Bill Hobba wrote:
"shevek" wrote in message
oups.com...

shevek wrote:
Bill Hobba wrote:
I have given the following many times -
http://www.mathpages.com/home/kmath528/kmath528.htm. If
accelerating
charges radiate would seem an open question. My understanding
is
it
is not
an open question if an accelerating mass radiates gravitational
waves
- GR
predicts it does - or does it? Do others have any thoughts?

Thanks
Bill.

Hi Bill.

I was fortunate to attend a talk recently about uniformly
acclerated
charges.


Also, I should mention many references about the topic are
available in
the slides from the talk:

http://plasma4.sr.unh.edu/ng/uac-unh.pdf

cheers -

Thanks for the links - great stuff. Now for my two cents worth - I
do not
think a uniformly accelerated charge radiates because of the EEP.
The usual
explanation of why charges on earth do not radiate (a charge is an
extended
thing including the field which due to tidal forces is not the same
as a
uniformly accelerated charge) I do not agree with.

Are you talking about because of gravitational acceleration? An object
sitting on the surface of the earth is not accelerating downwards at
9.8m/s/s.. that force is balanced by frictional electric forces from
the rock pushing up. Some more examples of accelerated charges
radiating: a fluorescent light, a cell phone antenna, synchrotron,
bremstrahlung..

Feynman and Wheeler
demonstrated that fields are not required to formulate classical EM -
thus
the answer looks unreasonable - at least to me. I suspect the true
resolution lies in QED.

I disagree. Do you mean there is no field, because the force is
carried by an exchange of mesons (in this case photons)? First, one
can argue that there still is a field - just defined as the force on a
particle divided by the charge. Second, how is a proton attracted to
an antiproton by a photon exchange? That mechanism explains repulsive
forces but not attractive.

shevek wrote:

I disagree. Do you mean there is no field, because the force is
carried by an exchange of mesons (in this case photons)?
First, one
can argue that there still is a field - just defined as the force on
a
particle divided by the charge. Second, how is a proton attracted to
an antiproton by a photon exchange?

From the perspective of GR the Energy Density,
is the product of two E-fields from charges "a"
and "b" i.e.,

T_00 = E(a)*E(b)

T_00 = - E^2 ,(attractive a=-b)

T_00 = + E^2 ,(repulsive a= +b)

These in turn set-up the G_00, (G_uv=T_uv) and
the resulting g_uv and from them the geodesics.

Energy and therefore Energy Density varies dis-
continuosly, hence the E-fields (E^2 above) also
vary discontinuously, that would be by photons
emitted by the pairing of protons and
anti-protons.

That mechanism explains repulsive
forces but not attractive.

Virtual photon exchange is a conceptualiztion
of mathematics. If you want you are entitled
to think in terms of the annilihation of
virtual antiphotons and photons, to create the
-E^2 above, and by doing so emit a photon.
I wouldn't argue with that.

Regards
Ken S. Tucker

"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox wrote in
message news:qRsZd.5192$uk7.4852@fed1read01...
Dear Bill Hobba:

"Bill Hobba" wrote in message
...

"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox
wrote in
message news:JTqZd.5164$uk7.3057@fed1read01...
Dear Penelope:

"Penelope" wrote in message
...
On Mon, 14 Mar 2005 "Bill Hobba"
wrote:
I have given the following many times -
URL:http://www.mathpages.com/home/kmath528/kmath528.htm
If accelerating charges radiate would seem an
open question. My understanding is it is not
an open question if an accelerating mass
. radiates gravitational waves - GR predicts
it does - or does it? Do others have any
thoughts?

The question discussed on that web page
is whether a UNIFORMLY accelerating
charge radiates. General relativity does not
predict that a uniformly accelerating mass
radiates. In fact, there can't be any such
thing as a simple uniformly accelerating
mass... what would be causing it to
accelerate?

Sitting on the surface of a planet, a test
body is being uniformly accelerated from its
free-fall elliptical orbit. And electrostatic
force is what is providing the acceleration.
Does this qualify?

AFAICS David it does not qualify. I should
have seen it from the beginning as well.
From the EEP if a uniformly accelerated mass
radiated then we would, theoretically, have a
free source of energy for particles at rest on
earth. When viewed from a frame in free fall
it would be acclerating so should radiate.

Frame dragging. "Boosting" of photons (either red or blue shift)
past a spinning object (which was only proposed, not observed).
I think you may have too narrow an expectation of "radiation".
And it wouldn't be "free energy", since conservation of momentum
would require that-which-spins to spin a little more slowly. And
I also see that I have the object spinning, which is not what you
had stated.

It obviously does not (or you would have a
free source of energy) so unformily accelerated
masses do not radiate grvitaitonal waves.

Waves, perhaps not. Energy, I believe so. Note anomalous boosts
of satellites (during slingshot maneuvers), and frame dragging.

This is the same argument the mathpage link
I gave about uniformly accelrated charges
radiating used -

'One of the most familiar propositions of
elementary classical electrodynamics is that
"an accelerating charge radiates". In fact, the
power (energy per time) of electromagnetic
radiation emitted by a charged particle is
often said to be strictly a function of the
acceleration of that particle. However, if we
accept the strong Equivalence Principle (i.e., the
equivalence between gravity and acceleration),
the simple idea that radiation is a function of
acceleration becomes problematic, because in
this context an object can be both stationary
and accelerating. For example, a charged
object at rest on the Earth's surface is
stationary, and yet it's also subject to a
(gravitational) acceleration of about
9.8 m/sec2. It seems safe to say (and it is
evidently a matter of fact) that such an object
does not radiate electromagnetic energy, at
least from the point of view of co-stationary
observers. If it did, we would have a perpetual
source of free energy. Since the upward force
holding the object in place at the Earth's
surface does not act through any distance,
the work done by this force is zero. Therefore,
no energy is being put into the object, so if the
object is radiating electromagnetic energy
(and assuming the internal energy of the
object remains constant) we have a violation
of energy conservation.'

Let me say this about that... ;)
Co-stationary observers perceive a different charge than an
observer that is free falling past the charge (say on the
charge's geodesic without the support of the Earth). The
difference in energy (not charge) would be the magnetic field
created by this current. I'm probably wrong, but I don't see a
conservation of energy problem. If we had access to the geodesic
that any particular sacrificial mass would normally follow, yes
we could get "free" energy... but we'd eventually fill the well.

I'm ready for my trouncing.

Remember the caveat - 'and assuming the internal energy of the object
remains constant'. If you have a process how it could tap into that then
you have an out. Also, as the mathpage link I gave explained, the issue is
controversial.

Thanks
Bill


David A. Smith

"Tom Roberts" wrote in message
om...
Bill Hobba wrote:
My understanding is it is not
an open question if an accelerating mass radiates gravitational waves -
GR
predicts it does - or does it? Do others have any thoughts?

First you must define what you mean by "accelerating" -- in Minkowski
spacetime an inertially-moving mass can appear to be "accelerating" if
one describes its postion using non-inertial coordinates. As GR is
coordinate independent, clearly one must not attempt to use such a
coordinate-dependent meaning of "accelerate".

So it seems the only useful meaning is: having a nonzero 4-acceleration.

But a test particle orbiting a large mass has zero 4-acceleration, and
yet we know that two large masses orbiting each other will definitely
generate gravitational radiation. So zero 4-acceleration does not
preclude gravitational radiation (insofar as 4-aceleration makes sense
for a non-negligible mass object).

Good point.


A spherical shell pushed symmetrically by many rockets into its center
could have non-zero 4-acceleration for each point of its surface, and
yet no gravitational radiation can be emitted (due to the spherical
symmetry -- Birkhof's theorem applies). So nonzero 4-acceleration does
not ensure gravitational radiation.

Conclusion: no simple but generally-valid statement can be made.

Yep.

Thanks
Bill



Tom Roberts

Ken S. Tucker wrote:
shevek wrote:

I disagree. Do you mean there is no field, because the force is
carried by an exchange of mesons (in this case photons)?
First, one
can argue that there still is a field - just defined as the force
on
a
particle divided by the charge. Second, how is a proton attracted
to
an antiproton by a photon exchange?

From the perspective of GR the Energy Density,
is the product of two E-fields from charges "a"
and "b" i.e.,

T_00 = E(a)*E(b)

T_00 = - E^2 ,(attractive a=-b)

T_00 = + E^2 ,(repulsive a= +b)

These in turn set-up the G_00, (G_uv=T_uv) and
the resulting g_uv and from them the geodesics.

Energy and therefore Energy Density varies dis-
continuosly, hence the E-fields (E^2 above) also
vary discontinuously, that would be by photons
emitted by the pairing of protons and
anti-protons.


Which direction are the photons emitted - if they are to be
"exchanged", then the photons are emitted toward the other particle..
in which case the force is repulsive. If the (virtual) photons are
emitted away from the opposite particle, than what force caused them to
be emitted? We are back to a field again.

That mechanism explains repulsive
forces but not attractive.

Virtual photon exchange is a conceptualiztion
of mathematics. If you want you are entitled
to think in terms of the annilihation of
virtual antiphotons and photons, to create the
-E^2 above, and by doing so emit a photon.
I wouldn't argue with that.


Thanks Ken, well that's an idea.. I don't know QED actually, so I was
looking for a definitive answer. All the quantum field theorists I
talk to give a different answer - one popular one is that the virtual
photon being exchanged has a momentum facing the other way from it's
wavevector! Which doesn't make sense to me. Anyway, a photon is it's
own antiparticle - and photons move through each other without effect
in a linear medium.

shevek wrote:
Ken S. Tucker wrote:
shevek wrote:

I disagree. Do you mean there is no field, because the force is
carried by an exchange of mesons (in this case photons)?
First, one
can argue that there still is a field - just defined as the force
on
a
particle divided by the charge. Second, how is a proton
attracted
to
an antiproton by a photon exchange?

From the perspective of GR the Energy Density,
is the product of two E-fields from charges "a"
and "b" i.e.,

T_00 = E(a)*E(b)

T_00 = - E^2 ,(attractive a=-b)

T_00 = + E^2 ,(repulsive a= +b)

These in turn set-up the G_00, (G_uv=T_uv) and
the resulting g_uv and from them the geodesics.

Energy and therefore Energy Density varies dis-
continuosly, hence the E-fields (E^2 above) also
vary discontinuously, that would be by photons
emitted by the pairing of protons and
anti-protons.


Which direction are the photons emitted - if they are to be
"exchanged", then the photons are emitted toward the other particle..
in which case the force is repulsive. If the (virtual) photons are
emitted away from the opposite particle, than what force caused them
to
be emitted? We are back to a field again.

Ok, but the field itself is physically caused
by a relative relation of charges. Above I
provided a GR solution for a charge couple
"a" and "b".
All the charges in the Earth and Moon are
electro-magnetically coupled. Assuming neither
is electrostatically charged then for every
repelling couple there is an attractive couple.
Do the arithmetric on that, it quite neat.

Repelling Couples = Attractive Couples
when gravitating bodies are neutral.

Incidentally, gravitation results because
attraction is slightly stronger than repulsion,
between charges, that we call a gravitational
field.

The field doesn't really have an independent
existance, it's a simplification of the force
equation by setting the test particle to a
unit mass (F =GMm/r^2 = GM/r^2 when m=1),
likewise for EM-force.

That mechanism explains repulsive
forces but not attractive.

Virtual photon exchange is a conceptualiztion
of mathematics. If you want you are entitled
to think in terms of the annilihation of
virtual antiphotons and photons, to create the
-E^2 above, and by doing so emit a photon.
I wouldn't argue with that.


Thanks Ken, well that's an idea.. I don't know QED actually, so I
was
looking for a definitive answer. All the quantum field theorists I
talk to give a different answer - one popular one is that the virtual
photon being exchanged has a momentum facing the other way from it's
wavevector! Which doesn't make sense to me. Anyway, a photon is
it's
own antiparticle - and photons move through each other without effect
in a linear medium.

Set E(a) to the Electric field of "a" etc,
GR provides this solution for a pair of
naked charges "a" and "b",

G_00 = E(a)*E(b)= T_00

On the LHS is the continuum, and
on the RHS is the quantum field.

((BTW use g_00 = 1 - a*b/r^2, and Nabla^2(a/r)=0.))

Is that the resolution we care for.
Thanks Shevek
Ken S. Tucker

"Ken S. Tucker" wrote in message oups.com...
shevek wrote:
Ken S. Tucker wrote:
shevek wrote:

I disagree. Do you mean there is no field, because the force is
carried by an exchange of mesons (in this case photons)?
First, one
can argue that there still is a field - just defined as the force
on
a
particle divided by the charge. Second, how is a proton
attracted
to
an antiproton by a photon exchange?

From the perspective of GR the Energy Density,
is the product of two E-fields from charges "a"
and "b" i.e.,

T_00 = E(a)*E(b)

T_00 = - E^2 ,(attractive a=-b)

T_00 = + E^2 ,(repulsive a= +b)

These in turn set-up the G_00, (G_uv=T_uv) and
the resulting g_uv and from them the geodesics.

Energy and therefore Energy Density varies dis-
continuosly, hence the E-fields (E^2 above) also
vary discontinuously, that would be by photons
emitted by the pairing of protons and
anti-protons.


Which direction are the photons emitted - if they are to be
"exchanged", then the photons are emitted toward the other particle..
in which case the force is repulsive. If the (virtual) photons are
emitted away from the opposite particle, than what force caused them
to
be emitted? We are back to a field again.

Ok, but the field itself is physically caused
by a relative relation of charges. Above I
provided a GR solution for a charge couple
"a" and "b".
All the charges in the Earth and Moon are
electro-magnetically coupled. Assuming neither
is electrostatically charged then for every
repelling couple there is an attractive couple.
Do the arithmetric on that, it quite neat.

Repelling Couples = Attractive Couples
when gravitating bodies are neutral.

Incidentally, gravitation results because
attraction is slightly stronger than repulsion,
between charges, that we call a gravitational
field.

The field doesn't really have an independent
existance, it's a simplification of the force
equation by setting the test particle to a
unit mass (F =GMm/r^2 = GM/r^2 when m=1),
likewise for EM-force.

That is an interesting notion. Two plausible
mechanisms come to mind that had never occured
to me before.

Either as a result of deformation or displacement,
the attractive components could *effectively* have
a path just slighly shorter than their repulsive partners.

That's tooooo simple. But just offhand I can't
offer argument. Maybe later LOL.

Sue...


That mechanism explains repulsive
forces but not attractive.

Virtual photon exchange is a conceptualiztion
of mathematics. If you want you are entitled
to think in terms of the annilihation of
virtual antiphotons and photons, to create the
-E^2 above, and by doing so emit a photon.
I wouldn't argue with that.


Thanks Ken, well that's an idea.. I don't know QED actually, so I
was
looking for a definitive answer. All the quantum field theorists I
talk to give a different answer - one popular one is that the virtual
photon being exchanged has a momentum facing the other way from it's
wavevector! Which doesn't make sense to me. Anyway, a photon is
it's
own antiparticle - and photons move through each other without effect
in a linear medium.

Set E(a) to the Electric field of "a" etc,
GR provides this solution for a pair of
naked charges "a" and "b",

G_00 = E(a)*E(b)= T_00

On the LHS is the continuum, and
on the RHS is the quantum field.

((BTW use g_00 = 1 - a*b/r^2, and Nabla^2(a/r)=0.))

Is that the resolution we care for.
Thanks Shevek
Ken S. Tucker

jahn wrote:
"Ken S. Tucker" wrote in message
oups.com...
shevek wrote:
Ken S. Tucker wrote:
shevek wrote:

I disagree. Do you mean there is no field, because the force
is
carried by an exchange of mesons (in this case photons)?
First, one
can argue that there still is a field - just defined as the
force
on
a
particle divided by the charge. Second, how is a proton
attracted
to
an antiproton by a photon exchange?

From the perspective of GR the Energy Density,
is the product of two E-fields from charges "a"
and "b" i.e.,

T_00 = E(a)*E(b)

T_00 = - E^2 ,(attractive a=-b)

T_00 = + E^2 ,(repulsive a= +b)

These in turn set-up the G_00, (G_uv=T_uv) and
the resulting g_uv and from them the geodesics.

Energy and therefore Energy Density varies dis-
continuosly, hence the E-fields (E^2 above) also
vary discontinuously, that would be by photons
emitted by the pairing of protons and
anti-protons.


Which direction are the photons emitted - if they are to be
"exchanged", then the photons are emitted toward the other
particle..
in which case the force is repulsive. If the (virtual) photons
are
emitted away from the opposite particle, than what force caused
them
to
be emitted? We are back to a field again.

Ok, but the field itself is physically caused
by a relative relation of charges. Above I
provided a GR solution for a charge couple
"a" and "b".
All the charges in the Earth and Moon are
electro-magnetically coupled. Assuming neither
is electrostatically charged then for every
repelling couple there is an attractive couple.
Do the arithmetric on that, it quite neat.

Repelling Couples = Attractive Couples
when gravitating bodies are neutral.

Incidentally, gravitation results because
attraction is slightly stronger than repulsion,
between charges, that we call a gravitational
field.

The field doesn't really have an independent
existance, it's a simplification of the force
equation by setting the test particle to a
unit mass (F =GMm/r^2 = GM/r^2 when m=1),
likewise for EM-force.

That is an interesting notion. Two plausible
mechanisms come to mind that had never occured
to me before.

Either as a result of deformation or displacement,
the attractive components could *effectively* have
a path just slighly shorter than their repulsive partners.

That's tooooo simple. But just offhand I can't
offer argument. Maybe later LOL.

Sue...

Well Sue, I think GR is inclined to agree
with you. The simple formula is,

S^2 = X^2 + a*b (1)

where S=Signal distance, X is fixed, then,

S^2 = X^2 - q^2 (when a=-b, Attractive)

S^2 = X^2 + q^2 (when a=b, Repulsive).

If you would like an explanation, I figure,
it's because "repelling" charges store
positive energy in the field they create but
"attractive" store negative energy.

Further analogy, it takes light "longer"
to go threw air or water or glass, than
a vacuum, each medium has positive energy.

The thing about GR is one can use a dozen
analogies and each is right, and Sue's
deformation analogy is as good as any.

I liken the way SR unified electrostatic
forces and magnetism, to the way GR
unifies electrostatics and gravitation.

Back in the 70's I met with Ray Vanstone
of the U of T about unification of EM and
gravitation. He pointed out that it was
done at least implicitly, by G_uv=T_uv,
((he used his own text BTW)).

I haven't found an explicit solution,
aside from the one below, (I call it
unitivity because in classical GR charge
is not a length).

By unitizing charge to a be length, and an
invariant one at that, we're able to use
Eq.(1) sensibly.

If you figure it out you'll find (I hope!),
charge^2 = action and action = length^2
when you convert mass to length as GR allows.

Because charge is invariant, it is orthogonal
to spacetime, so in Eq.(1) the charges are
orthogonal to X. (Recall lengths perpendicular
to motion are invariant).

The unitivity solution needs to be consistent
with QT, and I've argued GR predicts QT recently
with someone called "Non ame".

That mechanism explains repulsive
forc
es but not attractive.

Virtual photon exchange is a conceptualiztion
of mathematics. If you want you are entitled
to think in terms of the annilihation of
virtual antiphotons and photons, to create the
-E^2 above, and by doing so emit a photon.
I wouldn't argue with that.


Thanks Ken, well that's an idea.. I don't know QED actually, so
I
was
looking for a definitive answer. All the quantum field theorists
I
talk to give a different answer - one popular one is that the
virtual
photon being exchanged has a momentum facing the other way from
it's
wavevector! Which doesn't make sense to me. Anyway, a photon is
it's
own antiparticle - and photons move through each other without
effect
in a linear medium.

Set E(a) to the Electric field of "a" etc,
GR provides this solution for a pair of
naked charges "a" and "b",

G_00 = E(a)*E(b)= T_00

On the LHS is the continuum, and
on the RHS is the quantum field.

((BTW use g_00 = 1 - a*b/r^2, and Nabla^2(a/r)=0.))

Is that the resolution we care for.
Thanks Shevek
Ken S. Tucker

Regards
Ken S. Tucker

"Ken S. Tucker" wrote in message ups.com...
jahn wrote:
"Ken S. Tucker" wrote in message
oups.com...
shevek wrote:
Ken S. Tucker wrote:
shevek wrote:

I disagree. Do you mean there is no field, because the force
is
carried by an exchange of mesons (in this case photons)?
First, one
can argue that there still is a field - just defined as the
force
on
a
particle divided by the charge. Second, how is a proton
attracted
to
an antiproton by a photon exchange?

From the perspective of GR the Energy Density,
is the product of two E-fields from charges "a"
and "b" i.e.,

T_00 = E(a)*E(b)

T_00 = - E^2 ,(attractive a=-b)

T_00 = + E^2 ,(repulsive a= +b)

These in turn set-up the G_00, (G_uv=T_uv) and
the resulting g_uv and from them the geodesics.

Energy and therefore Energy Density varies dis-
continuosly, hence the E-fields (E^2 above) also
vary discontinuously, that would be by photons
emitted by the pairing of protons and
anti-protons.


Which direction are the photons emitted - if they are to be
"exchanged", then the photons are emitted toward the other
particle..
in which case the force is repulsive. If the (virtual) photons
are
emitted away from the opposite particle, than what force caused
them
to
be emitted? We are back to a field again.

Ok, but the field itself is physically caused
by a relative relation of charges. Above I
provided a GR solution for a charge couple
"a" and "b".
All the charges in the Earth and Moon are
electro-magnetically coupled. Assuming neither
is electrostatically charged then for every
repelling couple there is an attractive couple.
Do the arithmetric on that, it quite neat.

Repelling Couples = Attractive Couples
when gravitating bodies are neutral.

Incidentally, gravitation results because
attraction is slightly stronger than repulsion,
between charges, that we call a gravitational
field.

The field doesn't really have an independent
existance, it's a simplification of the force
equation by setting the test particle to a
unit mass (F =GMm/r^2 = GM/r^2 when m=1),
likewise for EM-force.

That is an interesting notion. Two plausible
mechanisms come to mind that had never occured
to me before.

Either as a result of deformation or displacement,
the attractive components could *effectively* have
a path just slighly shorter than their repulsive partners.

That's tooooo simple. But just offhand I can't
offer argument. Maybe later LOL.

Sue...

Well Sue, I think GR is inclined to agree
with you. The simple formula is,

What? They must be offering a prize for GR
salesperson of the month. It's a great technique
tho' and I'll put a good word in for ya if Al's Used
Car Lot calls for a reference. )


S^2 = X^2 + a*b (1)

where S=Signal distance, X is fixed, then,

S^2 = X^2 - q^2 (when a=-b, Attractive)

S^2 = X^2 + q^2 (when a=b, Repulsive).

If you would like an explanation, I figure,
it's because "repelling" charges store
positive energy in the field they create but
"attractive" store negative energy.

Further analogy, it takes light "longer"
to go threw air or water or glass, than
a vacuum, each medium has positive energy.

OK ... I'll add those to the list.



The thing about GR is one can use a dozen
analogies and each is right, and Sue's
deformation analogy is as good as any.

I liken the way SR unified electrostatic
forces and magnetism, to the way GR
unifies electrostatics and gravitation.

Back in the 70's I met with Ray Vanstone
of the U of T about unification of EM and
gravitation. He pointed out that it was
done at least implicitly, by G_uv=T_uv,
((he used his own text BTW)).

I haven't found an explicit solution,
aside from the one below, (I call it
unitivity because in classical GR charge
is not a length).

Probably only because it is a bit harder to
express.


By unitizing charge to a be length, and an
invariant one at that, we're able to use
Eq.(1) sensibly.

If you figure it out you'll find (I hope!),
charge^2 = action and action = length^2
when you convert mass to length as GR allows.

Because charge is invariant, it is orthogonal
to spacetime, so in Eq.(1) the charges are
orthogonal to X. (Recall lengths perpendicular
to motion are invariant).

OK ... I'll let ya get away with that since you
have declared your clocks to be funny. LOL


The unitivity solution needs to be consistent
with QT, and I've argued GR predicts QT recently
with someone called "Non ame".

OK... I'll dump my tinker-toys out this weekend
and see if we can include or exclude some of these
mechanisms by some other means.

Sue...



That mechanism explains repulsive
forc
es but not attractive.

Virtual photon exchange is a conceptualiztion
of mathematics. If you want you are entitled
to think in terms of the annilihation of
virtual antiphotons and photons, to create the
-E^2 above, and by doing so emit a photon.
I wouldn't argue with that.


Thanks Ken, well that's an idea.. I don't know QED actually, so
I
was
looking for a definitive answer. All the quantum field theorists
I
talk to give a different answer - one popular one is that the
virtual
photon being exchanged has a momentum facing the other way from
it's
wavevector! Which doesn't make sense to me. Anyway, a photon is
it's
own antiparticle - and photons move through each other without
effect
in a linear medium.

Set E(a) to the Electric field of "a" etc,
GR provides this solution for a pair of
naked charges "a" and "b",

G_00 = E(a)*E(b)= T_00

On the LHS is the continuum, and
on the RHS is the quantum field.

((BTW use g_00 = 1 - a*b/r^2, and Nabla^2(a/r)=0.))

Is that the resolution we care for.
Thanks Shevek
Ken S. Tucker

Regards
Ken S. Tucker

jahn wrote:
"Ken S. Tucker" wrote in message
ups.com...
jahn wrote:
"Ken S. Tucker" wrote in message
oups.com...
shevek wrote:
Ken S. Tucker wrote:
shevek wrote:

I disagree. Do you mean there is no field, because the
force
is
carried by an exchange of mesons (in this case photons)?
First, one
can argue that there still is a field - just defined as
the
force
on
a
particle divided by the charge. Second, how is a proton
attracted
to
an antiproton by a photon exchange?

From the perspective of GR the Energy Density,
is the product of two E-fields from charges "a"
and "b" i.e.,

T_00 = E(a)*E(b)

T_00 = - E^2 ,(attractive a=-b)

T_00 = + E^2 ,(repulsive a= +b)

These in turn set-up the G_00, (G_uv=T_uv) and
the resulting g_uv and from them the geodesics.

Energy and therefore Energy Density varies dis-
continuosly, hence the E-fields (E^2 above) also
vary discontinuously, that would be by photons
emitted by the pairing of protons and
anti-protons.


Which direction are the photons emitted - if they are to be
"exchanged", then the photons are emitted toward the other
particle..
in which case the force is repulsive. If the (virtual)
photons
are
emitted away from the opposite particle, than what force
caused
them
to
be emitted? We are back to a field again.

Ok, but the field itself is physically caused
by a relative relation of charges. Above I
provided a GR solution for a charge couple
"a" and "b".
All the charges in the Earth and Moon are
electro-magnetically coupled. Assuming neither
is electrostatically charged then for every
repelling couple there is an attractive couple.
Do the arithmetric on that, it quite neat.

Repelling Couples = Attractive Couples
when gravitating bodies are neutral.

Incidentally, gravitation results because
attraction is slightly stronger than repulsion,
between charges, that we call a gravitational
field.

The field doesn't really have an independent
existance, it's a simplification of the force
equation by setting the test particle to a
unit mass (F =GMm/r^2 = GM/r^2 when m=1),
likewise for EM-force.

That is an interesting notion. Two plausible
mechanisms come to mind that had never occured
to me before.

Either as a result of deformation or displacement,
the attractive components could *effectively* have
a path just slighly shorter than their repulsive partners.

That's tooooo simple. But just offhand I can't
offer argument. Maybe later LOL.

Sue...

Well Sue, I think GR is inclined to agree
with you. The simple formula is,

What? They must be offering a prize for GR
salesperson of the month. It's a great technique
tho' and I'll put a good word in for ya if Al's Used
Car Lot calls for a reference. )

LOL, classical Grist's have dumped me,
I couldn't get g-waves for LIGO or
gravitomagnetism for GP-b.

S^2 = X^2 + a*b (1)

where S=Signal distance, X is fixed, then,

S^2 = X^2 - q^2 (when a=-b, Attractive)

S^2 = X^2 + q^2 (when a=b, Repulsive).

If you would like an explanation, I figure,
it's because "repelling" charges store
positive energy in the field they create but
"attractive" store negative energy.

Further analogy, it takes light "longer"
to go threw air or water or glass, than
a vacuum, each medium has positive energy.

OK ... I'll add those to the list.

The thing about GR is one can use a dozen
analogies and each is right, and Sue's
deformation analogy is as good as any.

I liken the way SR unified electrostatic
forces and magnetism, to the way GR
unifies electrostatics and gravitation.

Back in the 70's I met with Ray Vanstone
of the U of T about unification of EM and
gravitation. He pointed out that it was
done at least implicitly, by G_uv=T_uv,
((he used his own text BTW)).

I haven't found an explicit solution,
aside from the one below, (I call it
unitivity because in classical GR charge
is not a length).

Probably only because it is a bit harder to
express.


By unitizing charge to a be length, and an
invariant one at that, we're able to use
Eq.(1) sensibly.

If you figure it out you'll find (I hope!),
charge^2 = action and action = length^2
when you convert mass to length as GR allows.

Because charge is invariant, it is orthogonal
to spacetime, so in Eq.(1) the charges are
orthogonal to X. (Recall lengths perpendicular
to motion are invariant).

OK ... I'll let ya get away with that since you
have declared your clocks to be funny. LOL

The unitivity solution needs to be consistent
with QT, and I've argued GR predicts QT recently
with someone called "Non ame".

OK... I'll dump my tinker-toys out this weekend
and see if we can include or exclude some of these
mechanisms by some other means.

Sue...
Ken

That mechanism explains repulsive
forc
es but not attractive.

Virtual photon exchange is a conceptualiztion
of mathematics. If you want you are entitled
to think in terms of the annilihation of
virtual antiphotons and photons, to create the
-E^2 above, and by doing so emit a photon.
I wouldn't argue with that.


Thanks Ken, well that's an idea.. I don't know QED actually,
so
I
was
looking for a definitive answer. All the quantum field
theorists
I
talk to give a different answer - one popular one is that the
virtual
photon being exchanged has a momentum facing the other way
from
it's
wavevector! Which doesn't make sense to me. Anyway, a
photon is
it's
own antiparticle - and photons move through each other
without
effect
in a linear medium.

Set E(a) to the Electric field of "a" etc,
GR provides this solution for a pair of
naked charges "a" and "b",

G_00 = E(a)*E(b)= T_00

On the LHS is the continuum, and
on the RHS is the quantum field.

((BTW use g_00 = 1 - a*b/r^2, and Nabla^2(a/r)=0.))

Is that the resolution we care for.
Thanks Shevek
Ken S. Tucker

Regards
Ken S. Tucker