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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Sep 14, 2023 at 17:33

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My view is similar to consider only one of the multiple solutions of a polynomial equations when we know that it has to fall within a physically given interval. Maxwell's equations do not define uniquely the solutions, they are necessary but not sufficient, so to speak. To get the physically true solution among the possibly many solutions of the system of those 4 vector differential equations we have to take into account the boundary conditions, the Sommerfeld radiation condition at infinity, the constitutive relationships, and demand that the solution also satisfy basic physical requirements, such as causality.

The retarded waves are generated by the motion of charges and time varying macroscopic currents ("continuous charges"). The advanced waves coming from infinity, if you wish to take them into account, must be absorbed properly by your local charges/currents, otherwise they are reflected and scattered, thus "misbehave". If you also impose the Sommerfeld radiation condition on the retarded solution then there is no funny business at infinity.

Unless you take care of this local absorption of the advanced wave you will not get the unique physical solution provided by the retarded wave itself.

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  • $\begingroup$ This answer contains no math, and does not attempt to answer my question as asked. Please see the bolded part. $\endgroup$
    – Jerome
    Commented Oct 17, 2022 at 19:03

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