I'm reading electrodynamics notes and come across that

$$\int_\text{all space} \mathrm{d}\vec r \; \nabla \cdot \left(\vec A \times \vec B \right)~=~0$$ in case of magnetostatics and$$ \int_\text{all space} \mathrm{d}\vec r \; \nabla \cdot \left(\phi \vec E \right)~=~0$$in case of electrostatics, where:

  • $A$ is the magnetostatic potential;

  • $B$ the magnetic field; and

  • $\phi$ the electrostatic potential.

Question: Why do both integrals equal $0$?

The above mentioned formulas are used to show that:$$ \begin{alignat}{7} W &=\frac{\epsilon_0}{2} && \int_\text{all space} \mathrm{d} \vec{r} \vec{E} ^2 \\[2.5px] W &=\frac{1}{2 \mu_0} && \int_\text{all space} \mathrm{d} \vec{r} \vec{B} ^2 \end{alignat} $$starting from$$ \begin{alignat}{7} W &= \frac{1}{2} && \int_\text{all space} \phi \left(\vec{r}\right) \rho\left(\vec{r}\right) \\[2.5px] W &= \frac{1}{2} && \int_\text{all space} \mathrm{d} \vec{r} \vec{j} \cdot \vec{A} \end{alignat} $$where $\rho$ the charge density, and $\vec{j}$ the current density.

I've tried using Gauss's Theorem,$$ \int_{\partial V} \mathrm{d}\vec r \; \left(\vec A \times \vec B \right)\cdot \mathrm{d}\vec{S}~=~0 \,,$$and $$ \int_{\partial V} \mathrm{d}\vec S\; \cdot \left(\phi \vec{E}\right)~=~0 \,,$$ but this doesn't bring me any further to solving my problem.

  • 1
    $\begingroup$ Can you provide a reference (name of text, page?) $\endgroup$ Commented Nov 2, 2013 at 17:05
  • $\begingroup$ There seems to be no a priori reason why those should be zero. Maybe an error? There are other ways to arrive at the expression for energy. See Griffiths for eg. $\endgroup$
    – guru
    Commented Nov 2, 2013 at 19:21
  • $\begingroup$ It really is just gauss's law. You have to assume the current and charges are localized (don't go to infinity). Otherwise it doesn't work. Of course I guess I can't officially convince you unless you give a source. $\endgroup$ Commented Nov 2, 2013 at 20:32
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs: I agree if you mean Gauss's theorem (rather than Gauss's law). $\endgroup$
    – Qmechanic
    Commented Nov 4, 2013 at 1:09
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Is there any way that I can extend my argument (Gauss's theorem) to ''see'' why it equals zero? Gauss's thereom itself is namely just a conversion from a volume integral to an area integral or vice versa (there is no 0 involved). $\endgroup$
    – user104662
    Commented Nov 4, 2013 at 7:05

2 Answers 2


Energy of static fields

The first two terms,$$ \int_{\text{all space}} \mathrm{d}\vec r \; \nabla \cdot \left(\vec A \times \vec B\right) $$and$$ \int_{\text{all space}} \mathrm{d}\vec r \; \nabla \cdot \left(\phi \vec E \right) $$become surface integrals of $\vec A\times \vec B$ and $\phi \vec E$, respectively, by Stoke's theorem. You can, by casting the static equations of motion for $E$ and $B$ in spherical coordinates, see that the Green's functions that result from these equations die off as $\frac{1}{r^2}$, so the products in the above terms should die off at least as $\frac{1}{r^3}$ as $r\to\infty$. When performing the surface integral, we have a area term $4\pi R^2$ which leaves a leading $\frac{1}{R}$ dependence in the integrand, which vanishes as $R \to \infty$.

Th calculation is sketched on Wikipedia here.

A note about point charges

(Jackson's Classical Electrodynamics, pg. 40). If you plug the electric field due to a point charge into$$ W=\frac{\epsilon_0}{2} \int \mathrm{d}V \vec E^2 \,,$$you get an integrand that behaves like $\tfrac{1}{r^4} 4\pi r^2 \, \mathrm{d}r ~ \tfrac{1}{r^2}{\mathrm{d}r}$. The integral therefore diverges at the $r=0$ limit, reflecting the fact that energy of the field configuration of a point charge diverges. Similarly, for the electric field due to two separated point charges, we get two divergent "self-energy" terms and a third term that is the familiar potential energy of point charges. It therefore makes sense to discard these self-energy terms as they only contribute a fixed amount that does not vary with the positions of the charges.

  • $\begingroup$ To provide some more info: When one wants to show the equivalence between: $ W=\frac{1}{2} \int_{all space} d \vec r\vec E ^2 $ and $W= \frac{1}{2} \int_{all space} d\vec r \phi (\vec r) \rho (\vec r)$ one needs that: $\int_{all space} d\vec r \; \nabla \cdot(\vec A \times \vec B)=0$ $\endgroup$
    – user104662
    Commented Nov 2, 2013 at 17:44
  • $\begingroup$ I've edited the question to clarify matters. $\endgroup$
    – user104662
    Commented Nov 2, 2013 at 18:03
  • 1
    $\begingroup$ You can't quite distribute the divergence over the cross product: the relationship is $\nabla \cdot(\vec A \times \vec B)= \vec{B}\cdot (\nabla \times \vec{A}) - \vec{A}\cdot (\nabla \times \vec{B}$) - you likely meant this but I just wanted to make clear to the OP. Actually, I can't see how this helps (in Coulomb gauge) - could you expand your answer a bit for the slow-on-the-uptake like me? $\endgroup$ Commented Nov 2, 2013 at 22:43

Hopefully Lionel will expand his answer because it looks elegant but I can't wholly make it work. In the meanwhile, here's a ham fisted approach:

From the standard identity $\nabla \cdot(\vec A \times \vec B)= \vec{B}\cdot (\nabla \times \vec{A}) - \vec{A}\cdot (\nabla \times \vec{B})$ and by Faraday's law to expand $\nabla \times \vec{B}$ on the rhs we get:

$$\mu_0^{-1} \nabla \cdot \left(\vec A \times \vec B \right)= \mu^{-1} {\left|\vec{B} \right|}^2 - \vec{A}\cdot\vec{J} - \epsilon_0 \vec{A}\cdot\partial_t \vec{A} \tag{1}$$

Expanding $\epsilon_0\,\nabla\cdot(\phi\,\vec{E}) = -\epsilon_0 \nabla\phi\cdot{E} + \epsilon_0 \phi \nabla\cdot\vec{E}$ and using Gauss's law for electricity as well as $\vec{E} = -\nabla \phi - \partial_t \vec{E}$ we get:

$$\epsilon_0 \nabla\cdot \left(\phi \vec{E} \right) = -\epsilon_0 \left|\vec{E}\right|^2 + \phi\,\rho + \epsilon_0 \vec{E}\cdot\partial_t\vec{A} \tag{2}$$

By applying the divergence theorem to a big spherical volume $V$ enclosing all of the current and charge to the left-hand side of both $\left(1\right)$ and $\left(2\right)$ you'll get surface integrals $\oint_{\partial V}\hat{n} \cdot \left(\vec{A}\times\vec{B} \right)\, {\rm d}S$ and $\oint_{\partial V}\hat{n}\cdot \left( \phi \vec{E} \right)\, {\rm d}S$. As in the comments, you need to assume decay behavior of $\vec{E}$, $\phi$, $\vec{B}$ and $\vec{A}$ their asymptotic dependencies of their magnitudes on the radius $R$ of the bign sphere for static conditions are at most $R^{-1}$ for $\phi$ and $\vec{A}$ and at most $R^{-2}$ for $\vec{E}$ and $\vec{B}$ (in general with dynamic problems with radiation, they all dwindle like $R^{-1}$). So in the static case, the surface integrals will vary like $R^{-1}$, hence vanish as the sphere grows big enough to enclose all space. On dropping the time varying terms on the right-hand side of $\left(1\right)$ and $\left(2\right)$ (for static conditions), you'll get the result you need.

As in the comments, there are easier – and more general – ways to study energy flows in the electromagnetic field. In the middle of my answer to the Physics SE question "How can Magnets be used to pick up pieces of metal when the force from a magnetic field does no work?" I show the standard method as in e.g. Griffiths or the literary horror of the symbol jungles of Jackson, "Classical Electrodynamics" (be sure to take a keenly sharpened machete with you to cut through all the equations).

The very best exposition I know of is in Chapter 27 of the second volume of the Feynman Lectures on Physics: the chapter called "Field Energy and Field Momentum". Feynman is every bit as mathematically rigorous as Jackson (probably better) and he studies the physics most carefully: discussing the often glossed over topics of the ambiguity in defining the field energy and flow as well as the locality of energy flows. The method in your question is interesting, though: I've not seen it before and it lets one split apart the $\frac{\epsilon}{2} \left| \vec{E}^2 \right|$ and $\frac{\mu}{2} \left|\vec{H}^2 \right|$ bits of the energy density to study the Lorentz-covariant $\left| \vec{E} \right|^2 - c^2 \left|\vec{B} \right|^2$.

Hope this all helps.

  • $\begingroup$ We are both in a bind, my friend. The integral of $\frac{1}{r}$ diverges as $\ln r$, which is to say the energy of a point charge is infinite. Griffiths gets by this by saying that the point charges are given to you, so and therefore we don't have to supply their energy nor worry about it. This is in essence a renormalization issue. $\endgroup$ Commented Nov 3, 2013 at 18:47
  • $\begingroup$ @lionelbrits Are you sure? If there are point charges, yes, but here we're thinking of them mathematically as densities $\rho$ and $J$ (which sweeps the self energy under the carpet!). Historically, people (Lorentz IIRC) tried this for the electron self energy - making it a nonzero radius. Anyhow, the surface integrands (as opposed to integrals) vary like $\phi\,|\vec{E}|\to R^{-1} R^{-2}$ so $|\oint_{\partial V}\hat{n}\cdot( \phi \vec{E})\, {\rm d}S|\approx 4\pi R^2 R^{-3} = 4\pi/R\to0$ for static fields. The method won't work with radiation (when you can only bound the integrand by .... $\endgroup$ Commented Nov 3, 2013 at 22:05
  • $\begingroup$ @lionelbrits ... $R^{-2}$). Or am I misunderstanding you? $\endgroup$ Commented Nov 3, 2013 at 22:06
  • $\begingroup$ Yes, you are right. The dangers of doing calculations in your head. Thanks for spotting my ^%$#@, I'll try to be more careful with what I type in the future :) At this point I'm not even sure I should leave my answer up, although I think I've ironed out all the bugs. $\endgroup$ Commented Nov 4, 2013 at 1:02
  • $\begingroup$ @lionelbrits No, please leave it up. Our answers are now very alike, but this (in my opinion) is quite OK for the simple reason that different writing styles work for different people. Neither of us know the OP and his / her way of thinking. Some would disagree with such "duplicate answers" but really the only problem I see is that plagiarism is unfair, but this is clearly not so here (within reason, clearly ten very alike answers could begin to get overwhelming for someone struggling with a concept). $\endgroup$ Commented Nov 4, 2013 at 1:07

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