On this page, it states:

"The only fields' couples (a,b) for which we can get a non-zero value of the Poynting vector for a large distance r₀ over the sphere are (R, R); (radiation, radiation) and (2, R); (relativistic, radiation). The reason is this. Spherical surface increases as r₀² and each field in these couples, onto this surface, decreases as 1/r₀ so, their product decreases as (1/r₀)². Consequently, even if r₀ goes to infinity, surface integral (eq.(33)) of the Poynting vector remains unchanged. In that case a well definite quantity of power goes out the sphere no matter the value of r₀ is. So, this energy is lost for ever."

Is there an error here, as surely each field decreases as $\frac{1}{r_{0}^{2}}$, meaning their product decreases as $\frac{1}{r_{0}^{4}}$?

For example, in my plasma physics notes, in the middle of a derivation of the conservation of energy (where $E^{f}$ is the energy of the EM fields) it states:

$$ \begin{aligned} \frac{\mathrm{d} E^{f}}{\mathrm{~d} t}=& \varepsilon_{0} \int \mathrm{d}^{3} x \mathbf{E} \cdot\left[\frac{1}{\varepsilon_{0} \mu_{0}} \nabla \times \mathbf{B}-\frac{1}{\varepsilon_{0}} \mathrm{j}\right]+\frac{1}{\mu_{0}} \int \mathrm{d}^{3} x \mathbf{B} \cdot(-\nabla \times \mathbf{E})=\\ &-\int \mathrm{d}^{3} x \mathbf{j} \cdot \mathbf{E}+\frac{1}{\mu_{0}} \int \mathrm{d}^{3} x \mathbf{E} \cdot(\nabla \times \mathbf{B})-\frac{1}{\mu_{0}} \int \mathrm{d}^{3} x \mathbf{B} \cdot(\nabla \times \mathbf{E}) \end{aligned} $$ Using the vector identity $$ \nabla \cdot(\mathbf{E} \times \mathbf{B})=\mathbf{B} \cdot(\nabla \times \mathbf{E})-\mathbb{E} \cdot(\nabla \times \mathbf{B}) $$ we can write: $$ \int \mathrm{d}^{3} x[\mathbf{E} \cdot(\nabla \times \mathbf{B})-\mathbf{B} \cdot(\nabla \times \mathbf{E})]=-\int \mathrm{d}^{3} x \nabla \cdot(\mathbf{E} \times \mathbf{B})=\int_{S}(\mathbf{E} \times \mathbf{B}) \cdot \mathrm{d} \boldsymbol{\sigma}=0 $$ as for $\mathbf{x} \rightarrow \infty$ the fields must go to zero, to keep energy finite.

This suggests that the fields decrease much faster than the surface enclosing them increases, meaning no flux traverses the surface at infinity, which contradicts the webpage quoted at the top of this question. I am quite confused by both of these explanations and would like to know the conditions/cases in which the surface integral of the Poynting vector does indeed vanish.

EDIT: Note, for energy conservation in this case, $\frac{\mathrm{d} E^{f}}{\mathrm{~d} t}= -\int \mathrm{d}^{3} x\ \mathbf{j} \cdot\mathbf{E}$ $ $ must be satisfied.


1 Answer 1


Surely each field decreases as $1/r_0^2$

No. The inverse-square dependence of Coulomb’s Law doesn’t apply to accelerating charges. Look at the dependence on distance of the Liénard-Wiechert fields of a point charge in arbitrary motion. The term proportional to the acceleration decreases as $1/r_0$, not $1/r_0^2$, and causes the Poynting flux to be nonzero at infinity.

  • $\begingroup$ Oh ok, I see... So, in a non-relativistic limit, the fields decrease as ~ $\frac{1}{\mathrm{r_{0}}}$ for accelerating charges? $\endgroup$ Commented Dec 22, 2020 at 20:49
  • $\begingroup$ Not just in the non-relativistic limit. The LW fields are valid for any velocity. $\endgroup$
    – G. Smith
    Commented Dec 22, 2020 at 20:51
  • $\begingroup$ BTW, I looked only briefly at that paper you’re reading, but it seemed like it might be non-mainstream. I recommend treating it with skepticism. $\endgroup$
    – G. Smith
    Commented Dec 22, 2020 at 20:54
  • $\begingroup$ The Wikipedia link in your answer has a $\frac{1}{\mathrm{r}}$ and a $\frac{1}{\mathrm{r^{2}}}$ dependence but the $\frac{1}{\mathrm{r^{2}}}$ term is multiplied by the Lorentz factor... $\endgroup$ Commented Dec 22, 2020 at 20:55
  • $\begingroup$ Yes... I just found that webpage while trying to figure out the reasoning in my plasma notes (quoted near the end of my question). I still can't work out why the integral vanishes here... $\endgroup$ Commented Dec 22, 2020 at 20:58

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