Equations of motion of $\mathcal{L}= - \frac{1}{4}F^{2}_{\mu \nu} - A_{\mu}J_{\mu}$ in momentum space

Question

I'm reading the Matthew D. Schwartz, Quantum field theory and the standard model, p.128 and some question arises.

Consider a lagrangian $\mathcal{L}= - \frac{1}{4}F^{2}_{\mu \nu} - A_{\mu}J_{\mu}$ ($J_{\mu}$ is current). It's equations of motion are $\partial_{\mu}F_{\mu \nu}= J_{\nu}$, so $\partial_{\mu} \partial_{\mu}A_{\nu}-\partial_{\mu}\partial_{\nu}A_{\mu}=J_{\nu}$.

Then, why $(-p^{2}g_{\mu \nu} + p_{\mu}p_{\nu})A_{\mu} = J_{\nu}$ in momentum space?

This question originates from next section in his book, p.128

Why the underlined statement is true?

Can anyone help?

Answer 1

@JeanbaptisteRoux is right. Schwartz should have used symbols less ambiguously to prevent your confusion. Denote the functions in Eq. (8.94) as $A_\mu(x),\,J_\mu(x)$, viz.$$\partial^2A_\nu(x)-\partial_\mu\partial_\nu A_\mu(x)=J_\nu(x).$$Define$$\tilde{A}_\mu(p):=\int_{\Bbb R^4}A_\mu(x)e^{ip\cdot x}d^4x,\,\tilde{J}_\mu(p):=\int_{\Bbb R^4}J_\mu(x)e^{ip\cdot x}d^4x.$$To (8.94) we now apply the operator $\int_{\Bbb R^4}e^{ip\cdot x}d^4x$, so $A_\mu(x)$ becomes $\tilde{A}_\mu(k)$, so by integration by parts twice$$(-p^2g_{\mu\nu}+p_\mu p_\nu)\tilde{A}_\nu(k)=\tilde{J}_\mu(k),$$which is how Eq. (8.95) should have been written for clarity. Schwartz also uses exclusively downstairs indices, so I'll leave repeating the above calculation with proper up/down contraction as an exercise.