In article ,
says...
The only force that comes out of the great density of a BH is great
force of gravity. If the magnetic attraction or repulsion can't escape
from a BH it is they are virtual photons,and they can't escape the
virtual horizon of a black hole. Bert


That makes sense.

Although surely electric forces must be able to come out of a black hole
as it can have a charge.



--
Mankind's future is in space.

Observations of Bernard - No 48

Bernardz:
In article ,
says...

Now if the black hole gurus are right, I should not be able to determine
the distribution of the charge inside a black hole. Why not?

Because the field propagates at the speed of light and no lightlike
or timelike trajectory inside the horizon will ever reach the horizon.


How then can I measure the total charge?

That's a more complex question. The simplest explanation is to
consider both the charge and the mass. We know we cannot locate
the mass, so that once it crosses the horizon, all we can measure
is a black hole with a larger mass, i.e., the mass appears to
be at the center of the hole. If the mass appears to be at the
center of the hole, then the charge must appear to be there too.

A slightly more complex answer is to consider the multipole expansion of
point charge. If we don't chose the point charge as the origin of our
coordinate system, it will, in general, have a dipole moment, quadrupole
moment, etc. Only the _lowest_ non-vanishing multipole moment is in-
dependent of the origin you choose.

(These multipole moments are, of course, artifacts of the coordinates,
but useful to make the point here.) For example, consider, the charge q
with a mass m below,

^y
| so that our origin is not where the charge is located.
| q The multipole expansion about the origin contains the
+---- the charge q plus higher order multipoles. In this case.
x it's a complicated way to write the field of a point
charge and say it isn't located at the origin.

Now, assume you are somewhere outside the horizon of the black hole, and
you take the center of your coordinate system to be the center of the
black hole. Now, even when the charge is outside the horizon, the monopole
term, which is q/r^2, is at the center of the hole. There's nothing
strange about this, because it's an artifact of choosing the origin of our
coordinate system in a strange way. Because we've chosen it in a strange
way, we also have all of the higher order multipole moments, but
regardless, we get the correct field once we add up all of the multipoles.

Now, let the charge cross the horizon. It has a mass m and we know
that we cannot locate the mass, so all of the position information
about the mass must vanish, so all of the multipole moments which
depend upon the location of the charge must vanish or else the charge
will appear to be someplace other than where the mass appears to
be. Since the only multipole moment which is independent of the origin,
is the monople moment, the charge itself, it must also appear to be
at the center.

Now consider the case of a dipole. The dipole moment for a pair
of charges, q and -q separated by a distance `a', is qa. However,
we can consider these as two separate charges. Once the charges
cross the horizon, they both must appear to be at the same exact
point, at the center of the black hole. Since all that's visible
outside is the sum of the two monopole momemts, the charge is zero,
the separation between them is zero and no dipole field is observable.


There's a slight catch here. For a pure dipole, the lowest non-vanishing
multipole moment _is_ the dipole moment, and therefore, the dipole moment
_is_ independent of origin. For an electric dipole, there is no problem,
because all electric dipoles are really systems of charges which then is
just the case described above. This isn't true for a magnetic dipole. The
most familiar magnetic dipoles are current loops, and since currents are
due to moving charges, we have some angular momentum. The angular momentum
can be transferred to the hole and a charged rotating black hole will have
a magnetic field. That won't hellp you locate the charges though. The
magnetic dipole moment of a current loop is equal to the current in the
loop x the area of the loop, independent of the shape of the loop (in a
plane - you can get an arbitrarily shaped loop by considering the
superposition of planar loops).

The electron differs somewhat. It has a magnetic moment, even when
treated as a point charge and it isn't rotating. The magnetic field comes
from the spin, which is not a mechanical rotation. I would think that this
requires the black hole to have a spin of 1/2 and a dipole moment, despite
not rotating, but I don't know precisely what (or if) there is any
consensus on how quantum mechanics figures in.

The above may not be very satisfying, but better arguments become
really complex. If you consider the field in the context of quantum
field theory, then the electric field will consist of the virtual
photons associated with the charge. Virtual photons can propagate
faster than light as they carry no information, and so there is no
problem in reaching the horizon.

Dear Bernardz:

"Bernardz" wrote in message
news:MPG.1aaaf80f61cde26d98994d@news...
In article ,
says...

Now if the black hole gurus are right, I should not be able to
determine
the distribution of the charge inside a black hole. Why not?

Because the field propagates at the speed of light and no lightlike
or timelike trajectory inside the horizon will ever reach the horizon.


How then can I measure the total charge?

It is written on the event horizon, remember? As if it were stuck there,
like a fly on flypaper.

David A. Smith

"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox wrote in
message news:Hl20c.4515$h23.389@fed1read06...
....
How then can I measure the total charge?

It is written on the event horizon, remember? As if it were stuck there,
like a fly on flypaper.

In light of Bilge's excellent response, I retract this.

David A. Smith

"Bill Vajk" wrote in message
...
Please reread the Baez html, I don't think you've understood it.


Which part do you think I've not understood, Bill, because I find
nothing remarkable in that link?

For instance,

"You will never quite see the fragments of grapefruit reach the horizon;
instead they will appear to slow down and get redshifted
to invisibility as they approach the horizon. The fragments themselves
however will pass the event horizon and hit the singularity in a finite
amount of proper time."

seem to says essentiall the same thing I tried to convey.

Black holes do *not* violate conservation laws so far as anyone
has been able to demonstrate. If you can detect magnetic force
coming from the magnet "forever," considering it doesn't even exist
any longer, you've discovered the equivalent of free energy. That's
not going to happen.

I agree that black holes do not violate conservation laws (at least not
at a classically level). However, the description of the magnetic effect
you would see (detect) would not be a violation of conservation law.

Besides, in order to "see" the virtual image
you are forced to introduce fresh new energy any time you
want to see it.

No "fresh" energy is required, and none was suggested. What is seen to
the distant observer (one at a greater radius than the remaining image
of the infalling magnet) is the light that was emitted or reflected from
the object in a direction away from the hole before the object arrived
at the horizon. Nothing more. No new energy required. However, this
image becomes fainter for being spread out in time (darkens and
redshifts).

The light is delayed because it has further to go. Shoot a flash light
off into space, at some great distance, would we be looking at it? or
would we understand we were looking at where it had been, since it took
time for the light to travel? It takes no additional energy to see where
it was, though. You just have to wait for that energy already "sent" to
arrive. In fact even if the flashlight batteries failed, you'd still see
the light arriving for a long time. No fresh energy required.

How long we can actually see the magnet and detect the magnetic effects
is more of a technical question. How faint a signal can we detect, and
how (low background noise level) dark the black hole background appears
to be (because of semiclassical thermodynamic issues like Hawking and
Gibbons radiation, etc.)?

--
Randy M. Dumse

Caution: Objects in mirror are more confused than they appear.

N:dlzc D:aol T:com \(dlzc\):

"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox wrote in
message news:Hl20c.4515$h23.389@fed1read06...
...
How then can I measure the total charge?

It is written on the event horizon, remember? As if it were stuck there,
like a fly on flypaper.

In light of Bilge's excellent response, I retract this.

I don't think it's necessary for you to retract that, actually.
Remember, that what I described was for an observer _outside_ the
black hole, in which case, all gauss' law says is that the field
is determined by any gaussian surface enclosing the charge.

Inside the black hole, my guess is that there might be something
counter-intuitive. Recall that for an observer _inside_ the black hole,
the radial direction, which is toward the singularity, is timelike, so
that it lies in the observer's future. It would seem that implies that the
observer should not measure an electric field originating at the
singularity, since that field would originate in the observer's future and
propagate toward the past. However, this is one of those things that would
be best answered by someone who has a better idea of what's happening
inside a black hole with respect to the sigularity and the interpretation
of the coordinates.

Randy M. Dumse wrote:

"Bill Vajk" wrote in message
...

Please reread the Baez html, I don't think you've understood it.

Which part do you think I've not understood, Bill, because I find
nothing remarkable in that link?

It is not remarkable, but it conflicts significantly with what
you wrote.

For instance,

"You will never quite see the fragments of grapefruit reach the horizon;
instead they will appear to slow down and get redshifted
to invisibility as they approach the horizon. The fragments themselves
however will pass the event horizon and hit the singularity in a finite
amount of proper time."

seem to says essentiall the same thing I tried to convey.

I'm glad you seem to agree with Baez. Just how small a
fragment of a magnet remains magnetic, espcially when
contemplating the high temperatures associated with its
being ripped apart?

Next, here's some of what this discussion has lost. You wrote:

Your detector won't see the magnet for where the magnet is, but for
where the magnet was at some time before crossing the horizon played
like a slowing movie out to time infinity (or you loose the signal in a
less theoretical and more "technically" manner of consideration).

Time infinity?????????????

which fed the next tidbit of mine:

Black holes do *not* violate conservation laws so far as anyone
has been able to demonstrate. If you can detect magnetic force
coming from the magnet "forever," considering it doesn't even exist
any longer, you've discovered the equivalent of free energy. That's
not going to happen.

I agree that black holes do not violate conservation laws (at least not
at a classically level). However, the description of the magnetic effect
you would see (detect) would not be a violation of conservation law.

Of course it would if you can detect a magnet forever that isn't
there any loner but was once there.

Besides, in order to "see" the virtual image
you are forced to introduce fresh new energy any time you
want to see it.

No "fresh" energy is required, and none was suggested.

I was talking about another snipped pertinent part of the
discussion which, for some reason, you've engaged in
goalpost moving. Hrere's what I was discussing:

You had written:
Just because the detector "sees" the magnet still approaching the
horizon doesn't mean the magnet is still where it is seen. This is true
in the same sense as seeing a movie of Einstein riding a bicycle in New
Jersey. Just because you now see it, doesn't mean if you went to New
Jersey you'd find Einstein on that bike. It only means he was once
there and it took a long time for the image to reach your eye by
the route they took. Einstein has long ago left New Jersey.

Of course viewing the movie reuires fresh energy, invalidating
your example which I labeled as follows: "The movie argument
separates men from boys where science is concerned. It was
designed for simple minds."

What is seen to
the distant observer (one at a greater radius than the remaining image
of the infalling magnet) is the light that was emitted or reflected from
the object in a direction away from the hole before the object arrived
at the horizon. Nothing more. No new energy required. However, this
image becomes fainter for being spread out in time (darkens and
redshifts).

Strawman argumentation.

How long we can actually see the magnet and detect the magnetic effects
is more of a technical question.

Wrong. It disintegrates and heats up sending no more of its original
photons, but rather begins to appear as xrays to the observer.

Are you Randy the programmer?

Dear Bilge:

"Bilge" wrote in message
...
N:dlzc D:aol T:com \(dlzc\):

"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox wrote in
message news:Hl20c.4515$h23.389@fed1read06...
...
How then can I measure the total charge?

It is written on the event horizon, remember? As if it were stuck
there,
like a fly on flypaper.

In light of Bilge's excellent response, I retract this.

I don't think it's necessary for you to retract that, actually.
Remember, that what I described was for an observer _outside_ the
black hole, in which case, all gauss' law says is that the field
is determined by any gaussian surface enclosing the charge.

But what I had been alluding to was that the charge would appear "stuck in
one spot" on the event horizon, and this is clearly wrong. One could
derive magnetically-induced energy from the BH, if the charge were "off
center". Since it is effectively at the center (as perceived by us
outsiders), any transmitted forces are directed at the center of mass.

Does a spinning Van deGraaf sphere (about an axis of symmetry) produce a
magnetic field? No.

Inside the black hole, my guess is that there might be something
counter-intuitive. Recall that for an observer _inside_ the black hole,
the radial direction, which is toward the singularity, is timelike, so
that it lies in the observer's future. It would seem that implies that
the
observer should not measure an electric field originating at the
singularity, since that field would originate in the observer's future
and
propagate toward the past. However, this is one of those things that
would
be best answered by someone who has a better idea of what's happening
inside a black hole with respect to the sigularity and the interpretation
of the coordinates.

I'm wondering if in fact we are not all "experts" already. If all matter
enters at the Big Bang (the horizon is time 0)... one wonders if a BH
cannot itself contain other BHs, and pocket philosophers and scientists to
wonder about the inside of a BH... ;}

David A. Smith

"Bill Vajk" wrote in message
...
I'm glad you seem to agree with Baez. Just how small a
fragment of a magnet remains magnetic, espcially when
contemplating the high temperatures associated with its
being ripped apart?

I don't accept "aprior" the magnet is being ripped apart. We haven't
established enough detail about its condition and motion to determine
that.

Next, here's some of what this discussion has lost. You wrote:

Your detector won't see the magnet for where the magnet is, but
for where the magnet was at some time before crossing the
horizon played like a slowing movie out to time infinity (or you
loose the signal in a less theoretical and more "technically"
manner of consideration).
Time infinity?????????????

which fed the next tidbit of mine:

Black holes do *not* violate conservation laws so far as
anyone has been able to demonstrate. If you can detect
magnetic force coming from the magnet "forever," considering it
doesn't even exist any longer, you've discovered the equivalent
of free energy. That's not going to happen.

I agree that black holes do not violate conservation laws (at least
not at a classically level). However, the description of the
magnetic effect you would see (detect) would not be a violation
of conservation law.

Of course it would if you can detect a magnet forever that isn't
there any loner but was once there.

Okay, I left your argument in tact there, so you don't have to repeat
it. I believe I may understand your premise: If there is infinity time,
there is infinite energy. But! I disagree.

Let's say I'm going to shine a watts worth of energy on you. In the
first day, I shine 1/2 a watt. On the second day, I shine 1/4 watt. On
the fourth day I shine 1/8th watt. Any limit to how long I can keep this
up in the classical limit?

Okay, at the quantum limit, when I get to my last billion or so photons,
I start hording them. I send one photon out each year the first year, I
send another photon out two years later. Then I send out another photon
out four years later. And so on.

And likewise, I cut the frequency of each remaining photon in half,
reducing the energy escaping even further. With each escaping photon, I
have enough energy left to let two more of the same go, but I again
split those down.

Mathematically, "the decreasing rate of transmission function does not
tend toward any limit of a fixed value on the time axis."

Infinite time does not mean infinite energy.

But do feel free to elaborate if I'm still not understanding what you're
meaning.

Besides, in order to "see" the virtual image
you are forced to introduce fresh new energy any time you
want to see it.

No "fresh" energy is required, and none was suggested.

I was talking about another snipped pertinent part of the
discussion which, for some reason, you've engaged in
goalpost moving.

Not intentionally engaging in goalpost moving. More likely, perhaps, we
are just talking past each other. In any case...

Hrere's what I was discussing:

You had written:
Just because the detector "sees" the magnet still approaching the
horizon doesn't mean the magnet is still where it is seen. This is
true in the same sense as seeing a movie of Einstein riding a
bicycle in New Jersey. Just because you now see it, doesn't
mean if you went to New Jersey you'd find Einstein on that bike.
It only means he was once there and it took a long time for the
image to reach your eye by the route they took. Einstein has long
ago left New Jersey.

Of course viewing the movie reuires fresh energy, invalidating
your example which I labeled as follows: "The movie argument
separates men from boys where science is concerned. It was
designed for simple minds."

Well, in every analogy or metaphor, there is the part that applies and
is hopefully most obvious to help understanding attempted to be
communicated, and there is the part that doesn't apply, because it is an
analogy.

The purpose in this movie example I offered was, what is seen later, is
not representative of what is happening there now. The image of Einstein
on a bike in NJ was created long ago. The image you see of Einstein on a
bike in NJ is currently arriving at your eye. The part of the analogy
which applies is you are seeing something that was delayed in
transmission and no longer exists in reality. The part that doesn't
apply is you had to shine new light through the film to see it. If
doesn't invalidate the example, in my opinion, it is just the
highlighting of the part that doesn't apply.

When we use apple to explain red, we hope the brown stem, and green
leaf, and the blackish bud residual be ignored. The abundance of red is
the message of the metaphor. The stem, etc., is the part that is not
intended in the metaphor. So you then use a tomato and a strawberry to
further communicate the idea of red. Stop there, and you've taught "red
fruit" where you only meant to imply red. Again, there is a part that
applies and a part that doesn't. So you might need a red fire truck...
since it doesn't have a stem... even though it has tires, etc, which may
distract from the central message of "red".

So I disagree the movie analogy was meant for simple minds. One could
easily assume that, if they focused on the part which wasn't meant to be
part of the metaphor, and ignored the part that was.

What is seen to
the distant observer (one at a greater radius than the remaining
image of the infalling magnet) is the light that was emitted or
reflected from the object in a direction away from the hole
before the object arrived at the horizon. Nothing more. No new
energy required. However, this image becomes fainter for being
spread out in time (darkens and redshifts).

Strawman argumentation.

Hummm... well strawman to me said by you here implies I've made a
"strawman" fabrication or the argument so I can then shoot down the
strawman. So I'm not able to tell much about what you're objecting to.
Maybe I got part of it above. Maybe not. Please be clearer for me, if
possible, so we can have a more useful discussion if you care to.

It disintegrates and heats up sending no more of its original
photons, but rather begins to appear as xrays to the observer.

Why do you think the magnet disintegrates and heats up and appears as
x-rays? What mechanism are you thinking does that? And what is the
situation of the magnet? freefall? constant r? orbit?

If it were orbiting a spinning black hole with lots of infalling gases
coming in from random trajectories, I could see it would be heated by
the gas collisions until it emitted x-rays, but in an, isolated settled
black hole without angular momentum or charge (Schwarzschild solution)
in free falling or orbit, I don't see where the heating would come from.
If it were accelerated, such as held at a constant r, such as jb's
grapefruit, yes, it would get ripped apart. (Grapefruits aren't all that
structurally strong, after all.) I don't think anyone far from the
horizon would see it emitting x-rays though.

Are you Randy the programmer?

I program... among many other things. [humor on]You know, there are so
many many people named Dumse out there, just like Vajk, it's such a
common name, you probably are thinking of someone else. [humor off]

--
Randy M. Dumse
www.newmicros.com
Caution: Objects in mirror are more confused than they appear.

In article V480c.5534$h23.2351@fed1read06, "N:dlzc D:aol T:com \(dlzc
\)" N: dlzc1 D:cox says...
one wonders if a BH
cannot itself contain other BHs,


No reason why not.





--
The bouncer causes more fights then anyone at the bar. Yet the roughest
bars are those without a bouncer.

Observations of Bernard - No 49