🚀 go-pugleaf

In article <1992Oct1.233347.16497@ichips.intel.com> bhoughto@sedona.intel.com (Blair P. Houghton) writes:
>In article <BvFCpL.Atx@well.sf.ca.us> sarfatti@well.sf.ca.us (Jack Sarfatti) writes:
>>Night Thoughts of a Quantum Mechanic in the style of Ludwig Wittgenstein. 1
>>
>>An electric charge is freely falling. A photon detector is freely falling.
>
>Okayyyy...
>
>>The freely falling detector will not detect real photons from the freely
>>falling charge.
>
>Sez who?

It's true.  The heuristic argument for why it should be true is just
the Principle of Equivalence.  "Being in free fall" is equivalent to
"being in an inertial reference frame", and if the charge and observer
are both in inertial reference frames, no radiation is seen.

But if that doesn't satisfy you, you can do the GR calculation rigorously.
Somebody wrote a nice paper on it a few years ago (Am. J. Phys., I think,
but I could be wrong).  He calculated the Poynting vector and hence
the energy flux from a charge in four cases: 

(i) Falling charge, falling observer,
(ii) Falling charge, supported observer,
(iii) Falling observer, supported charge,
(iv) Supported observer, supported charge.

(The background spacetime is Schwarzschild, far outside the horizon.
Here "falling" means "in free fall", and "supported" means at rest with respect
to the usual Schwarzschild coordinate frame.)

As one might expect, radiation is seen in cases (ii) and (iii) only.

When I first read this, I thought it made perfect sense from the classical
point of view, but I was bothered when I tried to look at it in the 
"photon picture".  After all, either the charge IS emitting photons
or it ISN'T, right?  Wrong.  I eventually learned enough physics to 
decide that everything was OK in the photon picture too, but if I
tried to explain why, I'd screw it up.  I'll leave that job to
someone else.

By the way, if you want the reference to the paper I described, send
me some mail.  I have the paper somewhere, and it shouldn't be too
hard to dig it up.

-Ted