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| Timo Nieminen |
 Posted: Wed Jun 09, 2004 5:09 pm |
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On Thu, 9 Jun 2004, John Schoenfeld wrote:
[quote:c3bb15ff31]Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Wed, 8 Jun 2004, John Schoenfeld wrote:
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?
I disagree. |B| = 0 implies no magnetic field.
[/quote:c3bb15ff31]
And are you claiming that in this case, div B != 0 ? If not, what is it
that you actually disagree about?
[quote:c3bb15ff31]You're the one who said
my statement is wrong - "electric charges in motion induce solenoidal
magnetic fields".
[/quote:c3bb15ff31]
I did not say this at all. I'm saying that div B = 0 is what has been
observed so far. Note that this agrees with your statement above.
[quote:c3bb15ff31]My statment implies |B| != 0 and div B = 0.
[/quote:c3bb15ff31]
Do note that the Maxwell equations are valid for B = 0.
[quote:c3bb15ff31]As I said before, neutrons *ARE* charged, and their motion induce
solenoidal magnetic fields.
[cut]
Yes, as I said, that's about the closest observed. What else would
you expect in the absence of magnetic charge?
The closest observed magnetic charge?
[/quote:c3bb15ff31]
No, as I wrote, it's the closest classically to a magnetic field produced
by something with zero charge. Since you don't appear to understand what I
wrote, I will rephrase:
Since no magnetic charge has ever been observed, one would not expect that
the magnetic field of a neutron was due to magnetic charge. Therefore,
experimental results showing non-zero charge densities should not be
surprising.
By the way, what is the lowest non-zero electric multipole moment of a
neutron?
[quote:c3bb15ff31]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?
As I said, whether this special case equates to all cases, one can
only abductively conclude.
[/quote:c3bb15ff31]
That's a non-answer.
Think about what div B = 0 actually means. 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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| John Schoenfeld |
| Posted: Thu Jun 10, 2004 12:09 pm |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.0406100851190.30717-100000@localhost>...
[quote:3c62343786]On Thu, 9 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Wed, 8 Jun 2004, John Schoenfeld wrote:
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?
I disagree. |B| = 0 implies no magnetic field.
And are you claiming that in this case, div B != 0 ? If not, what is it
that you actually disagree about?
[/quote:3c62343786]
Your proposition that the absence of electric charge induces
solenoidal magnetic fields. There is no B in the absence of electric
charge, so there is no magnetic field. It's like saying the absence of
mass causes uniform gravitational fields.
[quote:3c62343786]You're the one who said
my statement is wrong - "electric charges in motion induce solenoidal
magnetic fields".
I did not say this at all. I'm saying that div B = 0 is what has been
observed so far. Note that this agrees with your statement above.
My statment implies |B| != 0 and div B = 0.
Do note that the Maxwell equations are valid for B = 0.
[/quote:3c62343786]
Yes.
[quote:3c62343786]As I said before, neutrons *ARE* charged, and their motion induce
solenoidal magnetic fields.
[cut]
Yes, as I said, that's about the closest observed. What else would
you expect in the absence of magnetic charge?
The closest observed magnetic charge?
No, as I wrote, it's the closest classically to a magnetic field produced
by something with zero charge. Since you don't appear to understand what I
wrote, I will rephrase:
Since no magnetic charge has ever been observed, one would not expect that
the magnetic field of a neutron was due to magnetic charge. Therefore,
experimental results showing non-zero charge densities should not be
surprising.
[/quote:3c62343786]
Have experiments shown non-zero electric charge density?
[quote:3c62343786]By the way, what is the lowest non-zero electric multipole moment of a
neutron?
[/quote:3c62343786]
0.63E-25
[quote:3c62343786]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?
As I said, whether this special case equates to all cases, one can
only abductively conclude.
That's a non-answer.
Think about what div B = 0 actually means. How can it be valid if
contradicted by observation?
[/quote:3c62343786]
It is valid for the special case of electric charges in motion. That's
all Maxwell says, nothing more. |
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| Rob Woodside |
| Posted: Thu Jun 10, 2004 12:37 pm |
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Guest
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
snip
[quote:8420f5b831]
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.
Inelegant and misses the topologigcal nature of the Dirac monopole,[/quote:8420f5b831]
but the physics is right and sufficient for the purposes of this
thread. Thank you again for your efforts. |
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| Timo Nieminen |
| Posted: Thu Jun 10, 2004 5:33 pm |
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On Fri, 10 Jun 2004, John Schoenfeld wrote:
[quote:1f3ac1105b]Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 9 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Wed, 8 Jun 2004, John Schoenfeld wrote:
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?
I disagree. |B| = 0 implies no magnetic field.
And are you claiming that in this case, div B != 0 ? If not, what is it
that you actually disagree about?
Your proposition that the absence of electric charge induces
solenoidal magnetic fields. There is no B in the absence of electric
charge, so there is no magnetic field.
[/quote:1f3ac1105b]
I'm saying that, in the absence of charge, div B = 0, just as in the
presence of charge.
Do you disagree with this? If not, what do you disagree with?
Do you think that "B = 0" and "no magnetic field" are fundamentally
different statements? Why? Why is the difference important?
[quote:1f3ac1105b]It's like saying the absence of
mass causes uniform gravitational fields.
[/quote:1f3ac1105b]
Yes. Of course.
[quote:1f3ac1105b]Since no magnetic charge has ever been observed, one would not expect that
the magnetic field of a neutron was due to magnetic charge. Therefore,
experimental results showing non-zero charge densities should not be
surprising.
Have experiments shown non-zero electric charge density?
[/quote:1f3ac1105b]
You stated that they have. Why ask me, when you already know of them?
[quote:1f3ac1105b]By the way, what is the lowest non-zero electric multipole moment of a
neutron?
0.63E-25
[/quote:1f3ac1105b]
What n-pole moment is this?
[quote:1f3ac1105b]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?
As I said, whether this special case equates to all cases, one can
only abductively conclude.
That's a non-answer.
Think about what div B = 0 actually means. How can it be valid if
contradicted by observation?
It is valid for the special case of electric charges in motion. That's
all Maxwell says, nothing more.
[/quote:1f3ac1105b]
If magnetic charge is observed (pedant point: finite non-zero magnetic
charge density), then that would be a direct contradiction of div B = 0.
How could div B = 0 still 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: Fri Jun 11, 2004 10:17 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406090909420.24864-100000@localhost...
[quote:74ec372378]On Tue, 8 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
[/quote:74ec372378]
{snip higher levels}
[quote:74ec372378]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.
I did address the point that Maxwell's derivation is irrelevant.
[/quote:74ec372378]
Well, you did make that claim.
[quote:74ec372378]Addressed in other branch.
[cut remainder]
[/quote:74ec372378]
Let's drop this frayed end of the discussion........
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| greywolf42 |
| Posted: Fri Jun 11, 2004 11:21 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406090826540.24864-100000@localhost...
[quote:5a57d00eac]On Tue, 8 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
[/quote:5a57d00eac]
{snip higher levels}
[quote:5a57d00eac]Then Maxwell's ether model is not classical EM theory.
Non sequiteur.
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:5a57d00eac]
Because Maxwell derived Maxwell's equations with classical physical methods.
Thus, Maxwell's EM theory *is* classical. And it is *compatible* because
'modern' EM theory assumes that Maxwell did his work correctly.
[quote:5a57d00eac]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.
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.
[/quote:5a57d00eac]
??? Maxwell's model *is* a classical EM theory. According to you it is not
"Classical EM theory," only because you arbitrarily define the CEMT as
'everything except Maxwell's model.' Now, if you put quotations around the
term "Classical EM theory", and then defined it, you might have something
to gripe about.
[quote:5a57d00eac]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:5a57d00eac]
Let's examine the genesis of this side issue:
JS: "Is there anything in classical e&m that _prohibits_ a divergence in
magnetic fields?"
TN: "... div B = 0 is an experimental observation."
JS:" Yes an observation from steady currents."
TN: "It's simply the non-observation of magnetic charge. First noted, AFAIK,
by Gilbert back in around 1600."
gw42: "It's also a conclusion from Maxwell's physical derivation. It's not
'just' an observation."
Please note that John did not limit his question to classical EM *theory.*
Especially in light of the fact that your response was one of 'experimental
observation.'
[quote:5a57d00eac]That you wish to ignore Maxwell's derivation of his equations is merely
propaganda.
I note a deviation from the physics at hand. Propaganda to what end?
[/quote:5a57d00eac]
Trying to spike any awareness that classical EM is based on a fluid aether.
Like your attempt immediately below of dismissing Maxwell's derivation as
"historical curiosity."
[quote:5a57d00eac]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:5a57d00eac]
The 'derivations' that you list are mere mathematical symbol shuffling, with
no physical basis. And there is no benefit from 'exactness'. When I've
already pointed out that those 'exact' solutions failed almost immediately
(the "ultraviolet catastrophe"). While Maxwell's model passes the UC just
fine.
{snip higher levels}
[quote:5a57d00eac]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.
Do note the Lorentz symmetry inherent in the Maxwell equations.
[/quote:5a57d00eac]
Irrelevant to the question.
[quote:5a57d00eac]They are most certainly relativistic.
The Maxwell equations do meet the non-quantum
criterion for classicity, but not the non-SR criterion.
[/quote:5a57d00eac]
We aren't discussing "Maxwell's equations", but Maxwell's theory. Which is
purely non-relativistic. Regardless of what Maxwell's equations are.
{snip higher levels}
[quote:5a57d00eac]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.
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?
[/quote:5a57d00eac]
The fact that you claimed that Lorenz 'duplicated' Maxwell's derivation of
Maxwell's equations within a few years of Maxwell. Now you are backing off.
[quote:5a57d00eac]Are you going to deny that Lorenz derived a theory of EM equivalent to the
Maxwell equations etc., without even reading what he did?
[/quote:5a57d00eac]
How can I 'read what he did', when you refuse to provide a reference?
[quote:5a57d00eac]It's likely that Whittaker discusses Lorenz's work; this might be more
easily available to you.
[/quote:5a57d00eac]
LOL! I knew you were just trolling. You dug up a mention of Lorenz' name,
and confabulated that as rederiving Maxwell's equations.
[quote:5a57d00eac]The thread to which
you refer was with regard to the origin of the concept of gauge
invariance.
So what? Jackson & Okun give the references that you asked for. What are
you complaining about?
[/quote:5a57d00eac]
The fact that you made up the claim.
[quote:5a57d00eac]http://www.google.com/groups?selm=102id1b3b17soe5%40corp.supernews.com
You'll note that I caught Jackson being freely creative with history.
In your opinion.
[/quote:5a57d00eac]
Of course.
[quote:5a57d00eac]"(Jackson's) 'imprecision' with Maxwell does not make me confident of
his abilities with Lorenz, Hertz or Heaviside."
So what? Jackson & Okun give the references that you asked for. What are
you complaining about?
[/quote:5a57d00eac]
You didn't give references.
[quote:5a57d00eac]What does your confidence in Jackson's precision or lack thereof have to
do with the writings of Lorenz and Maxwell?
[/quote:5a57d00eac]
The fact that you have no references.
{snip higher levels}
[quote:5a57d00eac]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.)
There was no attempt to construct a mechanical model.
[/quote:5a57d00eac]
That's what I said. Therefore, they aren't physical derivations.
[quote:5a57d00eac]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:5a57d00eac]
I based my opinion upon reading Hertz' derivation and upon your definitive
statement about Lorenz (since I haven't seen the derivation that you claimed
that Lorenz made). Now that I know that you made it up, my assessment is
limited to Hertz.
[quote:5a57d00eac]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.
Coulomb's law is not a phenomenon?
[/quote:5a57d00eac]
That's right!!! Now you are getting it!
Coulomb's law is a mathematical equation. Nothing more.
[quote:5a57d00eac]Coulomb's law is devoid of physical meaning?
[/quote:5a57d00eac]
Any meaning arises from the derivation of the equation.
[quote:5a57d00eac][cut]
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.
You misunderstood the question. Sorry, I was not sufficiently
precise in my wording.
[/quote:5a57d00eac]
I wrote:
"... the only physical derivation of Maxwell's equations prohibits magnetic
monopoles."
You replied:
"But, as you have repeatedly said, the Maxwell equations are only an
approximation of the behaviour of Maxwell's mechanical model. .... If (div
B = 0) is exact, in what way are the Maxwell equations an approximation?"
You were quite precise, and quite clear. You are simply running again.
[quote:5a57d00eac](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".)
[/quote:5a57d00eac]
Would you settle for 'more than 10 times?' How many times do you need?
You know quite well in what ways Maxwell's equations are approximations.
However, you choose to repeatedly (and dishonestly) claim and/or imply that
I have not provided this many times before.
[quote:5a57d00eac]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?
[/quote:5a57d00eac]
Read the above list.
[quote:5a57d00eac]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?
[/quote:5a57d00eac]
Read the above list.
[quote:5a57d00eac]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?
[/quote:5a57d00eac]
Because that is an approximation that Maxwell made in his derivations.
[quote:5a57d00eac]Also, what is the energy density of the medium?
[/quote:5a57d00eac]
That wasn't calculated by Maxwell. And is irrelevant to the subject at
hand.
[quote:5a57d00eac]Is this a measurable property?
[/quote:5a57d00eac]
No, because energy is not a measurable property in any physical system. (We
always *calculate* energy from some combination of observables.)
[quote:5a57d00eac]Did Maxwell derive (or guess) a possible quantitative value?
[/quote:5a57d00eac]
Not to my knowledge. Why should he have? He did make some efforts along
the lines of measurable properties of his fluid.
{snip higher levels}
[quote:5a57d00eac]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.
You are the one ignoring the historical fact that Maxwell's mechanical
model is defunct.
[/quote:5a57d00eac]
The proof by repetition fallacy.
[quote:5a57d00eac]Hint: Lorentz's mechanical model is not the same as
Maxwell's mechanical model. Is this historical fact inconvenient to your
personal religion?
[/quote:5a57d00eac]
Lorentz didn't have a 'mechanical model'. However, Lorentz' model of
electrons (and charged particles) is consistent with Maxwell's model.
[quote:5a57d00eac]Anyway, it's nice to see that you are resorting to cheap debating tricks
so early in the thread.
[/quote:5a57d00eac]
The pure ad hominem diversion sophistry.
[quote:5a57d00eac]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.
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.
[/quote:5a57d00eac]
So did the Ptolemaic system. The claim of usefulness does not validate a
model.
[quote:5a57d00eac]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.).
[/quote:5a57d00eac]
:)
Because the equations co-opted by the meaningless social farce *were*
explicitly developed by a *real*, physical model. Then the farce took over,
and started worshipping the equations.
[quote:5a57d00eac]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.
[/quote:5a57d00eac]
LOL! The ad hominem (and flatly untrue) sophistry.
Rhetorical tricks include 'argument by definition' (i.e. defining a term
specifically and ad hoc to ignore the opposing viewpoint); diversion; ad
hominem; repetition; irrelevancy; tangential claims; etc. All efforts
performed by Timo.
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| John Schoenfeld |
| Posted: Fri Jun 11, 2004 11:45 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.0406110925410.4477-100000@localhost>...
[quote:64ac9c2e9f]On Fri, 10 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Thu, 9 Jun 2004, John Schoenfeld wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Wed, 8 Jun 2004, John Schoenfeld wrote:
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?
I disagree. |B| = 0 implies no magnetic field.
And are you claiming that in this case, div B != 0 ? If not, what is it
that you actually disagree about?
Your proposition that the absence of electric charge induces
solenoidal magnetic fields. There is no B in the absence of electric
charge, so there is no magnetic field.
I'm saying that, in the absence of charge, div B = 0, just as in the
presence of charge.
Do you disagree with this? If not, what do you disagree with?
Do you think that "B = 0" and "no magnetic field" are fundamentally
different statements? Why? Why is the difference important?
[/quote:64ac9c2e9f]
I said |B|!=0 and div B = 0, which is different than |B|=0 and div B =
0. I don't believe the latter is relevant.
[quote:64ac9c2e9f]It's like saying the absence of
mass causes uniform gravitational fields.
Yes. Of course.
[/quote:64ac9c2e9f]
There are no gravitational fields.
[quote:64ac9c2e9f]Since no magnetic charge has ever been observed, one would not expect that
the magnetic field of a neutron was due to magnetic charge. Therefore,
experimental results showing non-zero charge densities should not be
surprising.
Have experiments shown non-zero electric charge density?
You stated that they have. Why ask me, when you already know of them?
[/quote:64ac9c2e9f]
I didn't say that. I said neutron have charge density, you originally
implied that they were neutral.
[quote:64ac9c2e9f]By the way, what is the lowest non-zero electric multipole moment of a
neutron?
0.63E-25
What n-pole moment is this?
[/quote:64ac9c2e9f]
2
[quote:64ac9c2e9f]
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?
As I said, whether this special case equates to all cases, one can
only abductively conclude.
That's a non-answer.
Think about what div B = 0 actually means. How can it be valid if
contradicted by observation?
It is valid for the special case of electric charges in motion. That's
all Maxwell says, nothing more.
If magnetic charge is observed (pedant point: finite non-zero magnetic
charge density), then that would be a direct contradiction of div B = 0.
How could div B = 0 still be valid if contradicted by observation?[/quote:64ac9c2e9f] |
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| greywolf42 |
| Posted: Fri Jun 11, 2004 11:50 am |
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Guest
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greywolf42 <mingstb@marssim-ss.com> wrote in message news:...
[quote:479d86012e]Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406090912220.24864-100000@localhost...
[/quote:479d86012e]
{snip}
[quote:479d86012e]Physical fact: the Maxwell equations describe classical EM for all
observed velocities of sources, for inertial reference frames.
That is not a 'fact,' it is an assumption in SR. It is also untrue.
You claim it is untrue. Reference?
Einstein, 1905, "On the Electrodynamics of Moving Bodies". First
paragraph.[/quote:479d86012e]
Whups! I spent too much time reformatting, and not enough time reading.
Maxwell's equations (as applied by "classical EM theory") fail for the case
of the 'ultraviolet catastrophe.'
Assumptions are not 'facts.'
{snip}
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| Rob Woodside |
| Posted: Fri Jun 11, 2004 1:21 pm |
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Guest
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.0406110925410.4477-100000@localhost>...
[quote:8c9120ba1d]
If magnetic charge is observed (pedant point: finite non-zero magnetic
charge density), then that would be a direct contradiction of div B = 0.
How could div B = 0 still be valid if contradicted by observation?
[/quote:8c9120ba1d]
Pedantry is in the eyes of the beholder. Thank you for stating things
correctly. Cheers!!! |
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| Timo Nieminen |
| Posted: Mon Jun 14, 2004 5:54 pm |
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On Fri, 11 Jun 2004, greywolf42 wrote:
[quote:e853616248]Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Tue, 8 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
{snip higher levels}
Then Maxwell's ether model is not classical EM theory.
Non sequiteur.
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?
Because Maxwell derived Maxwell's equations with classical physical methods.
Thus, Maxwell's EM theory *is* classical. And it is *compatible* because
'modern' EM theory assumes that Maxwell did his work correctly.
[/quote:e853616248]
And yet, modern classical EM theory assumes that the Maxwell equations are
correct, while, as you say, they are only approximations in Maxwell's
model.
Surely, if one observes that the Maxwell equations are exact, this must
rule out Maxwell's model?
[cut]
[quote:e853616248]That you wish to ignore Maxwell's derivation of his equations is merely
propaganda.
I note a deviation from the physics at hand. Propaganda to what end?
Trying to spike any awareness that classical EM is based on a fluid aether.
[/quote:e853616248]
What do you mean by "based on a fluid aether"? Historically, yes, this was
Maxwell's foundation.
[quote:e853616248]Like your attempt immediately below of dismissing Maxwell's derivation as
"historical curiosity."
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.
The 'derivations' that you list are mere mathematical symbol shuffling, with
no physical basis.
[/quote:e853616248]
Care to explain why Lorenz's and Hertz's derivations are mere mathematical
symbol shuffling?
Are you claiming that Coulomb's law is not a physical basis?
[quote:e853616248]And there is no benefit from 'exactness'. When I've
already pointed out that those 'exact' solutions failed almost immediately
(the "ultraviolet catastrophe"). While Maxwell's model passes the UC just
fine.
[/quote:e853616248]
This is an entirely different matter. I see that you supplied a reference
in response to my request. Thanks.
[quote:e853616248]{snip higher levels}
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.
Do note the Lorentz symmetry inherent in the Maxwell equations.
Irrelevant to the question.
They are most certainly relativistic.
The Maxwell equations do meet the non-quantum
criterion for classicity, but not the non-SR criterion.
We aren't discussing "Maxwell's equations", but Maxwell's theory. Which is
purely non-relativistic. Regardless of what Maxwell's equations are.
[/quote:e853616248]
We were discussing the Maxwell *equations*. You stated that the Maxwell
*equations* were classical because they were derived from Maxwell's model.
All I did was point out that the Maxwell *equations* are relativistic.
[quote:e853616248]{snip higher levels}
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.
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?
The fact that you claimed that Lorenz 'duplicated' Maxwell's derivation of
Maxwell's equations within a few years of Maxwell. Now you are backing off.
[/quote:e853616248]
I note you are entering rhetoric mode. Are you evading the physics?
You asked for a reference. I provided one.
[quote:e853616248]Are you going to deny that Lorenz derived a theory of EM equivalent to the
Maxwell equations etc., without even reading what he did?
How can I 'read what he did', when you refuse to provide a reference?
[/quote:e853616248]
I provided a source for a reference. However, since you seem to not have
access to the reference list in Jackson & Okun:
Lorenz, L. V., 1867, "Ueber die Identität der Schwingungen des Lichts mit
den elektrischen Strömen," Ann. Phys. Chem. 131, 243-263
["On the Identity of the Vibrations of Light with Electrical Currents,"
Philos. Mag., Ser. 4, 34, 287-301 (1867)].
[quote:e853616248]It's likely that Whittaker discusses Lorenz's work; this might be more
easily available to you.
LOL! I knew you were just trolling. You dug up a mention of Lorenz' name,
and confabulated that as rederiving Maxwell's equations.
[/quote:e853616248]
It's trolling to suggest a source that you might have available?
[quote:e853616248]The thread to which
you refer was with regard to the origin of the concept of gauge
invariance.
So what? Jackson & Okun give the references that you asked for. What are
you complaining about?
The fact that you made up the claim.
[/quote:e853616248]
So you are stating that Lorenz did not derive a theory of EM equivalent to
the Maxwell equations etc.?
[cut]
[quote:e853616248]So what? Jackson & Okun give the references that you asked for. What are
you complaining about?
You didn't give references.
[/quote:e853616248]
You can't look up Jackson & Okun?
[quote:e853616248]What does your confidence in Jackson's precision or lack thereof have to
do with the writings of Lorenz and Maxwell?
The fact that you have no references.
[/quote:e853616248]
I fail to see what that has to do with the writings of Lorenz and Maxwell.
[quote:e853616248]{snip higher levels}
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.)
There was no attempt to construct a mechanical model.
That's what I said. Therefore, they aren't physical derivations.
[/quote:e853616248]
Why is a mechanical model needed for a physical derivation? Do you simply
use "physical" as a synonym for "mechanical"?
[quote:e853616248]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?
I based my opinion upon reading Hertz' derivation and upon your definitive
statement about Lorenz (since I haven't seen the derivation that you claimed
that Lorenz made). Now that I know that you made it up, my assessment is
limited to Hertz.
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.
Coulomb's law is not a phenomenon?
That's right!!! Now you are getting it!
Coulomb's law is a mathematical equation. Nothing more.
Coulomb's law is devoid of physical meaning?
Any meaning arises from the derivation of the equation.
[/quote:e853616248]
Can you answer with a yes or no? Is Coulomb's law devoid of physical
meaning?
[quote:e853616248][cut]
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.
You misunderstood the question. Sorry, I was not sufficiently
precise in my wording.
I wrote:
"... the only physical derivation of Maxwell's equations prohibits magnetic
monopoles."
You replied:
"But, as you have repeatedly said, the Maxwell equations are only an
approximation of the behaviour of Maxwell's mechanical model. .... If (div
B = 0) is exact, in what way are the Maxwell equations an approximation?"
You were quite precise, and quite clear. You are simply running again.
[/quote:e853616248]
No, I was asking for some specific details of the physics. I notice you do
not provide any below.
I ask for some specific details. You fail to provide any, and claim that I
am "simply running again". Such a nice rhetorical trick!
[quote:e853616248](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".)
Would you settle for 'more than 10 times?' How many times do you need?
[/quote:e853616248]
How about something that isn't a lie?
[quote:e853616248]You know quite well in what ways Maxwell's equations are approximations.
However, you choose to repeatedly (and dishonestly) claim and/or imply that
I have not provided this many times before.
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?
Read the above list.
[/quote:e853616248]
The answer is not contained in the above list.
Is div D = \rho exact? Can you answer yes or no?
Is curl H - dD/dt = J exact? Can you answer yes or no?
Is curl E + dB/dt = 0 exact? Can you answer yes or no?
Can you actually deal with the physics instead of blowing out a rhetorical
smokescreen?
[quote:e853616248]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?
Read the above list.
[/quote:e853616248]
The answer is not contained in the above list.
[quote:e853616248]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?
Because that is an approximation that Maxwell made in his derivations.
[/quote:e853616248]
Do you know why Maxwell made this approximation? Why is this assumption
necessary? Is this assumption necessary?
[cut]
[quote:e853616248]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.
You are the one ignoring the historical fact that Maxwell's mechanical
model is defunct.
The proof by repetition fallacy.
Hint: Lorentz's mechanical model is not the same as
Maxwell's mechanical model. Is this historical fact inconvenient to your
personal religion?
Lorentz didn't have a 'mechanical model'. However, Lorentz' model of
electrons (and charged particles) is consistent with Maxwell's model.
[/quote:e853616248]
So, is Lorentz's theory correct? Or, at least, does Lorentz's theory give
correct results?
[quote:e853616248]Anyway, it's nice to see that you are resorting to cheap debating tricks
so early in the thread.
The pure ad hominem diversion sophistry.
[/quote:e853616248]
... that you introduced: "That you want to ignore historical fact because
it is inconvenient to your personal religion is your problem.", along with
blatant lies: "The same way they have been described literally dozens of
times for you".
While evading questions of physics.
[quote:e853616248]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.
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.
So did the Ptolemaic system. The claim of usefulness does not validate a
model.
[/quote:e853616248]
No, it does not. Given that classical EM theory is known to be a classical
approximation of a non-classical reality, so what?
I wouldn't call the Ptolemaic system a meaningless social farce. Do you
think it's a meaningless social farce?
[quote:e853616248]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.).
:)
Because the equations co-opted by the meaningless social farce *were*
explicitly developed by a *real*, physical model. Then the farce took over,
and started worshipping the equations.
[/quote:e853616248]
I see. Can you define more precisely this term "meaningless social farce"?
[quote:e853616248]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.
LOL! The ad hominem (and flatly untrue) sophistry.
Rhetorical tricks include 'argument by definition' (i.e. defining a term
specifically and ad hoc to ignore the opposing viewpoint); diversion; ad
hominem; repetition; irrelevancy; tangential claims; etc. All efforts
performed by Timo.
[/quote:e853616248]
You give me too much credit. While all of these wonderful feats of
rhetoric have occured in this thread, I note that the primary source of
them all has been you. Surely you should take credit for your great skill
in debate?
--
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: Mon Jun 14, 2004 6:14 pm |
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Guest
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On Fri, 11 Jun 2004, greywolf42 wrote:
[quote:fa7566c057]greywolf42 <mingstb@marssim-ss.com> wrote in message news:...
Timo Nieminen <timo@physics.uq.edu.au> wrote:
{snip}
Physical fact: the Maxwell equations describe classical EM for all
observed velocities of sources, for inertial reference frames.
That is not a 'fact,' it is an assumption in SR. It is also untrue.
You claim it is untrue. Reference?
Einstein, 1905, "On the Electrodynamics of Moving Bodies". First
paragraph.
Whups! I spent too much time reformatting, and not enough time reading.
Maxwell's equations (as applied by "classical EM theory") fail for the case
of the 'ultraviolet catastrophe.'
[/quote:fa7566c057]
Which is usually regarded as non-classical. Do you have a reference for a
failure of the Maxwell equations in the classical regime?
[quote:fa7566c057]Assumptions are not 'facts.'
[/quote:fa7566c057]
Indeed. Which is why I asked for some evidence for your assumption that
the Maxwell equations don't describe classical EM for all observed
velocities of sources, for all intertial reference frames.
--
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: Thu Jun 17, 2004 1:37 pm |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406150900030.4991-100000@localhost...
[quote:0850c0f610]On Fri, 11 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
On Tue, 8 Jun 2004, greywolf42 wrote:
Timo Nieminen <timo@physics.uq.edu.au> wrote:
[/quote:0850c0f610]
{snip higher levels}
[quote:0850c0f610]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?
Because Maxwell derived Maxwell's equations with classical physical
methods. Thus, Maxwell's EM theory *is* classical. And it is
*compatible* because 'modern' EM theory assumes that Maxwell did
his work correctly.
And yet, modern classical EM theory assumes that the Maxwell equations are
correct, while, as you say, they are only approximations in Maxwell's
model.
[/quote:0850c0f610]
Yes.
[quote:0850c0f610]Surely, if one observes that the Maxwell equations are exact, this must
rule out Maxwell's model?
[/quote:0850c0f610]
A strange statement. One can never "observe" Maxwell's equations being
exact.
[quote:0850c0f610][cut]
That you wish to ignore Maxwell's derivation of his equations is
merely propaganda.
I note a deviation from the physics at hand. Propaganda to what end?
Trying to spike any awareness that classical EM is based on a fluid
aether.
What do you mean by "based on a fluid aether"? Historically, yes, this was
Maxwell's foundation.
[/quote:0850c0f610]
And the theoretical foundation of those equations remains viable.
[quote:0850c0f610]Like your attempt immediately below of dismissing Maxwell's derivation
as "historical curiosity."
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.
The 'derivations' that you list are mere mathematical symbol shuffling,
with no physical basis.
Care to explain why Lorenz's and Hertz's derivations are mere mathematical
symbol shuffling?
[/quote:0850c0f610]
Because they have no physical basis (no physical causation).
[quote:0850c0f610]Are you claiming that Coulomb's law is not a physical basis?
[/quote:0850c0f610]
Yes. Coulomb's law (pre-Maxwell) is an empirical math equation.
[quote:0850c0f610]And there is no benefit from 'exactness'. When I've
already pointed out that those 'exact' solutions failed almost
immediately (the "ultraviolet catastrophe"). While Maxwell's
model passes the UC just fine.
This is an entirely different matter.
[/quote:0850c0f610]
Not in the least! It is an explicit disproof of your claim of exactness and
perfection of Maxwell's equations. The 'ultraviolet catastrophe' was the
*explicit* result of using the ME sans any physical underpinning. And
they failed. Exactly as Maxwell's *physical model* would expect.
[quote:0850c0f610]I see that you supplied a reference
in response to my request. Thanks.
[/quote:0850c0f610]
You are welcome.
{snip repetitive arguments}
[quote:0850c0f610]Are you going to deny that Lorenz derived a theory of EM equivalent to
the Maxwell equations etc., without even reading what he did?
How can I 'read what he did', when you refuse to provide a reference?
I provided a source for a reference. However, since you seem to not have
access to the reference list in Jackson & Okun:
Lorenz, L. V., 1867, "Ueber die Identität der Schwingungen des Lichts mit
den elektrischen Strömen," Ann. Phys. Chem. 131, 243-263
["On the Identity of the Vibrations of Light with Electrical Currents,"
Philos. Mag., Ser. 4, 34, 287-301 (1867)].
[/quote:0850c0f610]
Thanks. I'll look it up soon. And I'll put into abeyance my suspicion that
you confabulated what's in that paper by reading Okun.
{snip higher levels}
[quote:0850c0f610]There was no attempt to construct a mechanical model.
That's what I said. Therefore, they aren't physical derivations.
Why is a mechanical model needed for a physical derivation? Do you simply
use "physical" as a synonym for "mechanical"?
[/quote:0850c0f610]
A mechanical model is always physical. The point is that a physical model
requires a direct, cause-and-effect model. Equations and assumed
Aristotlean 'principles' are not physical models.
{snip more on Lorenz derivation}
[quote:0850c0f610]Coulomb's law is not a phenomenon?
That's right!!! Now you are getting it!
Coulomb's law is a mathematical equation. Nothing more.
Coulomb's law is devoid of physical meaning?
Any meaning arises from the derivation of the equation.
Can you answer with a yes or no? Is Coulomb's law devoid of physical
meaning?
[/quote:0850c0f610]
Sorry, I can't answer with a yes or no. If you wish to derive Coulomb's
equation from a physical model, then the equation has physical meaning. If
you just write symbols on a page, then there is no physical meaning.
{snip higher levels}
[quote:0850c0f610]You know quite well in what ways Maxwell's equations are approximations.
However, you choose to repeatedly (and dishonestly) claim and/or imply
that I have not provided this many times before.
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?
Read the above list.
The answer is not contained in the above list.
[/quote:0850c0f610]
The answer *is* contained in the above list. But you have to read and
think.
[quote:0850c0f610]Is div D = \rho exact? Can you answer yes or no?
Is curl H - dD/dt = J exact? Can you answer yes or no?
Is curl E + dB/dt = 0 exact? Can you answer yes or no?
[/quote:0850c0f610]
As previously stated (and ignored by you), those three equations are the
result of physical approximations made by Maxwell's when using his physical
model.
[quote:0850c0f610]Can you actually deal with the physics instead of blowing out a rhetorical
smokescreen?
[/quote:0850c0f610]
I've dealt with the physics many times. You simply refuse to read the
references -- or my posts.
[quote:0850c0f610]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?
Read the above list.
The answer is not contained in the above list.
[/quote:0850c0f610]
The answer *is* contained in the list. The list gives you clear indication
of the types of terms that will need to be added.
I'm beginning to suspect that you are again confusing physics and
mathematics. These equations are the result of *physical* approximations
(specific physical conditions hold). Not a mathematical approximation where
we are dealing with higher order *terms* in a mathematical series.
[quote:0850c0f610]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?
Because that is an approximation that Maxwell made in his derivations.
Do you know why Maxwell made this approximation? Why is this assumption
necessary? Is this assumption necessary?
[/quote:0850c0f610]
The approximation is necessary if you want to return Faraday's law in its
'pure' (low-speed laboratory) form.
[quote:0850c0f610][cut]
[/quote:0850c0f610]
{snip higher levels}
[quote:0850c0f610]Hint: Lorentz's mechanical model is not the same as
Maxwell's mechanical model. Is this historical fact inconvenient to
your personal religion?
Lorentz didn't have a 'mechanical model'. However, Lorentz' model of
electrons (and charged particles) is consistent with Maxwell's model.
So, is Lorentz's theory correct?
[/quote:0850c0f610]
Who knows? One can't prove a theory in the scientific method.
[quote:0850c0f610]Or, at least, does Lorentz's theory give correct results?
[/quote:0850c0f610]
It does under certain conditions.
{snip}
[quote:0850c0f610]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.
So did the Ptolemaic system. The claim of usefulness does not validate
a model.
No, it does not.
[/quote:0850c0f610]
Good. Then we agree that your argument was specious.
[quote:0850c0f610]Given that classical EM theory is known to be a classical
approximation of a non-classical reality, so what?
[/quote:0850c0f610]
Your argument was specious. And your new argument about your personal
'knowledge' about 'non-classical reality' is just as specious. I see what
you call "non-classical reality" as simply a dodge to get out from under the
failure of mathemeticians blindly applying Maxwell's equations.
[quote:0850c0f610]I wouldn't call the Ptolemaic system a meaningless social farce. Do you
think it's a meaningless social farce?
[/quote:0850c0f610]
Yes.
[quote:0850c0f610]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.).
:)
Because the equations co-opted by the meaningless social farce *were*
explicitly developed by a *real*, physical model. Then the farce took
over, and started worshipping the equations.
I see. Can you define more precisely this term "meaningless social farce"?
[/quote:0850c0f610]
It's not a 'term.' It's a description.
{snip non physics}
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| greywolf42 |
| Posted: Thu Jun 17, 2004 1:38 pm |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406151011500.4991-100000@localhost...
[quote:1941cee364]On Fri, 11 Jun 2004, greywolf42 wrote:
greywolf42 <mingstb@marssim-ss.com> wrote in message news:...
Timo Nieminen <timo@physics.uq.edu.au> wrote:
Physical fact: the Maxwell equations describe classical EM for
all observed velocities of sources, for inertial reference
frames.
That is not a 'fact,' it is an assumption in SR. It is also
untrue.
You claim it is untrue. Reference?
Einstein, 1905, "On the Electrodynamics of Moving Bodies". First
paragraph.
Whups! I spent too much time reformatting, and not enough time reading.
Maxwell's equations (as applied by "classical EM theory") fail for the
case of the 'ultraviolet catastrophe.'
Which is usually regarded as non-classical. Do you have a reference for a
failure of the Maxwell equations in the classical regime?
[/quote:1941cee364]
Ummm. The 'ultraviolet catastrophe' was explicitly classical. Especially
at the time.
[quote:1941cee364]Assumptions are not 'facts.'
Indeed. Which is why I asked for some evidence for your assumption that
the Maxwell equations don't describe classical EM for all observed
velocities of sources, for all intertial reference frames.
[/quote:1941cee364]
I never made an assumption. I gave you the printed evidence that Maxwell
made those assumptions. ("On Physical Lines of Force")
--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail} |
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| Timo Nieminen |
| Posted: Thu Jun 17, 2004 5:40 pm |
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On Thu, 17 Jun 2004, greywolf42 wrote:
[quote:2a6434641d]greywolf42 <mingstb@marssim-ss.com> wrote:
So, I will presume that you want to see an identification of Planck's
constant as the action parameter of Maxwell's aether.
http://www.google.com/groups?selm=4tmm5m%24rp7%40sjx-ixn6.ix.netcom.com
Or, you can simply do a google search on 'Planck's Action'.
Now, since we have demonstrated the failure of the pure 'exact' Maxwell's
equations (you haven't challenged it), you will agree that the pure math
'equation mining' process is flawed. And now you will agree that there
*are* classical, theoretical EM reasons that magnetic monopoles have never
been seen.
Timo? Hello?
[/quote:2a6434641d]
Reference noted elsewhere in thread, and such reference replied to by you,
so you presumably mean the last paragraph.
Firstly, it is well-known that classical EM theory fails in the quantum
domain.
I don't think that this is due to a failure in what you call "the pure
math 'equation mining' process" at all, but more a failure of the physical
assumptions on which the whole thing is based.
I don't agree that magnetic monopoles are theoretically prohibited by
classical EM theory. I note that the *assumption* that magnetic monopoles
do not exist is one of the foundations of classical EM theory, and that
this assumption is well-supported by experiment and observation.
I think that a quite likely reason that magnetic monopoles have never been
seen is that they don't exist.
--
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: Fri Jun 18, 2004 11:25 am |
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Timo Nieminen <timo@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0406180935060.22535-100000@localhost...
[quote:c075da4685]On Thu, 17 Jun 2004, greywolf42 wrote:
greywolf42 <mingstb@marssim-ss.com> wrote:
So, I will presume that you want to see an identification of Planck's
constant as the action parameter of Maxwell's aether.
http://www.google.com/groups?selm=4tmm5m%24rp7%40sjx-ixn6.ix.netcom.com
Or, you can simply do a google search on 'Planck's Action'.
Now, since we have demonstrated the failure of the pure 'exact'
Maxwell's equations (you haven't challenged it), you will agree
that the pure math 'equation mining' process is flawed. And
now you will agree that there *are* classical, theoretical EM
reasons that magnetic monopoles have never been seen.
Timo? Hello?
Reference noted elsewhere in thread, and such reference replied to by you,
so you presumably mean the last paragraph.
[/quote:c075da4685]
Yes, I did notice that you thanked me for the reference. However, since
this evidence contradicted a position of yours, I was interested in seeing
the followup.
[quote:c075da4685]Firstly, it is well-known that classical EM theory fails in the quantum
domain.
[/quote:c075da4685]
That is a circular approach. *FIRST* the math-only EM theory was observed
to fail. Then -- eventually -- a new theory was created to account for what
happened when the EM math failed. Whether it is 'well-known' or not is
irrelevant.
[quote:c075da4685]I don't think that this is due to a failure in what you call "the pure
math 'equation mining' process" at all, but more a failure of the physical
assumptions on which the whole thing is based.
[/quote:c075da4685]
A view contrary to logic. The only 'physical assumptions' anywhere in
Maxwell's equations are the 'physical model' used by Maxwell do derive the
ME. And the 'physical model' predicts the observed failure of the ME. Yet
you claim that you think it is the 'failure of the physical assumptions'
that gives rise to the problem. When there *is* no problem if you use the
physcial model.
[quote:c075da4685]I don't agree that magnetic monopoles are theoretically prohibited by
classical EM theory.
[/quote:c075da4685]
I understand that this is your view. But you have explained that this is
your view *because* you define 'classical EM theory' as the math of
Maxwell's equations -- without Maxwell's actual theory.
[quote:c075da4685]I note that the *assumption* that magnetic monopoles
do not exist is one of the foundations of classical EM theory, and that
this assumption is well-supported by experiment and observation.
[/quote:c075da4685]
Of course, this is not an *assumption* in Maxwell's model. It is a
conclusion. Again, you are simply claiming that Maxwell's equations *are*
the origin of classical EM.
[quote:c075da4685]I think that a quite likely reason that magnetic monopoles have never been
seen is that they don't exist.
[/quote:c075da4685]
Well, at least we agree on this point.
So, I will sign off from Timo in this thread, after today. I think we've
managed to dance around all the various issues quite nicely.
Timo, feel free to have the last word, if you like.
--
greywolf42
ubi dubium ibi libertas
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