Re: [Phys-L] electromagnetism paradox, or not

[John Denker <jsd@av8n.com>]

You might be a redneck physicist if ... your supply of 
"charcoal-lighter fluid" consists entirely of liquid oxygen.

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

Once upon a time there was a "teacher" who really wanted to
impress the students with how smart he was.  He also wanted 
students to avoid making mistakes.  Therefore he made a point 
of explicitly dispelling every possible misconception and paradox.
For example, on the first day when the topic of kinetic energy 
came up, he discussed
     KE = 1/2 m a^2
and made it sound as plausible as it possibly could be, before
dispelling the misconception.  On the second day he discussed
     KE = 1/2 m b^2
and made it sound as plausible as it possibly could be, before
dispelling the misconception.  On the second day he discussed
     KE = 1/2 m c^2
which is somewhat of an interesting story...... Eventually on 
the 22nd day he got around to
     KE = 1/2 m v^2
which turns out to be the right answer.  He was proud of
the fact that the students knew explicitly how to avoid lots 
of wrong answers.  He took the same approach with other
topics.  He misstated the laws of physics in every possible
way, creating lots of paradoxes and misconceptions.  This
convinced the students that physics was exceedingly challenging
and counterintuitive, and the "teacher" must be a genius for
being able to overcome so many difficulties.

I mention this because on 09/06/2013 09:30 AM, William Maddox cited:

> ..... from
> "The Universe in the Rearview Mirror": Consider one of Einstein's
> free falling frames of reference.  Supposedly the physics inside
> would produce the same results as an SR inertial frame (constant
> speed). What happens if the reference frame is a space station in
> orbit and you turn on a Van de Graaff? From the point of view of an
> outside observer the space station and the charges on the globe of
> the VDG are accelerated and should radiate. From the point of view of
> an astronaut the charges are at rest and should not radiate. What
> happens?

First of all, I assume we are supposed to ignore the startup
transient that occurs when we "turn on the Van de Graaff".
This is supposed to be a puzzle, and the transient makes the
right answer even more obvious than it already is, so let's
assume the VdG has been on for a very long time, and we are
concerned with a more-or-less steady charge distribution.
While we're at it, we might as well dispense with the VdG
and assume that due to the solar wind or whatever, the entire
space station has picked up a "static" charge.  If I'm wrong
about these assumptions, please clarify the question.

But is this really a puzzle?  Is there anybody on this list 
who doesn't know the right answer????  You have a charged 
object in orbit.  Is it going to radiate?  Classical (19th
century) physics suffices to answer this question quantitatively.
Forsooth, dimensional analysis suffices to get the right answer 
within a factor of 2 or so.  Does anybody really think general
relativity is going to give a /less/ accurate answer?

As a teacher, getting the right answer is necessary, but it
is only the first step.  The next step is to figure out where 
the student is coming from, to find out what motivated him 
to ask such an off-the-wall question.  We have to find out 
who's feeding him misconceptions.

Implicit in the radiation "puzzle" is a gross misstatement 
of one of the laws of physics.  As is so often the case 
with students, we do not get a clear statement of the 
misconception;  we need to dig for it.

It's hard to be sure, because there are at least as many
ways of misstating the equivalence principle as there are
ways of misstating 1/2 m v^2.  So we have to consider a
few possibilities.  The leading hypothesis goes like this: 
"Einstein said free-fall is equivalent to no gravity at all."

Well ... it would be correct but pedagogically unhelpful 
to point out that Einstein never said any such thing.  It
would be better to clarify what he did say:
  A uniform straight-line acceleration of the reference
  frame is equivalent to a constant globally-uniform
  gravitational field.

That's clearly correct.  One can now begin to understand
where the student might be coming from:  A half-remembered 
snapshot of a generic elevator is not enough to distinguish
between the correct statement and innumerable incorrect
statements.  You need to actually /think/ about the
accelerating elevator.

And yes, to answer the obvious follow-up question, if you
had a charged source and an observer freely falling in 
the same globally-uniform field, the accelerating charge
would not radiate.  This should be obvious from the
structure of the Liénard-Wiechert potentials.  As with
so many other thingsk the Liénard-Wiechert potentials
are easier to understand if you draw the spacetime
diagram.
  http://www.av8n.com/physics/lienard-wiechert.htm
Relative to a uniformly accelerated reference frame,
the charge, the observer, /and the fields/ all fall
together.  The L-W integral depends on the relative
separation, so it doesn't care.

It must be emphasized that a uniform gravitational field
corresponds to *no curvature* of spacetime.

Again:  
  A uniform straight-line acceleration of the reference
  frame is equivalent to a constant globally-uniform
  gravitational field.

For a charged object in orbit around the earth, we do
not have anything resembling a straight-line acceleration,
nor do we have anything resembling a globally uniform
gravitational field.  As usual, if you state the laws
of physics properly, there is no paradox ... not even
a puzzle.

As a slightly different hypothesis, or perhaps a slightly
different way of expressing the previous hypothesis,
every first-year physics textbook I've ever seen abuses
the definition of "gravity" and "g".  There are two
significantly different things, both of which are
called "gravity" and denoted by "g".  I say that g
is frame-dependent.  I also say that in Newton's law
of universal gravitation,
   δg = G M / r^2               [1]
the RHS is frame-independent so the thing on the RHS
had better be frame-independent.  I say the RHS needs
to be written as δg (not plain g).  For details, see
  http://www.av8n.com/physics/weight.htm#sec-various-notions

Again t must be emphasized that a uniform gravitational 
field corresponds to *no curvature* of spacetime and
no δg.

You can use equation [1] and/or general relativity to
explain δg.  You will never understand the first thing
about GR -- or about gravity in general -- until you 
learn to distinguish g from δg.  One of them is frame-
dependent, while the other is not.

Does anybody have any better hypotheses as to what lies
behind this alleged paradox?

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

I hope nobody is fooled by the additional silliness 
of talking about "inside" versus "outside" the space
station.  According to the laws of physics, it is the
acceleration of the chosen /frame/ that matters.  The
acceleration of any particular object is irrelevant,
and being indoors or outdoors is even more irrelevant.

You can always find innumerable ways of misstating
the laws of physics ... but why would you want to?
Aren't there enough misconceptions running around 
already?