On Vibrating, Charged Objects

G.R.Dixon, September 25, 2014

 

When a charged object oscillates on the x-axis, it emits radiation.  This radiation carries momentum alternating in the positive and negative x-directions.  By momentum conservation the object recoils in the opposite directions.  Abraham and Lorentz called the force responsible for this recoil the “radiation reaction force.”  Its formula is

 

FRadReact = (q2 / 6peoc3)(d3x/dt3).                                                                  (1)

 

If there is also a driving force F driving the object to and fro, then the net acting force is F + FRadReact, and by Newton

 

F +  (q2 / 6peoc3)(d3x/dt3) = m d2x/dt2.                                                         (2)

 

Here m is the sum of the object’s “mechanical” mass plus the charge’s electromagnetic mass.

By way of an example, we might consider the case where the object is driven by the free end of an anchored spring.  The force exerted by the spring is

 

F = -kx,                                                                                                           (3)

 

and Eq. 2 becomes

 

(q2 / 6peoc3)(d3x/dt3) – m d2x/dt2 – kx = 0.                                      (4)

 

By energy conservation we expect the object’s oscillation amplitude to attenuate with time as energy is radiated away.  In cases where the attenuation is not very great from oscillation to oscillation, the motion is approximately sinusoidal and we can approximate that

 

d3x/dt3 = -w2 dx/t.                                                                                           (5)

 

Substituting into Eq. 4, we end up with the 2nd order differential equation

 

m d2x/dt2 + (w2q2 / 6peoc3) dx/dt + kx = 0.                                                  (6)

 

Defining constants b = w2q2 / 6peoc3 and w’ = [k/m – (b/2m)2]1/2 we have

 

m d2x/dt2 + b dx/dt + kx = 0.                                                             (7)

 

And the solution of this equation is

 

x = Ae-bt/2m sin(w’t + d),                                                                                 (8)

 

where d is determined by the initial conditions.

An interesting variation is when A does not attenuate with time, say x = A sin(wt).  Since the object is charged, it must emit a single-valued amount of radiant energy each cycle.  Presumably the driving force in Eq. 2 must be augmented and we have

 

-kx + Faug +  (q2 / 6peoc3)(d3x/dt3) = m d2x/dt2.                                           (9)

 

Since x = A sin(wt) is the solution of –kx = m d2x/dt2, we can see at once that

 

Faug = -(q2 / 6peoc3)(d3x/dt3).                                                                        (10)

 

The rate at which Faug does work is the rate at which radiant power is emitted:

 

Faug v = Prad.                                                                                                  (11)

 

And the definite integral of Prad over limits 0 and 2p/w will equal Erad, the radiant energy emitted per cycle:

 

Erad = w4q2A2 / 6eoc3.                                                                                    (12)

 

Note that the Larmor formula for radiant power is

 

PLarmor = q2a2 / 6peoc3.                                                                                  (13)

 

This also integrates to w4q2A2 / 6eoc3.  However, the Larmor power is greatest at x = +A, where v = 0.

It is interesting to speculate about the case where the spring agent force, on a charged object whose motion is a solution to Eq. 4, suddenly drops to zero.  From Eq. 4 the equation of motion becomes

 

(q2 / 6peoc3) (d3x/dt3) – m d2x/dt2 = 0.                                                         (14)

 

A solution is

 

d3x/dt3 = d2x/dt2 = 0.                                                                                     (15)

 

In this case the object ceases radiating and starts translating at a constant speed. 

It is noteworthy that Eq. 15 is the physically tenable solution, but not the only mathematical one.  For assuming that d2x/dt2 and d3x/dt3 are not zero leads to the solution

 

d2x/dt2 = ao et / t,                                                                                            (16)

 

where

 

t = q2 / 6peomc3                                                                                             (17)

 

and ao is a constant.  If ao is nonzero, then d2x/dt2 can “run away” to unlimited positive or negative values.  Such behaviors do not agree with experiment.