[Phys-L] Re: accelerating charge

[jsd at AV8N.COM (John Denker)]

Carl Mungan wrote:

> Ann is in an inertial frame. Bob is in a rocket accelerating relative
> to Ann. Bob is holding a plastic rod bearing a net positive charge.
> Is the charge radiating? Bob says no; Ann says yes. Who's right and
> why?

Ann's right.  Bob's wrong.  Accelerating charges radiate.

The only thing that makes this tricky, AFAICT, is that
ordinary rocket ships accelerate at only modest rates
(a few Gees) and plastic rods carry only very modest
amounts of charge, so that the magnitude of the radiation
will be small -- so small that our intuition tells us it
is "zero".

Also there may be some ambiguity about the counter-charge,
as discussed below.

It is highly amusing to estimate how much radiation there
will be.  It serves as a good exercise and example of
qualitative reasoning.  You can do quite a nice job using
little more than scaling arguments and dimensional analysis.
You can calculate it on half a sheet of paper.  Here goes:

*) We want to calculate the radiative intensity (P/A i.e.
power per unit area) that is observed a distance R from
the source.

*) If you remember anything about how radiation works, you
know that the leading source term involves the first moment
of the charge distribution, i.e. the dipole moment, D. (There
are good reasons to discount higher and lower moments, but I
won't go into them now.)

   Partial result:  P/A = D ...?

*) It has to involve time-derivatives of the dipole moment.
The lowest derivative that makes sense is the second derivative,
D dot dot, because two charges moving past each other in uniform
straight-line motion have a nonzero D dot, and we know from
special relativity that uniform motion cannot radiate, ergo
D dot cannot be a source term.  But D dot dot is open for
business.  Here "dot" is shorthand for a dot written over the
D, and means the time derivative.  We assume Bob's rocket ship
is not moving very fast, so we needn't fuss with the distinction
between Ann's time and Bob's time.

   P/A = D dot dot ...?

*) It has to involve the _square_ of D dot dot, because the
power must always be positive, and it must change sign if we
change the sign of the charges and/or change the direction
of acceleration.

   P/A = (D dot dot)^2 ...?

*) To get anything with units of energy or power, we have to
include a factor of Coulomb's constant, k.
(Recall 1/k = 4 pi eps0, and eps0 = 8.85 pF/m).

   P/A = k (D dot dot)^2 ...?

*) The intensity will have to fall off like one over R squared,
because of conservation of energy.

           (D dot dot)^2
   P/A = k ------------- ...?
               R^2

*) If you check the units in the previous expression, you find
you need to throw in three factors of the speed of light to
make the units come out right:

            (D dot dot)^2
   P/A =~ k -------------
               R^2 c^3

And that's the desired formula.  As I seem to recall, the
exact formula differs by a factor of two ... but if you want
the exact formula you'll have to do some actual work.

Let's plug in some numbers.  Suppose the capacitive energy
in Bob's rod is on the order of one Joule (probably an
overestimate) and his acceleration is a few Gees.  Then the
total radiated power will be something like 10^-20 watts.
It's nonzero, but Bob & Ann might decide it's not worth
worrying about.

When charging his rod, Bob must have gotten the charge from
somewhere, leaving behind an equal and opposite charge, i.e.
the counter-charge.  In the foregoing, I assumed the counter-
charge was stationary.  If instead Bob carries the counter-
charge with him, the dipole moment is unchanging and the
radiation will be even smaller (vastly smaller) than the
estimate given above.

===================

The foregoing line of reasoning works equally well if you
want to estimate the intensity of gravitational radiation.
You can parlay the most rudimentary understanding of GR,
plus basic physics and scaling laws, to get the radiation
formula, correct to within a factor of 2.  (There's a
funny story about that....)