1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Meaning of Sommerfeld radiation conditions

  1. Mar 1, 2013 #1
    Promising that I will not make other new questions in the next days :smile:, I have a doubt about the meaning of a pair of expressions.
    Sommerfeld's conditions for an electromagnetic field produced by a finite source bounded by a finite volume are:

    [itex]\lim_{r \to +\infty} r|\mathbf{E}| < q\\
    \lim_{r \to +\infty} r|\mathbf{H}| < q\\
    \lim_{r \to +\infty} r \left[\mathbf{E} - \eta \mathbf{H} \times \mathbf{\hat{u}}_k \right] = 0\\
    \lim_{r \to +\infty} r \left[\mathbf{H} - \displaystyle \frac{\mathbf{\hat{u}}_k \times \mathbf{E}}{\eta} \right] = 0[/itex]

    where [itex]q[/itex] are finite quantities, [itex]\eta[/itex] is the wave impedance in the considered medium (for example the free space), [itex]\mathbf{\hat{u}}_k[/itex] is the direction of propagation and [itex]r[/itex] is the distance from the source.
    The first two state that the fields' module must decrease at least as [itex]1/r[/itex].
    The last two state that the fields must be similar to those of a plane wave: they must be mutually orthogonal and also both orhogonal to the direction of propagation. Moreover, the "part" of [itex]\mathbf{E}[/itex] (in the first) and [itex]\mathbf{H}[/itex] (in the second) which does not contribute to that plane wave must decrease at least as [itex]1/r^2[/itex].

    Why these last two conditions are called "radiation conditions"? As a matter of fact, in the electric dipole non-radiative field components decrease as [itex]1/r^2[/itex] or [itex]1/r^3[/itex]. But why this is a necessary requirement to build a "radiation"? Couldn't we have a not-radiating component which decreases as [itex]1/r[/itex]? Why?
    Thank you, again, for having read.

  2. jcsd
  3. Mar 1, 2013 #2


    User Avatar
    2017 Award

    Staff: Mentor

    Every component which decreases with 1/r looks like radiation, and therefore it is called radiation.
  4. Sep 11, 2013 #3

    Instead of opening a new thread, I recover this old, poor one.
    The first two conditions select a field [itex]\mathbf{E}, \mathbf{H}[/itex] which asymptotically (that is, for [itex]r \to \infty[/itex]) is equal to a spherical wave, because

    [itex]|\mathbf{E}| = O \left(\displaystyle \frac{1}{r} \right)[/itex]
    [itex]|\mathbf{H}| = O \left(\displaystyle \frac{1}{r} \right)[/itex]

    if the limits are verified.
    But why a spherical wave? Why this is the only type of wave accepted by these conditions?

  5. Sep 11, 2013 #4


    User Avatar
    2017 Award

    Staff: Mentor

    The source is finite - every radiation (every 1/r source) has to come from the source, and for large radii the source is basically point-like (with higher orders accounting for the deviations from that).
  6. Sep 11, 2013 #5
    the higher orders that disappear as [itex]r \to \infty[/itex].
    Ok and thank you. But you are speaking about the source, not the field: who can say that a point source must generate a wave with a spherical (as [itex]r \to \infty[/itex]) wavefront? The wavefront could be an ellipsoid or something else and the radiation will transfer power to the infinity as well.
  7. Sep 11, 2013 #6
    Here http://www.math.wichita.edu/~deepak/IEMS.pdf
    (bottom of page 1) it is written what you said. The asymptotic behaviour of a field is showed as

    [itex]\mathbf{E} = \displaystyle \frac{e^{ikr}}{r} \left[ \mathbf{E}_{\infty} + O \left( \displaystyle \frac{1}{r} \right) \right][/itex]

    Even without this link, I knew you were reason, but I'm wondering: why?

  8. Sep 11, 2013 #7
    Sorry for the string of posts, but the answer may be: because of its symmetry, a point source must radiate the same field in all the possible directions.
  9. Sep 11, 2013 #8


    User Avatar
    2017 Award

    Staff: Mentor

    "Spherical" does not mean "spherical symmetry" (that is impossible). The radiation is emitted radially outwards.

    You can edit your own posts to avoid multiple posts.
  10. Sep 12, 2013 #9


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Have a look at the "multipole expansion". This is an expansion of the em. field from a charge-current distribution that is confined to a finite space region at points far away from this region. The expansion goes by using spherical coordinates in Maxwell's equations and expansion in powers of [itex]1/r[/itex].

    The general idea is to separate the wave equation in spherical coordinates, leading to an expansion in spherical harmonics [itex]\mathrm{Y}_{lm}(\vartheta,\varphi)[/itex]. You'll find that for the em. fields the spherically symmetric part, i.e. the term with [itex]l=m=0[/itex], is necessarily a static contribution and at far distances reduces to a Coulomb field of a point charge, i.e., a scalar [itex]1/r[/itex] potential.

    The wave solutions start at [itex]l=1[/itex], which is the dipole term. For wave fields there are electric and magnetic dipole contributions. For the wave fields they start with a spherical wave
    [itex]\vec{E} \propto \exp(\mathrm{i} k r)/r[/itex], where the temporal part of the wave is assumed as harmonic, [itex]\vec{E} \propto \exp(-\mathrm{i} \omega t)[/itex]. With this sign convention the spherical wave [itex]\propto \exp(\mathrm{i} k r)[/itex] is a spherical wave running outwards, i.e., away from the source, and that's the solution you are looking for. If you keep the time dependence general, these are precisely the retarded solutions, i.e., in Lorenz gauge (in Heaviside-Lorentz units):
    [tex]A^{\mu}(t,x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{J^{\mu}(t-r/c,\vec{x}')}{4 \pi c r} \quad \text{with} \quad r=|\vec{x}-\vec{x}'|.[/tex]

    The Sommerfeld radiation conditions are needed to impose these retardation conditions for the Helmholtz equation, where you already assumed the harmonic time dependence. They basically force the fields to run outwards from the sources when looking from a far distance from the sources, choosing the solution [itex]\propto \exp(+\mathrm{i} k r)[/itex] rather than [itex]\propto \exp(-\mathrm{i} k r)[/itex], which describe waves running towards the sources, being absorbed there.
  11. Sep 12, 2013 #10
    I had never seen the multipole expansion, but it seems very useful.
    Ok for the [itex]\mathrm{exp}(+ikr)[/itex]: so, maybe your refer to the original Sommerfeld condition

    [itex]\lim_{r \to \infty} r \left( \displaystyle \frac{\partial \phi}{\partial r} - ik \phi \right) = 0[/itex]

    where [itex]\phi[/itex] is proportional to [itex]\mathrm{exp}(+ikr)[/itex]. So, if we had inside the limit [itex]+ ik \phi[/itex] instead of [itex]- ik \phi[/itex], we would select "the solution running towards the sources".
    But this can work only if we assume [itex]\phi[/itex] proportional to [itex]\mathrm{exp}(+ikr)[/itex] and not otherwise. I considered this fact a loss of generality, but if the asymptotic behaviour of any field is like [itex]\mathrm{exp}(+ikr)[/itex], this is a general result.
    In the Silver-Muller conditions (the ones I wrote in the first post) you can't say the [itex]\mathrm{exp}(+ikr)[/itex] term, but I think the outgoing wave is given by the outgoing Poynting vector originated by such fields.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook