Another problem with Maxwells theory
John Schoenfeld
charge radiates. Since energy is conserved, this radiative emission
occurs at the expense of the particle's kinetic energy.
Therefore, under the influence of a separate force, a charged particle
accelerates LESS than a neutral particle of the same mass due to the
above effect.
Since the charged mass accelerated less but remained with constant
mass, one must draw the conclusion that the emitted radiation caused a
reaction force on the charged particle.
If you try to derive this recoil force caused by the radiation, you
end up with the formula:
Fr = (u0 q^2 /6 pi c) j
where
Fr = recoil force of radiation emission
u0 = permeability of free space
j = the rate of change of acceleration (the jerk)
c = light speed
m = mass or particle
From Newtons second law,
Frad = ma = (u0 q^2 / 6 pi c) j
From which it follows that
a(t) = a0 e^(t/mu)
where
mu = u0 q^2 / 6 pi m c
In the case of the eletron, mu = 6E-24.
Problem 1: the acceleration of the particle here would increase
exponentially with time, which is seemingly absurd.
Problem 2: If you insist that a0 = 0, then when you do apply an
external force, the particle begins to react BEFORE the external force
is applied (causality violation).
Maxwell was wrong.
JS
Ahmed Ouahi, Architect
............ ...There is no theory of the universe; events cannot be
predicted beyond a certain extent but occur in a random and arbitrary
manner!!!!!!!!!!!!!!!!.................... ...
--Stephen Hawking
--
Ahmed Ouahi, Architect
Best Regards!
"John Schoenfeld" <j.scho...@programmer.net> wrote in message
news:a98beaaa.04041...@posting.google.com...
Franz Heymann
"John Schoenfeld" <j.scho...@programmer.net> wrote in message
news:a98beaaa.04041...@posting.google.com...
[snip the shit]
Familiarise yourself with the radiation reaction term and stop spewing
crap.
Franz
Anonymous 2357
............ ...There is no theory of the universe; events
cannot be
> predicted beyond a certain extent but occur in a random and
arbitrary
> manner!!!!!!!!!!!!!!!!.................... ...
>
> --Stephen Hawking
>
>
> --
> Ahmed Ouahi, Architect
> Best Regards!
>
>
>
message
> news:a98beaaa.04041...@posting.google.com...
Do you always express yourself that way?
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Sam Wormley
>
> According to the laws of classical electrodynamics, an accelerating
> charge radiates. Since energy is conserved, this radiative emission
> occurs at the expense of the particle's kinetic energy.
>
Suggestion of self education
http://math.ucr.edu/home/baez/physics/Administrivia/booklist.html#quantum-mechanics
Usenet Physics FAQs
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Harold Ensle
"Franz Heymann" <notfranz...@btopenworld.com> wrote in message
news:c5gggc$ps0$1...@hercules.btinternet.com...
You are the one who needs an education. What he is talking about
is a well known paradox (look in Griffiths' Introduction to Electrodynamics
for example). The only _possible_ error in his post is to claim that
Maxwell's equations are wrong because of it. It could indeed mean
that Maxwell's equations need some kind of revision, though my
current take on the subject is that there is some non-general assumption
being made in the problem itself. It is an interesting problem and
is on my list of things to look into.
H.Ellis Ensle
Harold Ensle
"Sam Wormley" <swor...@mchsi.com> wrote in message
news:407BFB20...@mchsi.com...
His post, up to the last comment, is straight out of an E&M text,
apparently one you need to read.
H.Ellis Ensle
Uncle Al
>
> According to the laws of classical electrodynamics, an accelerating
> charge radiates. Since energy is conserved, this radiative emission
> occurs at the expense of the particle's kinetic energy.
>
> Therefore, under the influence of a separate force, a charged particle
> accelerates LESS than a neutral particle of the same mass due to the
> above effect.
How would you accelerate a neutral particle, jackass? A particle free
falling in a gravitational field is in Minkowski space and has no
in-frame acceleration at all. If you use an EM field to accelerate
it, Maxwell rules.
[snip muddled drivel]
> Maxwell was wrong.
You are an empirically self-demonstated voluble idiot.
Read Jackson for comprehension.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Robert J. Kolker
John Schoenfeld wrote:
> According to the laws of classical electrodynamics, an accelerating
> charge radiates. Since energy is conserved, this radiative emission
> occurs at the expense of the particle's kinetic energy.
>
> Therefore, under the influence of a separate force, a charged particle
> accelerates LESS than a neutral particle of the same mass due to the
> above effect.
Not if enough energy is applied to the particle to maintain its
accleration. That is why radio stations need electric power.
Bob Kolker
Bilge
>Problem 1: the acceleration of the particle here would increase
>exponentially with time, which is seemingly absurd.
>
>Problem 2: If you insist that a0 = 0, then when you do apply an
>external force, the particle begins to react BEFORE the external force
>is applied (causality violation).
See the chapter in jackson on radiation reaction.
Bilge
>"Franz Heymann" <notfranz...@btopenworld.com> wrote in message
>news:c5gggc$ps0$1...@hercules.btinternet.com...
>>
>> "John Schoenfeld" <j.scho...@programmer.net> wrote in message
>> news:a98beaaa.04041...@posting.google.com...
>>
>> [snip the shit]
>>
>> Familiarise yourself with the radiation reaction term and stop spewing
>> crap.
>>
>> Franz
>
>You are the one who needs an education. What he is talking about
>is a well known paradox (look in Griffiths' Introduction to Electrodynamics
>for example).
complete (non-quantum mechanical) discussion may be found in jackson,
who devotes a chapter to it.
>The only _possible_ error in his post is to claim that
>Maxwell's equations are wrong because of it. It could indeed mean
>that Maxwell's equations need some kind of revision, though my
>current take on the subject is that there is some non-general assumption
>being made in the problem itself. It is an interesting problem and
>is on my list of things to look into.
The ``non-general assumption'' is that charges can be treated classically.
The time scale at which any issue of causality occurs is on the order of
the time it takes light to cross the radius of the charged particle. Since
charged particles are known through experiment to be much smaller than
any classical model would allow, an analysis which is classical is likely
to make bizarre predictions when extrapolated beyond the regime in which
it is valid. To phrase it loosely, the radiation in question here, would
result in the instability of matter, since such a runaway reaction would
radiate away the energy of the charged particle (i.e., its mass). In that
case, we are talking about a massive photon, since only a massive photon
would conserve energy and momentum in such a reaction. There is nothing
particularly wrong with a massive photon, however, apart from leading
to non-conservation of charge, which would appear to be obvious, from
the premise. There is also nothing wrong with a massive photon if such
photons are ``reabsorbed'' by the charge that emits them, leaving only
massless photons to propagate as radiation. Loosely speaking, these
``massive'' photons are the self-energy corrections that renormalize the
mass of a charged particle.
Bilge
>charge radiates. Since energy is conserved, this radiative emission
>occurs at the expense of the particle's kinetic energy.
>
>Therefore, under the influence of a separate force, a charged particle
>accelerates LESS than a neutral particle of the same mass due to the
>above effect.
anything at all, in principle, (and hence exists but possibly hasn't
been discovered), has a field associated with it that corresponds to
a force that will accelerate that particle, the difference between
a neutral particle and a charged particle has nothing to do with
your argument, since by definition, your argument applies only to
charged particles. If you were referring to neutrinos, you would
have the same argument, except that the radiation in question would
consist of Z's rather than photons.
>Since the charged mass accelerated less but remained with constant
>mass, one must draw the conclusion that the emitted radiation caused a
>reaction force on the charged particle.
Not if the radiation emitted was reabsorbed by the charge that
emitted it. Since the timescale in question is on the order of
the time it takes for the radiation to propagate across the radius
of the charge, we aren't in a classical regime.
[...]
>Problem 1: the acceleration of the particle here would increase
>exponentially with time, which is seemingly absurd.
>
>Problem 2: If you insist that a0 = 0, then when you do apply an
>external force, the particle begins to react BEFORE the external force
>is applied (causality violation).
>
>Maxwell was wrong.
Maxwell wasn't complete, but maxwell's equations follow from a more
complete theory, i.e., qed.
John Schoenfeld
Correct. I wanted to probe the reaction of the so called "qualified"
posters. It is obvious a physics discussion with them approaches
pointlessness.
JS
John Schoenfeld
> John Schoenfeld wrote:
> >
> > According to the laws of classical electrodynamics, an accelerating
> > charge radiates. Since energy is conserved, this radiative emission
> > occurs at the expense of the particle's kinetic energy.
> >
> > Therefore, under the influence of a separate force, a charged particle
> > accelerates LESS than a neutral particle of the same mass due to the
> > above effect.
>
> How would you accelerate a neutral particle, jackass? A particle free
> falling in a gravitational field is in Minkowski space and has no
> in-frame acceleration at all. If you use an EM field to accelerate
> it, Maxwell rules.
It is clear that you demonstrate very little knowledge, only
obfuscated ignorance. In the post I explicitly stated "separate
force", which in classical physics could be a Newtonian gravitational
force or an electrostatic repulsive force of colliding particles
(assuming the neutral particle was internally charged). You don't even
need a neutral particle, all you need is a charged particle and an
equivalent less charged particle which have the same external force
vector acting on it. The more charged, the less acceleration, this is
clear from the Abraham-Lorentz derivation of radiation reaction. Much
to your dismay, this problem persists in Special Relativity by
deriving Abraham-Lorentz formula from Lienards formula as opposed to
Larmors.
> [snip muddled drivel]
>
> > Maxwell was wrong.
>
> You are an empirically self-demonstated voluble idiot.
Maxwell is wrong within the extremes of classical physics, both
empirically and theoretically, it seems. This is not so bad as it
leaves room for improvement and areas for research. Physics is not a
religion.
You might want to take your mindless attacks elsewhere.
JS
Timo Nieminen
> According to the laws of classical electrodynamics, an accelerating
> charge radiates. Since energy is conserved, this radiative emission
> occurs at the expense of the particle's kinetic energy.
[cut]
> From Newtons second law,
>
> Frad = ma = (u0 q^2 / 6 pi c) j
>
> From which it follows that
>
> a(t) = a0 e^(t/mu)
I note that somebody has been reading Griffiths.
> Problem 1: the acceleration of the particle here would increase
> exponentially with time, which is seemingly absurd.
Frad = ma only in the case of no external forces, in which case a0 = 0 is
an entirely reasonable assumption. No external force means none of the
causality violations noted below.
> Problem 2: If you insist that a0 = 0, then when you do apply an
> external force, the particle begins to react BEFORE the external force
> is applied (causality violation).
So Abraham--Lorentz equation of motion is wrong. See reference to Jackson
provided by Griffiths for details and an alternative. Even the next
section in Griffiths makes worthwhile reading. How does one apply a sudden
acceleration uniformly and instantly to an extended object?
> Maxwell was wrong.
Not a consequence of the above points, but yes, this is well-known. Is
reality classical?
Though one needs to be careful about what one means by "Maxwell was
wrong".
--
Timo Nieminen
John Schoenfeld
> John Schoenfeld:
> >According to the laws of classical electrodynamics, an accelerating
> >charge radiates. Since energy is conserved, this radiative emission
> >occurs at the expense of the particle's kinetic energy.
> >
> >Therefore, under the influence of a separate force, a charged particle
> >accelerates LESS than a neutral particle of the same mass due to the
> >above effect.
>
> An additonal comment. Since any particle for which we can measure
> anything at all, in principle, (and hence exists but possibly hasn't
> been discovered), has a field associated with it that corresponds to
> a force that will accelerate that particle, the difference between
> a neutral particle and a charged particle has nothing to do with
> your argument, since by definition, your argument applies only to
> charged particles. If you were referring to neutrinos, you would
> have the same argument, except that the radiation in question would
> consist of Z's rather than photons.
I should've stated then that the "neutral particle" is internally
charged. Either way, your point is moot as even if you use a less
charged particle, it will accelerate differently than an equivalent
more charged particle.
Here is something which is quite funny, although I hope I have not
made an error:
Given the right charge to mass ratio, a classically charged mass can
levitate in Earth gravity by simply counter-acting the gravitational
force with its radiation reaction force. When this occurs, energy
conservation is violated.
Here is why:
Since the charge emits an isotropic electric field, it only absorbs a
section of the field, not the entire field. Therefore, the charged
mass can sustain levitation by absorbing the component of the field it
requires, and letting the remainder of the field radiate away for
consumption elsewhere. Since the levitating charge goes nowhere, this
process can repeat itself infinitely. The obvious problem here is that
Energy is being created and radiated away by this self-levitating
charge which violates the Conservation of Energy law with extreme
prejudice.
What precisely is wrong with my argument here? (without invoking QED
or QFT).
Harold Ensle
"John Schoenfeld" <j.scho...@programmer.net> wrote in message
news:a98beaaa.04041...@posting.google.com...
Ha ha ...a very good trick. I used it once too. One time someone
asked about Einstein's derivation of E=mc^2 and I happened
to have his original paper on hand. So I posted the derivation
_Verbatim_ (without mentioning the source) You wouldn't
believe the absurd replies I got from "qualified" posters.
H.Ellis Ensle
Sam Wormley
See Chapter 16 of Jackson, "Radiation Damping, Classical Models of
Charged Particles"--specifically sec 16.2 "Radiative Reaction Force from
Conservation of Energy", pps 747-750.
Timo Nieminen
> Here is something which is quite funny, although I hope I have not
> made an error:
>
> Given the right charge to mass ratio, a classically charged mass can
> levitate in Earth gravity by simply counter-acting the gravitational
> force with its radiation reaction force.
How? Radiative reaction force is proportional to da/dt. Levitation implies
a = 0, thus radiative reaction force is zero.
> When this occurs, energy
> conservation is violated.
If one uses conservation of energy to derive an expression for radiative
reaction force, one does not expect application of said expression to
result in violation of energy conservation. Thus, actual application of
the theory should disagree with radiative self-levitation.
--
Timo Nieminen
Harry
IMO there was a little glitch in that, but without consequences (no further
comment because I want to publish it).
Now I like to have a laugh. Is that thread still available on internet? I
can't find it...
Harald
EjP
Leaving aside the problem that it's very hard to accelerate a
neutral particle, this issue is dealt with in Jackson.
-E
> Maxwell was wrong.
>
> JS
Harold Ensle
"Harry" <harald.v...@epfl.ch> wrote in message
news:407d4e8b$1...@epflnews.epfl.ch...
This should put you on the thread:
http://www.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&threadm=8orrpj%246n
j%241%40news.fsu.edu&rnum=8&prev=/groups%3Fas_q%3Densle%2520derivation%2520m
ass%2520energy%26safe%3Dimages%26ie%3DUTF-8%26oe%3DUTF-8%26lr%3D%26hl%3Den
H.Ellis Ensle
Alfred Einstead
> According to the laws of classical electrodynamics,
The laws are:
div E = rho/epsilon_0
div B = 0
curl E + dB/dt = 0
curl B - mu_0 epsilon_0 dE/dt = mu_0 J
with sources, for the case in question:
rho(r,t) = e delta(r - R(t))
J(r,t) = e R'(t) delta(r - R(t))
where (t |-> R(t)) descriibes the particle's trajectory.
> an accelerating
> charge radiates. Since energy is conserved, this radiative emission
> occurs at the expense of the particle's kinetic energy.
The equation of motion for the particle is given by:
d/dt (mR'(t)/sqrt(1-(|R'(t)|/c)^2)) = e (E0(R(t), t) + R'(t) x B0(R(t),t))
where the E0 and B0 are fields with the "particle's fields" subtracted
out.
> Therefore, under the influence of a separate force, a charged particle
> accelerates LESS than a neutral particle of the same mass due to the
> above effect.
The problem with saying this or anything about the particle is that
the force law does NOT determine the motion of the particle
because... there is no unambiguous definition of E0 and B0
because... there is no unambiguous field associated with this or any
source.
A field "associated with" a source is only given up to an arbitrary
unspecified free field. There are an infinite number of different
selections for fields to be associated with the source given above;
and therefore an infinite number of different E0's and B0's -- and
an infinite number of different INEQUIVALENT force laws.
All the conclusion you've drawn does is underscore this point;
particularly, that the "usual" selection of E0 and B0 breaks down.
The real problem is that the fields that are supposed to be in
the Lorentz force law, if it's being taken from its continuum
description, are E and B, themselves. But the force law stated
with these fields is mathematically meaningless, since
E(R(t),t) and B(R(t),t) are undefined.
So, the root of the problem is the assumption of point sources,
in the first place; or the extension of the vacuum constitutive
relations (D = epsilon_0 E; B = mu_0 H) to the vicinity of the
point sources. Take your pick.
This has been discussed a bit further in sci.physics.research, as
well.
Alfred Einstead
> Maxwell wasn't complete, but maxwell's equations follow from a more
> complete theory, i.e., qed.
QED is hardly complete. In particular, it is not known that the
interacting fields are even well-defined, and is widely believed
not to be; the interaction picture is known not to be well-defined
(in virtue of Haag's theorem), the S-matrix (even the renormalized
S-matrix) widely believed to be ill-defined.
Second, Maxwell's equations don't follow from QED, meaning specifically
the coupled Maxwell-Dirac equations for a spin 1 and spin 1/2 field
with the Heisenberg relations imposed on the Dirichlet data. They
ARE the (Maxwell part of the) equations in QED. Maxwell's equations
are kept intact. It's the objects which the equations are equations
of that is changed ... from c-numbers to q-numbers.
QED didn't provide a deeper foundation on which Maxwell rested. It
merely retained it intact. The only ingredient specific to Quantum
Field Theory is the posing of the Heisenberg relations on the
Dirichlet data. Were it classical, the spin 1/2 Dirac field would be
anti-commuting and the spin 1 field would be commuting and the
system would be a set of regular coupled non-linear partial
differential equations holding over regular numeric variables
(and fermion-number variables).
Unfortunately, since the entire coupled system (Maxwell + Dirac) is
non-linear, and since the Dirichlet data is singular, the system
is ill-defined, since there is no known way to mathematically
treat non-linear systems of partial differential equations with
singular (i.e. generalized function) boundary data ... unless
you posit a Colombeau theoretic structure or the like (and
even here, the resolution of the problem has never been clearly
made).
Tom Roberts
>>"John Schoenfeld" <j.scho...@programmer.net> wrote in message
>>news:a98beaaa.04041...@posting.google.com...
>>[snip the shit]
>
> is a well known paradox (look in Griffiths' Introduction to Electrodynamics
> for example). The only _possible_ error in his post is to claim that
> Maxwell's equations are wrong because of it.
Including the sign error?
According to his equation:
Fr = (u0 q2 /6 pi c) j
if I start pulling on a charge (so j points toward me), it will leap at
me due to this "radiation reaction" force Fr. In reality, of course, it
resists my pull more than an uncharged object of the same mass would do
-- that's why it is called a REACTION force. So his sign must be wrong;
that makes his other equation DECAY exponentially rather than "increse
exponentially" as he claims.
I don't have Griffiths' text. But I do know on basic physical grounds
that this is incorrect.
John Schoenfeld made several other errors in his post.
I posted an article on them, but it seems to not have
propagated fully. If it doesn't show up here soon I'll
repost it from this account.
Tom Roberts tjro...@lucent.com
Paul Stowe
>Harold Ensle wrote:
>>>"John Schoenfeld" <j.scho...@programmer.net> wrote in message
>>>news:a98beaaa.04041...@posting.google.com...
>>>[snip the shit]
>>
>> What he is talking about
>> is a well known paradox (look in Griffiths' Introduction to Electrodynamics
>> for example). The only _possible_ error in his post is to claim that
>> Maxwell's equations are wrong because of it.
>
> Including the sign error?
>
> According to his equation:
> Fr = (u0 q2 /6 pi c) j
Actually, Eq 9.148 [page 439] is given as, exactly,
2 q^2 .
F = ---- --- a
z4pi 3c^3
z = permittivity, u = permeability
Given 1/c^2 = zu, then this simplifies to,
uq^2 .
F = ------- *a*
6pic
We note that for Eq 9.126 [Page 434] is given as the same as 9.148 EXCEPT
that vector designator a is bold in 9.126 and not bold in 9.148. This
denotes the anti-direction of vector a to the vector *a*.
> if I start pulling on a charge (so j points toward me), it will leap at
> me due to this "radiation reaction" force Fr. In reality, of course, it
> resists my pull more than an uncharged object of the same mass would do
> -- that's why it is called a REACTION force. So his sign must be wrong;
> that makes his other equation DECAY exponentially rather than "increse
> exponentially" as he claims.
Could this, in conjunction with the virtual pairs, be related to,
if not the source of, inertia?
> I don't have Griffiths' text. But I do know on basic physical grounds
> that this is incorrect.
Only an incomplete description. The appropriate section in Griffiths
is Section 9.3 pages 432-439.
FrediFizzx
news:e58d56ae.04041...@posting.google.com...
I think Feynman said it very well,
"...we have allowed what is perhaps a silly thing, the possibility of the
'point' electron acting on itself."
FrediFizzx
Bilge
>> Maxwell wasn't complete, but maxwell's equations follow from a more
>> complete theory, i.e., qed.
>
>QED is hardly complete.
What I wrote was ``more complete'', vis-a-vis maxwell's equations.
I'm not really certain what it would mean to say ``complete'' without
qualification. [also, see below]
> In particular, it is not known that the
>interacting fields are even well-defined, and is widely believed
>not to be; the interaction picture is known not to be well-defined
>(in virtue of Haag's theorem), the S-matrix (even the renormalized
>S-matrix) widely believed to be ill-defined.
An example of a quantum field theory to which haag's theorem does
not apply and all of the objections above are addressed is,
http://arxiv.org/abs/hep-th/9703032
It's also possible that qed might be a ``complete'' theory of electro-
magnetic interactions. There are no particles which interact only via
the electromagnetic interaction (apart from the photon, which is itself,
not a source of the field), so there are no charged particles in nature
that require a description in terms of qed alone. Since, for example,
the electron interacts weakly, a complete theory of the electron would
(at least) require a description in terms of more than just qed. The
lack of particles in nature which interact only electromagnetically,
only precludes qed from being a complete description of particles that
don't exist.
Qed could still possibly be the most complete theory of electromagnetism
possible without having to ``exist'' as a theory indpendent of any other
interaction.
>Second, Maxwell's equations don't follow from QED, meaning specifically
>the coupled Maxwell-Dirac equations for a spin 1 and spin 1/2 field
>with the Heisenberg relations imposed on the Dirichlet data. They
>ARE the (Maxwell part of the) equations in QED. Maxwell's equations
>are kept intact.
I don't know why you say that. Once you've obtained the gauge covariant
derivative, you have the lagrangian,
_
L_qed = u(iD/ - m)u + L_free
with D/ == \gamma^u D_u = \gamma^u (d_u - ieA_u)
Assuming an abelian theory, the traditional faraday tensor comes from
the coupled fields through the commutator of the covariant derivatives,
F_vu = (1/ie)[D_u, D_v]
Since the covariant derivative appears as nothing more than
invariance under a local phase rotation, I don't see that maxwell's
equations were ever part of the qed lagrangian, a priori. Where do
I add maxwell's equations to the dirac equation in simply asserting
invariance under a local phase rotation to dirac lagrangian for
a free particle for which no assumption about being charged has been
made intially?
> It's the objects which the equations are equations
>of that is changed ... from c-numbers to q-numbers.
>
>QED didn't provide a deeper foundation on which Maxwell rested.
>It merely retained it intact. The only ingredient specific to Quantum
>Field Theory is the posing of the Heisenberg relations on the
>Dirichlet data.
Huh? If you mean to say that maxwell's equations ``rested'' on gauge
invariance all along and that qed didn't change that, I think you're
missing the point. Had gauge invariance been important to maxwell's
equations conceptually as a foundation, they would probably be called
weyl's equations, since it wasn't until after weyl failed to create a
geometric theory of E&M based on scale invariance, that phase invariance
was recognized as significant.
> Were it classical, the spin 1/2 Dirac field would be
>anti-commuting and the spin 1 field would be commuting and the
>system would be a set of regular coupled non-linear partial
>differential equations holding over regular numeric variables
>(and fermion-number variables).
>
>Unfortunately, since the entire coupled system (Maxwell + Dirac) is
>non-linear, and since the Dirichlet data is singular, the system
>is ill-defined, since there is no known way to mathematically
>treat non-linear systems of partial differential equations with
>singular (i.e. generalized function) boundary data ... unless
>you posit a Colombeau theoretic structure or the like (and
>even here, the resolution of the problem has never been clearly
>made).
However, that is a rather pessimistic view in that you only give reasons
that qed shouldn't work rather than say why it does work in the context of
a ``better'' theory that doesn't suffer those deficiencies (I'm being a
bit facetious here). It does work, so it might be more apt to ask why it
works so well. QED doesn't have to be (and almost certainly isn't) a
fundamental theory. It might be the most fundamental theory of
electromagnetism, though.
Bilge
>>news:<slrnc7opel....@radioactivex.lebesque-al.net>...
>> An additonal comment. Since any particle for which we can measure
>> anything at all, in principle, (and hence exists but possibly hasn't
>> been discovered), has a field associated with it that corresponds to
>> a force that will accelerate that particle, the difference between
>> a neutral particle and a charged particle has nothing to do with
>> your argument, since by definition, your argument applies only to
>> charged particles. If you were referring to neutrinos, you would
>> have the same argument, except that the radiation in question would
>> consist of Z's rather than photons.
>
>I should've stated then that the "neutral particle" is internally
>charged. Either way, your point is moot as even if you use a less
>charged particle, it will accelerate differently than an equivalent
>more charged particle.
So what? Since that acceleration never happens, what difference
does it make?
>Here is something which is quite funny, although I hope I have not
>made an error:
>
>Given the right charge to mass ratio, a classically charged mass can
>levitate in Earth gravity by simply counter-acting the gravitational
>force with its radiation reaction force. When this occurs, energy
>conservation is violated.
Are you sure you know what you mean by radiation reaction?
In particular, the radiation reaction to which you refer would
radiate the particle into oblivion by carrying off all of that
particles' mass in short order.
>
>Here is why:
>Since the charge emits an isotropic electric field, it only absorbs a
>section of the field, not the entire field. Therefore, the charged
>mass can sustain levitation by absorbing the component of the field it
>requires, and letting the remainder of the field radiate away for
>consumption elsewhere. Since the levitating charge goes nowhere, this
>process can repeat itself infinitely. The obvious problem here is that
>Energy is being created and radiated away by this self-levitating
>charge which violates the Conservation of Energy law with extreme
>prejudice.
>
>What precisely is wrong with my argument here? (without invoking QED
>or QFT).
If you consider only a classical argument, you won't violate
conservation of energy. You just won't have any stable matter,
as it will radiate its mass away.
John Schoenfeld
> > Given the right charge to mass ratio, a classically charged mass can
> > levitate in Earth gravity by simply counter-acting the gravitational
> > force with its radiation reaction force.
>
> How? Radiative reaction force is proportional to da/dt. Levitation implies
> a = 0, thus radiative reaction force is zero.
Not necessarily. If you use the Abraham-Lorentz radiation reaction
formula and Newtons law of gravity, you can model the radiating
self-levitating charge as follows:
Fg = -Frad
Gm1m2/r^2 = u0q^2/6pic da/dt
where
m1 = mass of gravitational source
m2 = mass of charge
Since da/dt is the jerk of the displacement field caused by the
gravitational field Gm1/r^2, da/dt evaluates to -Gm1/r^3.
Solving for q, you get
q = sqrt(m2 6 pi c r / u0 )
where
r = distance between gravitational source centre of gravity to
charge
m2 = mass of charge
u0 = permeability of free space
The interesting thing to note here is that charge self-levitation
occurs irrespective of the gravitational source mass, only the
distance between the source mass and the charge. This follows from the
fact that the radiation reaction is proportional to the rate of change
of acceleration, not the acceleration itself.
The charge required to levitate a particle with electron mass at earth
surface gravity is 1.61E-4 C, which is the charge of 1 000 000 000 000
000 electrons precisely.
> > When this occurs, energy
> > conservation is violated.
>
> If one uses conservation of energy to derive an expression for radiative
> reaction force, one does not expect application of said expression to
> result in violation of energy conservation. Thus, actual application of
> the theory should disagree with radiative self-levitation.
There is no reason why a formula derived from the Conservation of
Energy law cannot show it to be inconsistent.
Clearly, a self-levitating chare as I've showed above would radiate
infinite energy given infinite time.
The energy radiated away by the self-levitating charge is the energy
it radiates by accelerating due to gravity minus the energy required
to maintain it's position in the gravitational field.
NCE = E - LEV
where
NCE = non-conserved energy radiated away
E = energy radiated by accelerating mass
LEV = energy consumed by charge to levitate
E = Larmors formula
= q^2 a^2 / 6 pi e0 c^3
= q^2 g(r)^2 / 6 pi e0 c^3 (since acceleration is g(r)
LEV = Is simply the GPE
= m g(r) r
NCE = q^2 g(r)^2 / 6 pi e0 c^3 - mg(r)r
= g(r)(q^2 g(r) - 6 pi e0 c^3 m r) / 6 pi e0 c^3
where
m = mass of charge
q = electric charge of charge
g(r) = newtonian gravitational attraction at radius R
e0 = permitivitty of free space
c = light speed
Based on my NCE derivation, the non-conserved energy is greater when
there is high charge and high gravitational field strength.
It would seem logical that this non-conserved energy is is actually
conserved by a reduction in the mass of the charge, which is not
allowed in Maxwells theory.
JS
John Schoenfeld
> >I should've stated then that the "neutral particle" is internally
> >charged. Either way, your point is moot as even if you use a less
> >charged particle, it will accelerate differently than an equivalent
> >more charged particle.
>
> So what? Since that acceleration never happens, what difference
> does it make?
You are wrong, Abraham-Lorentz were right. Like I said before, Bilge,
even if you DON'T use a neutral particle (which can still be
accelerated despite your erroneous claims), you can use a charged
particle and a LESS charged particle to derive the Abraham-Lorentz
force. In fact, you just need a single charged particle.
Abraham-Lorentz were not wrong, you are.
> >Here is something which is quite funny, although I hope I have not
> >made an error:
> >
> >Given the right charge to mass ratio, a classically charged mass can
> >levitate in Earth gravity by simply counter-acting the gravitational
> >force with its radiation reaction force. When this occurs, energy
> >conservation is violated.
>
> Are you sure you know what you mean by radiation reaction?
> In particular, the radiation reaction to which you refer would
> radiate the particle into oblivion by carrying off all of that
> particles' mass in short order.
Why don't you state precisely where in classical physics mass can be
converted into pure em energy?
> >
> >Here is why:
> >Since the charge emits an isotropic electric field, it only absorbs a
> >section of the field, not the entire field. Therefore, the charged
> >mass can sustain levitation by absorbing the component of the field it
> >requires, and letting the remainder of the field radiate away for
> >consumption elsewhere. Since the levitating charge goes nowhere, this
> >process can repeat itself infinitely. The obvious problem here is that
> >Energy is being created and radiated away by this self-levitating
> >charge which violates the Conservation of Energy law with extreme
> >prejudice.
> >
> >What precisely is wrong with my argument here? (without invoking QED
> >or QFT).
>
> If you consider only a classical argument, you won't violate
> conservation of energy. You just won't have any stable matter,
> as it will radiate its mass away.
Wrong again. Matter is always stable in classical physics.
JS
Tom Roberts
> Given the right charge to mass ratio, a classically charged mass can
> levitate in Earth gravity by simply counter-acting the gravitational
> force with its radiation reaction force. When this occurs, energy
> conservation is violated.
Hmmm.
If you are assuming a uniform gravitational field, and imagine dropping
the charged particle, then you are implicitly accepting the sign error I
pointed out in an earlier post -- as you are dropping it, j is directed
down, and if Fr is in the direction of j it cannot possibly "levitate".
Ditto if you are assuming the 1/r^2 form of acceleration due to gravity
-- j is still directed down. As I pointed out before, after fixing the
sign error your "problem 1" disappears (i.e. Fr exponentially decays).
And as others have pointed out, for true "levitation" j=0 and Fr=0.
> Since the charge emits an isotropic electric field, it only absorbs a
> section of the field, not the entire field. Therefore, the charged
> mass can sustain levitation by absorbing the component of the field it
> requires, and letting the remainder of the field radiate away for
> consumption elsewhere. Since the levitating charge goes nowhere, this
> process can repeat itself infinitely. The obvious problem here is that
> Energy is being created and radiated away by this self-levitating
> charge which violates the Conservation of Energy law with extreme
> prejudice.
That is complete and utter gibberish.
Tom Roberts tjro...@lucent.com
Tom Roberts
> On Thu, 15 Apr 2004 00:41:38 GMT, Tom Roberts <tjro...@lucent.com> wrote:
>
> Actually, Eq 9.148 [page 439] is given as, exactly,
>
> 2 q^2 .
> F = ---- --- a
> z4pi 3c^3
>
> z = permittivity, u = permeability
>
> Given 1/c^2 = zu, then this simplifies to,
>
>
> uq^2 .
> F = ------- *a*
> 6pic
>
> We note that for Eq 9.126 [Page 434] is given as the same as 9.148 EXCEPT
> that vector designator a is bold in 9.126 and not bold in 9.148. This
> denotes the anti-direction of vector a to the vector *a*.
That is a VERY STRANGE nomenclature. In textbooks, normally a bold
letter deontes a 3-vector and the corresponding non-bold letter denotes
the norm of the 3-vector. Using bold to denote an "anti-direction" is
OUTRAGEOUSLY nonstandard -- why not simply use the standard minus sign?
[I do not have Griffith's textbook.]
I repeat: from basic physical grounds, Fr must be in the direction of
-da/dt (where a is the acceleration 3-vector of the charged particle and
t is the time coordinate of this inertial frame).
Tom Roberts tjro...@lucent.com
Harry
I see... you were criticized because you simplified it too much, not
noticing that he took the differerence of a difference at constant velocity.
So ... to my regret I had no laugh.
Harald
Harold Ensle
FrediFizzx
news:slrnc7soab....@radioactivex.lebesque-al.net...
It can't be. It is only a very good approximation at "normal" energies.
Electro-weak theory is probably a better approximation. But my research is
indicating that electromagnetism ought to even have strong components mixed
in.
FrediFizzx
Harold Ensle
This is not true. I simply put his last steps verbatim
>not
> noticing that he took the differerence of a difference at constant
velocity.
What are you talking about? I posted his last steps which resulted in
E=mc^2.
So I did it exactly as he did.
> So ... to my regret I had no laugh.
You must have missed the guy who said I was rewriting history because
Einstein never used the word "light" in the derivation.
You must have missed the guy who said Einstein could have never done
it that way because it was wrong.
Only Carr brought up a complaint that actually related. He complained
because I skipped the first step where Einstein sent two photons off at
oblique angles whose sum of momentum equaled the photon I started with
in my first step (Einsteins 2nd step)...the derivation then exactly matched
Einstein to the same result.
As for difference of a difference, I have no idea what you are talking
about.
H.Ellis Ensle
Ken S. Tucker
> Huh? If you mean to say that maxwell's equations ``rested'' on gauge
>invariance all along and that qed didn't change that, I think you're
>missing the point. Had gauge invariance been important to maxwell's
>equations conceptually as a foundation, they would probably be called
>weyl's equations, since it wasn't until after weyl failed to create a
>geometric theory of E&M based on scale invariance, that phase invariance
>was recognized as significant.
Sorry to interrupt, and not to change the topic,
but Weyl threw me a curve in High School,
in Dover's "Principle of Relativity", in his article
"Gravitation and Electricity". On pg 206 Eq.(7)
implies by
d (phi) = (phi)_u dx^u
that (phi)_u is a gradient.
But then his Eq.(10) on pg 207 curls that gradient.
We usually expect Curl Grad =0, so Weyl vanished
the EM-fields, but then vanished (phi),u at the end of
pg 207, and then at the top of pg 208, the F_uv
vanishes....wow!
At the bottom of page 208, where he re-institutes
a Maxwell equation based on
F_uv,w ...permetated =0
So F_uv is always zero but does change, he did
that relativistically. (read the ref).
Regrading Bilges, "phase terms" see below Weyl's
Eq. (7) and note the caboose term "- g_uv (phi)_p".
I think Einstien's nonsymmetrical approach renders
that phase term unnecessary, but I'm listening.
Ken S. Tucker
PS: snippable... Speaking for myself, I think both
Weyl and Einstein's unified field theories are reasonable,
so what I ended up doing is a merger, since both are
brighter than me. A merger occurs unless specific
contradictions occur, but I haven't found any yet.
kst
Androcles
"Tom Roberts" <tjro...@Lucent.com> wrote in message
news:407E99CF...@Lucent.com...
I repeat:
1-cos(pi/2).v/c
nu' = nu. ________________ is blueshift, illogical Roberts
sqrt(1-v^2/c^2)
Androcles
Timo Nieminen
> Timo Niemen wrote:
> > > Given the right charge to mass ratio, a classically charged mass can
> > > levitate in Earth gravity by simply counter-acting the gravitational
> > > force with its radiation reaction force.
> >
> > How? Radiative reaction force is proportional to da/dt. Levitation implies
> > a = 0, thus radiative reaction force is zero.
>
> Not necessarily. If you use the Abraham-Lorentz radiation reaction
> formula and Newtons law of gravity, you can model the radiating
> self-levitating charge as follows:
>
> Fg = -Frad
Yes, for levitation, you must have Fg = -Frad.
So a = 0. F = 0, F = ma, means a = 0.
a = 0 means that da/dt = 0.
But this means that Frad = 0.
So self-levitation by radiative reaction is not possible.
> Since da/dt is the jerk of the displacement field caused by the
> gravitational field Gm1/r^2, da/dt evaluates to -Gm1/r^3.
No, the charged particle has a = 0. Sure, you have Fg and Frad acting on
the particle, and Ftotal = Fg + Frad = 0. But you simply don't have any
meaningful a_g and a_rad being "components" of the acceleration of the
charge (with a_total = a_g + a_rad). You just have a = 0.
Note that the whole runaway solution problem depends on a being the actual
acceleration of the particle, not the acceleration that would result from
the external forces alone.
> It would seem logical that this non-conserved energy is is actually
> conserved by a reduction in the mass of the charge, which is not
> allowed in Maxwells theory.
Why is this not allowed in Maxwell's theory? Classical EM theory either
says nothing about mass and conservation of mass, unless special
relativity is included within classical EM theory, so how is it
disallowed?
You will note that classical models of charged particles usually (1) do
not assume a zero size point with infinite charge density, and (2) assume
that the mass is purely electromagnetic (ie due to the particle's own
field).
But that's beside the point, really, since self-levitation doesn't occur.
--
Timo Nieminen
John Schoenfeld
The radiation reaction force is proportional to the jerk, not the
acceleration. For example, if gravitational force was proportional to
the r^2, then Frad would be in the direction of the acceleration,
since it is inversly proportional, it is in the opposite direction..
i.e. a = r^2 da/dt = 2r^3,
a= 1/r^2 da/dt = -1/r^3
John Schoenfeld
i.e. a = r^2 da/dt = 2r,
Tom Roberts
> The radiation reaction force is proportional to the jerk,
You still have a sign error. The radiation reaction force Fr is
proportional to -jerk.
> For example, if gravitational force was proportional to
> the r^2, then Frad would be in the direction of the acceleration,
> since it is inversly proportional, it is in the opposite direction.
For a falling object dr/dt < 0, so it is falling into a region of
INCREASING force in the -r direction, which means increasing
acceleration DOWNWARD, so the jerk is pointing DOWNWARD. Fr must of
course point upward.
> a= 1/r^2 da/dt = -1/r^3
You have two separate errors here, both of which affect the overall
sign (but don't cancel each other):
1) a = -k/r^2 because gravitational force points DOWNWARD (k is
some constant > 0).
2) you took da/dr, not da/dt (and you omitted a factor of 2). As r
is strictly a function of t, we can use the chain rule:
j = da/dt = (da/dr) (dr/dt) = +(k/r^3)*(dr/dt)
From this we see that if the particle is falling down (dr/dt<0)
then j is negative (pointing down), so Fr MUST be in the -j
direction -- UPWARD. We also see that for an object thrown
upward (dr/dt>0), Fr is in the -j direction, and is therefore
downward (opposing the motion of the particle).
It is also QUITE CLEAR on basic physical grounds that Fr is in the -j
direction, not the +j direction: If I start pulling on a charge, j is
pointed toward me, and we know for sure that radiation REACTION does not
cause the particle to jump at me, it causes the charged particle to
resist my pull more than a neutral particle of the same mass would
resist. Note your formula depends on q^2, so this is true regardless of
the sign of the charge.
This sign error resolves your "problem 1" completely -- it's an
exponential decay, not increase. Your "problem 2" is essentially
unrelated, and I'll quote what I said before (in a post that aparently
never propagated very far in the newsgroup):
On 4/13/2004 5:15 AM, John Schoenfeld wrote:
>> If you try to derive this recoil force caused by the radiation, you
>> end up with the formula:
>> Fr = (u0 q2 /6 pi c) j
>> where
>> Fr = recoil force of radiation emission
>> u0 = permeability of free space
>> j = the rate of change of acceleration (the jerk)
>> c = light speed
>> m = mass or particle
Let me take this formula as given[#] (but see below). Then your
conclusions don't follow.
>> From Newtons second law,
>> Frad = ma = (u0 q2 / 6 pi c) j
>> From which it follows that
>> a(t) = a0 e^(t/mu)
>> where
>> mu = u0 q2 / 6 pi m c
>> In the case of the eletron, mu = 6E-24.
>> Problem 1: the acceleration of the particle here would increase
>> exponentially with time, which is seemingly absurd.
You have two errors (well, 3 -- see my response to your "Problem 2" below):
A) You used the wrong equation. a0=0 is the only physically-relevant
solution of your equation, because you omitted any external force
(there's no "startup" acceleration to create any radiation reaction
in the first place).
B) You have an overall sign error in Fr -- it is a reaction force, not
a driving force. This should be clear from the physical situation,
without a detailed examination of your derivation of Fr. That makes
a(t) exponentially decrease with time in your analysis, a MUCH more
sensible result (mu is manifestly > 0). Remember both a and j are
3-vectors.
An important aspect of successfully doing mathematical physics
is having enough physical intuition about the world so you can
detect such sign errors INDEPENDENT of your algebra.
Let me use the Newtonian approximation here for motion of the charged
particle, so the proper equation is:
Fext - (u0 q2/6pi c) da/dt = m a
Here's a simple solution that avoids your "problem":
Fext = constant
a = Fext/m (which is constant)
For a non-constant Fext things get more complicated, but never have the
pathology you claim....
>> Problem 2: If you insist that a0 = 0, then when you do apply an
>> external force, the particle begins to react BEFORE the external
>> force is applied (causality violation).
First, for a0=0 you never have any "external force" -- your analysis
omitted it.
But ignoring that, and your sign error, then you applied the wrong
boundary conditions. This is probably due to your usage of the Newtonian
approximation, which does not necessarily preserve the causality of
relativistic mechanics (remember that relativistic E&M is time-reversal
invariant, and you must explicitly select the retarded Green's function,
not the advanced one; the Newtonian approximation often corresponds to
the average of the retarded and advanced functions....).
[#] This is consistent with my recollection about this problem,
which is that a uniformly-accelerated charge does not radiate,
in the sense that \integral (ExB).dS = 0, with the integral
taken over any closed spatial 2-surface enclosing the charge.
That implies no radiation reaction for uniform acceleration
(Fr=0 for j=0). But it's been a long time since I studied
this....
Tom Roberts tjro...@lucent.com
Tom Roberts
> If you use the Abraham-Lorentz radiation reaction
> formula and Newtons law of gravity, you can model the radiating
> self-levitating charge as follows:
> Fg = -Frad
> Gm1m2/r^2 = u0q^2/6pic da/dt
> where
> m1 = mass of gravitational source
> m2 = mass of charge
> Since da/dt is the jerk of the displacement field caused by the
> gravitational field Gm1/r^2, da/dt evaluates to -Gm1/r^3.
As I pointed out before, this is wrong:
1) the gravitational acceleration is -Gm1/r^2
2) you need to compute da/dt, not da/dr (and you omitted a factor
of 2 in your irrelevant computation)
To compute da/dt you can use the chain rule:
da/dt = (da/dr) (dr/dt)
but the presence of dr/dt (velocity) screws up your entire analysis....
You have also omitted initial conditions in your analysis, and they can
change the conclusion completely (e.g. the particle could be thrown
UPWARD, in which case Frad points downward)....
Bottom line: with Frad proportional to either -da/dt or +da/dt, it is
not possible to use Frad to "levitate" a charge, because for levitation
v=0, a=0, da/dt=0, and Frad=0, implying that Fg is unopposed and pulls
it downward. Yes, Frad will then retard that falling (once you correct
the sign error), but it cannot prevent it -- the particle will
accelerate downward until Fg=-Frad, at which time it will fall with
constant velocity. By playing with q/m you can make that velocity
slower, but you cannot make it vanish.
Tom Roberts tjro...@lucent.com
Androcles
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:QpIfc.950$077...@newssvr32.news.prodigy.com...
[snip]
Consider the concept "moving clocks run slow"
Let a clock emit a frequency of 1 Hz be moving relative to
an observer with velocity 0.866c
t-vx/c^2
t' = _______________
sqrt(1-v^2/c^2)
ref (Electrodynamics, section 3)
= t * sqrt(1-v^2/c^2)
= 1 * 0.5
= 0.5 sconds
= 1 tick per 2 seconds
Consider Einstein's doppler shift, to be found in the
equation
1-cos(phi).v/c
nu' = nu. ________________
sqrt(1-v^2/c^2)
ref (Electrodynamics, section 7)
Let phi = pi/2
1-0
nu' = 1 . ___________
0.5
= 2 ticks per second.
So according to Einstein,
0.5 = 2
Androcles
Harry
Sorry, I first had the impression that you oversimpified the kinetic energy
change, but now I took a second look and I think it's OK like that, except
that it was essential for simplicity to keep v constant and you did not
understand that point.
> > So ... to my regret I had no laugh.
>
> You must have missed the guy who said I was rewriting history because
> Einstein never used the word "light" in the derivation.
> You must have missed the guy who said Einstein could have never done
> it that way because it was wrong.
Yes, I did miss them indeed! I just went down in the Google thread....
> Only Carr brought up a complaint that actually related. He complained
> because I skipped the first step where Einstein sent two photons off at
> oblique angles whose sum of momentum equaled the photon I started with
> in my first step (Einsteins 2nd step)...the derivation then exactly
matched
> Einstein to the same result.
I still wonder if you got that point... which was that v remain constant.
> As for difference of a difference, I have no idea what you are talking
> about.
Now I do agree with the way you simplified that part.
Harald
John Schoenfeld
> On Thu, 15 Apr 2004, John Schoenfeld wrote:
>
> > > > Given the right charge to mass ratio, a classically charged mass can
> > > > levitate in Earth gravity by simply counter-acting the gravitational
> > > > force with its radiation reaction force.
> > >
> > > How? Radiative reaction force is proportional to da/dt. Levitation implies
> > > a = 0, thus radiative reaction force is zero.
> >
> > Not necessarily. If you use the Abraham-Lorentz radiation reaction
> > formula and Newtons law of gravity, you can model the radiating
> > self-levitating charge as follows:
> >
> > Fg = -Frad
>
> Yes, for levitation, you must have Fg = -Frad.
>
> So a = 0. F = 0, F = ma, means a = 0.
The acceleration acted on the charged mass is GM/r^2. Your assertion
that a = 0 is true only if you are refering to the average
acceleration for the time interval r/c (where r is the radius of the
particle). The reason is that the radiation reaction force does not
act instantaneously with the gravitatioanl force as time is required
for the radiation to pass through the charge volume. Obviously, r/c is
approaches hbar for "charged particles" but I posted this thread under
the context of classical electrodynamics and only classical
electrodynamics I shall consider.
If you are making the argument that point charges could not
self-levitate then are correct, since r=0 and thus a = 0 at all times.
Whilst this argument is valid, for all intensive purposes I refer to
particles of radius > 0 which do self-levitate under gravitational
fields and seemingly violate conservation of energy.
> a = 0 means that da/dt = 0.
>
> But this means that Frad = 0.
>
> So self-levitation by radiative reaction is not possible.
Radiative self-levitation of POINT charges may not be possible, but I
iterate again that spherical charges can exhibit this behaviour since
a(t) != 0 for all 0 < t < r/c. Rather, int(t=0, r/c) v(t) . dt = 0
which does not preclude Fg=-Frad where a > 0.
> > Since da/dt is the jerk of the displacement field caused by the
> > gravitational field Gm1/r^2, da/dt evaluates to -Gm1/r^3.
>
> No, the charged particle has a = 0. Sure, you have Fg and Frad acting on
> the particle, and Ftotal = Fg + Frad = 0. But you simply don't have any
> meaningful a_g and a_rad being "components" of the acceleration of the
> charge (with a_total = a_g + a_rad). You just have a = 0.
>
> Note that the whole runaway solution problem depends on a being the actual
> acceleration of the particle, not the acceleration that would result from
> the external forces alone.
It seems the entirety of your argument is based upon the assumption
that the acceleration of the charged mass is always 0, which is false
as I have explained above.
An analysis of the "micromotion" of this levitating charge shows that
it oscillates along its vertical diameter with amplitude r and
frequency proportional to r and q.
> > It would seem logical that this non-conserved energy is is actually
> > conserved by a reduction in the mass of the charge, which is not
> > allowed in Maxwells theory.
>
> Why is this not allowed in Maxwell's theory? Classical EM theory either
> says nothing about mass and conservation of mass, unless special
> relativity is included within classical EM theory, so how is it
> disallowed?
Mass invariance is implicit in Newtonian mechanics by simply observing
Newtons 2nd Law, the rate of change of momentum.
F = d (mv) / dt
Despite Newton never assuming this himself, "Academia" decided to
assume mass invariance in the 2nd law and simply remove m from the
derivative using the product rule.
F = m dv/dt = m a
This is probably a fundamental mistake which has propagated through
all of Physics, but that is anothe issue. If you accept F=ma in
mechanics, then you must accept mass invariance in dynamics.
JS
John Schoenfeld
> John Schoenfeld wrote:
> > The radiation reaction force is proportional to the jerk,
>
> You still have a sign error. The radiation reaction force Fr is
> proportional to -jerk.
There is no error, Tom.
>
> > For example, if gravitational force was proportional to
> > the r^2, then Frad would be in the direction of the acceleration,
> > since it is inversly proportional, it is in the opposite direction.
>
> For a falling object dr/dt < 0, so it is falling into a region of
> INCREASING force in the -r direction, which means increasing
> acceleration DOWNWARD, so the jerk is pointing DOWNWARD. Fr must of
> course point upward.
Tom, you are wrong.
If a(r) = 1/r^2, then the jerk is in the OPPOSITE DIRECTION. Think
about it.
Or, do the math.
d a(r)/dr = - 1/r^3
The snap is in the same direction as the acceleration, and the crackle
the same as the jerk. But what about the pop?
[snip]
> An important aspect of successfully doing mathematical physics
> is having enough physical intuition about the world so you can
> detect such sign errors INDEPENDENT of your algebra.
>
> But ignoring that, and your sign error, then you applied the wrong
> boundary conditions. This is probably due to your usage of the Newtonian
> approximation, which does not necessarily preserve the causality of
> relativistic mechanics (remember that relativistic E&M is time-reversal
> invariant, and you must explicitly select the retarded Green's function,
> not the advanced one; the Newtonian approximation often corresponds to
> the average of the retarded and advanced functions....).
>
>
> [#] This is consistent with my recollection about this problem,
> which is that a uniformly-accelerated charge does not radiate,
> in the sense that \integral (ExB).dS = 0, with the integral
> taken over any closed spatial 2-surface enclosing the charge.
> That implies no radiation reaction for uniform acceleration
> (Fr=0 for j=0). But it's been a long time since I studied
> this....
>
>
> Tom Roberts tjro...@lucent.com
What is all this? Are you seriously mounting a challenge to the
validity of the Abraham-Lorentz formula?
The problems I refered to are well known in physics:
problem 1: "runaway solutions"
problem 2: "acausual preacceleration"
As for my "sign error", i think that the error is not mine.
Good day Sir,
JS
Tom Roberts
> Tom Roberts <tjro...@lucent.com> wrote in message news:<UQHfc.930$Nb7...@newssvr32.news.prodigy.com>...
>>John Schoenfeld wrote:
>>>The radiation reaction force is proportional to the jerk,
>>You still have a sign error. The radiation reaction force Fr is
>>proportional to -jerk.
>
> There is no error, Tom.
Yes there is. You simply repeat the error(s) I have already pointed out.
> If a(r) = 1/r^2, then the jerk is in the OPPOSITE DIRECTION. Think
> about it.
But for gravitation, a(r) = -K/r^2 THE MINUS SIGN IS IMPORTANT. Remember
that a is a 3-vector. And what you call "jerk" is not the jerk -- you
differentiate wrt r, not t.
The acceleration due to gravity is DOWNWARD; the radiation reaction for
a falling charged particle is in the direction opposite to the velocity
of the particle (see my previous posts for a derivation).
Think about that for a second, and you'll see that for a
charged particle in a gravitational field but supported
against gravity (i.e. at rest wrt the earth) there is no
radiation reaction, and hence no radiation.
Remember that a particle in freefall can be thrown upward -- your
analysis completely omitted the boundary conditions, and therefore this
case.
> Or, do the math.
> d a(r)/dr = - 1/r^3
But jerk is not "d a(r)/dr", it is d a(t)/dt. THE DIFFERENCE IS IMPORTANT.
> The snap is in the same direction as the acceleration, and the crackle
> the same as the jerk. But what about the pop?
Successive derivatives wrt r are IRRELEVANT -- you need to express a as
a function of t and take successive derivatives wrt t. THE DIFFERENCE IS
IMPORTANT. This makes the signs of these derivatives more complicated
than you think (you get additional powers of velocity).
> What is all this? Are you seriously mounting a challenge to the
> validity of the Abraham-Lorentz formula?
On clear physical grounds there is a sign error in your formula for Fr.
> The problems I refered to are well known in physics:
> problem 1: "runaway solutions"
> problem 2: "acausual preacceleration"
These are not "well known", these are errors. With reasons I discussed
in earlier posts.
> As for my "sign error", i think that the error is not mine.
You certainly make sign mistakes. And conceptual errors (r is not t).
You will never understand this until you correct your errors and
misconceptions. It is more complicated than you think.
Tom Roberts tjro...@lucent.com
Dirk Van de moortel
"John Schoenfeld" <j.scho...@programmer.net> wrote in message news:a98beaaa.04041...@posting.google.com...
> > John Schoenfeld wrote:
> > > The radiation reaction force is proportional to the jerk,
> >
> > You still have a sign error. The radiation reaction force Fr is
> > proportional to -jerk.
>
> There is no error, Tom.
>
> >
> > > For example, if gravitational force was proportional to
> > > the r^2, then Frad would be in the direction of the acceleration,
> > > since it is inversly proportional, it is in the opposite direction.
> >
> > For a falling object dr/dt < 0, so it is falling into a region of
> > INCREASING force in the -r direction, which means increasing
> > acceleration DOWNWARD, so the jerk is pointing DOWNWARD. Fr must of
> > course point upward.
>
> Tom, you are wrong.
>
> If a(r) = 1/r^2, then the jerk is in the OPPOSITE DIRECTION. Think
> about it.
>
> Or, do the math.
>
> d a(r)/dr = - 1/r^3
>
> The snap is in the same direction as the acceleration, and the crackle
> the same as the jerk. But what about the pop?
Not even looking at the missing factor 2, this is a very nice one:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Confusion.html
Dirk Vdm
John Schoenfeld
That was a typo, obviously.
Tom Roberts
> Not even looking at the missing factor 2, this is a very nice one:
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Confusion.html
Dirk:
Snap, crackle, and pop have been used semi-humorously to describe higher
derivatives of position (Newtonian mechanics):
dx/dt velocity
d^2x/dt^2 acceleration
d^3x/dt^3 jerk
d^4x/dt^4 snap
d^5x/dt^5 crackle
d^6x/dt^6 pop
I believe this came from someone having a real use for "snap", and the
others just followed naturally (:-). Their utility in real problems is
unlikely -- even using jerk is rare.
But John Schoenfeld has been differentiating wrt r, not t, so he did fumble.
Tom Robertts tjro...@lucent.com
FrediFizzx
| Dirk Van de moortel wrote:
| > Not even looking at the missing factor 2, this is a very nice one:
| >
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Confusion.html
|
| Dirk:
|
| Snap, crackle, and pop have been used semi-humorously to describe higher
| derivatives of position (Newtonian mechanics):
| dx/dt velocity
| d^2x/dt^2 acceleration
| d^3x/dt^3 jerk
| d^4x/dt^4 snap
| d^5x/dt^5 crackle
| d^6x/dt^6 pop
|
| I believe this came from someone having a real use for "snap", and the
| others just followed naturally (:-). Their utility in real problems is
| unlikely -- even using jerk is rare.
My wife calls me a jerk when I stomp on the accelerator. ;-0
FrediFizzx
Dirk Van de moortel
"Tom Roberts" <tjro...@lucent.com> wrote in message news:6VZfc.9206$ni3....@newssvr31.news.prodigy.com...
> Dirk Van de moortel wrote:
> > Not even looking at the missing factor 2, this is a very nice one:
> > http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Confusion.html
>
> Dirk:
>
> Snap, crackle, and pop have been used semi-humorously to describe higher
> derivatives of position (Newtonian mechanics):
> dx/dt velocity
> d^2x/dt^2 acceleration
> d^3x/dt^3 jerk
> d^4x/dt^4 snap
> d^5x/dt^5 crackle
> d^6x/dt^6 pop
I know.
Last year I posted this:
http://groups.google.com/groups?&as_umsgid=d2rGa.51399$1u5....@afrodite.telenet-ops.be
>
> I believe this came from someone having a real use for "snap", and the
> others just followed naturally (:-). Their utility in real problems is
> unlikely -- even using jerk is rare.
>
> But John Schoenfeld has been differentiating wrt r, not t, so he did fumble.
Indeed, that is why I highlighted the d/dr part and the 'jerk' word only.
Dirk Vdm
Dirk Van de moortel
If it was a typo, you wouldn't have come up with the r^3 in the
denominator.
Dirk Vdm
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:fX5gc.74383$yX2.4...@phobos.telenet-ops.be...
>
> "John Schoenfeld" <j.scho...@programmer.net> wrote in message
news:a98beaaa.04041...@posting.google.com...
...
> > That was a typo, obviously.
>
> If it was a typo, you wouldn't have come up with the r^3 in the
> denominator.
Would that make it a derivo?
David A. Smith
Dirk Van de moortel
"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:n...@nospam.com> wrote in message news:nvegc.14359$Yf6.7124@fed1read07...
A defumblo ;-)
Dirk Vdm
John Schoenfeld
>
> Yes there is. You simply repeat the error(s) I have already pointed out.
>
>
> > If a(r) = 1/r^2, then the jerk is in the OPPOSITE DIRECTION. Think
> > about it.
>
> > d a(r)/dr = - 1/r^3
>
> But jerk is not "d a(r)/dr", it is d a(t)/dt. THE DIFFERENCE IS IMPORTANT.
Looks like I made an error.
Yes, d a(r)/dr gives the gradient of the gravitational field in the
direction of r, not the jerk of a body in the gravitational field as I
erroneously thought.
To get the jerk:
a(r(t)) = -GM/r^2
using the chain rule
d(a o r)/dt = da/dr . dr/dt
da/dr = which is simply 2GM/r^3.
Since dr/dt gives the component of velocity parallel to r, it is
equivalent to the actual velocity projected onto r.
dr/dt = (v . r) / |r|
So the jerk of a body in a Newtonian gravitational field is:
j = 2GM(v . r)/r^4
where
M = gravitational source mass
v = velocity of body
r = displacement between body and M's center of gravity
>
> > The snap is in the same direction as the acceleration, and the crackle
> > the same as the jerk. But what about the pop?
>
> Successive derivatives wrt r are IRRELEVANT -- you need to express a as
> a function of t and take successive derivatives wrt t. THE DIFFERENCE IS
> IMPORTANT. This makes the signs of these derivatives more complicated
> than you think (you get additional powers of velocity).
I don't agree they are irrelevant in general, but irrelevant in this
scenario.
>
> > What is all this? Are you seriously mounting a challenge to the
> > validity of the Abraham-Lorentz formula?
>
> On clear physical grounds there is a sign error in your formula for Fr.
This formula is known as the "Abraham-Lorentz" formula, and it does
not have a sign error.
www.physics.ubc.ca/~mattison/Courses/ Phys454/lecture31.pdf
http://www.physics.uq.edu.au/people/jones/phys6040/tut04/node1.html
> > The problems I refered to are well known in physics:
> > problem 1: "runaway solutions"
> > problem 2: "acausual preacceleration"
>
> These are not "well known", these are errors. With reasons I discussed
> in earlier posts.
Runaway solutions and acausal preacceleration have been a topic of
dicussion for quite some time, I doubt they are "errors" as you put
it.
www.pas.rochester.edu/~dmw/phy218/ Lectures/Lect_70b.pdf
> > As for my "sign error", i think that the error is not mine.
>
> You certainly make sign mistakes. And conceptual errors (r is not t).
> You will never understand this until you correct your errors and
> misconceptions. It is more complicated than you think.
I did have a misconception about dr/dt, which I now resolved thanks in
large part to you pointing it out.
However, this does not invalidate the validity of Radiation-Reaction
and the problems in poses for Classical electrodynamics.
JS
John Schoenfeld
Yes it was a blunder. It should've been, jerk = 2GM(v.r)/r^4 taking
the direction of acceleration as negative.
> Dirk Vdm
John Schoenfeld
> > Since da/dt is the jerk of the displacement field caused by the
> > gravitational field Gm1/r^2, da/dt evaluates to -Gm1/r^3.
>
> As I pointed out before, this is wrong:
> 1) the gravitational acceleration is -Gm1/r^2
> 2) you need to compute da/dt, not da/dr (and you omitted a factor
> of 2 in your irrelevant computation)
>
> To compute da/dt you can use the chain rule:
> da/dt = (da/dr) (dr/dt)
> but the presence of dr/dt (velocity) screws up your entire analysis....
Not so fast. dr/dt is NOT the velocity as you erroneously put it. It
is the component of velocity parallel to r. dr/dt = (v.r)/|r|. "THE
DIFFERENCE IS IMPORTANT". I wonder if Dirk will put this on his
website (i doubt it).
> You have also omitted initial conditions in your analysis, and they can
> change the conclusion completely (e.g. the particle could be thrown
> UPWARD, in which case Frad points downward)....
If the particle is thrown upward, then it is decelerating because it's
velocity is approaching 0. If you look at my jerk formula =
2GM(v.r)/r^4, you can see that if the velocity makes an angle greater
than 90deg with r, then the sign gets flipped.
The initial conditions therefore do make a difference (as you
contend), a DECELERATING charge will experience a force downwards, and
accelerating charge a force upwards.
> Bottom line: with Frad proportional to either -da/dt or +da/dt, it is
> not possible to use Frad to "levitate" a charge, because for levitation
> v=0, a=0, da/dt=0, and Frad=0, implying that Fg is unopposed and pulls
> it downward. Yes, Frad will then retard that falling (once you correct
> the sign error), but it cannot prevent it -- the particle will
> accelerate downward until Fg=-Frad, at which time it will fall with
> constant velocity. By playing with q/m you can make that velocity
> slower, but you cannot make it vanish.
You are making a fundamental error in assuming that Fg and Frad act
simultaneously on the charge, which they don't.
It takes time for Frad to act on the entire charge as the emitted wave
needs to propagate the entire particle.
For all intensive purposes Fg acts on the entire body instantaneously
but Frad acts variably over the period of c/r where r is the radius of
the charge.
Since your conclusion was derived from this erroneous assumption, it
too is transistively wrong.
> Tom Roberts tjro...@lucent.com
JS
Dirk Van de moortel
"John Schoenfeld" <j.scho...@programmer.net> wrote in message news:a98beaaa.04041...@posting.google.com...
[snip]
>
> Since your conclusion was derived from this erroneous assumption, it
> too is transistively wrong.
And on top of everything else, you have a severe problem
with spelling *and* with logic as well.
Dirk Vdm
Timo Nieminen
> Timo Nieminen <ti...@physics.uq.edu.au> wrote:
> > On Thu, 15 Apr 2004, John Schoenfeld wrote:
> >
> > > > > Given the right charge to mass ratio, a classically charged mass can
> > > > > levitate in Earth gravity by simply counter-acting the gravitational
> > > > > force with its radiation reaction force.
> > > >
> > > > How? Radiative reaction force is proportional to da/dt. Levitation implies
> > > > a = 0, thus radiative reaction force is zero.
> > >
> > > Not necessarily. If you use the Abraham-Lorentz radiation reaction
> > > formula and Newtons law of gravity, you can model the radiating
> > > self-levitating charge as follows:
> > >
> > > Fg = -Frad
> >
> > Yes, for levitation, you must have Fg = -Frad.
> >
> > So a = 0. F = 0, F = ma, means a = 0.
>
> The acceleration acted on the charged mass is GM/r^2.
In the absence of forces other than gravitational attraction, yes.
Levitation requires a = 0.
> Your assertion
> that a = 0 is true only if you are refering to the average
> acceleration for the time interval r/c (where r is the radius of the
> particle).
No. Stable levitation requires r = constant, so dr/dt = v = 0, dv/dt = a =
0. And da/dt = 0.
> If you are making the argument that point charges could not
> self-levitate then are correct, since r=0 and thus a = 0 at all times.
Why r = 0? For levitation, you just need r = constant. r is not the radius
of the particle; it's the distance from the center of the nearby massive
(spherically symmetric) object.
> Radiative self-levitation of POINT charges may not be possible, but I
> iterate again that spherical charges can exhibit this behaviour since
> a(t) != 0 for all 0 < t < r/c. Rather, int(t=0, r/c) v(t) . dt = 0
> which does not preclude Fg=-Frad where a > 0.
If this is what you meant, you should have been more precise. OK, so you
have a charge distribution with accelerating charges therein. Self
leviation of such a thing might well be possible, at least for an instant
of time. A constant input power will be required for stable sustained
self-levitation. But, as you noted earlier, the minimum power required is
Fg/c.
In this case, the acceleration of a small volume element of the charge
will not be GM/r^2, since it will also depend on the internal forces
holding the charge distribution together.
> An analysis of the "micromotion" of this levitating charge shows that
> it oscillates along its vertical diameter with amplitude r and
> frequency proportional to r and q.
Then it isn't stably levitating. Is this the actual behaviour that
you'd expect, from Lorentz-Abraham? Do the maths and show it.
> > > It would seem logical that this non-conserved energy is is actually
> > > conserved by a reduction in the mass of the charge, which is not
> > > allowed in Maxwells theory.
> >
> > Why is this not allowed in Maxwell's theory? Classical EM theory either
> > says nothing about mass and conservation of mass, unless special
> > relativity is included within classical EM theory, so how is it
> > disallowed?
>
> Mass invariance is implicit in Newtonian mechanics
[... cut]
And Newtonian mechanics is not Maxwell's theory. Thus, irrelevant.
Consider the electromagnetic nature of mass in classical models of charged
particles.
--
Timo Nieminen
Bilge
I have no idea what you are talking about. I don't have that book
so where you come by the curl of a divergence, I don't know.
The phase here is the phase of the electron wavefunction (or
field - however you want to think of it).
Tom Roberts
> Tom Roberts <tjro...@lucent.com> wrote:
>>To compute da/dt you can use the chain rule:
>> da/dt = (da/dr) (dr/dt)
>>but the presence of dr/dt (velocity) screws up your entire analysis....
>
> Not so fast. dr/dt is NOT the velocity as you erroneously put it. It
> is the component of velocity parallel to r.
Yes. You were discussing "levitation", which involves purely radial
motion, and for that dr/dt is indeed velocity.
[In the common approach where vectors and components are
interchanged (this is really the radial COMPONENT of the
velocity 3-vector -- remember we are in the Newtonian
approximation here). I deliberately chose not to belabor
this.]
>>Bottom line: with Frad proportional to either -da/dt or +da/dt, it is
>>not possible to use Frad to "levitate" a charge, because for levitation
>>v=0, a=0, da/dt=0, and Frad=0, implying that Fg is unopposed and pulls
>>it downward. Yes, Frad will then retard that falling (once you correct
>>the sign error), but it cannot prevent it -- the particle will
>>accelerate downward until Fg=-Frad, at which time it will fall with
>>constant velocity. By playing with q/m you can make that velocity
>>slower, but you cannot make it vanish.
>
> You are making a fundamental error in assuming that Fg and Frad act
> simultaneously on the charge, which they don't.
Nonsense. They BOTH operate CONTINUOUSLY on the charge. And in the
Newtonian approximation any sort of "propagation delay" is ignored
(remember that approximation involves the average of the retarded and
advanced potentials of relativistic electrodynamics). I was merely
giving a simplified description of how I think the solution to the
differential equation behaves -- when dropped from rest, levitation is
NOT possible, and the charge will accelerate down until it reaches[#] a
terminal velocity (i.e. the velocity and jerk where Frad = -Fg).
[#] Actually I suspect it only asymptotically approaches the
"terminal velocity"....
> It takes time for Frad to act on the entire charge as the emitted wave
> needs to propagate the entire particle.
Nonsense. The charged particle is of negligible size for such
considerations. Certainly this is so for any elementary particle, but it
is true even for a beachball with a static charge falling on earth. The
Newtonian approximation includes this (for that beachball the
propagation delay is ~2 ns, which is FAR below your ability to measure,
so the approximation involved is quite good).
Tom Roberts tjro...@lucent.com
Ken S. Tucker
I wrote Curl Grad =0. I thought everybody had that
book, it's cheap, if you can't afford it, I'll buy you
a bithday gift, it's certainly dated but is a good read.
>The phase here is the phase of the electron wavefunction (or
>field - however you want to think of it).
Ok, Weyl is very reputable, so likely his article is
sensible, I can't believe he made such an obvious
error, so I was trolling for opinions.
I'll call the electromagnetic potential vector A_u.
This can't be a gradient otherwise it's Curl ==F_uv
would be zero. I now use,
A_u = A(x_u/s), A'^u = A(dx^u/ds), A,u =0
with "A" the invariant potential. It looks a bit queer,
especially when association is employed! So I prime
the velocity potential A'^u.
Comments welcome
Ken S. Tucker
John Schoenfeld
> > simultaneously on the charge, which they don't.
>
> Nonsense. They BOTH operate CONTINUOUSLY on the charge. And in the
> Newtonian approximation any sort of "propagation delay" is ignored
> (remember that approximation involves the average of the retarded and
> advanced potentials of relativistic electrodynamics). I was merely
> giving a simplified description of how I think the solution to the
> differential equation behaves -- when dropped from rest, levitation is
> NOT possible, and the charge will accelerate down until it reaches[#] a
> terminal velocity (i.e. the velocity and jerk where Frad = -Fg).
>
> [#] Actually I suspect it only asymptotically approaches the
> "terminal velocity"....
>
>
> > It takes time for Frad to act on the entire charge as the emitted wave
> > needs to propagate the entire particle.
>
> Nonsense. The charged particle is of negligible size for such
> considerations. Certainly this is so for any elementary particle, but it
> is true even for a beachball with a static charge falling on earth. The
> Newtonian approximation includes this (for that beachball the
> propagation delay is ~2 ns, which is FAR below your ability to measure,
> so the approximation involved is quite good).
Interesting, you have absolutely no mathematical basis for any of
these claims, but interesting. I am fascinated the acceleration of a
charge can be 0 despite the fact that gravitational force and the
caused radiation reaction force do not occur simultaneously. In fact
there is a delay of c/r, which gives the charge ample opportunity to
accelerate by gravity. But still, interesting.
JS
Bilge
Yeah, I know. I wrote divergence by mistake. Repeat what I wrote
using grad instead of divergence.
>I thought everybody had that book, it's cheap, if you can't afford it,
>I'll buy you a bithday gift, it's certainly dated but is a good read.
I can probably live without yet another book to explain what a
grad, curl and divergence are.
>
>>The phase here is the phase of the electron wavefunction (or
>>field - however you want to think of it).
>
>Ok, Weyl is very reputable, so likely his article is sensible, I
>can't believe he made such an obvious error, so I was trolling for
>opinions.
I seriously doubt that weyl made error, too. That still doesn't
explain what you are talking about.
>I'll call the electromagnetic potential vector A_u.
>
>This can't be a gradient otherwise it's Curl ==F_uv would be zero.
>I now use,
> A_u = A(x_u/s), A'^u = A(dx^u/ds), A,u =0
A curl is a 3-d operator. F_uv is a rank two tensor in 4 dimensions.
>with "A" the invariant potential. It looks a bit queer, especially when
>association is employed! So I prime the velocity potential A'^u.
I still don't know what you are talking about.
Ken S. Tucker
>Ken S. Tucker:
[chop]
> >This can't be a gradient otherwise it's Curl ==F_uv would be zero.
> >I now use,
> > A_u = A(x_u/s), A'^u = A(dx^u/ds), A,u =0
>
> A curl is a 3-d operator. F_uv is a rank two tensor in 4 dimensions.
This is about as close to plain english as we ever get!
Cut me a bit of slack and allow me to call the invariant
A =q/s for some definition....
with q=charge, and "s" an invariant spacetime interval.
Then Grad that to make A,u (partial diff), followed
by a Curl to make,
Curl A,u = A,u,v - A,v,u
I maintain that last equation is zero, in any number of
dimensions, is that my mistake?
Regards
Ken S. Tucker
PS: If true then Gradients of the potential
do not enter into EM-fields.
Bilge
>news:<slrnc8a8ho....@radioactivex.lebesque-al.net>...
>>Ken S. Tucker:
>[chop]
>
>> >This can't be a gradient otherwise it's Curl ==F_uv would be zero.
>> >I now use,
>> > A_u = A(x_u/s), A'^u = A(dx^u/ds), A,u =0
>>
>> A curl is a 3-d operator. F_uv is a rank two tensor in 4 dimensions.
>
>This is about as close to plain english as we ever get!
>
>Cut me a bit of slack and allow me to call the invariant
>
> A =q/s for some definition....
>
>with q=charge, and "s" an invariant spacetime interval.
>
> Then Grad that to make A,u (partial diff), followed
>by a Curl to make,
>
> Curl A,u = A,u,v - A,v,u
notation which is self-consistent). Your expression has 1 index on the lhs
and two indicies on the rhs. If you use indicies, make sure the indicies
match. If you don't use indicies, you'll still have to define what you
mean by ``Curl'' in four dimensions. In three dimensions, it's defined
by,
curl A = \epsilon_ijk \nabla_i A_j
If by ``Curl'', you mean:
\epsilon_abc d^a d^b A^c = d^a d^b A^c + d^c d^a A^b + d^b d^c A^a
= d^a F^bc + d^c F^ab + d^b F^ca
= 0
Then, yes, that expression is zero. It's the equivalent to a ``bianchi
identity'' for the faraday tensor, which is the equivalent of
``curvature'' for the electromagnetic field.
>I maintain that last equation is zero, in any number of
>dimensions, is that my mistake?
Hard to say. I don't know what your equation means.
>PS: If true then Gradients of the potential do not enter into EM-fields.
The gradients don't enter into the fields. Whether or not the reason
is what you think is the reason, depends upon what you think is the reason.
To make a long story short, the reason is gauge invariance. Lorentz
invariance requires that it be possible to make a gauge transformation
such that A^u satisfies, d_u A^u = 0. That means, the four vector,
A^u = (\phi, A) has (at most) 3 degrees of freedom, so you can always
choose the time component to be zero, leaving you free to choose the
other three components, A^i, such that div . A = 0. If you require that
curl A = 0, too, you have no E or B field and you have to resort to
quantum mechanics to tell what happens.
Ken S. Tucker
Thanks Bilge, really nice response...
>Ken S. Tucker:
> >dub...@radioactivex.lebesque-al.net (Bilge) wrote in message
> >news:<slrnc8a8ho....@radioactivex.lebesque-al.net>...
> >>Ken S. Tucker:
> >[chop]
> >
> >> >This can't be a gradient otherwise it's Curl ==F_uv would be zero.
> >> >I now use,
> >> > A_u = A(x_u/s), A'^u = A(dx^u/ds), A,u =0
> >>
> >> A curl is a 3-d operator. F_uv is a rank two tensor in 4 dimensions.
> >
> >This is about as close to plain english as we ever get!
> >Cut me a bit of slack and allow me to call the invariant
> > A =q/s for some definition....
> >with q=charge, and "s" an invariant spacetime interval.
> > Then Grad that to make A,u (partial diff), followed
> >by a Curl to make,
> >
> > Curl A,u = A,u,v - A,v,u
>
> Could you please stick to notation that makes sense (i.e., _one_ standard
>notation which is self-consistent). Your expression has 1 index on the lhs
>and two indicies on the rhs.
Bilge, the equation is facetious, but not sarcastic,
I'm Curling a Gradient in any number of dimensions
summed over "u", to show Grad A,u =0.
>If you use indicies, make sure the indicies
>match. If you don't use indicies, you'll still have to define what you
>mean by ``Curl'' in four dimensions. In three dimensions, it's defined
>by,
>
> curl A = \epsilon_ijk \nabla_i A_j
understood, ((epslion permeates))...
> If by ``Curl'', you mean:
>
>\epsilon_abc d^a d^b A^c = d^a d^b A^c + d^c d^a A^b + d^b d^c A^a
>
> = d^a F^bc + d^c F^ab + d^b F^ca
>
> = 0
>
> Then, yes, that expression is zero. It's the equivalent to a ``bianchi
>identity'' for the faraday tensor, which is the equivalent of
>``curvature'' for the electromagnetic field.
Wow, if I understand correctly, you did a 4D permeatation
on the epsilon tensor fucking over the field tensor to confirm
Maxwell's second set to be zero in a vacuum., ok...
But I will hesitate, about agreeing on the zero
curvature, when variations of the Field tensor
"F^ab" occur.
I think this because, we usually use the propagation
of light waves and thus the variations of "F_ab" to map
our gravitational fields. For instance, when light is deflected
by a g-field,
> >I maintain that last equation is zero, in any number of
> >dimensions, is that my mistake?
>
> Hard to say. I don't know what your equation means.
>
> >PS: If true then Gradients of the potential do not enter into EM-fields.
>
> The gradients don't enter into the fields. Whether or not the reason
>is what you think is the reason, depends upon what you think is the reason.
>To make a long story short, the reason is gauge invariance. Lorentz
>invariance requires that it be possible to make a gauge transformation
>such that A^u satisfies, d_u A^u = 0.
>That means, the four vector,
>A^u = (\phi, A) has (at most) 3 degrees of freedom, so you can always
>choose the time component to be zero, leaving you free to choose the
>other three components, A^i, such that div . A = 0. If you require that
>curl A = 0, too, you have no E or B field and you have to resort to
>quantum mechanics to tell what happens.
Thank you, I think that too.
I'm captivated by d_u A^u=0
((I'm uncertain about your notion "d_u"))
Tanks Bilge...
Ken S. Tucker