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Covariance of the Larmor-Lienard Formula for Radiated Power, Oscillation Along the x-axis

G.R.Dixon, 6/30/2006

The Larmor-Lienard formula for the power radiated by a charge moving along the x-axis of inertial frame K is

. (1)

For purposes of numerical integration of PRad(t), we can build a table with the following columns:






If the charges motion is periodic, say

, (2)

then consecutive values of ti can be separated by increments dt, where

. (3)

Having populated the table, the energy radiated per cycle can be computed:

. (4)

For q=1 coul, A=1E-6 meters, w=.0001c/A, Eq. 4 yields

. (5)

Given the motion in Eq. 2, q radiates a spherical wave. In frame K the net momentum in one complete wave is zero. Thus from the perspective of inertial frame K (moving in the positive x-direction of K at speed v),

. (6)

We can determine whether Larmor-Lienard (Eq. 1) produces this result in K by transforming the quantities in the table into K quantities. Since the ti will not be evenly spaced in K time, we define dti to be

. (7)

Fig. 1a plots P(t), and Fig. 1b plots P(t), with v=.95c. Note the identical shapes, but the different scales on the time axes. When applied to the K values, the sum in Eq. 4 produces

, (8)


. (9)

Figure 1a


Figure 1b


If the Lorentz transformations of ux and ax are substituted in Eq. 1, then it is readily demonstrated that


and thus that

. (11)

Like Newtons 2nd law, then, the Larmor-Lienard formula has the same form and works equally well in all inertial frames. As discussed in a previous article, frame K might be said to be the "motional rest frame." And ERad has its least value in the motional rest frame. Although the transformations for E and p (momentum) ordinarily pertain to material particles, they evidently also apply to spherical waves that are distributed in space.

In the motional rest frame the energy flux per cycle, through a spherical surface centered on the origin, can be computed using the solutions for E and B of an oscillating point charge. When this is done it is found that

, (12)

where dA is an increment of surface area. This same exercise cannot be successfully executed in K, however, since part of the energy flux through a (fixed) surface in K corresponds to field energy that moves along with the translating oscillator.

The Larmor-Lienard formula is enigmatic in the following sense. In mechanics, the power expended by a driving force is Fxux. Since ux is zero at x=+A (Eq. 2), a driving agent expends no power at these positions. Yet this is precisely where the Larmor-Lienard formula for P has its maximum. It seems that there is a delay between (a) work increments done by the driving agent, and (b) the appearance of increments of radiant energy in space. To quote R. Feynman, "The exact time when the energy is liberated cannot be defined precisely."(1) The "materialization" of such radiant energy increments is explored in an article about a constant-velocity charge that is, during some short time interval, decelerated and brought to a halt.


1. The Feynman Lectures on Physics, V1, Sect. 32-2, "The rate of radiation of energy."