Author’s Note: Following is a section from a work now in progress. The results are interesting enough to prompt publication without delay. Following the section’s text is the Visual Basic program used to compute the results. GRD, 11/8/2009

Let us suppose that, viewed from the positive z-axis, two equal charges travel in a CCW circle of radius A in the xy-plane. The (centrifugal) inertial reaction forces (and the driving agent’s centripetal counteractions) point normal to the velocity vectors at all time, and thus do no work. (Any magnetic field will have only a z-component. Hence any magnetic force on a subject charge will also point normal to the charge’s velocity.)

The radiation reaction forces point opposite to the velocities. The work per cycle, expended to counteract each radiation reaction force, is q^{2}w^{3}A^{2}/(3e_{0}c^{3}). The work per cycle expended to counteract both radiation reaction forces is twice this amount.

Each charge also experiences an interactive force attributable to the other charge. One of the components of this force acts normal to the subject charge’s velocity, and thus only serves to modify the inertial reaction force and/or the magnetic force. Again, the work per cycle expended by the driving agent to counteract this component of the interactive force is zero. The other component of the interactive force acts *tangentially* to the subject charge’s circular orbit. It therefore modifies the radiation reaction force.

Program 5 computes the tangential components of the interactive forces and the work per orbit expended to counteract these components. Perhaps surprisingly, *the interactive force acting upon a charge points in the same direction as the charge’s velocity. *This *decreases* the work per cycle that must be expended to counteract the sums of the radiation reaction and the interactive forces. Indeed for values used in the program, the work per orbit expended to counteract the radiation reaction forces if 75329 joules, whereas the work to counteract the tangential components of the interactive forces if 75308 joules. The net work per cycle, required to drive both charges in tandem around the circular orbit, is thus only 21 joules!

It might be expected that by adding more charges, evenly distributed around a circular orbit, the attenuating effect of the interactive forces would be even more pronounced. Indeed if an infinite number infinitesimal point charges are evenly distributed around a circular orbit, then we have what amounts to a circular current loop. And it takes no power at all to drive a continuous, circular line charge at constant speed around a circle. Of course a corollary to this result is that a constant, circular current emits no radiant energy.

Private Sub cmdProgram5_Click()

'PROGRAM 5

'*****************

'Compute the work per cycle that must be done

'to drive 2 diametrically opposed charges around a circle.

'*****************

Const c As Double = 300000000#

Const epsilon0 As Double = 0.00000000000885

Const pi As Double = 3.141592654

Const q As Double = 1

Const A As Double = 1

Const omega As Double = 0.01 * c / A

Dim x2, y2 As Double 'right (subject)charge

Dim v2 As Double

Dim dtmin, dtmax As Double

Dim dt As Double

Dim t, tr As Double

Dim x1r, y1r As Double 'left (source) charge

Dim drx, dry, dr As Double

Dim v1xr, v1yr, a1xr, a1yr, v1r As Double

Dim ux, uy As Double

Dim Ey As Double

Dim Fy As Double

Dim Work, WorkRR, WorkInt, WorkTotal As Double

'Compute the work per cycle to counteract the radiation

'reaction forces

WorkRR = q ^ 2 * omega ^ 3 * A ^ 2 / (3 * epsilon0 * c ^ 3)

WorkRR = 2 * WorkRR

MsgBox ("WorkRR = " & WorkRR)

'Compute Ey at q2

t = 0

x2 = A

y2 = 0

dtmin = 0

dtmax = 5 * A / c

Do

dt = (dtmin + dtmax) / 2

tr = t - dt

x1r = -A * Cos(omega * tr)

y1r = -A * Sin(omega * tr)

drx = x2 - x1r

dry = y2 - y1r

dr = Sqr(drx ^ 2 + dry ^ 2)

If Abs(c * dt - dr) < 2 ^ (-30) Then Exit Do

If c * dt - dr > 0 Then

dtmax = dt

Else

dtmin = dt

End If

Loop

v1xr = omega * A * Sin(omega * tr)

v1yr = -omega * A * Cos(omega * tr)

v1r = Sqr(v1xr ^ 2 + v1yr ^ 2)

a1xr = omega ^ 2 * A * Cos(omega * tr)

a1yr = omega ^ 2 * A * Sin(omega * tr)

ux = c * drx / dr - v1xr

uy = c * dry / dr - v1yr

Ey = q / (4 * pi * epsilon0) * dr / (drx * ux + dry * uy) ^ 3 * (uy * (c ^ 2 - v1r ^ 2) - drx * (ux * a1yr - uy * a1xr))

MsgBox ("Ey = " & Ey)

Fy = -q * Ey

WorkInt = Fy * 2 * pi * A

WorkInt = 2 * WorkInt

MsgBox ("WorkInt = " & WorkInt)

WorkTotal = WorkRR + WorkInt

MsgBox ("WorkTotal = " & WorkTotal)

Stop

End Sub