[Phys-L] Re: accelerating charge

[jsd at AV8N.COM (John Denker)]

John Mallinckrodt wrote:
> <http://www.mathpages.com/home/kmath528/kmath528.htm>
>
> I won't claim to have thought enough about all of this, but I did
> find the arguments there somewhat compelling.

There is a wide assortment of arguments there, some of
them better than others.

1) I find the equivalence-principle arguments to be not
compelling at all.

Suppose we put into orbit a charged object.  Put the
counter-charge far, far away so we don't need to worry
about it.  Similarly ignore any effects arising from the
dielectric constant of the planet.  The object is in
free-fall, i.e. according to a comoving colocated observer
it is unaccelerated, yet it seems entirely obvious to me
that it radiates.  You can use Newtonian laws of gravity
to calculate the orbit, and you can use classical
electromagnetism to calculate the radiation.

Maybe I'm missing something, but I just cannot imagine any
other result.

This does not conflict with my understanding of the equivalence
principle.  It speaks of a very *local* equivalence between
gravitation and acceleration ... but radiation is something
that involves the far field.  The radiation field has a
memory of other places and other times.

It is possible to rederive much of general relativity starting
from special relativity.  To deal with accelerated frames, use
a succession of instantaneously comoving unaccelerated frames.
But -- as always -- you have to be careful when switching from
one frame to another.  By way of analogy, in the frame comoving
with a baseball bat, the ball does not gain any energy when it
collides with the bat, yet when we switch to the frame of a
baseball player or spectator, the ball gains a great deal of
energy from the collision.

2) On another point, I stand corrected:  My scaling argument
is not powerful enough to distinguish
   (D dot) × (D dot dot dot)
versus
   (D dot dot) × (D dot dot)    i.e. (D dot dot)^2

where as before, D is the dipole moment and dot means
time derivative.

In the case of bounded motion (including periodic motion)
you can freely convert one to the other by integrating by
parts, but not in the case of uniform acceleration aka
hyperbolic motion.

I note that the original question did not specify uniform
acceleration or even straight-line acceleration, so I will
retroactively make my analysis correct by restricting it to
the following scenario:  Bob is using his rocket to drive
himself around a closed path somewhere in outer space.  Ann
is stationary relative to the path.  You can optionally
simplify the scenario by assuming Bob is undergoing uniform
circular motion as observed by Ann, i.e. constant |v|, constant
|a|, and always v perpendicular to a.

I actually had the circular picture in mind when doing my
calculation, but I forgot to mention it, perhaps because
I thought my analysis was more general than it really was.
Sorry.

As for the case of hyperbolic motion, I will need to think
about that some more.  I would expect classical electrodynamics
to have an unambiguous prediction, one way or the other.

rlamont wrote:
> I find the concept of a single stationary charge in a
> gravitational field spontaneously radiating energy rather unnerving.

Me too.

This is complementary to my argument in part (1) above.

The GR curvature of space doesn't prevent you from building
Gaussian pillboxes.  For the charge sitting stationary
relative to the planet, the charge in each pillbox stays
the same, so I just cannot imagine how there could be
radiation.

> Appealing to a phantom charge
> to complete a "dipole" seems too easy a way out of the dilemma since
> the history of the separation of charges is not part of the argument.

I agree it isn't essential to the argument.  Under mild
assumptions, the multipole expansion is a complete basis.
If the dipole contribution turns out to be zero, you might
need to consider higher-order terms and/or other arguments
to decide whether the grand-total result is zero, but if the
dipole contribution is nonzero you should be able to rely
on it.