VELOCITY OF GRAVITY
allen c goodrich
A GRAND UNIFIED THEORY
Newton's Laws of Motion work very well, and certainly
there is a Galileo acceleration effect of gravity,but,
we must understand why these effects exist if we,
like Newton and Galileo, want to discover new worlds
of knowledge and progress.
WHY a grand unified theory of the universe?
Einstein's General Relativity equation does not
include relative mass-energy (kinetic energy). Gravity,
relative volumetric acceleration, is a function of relative
mass-energy density! THE LAW OF MOTION,
THE BERNOULLI PRINCIPLE AND TORRICELLI
THEOREM can all be derived directly from the
FUNDAMENTAL EQUATION OF THE UNIVERSE.
Current physics theories do not properly
explain the ocean tides, the photon (particle or wave),
gravity, time, mass, or the electromagnetic field.
The low tide, not the high tide, is observed in the deep
ocean directly under the full moon.( U.S. Coast and
Geodetic Survey Tables ). This observed fact, contradicts
physics texts, the dictionary and encyclopedia definition
of tide, which shows a picture of the earth with a bulge of
water on the side facing the full moon, and states that the
high tide tends to occur directly under the full moon.
This is an error! The tidal wave would be 12000 miles long.
It is not conceivable that the moon could pull several feet
of ocean water around the earth at better than 1000 mph to
generate this wave.This would wash away the continents
and humanity in a day.
The copyrighted theory 1988 A.C.Goodrich; explains the
tides as a decrease of kinetic energy and volume of the
ocean water with the increase of potential energy as the
moon direction changes and distance decreases relative
to a particular side of the earth's ocean, to maintain a
constant total energy of the effective universe.
THE FUNDAMENTAL EQUATION AND PRINCIPLE
of the universe is one of constant total energy expressed
by the (modified Galileo pendulum-Kepler-Newton-
equation by Goodrich) equation:
L ^3 / T ^2 = / K(M-m) ,relative volumetric acceleration,
GRAVITY, is a function of M-m, where M is the total
energy of the universe,m is the mass-energy in question
and T and L are time and distance.
This equation is derived from the
FUNDAMENTAL EQUATION AND PRINCIPLE of the
universe (by Goodrich)
mL ^2 / T ^2 + K(M-m)m/L = a constant M.
The total of kinetic and potential energy of the universe is a
constant M. The gravitational field is the potential energy
difference. Kinetic and potential energies are strictly
relative and directional..
This grand unified theory defines time, mass,
energy, gravity,the photon, other forces, and the
electromagnetic field as geometric properties of the
universe.
If the kinetic energy of a mass m, relative to the rest
of the effective universe, is equal to the negative potential
energy of m, relative to the rest of the effective universe,
any positive change of kinetic energy must equal a
negative change of potential energy. All of the planets
around the sun and the moon around the earth, in orbital
motion, were found to obey this fundamental law:
m(2pi L)^2/T^2 = -G (M-m) m/L ,where m is the planet ,
(M) is the effective universe, G is the gravitational
constant 6.67 x10^-8 dynes cms^2/gms^2, L is the radial
distance in centimeters and T is the orbital time in
seconds.^ means to the power of. The planetary orbits
are stable because it takes an energy change to change
the orbit.The FUNDAMENTAL EQUATION OF THE
UNIVERSE controls.Mass-energy, kinetic
energy, is created as the universe expands and
relative potential energy decreases, to maintain a
constant total energy of the universe.
See Library of Congress Card Catalog No.94-90554
THE UNIVERSE- A UNIFIED THEORY-GOODRICH
and ISBN 0-9644267.( Hard cover 396 pg. $55 +tax.& shp.
A summary 10 pg.for $13.00 ) from
NEW ALLEN GOODRICH ENT. INC.,1620 GROVER RD.
EAST AURORA, N.Y. 14052-9721
###
Uncle Al
[nothing useful]
Gravity propagates at lightspeed. Go away.
--
Uncle Al Schwartz
http://www.mazepath.com/uncleal/
http://www.ultra.net.au/~wisby/uncleal/
http://www.guyy.demon.co.uk/uncleal/
http://uncleal.within.net/
(Toxic URLs! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Jeremy Boden
<uncl...@hate.spam.net> writes
>allen c goodrich wrote:
>
>[nothing useful]
>
>Gravity propagates at lightspeed. Go away.
>
--
Jeremy Boden mailto:jer...@jboden.demon.co.uk
Joe Fischer
: I know - but - How does gravity escape from a black hole?
Gravity does not propagate.
Joe Fischer
Chris Hillman
On Mon, 28 Jun 1999, Jeremy Boden wrote:
> In article <3777B2C9...@hate.spam.net>, Uncle Al
> <uncl...@hate.spam.net> writes
> >allen c goodrich wrote:
> >
> >[nothing useful]
> >
> >Gravity propagates at lightspeed.
http://www.corepower.com/~relfaq/grav_speed.html
http://www.math.washington.edu/~hillman/PUB/Speed
> I know - but - How does gravity escape from a black hole?
http://www.corepower.com/~relfaq/black_gravity.html
http://www.math.washington.edu/~hillman/PUB/Escape
Chris Hillman
Home Page: http://www.math.washington.edu/~hillman/personal.html
Tom Roberts
> How does gravity escape from a black hole?
The word "escape" implies movement. The gravitational fields do not
move in any usual sense of the word. As I said before, _changes_ to
the field do indeed move with velocity c, but that is a different
aspect of gravitation.
In short, gravity has no need to "escape" from a black hole. The initial
conditions which "created" the black hole are required by the field
equations of GR to generate fields outside the horizon which are
consistent with the initial mass/energy content of the black hole. And
these external fields are "frozen" and continue to mimic the
mass/energy content of the hole even as additional mass/energy enters
it.
Tom Roberts tjro...@lucent.com
Paul Alan Cardinale
Doesn't the fact that the external fields are "frozen" go hand in hand
with that fact that from outside the black hole, additional mass/energy
is never observed to enter it, but approach it asymtotically?
Paul Cardinale
Jeremy Boden
<tjro...@lucent.com> writes
>Jeremy Boden wrote:
>> How does gravity escape from a black hole?
>
>The word "escape" implies movement. The gravitational fields do not
>move in any usual sense of the word. As I said before, _changes_ to
>the field do indeed move with velocity c, but that is a different
>aspect of gravitation.
>
>In short, gravity has no need to "escape" from a black hole. The initial
>conditions which "created" the black hole are required by the field
>equations of GR to generate fields outside the horizon which are
>consistent with the initial mass/energy content of the black hole. And
>these external fields are "frozen" and continue to mimic the
>mass/energy content of the hole even as additional mass/energy enters
>it.
>
That sounds like a good answer.
I should of course have phrased my original question in terms of the
apparent parallel between light waves (photons) and the yet to be
experimentally verified gravity waves (gravitons(?)).
Tom Roberts
> [about fields outside a black hole]
> Doesn't the fact that the external fields are "frozen" go hand in hand
> with that fact that from outside the black hole, additional mass/energy
> is never observed to enter it, but approach it asymtotically?
Yes. "Go hand in hand" is a good phrase, as neither "causes" the other, in
_either_ the naive "causality" advocated by some people around here _or_
by the causal structure of GR itself.
Note, however, that a hovering observer just outside the horizon can be
engulfed by it if sufficient additional mass/energy enters the black hole.
As more mass/energy enters, the horizon expands outward, and as the "mass"
of the black hole increases it also requires a hovering observer to
expend more thrust to remain hovering; the horizon can expand outward
past the observer, in which case no finite thrust can keep her hovering
and she will intersect the singularity inside the black hole (Schw.
case). This luckless observer can observe additional mass/energy
entering the horizon (for a while (:-)).
Tom Roberts tjro...@lucent.com
Steven B. Harris
writes:
>
>allen c goodrich wrote:
>
>[nothing useful]
>
Useful information ("information" here meaning arbitrariness and
unpredictability-- which means being able to encode any message you
like), if carried by gravity, propagates at light speed. In other
words, gravity WAVES propagate at light speed. But "gravity"
colloquially refers to the near field as well as the far field.
Near-field gravity (virtual gravitons, if there was such a thing), such
as the "static" force emanating from a mass moving past you, or a mass
about which you orbit in a nice free fall, like a planet about a sun,
doesn't "propagate" in the usual sense, but such a field certainly
changes in direction and magnitude with time, *as seen by you.* And
that change, so long as no accelerations of the source as seen by the
observer, are involved, is such that there is no parallax. This is
true for "static" electromagnetic fields or gravity fields, as it is
NOT true for light waves or gravity waves (which are caused by
accelerations of the sources, as seen by observers). For a field
emitted by a non-accelerated object (as seen by the observer) the pull
is exactly in the direction of the object that emits the field, just as
if the transmission was instantaneous (though it may change in
MAGNITUDE for different observers, the pull doesn't change in DIRECTION
at a given point--- if it did, the force between objects wouldn't point
exactly at the objects, and Newton's 3rd law and conservation of
momentum would go down the drain). And though such a force may change
in direction over time at a given point, that direction is the same for
anyone AT that point, in any inertial frame).
Alas, if you try to quickly or arbitrarily change the position of
said emitting or "source" object by accelerating it (in order to send
information on the static field) the *changes* in the field caused by
the acceleration (which are now waves composed of "real" particles) DO
travel at the speed of light, and DO show parallax. And they are all
that can carry information. Objects intercepting them "see" them (the
waves carrying information) come from a place where the object emitting
them may no longer be. From where it WAS when the wave was emitted.
Uncle Al understands this, but not all those reading may have been
brought up to speed yet. So to speak.
This all leads to some decidedly odd effects, which are the subject
of endless debate on sci.physics and elsewhere (see the physics faq).
For one thing, if you have an electron go past you, the total
electromagnetic field (as defined by the Lorentz force you feel from a
test charge) points exactly at the position of the electron in the
observer's inertial frame. If the observer is moving, this remains
true (though that direction may change for any given observation
point-- from Maxwell's equations and conservation of momentum, you can
derive the inertial frame coordinate transformations of special
relativity, just as Lorentz did). But it is the direction to the
electron which changes, not where the electron appears to be, from the
force it exerts. The electron is always where it appears to be, as
seen from the Lorentz force. It is simply in different places for
different observers, even if the observers at at the same place but
moving at different velocities. It took the genius of Einstein to say
that was okay, since there is no telling which observer is right, and
it doesn't really matter. It's all relative.
HOWEVER, if somebody waggles the electron as it goes past you, the
acceleration of this will produce electromagnetic waves which will
propagate only at the speed of light (since they ARE light, of a
certain kind, or EM radiation) and they will reach the observer from a
different direction: to wit, the direction to the place (coordinates)
where the electron was when it was accelerated, in the inertial frame
of the observer. Again, if the observer changes to a different
inertial frame, but at the same point, this direction may change (due
to parallax). Now, however, the change in direction never does point
in the direction where the electron actually is in this frame at the
time the transmission is intercepted (except by accident or
coincidence, as for instance if somebody moves the electron back to
where it was in the meantime, while the signal travels). It doesn't
even point to the 3-space direction where the electron WAS when the
radiation was emitted, necessarily. It's arbitrary, and only points to
the direction the electron was, if the observer happens to occupy more
or less the same inertial frame the electron did at the time (example:
consider an observer at rest with regard to the antenna of the
spaceship that sent a radio signal, at the time the signal went out).
That observer will see the signal come from where the spaceship was in
the observer's frame when the signal was sent. All other inertial
observers at that point will see the signal from from some other
direction).
Now what's odd here? We sometimes think of a radio wave as being a
change in the electric field seen by an observer, when the charge is
wiggled. This is incorrect. An observer will feel a force from an
electron moving by which points to the true position of the electron,
but will see radio waves (or any other EM waves) emitted by
acceleration of the electron at a position RETARDED, or behind the
position from which the force eminates. The Lorentz force does not
come from the same place the EM waves come from! EM waves are not just
changes in the Lorentz force when a static charge is wiggled; they are
something entirely different and wierd. We think of "seeing" an
electron as it goes past, and "seeing" it wiggle in the same way. This
is a bad way to look at it. If the electron is not accelerated, we
FEEL where it is (Lorentz force), no matter the distance, but if it is
accelerated we SEE where it was (EM radiation), with a speed of light
lag. One is NOT just a change in the other.
Gravity is exactly the same way, but here things are harder to
visualize. If a massive object goes past without accelerating, we FEEL
where it is (gravitational force), but if someone waggles it a bit on
the way, we "SEE" the waggle from the retarded position, in the
direction of where it WAS at a time d/c, where d is the distance and c
is the speed of light. Such waves do not have to eminate from the
direction of the source to conserve force and momentum, because the
waves themselves carry momentum, which was lost from the object when
they were emitted. That is WHY they carry momentum. Momentum carried
by a light wave or gravity wave is simply a force, retarded by the
speed of light lag time. If Newton had thought about action at a
distance, he would have seen that it is logically possible in terms of
his 3rd law only when there is no lag time. Otherwise, it must be
carred by an intermediate body which itself exerts force.
When the Earth circles the sun, therefore, there is a certain amount
of parallax, and telescopes need to point ahead of the sun's true
position, so that the image of the sun seen against the starts is the
one it actually occupied 8 minutes before, when the light was emitted
(this amounts to a tiny fraction of the Sun's apparently diameter in
the sky, but that doesn't matter for purposes of this discussion). And
this "c-retarded point" is also the point that gravity waves would come
from, if a God wiggled the sun up and down a tiny bit. However, the
direction from which the constant force of gravity on the Earth from
the non-accelerated Sun comes, is not subject to parallax, and points
in the direction where the sun actually is at any given time, without
correction for the 8 minute lag time. The sun's light and the sun's
gravity pull do not come from the same place! If this were not true of
the gravitational force, the force between Earth and Sun would not act
exactly between them, and the Earth's orbit would be unstable. What is
not generally appreciatiated, is that if the sun's static
electromagnetic field could be detected, the force from THAT ALSO would
point toward the Sun's true position, and not the place from which
light (EM waves) emanate. And in neither case could the Sun's static
gravity field, or static electromagnetic field, be used to send a
message. Relativity does not say no influence moves faster than
light-- it only provides that to avoid paradox, no information
(arbitrary wiggles) can move faster than light. The sun's constant
gravity does not qualify.
Steven B. Harris
<jer...@jboden.demon.co.uk> writes:
>
>In article <3777B2C9...@hate.spam.net>, Uncle Al
><uncl...@hate.spam.net> writes
>>allen c goodrich wrote:
>>
>>[nothing useful]
>>
>>Gravity propagates at lightspeed. Go away.
>>
The outside gravity is never IN the black hole. In any case, it's a
myth that NOTHING gets out of a black hole. Rather, no information (no
signal) gets out.
Yannopoulos Ruquist
of a distant observer, yet space itself is crossing this surface at
the speed of light, any observer "hovering" just outside the horizon
would essentially be traveling at the speed of light in its local
frame. Hovering is a very misleading term.
Charles W. Shults III
>
> In article <3777B2C9...@hate.spam.net>, Uncle Al
> <uncl...@hate.spam.net> writes
> >allen c goodrich wrote:
> >
> >[nothing useful]
> >
> >Gravity propagates at lightspeed. Go away.
> >
> I know - but - How does gravity escape from a black hole?
>
> Jeremy Boden mailto:jer...@jboden.demon.co.uk
Take a rubber sheet. Push your finger into it. Now,
hold still and ask yourself how the curvature gets away
from your finger- once the curvature is established, it
is a part of the structure of the sheet, a geometry feature.
Gravitation is a geometry feature of spacetime. All mass
has the innate ability (still not quite understood) to curve
the spacetime it is in. While some models predict a sort
of particle for the propagation of gravitational fields
(the hypothetical graviton), it's probably not from the
mass so much as from the curvature of spacetime that these
particles would be emitted. In any case, it isn't a
problem of how gravitation escapes from a black hole, because
gravitation is what -causes- a black hole.
Think in terms of geometry, and the answer is obvious.
If a black hole could swallow gravity, the matter within it
would simply disappear, leaving a larger hole in our
understanding of physics!
Cheers!
Chip Shults
Paul Alan Cardinale
Tom Roberts wrote:
>
<snip>
>
> Note, however, that a hovering observer just outside the horizon can be
> engulfed by it if sufficient additional mass/energy enters the black hole.
> As more mass/energy enters, the horizon expands outward, and as the "mass"
> of the black hole increases it also requires a hovering observer to
> expend more thrust to remain hovering; the horizon can expand outward
> past the observer, in which case no finite thrust can keep her hovering
> and she will intersect the singularity inside the black hole (Schw.
> case). This luckless observer can observe additional mass/energy
> entering the horizon (for a while (:-)).
>
I don't understand that.
Does the hovering observer not see the external fields of the black hole
as "frozen"? And does that not mean that she will not observe the event
horizon
to remain in the same place?
Paul Cardinale
deja
But to what...
Some numerical results show:
From x=2 to x=1000:
SUM 1/[ln(x)^x] = 3.24260941093
Which is the same result that I got for:
From x=2 to x=50
So it converges mighty quick. :)
deja
:P
-Dej
deja wrote in message <7le7fo$5ch$1...@news-02.meganews.com>...
zoo
If it disappears is that a singularity then that Hawking and many others
are speaking about? And if blackholes do disappear then would not the
gravity actually be a force that is exerted but at much higher speeds
than c?
>
> Cheers!
>
> Chip Shults
ystein Olsen
A similar question. Black holes can have charge, I've been told. Some
theories claim that exchange(?) of virtual photons is the cause of
electric charge. How can a black hole then have charge?
--
Oeystein Olsen, oeys...@astro.uio.no
http://www.astro.uio.no/~oeysteio/
Bruce Bowen
sbha...@ix.netcom.com(Steven B. Harris) wrote:
> Now what's odd here? We sometimes think of a radio wave as being a
> change in the electric field seen by an observer, when the charge is
> wiggled. This is incorrect. An observer will feel a force from an
> electron moving by which points to the true position of the electron,
> but will see radio waves (or any other EM waves) emitted by
> acceleration of the electron at a position RETARDED, or behind the
> position from which the force eminates. The Lorentz force does not
> come from the same place the EM waves come from! EM waves are not
just
> changes in the Lorentz force when a static charge is wiggled; they are
> something entirely different and wierd. We think of "seeing" an
> electron as it goes past, and "seeing" it wiggle in the same way.
This
> is a bad way to look at it. If the electron is not accelerated, we
> FEEL where it is (Lorentz force), no matter the distance, but if it is
> accelerated we SEE where it was (EM radiation), with a speed of light
> lag. One is NOT just a change in the other.
Well yes and no. Assume you have a stationary electron off at a
distance. You have your lab assistant, who is sitting next to the
electron, give it a sharp kick. You are not aware, by observation, of
this kick until the appropriate light delay, then the field changes to
that of the now moving electron at its current new position. But the
rub is the field must change continuously and the boundary conditions
must be the field of the stationary electron, and the field of the
moving electron at its new contemporaneous position. This is
information transfer, and it comes to you at the speed of light from
the spacetime postion of the kick, but it can still be thought of as a
"change in the field".
-Bruce bbo...@ppppppacbell.net (remove the stutter)
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
Steven B. Harris
writes:
>
>A similar question. Black holes can have charge, I've been told. Some
>theories claim that exchange(?) of virtual photons is the cause of
>electric charge. How can a black hole then have charge?
Because virtual photons emanate from black holes like virtual
gravitons. Alas, they cannot carry information.
Tom Roberts
> Tom Roberts wrote:
> > Note, however, that a hovering observer just outside the horizon can be
> > engulfed by it if sufficient additional mass/energy enters the black hole.
> > As more mass/energy enters, the horizon expands outward, and as the "mass"
> > of the black hole increases it also requires a hovering observer to
> > expend more thrust to remain hovering; the horizon can expand outward
> > past the observer, in which case no finite thrust can keep her hovering
> > and she will intersect the singularity inside the black hole (Schw.
> > case). This luckless observer can observe additional mass/energy
> > entering the horizon (for a while (:-)).
> as "frozen"? And does that not mean that she will not observe the event
> horizon
> to remain in the same place?
Only as long as the mass/energy contained in the black hole remains
constant. As additional mass/energy enters the hole, its event horizon
expands, as seen by any observer.
I used that as a way an outside observer can "observe" the infall of
something into the hole. An observer a finite distance away from
the horizon cannot directly observe something enter the horizon (loosely
because the fields are "frozen"). In the case where the horizon expands
to engulf the observer she can do so (because it is not possible to hover
_inside_ the horizon, and the observer is no longer a finite distance
outside the horizon). Note that in any case, an infalling observer can
observe her own crossing of the horizon, and can also observe the
crossing of approximately-comoving objects which are sufficiently close
to her.
Note that this situation converts the initially-hovering outside observer
into an infalling observer. I should have mentioned that more explicitly.
The above claims can be proven for a spherically-symmetric
situation in which a thin spherical shell falls into a Schwarzschild black
hole. For
non-spherical cases things are expected to be similar.
Tom Roberts tjro...@lucent.com
Tom Roberts
> Black holes can have charge, I've been told. Some
> theories claim that exchange(?) of virtual photons is the cause of
> electric charge. How can a black hole then have charge?
The theory which describes charge and its relationship to virtual
photons is QED. But it does not really claim that charge "causes"
photons or vice-versa; like most physical theories it merely
expresses their relationship (Maxwell's equations are similar:
charge does not "cause" electric and magnetic fields, it is merely
related to them in a specific way).
Virtual photons are "off the mass shell", meaning that they need not
have 0 mass. This permits spacelike virtual photons (i.e. virtual
photons which can "go faster than light"). These can escape from the
black hole. Due to the nature of the event horizon and the properties
of these virtual photons, the only information they can carry outside
the event horizon is how much charge there is inside the horizon.
This is analgous to expectations of how gravitons will behave in
a theory of quantum gravity: spacelike virtual gravitons will escape
a black hole, but the only information they will be able to carry
is how much mass and angular momentum there is inside the horizon.
Loosely, they can carry two pieces of information because they are
tensor particles rather than vector particles like photons.
Tom Roberts tjro...@lucent.com
Christian A. Ross
If that were true, then it would be possible to arrange a system of masses
to send gravitational messages into the past. And we would be feeling the
effects of the motion of the planets before they actually occurred. The
speed of light, as I understand it, is not just the velocity of photons,
but the maximum rate that information can be exchanged between objects in
different reference frames. And why would gravitons have to interact with
gravitons anyway. I had a physics professor who said black holes didn't
exist because they would swallow their own gravitational emission. He was a
kook, who poohpoohed the likes of Steve Weinberg and Hawking ... but I
don't see his name in the journals.
Chris Ross
(Nobody in particular).
Douglas W Bass
suggesting that gravity propagated at a much greater speed than light. I
must confess that I didn't understand much of the article, but the
engineering of the Global Positioning System was part of the argument.
Is Van Flandern known in the area of this conversation?
If this is actually the case, then gravity waves would be the perfect
medium for sending messages extremely long distances.
Best wishes,
Douglas Bass
-------------------------------------------------------------------------
Douglas Bass
Ph.D Student, Computer Science, University of Texas at Dallas
Phone: (972)-883-6444
email: db...@utdallas.edu
URL: http://www.utdallas.edu/~dbass
The opinions expressed are strictly my own, and are not necessarily those
of the faculty or administration of the University of Texas at Dallas.
Charles W. Shults III
>
> A similar question. Black holes can have charge, I've been told. Some
> theories claim that exchange(?) of virtual photons is the cause of
> electric charge. How can a black hole then have charge?
The virtual photon field will appear at the event horizon,
and we will still measure a charge present at that zone.
And, if you consider that dumping a bunch of electrons into
the hole creates a situation where the mass is not vanishing
into the hole, but being distributed evenly about it, you
can visualize a process that does the same thing to charges.
It becomes easier to visualize if you imagine that both
mass and charge are assigned specific vectors in a multi-
dimensional spacetime. Each charge unit adds a little twist
to the spacetime of the whole, resulting in a larger overall
charge. Mass does likewise, but in an axis that is orthogonal
to the three dimensions we can directly observe.
Some of the manifold theories (derivatives of the Kaluza-
Klein theory) assign dimensions to gravitation, electromagnetism,
strong and weak nuclear forces. Those add up and still are
measureable at the hole's "surface".
Cheers!
Chip Shults
Charles W. Shults III
>
> "Charles W. Shults III" wrote:
>
> > Jeremy Boden wrote:
> > >
> > > In article <3777B2C9...@hate.spam.net>, Uncle Al
> > > <uncl...@hate.spam.net> writes
> > > >allen c goodrich wrote:
> > > >
> > > >[nothing useful]
> > > >
> > > >Gravity propagates at lightspeed. Go away.
> > > >
> > > I know - but - How does gravity escape from a black hole?
> > >
> > > --
> > > Jeremy Boden mailto:jer...@jboden.demon.co.uk
> >
> > Take a rubber sheet. Push your finger into it. Now,
> > hold still and ask yourself how the curvature gets away
> > from your finger- once the curvature is established, it
> > is a part of the structure of the sheet, a geometry feature.
> > Think in terms of geometry, and the answer is obvious.
> > If a black hole could swallow gravity, the matter within it
> > would simply disappear, leaving a larger hole in our
> > understanding of physics!
> >
>
> If it disappears is that a singularity then that Hawking and many others
> are speaking about?
No. The singularity is the location where our known
laws of physics must be unable to apply. Consider that
we have numerous regimes for each type of physics that
we know about. For instance, macro scale objects (larger
than particles) at pedestrian velocities are quite easily
and accurately handled by plain Newtonian physics.
At high velocities, such as near lightspeed, Newtonian
physics breaks down because it isn't a truly accurate model
of the processes, and at this new regime of velocities, we
need to use relativity to handle the ever increasing errors
that show up. Newton is a subset of GR, GR just has a more
accurate representation of the physical world.
At the microscopic scale, Newtonian rules fail because
quantum effects begin to dominate. Here, we have a new set
of rules called quantum physics. It more accurately handles
the processes and observations at that scale. The boundary
is at about 10^-10 meters, about the scale of an atom.
These rules work down to about 10^-14 meters, the scale of
the nucleus.
For small scale and relativistic velocities, such as encountered
in particle accelerators, there is a new set of hybrid rules
called relativistic quantum physics. It works well at the
overlap of relativity physics and quantum physics.
Now, below 10^-14 meters, smaller than the nucleus, what
happens is pretty much anyone's guess. Singularities combine
immense masses, severely distorted spacetime geometry, tiny
scales, and high velocities. Physicists are still trying to
formulate the rules for this regime, but they are stymied by
problems with the uncertainty principle and other things as
well.
Given time, a merging of these rules may be possible. Only
then could we begin to explain what happens in a singularity.
> And if blackholes do disappear then would not the
> gravity actually be a force that is exerted but at much higher speeds
> than c?
No. They can evaporate through Hawking radiation, but then
the mass is distributed, and the dimple in spacetime is
smoothed out.
Cheers!
Chip Shults
Jeremy Boden
<sbha...@ix.netcom.com> writes
>>theories claim that exchange(?) of virtual photons is the cause of
>>electric charge. How can a black hole then have charge?
>
>gravitons. Alas, they cannot carry information.
This may be drifting off-topic here... but another similar question:-
Suppose a particle (photon?) is sitting inside a black hole near the
event horizon. Due to Heisenberg's uncertainty principle its position is
a little *uncertain*. This enables it to tunnel through the event
horizon and escape from the black hole...
Is this a possibility?
--
Jeremy Boden
Matthew Montchalin
In a previous article, oeys...@ulrik.uio.no (\ystein Olsen) says:
| A similar question. Black holes can have charge, I've been told. Some
| theories claim that exchange(?) of virtual photons is the cause of
| electric charge. How can a black hole then have charge?
How can *any* pointlike particle have charge? ;) And a similar question
arises, namely, whether a pointlike particle can have spin. Most people
suppose the possibility that black holes have spin, mass, and charge, but
these are merely attributes that can be conferred to any pointlike
particle moving through 3D space.
--
Magnus Nyborg
Clear Skies,
Magnus
Jeremy Boden <jer...@jboden.demon.co.uk> wrote in message
news:1YGYaIAR...@jboden.demon.co.uk...
> In article <7lg7qa$j...@dfw-ixnews15.ix.netcom.com>, Steven B. Harris
> <sbha...@ix.netcom.com> writes
> >In <7lfue4$6b7$1...@readme.uio.no> oeys...@ulrik.uio.no (\ystein Olsen)
> >writes:
> >>
> >>A similar question. Black holes can have charge, I've been told. Some
> >>theories claim that exchange(?) of virtual photons is the cause of
> >>electric charge. How can a black hole then have charge?
> >
Steven B. Harris
<jer...@jboden.demon.co.uk> writes:
>This may be drifting off-topic here... but another similar question:-
>
>Suppose a particle (photon?) is sitting inside a black hole near the
>event horizon. Due to Heisenberg's uncertainty principle its position
is
>a little *uncertain*. This enables it to tunnel through the event
>horizon and escape from the black hole...
>
>Is this a possibility?
>
>--
>Jeremy Boden
Not according to current theory. You are thinking of the event
horizon as a thing, a wall, a membrane. It's not any of these. It
simply marks a boundary. On one side is normal space, albeit space
with exceedingly high gravity fields, slowed time, and other nastiness.
On the other side of the boundary is "who knows?" Light cannot get
out, because space in that direction is infinitely long, the gravity
fields are infinitely large, and the flow of time is entirely stopped.
If it was a hole, it would be infinitely deep.
In quantum mechanics you can calculate the probability of things
"tunneling" though a region of space with a potential barrier (a field
of force opposing the motion, like a gravity field or electric field).
But you have to know the length of the barrier, the wavelength of the
particle (energy or momentum-- it's all related) and you have to know
the "height" of the barrier --- that is, how much energy is involved in
penetrating it by a certain distance. There is a probability of
tunneling to this distance for any particle by the Heisenberg principle
if you know the values of all of these things. If the barrier is
infinite in length and height, however, the probabilty of tunneling is
zero. So, no, nothing gets out from beyond the even horizon of a black
hole, as defined by general relativity (GR).
Remember, however, that it is only the *theory* of GR which predicts
these infinities. They are mathematical, and whether they are "real"
or not is up for grabs. Since something very like GR's black holes with
"infinite" gravity fields near them apparently exist, the actual fields
near the actual objects must indeed be very, very, very.... large.
Perhaps large enough to be infinity for all practical purposes,
including tunneling purposes, if you get my meaning. Whether they are
actually "infinitely large", however, it's another question. Infinity
is a big proposition. A collapsed star with a gravity field of a
trillion trillion g's would act no differently than the ones we think
we can "see" by their secondary effects, but such a field would be no
closer to infinity than than the one you're standing in. Nobody knows
if GR is exactly correct. It would be surprising if it were, IMHO. So
the answers given above are only according to this theory, and in
reality, something else might be true for very large ( > 3 times the
sun) masses.
Chris Hillman
On Thu, 1 Jul 1999, Tom Roberts wrote:
> As additional mass/energy enters the hole, its event horizon expands,
> as seen by any observer.
Just thought I'd break in to stress that the expansion of the event
horizon can -precede- the infall of a packet of mass-energy, and that this
is not paradoxical.
For concreteness, and for ease of modeling, let's say this mass-energy
consists of a radially infalling spherical shell of incoherent radiation
moving at the speed of light. It carries energy, and as it passes by an
observer using his rocket engines to hover outside a Schwarzschild hole,
this mass-energy will temporarily make the Ricci curvature go nonzero;
after it passes, the Ricci curvature near our observer goes back to zero,
but the Weyl curvature will have increased to take account of the
increased amount of mass-energy; if you like, of the increased mass of the
hole. In particular, if the shell contains enough energy, it will have
expanded to enclose our observer, -before- he finds he can no longer
maintain his position and is falling into the hole. This is no paradox,
but merely a reflection of the global, abstract definition of the horizon.
Interested readers are referred to my archived post
http://www.math.washington.edu/~hillman/PUB/vaidya
See also Frolov & Novikov, Black Hole Physics, Kluwer, 1998 for another
detailed discussion of infalling radiation as studied using the Vaidya
null dust solution (1951). This a spherically symmetric, not necessarily
static generalization of the Schwarzschild vacuum with one freely chosen
function, m(r), which describes a time varying amount of ingoing (resp.
outgoing) massless radiation with spherical wavefronts concentric with a
nonrotating star or black hole.
> Note that this situation converts the initially-hovering outside observer
> into an infalling observer. I should have mentioned that more explicitly.
Every now and again it is worth attempting to make interested students
aware of the existence of a great variety of alternative coordinate
systems for the Schwarzschild vacuum, each having a simple, explicit
metric, and each useful for different purposes. Everyone knows the
(exterior) "static coordinates" of Schwarzschild 1915:
ds^2 = -(1-2m/r) dt^2 + dr^2/(1-2m/r) + r^2 (du^2 + sin^2 u dv^2), r > 2m
which are appropriate for studying the experience of static observers, the
"Schwarzschild observers", who use rocket engines to hover motionless
above a black hole (or a star, if we use the Schwarzschild vacuum solution
for r > R, some sufficiently large "surface radius", and match to an
interior perfect fluid solution to model the matter of the star). Here,
note that every slice t = t0 is isometric to any other such slice, and can
be embedded as (the upper half of) a "Flamm paraboloid".
Many students are also aware of the "interior" coordinates
ds^2 = -dt/(2m/t - 1) + (2m/t - 1) dz^2 + t^2 (du^2 + sin^2 u dv^2),
0 < t < 2m, -infty < z < infty
These two patches turn out to describe -disjoint- portions of the -same-
spacetime, but to verify that you must find other coordinate patches which
overlap both of these. (These particular "interior coordinates" cover one
of two "interior" regions in the largest spacetime which contains pieces
which can be described by either of the above metrics; it turns out that
subject to smoothness assumptions there is only one such "maximal
extension" of the Schwarzschild vacuum; it is sometimes called the Kruskal
vacuum.)
There are a number of patches which extend the exterior coordinates into
the interior of a black hole, and which also happen to suitable for
studying the physical experience of freely falling observers who fall
radially into the hole (or star). They include the Eddington coordinates
(1924)
ds^2 = -(1-2m/r) dt^2 + 2 dt dr + r^2 (du^2 + sin^2 u dv^2), r > 0
(the Vaidya null dust is given by the same metric, but with m now taken to
be a function of t), the Lemaitre coordinates (1929)
ds^2 = -dT^2 - 2m/g(R-T) dR^2 - g(R-T)^2 (du^2 + sin^2 u dv^2),
R - T > 0, g(R-T) = (9m/2)^(1/3) (R-T)^(2/3),
and the "exponential" coordinates
ds^2 = dT^2 - e^(-2z) [(dz - sqrt(2m e^z) dt)^2 + du^2 + sin^2 u dv^2],
-infty < z < infty
The ONB (orthonormal basis, aka "local Lorentz frame", c.f. the
equivalence principle and how gtr generalizes str) corresponding to the
Lemaitre observers is not at all obvious in the Eddington metric, although
it turns out to be very simple; the required ONB's in the other two
patches are obvious, and the Eddington ONB is easily found from either of
these.
For studying phenomena such as "light bending", the simplest approach
involves finding coordinates which correctly represent angles in the
slices t = t0. These include
[ 1-m/(2R)]^2
ds^2 = -[ ------- ] dt^2
[ 1+m/(2R)]
+ (1+m/(2R))^4 [dR^2 + R^2(du^2 + sin^2 u dv^2)],
R > m/2
and also the Costa coordinates (1995 or so)
ds^2 = -tanh(z/2)^2 dt^2
+ 4m^2 cosh^4(z/2) [dz^2 + du^2 + sin^2 u dv^2],
-infty < z < infty
The first of these patches shows that each slice t = t0, r > 2m through
the Schwarzschild exterior patch is conformally equivalent to E^3 with a
ball subtracted. The second shows they are also conformally equivalent to
R x S^2 with the obvious euclidean metric, or rather the half cylinder z >
0. Here, z = 0 corresponds to r = 2m and represents the event horizon.
The cylinders are of course homeomorphic to the Flamm paraboloid; the
Costa coordinates provide a particularly insightful approach to
understanding the "Einstein-Rosen bridge".
For studying the nature of the event horizon at r = 2m, the "redshift"
coordinates
ds^2 = -(w dt)^2 + 16 m^2 dw^2/(1-w^2)^4
+ 4m^2/(1-w^2)^2 (du^2 + sin^2 u dv^2),
-1 < w < 1
are useful, since these have the same form as the Rindler coordinates
which describe the physical experience of uniformly accelerating
observers in the Minkowksi vacuum (1908):
ds^2 = -(x dt)^2 - dx^2 - dy^2 - dz^2, -infty < x < infty
To understand the causal structure, the Kruskal coordinates
16 m^2
ds^2 = ------------- (dR^2 - dT^2) - g (du^2 + sin^2 u dv^2)
g exp[g/(2m)]
are convenient; here g(R^2-T^2) is defined by the relation
(g(R^2-T^2) - 2m) exp[g(R^2-T^2)/(2m)] = R^2 - T^2
The patch obtained by restriction to the exterior region may be written
16 m^2 plog[(R^2-T^2)/(2me)] dR^2 - dT^2
ds^2 = ---------------------------- -----------
1 + plog[(R^2-T^2)/(2me)] R^2 - T^2
+ 4m^2 (1 + plog[(R^2-T^2)/(2me)])^2 (du^2 + sin^2 u dv^2)
where plog(w) = z is the principal branch of the inverse function of w = z
exp(z).
The Penrose coordinates are even more incisive, but rather than discussing
them in detail, I'll just give this ASCII sketch:
____
/\ /\
/ \/ \
\ /\ /
\/__\/
Here, the two exterior regions are the left and right diamonds and the two
interior regions are the bottom and top triangles. The horizon is the
"X", and every point is really a sphere S^2, with radius r varying from
place to place; r - > infty near the outer flanks of the two diamonds, and
r -> 2m near the inner flanks, i.e. the horizon, and r -> 0 near the top
or bottom segments, which represent the curvature singularities. The
slice t = 0 (-infty < z < infty) corresponds to a horizontal segment
through the central sphere (radius 2m), and as I already said, can be
visualized as a Flamm paraboloid or, up to conformal equivalence, as a
cylinder R x S^2.
Novikov's coordinates are also very interesting (they are comoving with a
different family of free falling observers, the Novikov observers), but
too complex to write out here; see, however, the pictures in MTW,
Gravitation, Freeman, 1973, and in Frolov & Novikov, Black Hole Physics,
Kluwer, 1998.
Note well: in the above, I ran out letters, so not all coordinates called
R (say) are necessarily the same.
Remark: the Schwarzschild solution, or rather family of solutions
parametrized by m, is contained not only in the Vaidya family of null dust
solutions but also in the Kerr family of vacuum solutions, and in families
of "cosmic string interior" solutions, and in still larger families of
exact solutions. Too understand one reason why this is interesting, see
the literature on generic singularities.
Chris Hillman
Home Page: http://www.math.washington.edu/~hillman/personal.html
Steven B. Harris
writes:
>I recently read an article by a gentleman named Tom Van Flandern,
>suggesting that gravity propagated at a much greater speed than light.
I
>must confess that I didn't understand much of the article, but the
>engineering of the Global Positioning System was part of the argument.
>
>Is Van Flandern known in the area of this conversation?
It's been explained to the man again and again that "gravity" (what
the sun holds the Earth with) and "gravity waves" have entirely
different properties. The last could carry information, but it would
also travel at the speed of light, and there's NO evidence that it
doesn't (gravity waves haven't even been detected, at this point,
though there's little doubt from observation of binary stars that they
exist). As for ordinary gravity of the kind we're used to, it is
static and does not "propagate," nor does it carry information. The
fact that it's not subject to parallax is irrelevent, and means nothing
about how "fast" it moves. To everyone but Van Flandern.
Chris Hillman
On 2 Jul 1999, Steven B. Harris wrote:
> Not according to current theory. You are thinking of the event
> horizon as a thing, a wall, a membrane. It's not any of these. It
> simply marks a boundary. On one side is normal space, albeit space
> with exceedingly high gravity fields, slowed time, and other nastiness.
> On the other side of the boundary is "who knows?" Light cannot get
> out, because space in that direction is infinitely long, the gravity
> fields are infinitely large, and the flow of time is entirely stopped.
> If it was a hole, it would be infinitely deep.
I'd have to take some exception to this description, which I think is
misleading.
First of all, time doesn't really "slow" anywhere; rather, light signals
from observers hovering close a massive object will be received with a
redshift by observers hovering further away; this is the same thing as
saying that clocks held stationary close to a massive object will run slow
compared (by light or radio signals) to clocks held stationary further
away.
Second, "time" and "space" are equally "normal" just inside and just
outside the event horizon.
Third, when you say "space is infinitely long in that direction", you
might be thinking of the fact that slices t = t0 in the Schwarzschild
coordinates correspond to paraboloids, so that these slices "never reach
inside the horizon", a fact which has misled many into thinking that the
"interior" either does not exist or that r = 0 is somehow "infinitely"
distant from r = 5m (say). MTW, Gravitation, Freeman, 1973, offer a very
good discussion of the geometry of the Schwarzschild solution.
> Remember, however, that it is only the *theory* of GR which
> predicts these infinities. They are mathematical, and whether they
> are "real" or not is up for grabs. Since something very like GR's
> black holes with "infinite" gravity fields near them apparently exist,
The components of the curvature tensors scale like m/r^3, just as in
Newtonian theory, so they are finite at the event horizon, indeed in
geometric units they are on the order 1/(8 m^2). (Thus, in the case of a
supermassive black hole, humans could survive the tidal forces long enough
to still be alive when they enter such a hole, but they could not survive
a close approach to the horizon of the typical ten solar mass supernova
remnant black hole.)
Gtr is expected to hold to a very high degree of accuracy not only just
outside the event horizon, but also deep inside real black holes.
However, ultimately quantum effects should become important, and then all
bets are off. And as you say, we have no way of observing the interesting
new physics which we think must be going on inside these objects.
Steven B. Harris
fact that it's not subject to aberration is irrelevent, and means
Chris Hillman
On 2 Jul 1999, Steven B. Harris wrote:
Anyone interested in learning what Flandern hasn't should read very
carefully this compilation of posts from sci.math.research, sci.physics,
and sci.physics.relativity (posts by experts like Steve Carlip):
http://www.math.washington.edu/~hillman/PUB/debate
Paul Alan Cardinale
Tom Roberts wrote:
>
> Only as long as the mass/energy contained in the black hole remains
> constant. As additional mass/energy enters the hole, its event horizon
> expands, as seen by any observer.
>
> I used that as a way an outside observer can "observe" the infall of
> something into the hole. An observer a finite distance away from
> the horizon cannot directly observe something enter the horizon (loosely
> because the fields are "frozen"). In the case where the horizon expands
> to engulf the observer she can do so (because it is not possible to hover
> _inside_ the horizon, and the observer is no longer a finite distance
> outside the horizon). Note that in any case, an infalling observer can
> observe her own crossing of the horizon, and can also observe the
> crossing of approximately-comoving objects which are sufficiently close
> to her.
>
How can an someone outside the event horizon observe the event horizon
to expand?
Wouldn't she have to observe matter fall into the black hole in order to
observe
expansion of the event horizon?
Paul Cardinale
Larry Mead
: A similar question. Black holes can have charge, I've been told. Some
: theories claim that exchange(?) of virtual photons is the cause of
: electric charge. How can a black hole then have charge?
No. An overall macroscopic charge is the cause of the net charge.
--
Lawrence R. Mead Ph.D. (Lawren...@usm.edu)
Eschew Obfuscation! Espouse Elucidation!
www-dept.usm.edu/~physics/mead.html
Larry Mead
: I recently read an article by a gentleman named Tom Van Flandern,
: suggesting that gravity propagated at a much greater speed than light. I
: must confess that I didn't understand much of the article, but the
: engineering of the Global Positioning System was part of the argument.
: Is Van Flandern known in the area of this conversation?
Mr. Van Flandern was once a respected and active astronomer. His
understanding of the GPS is very much flawed. *All* gravitational effects
are taken into account in the GPS. It was very surprising that the
article got through the editors of Physics Letters. The reviewers screwed
up. For a thorough account of how this is done and why there is no
conflict with GR, see "Was Einstein Right? " by Clifford Will. I just
spoke to Cliff about this [he is chair of physics at Washington University
in St. Louis], and he explained that Van Flandern is apparently not aware
that all relativistic effects are accounted for by the physics people who
designed the GPS; the computer program used to calculate where you are
takes GR automatically into account in the correct way. That is why it
in fact works.
[snip]
Tom Roberts
> How can an someone outside the event horizon observe the event horizon
> to expand?
By being engulfed by it, and therefore being unable to hover. Or by observing
that the mass of the black hole has increased (e.g. by measuring the period of
a small pendulum); the area of the event horizon is directly related to the
total mass of the black hole -- yes, this is an inference and not a direct
observation.
> Wouldn't she have to observe matter fall into the black hole in order to
> observe
> expansion of the event horizon?
No. I was discussing a spherical case, and the observer certainly would see the
thin spherical shell go by. But in general, even a non-symmetric influx of
(uncharged) mass/energy into a black hole will "damp out" into a Kerr solution in
a finite time (to any outside observer); the mass of the hole is always observable
to an outside observer (hovering or not). So matter falling in on the other side
of the hole (and not visible to the observer) would still be felt as an increase
of the hole's mass by that observer, in a finite time. In general there would also
be gravitational radiation, which is the mechanism by which asymmetries "damp out".
You might be interested in Kip Thorne's book, _Black_Holes_and_Time_Warps_, in which
he discusses this in detail, non-mathematically (!).
Tom Roberts tjro...@ucent.com
Steven B. Harris
writes:
>\ystein Olsen (oeys...@ulrik.uio.no) wrote:
>
>: A similar question. Black holes can have charge, I've been told.
Some
>: theories claim that exchange(?) of virtual photons is the cause of
>: electric charge. How can a black hole then have charge?
>
>No. An overall macroscopic charge is the cause of the net charge.
Reminds me of Moliere's physician who said that morphine puts people
to sleep by virtue of its "dormative properties." <g>
What in the world is an "overall macroscopic charge"? If not a net
charge? I think black holes are charged by virtue of having eaten an
excess of either positive or negative particles. Same as any other
large object.
John R Ramsden
>> has the innate ability (still not quite understood) to curve
>> the spacetime it is in.
Presumably that must be because, at some level, matter and energy
themselves are highly curved knots of space time. Their gravity
can then be interpreted as a secondary effect to smooth out (or
minimize) the overall curvature. If you sew a few stitches in a
sheet and then tug the cotton the same phenomenon occurs - the
result is a localized pucker surrounded by a gentler distortion.
>> Think in terms of geometry, and the answer is obvious.
>> If a black hole could swallow gravity, the matter within it
>> would simply disappear, leaving a larger hole in our
>> understanding of physics!
>
>If it disappears is that a singularity then that Hawking and many others
>are speaking about? And if blackholes do disappear then would not the
>gravity actually be a force that is exerted but at much higher speeds
>than c?
Black holes _do_ disappear, well small ones anyway. They "evaporate" at a
rate inversely related to their mass. (That word "related is deliberately
vague, as I don't remember the exact rate; it may be 1 / mass^3.)
I suppose one could relate this to the point I mentioned above: As a black
hole becomes smaller its curvature increases, and at some point it "thinks"
(so to speak) "Well I'm getting almost as curved as ordinary mass/energy,
I may as well turn back into it!". Don't take this analogy too seriously
though; it is only a passing thought, and I'm not a physicist as you may
have gathered.
Cheers
John R Ramsden (j...@redmink.demon.co.uk)
Steven B. Harris
Bruce Bowen <bru...@my-deja.com>
Thu, 01 Jul 1999 18:09:04 GMT
in Message-ID: <7lgarm$kj1$1...@nnrp1.deja.com>
notes:
>In article <7lcsc5$m...@dfw-ixnews4.ix.netcom.com>,
> sbha...@ix.netcom.com(Steven B. Harris) wrote:
>Now what's odd here? We sometimes think of a radio wave as
>being a change in the electric field seen by an observer, when
>the charge is wiggled. This is incorrect. An observer will
>feel a force from an electron moving by which points to the true
>position of the electron, but will see radio waves (or any
>other EM waves) emitted by acceleration of the electron at a
>position RETARDED, or behind the position from which the force
>emanates. The Lorentz force does not come from the same place
>the EM waves come from! EM waves are not just changes in the
>Lorentz force when a static charge is wiggled; they are
>something entirely different and wierd. We think of "seeing" an
>electron as it goes past, and "seeing" it wiggle in the same
>way. This is a bad way to look at it. If the electron is not
>accelerated, we FEEL where it is (Lorentz force), no matter the
>distance, but if it is accelerated we SEE where it was (EM
>radiation), with a speed of light lag. One is NOT just a change
>in the other.
Bruce writes:
>Well yes and no. Assume you have a stationary electron off at a
>distance. You have your lab assistant, who is sitting next to
>the electron, give it a sharp kick. You are not aware, by
>observation, of this kick until the appropriate light delay,
>then the field changes to that of the now moving electron at its
>current new position. But the rub is the field must change
>continuously and the boundary conditions must be the field of
>the stationary electron, and the field of the moving electron at
>its new contemporaneous position. This is information transfer,
>and it comes to you at the speed of light from the spacetime
>position of the kick, but it can still be thought of as a
>"change in the field".
Comment:
Agree. But note that this kind of c-retarded information lag
is more or less during accelerations, which is what I noted
above. If you make your accelerations delta function kicks, all
this does is give you boundary problems and discontinuities in
your apparent position answer (in which discontinuous changes in
velocity are transformed into apparently discontinuous changes in
position, as you say). But that's not a physical problem, just a
problem introduced by sticking in simplifying math. In a real
situation you'd have to do the problem in pieces (with an
acceleration piece) and then, to insure physicality, make sure
that apparent position points connect up for accelerated and
constant velocity segments.
Interestingly, if the electron is kicked directly toward you,
you see no EM radiation from the kick, and yet there is still a
c-delay time for the field at your position to change. If that
was your point, it is well-taken. Interesting.
(For those just entering this thread: if the electron is a ways
off at d1, and is kicked suddenly directly toward you at speed v,
it's true that the sudden and UNEXPECTED position change is
information, and you're not going to notice the kick until a time
t= d1/c elapses. By that time the electron's position has
changed to d2 (closer to you) = d1-vt = d1-v = d1(1-v/c). If v/c
is small, it turns out that the new electron E field strength
when you finally notice the kick is:
E = q (1/d2)^2 * [1 - 4v^2/c^2]
(I think the 4 is in there-- it appeared when I did the calcula-
tion last, but not the time before. I am sure that the
correction to instantaneous position only of order v^2/c^2 for
small v).
Note that the expression above is much closer to the field if
it came from the "instantaneous" position d2 than it is to the
"speed of light lag" position d1, which is where you'd still be
"seeing" the field emanating (feeling it emanate) if all changes
in the field (both expected and unexpected) simply traveled at
the speed of light (ie, if you had just seen the thing kicked at
d1, and therefore still saw it at d1, more or less, but now
moving toward you). So it looks (field-wise) as if the electron
had disappeared at d1 and reappeared closer, at d2. If the
electron keeps moving toward you at constant v, after the initial
delay associated with the kick, the changing field corresponds
closely to the electron's "real" distance from you (which is
closer to you than the one you would "see", if there was light
(EM radiation) coming from the electron). This is also true of
transverse kicks, so long as velocity doesn't change afterwards.
This is true because there is a term in the field equation of a
moving charge which adds a linear correction to the static field,
according to how fast the field strength is changing in the
direction of the observer at the "speed of light lag time"
position, multiplied by time for light to get to the observer
from that position. So all strictly linear changes in the fields
of charges with time, for an observer, are corrected for, speed
of light lag-wise.
Of course, since fields vary as 1/r^2, even constant
speeds don't give totally linear changes in the field strength
with time, so the correction isn't perfect, usually-- just very
good. For example, people may think that the magnetic field of
one coil of a transformer or motor is seen by the other coil with
a speed of light delay of d/c, according to how far apart the
coils are. But actually the delay is much, much smaller than
that, so long as the frequency is low enough and distances
between coils is small enough that little tiny segments of sine
wave change during t= d/c can no longer be approximated linearly.
The point there being that so long as fields change in a
predictable and approximately linear way which carries NO
information, the direction and distance to the field source
appears to vary much less than speed of light lag would predict.
Aberration of linear changes in static fields is not the same as
the normal aberration in EM radiation.
I think the only time the speed of light lag "update"
approximation for static fields by nature is perfect, so there is
no aberration, is when charges are at constant distance, and the
change in field vector with time is due only to change in
velocity which has a constant time derivative (as when charges
orbit each other in circular motion). Then the correction is
perfect, and orbiting charges see no aberration for each other's
position at all even when they move at relativistic speeds. That
was merely the point I wanted to make for the gravity people.
There's been a lot of talk about the reason for the lack of
aberration for the sun's static G field, as seen by the earth, is
because it is a property of space-time itself at the Earth's
position, blah, blah. But the static electric and magnetic
fields of the sun aren't subject to aberration, either, according
to Maxwell (and of course, Einstein). Does this also mean they
too are properties of "space itself"? Your call.
I don't know for sure whether GR makes a similar prediction
about correction for aberration in the static G fields of a
particle which moves at constant velocity (ie, one that predicts
that this should be good, but not perfect, except for circular or
orbital motion). Anybody out there good enough with the tensor
analysis to be able to say?
Steven B. Harris
Comment:
you see no standard 1/r^2 E dropoff farfield type EM radiation from the
kick (real photons), and yet there is a kind of EM pulse, and still a
c-delay time for the field at your position to change. If that
was your point, it is well-taken. Interesting.
(For those just entering this thread: if the electron is a ways
off at distance d1, and is kicked suddenly directly toward you at speed
v, it's true that the sudden and UNEXPECTED position change is
information, and you're not going to notice the kick until a time
t = d1/c elapses. By that time the electron's position has
changed to d2 (closer to you) = d1-vt = d1-(v*dl/c) = d1(1-v/c). If
v/c is small, it turns out that the new electron E field strength at
your position due to the electron moving toward you, when you finally
notice the kick is:
E = q (1/d2)^2 * [1 - 4v^2/c^2]
(I think the 4 is in there-- it appeared when I did the calcula-
tion last, but not the time before. I am sure that the
correction to instantaneous position only of order v^2/c^2 for
small v).
Note that the expression above is much closer to the field if
it came from the "instantaneous" closer position d2, where it is the
instant you get the signal from d1, than it is to what it would be from
the "speed of light lag" position d1, which is where you'd still be
"seeing" the field emanating (or feeling it emanate) if all changes in
the field (both expected and unexpected) simply traveled at the speed
of light (i.e., if you had just seen the thing kicked at
d1, and therefore still *saw* it at the d1 position, more or less, but
now moving toward you). So it looks (field-wise) as if the electron
had disappeared at d1, and reappeared closer, at d2. If the
electron keeps moving toward you at constant v, after the initial
delay associated with the kick, the changing field corresponds
closely to in magnitude the electron's "real" or simultaneous d2
distance from you (which is closer to you than the position d1 at which
you would "see" it, if there was light (EM radiation) coming from the
electron). This is also true of transverse kicks, so long as velocity
doesn't change afterwards.
This is true because there is a term in the field equation of a
moving charge which adds a linear correction to the static field,
according to how fast the field strength is changing in the
direction of the observer at the "speed of light lag time"
position, multiplied by time for light to get to the observer
from that position. So all strictly linear changes in the observed
distant fields of moving charges with time, are corrected for, speed of
light lag-wise.
Of course, since fields vary as 1/r^2, even constant
speeds don't give totally linear changes in the field strength
at distance with time, so the correction isn't perfect, usually-- just
very good. For example, people may think that the magnetic field of
one coil of a transformer or motor is seen by the other coil d distance
away, with a simple speed of light delay of d/c, according to how far
apart the coils are. But actually the delay is much, much smaller than
that, so long as the frequency is low enough, and distance
between coils is small enough, that little tiny segments of sine
wave change during t= d/c can be approximated linearly. Which
certainly is the case at 60 Hz in a standard AC transformer.
The point there being that so long as fields change in a
predictable and approximately linear way, which carries NO
information, the direction and distance to the field source
appears to vary much less than speed of light lag would predict.
Aberration of approximately linear changes in static fields is
not the same as the normal aberration in farfield EM radiation.
I think the only time the speed of light lag "update"
approximation by nature for far away static fields of moving
objects is perfect, is when charges are at constant distance, and
the change in field vector with time is due only to change in
velocity which has a constant time derivative (as when charges
orbit each other in circular motion). Then the correction is
perfect, and orbiting charges see no aberration for each other's
position at all, even when they move at relativistic speeds. That
was merely the point I wanted to make for the gravity people.
There's been a lot of talk that the reason for the lack of
aberration for the sun's static G field, as seen by the earth, is
because it is a property of space-time itself at the Earth's
position, blah, blah. But the static electric and magnetic
fields of the sun aren't subject to aberration, either, according
to Maxwell (and of course, Einstein). Does this also mean they
too are properties of "space itself"? Your call.
I don't know for sure whether GR makes a similar prediction
about correction for aberration in the far-away observed static G
Joe Fischer
:[snip things other than gravity]
: I don't know for sure whether GR makes a similar prediction
: about correction for aberration in the far-away observed static G
: fields of a particle which moves at constant velocity (ie, one that
: predicts that this should be good, but not perfect, except for circular
: or orbital motion). Anybody out there good enough with the tensor
: analysis to be able to say?
Start all over please, gravitation is not
a propagated field, it is a geometric field (pseudo-field).
As long as gravity is considered to be an attractive
"force" caused by particle propagation, knowledge is at
a dead end!
Newtonian gravitation _is_ a model based on
attraction, but not necessarily caused by propagated
particles.
General Relativity is not based on attraction,
it describes relative accelerations, and gravitational
"force" is merely assumed from the accelerations and
the masses of the accelerated objects.
It is very important for students to learn
Newtonian gravitation, both because it is extremely
useful due to the simplicity, and because it was
designed to work with Newtonian mechanics, and
not everybody is going to be a gravity theorist. :-)
But a good textbook will say what lies
beyond it's scope, even if all the answers are
not known. The continued perception of gravity
as "attractive", and caused by propagated particles
results from people thinking about gravity at the
high school level, expecting to find some profound
answer to the riddle of gravitation.
After Newtonian gravitation is learned well,
and General Relativity is understood to be geometry
od space-time, then thinking takes a completely
different direction, away from attraction, and
toward geometry of relative accelerations predicted
by 4-D space-time geometry.
What is left to do is understand how space
and time dimensions can, or would, change in a way
to convince surface dwellers that they are "attracted".
Regards,
Joe Fischer
Steven B. Harris
>
>Steven B. Harris (sbha...@ix.netcom.com) wrote:
>:[snip things other than gravity]
>: about correction for aberration in the far-away observed static G
>: fields of a particle which moves at constant velocity (ie, one that
>: predicts that this should be good, but not perfect, except for
circular
>: or orbital motion). Anybody out there good enough with the tensor
>: analysis to be able to say?
>
Try it one somebody who is easier to B.S., Joe. Gravity is field, it
is a field which produces waves which can carry information, and these
waves propagate at c, carry linear momentum, energy, angular momentum,
and are expected to have a spin of 2. There is no reason why a quantum
field theory of gravity, just as with any attractive force, cannot be
developed.
Francis Rey
Stephen Harris wrote
------snip----
Try it one somebody who is easier to B.S., Joe. Gravity is field, it
is a field which produces waves which can carry information, and these
waves propagate at c, carry linear momentum, energy, angular momentum,
and are expected to have a spin of 2. There is no reason why a quantum
field theory of gravity, just as with any attractive force, cannot be
developed.
Francis Rey
What a strong self assurance, and strong knowledge. Necessarily Stephen is
able to tell us where have been done the experiments which prove its
statements, and the data obtained. Words without verified numerical values
is not enough.
--
francis Rey
The fact much to learn don't teach intelligence
Heraclite of Ephesus.
Jason Carreiro
: Francis Rey writes
: Stephen Harris wrote
: ------snip----
: Try it one somebody who is easier to B.S., Joe. Gravity is field, it
: is a field which produces waves which can carry information, and these
: waves propagate at c, carry linear momentum, energy, angular momentum,
: and are expected to have a spin of 2. There is no reason why a quantum
: field theory of gravity, just as with any attractive force, cannot be
: developed.
: Francis Rey
: What a strong self assurance, and strong knowledge. Necessarily Stephen is
: able to tell us where have been done the experiments which prove its
: statements, and the data obtained. Words without verified numerical values
: is not enough.
: --
: francis Rey
Try looking at:
Which is the home page of the LIGO experiment at caltech. You can also
look at:
http://www.ligo.caltech.edu/LIGO_web/other_gw/gw_projects.html
Which is a page with links to various other projects to detect gravitational
waves, which may have originated from:
violent events in the distant universe, for example by the collision
of two black holes or by the cores of supernova [1]
In addition, you may find the following interesting:
influence of gravitational waves on a binary pulsar (two neutron
stars orbiting each other) has been measured accurately [1]
Jason M. Carreiro "Let me control a planet's oxygen supply and I
<mer...@bu.edu> don't care who makes the laws."
http://www.cs.bu.edu/staff/TA/merlinz/ -Great Cthulu's Starry Wisdom Band
[1] http://www.ligo.caltech.edu/LIGO_web/about/factsheet.html
Francis Rey
Francis Rey writes
Josian Carriego wrote
: Stephen Harris wrote
: ------snip----
: Francis Rey
Try looking at:
http://www.ligo.caltech.edu/LIGO_web/other_gw/gw_projects.html
Francis Rey
I have not paid attention to the fact the post of Stephen Harris was cross
posted to sci.physics, and astro.
Carriego is answering in place of Stephen, but he is not impressive.
Perhaps is I have time I will look at LIGO some day.
I certainly does not lose time to read about the fictitious dreams of some
astronomers. I have read the Stephen Hawking divagations and it is enough.
I will read when astronomers will be knowing the difference between MASS
and DENSITY. To understand that it is necessary to be physicist, what are
not astronomers.
The influence of gravitational waves would be very interesting if it
existed such waves.
In old times it was said :
"it is easy to lie to who is coming from far country".
Now we can replace the end by :
" to who says he is looking far in the sky ""
The best astronomer joke I likes is the "Velocity of Gravity".
francis Rey
The fact much to learn don't teach intelligence
Heraclite of Ephesus.
franc...@wanadoo.fr
http://perso.wanadoo.fr/francis.rey/
Joe Fischer
: : Try it one somebody who is easier to B.S., Joe. Gravity is field, it
: : is a field which produces waves which can carry information, and these
: : waves propagate at c, carry linear momentum, energy, angular momentum,
: : and are expected to have a spin of 2. There is no reason why a quantum
: : field theory of gravity, just as with any attractive force, cannot be
: : developed.
I am afraid this is a jumble of speculations
taken from several different theories or models of
gravitation, and _none_ of it is factual.
In Newtonian gravitation, as used by most
astronomers who do not really need General Relativity,
there is a relative acceleration supposedly caused
by a universal mutual attraction according to Newtonian
formulae.
But there are no experiments that show this
apparent attraction to be the result of a "field",
and absolutely nothing to suggest what such a
"field" would be composed of.
_ANY_ "attractive force" would be contrary
to General Relativity, and could not truly represent
gravitation, which is different from all other
processes.
: Try looking at:
:
: http://www.ligo.caltech.edu/
:
: Which is the home page of the LIGO experiment at caltech.
I did (again), and thank you very much,
the progress of LIGO is very interesting.
: You can also
: look at:
: http://www.ligo.caltech.edu/LIGO_web/other_gw/gw_projects.html
: Which is a page with links to various other projects to detect
: gravitational
: waves, which may have originated from:
:
: violent events in the distant universe, for example by the collision
: of two black holes or by the cores of supernova [1]
Thanks again, but I am aware of the history
of gravitational rradiation, having read every paper
I could find on the subject beginning with the prize
winning paper by Joe Weber in 1958.
: In addition, you may find the following interesting:
:
: influence of gravitational waves on a binary pulsar (two neutron
: stars orbiting each other) has been measured accurately [1]
Interesting, but it will take positive results
from LIGO to add real meaning to the subject, as it
is still not certain gravity waves exist.
Regards,
Joe Fischer