dlzc

Roughly speaking, there are no geodesics that extend from "inside" the
event horizon into the Universe at large.

If I selectively drop 10^30 electrons into a black hole, and somehow keep
an attendant number of protons from doing a similar dive, does the hole end
up with a net charge?

If so, where does the Universe perceive this charge to reside?

David A. Smith

starblade13

It resides in the electrons right before they fall into a black hole.

What we should be asking is what happens to the color charges if a
nucleon if it falls into a black hole, at least, the color charges at
the event horizon. The same could be asked of the weak charges of
neutrinos or antineutrinos.

(...Starblade Riven Darksquall...)

dlzc

Dear Starblade Darksquall:


news:<qy2jb.77970$gv5.57278@fed1read05>...


I have heard people discuss how objects appear to always be stuck at the
event horizon (like klingons), and they just get dimmer and dimmer. I
wonder what "dimmer" means to a charge?

David A. Smith

dubious

dlzc@aol.com \(formerly\):

Yes, but there is a limit to the charge a black hole can have which
depends upon its mass and total angular momentum.

I assume that it "resides" in the the same place the mass "resides" -
the singularity. I don't think the word "resides" makes a lot of sense
when referring to the singularity, though. Personally, my inclination
would be to give the singularity the bare charge on the electrons and
take the charge of the black hole outside the horizon to be the charge
as measured once one includes the vacuum polarization.

zed

Do charge potentials propagate ?
Why would they be confined by an event horizon ?

dlzc

Dear Bilge:

Thanks. I was just wondering if the E-field of a charge was written into
the Universe way back when, and the particle entered the hole, how its
virtual photon(s) could express outside (as in the Universe side of) the
event horizon. Mixed metaphors I guess...

Answers accepted as I'll understand your answers about the same time as we
actually get to visit the neighborhood of a black hole.

David A. Smith

clade

Hello,

You´re right: they *appear*. An electric charge does not become
"dimmer". You will end up with a black hole that carries the sum of
the charges of the electrons you have dropped in (because electric
charge is conserved). You can measure this charge by measuring the
black hole´s electric field.

regards,
Jürgen

dubious

Starblade Darksquall:

Nucleons are colorless, so an infalling nucleon has a net color
charge of zero. Ditto for mesons. The only way to give a black hole
a net color charge is with a quark or gluon. It's pretty hard to
come up with an example of how to do that.

This isn't as hard to address. The charge Q is related to the weak
hypercharge (current) through the relation:

Q = T_3 + Y/2

For a neutrino, T_3 = 1/2 and Y = -1, so T_3 + Y/2 is also 0.
In other words, the electric charge and weak charge are different
manifestations of the same thing. It might be more interesting to
ask if a black hole could beta decay. If all neutrinos are massive,
then I think it's not possible. If one of the neutrinos is
massless, then it might be possible.

dlzc

Dear Bilge:


Not sure what you mean by "beta decay". If no geodesics extend from inside
to the Universe side, I am assuming you don't mean an electron would be
ejected from inside the event horizon...

David A. Smith

dubious

dlzc@aol.com \(formerly\):

No. I was imagining something like the following. Hawking radiation
results involves photons from vacuum fluctuations at the horizon.
When a virtual photon pair is produced, if one of the pair is
absorbed by the black hole, and satisfies the "on mass shell[1]"
condition (i.e., becomes real and propagates inside the horizon),
then the second photon must go "on mass-shell" and propagate as
well. If it's trajectory doesn't take it inside the horizon,
then it escapes as hawking radiation. Now, instead of photons,
picture the fluctuation of a W+/W- pair, such that the W+ ends
up propagating inside the horizon and satisfies the on mass-shell
condition. The W- must then go on mass shell and propagate. Assuming
it decays before its trajectory takes it inside the horizon, the
decay looks like this:
e-, \mu- or \tau- /
W- ---->
\ - - -
\nu_e, \nu_mu, or \nu_tau


If one on the neutrinos is massless, I would assume it
could escape, just as a photon does. The lepton would
fall back into the hole. This is all pretty much a guess
and doing the actual calculation to see how probable
such a process might be, would be difficult (for me,
at least).