On the Fields of Radiant Pulses Emitted by a Relativistically Oscillating Point Charge G.R.Dixon, 7/4/04
Programs that can be viewed by clicking on this link created the figures and field fluxes in this article. In this article the fields of a 1-coulomb point charge with the following motion are investigated: (1) Such a charge emits radiation in Figure 1a
Energy Flux Through Surrounding Surface vs. Time Figure 1b
Energy Flux per Cycle vs. Azimuth Angle The relativistic formulas for radiated power (both the Larmor and Abraham-Lorentz formulas) indicate that these pulses are emitted each time the charge passes through the origin. Thus at time t=0 the center of one of the pulses is presumably at x = nl, n=1,2,3,… We shall begin by using the point charge field solutions to solve for the fields in the immediate vicinity of x=10l at time t=0. In particular the electric field is found at points on a small cylindrical surface, concentric to the x-axis and centered on x=10l. The cylinder’s length, L, is set to slightly more than a pulse width of approximately 3.377E-5 meters. It turns out that E Figure 2
The general goal will be to plot the components of _{x}) and sidewall (E_{y}). There are other components, but for present the focus is upon the transverse components on the cylinder wall, and upon the longitudinal components on the end caps. Secondarily, the electric field flux through part of the cylinder will be computed and shown to be zero (as, according to Maxwell, it must be).Figs. 3a and 3b plot E tangent to the cylinder wall. Note that the x-component of the Poynting vector, SB_{x}=e_{o}oc^{2}(E_{perpendicular}B_{tangential}) is positive at all points on the cylinder sidewall. Similar remarks presumably apply for all y<R.Figure 3a
E Figure 3b
B It is interesting to view this transverse component of the electric field in time, say in the plane that perpendicularly intersects the x-axis at x=10l. Viewing things at distances from the x-axis equal to R (the cylinder radius) and starting at some time slightly before t=0, the perpendicular component of Again considering the "snapshot" of E
F F F Thus the total flux through the surface of the half cylinder is:
F It is interesting to compute the fields on the cylinder’s side when
wA=c. (The ad hoc rationale for doing this is that, in any given half cycle, the charge moves at the speed of light only for an instant.) Figs. 4a and 4b plot E
Figure 4a
E
Figure 4b
B The electric and magnetic fields at points off the x-axis, at distances greater than R (the small cylinder radius discussed above), are certainly not zero. For example, Figs 5a - c plot E Figure 5a
E Figure 5b
E Figure 5c
B |