This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 nonradiative. We see this argument also in Newtonian gravity: because of the inverse square force law (common to both Newtons and Coulombs 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 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 Gausss law, V E d A = q , where q is the total charge inside the gaussian surface, and 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 x . Notice that there is no flux of E x through the faces of the gaussian surfaces in the y or z directions because of the orthogonality of E x to the surfaces normals. This immediately implies E x / x = 0, or E x does not obey a nontrivial wave equation. Therefore, the most general electric field isdoes not obey a nontrivial wave equation....
View
Full
Document
This note was uploaded on 07/19/2011 for the course PH 113 taught by Professor Liorburko during the Spring '09 term at University of Alabama  Huntsville.
 Spring '09
 LIORBURKO
 Physics, Charge, Force

Click to edit the document details