Why do charged particles behave strangely near black holes?

"BO" == Barry OGrady writes:

BO Its all conjecture anyway since black holes by their very nature
BO can't be detected, So they remain a crazy and unprovable theory.

What's Sgr A* then?

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Neil:

"Bilge" wrote in message
e-al.net...
Neil:
"Bilge" wrote in message

Well really, I do, if the reference to the article was important
to
your point.

Well, it isn't. What I mean is, my point stands on its own as an issue
about how charged particles behave in free fall in tidal fields.

That would be true if I knew what was your point [see below].

My point was, what is the self-force on a charge falling through a tidal
gravitational field, and what are the implications of that force, in
general. Well?

I thought I made that clear. (1) There is no tidal force on a charged
particle. A charged particle is a point, (2) The bottom line is the
charge doesn't radiate.


That is what I'm talking about, too. If you are talking about some
other effect, then I don't know what it might be. That is why I wanted
to see the article.

OK, I realize that we have to be careful about "radiation" because of the
comingling of what would happen with no charge present (intrinsic to space

There's no ``comingling'' here to consinder. That was the point.

[...]
It really doesn't make any difference. For example, the exterior
schwarzchild metric for a black hole applies to the earth for a radius
(r R_earth). The analysis is the same.

"The analysis", but not the relative degrees of involvement.

Yes, it is. Plug in the coordinate acceleration into the
larmour formula.

In any case,
why not try your hand at the direct points I made, rather than just going
around tangentially so much. Look at my argument about self-force, and
forget the NS material until you can look at it more.

I did. You just weren't reading what I wrote. If you want a different
answer, calculate it yourself. It's not that hard.

"Bilge" wrote in message
...
Neil:

"Bilge" wrote in message
e-al.net...
Neil:
"Bilge" wrote in message

Well really, I do, if the reference to the article was
important
to
your point.

Well, it isn't. What I mean is, my point stands on its own as an
issue
about how charged particles behave in free fall in tidal fields.

That would be true if I knew what was your point [see below].

My point was, what is the self-force on a charge falling through a
tidal
gravitational field, and what are the implications of that force, in
general. Well?

I thought I made that clear. (1) There is no tidal force on a charged
particle. A charged particle is a point, (2) The bottom line is the
charge doesn't radiate.

As I explained, it *has* to radiate if it undergoes oscillatory motion!
Don't confuse that with the constant acceleration case. Think of the
airless tunnel through the earth. The electromagnetic field changes with
time near any point on the earth, and that has to propagate and carry off
energy (consider that the Poynting vector is very close to any other
radiant source's) Are you saying that in the case of the tunnel, the
earth's gravity is strong enough to *nullify* the field - it can't. Yes,
ithe radiation would be weak, but the oscillations must be measured at a
distance, and that entails energy (Just consider: start the oscillation at
a certain time, and the EM wiggles must continue to move throughout space
like any other EM wave. You have made the mistake of thinking in terms of
certain parameters that seem intuitively critical to you, but ignoring
other conditions that *must* be satisfied. Our experience in history of
physics is that nature does what it must to preserve conservation laws,
even if that mucks up other nice expectations.

That is what I'm talking about, too. If you are talking about
some
other effect, then I don't know what it might be. That is why I
wanted
to see the article.

OK, I realize that we have to be careful about "radiation" because of
the
comingling of what would happen with no charge present (intrinsic to
space

There's no ``comingling'' here to consinder. That was the point.

[...]
It really doesn't make any difference. For example, the exterior
schwarzchild metric for a black hole applies to the earth for a
radius
(r R_earth). The analysis is the same.

"The analysis", but not the relative degrees of involvement.

Yes, it is. Plug in the coordinate acceleration into the
larmour formula.

But if you use "coordinate acceleration" that supports my own contention
that the charge in the diameter tunnel radiates and experiences the
Lorentz-Abraham self-force! (BTW do you really appreciate what that
means?) : even though in free fall, it's *coordinate* acceleration changes
as I described, although of course an inertial accelerometer registers
zero acceleration at all times (regardless of tidal field per se.)

In any case,
why not try your hand at the direct points I made, rather than just
going
around tangentially so much. Look at my argument about self-force, and
forget the NS material until you can look at it more.

I did. You just weren't reading what I wrote. If you want a different
answer, calculate it yourself. It's not that hard.

I did read what you wrote, and noted that it was tangential to the crucial
issues. I did calculate the relevant implications of the changing
coordinate acceleration that you seem to accept some places, but you still
don't get what that entails: radiation and self-force pertaining to
charges falling in tidal fields. Your referencing the heat bath etc. could
be relevant, but you fail to show how that undermines my point based on
other criteria, what degree of correction it would have on the basic
classical formulas based on v dotdot, etc.

Neil:
"Bilge" wrote in message

I thought I made that clear. (1) There is no tidal force on a charged
particle. A charged particle is a point, (2) The bottom line is the
charge doesn't radiate.

As I explained, it *has* to radiate if it undergoes oscillatory motion!

You said ``freely falling''.

Don't confuse that with the constant acceleration case. Think of the
airless tunnel through the earth. The electromagnetic field changes with
time near any point on the earth, and that has to propagate and carry off
energy

Huh? First, the fields which ``propagate'' are the ones which escape
to infinity, i.e., the radiation fields. Second, I clearly stated the
context of my response, which was for r R_earth, in order to have a
simple example to go with the physically intuitive argument. The ``tunnel
through the earth'' is a different question and doesn't negate my response
as I gave it. Although I think the same physical argument is applicable,
the mathematical problem is harder.

(consider that the Poynting vector is very close to any other
radiant source's)

You're mixing different things. Radiation and static fields are
different things. If you want to include the static fields, the
expression which defines the poynting vector has an additional term,
div S + du/dt = -J.E.

Are you saying that in the case of the tunnel, the earth's gravity is
strong enough to *nullify* the field - it can't.

Who said anything about nullifying anything? When the charge is moving
inertially, i.e., ``at rest'', it still has an electric field, which
one may attribute to virtual photons, not propagating photons. An observer
in free fall with the charge will still interact with that field, but
won't see the photons as propagating, i.e., the observer sees the same
coulomb field described by the coulomb potential.

Yes, ithe radiation would be weak, but the oscillations must be measured
at a distance, and that entails energy (Just consider: start the oscillation
at a certain time, and the EM wiggles must continue to move throughout space
like any other EM wave. You have made the mistake of thinking in terms of
certain parameters that seem intuitively critical to you, but ignoring
other conditions that *must* be satisfied.

No, I haven't. Those ``EM wiggles'' to which you refer are _not_
radiation fields. Those are the quasi-static fields, i.e., the near
field which is just a time dependent static field. Radiation fields
exist independent of the charge from which they originate.

[...]
Yes, it is. Plug in the coordinate acceleration into the
larmour formula.

But if you use "coordinate acceleration" that supports my own contention
that the charge in the diameter tunnel radiates and experiences the
Lorentz-Abraham self-force!

OK.

(BTW do you really appreciate what that means?):

Forget the gravitational field for a moment and consider an electron
in free space. It has a self-energy, which I can depict in the form
of a feynman diagram:

. .
. B .
---+-+-+--+--- etc.,
A . .
. .


So, I can interpret that as the continual exchange of virtual photons
with itself. Since energy and momentum have to be conserved at the
verticies, the electron has to be off mass shell and the photon has to
have a small mass. This is perfectly consistent with the idea that
energy is stored in the fields and the fields carry momentum. If the
average momentum is zero, then we are in the ``rest frame'' of the
charge. Note that if the virtual photon at point B is exchanged with
another charge rather than reabsorbed, the resulting diagram is the
first order correction to magnetic moment.

If the photons are radiated rather than reabsorbed, the result would be
the type of ``runaway acceleration'' of the abraham-lorentz force, in
which case the electron would be radiating itself away into oblivion.

You could reason why that doesn't happen in a couple of ways. First,
if the photon is massless, it can't happen without violating conservation
of energy and momentum. If it had a mass the result wouldn't be a real
mystery as that sort of process happens all the time. The Delta+ radiates
a pi_0 and comes a proton.

In any case, I can include the vacuum to improve on the picture.
The vacuum allows for the spontaneous creation and anihilation of
particle anti-particle pairs, for example,

/~\ Represents a virtual photon created from the anihilation of
/ ~ \ the e+/e- pair it produced. One could instead imagine the
| ~ | If you were to include the diagram above, you could inter-
\ ~ / change any of the photon lines with no effect, so this sort
\~/ of diagram is disregarded and doesn't contribute to any
~ interaction.

I'm going to go out on limb here, since I have no experience doing
quantum mechanics in curved spacetime, but if I take the gravitational
field to be modification to the vaccum, then I could easily picture
the different perspectives beween inequivalent observers of a charge
in a gravitational field to be a disagreement over which photons are
virtual. The vacuum the electron sees is locally the flat quantum
vacuum. The vacuum an observer in some other frame is also the
flat quantum vacuum _local_ to that observer. That doesn't mean the
two electron and the observer agree on the vacuum as the same vacuum
in the global sense.

even though in free fall, it's *coordinate* acceleration changes
as I described, although of course an inertial accelerometer registers
zero acceleration at all times (regardless of tidal field per se.)

Actually, it's the observer who is sitting a fixed point in the
gravitational field who is accelerated. It's when that observer
treats himself as inertial and the particle as accelerated that
causes the difficulty in interpretation.

I did. You just weren't reading what I wrote. If you want a different
answer, calculate it yourself. It's not that hard.

I did read what you wrote, and noted that it was tangential to the crucial
issues.

Since I think what I wrote is the entire issue, and you've given
me no reference to anything which would explain that you are talking
about some completely different phenomena rather than the one it appears,
you are free to pursue whatever that happens to be. I think you just
aren't getting the point, since I've used what you've told me as the
basis for my responses. If there is something you've left out, then
that isn't my problem.

I did calculate the relevant implications of the changing
coordinate acceleration that you seem to accept some places, but you still
don't get what that entails: radiation and self-force pertaining to
charges falling in tidal fields. Your referencing the heat bath etc. could
be relevant, but you fail to show how that undermines my point based on
other criteria, what degree of correction it would have on the basic
classical formulas based on v dotdot, etc.

As far as I can tell, your point doesn't involve anything other than
what is contained in my response. From your response above, it appears
you are confusing the radiation and quasi-static fields as being the
same thing.