Answer to the Question 08/00
RADIATING SPHEREThe question was:
Consider a thin charged spherical shell with time dependent radius R(t)=Ro+A*cos(wt) and total charge Q. How much power will be radiated by this shell?
(3/01) We received various solutions of the problem from Jürgen Holetzeck (14/9/2000) from Wiesbaden, Germany (e-mail j.holetzeck@germanynet.de), Nadav Shnerb (20/9/00) from Department of Physics in Judea and Samaria College, Israel (e-mail nadav@ycariel.yosh.ac.il), Michael Cohen (2/10/00) from Department of Physics & Astronomy at University of Pennsylvania (e-mail mcohen@dept.physics.upenn.edu), Kirk T. McDonald (5/10/00) from Princeton University (e-mail mcdonald@puphed.Princeton.edu), Zoran Hadzibabic (8/12/00) from MIT (e-mail zoran@mit.edu), Luis F. Rodriguez (27/1/01) from Institute of Astronomy of the National Autonomuos Institute of Mexico (e-mail l.rodriguez@astrosmo.unam.mx), Sylvain Wolf (5/3/01) from University of Lausanne, Switzerland (e-mail Sylvain.Wolf@ipt.unil.ch), and Itzhak Shapir (18/3/01) (e-mail ishapir1@netvision.net.il).
The answer: There will be no radiation.
The solution:
There are several ways to see that there will be no radiation from the sphere:
(a) The radiation consists of multipoles (dipole (magnetic or electric) radiation, quadrupole radiation, etc.) The symmetry of the source is "monopole", since clearly there is no directional dependence. Since there is no such thing as "monopole radiation", the spherical shell will not radiate. One can also say that by symmetry both electric and magnetic fields must be radial, while far away from the source the electromagnetic wave must be transverse!
(b) Let us to calculate the electromagnetic fields more carefully. The (particular Fourier component (frequency) of the) vector potential A at position x can be found by integrating the expression (1/c)J(x')exp[(w/c)|x-x'|]/|x-x'| over the positions x' of the source of radiation which is the (Fourier component of the) current density J(x'). Due to symmetry of the source the results of this integration will be of the form f(r)r where f(r) is some function of the distance from the center of the sphere and r is the radius-vector. [See, e.g.., Ch.9 in J.D. Jackson, Classical electrodynamics, Wiley, NY, 2nd edition (1975). The equations above are written in Gaussian units.] Since curl of radial vector A vanishes, the magnetic field will be zero everywhere and therefore there will be no radiation.
(c) Gauss law is applicable also in dynamic circumstances. If we draw a large sphere around our source charges, and assume that the electric field is isotropic, i.e. is in radial direction, and depends only on the distance from the center, then we reach a conclusion that the electric field is is constant in time. Consequently, we do not expect any radiation.
(d) Because of the symmetry the magnetic field B must point in radial direction and depend only on the distance from the center. This, together with equation div B=0, leads to the conclusion that B=0 everywhere. Consequently, there is no radiation.
Comment: Michael Cohen told us that this problem appears as a problem in Sec. 14 of the Classical Electricity and Magnetism by W.K.H. Panofsky and M. Phillips.
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