Re: Acceleration of charge

On Wed, 22 Feb 2006 08:31:54 +0000, Bilge wrote:

vanep@xxxxxxx:

>> But this suggests that the virtual pair, localized as it is in
>> its own free-falling coordinate system, should also be either
>> outside or inside ... not one particle out, one particle in.
>
>That is similar to what Thorne was saying.


Diagramatically, then you are picturing a closed loop that looks
like,

e- ^ r
-->-+ |
| | +--> t
-->-
e+

But, recall, that in a feynman diagram, positrons are depicted as
negative energy electrons propagating backwards in time, so you can
draw the same loop as,

-->-
| |
-<--

(In a feynman diagram, _all_ lines are particle lines by convention.
The difference between a particle and anti-particle in a feynman
diagram is the direction in time the arrow points, and the time axis
is generally the horizontal axis, which is backwards from the usual
spacetime diagram. So, for example, the following diagram
represents electron compton scattering:



~ ~
~ ~
+-->--+
/ \
/ \


because the arrow on the particle line line points in the positive t
direction. (actually, there is a second in which the photon lines
are crossed). If the arrow pointed the other way, it would represent
compton scattering by a positron. If you rotate it 90 degrees so
that the arrow points along the spatial direction and the solid
lines are on the left, you have pair anihilation - the incoming
electron emits a pair of photons and scatters backwards in
time. Rotated the other way, you have pair creation.

So, applying that to pair creation at the horizon, you would picture
the process as an electron falling into the hole, emitting a virtual
photon, then scattering backward in time and destroying _itself_
outside the horizon by emitting a photon. Since a particle which
propagates forward in time falls into the black hole, it will cross
the horizon the other way by propagating backward in time as an
anti-particle. Try to figure out the plot from the compton
scattering diagram and the closed loop will make more sense. It's
completely equivalent to having the virtual pair fall across the
horizon.

(It's sometimes said that the only dumb question is the one you didn't
ask. Let's test that a bit.)

How literally can I take the notion of the electron propagating
forward in time across the horizon, and then propagating back across
horizon, going backward in time?

If that's literally correct then it seems to lead to a problem. From
the point of view of an outside observer, it takes infinite time for
something to cross the horizon going in. If the electron does that,
and then re-crosses going "the other way", it would seem that the
visible radiation on the outside would be paired with "negative
radiation" on the inside _but_ the "negative" radiation should happen
an infinite time _later_ than the external "positive" radiation.

The consequence would be that, while black holes could radiate, they
could never (observably) evaporate as a result, because the
evaporation is due to the hole swallowing negative energy -- and that
won't happen until the end of time.


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