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2.10.9   Nonradiating sources
                          Not all time-dependent sources produce electromagnetic waves. In fact, certain local-
                        ized source distributions produce no fields external to the region containing the sources.
                        Such distributions are said to be nonradiating, and the fields they produce (within their
                        source regions) lack wave characteristics.
                          Let us consider a specific example involving two concentric spheres. The inner sphere,
                        carrying a uniformly distributed total charge −Q, is rigid and has a fixed radius a; the
                        outer sphere, carrying uniform charge +Q, is a flexible balloon that can be stretched to
                        any radius b = b(t). The two surfaces are initially stationary, some external force being
                        required to hold them in place. Now suppose we apply a time-varying force that results
                        in b(t) changing from b(t 1 ) = b 1 to b(t 2 ) = b 2 > b 1 . This creates a radially directed
                        time-varying current ˆ rJ r (r, t). By symmetry J r depends only on r and produces a field
                        E that depends only on r and is directed radially. An application of Gauss’s law over a
                        sphere of radius r 0 > b 2 , which contains zero total charge, gives
                                                          2
                                                      4πr E r (r 0 , t) = 0,
                                                          0
                        hence E(r, t) = 0 for r > r 0 and all time t.So E = 0 external to the current distribution
                        and no outward traveling wave is produced. Gauss’s law also shows that E = 0 inside
                        the rigid sphere, while between the spheres
                                                                  Q
                                                     E(r, t) =−ˆ r    .
                                                                4π
 0 r 2
                        Now work is certainly required to stretch the balloon and overcome the Lorentz force
                        between the two charged surfaces. But an application of Poynting’s theorem over a
                        surface enclosing both spheres shows that no energy is carried away by an electromagnetic
                        wave. Where does the expended energy go? The presence of only two nonzero terms in

                        Poynting’s theorem clearly indicates that the power term  E · J dV corresponding to
                                                                            V
                        the external work must be balanced exactly by a change in stored energy. As the radius
                        of the balloon increases, so does the region of nonzero field as well as the stored energy.
                          In free space any current source expressible in the form
                                                                ∂ψ(r, t)

                                                    J(r, t) =∇                                (2.359)
                                                                  ∂t
                        and localized to a volume region V , such as the current in the example above, is nonra-
                        diating. Indeed, Ampere’s law states that

                                                           ∂E      ∂ψ(r, t)
                                                 ∇× H = 
 0   +∇                              (2.360)
                                                           ∂t         ∂t
                        for r ∈ V ; taking the curl we have

                                                         ∂∇× E            ∂ψ(r, t)
                                         ∇× (∇× H) = 
 0        +∇ ×∇             .
                                                           ∂t               ∂t
                        But the second term on the right is zero, so
                                                                   ∂∇× E
                                                   ∇× (∇× H) = 
 0
                                                                     ∂t
                        and this equation holds for all r. By Faraday’s law we can rewrite it as

                                                                2
                                                             1 ∂
                                                 (∇× ∇×) +         H(r, t) = 0.
                                                             2
                                                            c ∂t 2

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