[net.astro] Equivalence Principle & Electric Charge

sharp@noao.UUCP (Nigel Sharp) (12/11/84)

The following question was recently posed:

An accelerated charge radiates electromagnetic waves.  A charge placed at
rest in the laboratory does not.  But, the equivalence principle states
that a gravitational field is locally equivalent to an acceleration.  Why
then does the stationary charge not radiate ?

And now, a two-part answer:

Everything is relative, including acceleration.  Thus, if the charge is
accelerated with respect to the measuring apparatus, the apparatus will
detect emitted EM radiation.  This is true whether it is the charge that is
being accelerated, or the apparatus.  Conversely, if both are being accelerated
together, there is no relative acceleration, and no detected radiation.
Thus, Doug Gwyn's earlier answer is correct:  apparatus in free fall detects
radiation from a charge supported in a gravitational field (and conversely,
apparatus supported in a gravitational field detects radiation from a freely-
falling charge).

[I have voluminous notes on this topic, which, I'm happy to say, was solved by
someone else (S.Teukolsky, if you want to look it up).  It does lead to an
interesting conclusion, which is that the definition of a particle is frame
dependent - i.e. the photons detected by an accelerated detector are not
there for a non-accelerated detector.  More details if people request them.]

The second part is something of an aside.  Electromagnetism actually violates
the equivalence principle, simply because it cannot be localised.  EM is a
long range, inverse square force.  It follows that charge distributions must
be sensitive to non-local structure, and particularly to the curvature of
spacetime.  This leads to some interesting effects in the consideration of
radiation reaction and problems involving gravity and acceleration.

Ah, isn't it amazing when it turns out that someone clever has actually solved
the problems ?  I was quite happy - I was having trouble with it myself,
although I was pleased to see that I at least understood the solution.
-- 
	Nigel Sharp   [noao!sharp  National Optical Astronomy Observatories]