Re: Gaussian statics law
Jimmie D
"art" <aun...@insightbb.com> wrote in message
news:1173413632.1...@j27g2000cwj.googlegroups.com...
> Gentlemen from outside of America. Gauss's law with respect to statics
> is quite specific and easy to understand. What is so wrong in
> mathematical terms by adding the metric of time to the law so that
> curl can be accomodated? i.e. change from a conservative field where
> all vectors have zero length,
> to a electro magnetic equation by adding the words " the addition of
> time" which by providing a three dimensional field has the true
> inclusion of curl i.e. all vectors have value in length and direction.
> America denies the feasability of such an addition to an existing law
> which in essence is regarded as a new law without basis on this side
> of the pond.Are all countries of this mentallity?
> Art
>
Because a static field does not produce an EM field(curl) only if that
static charge is in motion. Motion would even include taking a charged body,
say a pith ball and waving it back and forth. Electrons have a static charge
but when they are in motion in a conductor they produce fields(curl).
Electrons moving about an atom also produces fields but the net result of
all the aoms moving about is zero. PLEASE REFERENCE THE GUASSIAN LAW ON
STATICS. I still think you are confusing static with statistics.
ji...@specsol.spam.sux.com
> But Jimmie my friend, now you have an understanding of Gaussian law
> what is preventing you adding the metric of time or a length of time
> to the statics law?
Because by the definition of "static field" nothing changes over time.
<snip remaining babbling nonsense>
--
Jim Pennino
Remove .spam.sux to reply.
Gene Fuller
>
> But Jimmie my friend, now you have an understanding of Gaussian law
> what is preventing you adding the metric of time or a length of time
> to the statics law?
Art,
Adding the "metric of time" is exactly what J.C. Maxwell did, in 1865.
The detailed hard work surrounding Maxwell's Equations, as we know them
today, was probably more attributable to Oliver Heaviside. However,
Maxwell gets the credit for adding the time contribution.
73,
Gene
W4SZ
Dave
"Gene Fuller" <W4SZz...@att.net> wrote in message
news:ZteIh.48733$as2....@bgtnsc05-news.ops.worldnet.att.net...
unfortunately art is stuck on one of the 4 equations and is ignoring all the
others. if he really understood maxwell's work he would know:
Gauss' Law is for static electric charges and fields.
Ampere's Law is for static magnetic fields, that is fields set up by
constant (read non-time varying) currents.
Faraday's Law introduced the time varying part of the relation between
magnetic fields and currents.
Then Maxwell tied them together with the displacement current into the 4
equations that we have been using and which have successfully been used to
calculate all kinds of electromagnetic phenomena for many years.
By talking about curl of electric fields art is forgetting that this is one
of the representations of Faraday's law:
curl(E)= -dB/dt (E and B are vectors of course) which automatically adds
the time relationship that he is trying to force into Gauss's law where it
has no place.
personally i recommend ignoring him until he goes back to fields and waves
101 and gets the equations straight.
ji...@specsol.spam.sux.com
> On 9 Mar, 06:45, j...@specsol.spam.sux.com wrote:
<snip old crap>
> > Because by the definition of "static field" nothing changes over time.
> >
> > <snip remaining babbling nonsense>
> >
> > --
> > Jim Pennino
> >
> > Remove .spam.sux to reply.- Hide quoted text -
> >
> > - Show quoted text -
> Jim, it is the logic applied that produces the law is what you should
> be concentrating on since that same logic can be applied elsewher.
Is this babble supposed to mean something?
If something changes over time, it isn't static.
If it isn't static, static laws don't apply.
See Maxwell and friends for what applies when things are not static.
<snip rambling babble>
ji...@specsol.spam.sux.com
> But he did not associate it with antennas period.
Correct, he described EM fields in general.
> In the previous post
> I applied the same logic to an antenna array and using the initial
> logic that Gauss used and which Maxwell enlarged upon for other
> reasons. And to follow the logic applied by Gauss one must focus on
> equilibrium such that the static particles on the enclosed antenna
> array MUST be in equilibrium or else all falls apart inside the
> enclosed border. Remember that static particles reside on the surface
> of a radiator when energy is not applied.
Babbling nonsense.
The existance of static particles (whatever the hell they are, I presume
you mean electrons) on the surface of a radiator has nothing to do with
applied energy (other than maybe wind energy).
> It departes from the SURFACE when energy is applied and continues to
> do so as time passes by in a time varying form until time stops where
> at that time it must be in a state of equilibrium in static form Q.E.D
More babbling nonsense.
EM waves depart when energy is applied, not particles.
Cecil Moore
> EM waves depart when energy is applied, not particles.
Quantum Electrodynamics tells us that EM waves consist
of photons which are particles.
--
73, Cecil, w5dxp.com
John Smith I
So, which is the real question:
1) Why do waves act like particles?
--OR--
2) Why do particles act like waves?
Cecil Moore
> So, which is the real question:
How can a single photon go through two slits
at the same time and interfere with itself
on the other side?
--
73, Cecil http://www.w5dxp.com
ji...@specsol.spam.sux.com
> ji...@specsol.spam.sux.com wrote:
> > EM waves depart when energy is applied, not particles.
> Quantum Electrodynamics tells us that EM waves consist
> of photons which are particles.
Like other quanta the photon has both wave and particle
properties, but has zero rest mass and is not a particle like
an electron
Cecil Moore
> Same way with regard to free particles residing on the surface of a
> conductor,
Are you referring to free electrons?
John Smith I
> ...
> properties, but has zero rest mass and is not a particle like
> an electron
>
>
Really? I am going to have to see that one to believe it (not saying it
is incorrect though), a "something" with no mass--kinda like a "ghost
particle!"
ji...@specsol.spam.sux.com
> > ...
> > properties, but has zero rest mass and is not a particle like
> > an electron
> >
> >
> Really? I am going to have to see that one to believe it (not saying it
> is incorrect though), a "something" with no mass--kinda like a "ghost
> particle!"
You see it every day; it is called light.
Photons are not particles because they have no rest mass; particles
by definition do.
If photons had mass, they couldn't travel at the speed of light.
Or didn't you know that nothing with mass can travel at the speed of
light?
Cecil Moore
> Really? I am going to have to see that one to believe it (not saying it
> is incorrect though), a "something" with no mass--kinda like a "ghost
> particle!"
It's old hat knowledge, John, and one of the reasons
why standing wave energy doesn't just stand there or
just "slosh around" as one guru asserted. If a photon
is slowed to zero velocity, its mass vanishes and
it ceases to be detectable. A photon's mass derives
from its speed-of-light velocity, i.e. it is 100%
kinetic. Any particle with a resting mass would necessarily
have infinite mass at the speed of light. Therefore, any
particle with a finite mass at the speed of light must
necessarily have a zero rest mass.
Cecil Moore
> Photons are not particles because they have no rest mass; particles
> by definition do.
Quantum ElectroDynamics tells us that nothing except
particles exist. Photons are *particles* with zero
rest mass. QED! (Get it? :-)
Would you like to hear about virtual particles?
Dave
"Gene Fuller" <W4SZz...@att.net> wrote in message
news:ZteIh.48733$as2....@bgtnsc05-news.ops.worldnet.att.net...
>
>>
>> But Jimmie my friend, now you have an understanding of Gaussian law
>> what is preventing you adding the metric of time or a length of time
>> to the statics law?
>
>
> Art,
>
> Adding the "metric of time" is exactly what J.C. Maxwell did, in 1865. The
> detailed hard work surrounding Maxwell's Equations, as we know them today,
> was probably more attributable to Oliver Heaviside. However, Maxwell gets
> the credit for adding the time contribution.
unfortunately art is stuck on one of the 4 equations and is ignoring all the
ji...@specsol.spam.sux.com
> ji...@specsol.spam.sux.com wrote:
> > Photons are not particles because they have no rest mass; particles
> > by definition do.
> Quantum ElectroDynamics tells us that nothing except
> particles exist. Photons are *particles* with zero
> rest mass. QED! (Get it? :-)
Cecil, do you ever tire of playing semantic games?
Cecil Moore
> Cecil Moore <myc...@hotmail.com> wrote:
>> Quantum ElectroDynamics tells us that nothing except
>> particles exist. Photons are *particles* with zero
>> rest mass. QED! (Get it? :-)
>
> Cecil, do you ever tire of playing semantic games?
I never tire of semantic humor. A double meaning or,
even better, a triple meaning, is one of the things
that makes English so enjoyable. (And I really enjoy
the "wench for sell" over on rec.radio.swap.)
But seriously, QED indicates that everything that
exists is a particle, even if it has no rest mass,
even if it is only virtual.
John Smith I
> ...
No, I am far from thinking light is actually "something." (at least not
a "something" we are familiar with or have "true" examples of ...)
It is unthinkable that any object/particle can exist without mass ...
the discovery and absolute proof of that being possible is in our
future; presently we only have theories ...
I don't argue that it is impossible, rather only improbable. It is more
than likely, like has happened so many times, when we know why rf waves
appear to be both wave and particle, that physicists and mathematicians
will go scurrying to their dens and emerge with new "laws." And,
finally we will have a more complete picture of the phenomenon.
We only see a puzzle, although we can "work with the puzzle", although
we can "seem" to get meaningful data from this puzzle, or manipulate it
to do useful things for us, although we "seem" to have laws, equations
and formulas to describe this puzzle--we have been there and done that
before--that is, we have rewritten those laws, equations and formulas to
fit our new findings and started pretending we have reached the final
conclusions and "know" the phenomenon--but then, at some future date, we
do it all over again ...
John Smith I
> ...
> the "wench for sell" over on rec.radio.swap.)
> ...
How much is the wench? What does that wench look like, there a pic? <grin>
John Smith I
Cecil:
I know that is argued, and I suspect it all hogwash.
If "they" have to create theories depending on disappearing particles,
when you get one where you can take a look at one of them dern
"particles", that is just TOO desperate.
However, it does, in my personal opinion, suggest a STRONG relationship
of the "particles" to the ether--the ether cannot be seen nor detected
either ... (that is, IF it really exists, as I suspect it does ... )
ji...@specsol.spam.sux.com
> > ...
Ignorant nonsense.
John Smith I
> ...
Jim:
I do believe much of higher academia, and the subjects which drives it,
is above you, and confuses you--frankly, I think it all or most appears
as BS to you ...
That is too bad man. Perhaps a group centered around appliance usage
would better suit you ...
John Smith I
> John Smith I wrote:
>> Really? I am going to have to see that one to believe it (not saying
>> it is incorrect though), a "something" with no mass--kinda like a
>> "ghost particle!"
>
> It's old hat knowledge, John, and one of the reasons
> why standing wave energy doesn't just stand there or
> just "slosh around" as one guru asserted. If a photon
> is slowed to zero velocity, its mass vanishes and
Cecil:
One more thing ... On those those frisky, frolicking photons.
What would you attribute the fact the "photons" in HF behave much
differently then then the photons of VHF/UHF/SHF/LIGHT to?
JS
John E. Davis
wrote:
>Gauss' Law is for static electric charges and fields.
It is usually used for problems in electrostatics, but it is not
confined to such problems. The differential form of it is just one of
the Maxwell equations:
div E(x,t) = 4\pi\rho(x,t)
Integrate it over a fixed surface and you get the integral form, which
is Gauss's law. It is valid with time-dependent charge densities and
time-dependent electric fields.
--John
Ian White GM3SEK
That isn't quite true. The big gap is between anything we'd normally
call "light", and anything we'd normally call "RF".
Long explanation coming up... Executive Summary: at normal radio
frequencies, quantum theory is totally irrelevant.
Quantum theory describes electromagnetic energy as being divided into a
series of packets called photons, so (total energy in a stream of
photons) = (number of photons/second) x (energy of individual photons).
This also means that EM energy doesn't exist in pure sine-waves - the
waveform is actually built up in steps, very much like digitized audio.
The step size is the energy content of one quantum.
The question is: are those steps noticeable enough to be important?
For light and shorter wavelengths, the answer is often Yes. Quantum
theory was developed to explain observations like some kinds of light
being emitted in a series of sharp spectral lines, which cannot be
explained by a wave-only theory. Instead, it has to be thought of as
being built up of individual photons/quanta which can only have certain
"allowed" energy levels.
It turned out that the energy content of a single photon is uniquely
related to the wavelength of the radiation. Any given wavelength has
only one quantum energy. More energy can only be made up from larger
numbers of the same identical quanta. That unique quantum energy is
inversely proportional to the wavelength, and directly proportional to
the frequency:
E = hf
where
E is the energy content of a single quantum/photon (joules),
f is in Hz
h is Planck's constant which has a value of 6.6 x 10^-34 joule seconds.
This applies to all forms of EM energy, so let's calculate the 'step
size' in an RF waveform at 10MHz. That will be the energy content of a
single quantum, which turns out to be 0.000 000 000 000 000 000 000 000
0066 joules - which is unimaginably small.
It means that an RF waveform must be quantized into unimaginably large
numbers of tiny steps. Those steps will NEVER be observable, and there
will NEVER be any noticeable quantum effects at RF.
That holds true for all frequencies up to at least 100 gigahertz
(millimetre wavelengths), where quantum effects just begin to be
noticeable in precision measurements of very low noise levels - but even
way up there, quantum effects are still only a small correction.
So after all that, we come back to the plain fact that normal RF
radiation behaves purely as waves, just like we always thought it did.
The only new information from quantum theory is to *confirm* that
classical EM theory is all you'll ever need.
--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
Jimmie D
"art" <aun...@insightbb.com> wrote in message
news:1173448470....@q40g2000cwq.googlegroups.com...
> On 9 Mar, 02:33, "Jimmie D" <GFEN...@carolina.rr.com> wrote:
>> "art" <aun...@insightbb.com> wrote in message
>>
>> news:1173413632.1...@j27g2000cwj.googlegroups.com...
>>
>> > Gentlemen from outside of America. Gauss's law with respect to statics
>> > is quite specific and easy to understand. What is so wrong in
>> > mathematical terms by adding the metric of time to the law so that
>> > curl can be accomodated? i.e. change from a conservative field where
>> > all vectors have zero length,
>> > to a electro magnetic equation by adding the words " the addition of
>> > time" which by providing a three dimensional field has the true
>> > inclusion of curl i.e. all vectors have value in length and direction.
>> > America denies the feasability of such an addition to an existing law
>> > which in essence is regarded as a new law without basis on this side
>> > of the pond.Are all countries of this mentallity?
>> > Art
>>
>> Because a static field does not produce an EM field(curl) only if that
>> static charge is in motion. Motion would even include taking a charged
>> body,
>> say a pith ball and waving it back and forth. Electrons have a static
>> charge
>> but when they are in motion in a conductor they produce fields(curl).
>> Electrons moving about an atom also produces fields but the net result of
>> all the aoms moving about is zero. PLEASE REFERENCE THE GUASSIAN LAW ON
>> STATICS. I still think you are confusing static with statistics.
>
> But Jimmie my friend, now you have an understanding of Gaussian law
> what is preventing you adding the metric of time or a length of time
> about that
> little step? Start off my looking at it in pure mathematical terms and
> determine if the intent of the law is still not violated. Don't go
> beyond that at this time just consider the mathematics and get
> comfortable with it
> Art
>
Because it is meaningless
Dave
"John E. Davis" <da...@space.mit.edu> wrote in message
news:slrnev4j3i...@aluche.mit.edu...
in the general case. time varying charge implies a current, a current
implies a magnetic field, then you have to include Ampere's law and add
curl(E)=-dB/dt to the mix. while you may be able to constrain the changes
in rho(t) to some short time or constant current and eliminate the dB/dt
part of the problem, that would only apply in specific conditions, not to
the general case.
Cecil Moore
> What would you attribute the fact the "photons" in HF behave much
> differently then then the photons of VHF/UHF/SHF/LIGHT to?
In a free space vacuum, they all behave
essentially the same way. Their different
quantum energy levels causes them to interact
differently with the atomic structure of matter.
At 1 MHz, a photon has an energy level of
4x10^-9 eV while a gamma ray photon might have
an energy level of 4x10^+9 eV or higher.
Cecil Moore
> This applies to all forms of EM energy, so let's calculate the 'step
> size' in an RF waveform at 10MHz. That will be the energy content of a
> single quantum, which turns out to be 0.000 000 000 000 000 000 000 000
> 0066 joules - which is unimaginably small.
Maybe 40 nano-eV is more imaginable? :-)
For me, it is a lot easier to visualize a cloud
of photons leaving an antenna than it is to visualize
the field lines closing upon themselves (like a soap
bubble) and breaking free of the antenna.
Incidentally, your above number is off by 0.3% :-)
Dave
> On 9 Mar, 22:13, d...@space.mit.edu (John E. Davis) wrote:
>> On Fri, 09 Mar 2007 16:45:31 GMT, Dave <n...@nowhere.com>
> and that logic applies for a resonant array in situ
> inside a closed border whether time is variant or otherwise.
> The importantant point of the underlying logic that all inside the
> arbitary border must be in equilibrium at the cessation of time
> because the issue is not the static particles but of the flux. Period
> Thus the very reason for a conservative field in that
> it is able to project static particles in terms of time if time was
> added. For static particles time is not involved therefore
> ALL vectors are of ZERO length and direction is an asumption based on
> the action if and when time is added.
> John, you included time but did not mention time variant, was this for
> a reason? I have specifically use time variance since that enclosed
> within the border is an array in equilibrium
> from which the conservative field is drawn from.
> I am so pleased that some one came along that concentrated on the
> logic and not the retoric and abuse.
> Art
>
he may have hit what you believe correctly.. but unfortunately it is not a
valid generalization. as i stated in my other message:
John Smith I
> ...
> At 1 MHz, a photon has an energy level of
> 4x10^-9 eV while a gamma ray photon might have
> an energy level of 4x10^+9 eV or higher.
So, do I get you right here?
Like a high power hunting rifle, the energy that the photon is "shot"
from the antenna at guarantees a far and straight course of projection
(at vhf+ freqs)--as opposed to the lowly bb gun where the bb with low
energy is forced to fall to the forces of gravity (on in the photons
case, the earths magnetosphere) and come to earth much sooner?
John Smith I
The above is meant to be a bit "humorous", but you are stating HF is
much more prone to obey magnetic forces in the magnetosphere as opposed
to high freqs where photons are endowed with much more "kinetic energy"
in the form of the voltage(E) charge they have?
Since the law of conservation of energy exists, I am assuming you
consider some relationship of E/I to have changed in the VHF photon as
opposed to the HF photon--since there is no way for the 5 watts HF to
have different power levels than 5 watts VHF?
Dave
>>
>> - Show quoted text -
>
> Thats O.K. David,
> The appeal made for this thread was for people outside of America
> since eamericans were more interested in other things and I am
> assuming the Gentleman is from outside America. This discussion in the
> past has been bedeviled with arraogance and abuse to the neglect of
> logic, this has been the mode of this group for a very long time. If
> there was not such derision you could have looked up Gaussian law on
> the web where you would have found the mathematics behind the logic.
> If you had done this you would have found that curl is a part of the
> mathematical underpinning that in the event of time that part of the
> equation is zero. If time was part o0f the logic then you insert the
> value of curl in the equation, look up curl for your self and place it
> in the original equation which you are not changing i.e. concentrate
> on the mathematics and the underlying logic and the result becomes
> apparent.( and I have stated as such in past threads)
>
you obviously have not read and understood my recent posts. when you do
curl of electric fields you get ampere's law which takes into account the
time varying electric field... but of course also brings in the magnetic
field that is related to it. unless you add that part into the equation you
are ignoring half of the effects and will never get the proper understanding
of the equations. simply adding time to gauss's equation, written either in
differential or integral form simply ignores the magnetic field part and the
effects of curl and the resulting field and wave propagation effects that
must be taken into account when you start talking about time varying fields.
Cecil Moore
> Like a high power hunting rifle, the energy that the photon is "shot"
> from the antenna at guarantees a far and straight course of projection
> (at vhf+ freqs)--as opposed to the lowly bb gun where the bb with low
> energy is forced to fall to the forces of gravity (on in the photons
> case, the earths magnetosphere) and come to earth much sooner?
Not exactly. All photons, regardless of energy content,
travel at the speed of light (modified by VF, of course).
Your above example assumes most of the difference in energy
level is associated with the square of different velocities.
Cecil Moore
> Since the law of conservation of energy exists, I am assuming you
> consider some relationship of E/I to have changed in the VHF photon as
> opposed to the HF photon--since there is no way for the 5 watts HF to
> have different power levels than 5 watts VHF?
Five joules of HF (10 MHz) requires ten times as many
photons as five joules of VHF (100 MHz). What HF photons
lack in energy, they make up for in quantity.
I get ~7.553x10^28 photons in 5 joules of 10 MHz RF
energy and ~7.553x10^27 photons in 5 joules of 100
MHz RF energy.
The E-field/B-field ratio is the same for both in
free space.
John Smith I
> ...
> Five joules of HF (10 MHz) requires ten times as many
> photons as five joules of VHF (100 MHz). What HF photons
> lack in energy, they make up for in quantity.
> ...
Like I said, my original post in response to you was just "a joke", of
course the velocity of all photons is assumed constant.
However, the fact we fire a shotguy (HF) or a single bullet (VHF) makes
the photons in HF assume different charastistics than that of the fewer
photons of VHF?
I mean, I may be rather dense here, but I am attempting to put the model
you are presenting here together--obviously, I am missing something ...
Regards,
JS
--
http://assemblywizard.tekcities.com
John Smith I
> ...
> However, the fact we fire a shotguy (HF) or a single bullet (VHF) makes
> ...
Of course, in the above, "shotguy" = shotgun!
John E. Davis
wrote:
>no, i'm afraid you can't just put a 't' on each side and have it make sense
>in the general case. time varying charge implies a current, a current
>implies a magnetic field, then you have to include Ampere's law and add
>curl(E)=-dB/dt to the mix. while you may be able to constrain the changes
>in rho(t) to some short time or constant current and eliminate the dB/dt
>part of the problem, that would only apply in specific conditions, not to
>the general case.
I encourage you to review the Maxwell equations in a book on
electrodynamics. I personally like the book by Jackson, which is
oriented more towards physicists. In any case, the equation that I
wrote is one of the 4 Maxwell equations. It is valid for arbitrary
time-dependent electric fields. All it says is that the divergence of
the electric field at a point is proportional to the charge density at
that point:
div E(x,t) = 4\pi\rho(x,t) (Gaussian Units)
If you integrate this over a closed surface, and then use the
divergence theorem you get
\integral dA.E(x,t) = 4\pi \integral dV \rho(x,t)
The integral on the right-hand side is 4\pi times the total
(time-varying) charge enclosed by the surface. The other equations
are also valid, including the one you wrote.
Coincidently earlier this morning I was reviewing the derivation of
the energy loss of a heavy charged particle as it passes through
matter. The derivation made use of a very long cylinder with the
charged particle traveling along the axis of the cylinder. One point
in the calculation required the integral of the normal component
electric field (dA.E) produced by the charged particle over the
surface of the cylinder. That is, the left hand side of the above
equation. The answer is given by the right hand side of the above
equation. In this case, the charge density \rho(x,t) was created by
the moving charged particle.
--John
Cecil Moore
> --obviously, I am missing something ...
Maybe cause and effect? Cause and effect is indeed
missing in a lot of QED stuff. Not only do some
virtual particles move faster than the speed of
light but also apparently necessarily jump
backwards in time.
John E. Davis
wrote:
>Maybe cause and effect? Cause and effect is indeed
>missing in a lot of QED stuff. Not only do some
>virtual particles move faster than the speed of
>light but also apparently necessarily jump
>backwards in time.
So-called Feynman diagrams represent antimatter particles (positrons,
anti-quarks, etc) as the corresponding "matter" particles going
backward in time. Of course no physicist actually takes this
interpretation seriously. Nor do I believe particle theorists take
virtual particles seriously. They are just a convenient
representation of the terms in a perturbation expansion. It is unclear
to me that virtual particles play a role in non-perturbative theories.
--John
Dave
Gauss's law in Jackson's 'Classical Electrodynamics' 2nd edition, ppg
30-32,33 has NO 't'. nor does it in Ramo-Whinnery-VanDuzer 'Fields and
Waves in Communications Electronics' ppg 70-72(differential form),
75-76(integral form)
your final statement means that you are obviously outside the applicability
of Gauss's law since you have a moving charged particle, which can not be
described by a static field. i would guess that whatever derivation you are
looking at placed some other restrictions on the conditions such that you
could approximate the field by that type of equation. possibly a small
velocity or short distance or very short time period.
John E. Davis
wrote:
>Gauss's law in Jackson's 'Classical Electrodynamics' 2nd edition, ppg
>30-32,33 has NO 't'. nor does it in Ramo-Whinnery-VanDuzer 'Fields and
>Waves in Communications Electronics' ppg 70-72(differential form),
>75-76(integral form)
This is not surprising since that chapter in Jackson deals with
electrostatics. Look at section 1.5 on page 17. The section states:
The Maxwell equations are differential equations applying locally
at each point in space-time (x,t). By means of the divergence
theorem and Stoke's theorem they can be cast in integral form. [...
a few sentences later...] Then the divergence theorem applied to
the first and last [Maxwell] equations yields the integral
statements... The first is just Gauss's law...
--John
Cecil Moore
> It is unclear
> to me that virtual particles play a role in non-perturbative theories.
How about the static magnetic field from a permanent
magnet?
Richard Clark
<gm3...@ifwtech.co.uk> wrote:
>Quantum theory describes electromagnetic energy as being divided into a
>series of packets called photons, so (total energy in a stream of
>photons) = (number of photons/second) x (energy of individual photons).
Energy is not expressed with a time in the denominator. The standard
quantum theory expression for energy is eV - time is wholly absent as
it should be.
>This also means that EM energy doesn't exist in pure sine-waves
EM theory does not exclude the classic description of pure sine waves.
This is not a neither/nor situation.
> - the
>waveform is actually built up in steps, very much like digitized audio.
This appears to be the beginnings of a description about to fall off
the edge. What waveform? This is a conceit of time.
>The step size is the energy content of one quantum.
No, the step size as you describe is the potential difference of
quantization, an engineering term, not a quantum mechanics term. It
is quantisizing an amplitude to construct the wave in a time domain.
Quanta is the complete wave in a frequency domain.
The transform from one domain to the other is classic Fourier. His
analysis revealed that one unique energy (a single frequency) can be
decimated into many components (amplitudes over time). The
transformation is fully reversible (many amplitudes over time turned
back into one single frequency) without any information loss.
Mixing the two as being one analysis, corrupts it. Time is not a
factor in energy and cannot be drawn into its discussion through
transforms that mix topics.
>The question is: are those steps noticeable enough to be important?
>
>For light and shorter wavelengths, the answer is often Yes. Quantum
>theory was developed to explain observations like some kinds of light
>being emitted in a series of sharp spectral lines, which cannot be
>explained by a wave-only theory. Instead, it has to be thought of as
>being built up of individual photons/quanta which can only have certain
>"allowed" energy levels.
There is absolutely no distinction between "allowed" energy levels and
resonance which allows only certain frequencies. Resonance has been
historically correlated with circularity, and every instance you site
above is found in the change of orbitals - circular, harmonic motion.
A photon is emitted in the cM band when an electron orbiting a
Hydrogen atom flips its magnetic pole. This event is vastly below the
short wavelengths you describe by a million-fold. A good number of
correspondents here are fully capable of detecting this event with
commercial gear already suitable for the Ham market. They could have
done it 50 years ago too.
This is a quantum event, it is expressed with a classic quantum
energy, and it is resolvable as being important (insofar as
"importance" is a subjective, not quantitative quality). It is
certainly noticeable and is not artificially constrained by scale.
73's
Richard Clark, KB7QHC
Dave
yes, referring to all 4 Maxwell equations you do have a 't' dependency.
however, even equations 1.13 and 1.14 referred to by your quote have NO time
dependency in them. the equations on the next page,1.15 and 1.16 have the
time dependency that the 't' in your quote refers to. remember, those
integrals are NOT integrals over time, they are over the surface or volume.
John Smith I
> ...
> 73's
> Richard Clark, KB7QHC
I don't know ...
A digitized/stepped wave sorta' makes sense.
If you have an "amplitude" of one photon, then you add another, and yet
another, etc. it would tend to look "stepped" at some point--and, since
no one really knows, let's reserve the final determination ...
Right now, "quantums" are kinda like "nauga hide" (or, naugahyde) to me ...
You DO know about naugas' right?
My couch and chair and covered with their hides ... <grin>
Regards,
JS
John E. Davis
wrote:
>yes, referring to all 4 Maxwell equations you do have a 't' dependency.
>however, even equations 1.13 and 1.14 referred to by your quote have NO time
>dependency in them. the equations on the next page,1.15 and 1.16 have the
>time dependency that the 't' in your quote refers to. remember, those
>integrals are NOT integrals over time, they are over the surface or volume.
As I stated before, this chapter deals with electrostatics so the time
dependence has been dropped. However, the fact remains that Gauss's
law is the integral form of the first Maxwell equation, which holds
for an arbitrary space-time point. Unless you reject the first
equation, namely
div E(x,t) = 4 \pi \rho(x,t)
or the divergence thereom, you have to accept the fact that
\integral_S E(x,t).dA = 4 \pi \integral_V dV \rho(x,t)
which is Gauss's theorem.
This is my last post regarding this subject.
--John
Dave
the later chapter i quoted first is not based on electrostatics, and the
formula for gauss's law is always the same. it is not dependent on time in
any form. thank you for not continuing to prolong the misinformation in
this thread.
John E. Davis
wrote:
>the later chapter i quoted first is not based on electrostatics, and the
>formula for gauss's law is always the same. it is not dependent on time in
>any form.
I will make one last effort to to set the record straight. In volume
II of the Feynman Lectures on Physics, the title of chapter 15,
section 6 is "What is true for statics is false for dynamics". The
5th paragraph of that section states "Gauss' law, [eq omitted]
remains...". Also in that section, he has a table (Table 15-1) that
contains two columns:
FALSE IN GENERAL | TRUE ALWAYS
(true only for statics) |
------------------------------------------------------
Coulomb's Law | Gauss' law
[...] | [...]
Then in chapter 18, section 1 paragraph 3 you will find the statement:
"In dynamic as well as static fields, Gauss' law is always valid".
I do not think I can make it any more clear than this.
> thank you for not continuing to prolong the misinformation in this
> thread.
Do you also accuse Feynman of spreading misinformation? Unfortunately
he died a few years ago.
--John
Richard Clark
Davis) wrote:
>On Sat, 10 Mar 2007 23:21:35 GMT, Dave <no...@nowhere.com>
>wrote:
>>the later chapter i quoted first is not based on electrostatics, and the
>>formula for gauss's law is always the same. it is not dependent on time in
>>any form.
>
>I will make one last effort to to set the record straight. In volume
>II of the Feynman Lectures on Physics, the title of chapter 15,
>section 6 is "What is true for statics is false for dynamics". The
>5th paragraph of that section states "Gauss' law, [eq omitted]
>remains...". Also in that section, he has a table (Table 15-1) that
>contains two columns:
>
> FALSE IN GENERAL | TRUE ALWAYS
> (true only for statics) |
> ------------------------------------------------------
> Coulomb's Law | Gauss' law
> [...] | [...]
At the bottom of that Table is a footnote explaining the bold arrow of
your Gauss' law. It reads:
"The equations marked by an arrow (-ยป) are Maxwell's equations."
The table equation, and the one you reference in the text are both
Maxwell's.
>Then in chapter 18, section 1 paragraph 3 you will find the statement:
>
> "In dynamic as well as static fields, Gauss' law is always valid".
That chapter, too, clearly defines the same equation you are making an
appeal to as "Maxwell's equations." Observe Table 18-1 "Classical
Physics"
It is explicitly derived from the treatment as equation 4.1 - also
denoted Maxwell's equations.
"All charges are permanently fixed in space, or if they do move,
they move as a steady flow in the circuit ( so rho and j are
constant in time). In these circumstances, all of the terms in
the Maxwell equations which are time derivatives of the field are
zero."
Equations 4.6 and 4.8, the cross and dot products resolve to zero.
If you crank up the clock, Feynman concludes
"Only when there are sufficiently rapid changes, so that the time
derivatives in Maxwell's equations become significant, will E and
B depend on each other."
We will, of course, recognize this EB relationship as the field of
radiation and further recognize there is no field of radiation without
a significant time factor.
The grad operator, an inverted, enbolded del, is discussed by Feynman
in Chapter 2-4 is a significant element of these equations. The grad
operator obeys the same convention as the derivative notation.
Feynman's instruction clearly shows that Maxwell's treatment (actually
Heaviside's work before him) is a generalization of Gauss to include
time (sorry Art, he got there two centuries ago) and hence describes
Gauss equations as special (zero-time) instances of the generality.
Yuri Blanarovich
"art" <aun...@insightbb.com> wrote
>
> John
> I thank you for your input and continued attempts to overcome the
> barriers placed before you. I don't know what Country you come from
> but I apologise that the baggage associated with me was then dumped
> upon you and I fully understand your action of withdrawal.
> Thanks again for your efforts and I trust that you will not see
> this as sign of a new America emerging. It is certainly not the
> America I envisaged some forty years ago when I arrived. I think it is
> best that I to withdraw . Sometime it takes a hundred years
> before science is permitted to move on. A similar thing happened to
> George Green of Nottingham U.K. and only reemerged in full by the
> presentation by some body else who received the acreditation.
> Best regards
> Art Unwin KB9MZ...XG
Yo XG man!
While most of us sympatize with your condition, but your drivell is getting
beyong pathetic, you dumping on America is picture of your messed up
judgement and your "evaluation" of people here is just reflection of who is
messed up.
If you can't get over losing your colonies, or superiority of colonist
inbreds, you are free to go back, Eurabia is waiting for you and will
undoubtly recognize your genius (of calling reflector - director, and having
patent to prove it) and award you cross of the empire or something.
Just what the heck is your "Gausian" contraption suppose to get me that all
other known antennas or my designs don't? Lousy pattern with three lobes
over perfect ground and 6 dB F/B at 200 MHz??? Whopeeeee!!!
God bless America, the last bastion of freedom and the greatest country on
Earth!
Love it, or leave it!
73, cut the crap and get well!
Yuri, ex OK3BU
Ian White GM3SEK
>Ian White GM3SEK wrote:
>> This applies to all forms of EM energy, so let's calculate the 'step
>>size' in an RF waveform at 10MHz. That will be the energy content of a
>>single quantum, which turns out to be 0.000 000 000 000 000 000 000
>>000 0066 joules - which is unimaginably small.
>
>Maybe 40 nano-eV is more imaginable? :-)
>For me, it is a lot easier to visualize a cloud
>of photons leaving an antenna than it is to visualize
>the field lines closing upon themselves (like a soap
>bubble) and breaking free of the antenna.
>
Each to his own imagination. I'd find it hard to keep track of all
10^(30) photons emitted from here this weekend... though some of them
will probably send postcards :-)
>Incidentally, your above number is off by 0.3% :-)
Well, imagine that!
--
73 from Ian GM3SEK
John Smith I
> Ian White GM3SEK wrote:
>> This applies to all forms of EM energy, so let's calculate the 'step
>> size' in an RF waveform at 10MHz. That will be the energy content of a
>> single quantum, which turns out to be 0.000 000 000 000 000 000 000
>> 000 0066 joules - which is unimaginably small.
>
> Maybe 40 nano-eV is more imaginable? :-)
> For me, it is a lot easier to visualize a cloud
> of photons leaving an antenna than it is to visualize
> the field lines closing upon themselves (like a soap
> bubble) and breaking free of the antenna.
>
Well, I am just sitting here with this 120,000 gauss magnet in my hand
and a magnifying glass in the other--attempting to see the photons
shooting from one side of the magnet to the other ... I'd at least think
the dern photons excite the gas and would light up the neon bulb I am
holding next to one side!
Cecil Moore
> Well, I am just sitting here with this 120,000 gauss magnet in my hand
> and a magnifying glass in the other--attempting to see the photons
> shooting from one side of the magnet to the other ... I'd at least think
> the dern photons excite the gas and would light up the neon bulb I am
> holding next to one side!
Maybe a quote from "Optics", by Hecht is in order:
"Virtual photons can never escape to be detected
directly by some instrument." Sorry about that.
Ian White GM3SEK
>On Sat, 10 Mar 2007 09:33:46 +0000, Ian White GM3SEK
><gm3...@ifwtech.co.uk> wrote:
>
>>Quantum theory describes electromagnetic energy as being divided into a
>>series of packets called photons, so (total energy in a stream of
>>photons) = (number of photons/second) x (energy of individual photons).
>
>Energy is not expressed with a time in the denominator. The standard
>quantum theory expression for energy is eV - time is wholly absent as
>it should be.
>
Sorry, that was an editing mistake: obviously it's power that includes
the time dimension, energy does not.
>>This also means that EM energy doesn't exist in pure sine-waves
>
>EM theory does not exclude the classic description of pure sine waves.
>This is not a neither/nor situation.
>
>> - the
>>waveform is actually built up in steps, very much like digitized audio.
>
>This appears to be the beginnings of a description about to fall off
>the edge. What waveform? This is a conceit of time.
>
>>The step size is the energy content of one quantum.
>
>No, the step size as you describe is the potential difference of
>quantization, an engineering term, not a quantum mechanics term. It
>is quantisizing an amplitude to construct the wave in a time domain.
>Quanta is the complete wave in a frequency domain.
>
Richard is right about that. Please ignore my remarks about quantization
within the waveform itself. The quantization only affects the total
quantity of energy; or the power level if you wish to consider the rate
of energy transfer.
However, that doesn't affect the main point: at normal radio
frequencies, RF energy is composed of unimaginably small packets, in
unimaginably large numbers. This means that quantization effects in
energy or power levels are utterly negligible, and we can always think
of RF energy or power as a continuous stream.
It's rather similar to being aware that electric current is actually
made up of individual electrons - it's interesting information, but
electronic engineering very rarely needs to acknowledge the existence of
individual electrons. Even less does antenna engineering need to
acknowledge the existence of individual RF photons.
As I said in the previous posting, the energy of EM photons is
proportional to the frequency, so quantization effects only begin to be
noticeable at frequencies of hundreds of gigahertz, and still only as a
small correction in measurements of the very lowest power levels we can
detect.
Richard cited the following as a claimed exception:
>A photon is emitted in the cM band when an electron orbiting a
>Hydrogen atom flips its magnetic pole. This event is vastly below the
>short wavelengths you describe by a million-fold. A good number of
>correspondents here are fully capable of detecting this event with
>commercial gear already suitable for the Ham market. They could have
>done it 50 years ago too.
>
That is an example of a quantum effect determining the *frequency* of an
RF emission... but the origin of the RF energy doesn't change its
character. If a signal generator is tuned to that frequency, it will
produce exactly the same kind of RF energy - a torrent of quanta so tiny
that their individual existence is irrelevant.
--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
John Smith I
> ...
> Maybe a quote from "Optics", by Hecht is in order:
> "Virtual photons can never escape to be detected
> directly by some instrument." Sorry about that.
Cecil:
You know me by now, I jest.
If we knew as much (in quantity) as we DON'T know--we'd know enough to
be of importance ... :(
Richard Clark
Hi Ian
Determining the *frequency*? That has to be the most obscure
contribution I've ever seen.
The origin in fact does change its character - that is the whole point
and it is an elemental point of sub-atomic physics at that. Can you
offer a counter-example or any example of Photon generation that
leaves no trace of change? I can offer one that doesn't, I can offer
one that does leave a trace (already done), and examples where the
photon originates from non-electron interaction.
For this last, they too inhabit wavelengths that are orders of
magnitude below the visible. Phononic-Photonic interactions may not
fill library shelves, but their several volumes that do fill at least
one shelf are quite thick. Beyond their contributions we find those
from Excitons, Polarons, Polaritons, Plasmons and so on down the
energy band into the less than milli-eV range. Such photon generation
is in a continuum of wavelengths that challenges the simple Lyman
series of discrete resonances (much less all the other series) you
alluded to previously. That continuum extends over great swaths of
the RF spectrum. To your credit, you allude to this spectrum but
underplay the consequences:
>noticeable at frequencies of hundreds of gigahertz, and still only as a
>small correction in measurements of the very lowest power levels we can
>detect.
The effects are not marginally detectable and are the basis of a new
industry called Nanotechnology.
As for a torrent of quanta so tiny that their individual existence is
irrelevant, this distinction could be as easily lost on sunshine, much
less HF wavelengths. One can certainly find a power density from an
HF antenna that equals that of sunshine. The scale of comparing the
number of photons would be the difference between drinking out of a
firehose or a tidal wave. Clearly both have long escaped the
magnitudes of a gulp and to a drowning man, the comparison would be
ironically trivial.
Cecil Moore
> Determining the *frequency*? That has to be the most obscure
> contribution I've ever seen.
Richard, do you understand that free electrons can
emit photons of any frequency? i.e. no orbital change
necessary?
bluey
> On Sun, 11 Mar 2007 06:00:27 +0000 (UTC), d...@space.mit.edu (John E.
>
> Davis) wrote:
> >On Sat, 10 Mar 2007 23:21:35 GMT, Dave <n...@nowhere.com>
> >wrote:
I'm easily impressed, but none the less I'm still impressed.
Derek.
Richard Clark
>Cecil, you should look up the background of JOHN E DAVIS from MIT that
>the group just dissed. He is not just a nobody, he has credentials
>that should be respected
Hi Art,
Unfortunately John has proven that even (what you assume to be) an MIT
Don can get it wrong. He is not (just a software monkey like a lot of
us.) His references clearly show in exactly the same areas (chapter
and verse) how it is Maxwell's (actually Heavisides') equations that
add time to Gauss. In this case Feynman (the winner of the Nobel
prize in Physics) clearly states the time relationship introduced by
Maxwell two centuries ago.
I am quite sure the very book under your nose says exactly the same
thing, using exactly the same equations.
The long and short of it is exactly the same thing you've been told ad
infinitum, but have spit on those correspondents. The only thing
you've added is shooting sparks (must be related to that spit thing)
and pixie sticks.
Haven't you left yet?
Ian White GM3SEK
>On Sun, 11 Mar 2007 21:23:28 +0000, Ian White GM3SEK
><gm3...@ifwtech.co.uk> wrote:
>
>>Richard cited the following as a claimed exception:
>>
>>>A photon is emitted in the cM band when an electron orbiting a
>>>Hydrogen atom flips its magnetic pole. This event is vastly below the
>>>short wavelengths you describe by a million-fold. A good number of
>>>correspondents here are fully capable of detecting this event with
>>>commercial gear already suitable for the Ham market. They could have
>>>done it 50 years ago too.
>>>
>>That is an example of a quantum effect determining the *frequency* of an
>>RF emission... but the origin of the RF energy doesn't change its
>>character. If a signal generator is tuned to that frequency, it will
>>produce exactly the same kind of RF energy - a torrent of quanta so tiny
>>that their individual existence is irrelevant.
>
>Hi Ian
>
>Determining the *frequency*? That has to be the most obscure
>contribution I've ever seen.
>
most basic equation of quantum mechanics:
E = hf
It means that whenever there is a transition between two energy levels,
a photon is emitted whose frequency is uniquely determined by the
difference between those energy levels.
The case you quoted was the so-called "hydrogen line". A hydrogen atom
can have the spin of its single electron aligned in the same direction
as that of its single proton; or in the opposite direction. The former
state has slightly more energy, and when the spin of one atom flips to
the lower-energy state, one quantum of EM radiation is emitted. The
frequency of that radiation is determined by the difference in energy
levels between the two states, and is 1.42GHz.
The hydrogen line is like any other spectral line, except that the
difference in energy levels is unusually small (optical spectroscopists
would call it "hyperfine splitting") so the energy comes out as
microwaves rather than light.
The point I was making was that 1.42GHz radiation generated in this
particular manner has no special properties other than its frequency. It
is exactly the same kind of RF energy as you'd get from a signal
generator tuned to the same frequency.
Derek
> the group just dissed. He is not just a nobody, he has credentials
> that should be respected
Richard Clark
Hi Ian,
So, the photon thus emitted is indistinguishable from any signal
generator's output. Neither of us is surprised, granted.
What distinction are you trying to make that is not already obvious?
>>If a signal generator is tuned to that frequency, it will
>>produce exactly the same kind of RF energy - a torrent of quanta so tiny
>>that their individual existence is irrelevant.
This is nothing more than a tautology. No one is going to be
surprised by this event either.
Can you give us an example of "special properties" that differentiates
a photon from an EM wave? On the face of it, that question is absurd,
but I see nothing distinctive about your comments except in this fine
parsing of "special properties" that seem to vanish (no pun) for
longer wavelengths.
>It means that whenever there is a transition between two energy levels,
>a photon is emitted whose frequency is uniquely determined by the
>difference between those energy levels.
If this is the "special property" and hyperfine are not, then I
suppose it could as easily be called "very special property" to no
less acclaim. The production of photons through a myriad of other
interactions that I offered rather makes this "special property"
rather banal, because those interactions also present harmonic
relationships and are not uniquely determined by transitions - and yet
they remain photons none the less. Would I be overstepping to call
them "extraordinary properties?"
I must presume this casts back to your comment:
>>Quantum
>>theory was developed to explain observations like some kinds of light
>>being emitted in a series of sharp spectral lines, which cannot be
>>explained by a wave-only theory. Instead, it has to be thought of as
>>being built up of individual photons/quanta which can only have certain
>>"allowed" energy levels.
My presumption is that "special properties" and "some kinds of light"
are congruent. Again, it appears to be as tautological as might my
examples of "extraordinary properties."
Where does this lead us?
Derek
West Coast against M.I.T. space engineering on the East Coast.
The Patent Office may have to invite the IEEE to adjudicate on Art's
patent or else dig up Gauss's grave.
Looks like this could run a bit longer yet.
Derek
ji...@specsol.spam.sux.com
<snip>
> My presumption is that "special properties" and "some kinds of light"
> are congruent. Again, it appears to be as tautological as might my
> examples of "extraordinary properties."
> Where does this lead us?
To the conclusion that a photon is a photon.
All photons propagate the same way.
Electromagnetic radiation is electromagnetic radiation.
All electromagnetic radiation propagates the same way.
The interaction of electromagnetic radiation and matter is a function
of the photon energy, which is a function of the photon frequency.
There is no hard and fast dividing line between antennas and optics.
And probably a few more if I thought about it for a while.
--
Jim Pennino
Remove .spam.sux to reply.
Derek
Who's next?
D
Ian White GM3SEK
>Richard Clark <kb7...@comcast.net> wrote:
>
><snip>
>
>> My presumption is that "special properties" and "some kinds of light"
>> are congruent. Again, it appears to be as tautological as might my
>> examples of "extraordinary properties."
>
>> Where does this lead us?
>
>To the conclusion that a photon is a photon.
>
>All photons propagate the same way.
>
>Electromagnetic radiation is electromagnetic radiation.
>
>All electromagnetic radiation propagates the same way.
>
>The interaction of electromagnetic radiation and matter is a function
>of the photon energy, which is a function of the photon frequency.
>
Agreed with all the above, but...
>There is no hard and fast dividing line between antennas and optics.
Certainly no hard and fast line, because the same basic physics applies
at all frequencies and wavelengths. But there are HUGE differences in
the size and importance of some effects, for RF and for light.
We cannot tell how big those differences are by just talking about them.
We need to rub a few numbers together, and then see what comes out.
So the very first step into quantum mechanics is to put some numbers
into the E=hf equation. This immediately proves that quantum effects
(although still theoretically present) are too small to be of any
practical importance in antenna engineering. At that point, any sensible
person would realise they had taken a wrong turning, and get straight
back on the road.
For antenna engineering, that road is ENTIRELY built on the classical
physics of the 18th-19th century. It can be a hard road to travel, but
it's a reliably straight one. Any side turnings are NOT going to be
short-cuts to a better understanding.
Gene Fuller
[snip]
>
> For antenna engineering, that road is ENTIRELY built on the classical
> physics of the 18th-19th century. It can be a hard road to travel, but
> it's a reliably straight one. Any side turnings are NOT going to be
> short-cuts to a better understanding.
Ian,
For the misunderstood and unappreciated "inventor", hope springs eternal.
It's all for the good, however. RRAA would simply fade away without
fractals, crossed-fields, RoomCaps, unmodelable structures, traveling
waves, one-second long transmission lines, Poynting vectors, etc.
73,
Gene
W4SZ
Gene Fuller
> On 13 Mar, 08:02, Gene Fuller <W4SZzzz...@att.net> wrote:
>
> Gene, I was just reading the archives of 2004 where you fought with
> everybody in ham radio,QEX as well as on this newsgroup as to how
> everybody was inerpretating Maxwells laws plus used a lot of
> accusatory words against Walt and many others. You couldn't push any
> of them away then so what makes you think that all are going to line
> up behind you to get rid of me? Now you are lining up with the amateur
> group and the West Coast without resolving your past disagrements with
> every body about your disagreements with Maxwell resolved . Are you
> going to start a third front about what Maxwell really meant? NASA has
> been in error before, remember the "O"
> ring saga . They then dug a hole for themselves thinking that the
> deeper they dug the closer they were to escaping, maybe you are of the
> same thinking.Think about all those clever guys that were part of MIT
> and you are going to take them on with respect to Maxwell's teachings
> or at least what you thinl he meant? I'll back MIT anyday against you
> and others with respect to electrical laws.He gave the mathematical
> analysis which all have been craving for and he gets accused of
> spreading mis information. What is it that this group and the West
> coast NASA want with respect to Gaussian arrays, remove him from all
> the text books and replace him by Stokes?
>
> Art
>
Art,
You need to learn to read more carefully. My one and only argument with
Walt Maxwell was about the fuss between him and Steve Best. My position
then, and still today, was that both of these experts were correct in
their technical analysis.
Walt chose a novel approach involving "virtual short circuits", and
Steve chose a more traditional wave model. The physical, measurable
results were identical, and there would have been no way that anyone
could test the difference in the two analyses by any sort of measurement.
I believe there were some harsh words in addition to the technical
analysis, but I was not part of that. There was also a huge amount of
chatter along the lines of 2 + 2 is not equal to 7, from our favorite
nit-picker.
I have no idea why you have lumped me into something to do with MIT. I
have been there a few times over the years, but I don't think that would
have any connection to RRAA.
73,
Gene
W4SZ
Richard Clark
> My sincere thanks to M.I.T for supplying another avenue of
>authentification
Hi Art,
Was this thanks for his misreading Gauss where it should have been
Maxwell?
John E. Davis
wrote:
>Was this thanks for his misreading Gauss where it should have been
>Maxwell?
I do not understand your comment. If you go back and look at my first
post on this subject (Message-ID <slrnev4j3i...@aluche.mit.edu>),
you will see that I equated Gauss's law with the first Maxwell equation.
Gauss's law is commonly stated as:
The electric flux through a closed surface is proportional to the
amount of charge enclosed by the surface.
As I wrote before, this also happens to be the integral form of the
first Maxwell equation:
div E(x,t) = 4\pi\rho(x,t)
While Gauss may have stated this law in terms of static charges, and
it finds most applications in the static case, the law also holds for
the dynamic case. This is why physicists equate Gauss's law with the
integral form of the first Maxwell equation. And as evidence of this
association, you indirectly pointed out in Message-ID
<tma7v2dqk0b3vpt1u...@4ax.com> that Feynman equated the
two in the table 15-1 of volume II of his lectures.
--John
Richard Clark
wrote:
>I do not understand your comment.
Hi John,
It was rather explicit. To have disputed Dave's assertion with
additional material that substantiated him, makes for a rather strange
reading of Feynman.
As Art is presenting material that he has drawn time into Gauss' work
through his own invention exclusive of all other's, you have become
his touchstone as vouchsafing his claim. You can bow out once again,
of course, and become a martyr instead. Or you can indulge us with a
dialog with Art.
Tom Donaly
Using the MKSA system, Gauss' law is expressed as div D = rho. Art can
take the time derivative of both sides, if he wants to, in which case he
gets div d(D)/dt = d(rho)/dt. This doesn't mean much except that it's
what you end up with when you take the divergence of both sides of
the Maxwell equation curl H = j + d(D)/dt, and then apply the equation
of continuity where it fits. (You have to pretend the 'd's'
in each equation are the funny little Greek letters that signify
partial differentiation.) Feynman didn't like to use the magnetic field
intensity vector H or the electric flux density vector D
so he used their B and E equivalents in his presentation of Maxwell's
equations in his _Lectures on Physics_. I guess you could start an
argument over whether or not H and D have physical significance, but
don't ask me to join in.
John, I think you might want to re-think your equation div
E(x,t)=4\pi\rho(x,t).
73,
Tom Donaly, KA6RUH
John E. Davis
wrote:
> John, I think you might want to re-think your equation div
>E(x,t)=4\pi\rho(x,t).
It is not my equation--- it is the first Maxwell equation (expressed
using Gaussian units). I did not make it up, nor did I add the
time-dependence as another poster suggested.
--John
Tom Donaly
Different texts have Maxwell's equations in different order. What text
did you get this from? Becker has it (in Gaussian CGS units) as
div D = 4\pi\rho (where the backslash indicates multiplication, and D
and rho have the usual meanings. You can add the 't' if you want to, but
it's unnecessary. Also, since you're dealing in 3 dimensions, why not
indicate them as in E(x,y,z), or E(x,y,z,t) (if the time means something
to you)?
73,
Tom Donaly, KA6RUH
John E. Davis
wrote:
>Different texts have Maxwell's equations in different order. What text
>did you get this from? Becker has it (in Gaussian CGS units) as
>div D = 4\pi\rho (where the backslash indicates multiplication, and D
>and rho have the usual meanings. You can add the 't' if you want to, but
>it's unnecessary. Also, since you're dealing in 3 dimensions, why not
>indicate them as in E(x,y,z), or E(x,y,z,t) (if the time means something
>to you)?
I tend to write equations in LaTeX form as most people I exchange
emails with mathematical equations use that for formatting mathematics.
Here, \pi represents the greek letter pi, and \rho is the greek letter
rho. I used x to represent a spatial 3-vector. I could have written
it as (x,y,z) but I did not think this shorthand would cause any
confusion given the context.
The difference between E and D is not important here. If you use D,
then \rho must be interpreted as the so-called "free" charge density.
However, the fundamental field is E, and if you use it the \rho must
be interpreted as the _full_ charge density. The relationship between
E and D can be very complex and may well depend upon the strength of
the applied field E. For simple materials a linear relationship is
usually assumed, e.g., D = \epsilon E, where \epsilon is the
dielectric constant of the medium. Also even here in this linear
relationship, \epsilon need not be a scalar (a number). It could be a
tensor (a 3x3 matrix), in which case D and E would not have the same
direction.
--John
Derek
> On 13 Mar 2007 22:15:37 GMT, d...@space.mit.edu (John E. Davis)
> wrote:
>
> >I do not understand your comment.
>
> Hi John,
>
> It was rather explicit. To have disputed Dave's assertion with
> additional material that substantiated him, makes for a rather strange
> reading of Feynman.
. You can bow out once again,
> of course, and become a martyr instead. Or you can indulge us with a
> dialog with Art.
>
I would think this comment applied equally to Dave.
Derek
Tom Donaly
Thanks for explaining that, John. I am unfamiliar with the conventions
of LaTex, obviously (I get my information from books that are generally
older than I am, and I'm not young). I don't have any problem with
Gauss' law being used in a non-static context. It applies, regardless.
That's as far as I go in agreeing with Art, though, since I can't
understand the rest of his theory, at all (but might if I could turn
off the left side of my brain - maybe).
73,
Tom Donaly, KA6RUH
Richard Clark
>But that still leaves the question why haven't
>ham
>radio users not picked up the slack and tried them?
Hi Art,
That was done an hundred years ago, and people found better ways.
Dave
"Derek" <blue...@yahoo.com.au> wrote in message
news:1173900412....@b75g2000hsg.googlegroups.com...
equation... well done. sri i didn't state that myself, but i have had
better things to do than try to argue with art.
Derek
>
> i like the post that points out the unnecessary t in the Gauss's law
> equation... well done. sri i didn't state that myself, but i have had
> better things to do than try to argue with art.
As I remember it you were arguing with John.
Derek
Dave
same after a while.
JIMMIE
> Cecil Moore wrote:
> > j...@specsol.spam.sux.com wrote:
> >> EM waves depart when energy is applied, not particles.
>
> > Quantum Electrodynamics tells us that EM waves consist
> > of photons which are particles.
> > --
> > 73, Cecil, w5dxp.com
>
> So, which is the real question:
>
> 1) Why do waves act like particles?
>
> --OR--
>
> 2) Why do particles act like waves?
>
> JS
> --http://assemblywizard.tekcities.com
When a field is traveling at/near the speed of light it has mass(acts
as a particle) slower and it is a wave. EM lives on the hairy edge of
both worlds. vAt least thats what my Phd girlfriend told me once. Who
knows though, she was pretty weird.
JIMMIE