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# Lec9 - physics are symmetrical under time reversal(t t an...

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physics are symmetrical under time reversal ( t → - t ), an incoming imploding spherical monopole wave would make an otherwise static spherical charge distribution pulsate radi- ally. Each charge pulsates radially because of a radial force acting on it. The form of the Lorentz force law then implies there were a radial E field acting on the charges. But a radial E field is perpendicular to the imploding spherical wave front and therefore in the same direc- tion of its motion, so that the E field would have a longitudinal component. This qualitative argument implies that a monopole electromagnetic wave is necessarily longitudinal. We do know, however, that outside a spherical charge distribution the electromagnetic field is static, as the monopole piece of the electromagnetic field is non-radiative. We see this argument also in Newtonian gravity: because of the inverse square force law (common to both Newton’s and Coulomb’s laws) the field strength of any spherical charge (or mass) distribu- tion is the same as if all the charge (or mass) were concentrated at the center. Specifically, the electric field outside a radially pulsating spherical charge distribution is static. Therefore, the longitudinal component of the E field cannot exist, as such a longitudinal component would have to be radiated by a pulsating spherical charge distribution. Show that E does not have a longitudinal component : Consider a plane electromagnetic wave traveling in vacuum in the direction of ˆ x . (Our coordinates may be rotated to this orientation.) Because of the planar symmetry the most general electric field strength is E = E ( t, x ). In what follows we omit the explicit time dependence of fields. We first show there can be no component E x : construct a gaussian surface enclosing a volume V as in Fig. 1. According to Gauss’s law, V E · d A = q 0 , where q is the total charge inside the gaussian surface, and 0 is the permittivity of vacuum. Here, d A is a surface element normal to the (orientable) surface V , defined conventionally such that it is positive when pointing outward. Calculate the integral on the LHS: take E = E x ( x x + E y ( x y + E z ( x z . Then V E · d A = [ E x ( x + dx ) - E x ( x )] dy dz = 0 as there are no charges. The components E y and E z identically do not contribute to the LHS integral, as these components are not functions of y or z , respectively, and as they are in the plane of the faces of the surface whose normals are in the direction of ˆ x . Notice that there is no flux of E x through the faces of the gaussian surfaces in the ˆ y or ˆ z directions because of the orthogonality of E x to the surface’s normals. This immediately implies E x / x = 0, or E x does not obey a non-trivial wave equation. Therefore, the most general electric field is E = E y ( x y

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