# Vanishing boundary terms of the Inhomogeneous electromagnetic wave equation using Green functions

I'm trying to follow Zangwill's derivation of the solution to the inhomogeneous wave equation $$\bigg[ \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \bigg] \psi(\vec{r},t) = - f(\vec{r},t),$$ where he arrives at the complete solution given by $$\psi(\vec{r},t) = \int_{t_1}^{t_2} \text{d}t' \int_V \text{d}^3r' G(\vec{r},t|\vec{r}',t')f(\vec{r}',t') \\ + \int_{t_1}^{t_2} \text{d}t' \int_{\partial V} \text{d}\vec{S}' \cdot[G(\vec{r},t|\vec{r}',t')\nabla\psi(\vec{r}',t') - \nabla G(\vec{r},t|\vec{r}',t')\psi(\vec{r}',t')] \\ + \frac{1}{c^2} \int_V \text{d}^3r' \bigg[\frac{\partial}{\partial t}G(\vec{r},t|\vec{r}',t') \psi(\vec{r}',t') - G(\vec{r},t|\vec{r}',t') \frac{\partial}{\partial t} \psi(\vec{r}',t')\bigg]_{t'=t_1}^{t'=t_2},$$ with two Green functions satisfying the wave equation, $$G_\pm (\vec{r},t|\vec{r}',t') = \delta(t - t' \pm |\vec{r} - \vec{r}'|/c) / 4 \pi |\vec{r} - \vec{r}'|.$$ He then asserts that the spatial boundary term is identically zero, with the temporal boundary term representing the ingoing and outgoing waves in the interval $$[t_1 \; t_2].$$

The problem is, he then goes on to assume that the nobody in their right mind would ever be interested in scenarios other than those where $$V \rightarrow \infty$$, and tosses the advanced wave into the garbage bin for the crime of the third term (or $$\psi_{\text{out}}(\vec{r},t)$$) being undefined. With a wave of the hand, he intuits for the retarded solution that the boundary $$\partial V$$ at infinity kills the term on the second line of the solution on physical grounds, as the wave will never reach it in finite time. But this logic only applies to retarded waves integrated over all space!

... but what if I want to consider finite volumes, and the advanced wave is still perfectly valid? I'd like a purely mathematical proof that the spatial boundary integral disappears, but am having trouble making headway: I've tried massaging the integrand with identities, and calculating terms explicitly, but can't justify making it zero.

Surely, this integral must always be zero, because the advanced and retarded solutions $$\psi(\vec{r},t)_{\text{ret/adv}} = \psi(\vec{r},t)_{\text{in/out}} + \frac{1}{4 \pi} \int_V \text{d}^3r' \frac{f(\vec{r}',t \pm |\vec{r}-\vec{r}'|/c)}{|\vec{r} - \vec{r}'|}$$ explicitly assume the spatial boundary term is gone, and Zangwill presents these as the general solutions of the differential equation under all scenarios. Is there a trick to this? And to be clear, I'm only really interested in the advanced-wave solution.

• If the boundary is not in infinity, then in general, the boundary integrals can't be dropped. The final equation is not valid for arbitrary region $V$, it is meant only for the case where integration goes over the whole infinite space. Commented Sep 14, 2023 at 17:33