On the Fields of Radiant Pulses Emitted by a Relativistically Oscillating Point Charge
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:
Such a charge emits radiation in pulses that are highly concentrated around the x-axis. Fig. 1a plots the energy that fluxes through a surrounding spherical surface, of radius 10l, over the course of an oscillation time. Note how virtually all of the energy flux through the surface occurs in sharply defined pulses. Fig. 1b plots the energy per cycle that fluxes through the same surface, as a function of azimuth angle q. Note in this plot how virtually all of the energy flow is confined to narrow solid angles around the x-axis.
Energy Flux Through Surrounding Surface vs. Time
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 Ey is negative just above the pulse’s midpoint. But at y=2.64535E-4 meters Ey shoots off to large, positive values. We shall denote this value of y as "R" and use it for the cylinder radius. Fig. 2 illustrates the negative values of Ey for values of y<R.
The general goal will be to plot the components of E that are perpendicular to the cylinder’s end caps (Ex) and sidewall (Ey). 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 Ey(x,R,0,0) and Bz(x,R,0,0) respectively, with x ranging from 10l-L/2 (L/2 being half a pulse width) to 10l+L/2. More generally (by symmetry) Fig. 3a plots the component of E that is perpendicular to the cylinder wall at each value of x in the stated range. Similarly, Fig. 2b more generally plots the component of B tangent to the cylinder wall. Note that the x-component of the Poynting vector, Sx=eooc2(EperpendicularBtangential) is positive at all points on the cylinder sidewall. Similar remarks presumably apply for all y<R.
Eperpendicular at Points on the Cylinder Wall, wA=.9999c
Btangential at Points on the Cylinder Wall, wA=.9999c
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 E (initially zero) grows increasingly positive to some maximum value. It then collapses rapidly and as quickly shoots down to a maximum negative value. It then decays practically back to zero as the pulse completes its transit through the plane.
Again considering the "snapshot" of Eperpendicular at time t=0, the E field flux through the left half of the cylinder wall is clearly negative, and that through the right half is positive. If we consider only the flux through the surface of the left half of the cylinder, an interesting question arises. Since there is no charge within that left half, is Gauss’ law satisfied? That is, does the sum of the fluxes through the half cylinder’s two end caps cancel out the negative flux through its sidewall? In order to compute these fluxes, we compute Ex(10l - Pulse Width/2,y,0,0) and Ex(10l,y,0,0) over the ranges 0<y<R. Denoting the fluxes through the left and right end-caps and the half cylinder’s side as Fleftleft, Frightright, and Fside side respectively, it is found that
Fleft left = -1006.0 Volt-meters, (2a)
Fright right = 10012.5 Volt-meters, (2b)
Fside side = -9006.4 Volt-meters. (2c)
Thus the total flux through the surface of the half cylinder is:
Fcylinder cylinder cylinder = 2.7E-2 Volt-meters (or practically zero). (2d)
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 Eperpendicular and Btangential in this case. The small change in wA produces dramatic results. Each field component has now become totally compressed.
Eperpendicular at Points on the Cylinder Wall, wA=c
Btangential at Points on the Cylinder Wall, wA=c
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 Ex, Ey, and Bz at (10l,y,0,0) with y ranging from 0 up to 1/300th of a wavelength. The magnitude of the Poynting vector is generally much greater than zero. However, as indicated in Fig. 1a, the total power flux through an enveloping surface (say a spherical one of radius 10l), at any time other than pulse time, is practically zero. And (see Fig. 1b), the net energy flux per cycle, through any surface increment other than a small disc that is normal to and centered on the x-axis, is practically zero. These results are all consequences of the relativistically correct point charge field solutions.
Ex(10l,y,0,0), Range of y >> Cylinder Radius
Ey(10l,y,0,0), Range of y >> Cylinder Radius
Bz(10l,y,0,0), Range of y >> Cylinder Radius