__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

F_{RadReact} = (q^{2}
/ 6pe_{o}c^{3})(d^{3}x/dt^{3}). (1)

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

F + (q^{2} / 6pe_{o}c^{3})(d^{3}x/dt^{3}) = m d^{2}x/dt^{2}. (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

(q^{2} / 6pe_{o}c^{3})(d^{3}x/dt^{3}) – m d^{2}x/dt^{2}
– 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

d^{3}x/dt^{3}
= -w^{2} dx/t. (5)

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

m d^{2}x/dt^{2}
+ (w^{2}q^{2}
/ 6pe_{o}c^{3}) dx/dt + kx = 0. (6)

Defining
constants b = w^{2}q^{2}
/ 6pe_{o}c^{3} and w’ = [k/m –
(b/2m)^{2}]^{1/2} we have

m d^{2}x/dt^{2}
+ 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 + F_{aug} + (q^{2}
/ 6pe_{o}c^{3})(d^{3}x/dt^{3}) = m d^{2}x/dt^{2}. (9)

Since
x = A sin(wt) is the solution of –kx = m d^{2}x/dt^{2},
we can see at once that

F_{aug} = -(q^{2}
/ 6pe_{o}c^{3})(d^{3}x/dt^{3}). (10)

The rate at which F_{aug}
does work is the rate at which radiant power is emitted:

F_{aug} v = P_{rad}. (11)

And
the definite integral of P_{rad} over limits 0 and 2p/w will equal E_{rad}, the radiant energy emitted per cycle:

E_{rad} = w^{4}q^{2}A^{2}
/ 6e_{o}c^{3}. (12)

Note
that the __Larmor__ formula for radiant power is

P_{Larmor} = q^{2}a^{2}
/ 6pe_{o}c^{3}. (13)

This
also integrates to w^{4}q^{2}A^{2}
/ 6e_{o}c^{3}. 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

(q^{2} / 6pe_{o}c^{3}) (d^{3}x/dt^{3}) – m d^{2}x/dt^{2}
= 0. (14)

A solution is

d^{3}x/dt^{3}
= d^{2}x/dt^{2} = 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 d^{2}x/dt^{2} and d^{3}x/dt^{3}
are __not__ zero leads to the solution

d^{2}x/dt^{2}
= a_{o} e^{t / }^{t}, (16)

where

t = q^{2} / 6pe_{o}mc^{3} (17)

and a_{o}
is a constant. If a_{o} is nonzero, then d^{2}x/dt^{2}
can “run away” to unlimited positive or negative values. Such behaviors do not
agree with experiment.