Why is radiation power defined at infinity?

Question

According to Griffiths in chapter 11 , given a source of radiation enclosed by a sphere, the power passing through the sphere is $$ P(r) = \oint \mathbf{S} \cdot d\mathbf{a} = \frac{1}{\mu_0} \oint (\mathbf{E} \times \mathbf{B}) \cdot d\mathbf{a}, $$ and the power radiated is: $$ P_{rad} = \lim_{r\to\infty} P(r). $$ What I don't understand is why $r$ has to go to infinity for the power to be defined? I understand that Griffiths defines the radiation as the energy transported to infinity, but isn't the power the same at every r?

Answer 1

There is usually a "near field" part of the field energy that falls off faster than $1/r^2$. I suppose the $r\to \infty$ in Griffiths is designed so as not to count this part of the field.

Answer 2

This is one of the places where Griffiths is less conceptually correct and clear than elsewhere in the book.

Griffiths, in that section, is saying what a litmus test for radiation is, but his test actually is for radiation in infinity, an interesting and important thing, but not for radiation in general.

He says it is a signature of radiation that there is an irreversible flow of energy. But this is not true for radiation in general; there are counterexamples which are radiation, and are not associated with irreversible energy transfer.

Radiation in general is primarily the presence of waves (a special kind of field), not a transfer of energy from one place to another these waves can be associated with. A transfer of energy is a common coinciding phenomenon with non-equilibrium radiation, but it is not a necessary feature of all radiation. For example, consider a sphere which interacts via equilibrium thermal radiation with a smaller sphere of the same temperature in its insides. The Griffiths litmus test would say there is no radiation, because there is no irreversible flow of energy. Of course, the proper understanding of this litmus test is that there is no radiation energy going to infinity - but that there is radiation field in between the spheres.

You are correct that in the class of scenarios he considers (a finite-sized system in vacuum), explicit focus on the limit of $r$ going to infinity is weird. First of all, we can't ever check this in practice on any real field, it is only a formal condition on imagined EM fields, which are defined all the way to infinite distances and times. But OK, let's use it on fields on paper only. Griffiths shows some examples of fields which do not depend on time, for which the limit $\lim_{r\to\infty} P(r)$ has obvious meaning, and its value is zero. However, if we talk about radiation, then the interesting cases are those which have fields depending on time, but then $P$ depends on time $t$, and then the meaning of the expression $\lim_{r\to\infty} P(r,t)$ is not so clear; $P(r,t)$ in infinity depends also on value of the parameter $t$, or the way it changes with $r$. For some values of $t$, the limit value may be zero, and for others, it may not. For example, already in 11.20, in the formula for the Poynting vector for dipole radiation, we can see it strongly depends on time, and net Poynting flux does as well, so the limit for $r\to\infty$ does not exist independently of $t$; if we are to get the limit value, we need to know for which $t$. The litmus test is not so easy to apply.

Griffiths could say that well, in such cases, let's redefine the litmus test to work with time average value of the Poynting flux instead, to get rid of this problem with time. But then we get back to what you point out - that the limit of infinite $r$ is not actually needed.

For systems surrounded by vacuum all the way to infinity, time average of $P$ ($\overline{P(r)}$) being non-zero at any $r$ indicates the same thing as value of $\lim_{r\to\infty} \overline{P(r)}$ being non-zero - the presence of radiation going to infinity. If $\overline{P(r)}$ is non-zero at some $r$, then it is so for all greater $r$. We do not need to additionally check the limit $r\to\infty$; it is sufficient to find the average flux for finite $r$, large enough to contain the material system, and if it is non-zero, then Poynting energy moves out to infinity.

One way to interpret the focus on the limit is that this is a formal laconic way to enforce the condition that the system is contained inside some finite imagined sphere, so we're talking about Poynting flux at a boundary that is very far from the system. But if so, that should be stated explicitly in words anyway, because we can also talk about radiation through a boundary very close to the system. And the limit to infinity is not needed in any case.

Answer 3

Just adding to @mike stone

From: https://en.wikipedia.org/wiki/Dipole#Dipole_radiation, the fields from an oscillating ($\omega$) electric dipole ($\vec p$) is:

$$ \vec E = \frac 1 {4\pi\epsilon_0}\Big( \frac{\omega^2}{c^2 r}(\hat r \times \vec p)\times\hat r + \big(\frac 1 {r^3}-\frac{i\omega}{c r^2}\big) $$ $$ (3\hat r[\hat r \cdot\vec p]-\vec p) \Big) e^{\frac{i\omega r} c} e^{-i\omega t} $$

$$ \vec B = \frac{\omega^2}{4\pi\epsilon_0 c^3} (\hat r \times \vec p) \big( 1-\frac c {i\omega r} \big) \frac{e^{i\omega r/c}} r e^{-i\omega t} $$

Note: when just look at maximum dipole as the charges oscillate compared with the maximum current--they are out of phase, however, the $1/r$ terms in the above equation are in-phase, and represent the radiation...at infinity.