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| Timo Nieminen |
 Posted: Sun Jun 06, 2004 6:03 pm |
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On Sat, 4 Jun 2004, John Schoenfeld wrote:
[quote:d3e3bd6b81]Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Fri, 3 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 2 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 2 Jun 2004, John Schoenfeld wrote:
Is there anything in classical e&m that _prohibits_ a divergence in
magnetic fields?
div B = 0 is an experimental observation.
Yes an observation from steady currents.
No. It's from observations of steady currents, slowly-varying currents,
rapidly varying currents, and zero currents.
I meant "charges in motion".
You wrote "steady currents". Exactly what do you mean? In any case,
div B = 0 is observed with moving charges, stationary charges,
accelerating charges, and in the absence of charge.
Yes, hence "charges in motion".
[/quote:d3e3bd6b81]
Note "absence of charge" in the list above.
[quote:d3e3bd6b81]And Magnetic fields occur in the
absence of charge?
[/quote:d3e3bd6b81]
What is the magnetic moment of a neutron? What is the charge of a neutron?
That's about the closest to magnetic field in the absence of a charge
that's been observed.
[quote:d3e3bd6b81]Or are you trying to say that the divergence of a
non-existent field is 0?
[/quote:d3e3bd6b81]
No, I'm saying that div B = 0 for all magnetic fields that have been
observed.
[quote:d3e3bd6b81]Note also that the necessary modifications to the Maxwell equations if
magnetic monopoles are found have been known for a long time. 1931 if not
earlier (possibly much earlier).
My question is _why_ would the Maxwell equations require
modifications? If a premise of a theory is that charges in motion
cause solenoidal magnetic fields, then even if magnetic charge can
exist in some other way, why would this the theory require alteration
since the original premise remains valid?
If magnetic charge is found, then you no longer have div B = 0.
Since div B = 0 is one of the Maxwell equations, then the Maxwell
equations require modification.
Read properly. Why does magnetic charge change div B = 0 for magnetic
fields of charges in motion?
[/quote:d3e3bd6b81]
Why don't you read properly? The Maxwell equations are not restricted to
fields of charges in motion; they describe the behaviour of electric and
magnetic fields, irrespective of their sources. Since this behaviour does
depend on sources, all known sources of electric and magnetic fields must
appear in the Maxwell equations. At the moment, only electric charge and
electric current are included, since these are the only known sources. If
magnetic charge is found, then magnetic charge and current will need to be
included. Until then, they're not needed. (But you might note that
magnetic currents are sometimes used as equivalent sources.)
[quote:d3e3bd6b81]The original premise is div B = 0, and this would no longer be valid.
No, it would still be valid.
[/quote:d3e3bd6b81]
No, it would not. Think about what div B = 0 actually means.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Richard Herring |
| Posted: Mon Jun 07, 2004 2:54 am |
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In message <a98beaaa.0406040750.4dffead5@posting.google.com>, John
Schoenfeld <j.schoenfeld@programmer.net> writes
[quote:3412e5e5a9]Richard Herring <junk@[127.0.0.1]> wrote in message
news:<5sxmXUCc5DwAFwuq@baesystems.com>...
In message <a98beaaa.0406031501.6f8f9de9@posting.google.com>, John
Schoenfeld <j.schoenfeld@programmer.net> writes
Richard Herring <junk@[127.0.0.1]> wrote in message
news:<po3i2pa7mxvAFwUB@baesystems.com>...
In message <a98beaaa.0406021802.20b74de4@posting.google.com>, John
Schoenfeld <j.schoenfeld@programmer.net> writes
"Robert J. Kolker" <robert_kolker@hotmail.com> wrote in message
news:<aSrvc.37813$eY2.10359@attbi_s02>...
John Schoenfeld wrote:
Is there anything in classical e&m that _prohibits_ a divergence in
magnetic fields? If so, what precisely is it that forces _all_
magnetic fields to be solenoidal (abductive reasoning is not good
enough)?
It will have to do. The only magenetic fields that have been seen are
solenoidal and at right angles to the current. If someone does come up
with a magnetic monople Maxwells's Equations will have to be altered.
I don't see why they would require alteration since the equations are
considering only the special case of steady currents
Really? Which part of Maxwell's equations implies steady current?
I meant charges in motion, forgetting the reserved words steady current.
Sense no makes that.
Having trouble reading again?
[/quote:3412e5e5a9]
Yes, but only when you write it. Odd, that.
--
Richard Herring |
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| John Schoenfeld |
| Posted: Mon Jun 07, 2004 11:58 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.0406070956050.13539-100000@localhost>...
[quote:1ac70a72fa]On Sat, 4 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Fri, 3 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 2 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 2 Jun 2004, John Schoenfeld wrote:
Is there anything in classical e&m that _prohibits_ a divergence in
magnetic fields?
div B = 0 is an experimental observation.
Yes an observation from steady currents.
No. It's from observations of steady currents, slowly-varying currents,
rapidly varying currents, and zero currents.
I meant "charges in motion".
You wrote "steady currents". Exactly what do you mean? In any case,
div B = 0 is observed with moving charges, stationary charges,
accelerating charges, and in the absence of charge.
Yes, hence "charges in motion".
Note "absence of charge" in the list above.
[/quote:1ac70a72fa]
That is an extraordinary claim, which I hope you can back up with
references. The only known cause of magnetic fields are electric
charges in motion.
[quote:1ac70a72fa]And Magnetic fields occur in the
absence of charge?
What is the magnetic moment of a neutron? What is the charge of a neutron?
That's about the closest to magnetic field in the absence of a charge
that's been observed.
[/quote:1ac70a72fa]
Quantum Theory and scattering experiments both show neutron's are
internally charged. A neutron consists of 1 up quark (charge: 2/3) and
2 down quarks (charge: -1/3) which sum to the expected 0. However, the
charge distribution of a neutron is certainly not uniformly 0, and
thus a neutron in motion would indeed induce a solenoidal magnetic
field, as predicted by classical E&M. The problem is in the definition
of a classical neutron as having 0 electric charge, which is false.
[quote:1ac70a72fa]Or are you trying to say that the divergence of a
non-existent field is 0?
No, I'm saying that div B = 0 for all magnetic fields that have been
observed.
[/quote:1ac70a72fa]
I do not disagree with that.
[quote:1ac70a72fa]Note also that the necessary modifications to the Maxwell equations if
magnetic monopoles are found have been known for a long time. 1931 if not
earlier (possibly much earlier).
My question is _why_ would the Maxwell equations require
modifications? If a premise of a theory is that charges in motion
cause solenoidal magnetic fields, then even if magnetic charge can
exist in some other way, why would this the theory require alteration
since the original premise remains valid?
If magnetic charge is found, then you no longer have div B = 0.
Since div B = 0 is one of the Maxwell equations, then the Maxwell
equations require modification.
Read properly. Why does magnetic charge change div B = 0 for magnetic
fields of charges in motion?
Why don't you read properly? The Maxwell equations are not restricted to
fields of charges in motion; they describe the behaviour of electric and
magnetic fields, irrespective of their sources. Since this behaviour does
depend on sources, all known sources of electric and magnetic fields must
appear in the Maxwell equations. At the moment, only electric charge and
electric current are included, since these are the only known sources. If
magnetic charge is found, then magnetic charge and current will need to be
included. Until then, they're not needed. (But you might note that
magnetic currents are sometimes used as equivalent sources.)
[/quote:1ac70a72fa]
This is where I disagree. By design, Maxwell's theory is constrained
to the special case of electric charges in motion. Therefore, an
observation of magnetic charge does not suddenly disprove Maxwell's
theory, it just shows that Maxwell's theory does not account for all
classes of electromagnetic phenomena. Obviously, if one was to
describe all known cases of e&m in the presence of magnetic charge,
the equations require modification; not so for the special case of
electric charges in motion.
[quote:1ac70a72fa]The original premise is div B = 0, and this would no longer be valid.
No, it would still be valid.
No, it would not. Think about what div B = 0 actually means.
[/quote:1ac70a72fa]
div B = 0 is an experimentally verified premise which the remainder of
the equations are based off. Maxwell's equations are true for div B =
0. |
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| Alfred Einstead |
| Posted: Mon Jun 07, 2004 3:24 pm |
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j.schoenfeld@programmer.net (John Schoenfeld) wrote:
[quote:8269297640]Is there anything in classical e&m that _prohibits_ a divergence in
magnetic fields?
[/quote:8269297640]
The opposite is true.
The classical electromagnetic field is part of a larger
classical Yang-Mills field (the electroweak field) whose
Maxwell equations are non-linear and DO have magnetic current
terms on the righthand side. There's a contribution from
the W+ and W- fields on the right hand side of the Maxwell
equations directly related to the weak mixing angle. |
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| Timo Nieminen |
| Posted: Mon Jun 07, 2004 5:15 pm |
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On Tue, 7 Jun 2004, John Schoenfeld wrote:
[quote:3e0326d00f]Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Sat, 4 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Fri, 3 Jun 2004, John Schoenfeld wrote:
I meant "charges in motion".
You wrote "steady currents". Exactly what do you mean? In any case,
div B = 0 is observed with moving charges, stationary charges,
accelerating charges, and in the absence of charge.
Yes, hence "charges in motion".
Note "absence of charge" in the list above.
That is an extraordinary claim, which I hope you can back up with
references. The only known cause of magnetic fields are electric
charges in motion.
[/quote:3e0326d00f]
.... with a possible exception of free radiation fields. But this is a
matter of definition.
I don't see why you believe the claim above to be extraordinary. Do you
know of an observation of div B != 0 in the absence of charge?
Granted, if one excludes free radiation fields and fields of particles
such as neutrons, div B = 0 in the absence of charge reduces to the rather
tautologous div 0 = 0, but this certainly doesn't harm the correctness of
the claim.
[quote:3e0326d00f]And Magnetic fields occur in the
absence of charge?
What is the magnetic moment of a neutron? What is the charge of a neutron?
That's about the closest to magnetic field in the absence of a charge
that's been observed.
Quantum Theory and scattering experiments both show neutron's are
internally charged.
[/quote:3e0326d00f]
As I said, that's about the closest observed. What else would you expect
in the absence of magnetic charge?
But note that your original question restricted the matter to classical EM
theory, and bringing in the question of particulate matter is somewhat
beside the point, especially elements of quantum theory.
[quote:3e0326d00f]No, I'm saying that div B = 0 for all magnetic fields that have been
observed.
I do not disagree with that.
[/quote:3e0326d00f]
So what is your point?
[quote:3e0326d00f]My question is _why_ would the Maxwell equations require
modifications? If a premise of a theory is that charges in motion
cause solenoidal magnetic fields, then even if magnetic charge can
exist in some other way, why would this the theory require alteration
since the original premise remains valid?
If magnetic charge is found, then you no longer have div B = 0.
Since div B = 0 is one of the Maxwell equations, then the Maxwell
equations require modification.
Read properly. Why does magnetic charge change div B = 0 for magnetic
fields of charges in motion?
Why don't you read properly? The Maxwell equations are not restricted to
fields of charges in motion; they describe the behaviour of electric and
magnetic fields, irrespective of their sources. Since this behaviour does
depend on sources, all known sources of electric and magnetic fields must
appear in the Maxwell equations. At the moment, only electric charge and
electric current are included, since these are the only known sources. If
magnetic charge is found, then magnetic charge and current will need to be
included. Until then, they're not needed. (But you might note that
magnetic currents are sometimes used as equivalent sources.)
This is where I disagree. By design, Maxwell's theory is constrained
to the special case of electric charges in motion.
[/quote:3e0326d00f]
Then you're quibbling about a philosophical matter of opinion, and not
physics.
If you really, truly, believe that the Maxwell equations are intended to
be restricted only to classical electromagnetic phenomena produced by
electric charges in motion (presumably including the special case of
stationary charges), then what was the point of your original question,
since in that case, classical EM theory can have nothing at all to say
about magnetic charge?
Perhaps you would be more content if I rephrase the statement on
modification of the Maxwell equations to:
The modifications that would be required for a classical EM theory
desribing all classical EM phenomena, were magnetic charge to be
discovered, are simple and well-known.
[quote:3e0326d00f]The original premise is div B = 0, and this would no longer be valid.
No, it would still be valid.
No, it would not. Think about what div B = 0 actually means.
div B = 0 is an experimentally verified premise which the remainder of
the equations are based off. Maxwell's equations are true for div B =
0.
[/quote:3e0326d00f]
If magnetic charge is observed, div B = 0 will no longer be an
experimentally verified premise. And therefore will require modification.
How can div B = 0 be valid if contradicted by observation?
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| greywolf42 |
| Posted: Tue Jun 08, 2004 10:58 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406070828470.13486-100000@localhost...
[quote:8fc7465d48]On Fri, 4 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
[/quote:8fc7465d48]
{snip higher levels}
[quote:8fc7465d48]
Sure, Maxwell is definitely classical, and it may well be that
Maxwell's ether model prohibits divergence in magnetic fields.
However, Maxwell's ether model is not classical EM theory.
ROTFLMAO!!!!
Have you not stated in the past that in Maxwell's theory, the Maxwell
equations are only approximations valid in the case of low speed
relative to the ether of both observer and source?
I have indeed so stated.
Then Maxwell's ether model is not classical EM theory.
[/quote:8fc7465d48]
Non sequiteur.
[quote:8fc7465d48]Do note that
"classical electromagnetic theory" is a technical term with a meaning more
precise than "classical" "electromagnetic theory". Ritz's and Weber's
electromagnetic theories are classical as well, but they're not "classical
electromagnetic theory", either.
[/quote:8fc7465d48]
Then why don't you avoid silly quibbling about what is contained in the
definition of the three words 'classical electromagnetic theory' and address
the substantive issue?
The point is, that a boldfaced lie was made, when it was claimed that there
was no theoretical argument against the existence of magnetic monopoles.
That you wish to ignore Maxwell's derivation of his equations is merely
propaganda.
[quote:8fc7465d48]What bearing, then, does
Maxwell's model have on classical EM theory?
Well, Maxwell's equations were derived from Maxwell's classical model.
That makes them classical.
No, that isn't what makes them classical.
[/quote:8fc7465d48]
What does? It doesn't use quantum mechanics. It doesn't use SR. It is
entirely based on Newtonian mechanics. That qualifies for 'classical' in my
book.
[quote:8fc7465d48]Note well that Lorenz obtained the Maxwell equations almost
simultaneously without recourse to an ether or a mechanical
model,
Well, this is news to me. Do you have a reference for this work of
Lorenz?
Given that the paper by Jackson & Okun (RMP 2001) you have been previously
referred to, and the content of which you did discuss over the course of
many posts in a previous thread, discusses the work of Lorenz, and gives
references to Lorenz's original publication, I am somewhat surprised that
this is news to you.
Jackson & Okun also note that Maxwell refers to Lorenz's work in his
Treatise.
[/quote:8fc7465d48]
That Maxwell's Treatise (and Jackson & Okun) refer to Lorenz does not equate
with Lorenz having derived all of Maxwell's equations. The thread to which
you refer was with regard to the origin of the concept of gauge invariance.
http://www.google.com/groups?selm=102id1b3b17soe5%40corp.supernews.com
You'll note that I caught Jackson being freely creative with history.
"(Jackson's) 'imprecision' with Maxwell does not make me confident of his
abilities
with Lorenz, Hertz or Heaviside."
[quote:8fc7465d48]and Hertz did the same later.
And then there are the modern derivations.
But none of these are physical derivations. They are merely mathematical
manipulations.
Why? Both Lorenz and Hertz made use of the physical behaviour of
electromagnetic systems in their derivations. How they merely mathematical
manipulations?
[/quote:8fc7465d48]
Because there was never any attempt to handle the symbols as anything
physical. (Hint: "Electromagnetic systems" begs the question.)
[quote:8fc7465d48]The modern derivations are very firmly based on a small set of
well-observed physical phenomena.
[/quote:8fc7465d48]
But there are no phenomena at all used in these 'modern' derivations. It's
all pure symbolism. Arbitrary constants. Nothing physical.
[quote:8fc7465d48]All more physical than Maxwell's model, in that they don't invoke the
mechanical properties of a non-existent ether.
[/quote:8fc7465d48]
Without physical properties, they aren't physical.
[quote:8fc7465d48]Irrelevant to the fact that the only physical derivation of Maxwell's
equations prohibits magnetic monopoles.
But, as you have repeatedly said, the Maxwell equations are only an
approximation of the behaviour of Maxwell's mechanical model.
[/quote:8fc7465d48]
But a very good approximation for most laboratory work.
[quote:8fc7465d48]Is div B = 0 an approximation, or is it actually exact?
[/quote:8fc7465d48]
That part is exact.
[quote:8fc7465d48]If it is exact, in
what way are the Maxwell equations an approximation?
[/quote:8fc7465d48]
The same way they have been described literally dozens of times for you,
Timo.
http://www.google.com/groups?selm=101vu9m7c668ua4%40corp.supernews.com
1) Green's identities hold (the continuum approximation for fluid
mechanics).
2) The motion of the source and/or the motion of the receiver through the
aether is small relative to the speed of light.
3) The energy density of the effects are small relative to the energy
density of the medium.
4) Maxwell's assumption of incompressibility is not violated by attempting
to get too "close" to a charge.
5) The distances involved do not exceed the limits of the dynamic viscosity
of the fluid.
[quote:8fc7465d48]div B = 0 is simply an experimental observation.
That would imply that there are *no* theoretical explanations. And
Maxwell's derivation is clearly theoretical. That Lorenz, Hertz, and
modern 'derivations' do not have this same theoretical limitation
does not mean that there are *no* theoretical derivations.
Of course, the theory
would need to be modified if a counter-observation is made,
I believe that would crisp Maxwell's theory completely. I'm not sure
how one could salvage it.
Maybe it would crisp Maxwell's model completely. But so what? Maxwell's
mechanical model is already defunct.
[/quote:8fc7465d48]
That you want to ignore historical fact because it is inconvenient to your
personal religion is your problem.
[quote:8fc7465d48]Classical EM theory, as the term is
currently used in physics, would have no difficulty accomodating it.
[/quote:8fc7465d48]
But the term is a meaningless social farce. If you simply scrawl marks on
paper, anything is possible. However, teh real universe requires physical
action.
[quote:8fc7465d48][cut remainder]
[/quote:8fc7465d48]
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| greywolf42 |
| Posted: Tue Jun 08, 2004 11:09 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406070903300.13527-100000@localhost...
[quote:34c44f69b3]On Fri, 4 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 3 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
It's simply the non-observation of magnetic charge. First noted,
AFAIK, by Gilbert back in around 1600.
It's also a conclusion from Maxwell's physical derivation. It's not
'just' an observation.
[cut]
These people do so because there is no fundamental reason why they
should not exist according to classical EM theory.
A false statement. You can't get more classical than Maxwell. And
Maxwell when explicitly derived his equations (On Physical Lines
of Force, 1861), the derivation found that magnetic monopoles
were theoretically impossible. Which is why they don't show up
in the equations.
Addressed in detail elsewhere in the thread.
Apparently you mean to do this later. Or my newsreader is acting up
again.....
Odd. Your reply to the post where this was addressed in more detail shows
up as posted earlier than this one on both my local newsserver and google.
[/quote:34c44f69b3]
Oh, you meant your statements about what you would include in the term
'classical EM theory.' Your words implied that you were going to address
Maxwell's derivation. Which you didn't.
[quote:34c44f69b3]To summarise:
Maxwell's ether model is not classical EM theory, as the term
"classical EM theory" is used today.
Ahhh, so!! The 'exclusion by definition' trick. This is as useful a
statement as:
A poodle is not a dog, as the term 'dog' is used today.
Are you wanting to discuss physics, or are you just trolling? First, you
introduce a misleading and irrelevant point into an already existing
discussion, and then you start with cheap debating tricks like the one
above.
[/quote:34c44f69b3]
LOL! Your 'cutting' of the text in no way means that it didn't exist!
The historical fact that Maxwell has a very explicit theoretical prohibition
on magnetic monopoles, is by no means 'misleading' or 'irrelevant' to the
false claim that there exists no theoretical reason why MMs should not
exist.
[quote:34c44f69b3]Do note that the statement:
Maxwell's ether model is not classical EM theory, as the term
"classical EM theory" is used today.
is actually true. If you doubt this, try reading some classical EM theory
textbooks.
[/quote:34c44f69b3]
That classical EM textbooks fail to mention Maxwell's original derivation
does not make Maxwell's original derivation nonclassical.
[quote:34c44f69b3]Your statement that the above statement is as useful as:
A poodle is not a dog, as the term 'dog' is used today.
which is (1) off-topic and (2) wrong is simply incorrect, since the former
is (1) relevant to your intended diversion into an irrelevant discussion
of Maxwell's ether model, and (2) correct.
So away with your cheap debating tricks!
[/quote:34c44f69b3]
LOL! Says the sophist.
[quote:34c44f69b3]Do note that I did not invent any new definitions above, just reminded you
of the industry standard definition that is in common use. Your claim to
be a victim of "exclusion by definition" is rather flimsy.
[/quote:34c44f69b3]
Your attempt to exclude historical fact by social fiat is laughable.
Maxwell used an explicitly classical method to derive Maxwell's equations.
Pure Newtonian mechanics. No QM. No relativity.
{snip higher levels}
[quote:34c44f69b3]Well, I don't think that magnetic monopoles exist. But then, the
universe is what the universe is, and can surprise us.
I like surprises. I just don't care for callous disregard for history,
or attempts to shade the truth within a supposedly scientific
discourse.
Neither do I. That's why I think cheap debating tricks should be avoided.
[/quote:34c44f69b3]
So why don't you avoid the cheap debating tricks -- like trying to define
away historical fact?
[quote:34c44f69b3]Stick to the physics.
[/quote:34c44f69b3]
Then why is your argument entirely social convention?
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| greywolf42 |
| Posted: Tue Jun 08, 2004 12:26 pm |
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greywolf42 <mingstb@marssim-ss.com> wrote in message
news:10cbua5g0igt863@corp.supernews.com...
[quote:ea5e648f16]Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406070853280.13486-100000@localhost...
On Fri, 4 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
{snip higher levels}
Maxwell's derivation of Maxwell's equations (On
Physical Lines of Force, 1861) explicitly prohibits magnetic
monopoles. Find a magnetic monopole, and disprove
Maxwell's aether!
Maxwell's ether is already disproved,
Really? When did Maxwell's equations get disproved?
As you have repeatedly written, the Maxwell equations are only an
approximate description of Maxwell's ether.
Not quite. What I've pointed out is that Maxwell's equations are
approximations that hold only under certain conditions.
One can therefore disprove
Maxwell's ether without disproving the Maxwell equations. In fact, one
can disprove Maxwell's ether by failing to disprove the Maxwell
equations.
Your logic is inverted. And incorrect.
One cannot claim that because Maxwell's equations have not yet been
disproven, that they are proved true.
One must set up experiments that can distinguish where Maxwell's equations
should break down -- according to Maxwell's theory.
[/quote:ea5e648f16]
{snip}
I should probably point out that the "pure" mathematicians' equation mining
(what you call 'classical EM theory') punted it's first derivative test
almost immediately. It was called the 'ultraviolet catastrophe.' And you
won't find it mentioned in many 'classical EM' books.
This little problem arose when the mathematicians worked out the
'equipartition theorem.' Fully mathematically correct, and derived from
Maxwell's equations (sans Maxwell's physical model). The pure-math
("classical EM") Rayleigh-Jeans law predicts that the energy of a blackbody
radiator will increase without limit as wavelength approaches zero.
Planck found an empirical curve fit for the actual curve, in 1900. This
curve fit contained an arbitrary constant (Planck's constant). As it turns
out, Maxwell's physical model provides the same curve -- where the physical
action parameter of Maxwell's aether is equal to Planck's constant.
So here is an area where "Maxwell's equations" (as interpreted by
mathematicians) broke down. And "Maxwell's model" provides the correct
answer. This is a classic case of failure of Maxwell's first approximation
(as noted in the parallel thread):
1) Green's identities hold (the continuum approximation for fluid
mechanics).
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| John Schoenfeld |
| Posted: Tue Jun 08, 2004 12:46 pm |
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Timo Nieminen <timo@physics.uq.edu.au> wrote:
[quote:9f76b739ae]Note "absence of charge" in the list above.
That is an extraordinary claim, which I hope you can back up with
references. The only known cause of magnetic fields are electric
charges in motion.
... with a possible exception of free radiation fields. But this is a
matter of definition.
I don't see why you believe the claim above to be extraordinary. Do you
know of an observation of div B != 0 in the absence of charge?
Granted, if one excludes free radiation fields and fields of particles
such as neutrons, div B = 0 in the absence of charge reduces to the rather
tautologous div 0 = 0, but this certainly doesn't harm the correctness of
the claim.
[/quote:9f76b739ae]
Saying an absence of charge gives rise to a solenoidal magnetic field
is simply false. There are no magnetic fields in the absence of
electric charges in motion.
[quote:9f76b739ae]And Magnetic fields occur in the
absence of charge?
What is the magnetic moment of a neutron? What is the charge of a neutron?
That's about the closest to magnetic field in the absence of a charge
that's been observed.
Quantum Theory and scattering experiments both show neutron's are
internally charged.
As I said, that's about the closest observed. What else would you expect
in the absence of magnetic charge?
[/quote:9f76b739ae]
As I said before, neutrons *ARE* charged, and their motion induce
solenoidal magnetic fields. The problem is with the classical
definition of a neutron as being electrically neutral which is
empirically false. Neutron's are composed of 3 quarks, 1 up quark
(charge: +2/3) and 2 down quarks (charge: -1/3) which have a combined
charge of 0. This does not mean that the charge density distribution
of a neutron is uniformly 0. Neutrons in motion induces magnetic
fields for precisely this reason; as do all other electric charges.
[quote:9f76b739ae]But note that your original question restricted the matter to classical EM
theory, and bringing in the question of particulate matter is somewhat
beside the point, especially elements of quantum theory.
[/quote:9f76b739ae]
The problem is with the false notion of electrically neutral neutrons.
This is simply empirically false, irrespective of your theoretical
context.
[quote:9f76b739ae]My question is _why_ would the Maxwell equations require
modifications? If a premise of a theory is that charges in motion
cause solenoidal magnetic fields, then even if magnetic charge can
exist in some other way, why would this the theory require alteration
since the original premise remains valid?
If magnetic charge is found, then you no longer have div B = 0.
Since div B = 0 is one of the Maxwell equations, then the Maxwell
equations require modification.
Read properly. Why does magnetic charge change div B = 0 for magnetic
fields of charges in motion?
Why don't you read properly? The Maxwell equations are not restricted to
fields of charges in motion; they describe the behaviour of electric and
magnetic fields, irrespective of their sources. Since this behaviour does
depend on sources, all known sources of electric and magnetic fields must
appear in the Maxwell equations. At the moment, only electric charge and
electric current are included, since these are the only known sources. If
magnetic charge is found, then magnetic charge and current will need to be
included. Until then, they're not needed. (But you might note that
magnetic currents are sometimes used as equivalent sources.)
This is where I disagree. By design, Maxwell's theory is constrained
to the special case of electric charges in motion.
Then you're quibbling about a philosophical matter of opinion, and not
physics.
If you really, truly, believe that the Maxwell equations are intended to
be restricted only to classical electromagnetic phenomena produced by
electric charges in motion (presumably including the special case of
stationary charges), then what was the point of your original question,
since in that case, classical EM theory can have nothing at all to say
about magnetic charge?
Perhaps you would be more content if I rephrase the statement on
modification of the Maxwell equations to:
The modifications that would be required for a classical EM theory
desribing all classical EM phenomena, were magnetic charge to be
discovered, are simple and well-known.
[/quote:9f76b739ae]
Yes, that is more accurate.
[quote:9f76b739ae]The original premise is div B = 0, and this would no longer be valid.
No, it would still be valid.
No, it would not. Think about what div B = 0 actually means.
div B = 0 is an experimentally verified premise which the remainder of
the equations are based off. Maxwell's equations are true for div B =
0.
If magnetic charge is observed, div B = 0 will no longer be an
experimentally verified premise. And therefore will require modification.
How can div B = 0 be valid if contradicted by observation?
[/quote:9f76b739ae]
Because div B = 0 is valid for the special case of electric charges in
motion, that's why. Whether this special case equates to all cases,
one can only abductively conclude. |
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| Timo Nieminen |
| Posted: Tue Jun 08, 2004 5:09 pm |
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On Tue, 8 Jun 2004, greywolf42 wrote:
[quote:c3d36b1a3d]Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Fri, 4 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
{snip higher levels}
Sure, Maxwell is definitely classical, and it may well be that
Maxwell's ether model prohibits divergence in magnetic fields.
However, Maxwell's ether model is not classical EM theory.
ROTFLMAO!!!!
Have you not stated in the past that in Maxwell's theory, the Maxwell
equations are only approximations valid in the case of low speed
relative to the ether of both observer and source?
I have indeed so stated.
Then Maxwell's ether model is not classical EM theory.
Non sequiteur.
[/quote:c3d36b1a3d]
No, you are simply incorrect. Classical EM theory assumes that the Maxwell
equations are correct. You say that Maxwell's ether theory assumes that
the Maxwell equations are only low-speed approximations.
How then can Maxwell's ether model be compatible with (modern) classical
EM theory?
[quote:c3d36b1a3d]Do note that
"classical electromagnetic theory" is a technical term with a meaning more
precise than "classical" "electromagnetic theory". Ritz's and Weber's
electromagnetic theories are classical as well, but they're not "classical
electromagnetic theory", either.
Then why don't you avoid silly quibbling about what is contained in the
definition of the three words 'classical electromagnetic theory' and address
the substantive issue?
The point is, that a boldfaced lie was made, when it was claimed that there
was no theoretical argument against the existence of magnetic monopoles.
[/quote:c3d36b1a3d]
The point is that a bold-faced lie was made when it was claimed that there
was a prohibition against magnetic monopoles in classical EM theory.
It may well be that Maxwell's ether model prohibits such, but this does
not imply any theoretical prohibition in classical EM theory against such.
[quote:c3d36b1a3d]That you wish to ignore Maxwell's derivation of his equations is merely
propaganda.
[/quote:c3d36b1a3d]
I note a deviation from the physics at hand. Propaganda to what end? Let
me try to bring this back to the physics:
Maxwell's derivation is a historical curiosity. Consider that 3 of the
modern Maxwell equations preceded Maxwell, and the 4th also did for the
limiting case of static fields.
Consider that at least 4 derivations (Lorenz, Hertz, Coulomb's law +
Lorentz symmetry, photons with spin 1) yield the Maxwell equations
exactly, while Maxwell's ether model does not.
[quote:c3d36b1a3d]What bearing, then, does
Maxwell's model have on classical EM theory?
Well, Maxwell's equations were derived from Maxwell's classical model.
That makes them classical.
No, that isn't what makes them classical.
What does? It doesn't use quantum mechanics. It doesn't use SR. It is
entirely based on Newtonian mechanics. That qualifies for 'classical' in my
book.
[/quote:c3d36b1a3d]
Do note the Lorentz symmetry inherent in the Maxwell equations. They are
most certainly relativistic. The Maxwell equations do meet the non-quantum
criterion for classicity, but not the non-SR criterion.
[quote:c3d36b1a3d]Note well that Lorenz obtained the Maxwell equations almost
simultaneously without recourse to an ether or a mechanical
model,
Well, this is news to me. Do you have a reference for this work of
Lorenz?
Given that the paper by Jackson & Okun (RMP 2001) you have been previously
referred to, and the content of which you did discuss over the course of
many posts in a previous thread, discusses the work of Lorenz, and gives
references to Lorenz's original publication, I am somewhat surprised that
this is news to you.
Jackson & Okun also note that Maxwell refers to Lorenz's work in his
Treatise.
That Maxwell's Treatise (and Jackson & Okun) refer to Lorenz does not equate
with Lorenz having derived all of Maxwell's equations.
[/quote:c3d36b1a3d]
You asked for a reference. I noted that references to Lorenz's work, and
references to Maxwell's recognition of Lorenz's work are contained in
Jackson & Okun. What are you complaining about?
Are you going to deny that Lorenz derived a theory of EM equivalent to the
Maxwell equations etc., without even reading what he did?
It's likely that Whittaker discusses Lorenz's work; this might be more
easily available to you.
[quote:c3d36b1a3d]The thread to which
you refer was with regard to the origin of the concept of gauge invariance.
[/quote:c3d36b1a3d]
So what? Jackson & Okun give the references that you asked for. What are
you complaining about?
[quote:c3d36b1a3d]http://www.google.com/groups?selm=102id1b3b17soe5%40corp.supernews.com
You'll note that I caught Jackson being freely creative with history.
[/quote:c3d36b1a3d]
In your opinion.
[quote:c3d36b1a3d]"(Jackson's) 'imprecision' with Maxwell does not make me confident of his
abilities
with Lorenz, Hertz or Heaviside."
[/quote:c3d36b1a3d]
So what? Jackson & Okun give the references that you asked for. What are
you complaining about?
What does your confidence in Jackson's precision or lack thereof have to
do with the writings of Lorenz and Maxwell?
[quote:c3d36b1a3d]and Hertz did the same later.
And then there are the modern derivations.
But none of these are physical derivations. They are merely mathematical
manipulations.
Why? Both Lorenz and Hertz made use of the physical behaviour of
electromagnetic systems in their derivations. How they merely mathematical
manipulations?
Because there was never any attempt to handle the symbols as anything
physical. (Hint: "Electromagnetic systems" begs the question.)
[/quote:c3d36b1a3d]
There was no attempt to construct a mechanical model.
Anyway, that seems like a very definitive statement on Lorenz's and
Hertz's derivations. Have you read the original derivations, or a
secondary source?
[quote:c3d36b1a3d]The modern derivations are very firmly based on a small set of
well-observed physical phenomena.
But there are no phenomena at all used in these 'modern' derivations. It's
all pure symbolism. Arbitrary constants. Nothing physical.
[/quote:c3d36b1a3d]
Coulomb's law is not a phenomenon? Coulomb's law is devoid of physical
meaning?
[cut]
[quote:c3d36b1a3d]
All more physical than Maxwell's model, in that they don't invoke the
mechanical properties of a non-existent ether.
Without physical properties, they aren't physical.
Irrelevant to the fact that the only physical derivation of Maxwell's
equations prohibits magnetic monopoles.
But, as you have repeatedly said, the Maxwell equations are only an
approximation of the behaviour of Maxwell's mechanical model.
But a very good approximation for most laboratory work.
Is div B = 0 an approximation, or is it actually exact?
That part is exact.
If it is exact, in
what way are the Maxwell equations an approximation?
The same way they have been described literally dozens of times for you,
Timo.
http://www.google.com/groups?selm=101vu9m7c668ua4%40corp.supernews.com
1) Green's identities hold (the continuum approximation for fluid
mechanics).
2) The motion of the source and/or the motion of the receiver through the
aether is small relative to the speed of light.
3) The energy density of the effects are small relative to the energy
density of the medium.
4) Maxwell's assumption of incompressibility is not violated by attempting
to get too "close" to a charge.
5) The distances involved do not exceed the limits of the dynamic viscosity
of the fluid.
[/quote:c3d36b1a3d]
You misunderstood the question. Sorry, I was not sufficiently precise in
my wording. (I don't recall you describing the above for me "literally
dozens of times". A quick check on google fails to reveal "literally
dozens of times".)
Let me repeat:
Noting that div B = 0 is exact in Maxwell's theory:
1) Are any of the other 3 modern Maxwell equations exact in Maxwell's
theory? Which ones?
2) Can you write the actual exact equations corresponding to the
approximate ones, or at least give some indication of the type of terms
required to make them exact?
But since you reposted the above list, why is it required that the speed
of the source relative to the medium be small compared to the speed of
light?
Also, what is the energy density of the medium? Is this a measurable
property? Did Maxwell derive (or guess) a possible quantitative value?
[quote:c3d36b1a3d]div B = 0 is simply an experimental observation.
That would imply that there are *no* theoretical explanations. And
Maxwell's derivation is clearly theoretical. That Lorenz, Hertz, and
modern 'derivations' do not have this same theoretical limitation
does not mean that there are *no* theoretical derivations.
Of course, the theory
would need to be modified if a counter-observation is made,
I believe that would crisp Maxwell's theory completely. I'm not sure
how one could salvage it.
Maybe it would crisp Maxwell's model completely. But so what? Maxwell's
mechanical model is already defunct.
That you want to ignore historical fact because it is inconvenient to your
personal religion is your problem.
[/quote:c3d36b1a3d]
You are the one ignoring the historical fact that Maxwell's mechanical
model is defunct. Hint: Lorentz's mechanical model is not the same as
Maxwell's mechanical model. Is this historical fact inconvenient to your
personal religion?
Anyway, it's nice to see that you are resorting to cheap debating tricks
so early in the thread.
[quote:c3d36b1a3d]Classical EM theory, as the term is
currently used in physics, would have no difficulty accomodating it.
But the term is a meaningless social farce. If you simply scrawl marks on
paper, anything is possible. However, teh real universe requires physical
action.
[/quote:c3d36b1a3d]
Utter claptrap! How can a theory that describes EM phenomena in the
non-quantum regime be a meaningless social farce? Do note that it works,
and works well.
I'd like to know why you think that a meaningless social farce can be used
to design working physical devices (hint: it works for antennas,
waveguides, circuits, scattering, etc.).
Anyway, are we here to discuss physics or play cheap debating games? The
only physics content in your post was the 1)-5) list, and you even
accompanied that with a rhetorical lie.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Timo Nieminen |
| Posted: Tue Jun 08, 2004 5:12 pm |
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On Tue, 8 Jun 2004, greywolf42 wrote:
[quote:3430f12d9a]Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Fri, 4 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 3 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
It's simply the non-observation of magnetic charge. First noted,
AFAIK, by Gilbert back in around 1600.
It's also a conclusion from Maxwell's physical derivation. It's not
'just' an observation.
[cut]
These people do so because there is no fundamental reason why they
should not exist according to classical EM theory.
A false statement. You can't get more classical than Maxwell. And
Maxwell when explicitly derived his equations (On Physical Lines
of Force, 1861), the derivation found that magnetic monopoles
were theoretically impossible. Which is why they don't show up
in the equations.
Addressed in detail elsewhere in the thread.
Apparently you mean to do this later. Or my newsreader is acting up
again.....
Odd. Your reply to the post where this was addressed in more detail shows
up as posted earlier than this one on both my local newsserver and google.
Oh, you meant your statements about what you would include in the term
'classical EM theory.' Your words implied that you were going to address
Maxwell's derivation. Which you didn't.
[/quote:3430f12d9a]
I did address the point that Maxwell's derivation is irrelevant.
[quote:3430f12d9a]To summarise:
Maxwell's ether model is not classical EM theory, as the term
"classical EM theory" is used today.
Ahhh, so!! The 'exclusion by definition' trick. This is as useful a
statement as:
A poodle is not a dog, as the term 'dog' is used today.
Are you wanting to discuss physics, or are you just trolling? First, you
introduce a misleading and irrelevant point into an already existing
discussion, and then you start with cheap debating tricks like the one
above.
LOL! Your 'cutting' of the text in no way means that it didn't exist!
The historical fact that Maxwell has a very explicit theoretical prohibition
on magnetic monopoles, is by no means 'misleading' or 'irrelevant' to the
false claim that there exists no theoretical reason why MMs should not
exist.
[/quote:3430f12d9a]
Addressed in other branch.
[cut remainder]
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Timo Nieminen |
| Posted: Tue Jun 08, 2004 5:31 pm |
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On Tue, 8 Jun 2004, greywolf42 wrote:
[quote:b1514bc1dd]greywolf42 <mingstb@marssim-ss.com> wrote:
One must set up experiments that can distinguish where Maxwell's equations
should break down -- according to Maxwell's theory.
{snip}
I should probably point out that the "pure" mathematicians' equation mining
(what you call 'classical EM theory') punted it's first derivative test
almost immediately. It was called the 'ultraviolet catastrophe.' And you
won't find it mentioned in many 'classical EM' books.
This little problem arose when the mathematicians worked out the
'equipartition theorem.' Fully mathematically correct, and derived from
Maxwell's equations (sans Maxwell's physical model). The pure-math
("classical EM") Rayleigh-Jeans law predicts that the energy of a blackbody
radiator will increase without limit as wavelength approaches zero.
Planck found an empirical curve fit for the actual curve, in 1900. This
curve fit contained an arbitrary constant (Planck's constant). As it turns
out, Maxwell's physical model provides the same curve -- where the physical
action parameter of Maxwell's aether is equal to Planck's constant.
[/quote:b1514bc1dd]
Reference?
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Timo Nieminen |
| Posted: Tue Jun 08, 2004 5:37 pm |
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On Wed, 8 Jun 2004, John Schoenfeld wrote:
[quote:b5f27d8cf4]Timo Nieminen <timo@physics.uq.edu.au> wrote:
Note "absence of charge" in the list above.
That is an extraordinary claim, which I hope you can back up with
references. The only known cause of magnetic fields are electric
charges in motion.
... with a possible exception of free radiation fields. But this is a
matter of definition.
I don't see why you believe the claim above to be extraordinary. Do you
know of an observation of div B != 0 in the absence of charge?
Granted, if one excludes free radiation fields and fields of particles
such as neutrons, div B = 0 in the absence of charge reduces to the rather
tautologous div 0 = 0, but this certainly doesn't harm the correctness of
the claim.
Saying an absence of charge gives rise to a solenoidal magnetic field
is simply false. There are no magnetic fields in the absence of
electric charges in motion.
[/quote:b5f27d8cf4]
Better to say that B = 0 everywhere. How is it false to say that div B = 0
if B = 0 everywhere?
[quote:b5f27d8cf4]And Magnetic fields occur in the
absence of charge?
What is the magnetic moment of a neutron? What is the charge of a neutron?
That's about the closest to magnetic field in the absence of a charge
that's been observed.
Quantum Theory and scattering experiments both show neutron's are
internally charged.
As I said, that's about the closest observed. What else would you expect
in the absence of magnetic charge?
As I said before, neutrons *ARE* charged, and their motion induce
solenoidal magnetic fields. The problem is with the classical
definition of a neutron as being electrically neutral which is
empirically false. Neutron's are composed of 3 quarks, 1 up quark
(charge: +2/3) and 2 down quarks (charge: -1/3) which have a combined
charge of 0. This does not mean that the charge density distribution
of a neutron is uniformly 0. Neutrons in motion induces magnetic
fields for precisely this reason; as do all other electric charges.
[/quote:b5f27d8cf4]
Yes, as I said, that's about the closest observed. What else would
you expect in the absence of magnetic charge?
[quote:b5f27d8cf4]But note that your original question restricted the matter to classical EM
theory, and bringing in the question of particulate matter is somewhat
beside the point, especially elements of quantum theory.
The problem is with the false notion of electrically neutral neutrons.
This is simply empirically false, irrespective of your theoretical
context.
[/quote:b5f27d8cf4]
So what? You asked about any prohibitions in classical EM theory on the
existence of magnetic monopoles.
[cut]
[quote:b5f27d8cf4]The original premise is div B = 0, and this would no longer be valid.
No, it would still be valid.
No, it would not. Think about what div B = 0 actually means.
div B = 0 is an experimentally verified premise which the remainder of
the equations are based off. Maxwell's equations are true for div B =
0.
If magnetic charge is observed, div B = 0 will no longer be an
experimentally verified premise. And therefore will require modification.
How can div B = 0 be valid if contradicted by observation?
Because div B = 0 is valid for the special case of electric charges in
motion, that's why. Whether this special case equates to all cases,
one can only abductively conclude.
[/quote:b5f27d8cf4]
No, div B = 0 is a statement about magnetic fields in general. How can it
be valid if contradicted by observation?
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Timo Nieminen |
| Posted: Wed Jun 09, 2004 1:48 am |
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On Wed, 8 Jun 2004, Rob Woodside wrote:
[quote:9aaa95433a]Timo Nieminen <timo@physics.uq.edu.au> wrote:
If magnetic charge is observed, div B = 0 will no longer be an
experimentally verified premise. And therefore will require modification.
Timo, I appreciate your good efforts and don't mean to detract from
them. However, you can have topological magnetic monopoles and still
have Div B=0 "everywhere". This is what Dirac did and paid for it with
the "Dirac strings". Dirac realized that these strings were phony, but
it wasn't until Wu & Yang that the strings were removed and the theory
laid bare. (see my other post on this thread)
[/quote:9aaa95433a]
An unnecessary complication at the moment, I think. But I think it isn't
necessary to go to that much effort. Surely it is sufficient to simply
say that all charges are point charges, the field is discontinuous at
point charges, and therefore div B = 0 and div D = 0 everywhere where
div B and div D are defined. Inelegant, of course, but simple. The
difficulties with infinite energies are left as an exercise for the
reader.
But your point is well-made, and needs no real further comment. Thanks.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| John Schoenfeld |
| Posted: Wed Jun 09, 2004 9:49 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.0406090931590.24864-100000@localhost>...
[quote:910ae0bb7c]On Wed, 8 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
Note "absence of charge" in the list above.
That is an extraordinary claim, which I hope you can back up with
references. The only known cause of magnetic fields are electric
charges in motion.
... with a possible exception of free radiation fields. But this is a
matter of definition.
I don't see why you believe the claim above to be extraordinary. Do you
know of an observation of div B != 0 in the absence of charge?
Granted, if one excludes free radiation fields and fields of particles
such as neutrons, div B = 0 in the absence of charge reduces to the rather
tautologous div 0 = 0, but this certainly doesn't harm the correctness of
the claim.
Saying an absence of charge gives rise to a solenoidal magnetic field
is simply false. There are no magnetic fields in the absence of
electric charges in motion.
Better to say that B = 0 everywhere. How is it false to say that div B = 0
if B = 0 everywhere?
[/quote:910ae0bb7c]
I disagree. |B| = 0 implies no magnetic field. You're the one who said
my statement is wrong - "electric charges in motion induce solenoidal
magnetic fields". My statment implies |B| != 0 and div B = 0.
[quote:910ae0bb7c]And Magnetic fields occur in the
absence of charge?
What is the magnetic moment of a neutron? What is the charge of a neutron?
That's about the closest to magnetic field in the absence of a charge
that's been observed.
Quantum Theory and scattering experiments both show neutron's are
internally charged.
As I said, that's about the closest observed. What else would you expect
in the absence of magnetic charge?
As I said before, neutrons *ARE* charged, and their motion induce
solenoidal magnetic fields. The problem is with the classical
definition of a neutron as being electrically neutral which is
empirically false. Neutron's are composed of 3 quarks, 1 up quark
(charge: +2/3) and 2 down quarks (charge: -1/3) which have a combined
charge of 0. This does not mean that the charge density distribution
of a neutron is uniformly 0. Neutrons in motion induces magnetic
fields for precisely this reason; as do all other electric charges.
Yes, as I said, that's about the closest observed. What else would
you expect in the absence of magnetic charge?
[/quote:910ae0bb7c]
The closest observed magnetic charge?
[quote:910ae0bb7c]But note that your original question restricted the matter to classical EM
theory, and bringing in the question of particulate matter is somewhat
beside the point, especially elements of quantum theory.
The problem is with the false notion of electrically neutral neutrons.
This is simply empirically false, irrespective of your theoretical
context.
So what? You asked about any prohibitions in classical EM theory on the
existence of magnetic monopoles.
[/quote:910ae0bb7c]
And there are none. Maxwell describes electric charges in motion and
their fields, that's all.
[quote:910ae0bb7c][cut]
The original premise is div B = 0, and this would no longer be valid.
No, it would still be valid.
No, it would not. Think about what div B = 0 actually means.
div B = 0 is an experimentally verified premise which the remainder of
the equations are based off. Maxwell's equations are true for div B =
0.
If magnetic charge is observed, div B = 0 will no longer be an
experimentally verified premise. And therefore will require modification.
How can div B = 0 be valid if contradicted by observation?
Because div B = 0 is valid for the special case of electric charges in
motion, that's why. Whether this special case equates to all cases,
one can only abductively conclude.
No, div B = 0 is a statement about magnetic fields in general. How can it
be valid if contradicted by observation?
[/quote:910ae0bb7c]
As I said, whether this special case equates to all cases, one can
only abductively conclude. |
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