$$
The next step is where I lose our author and go from "Oh ok I get this" to "completely clueless". He rewrites $\delta S$ as:
$$\delta S = \int dt \int d^3 (\delta x) \vec{A} \mathcal{F(\vec{A}_0)}+ \mathcal{O}(A^2)$$
where $\mathcal{F} = \frac{\delta S}{\delta A}$. What is the purpose of $x\to \delta x$ in the above expression? Is the functional derivative $\mathcal{F}$ a vector quantity? He has it bolded with a bolded $\vec{A}$ in the above expression. One last question takes the above expression further. We assume $\delta \vec{A}$ vanishes at infinity, as it should, and integration by parts yields:
$$\delta S = -\int dt \int d^3 x \left( \partial^2_0 \vec{A}_0 – (\vec{\nabla} \times \vec{\nabla})\times \vec{A}_0 \right)\delta \vec{A}\,\,\,\,\,\,\,\, (2)$$
How exactly did we go from (1) to (2)? Thanks I appreciate your time!
Best Answer
Author gives a clue on the transition:
This is the usual step in the Lagrangian theory of field (actually, of anything). At first, we have the variation of action written in an awkward form: $$\delta S=\int_{\substack{\text{domain of least}\\\text{action problem}}}(\text{something})\cdot(\text{derivative of }(\text{variation}))$$ Since we want the variation of action to be in form $\delta S=(\text{something})\cdot(\text{variation})$, we have to "pull out" the variation from under the derivative. This is done by integration by parts: $$\delta S=\Bigl[(\text{something})\cdot(\text{variation})\Bigr]_{\text{boundary of domain}}\\-\int_{\text{domain}}(\text{variation})\cdot(\text{derivative of }(\text{that something}))$$ And then we use the fact that the variation is put to be zero on the boundary of the domain (in this case, at infinity). That means that the first term cancels out, and we have finally $$\delta S=-\int_{\text{domain}}(\text{variation})\cdot(\text{derivative of }(\text{something}))$$ Exactly what we need to proceed to $(\text{derivative of }(\text{something}))=0$.
A side note: I found a typo in the book, comparing the Russian and English editions. In the English edition the formula (1.6) is typesetted as $$\delta S=\int \mathrm{d}t\,\mathrm{d}^3\delta x\mathbf{A}(t,\mathbf{x})\mathbf{F}[\mathbf{A}_0(t,\mathbf{x})]+\mathcal{O}(\delta A^2)$$ which for me hardly makes any sense (what is the differential of the variation, and what is $\delta x$ in the field case in the first place?). Actually, in the Russian edition this formula looks like $$\delta S=\int dt\,d^3x\,\delta\mathbf{A}(t,\mathbf{x})\mathbf{F}[\mathbf{A}_0(t,\mathbf{x})]+O(\delta A^2)$$ which is rather more comprehensible. No wonder you stumbled over this. My condolences.