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Science Forum Index » Physics - Electromagnetic Forum » Is B just that part of E that does no work?
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| blackhead |
Posted: Wed Jul 25, 2007 4:51 pm |
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Guest
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B and E are two fields that act on charge, with the former doing no
work. So isn't B just that part of E which does no work? i.e. that
part of E which is normal to the velocity, v, of a charge and so could
be defined as E x v? So every E will have a B that is dependent upon
the path taken by a charge through it.
I want to calculate B for a straight wire using the above idea. I will
calculate the delayed E at a point due to the moving electrons, and
subtract that from E due to the stationary ions which should give zero
since a stationary charge experiences no force in practise. The
problem is why a moving charge, Q, experiences a force if E is zero
there? I suspect it's because it's E will have an effect on the
electrons in the wire because of the finite time it takes to effect
them, which in turn generate an E that effects Q so that an E normal
to v of Q is the net result.
Is this going to be a waste of time, so that I will end up with
something that conflicts with Maxwell's equations?
Thanks in advance. |
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| John C. Polasek |
Posted: Thu Jul 26, 2007 8:30 pm |
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Guest
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On Wed, 25 Jul 2007 14:51:20 -0700, blackhead
<larryharson@softhome.net> wrote:
Quote: B and E are two fields that act on charge, with the former doing no
work. So isn't B just that part of E which does no work? i.e. that
part of E which is normal to the velocity, v, of a charge and so could
be defined as E x v? So every E will have a B that is dependent upon
the path taken by a charge through it.
I want to calculate B for a straight wire using the above idea. I will
calculate the delayed E at a point due to the moving electrons, and
subtract that from E due to the stationary ions which should give zero
since a stationary charge experiences no force in practise. The
problem is why a moving charge, Q, experiences a force if E is zero
there? I suspect it's because it's E will have an effect on the
electrons in the wire because of the finite time it takes to effect
them, which in turn generate an E that effects Q so that an E normal
to v of Q is the net result.
Is this going to be a waste of time, so that I will end up with
something that conflicts with Maxwell's equations?
Thanks in advance.
(??"B is an E"??).
It would help your case if you attach units to B & E.
That's what distinguishes physics from mathematics. Mistakes become
glaringly obvious. Unless of course you use cgs units then write
anything you want. |
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| Autymn D. C. |
Posted: Mon Aug 13, 2007 8:24 pm |
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it's E will?
No, start with the expression for F, then isolat your terms
therefrom. It's the current, not the charge, that is affected/
effected by B. |
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| Benj |
Posted: Wed Aug 15, 2007 2:29 am |
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Guest
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Autymn D. C. wrote:
Hey Autymn Womyn!
You forgot to yell: its!!!!! SHITHEAD!!!!!!
Quote: No, start with the expression for F, then isolat your terms
therefrom. It's the current, not the charge, that is affected/
effected by B.
Totally wrong as usual.
But since Autymn has entered the fray let's stray into never-never
land.
First off. E and B are not one thing. They are two different things.
E (electrostatic) is a force developed between charges (coulomb) which
is a result of a connection between charges in higher dimensions. B
on the other hand is a flow of Aether. When aether flows past a
spinning charge, it creates a "curveball" effect that causes the right
angle forces observed in qV x B. Note that this force which which is
widely held to be an E is in fact nothing of the sort. It has very
little in common with Electrostatic E. This is easily seen by an
application of the Bridgman "operational" approach to identifying
fields.
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work? |
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| FrediFizzx |
Posted: Thu Aug 16, 2007 7:13 pm |
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"Benj" <bjacoby@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
Quote: So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations
are wrong for Classical EM has an automatic trip to Stockholm. Don't ya
think lots of people have tried and failed?
Best,
Fred Diether
Moderator sci.physics.foundations |
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| Benj |
Posted: Fri Aug 17, 2007 6:34 am |
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FrediFizzx wrote:
Quote: "Benj" <bjacoby@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations
are wrong for Classical EM has an automatic trip to Stockholm. Don't ya
think lots of people have tried and failed?
Say, wait a minute! This sounds rather suspiciously like one of those
"Everything has already been invented, so don't bother trying anything
new" statements!  |
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| Szczepan Bialek |
Posted: Fri Aug 17, 2007 1:07 pm |
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Guest
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"FrediFizzx" <fredifizzx@hotmail.com> wrote
news:5ik7guF3qjtfiU1@mid.individual.net...
Quote: "Benj" <bjacoby@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations are
wrong for Classical EM has an automatic trip to Stockholm. Don't ya think
lots of people have tried and failed?
It was already done. Displacement current do not exist. But it is not enough
for the trip. Before such trip the new Treatise must be written. Maxwell's
one is based on the "hydraulic analogy" (current is an incompressible
liquid). The new Treatise must be based on the "gas (compressible) analogy".
S* |
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| FrediFizzx |
Posted: Fri Aug 17, 2007 9:30 pm |
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Guest
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"Benj" <bjacoby@iwaynet.net> wrote in message
news:1187350447.161703.201480@e9g2000prf.googlegroups.com...
Quote:
FrediFizzx wrote:
"Benj" <bjacoby@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad?
Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations
are wrong for Classical EM has an automatic trip to Stockholm. Don't
ya
think lots of people have tried and failed?
Say, wait a minute! This sounds rather suspiciously like one of those
"Everything has already been invented, so don't bother trying anything
new" statements!
It is not even close to a statement like that. You are quite free to
try to disprove them experimentally. The Maxwell Equations are too
simple for there to be anything wrong with them for classical EM. Now
microscopically at the quantum level, it's a somewhat different ball
game and they don't hold up. We use QED for that.
Best,
Fred Diether
Moderator sci.physics.foundations |
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| FrediFizzx |
Posted: Fri Aug 17, 2007 9:33 pm |
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"Szczepan Bialek" <sz.bialek@wp.pl> wrote in message
news:fa4o5l$mcs$1@node1.news.atman.pl...
Quote:
"FrediFizzx" <fredifizzx@hotmail.com> wrote
news:5ik7guF3qjtfiU1@mid.individual.net...
"Benj" <bjacoby@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations
are wrong for Classical EM has an automatic trip to Stockholm. Don't
ya think lots of people have tried and failed?
It was already done. Displacement current do not exist. But it is not
enough for the trip. Before such trip the new Treatise must be
written. Maxwell's one is based on the "hydraulic analogy" (current is
an incompressible liquid). The new Treatise must be based on the "gas
(compressible) analogy".
S*
No, before the trip an experiment must be devised and performed that
shows that there is no displacement current.
Best,
Fred Diether
Moderator sci.physics.foundations |
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| blackhead |
Posted: Fri Aug 17, 2007 9:42 pm |
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On 15 Aug, 08:29, Benj <bjac...@iwaynet.net> wrote:
Quote: Autymn D. C. wrote:
it's E will?
Hey Autymn Womyn!
You forgot to yell: its!!!!! SHITHEAD!!!!!!
No, start with the expression for F, then isolat your terms
therefrom. It's the current, not the charge, that is affected/
effected by B.
Totally wrong as usual.
But since Autymn has entered the fray let's stray into never-never
land.
First off. E and B are not one thing. They are two different things.
E (electrostatic) is a force developed between charges (coulomb) which
is a result of a connection between charges in higher dimensions.
I was playing around with this idea, but worded it badly. I meant that
E and B are those parts of the E-M field that do and don't do work on
charge respectively. I have always viewed the static E field of an
electron say, as a region where B does not exist. Yet if B is that
part of the E-M field that does no work, then B has a magnitude 0, but
a direction still normal to the E field.
The force on a moving charge in a static E is always eE regardless of
it's v. So if the force eE is resolved into an E1 along v and B1
orthogonal to v, they may both satisfy Maxwell's equations, while at
the same time making some problems easier to solve. I don't see how
it's possible, but is the scalar product, E dot B, ever non-zero,
meaning they are not always orthogonal?
Quote: B
on the other hand is a flow of Aether. When aether flows past a
spinning charge, it creates a "curveball" effect that causes the right
angle forces observed in qV x B.
Sounds crackpot but interesting, where did you get the idea from? :)
Quote: Note that this force which which is
widely held to be an E is in fact nothing of the sort. It has very
little in common with Electrostatic E. This is easily seen by an
application of the Bridgman "operational" approach to identifying
fields.
I don't think it's widely believed to be an E, since that's part of
the Lorentz force equation F = e( E + v x B).
Quote: So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
Check his work? Mine more like since his has been so successful.
Regards Blackhead. |
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| Szczepan Bialek |
Posted: Sat Aug 18, 2007 4:06 am |
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Guest
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"FrediFizzx" <fredifizzx@hotmail.com> wrote
news:5in422F3qovu0U1@mid.individual.net...
Quote: "Szczepan Bialek" <sz.bialek@wp.pl> wrote in message
news:fa4o5l$mcs$1@node1.news.atman.pl...
"FrediFizzx" <fredifizzx@hotmail.com> wrote
news:5ik7guF3qjtfiU1@mid.individual.net...
"Benj" <bjacoby@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations
are wrong for Classical EM has an automatic trip to Stockholm. Don't ya
think lots of people have tried and failed?
It was already done. Displacement current do not exist. But it is not
enough for the trip. Before such trip the new Treatise must be written.
Maxwell's one is based on the "hydraulic analogy" (current is an
incompressible liquid). The new Treatise must be based on the "gas
(compressible) analogy".
S*
No, before the trip an experiment must be devised and performed that shows
that there is no displacement current.
Iam sure that this was wrote Bill Miller is thruth:
"The simple answer seems to be, no. There have been many attempts to
demonstrate the interpretation that Displacement Current leads directly to a
magnetic field. None have proven it. The most exhaustive used cryogenics to
attempt to prove the Displacement Current--->H idea. It failed to do so.
(Sorry, I don't have the reference handy.)
An excellent overview of Displacement Current is Roche's "The presen status
of Maxwell's displacement current." Eur. J. Physics 19(1998) The
bibliography is rich. Also see WGV Rosser's book, "Classical
Electromagnetism via relativity." On page 81, after analyzing the magnetic
field of a moving "point" charge, he says, "The Maxwell term (What he calls
Displacement Current) the magnetic field at P... both the electric and
magnetic fileds ... arise from the moving charge."
Rosser says lots more.
Unfortunately, textbooks have ignored all of this, and every online EM
syllabus that I have seen persists with the myth of Displacement Current.
I hope this helps.
Bill "
S*
Quote:
Best,
Fred Diether
Moderator sci.physics.foundations |
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| blackhead |
Posted: Sat Aug 18, 2007 7:46 am |
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On Aug 18, 10:06 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
Quote: "FrediFizzx" <fredifi...@hotmail.com> wrotenews:5in422F3qovu0U1@mid.individual.net...
"Szczepan Bialek" <sz.bia...@wp.pl> wrote in message
news:fa4o5l$mcs$1@node1.news.atman.pl...
"FrediFizzx" <fredifi...@hotmail.com> wrote
news:5ik7guF3qjtfiU1@mid.individual.net...
"Benj" <bjac...@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations
are wrong for Classical EM has an automatic trip to Stockholm. Don't ya
think lots of people have tried and failed?
It was already done. Displacement current do not exist. But it is not
enough for the trip. Before such trip the new Treatise must be written.
Maxwell's one is based on the "hydraulic analogy" (current is an
incompressible liquid). The new Treatise must be based on the "gas
(compressible) analogy".
S*
No, before the trip an experiment must be devised and performed that shows
that there is no displacement current.
Iam sure that this was wrote Bill Miller is thruth:
"The simple answer seems to be, no. There have been many attempts to
demonstrate the interpretation that Displacement Current leads directly to a
magnetic field. None have proven it. The most exhaustive used cryogenics to
attempt to prove the Displacement Current--->H idea. It failed to do so.
(Sorry, I don't have the reference handy.)
An excellent overview of Displacement Current is Roche's "The presen status
of Maxwell's displacement current." Eur. J. Physics 19(1998) The
bibliography is rich. Also see WGV Rosser's book, "Classical
Electromagnetism via relativity." On page 81, after analyzing the magnetic
field of a moving "point" charge, he says, "The Maxwell term (What he calls
Displacement Current) the magnetic field at P... both the electric and
magnetic fileds ... arise from the moving charge."
Rosser says lots more.
Unfortunately, textbooks have ignored all of this, and every online EM
syllabus that I have seen persists with the myth of Displacement Current.
Displacement current exists as a mathematical term @D/@dt =
epsilon@E@dt. The idea that it generates a magnetic field is the myth,
right?
Quote: I hope this helps.
Bill "
S*
Best,
Fred Diether
Moderator sci.physics.foundations- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text - |
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| maxwell |
Posted: Sat Aug 18, 2007 10:09 am |
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Guest
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On Aug 17, 11:07 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
Quote: "FrediFizzx" <fredifi...@hotmail.com> wrotenews:5ik7guF3qjtfiU1@mid.individual.net...
"Benj" <bjac...@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations are
wrong for Classical EM has an automatic trip to Stockholm. Don't ya think
lots of people have tried and failed?
It was already done. Displacement current do not exist. But it is not enough
for the trip. Before such trip the new Treatise must be written. Maxwell's
one is based on the "hydraulic analogy" (current is an incompressible
liquid). The new Treatise must be based on the "gas (compressible) analogy".
S*
Wrong. Only Maxwell's first (1859) EM paper was based on the analogy
with incompressible fluid. His second EM paper (1861) was based on
his fantastic 'gears & wheels' model - that got the 'right' answer.
Maxwell's masterpiece was his 1864/1865 paper that adopted a
Lagrangian model for the aether (a 3D elastic solid with some weird
properties that would give him the 1861 answers). The 'displacement
current' was a distortion of this aether. Maxwell rejected the idea
that electricity was some kind of 'stuff' (like electrons) that was in
motion through space. Since all of his equations (now viewed through
the Heaviside lens) were the differential result of macroscopic
experiments one will never prove these equations 'wrong' at the
macroscopic level as this is where they came from in the first place
and these correspond to reality e.g. Faraday, Oersted, etc. The real
problem with this approach (that is still followed by modern field
theorists) is that it is based on the 'continuum hypothesis' (so that
the mathematical differential limit is assumed to correspond to
reality). Unfortunately for all these math-mad Pythagoreans,
experimental physicists have demonstrated that the world is discrete
(electrons, protons etc) so non-continuum math is needed at the micro
level, not all this lovely calculus that has been the focus for over
350 years. |
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| Bill Miller |
Posted: Sat Aug 18, 2007 11:06 am |
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Guest
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"<SNIP>
Displacement current exists as a mathematical term @D/@dt =
Quote: epsilon@E@dt. The idea that it generates a magnetic field is the myth,
right?
Of course. Displacement Current as a mathematical entity exists. It is
nothing more nor less than the rate of change of the E field WRT time.
The physical meaning is that -- as the E field changes -- the velocity and
position of (nearby) charges will also vary. It is the movement of those
charges that causes the magnetic field.
Bill
<SNIP> |
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| Szczepan Bialek |
Posted: Sat Aug 18, 2007 11:46 am |
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Guest
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"maxwell" <spsi@shaw.ca> wrote
news:1187449795.389881.220960@x40g2000prg.googlegroups.com...
Quote: On Aug 17, 11:07 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
"FrediFizzx" <fredifi...@hotmail.com
wrotenews:5ik7guF3qjtfiU1@mid.individual.net...
"Benj" <bjac...@iwaynet.net> wrote in message
news:1187162958.426098.305710@j4g2000prf.googlegroups.com...
So will your little exercise be a wast of time? Who knows? Will it
conflict with Maxwell? Who cares? And why would that be bad? Maxwell
was from the 19th century for crying out loud! Don't you think it's
past time to go back and check his work?
LOL! Anyone that can experimentally prove that the Maxwell equations
are
wrong for Classical EM has an automatic trip to Stockholm. Don't ya
think
lots of people have tried and failed?
It was already done. Displacement current do not exist. But it is not
enough
for the trip. Before such trip the new Treatise must be written.
Maxwell's
one is based on the "hydraulic analogy" (current is an incompressible
liquid). The new Treatise must be based on the "gas (compressible)
analogy".
S*
Wrong. Only Maxwell's first (1859) EM paper was based on the analogy
with incompressible fluid. His second EM paper (1861) was based on
his fantastic 'gears & wheels' model - that got the 'right' answer.
Maxwell's masterpiece was his 1864/1865 paper that adopted a
Lagrangian model for the aether (a 3D elastic solid with some weird
properties that would give him the 1861 answers). The 'displacement
current' was a distortion of this aether. Maxwell rejected the idea
that electricity was some kind of 'stuff' (like electrons) that was in
motion through space. Since all of his equations (now viewed through
the Heaviside lens) were the differential result of macroscopic
experiments one will never prove these equations 'wrong' at the
macroscopic level as this is where they came from in the first place
and these correspond to reality e.g. Faraday, Oersted, etc. The real
problem with this approach (that is still followed by modern field
theorists) is that it is based on the 'continuum hypothesis' (so that
the mathematical differential limit is assumed to correspond to
reality). Unfortunately for all these math-mad Pythagoreans,
experimental physicists have demonstrated that the world is discrete
(electrons, protons etc) so non-continuum math is needed at the micro
level, not all this lovely calculus that has been the focus for over
350 years.
So we have the two displacement currents: (See:
http://www.nyas.org/publications/readersWritersExcerpt.asp?excerptId=42 -
my English is not such as should be so I use citations)
1. "These waves were theoretically discovered by Maxwell after he added a
single new mathematical term, called the "displacement current," to the
equations of electricity and magnetism previously worked out by others. How
did Maxwell deduce his hypothetical displacement current? When I first
learned the theory of electricity and magnetism as a college physics major,
I thought that the displacement current had been derived by requiring
mathematical consistency and the conservation of electric charge."
This one still exist in the math phisycs for the currents in circuits. This
should be replaced by the gas analogy.
2. "When an electric force oscillated in time, the electric particles in the
elastic medium oscillated in response, giving rise to the so-called
displacement current. The underlying elastic medium that allowed all of this
to happen was called "ether." The ether was the material substance through
which electromagnetic waves propagated, just as air is the substance through
which sound waves propagate, by bumping one air molecule into the next.
Again, in Maxwell's words, "We have therefore some reason to believe, from
the phenomena of light and heat, that there is an aetheral medium filling
space and permeating bodies, capable of being set in motion and of
transmitting that motion from one part to another." Maxwell's mechanical
model for electricity and magnetism, including the ether, was completely
fictitious, but it led him to the correct equations."
This one is for space. It is too early to discuse eather.
"Don't you think it's past time to go back and check his work?"
S* |
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