Re: Acceleration of charge

sal:
On Thu, 09 Feb 2006 08:26:42 +0000, Bilge wrote:

sal:
>On Wed, 08 Feb 2006 12:39:10 -0800, shiv wrote:

>Finally, if the charge on the rocket radiates, then the energy
>must come from the rocket engine -- the rocket must accelerate
>more slowly that it would if it had no loose charges on it. So,
>if I run a Van de Graaf generator in a rocket (using an onboard
>power supply, so the net mass/energy of the rocket isn't affected)
>does the rocket's acceleration decrease as the generator separates
>positive and negative charges ... since the separated charges
>must be radiating away energy?
>
>One point: You can't transform away EM radiation by a frame shift.
>So, if one observer sees it, everybody sees it.

Surprise - the answer is not quite that straight
forward. Accelerated oberserver will see themselves surrounded by a
heat bath which is proportional to their acceleration, so the while
the charge in the frame of the accelerated oserver radiates, that
radiation will be in equilibrium with the thermal bath. For an
electron sitting on the earth's surface, the temperture of the
thermal at is,


T = \har g / 2 \pi ck = 3.9 x 10^-20 K

where k is boltzmann's constant. Look up unruh radiation for more
details.

OK ...

There exists evidence for unruh radiation for accelerated
circular motion from synchrotron radiation, which is polarized but
the electron polarization remains less than 100%. In the lab frame,
the ``depolarization'' is interpreted as scattering of the electrons
from the photons in the thermal bath seen by the accelerated
electron.

Would it be fair to say that Unruh radiation is a direct manifestation
of the "zero-point energy"?

It would be fair to say that the thermal background due to unruh
radiation in an accelerated frame is the zero-point energy in that
frame.

Speaking of Hawking radiation, I haven't (yet) read a mathematical
treatment of it, and the plain English treatments leave me feeling
extremely puzzled. The standard "English" explanation of Hawking

That is because the explanations fall into two basic categories:
(1) Overly simplistic, (2) Full blown quantum field theory.

I believe I have an alternative which is conceptually simple without
being really involved, in terms of feynman diagrams, and vacuum
fluctuations, but before I post it, I need to work out some of the details
so I don't try to convince you of something that sounds good but is
totally wrong. I sketched it out below, but take it with a grain of
salt until I get a chance to work out the details and check it.

radiation is that electron/positron pair production proceeds apace
just outside the horizon, and some of the time the black hole swallows
one member of the pair (the one with "negative energy"... or says
Hawking in "Brief History of Time"). So far so good. But here's the
sticking point: The black hole's mass can actually be radiated away in
finite (Schwarzschild-coordinate) time by Hawking radiation. The
actual mechanism is stated as being the addition of "negative mass" to
the black hole by the "negative mass" particles that fall into
it. _BUT_ it takes infinite (Schwarzschild-coordinate) time for
anything _outside_ the event horizon to cross the horizon -- so how
can an outside observer see a black hole evaporate due to Hawking
radiation? The "swallowed" member of each pair should fall forever
toward the horizon, and never cross it. (My speculation on the matter
is that the "swallowed" member tunnels past the horizon and hence
never "really" crosses it, but I'm not sure that's even legit -- it
would involve traversing a distance faster than a photon.)

As a hint of the picture I mentioned above, consider the feynman
diagram for a vacuum fluctuation,


. ->. Which you can interpret a couple of ways. One,
. .B at point A, a virtual e+/e- pair anihilates into
A.~~~~~. the vacuum, producing a virtual photon which pair
. . produces the e+/e- pair at B. Two, an electron
<- propagates forward in time from point A to point B,
interacts with a virtual photon at point B, then scatters backward
in time and destroys itself at point A. None of the particles are
on mass shell, so none can propagate. You could equivalently
write,

. ->.
. .B
~~~~A. .~~~~
. .
<-

allowing the virtual lines to terminate on real particles by crossing
with the virtual lines in the loops of the real particle, however,
such diagrams never contribute to physical effects and so are not
included. However, the situation is different at the horizon. Since
the photon lines are both virtual (and hence of negative energy),
it's possible that such a loop allows the black hole to put one of
the virtual photons on mass shell absoring the other. Don't take
this too seriously until I figure out whether or not I can get the
right thermal distribution this way.

.

Relevant Pages

  • Radiation reaction for a uniformly accelerated charge
    ... "Radiation from a uniformly accelerated charge ... can distinguish a charge in uniform acceleration ... radiates energy ... Boulware, "Radiation from a uniformly accelerated charge", ...
    (sci.physics.research)
  • Re: Q. about max acceleration, Plancks constant, PLUS: help locate Sci Am column (Amtr. Sci)
    ... just that the extra energy will be wasted and have no additional ... velocity increases to increase the velocity of the increasing mass. ... I doubt that such extreme acceleration as he would have been ... Beyond certain frequencies, radiation is ...
    (sci.physics)
  • Re: Acceleration of charge
    ... >Finally, if the charge on the rocket radiates, then the energy must come ... >would if it had no loose charges on it. ... You can't transform away EM radiation by a frame shift. ... proportional to their acceleration, so the while the charge in the frame ...
    (sci.physics.relativity)
  • Re: Acceleration of charge
    ... To an observer in the rocket ... Acceleration of *macroscopic* charge doesn't cause radiation, it only brings into phase to some degree the radiation that was already occurring wrt our frame of reference. ...
    (sci.physics.relativity)
  • Re: Acceleration of charge
    ... >Finally, if the charge on the rocket radiates, then the energy ... >more slowly that it would if it had no loose charges on it. ... >does the rocket's acceleration decrease as the generator separates ... You can't transform away EM radiation by a frame shift. ...
    (sci.physics.relativity)