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There is an integral that I stumbled upon when I saw a calculation related to magnetic field energy (in static current density case) $$ U_B= \int_\text{whole space} \mathbf j \cdot \mathbf A \:\mathrm dV . $$ The integral is expanded using Maxwell's Laws and we reach to something like this $$ U_B = C\left( \int_\text{whole space} \mathbf B\times \mathbf A \cdot \mathbf n \:\mathrm dS + \int_\text{whole space} \mathbf B\cdot \mathbf B \:\mathrm dV \right), $$ where $C$ is a constant.

Then the first integral is taken to be zero. There was no logic stated behind this. Is there is some trivial fact that I am missing because I cannot figure any reason for the first integral to be $0$.

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    $\begingroup$ I think there is a mistake (typo?) in the 1st integral; it should be written as a surface integral $\int \textbf{B} \times \textbf{A} \cdot \textbf{n} dS$, and then it follows from the regularity of the fields at infinity that it has to be zero $\endgroup$
    – hyportnex
    Jul 21, 2017 at 16:33
  • $\begingroup$ @hyportnex: Sorry that is a typo. I will correct it. But could you explain me a bit more about the "regularity of fields at infinity" $\endgroup$
    – Icchyamoy
    Jul 21, 2017 at 16:40
  • $\begingroup$ the regularity condition means that at infinity the fields go to zero sufficiently fast. One usually assumes that $r^2|\textbf{B}|$ is bounded, consequently $r|\textbf{A}|$ is also bounded, as $r \to \infty $ $\endgroup$
    – hyportnex
    Jul 21, 2017 at 16:54
  • $\begingroup$ But why is such condition taken? It might not always hold. Say when $\mathbf j $ is proportional to $r^3$ then obviously, $ \mathbf A=\int \mathbf j/|r| dV$ will not be 0 at infinity right? $\endgroup$
    – Icchyamoy
    Jul 21, 2017 at 17:04
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    $\begingroup$ I think that in first equation a factor "1/2" is missing $$ U_B= \frac{1}{2}\int_\text{whole space} \mathbf j \cdot \mathbf A \:\mathrm dV $$ "...If the currents are all localized, then both $\mathbf{A}$ and $\mathbf{B}$ tend to zero at infinity..." see : Energy in a magnetic field $\endgroup$
    – Frobenius
    Jul 21, 2017 at 18:21

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This condition essentially needs to be imposed 'on faith', to some degree, because we intuitively feel that physical sources should be localized, and for those we can work out that the fields decay fast enough that the surface term vanishes.

This is not to say that we're just wishing away terms that we find inconvenient: instead, we explicitly make the caveat that what follows is valid for localized sources only ─ and we make the slightly circular definition that a 'localized source' (or a localized field) is one for which the relevant surface terms vanish.

In the majority of cases, it is very hard to come up with a (useful) necessary condition on the fields and sources which is equivalent to the vanishing of that surface term. We know plenty of sufficient conditions (of the form 'for such-and-such class of localized sources, the surface term vanishes'), and those are broad enough to cover most cases of interest, but it's hard to make more general statements.

In a sense, this is more of a "credit" view of mathematical rigour, as opposed to the "debit" view that mathematicians tend to hold: where they say "OK, my fields satisfy X and Y niceness hypotheses, let's work out what results I can prove from those", physicists tend to tell rigour "well, I want that term to go away, so you can just bill me later by telling me what conditions my fields need to satisfy for my formalism to hold".

And indeed, this kind of assumption can indeed come back and bite us. One famous example is the separation of optical angular momentum into orbital and spin contributions, as described here, via an integration by parts of the form $$ \mathbf J = \frac{1}{\mu_0c^2} \int\mathrm d\mathbf r \: \mathbf r \times (\mathbf E\times\mathbf B) = \frac{1}{\mu_0c^2} \int\mathrm d\mathbf r \: (E_i (\mathbf r \times\nabla) A_i ) + \frac{1}{\mu_0c^2} \int\mathrm d\mathbf r \: \mathbf E\times \mathbf A =\mathbf L + \mathbf S, $$ where you get this weird paradox: if you plug in a circularly-polarized plane wave, then $\mathbf L$ is zero but $\mathbf S$ is not, but the initial $\mathbf J$ also seems to vanish, so something is off: the problem here is that the boundary terms (or the regions where the beam tapers off) carry nonzero angular momentum, and cannot be neglected. This apparent paradox has puzzled many an unsuspecting author.

Nevertheless, we still use the equation (as we do with the one you mentioned) because it holds for what we think are physical fields, and we're prepared to say "it's zero because I say so" and accept that if it's not zero then the fields are probably not physical and we shouldn't be using them as such.

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  • $\begingroup$ Is it for the same reason that dipole moment $\mathbf p$ is taken $0$ at infinity? I came across another integral where the same volume to surface integral transformation was made and the integrand was reduced to 0 as "$\mathbf p$ is 0 at infinity". $\endgroup$
    – Icchyamoy
    Jul 22, 2017 at 3:43
  • $\begingroup$ That's not enough context to tell either way. $\endgroup$ Jul 22, 2017 at 9:49
  • $\begingroup$ Well while evaluating electric potential$\phi$ from free charge density $\rho_F$ and dipole moment per unit volume $\mathbf P$ it was stated $\phi=\displaystyle \int (\rho_F /|r|)dV - \int \mathbf P\cdot \nabla(1/|r|)dV$, they expanded the integral and made $\displaystyle \int \partial \cdot (\mathbf P/|r|)dV=0$ since $\mathbf P=0$ at infinity. $\endgroup$
    – Icchyamoy
    Jul 22, 2017 at 11:26
  • $\begingroup$ Well, you can only have dielectric polarization where you have a dielectric medium, so unless you have an infinite medium then that one is particularly obvious. Where it gets hard is when you need to quantify the decay properties of fields (where you control the sources but then the response is what it is) but if it's just about the sources then it's pretty much trivial. $\endgroup$ Jul 22, 2017 at 12:25

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