Gaussian statics law
 

"art" wrote in message
s.com...
On 10 Mar, 06:41, "Dave" wrote:
"art" wrote in message

oups.com...



On 9 Mar, 22:13, (John E. Davis) wrote:
On Fri, 09 Mar 2007 16:45:31 GMT, Dave
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

John, you have hit it on the nose. It is the logic that is important
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:

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.- Hide quoted text -

- 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.

Gaussian statics law
 

John Smith I wrote:
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.
--
73, Cecil http://www.w5dxp.com

Gaussian statics law
 

John Smith I wrote:
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.
--
73, Cecil http://www.w5dxp.com

Gaussian statics law
 

Cecil Moore wrote:

...
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

Gaussian statics law
 

John Smith I wrote:

...
However, the fact we fire a shotguy (HF) or a single bullet (VHF) makes
...

Of course, in the above, "shotguy" = shotgun!

JS
--
http://assemblywizard.tekcities.com

Gaussian statics law
 

On Sat, 10 Mar 2007 13:08:39 GMT, Dave
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

Gaussian statics law
 

John Smith I wrote:
--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.
--
73, Cecil http://www.w5dxp.com

Gaussian statics law
 

On Sat, 10 Mar 2007 11:34:36 -0600, Cecil Moore
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

Gaussian statics law
 

"John E. Davis" wrote in message
...
On Sat, 10 Mar 2007 13:08:39 GMT, Dave
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

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.

Gaussian statics law
 

On Sat, 10 Mar 2007 18:38:08 GMT, Dave
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

Gaussian statics law
 

John E. Davis wrote:
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?
--
73, Cecil http://www.w5dxp.com

Gaussian statics law
 

On Sat, 10 Mar 2007 09:33:46 +0000, Ian White GM3SEK
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

Gaussian statics law
 

"John E. Davis" wrote in message
...
On Sat, 10 Mar 2007 18:38:08 GMT, Dave
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

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.

Gaussian statics law
 

Richard Clark wrote:

...
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

Gaussian statics law
 

On Sat, 10 Mar 2007 20:59:59 GMT, Dave
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

Gaussian statics law
 

"John E. Davis" wrote in message
...
On Sat, 10 Mar 2007 20:59:59 GMT, Dave
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

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.

Gaussian statics law
 

On Sat, 10 Mar 2007 23:21:35 GMT, Dave
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

Gaussian statics law
 

On Sun, 11 Mar 2007 06:00:27 +0000 (UTC), (John E.
Davis) wrote:

On Sat, 10 Mar 2007 23:21:35 GMT, Dave
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 (-») 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.

73's
Richard Clark, KB7QHC

Gaussian statics law
 

On 10 Mar, 23:00, (John E. Davis) wrote:
On Sat, 10 Mar 2007 23:21:35 GMT, Dave
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

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

Gaussian statics law
 

"art" 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

Gaussian statics law
 

Cecil Moore wrote:
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

Gaussian statics law
 

Cecil Moore wrote:
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.

Incidentally, your above number is off by 0.3% :-)

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!

JS
--
http://assemblywizard.tekcities.com

Gaussian statics law
 

John Smith I wrote:
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.
--
73, Cecil http://www.w5dxp.com

Gaussian statics law
 

Richard Clark wrote:
On Sat, 10 Mar 2007 09:33:46 +0000, Ian White GM3SEK
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

Gaussian statics law
 

Cecil Moore wrote:

...
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 ... :(

JS
--
http://assemblywizard.tekcities.com

Gaussian statics law
 

On Sun, 11 Mar 2007 21:23:28 +0000, Ian White GM3SEK
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.

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.

73's
Richard Clark, KB7QHC

Gaussian statics law
 

Richard Clark wrote:
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?
--
73, Cecil http://www.w5dxp.com

Gaussian statics law
 

On Mar 11, 12:38 am, Richard Clark wrote:
On Sun, 11 Mar 2007 06:00:27 +0000 (UTC), (John E.


Davis) wrote:
On Sat, 10 Mar 2007 23:21:35 GMT, Dave
wrote:


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 (-») 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.

73's
Richard Clark, KB7QHC



I'm easily impressed, but none the less I'm still impressed.

Derek.

Gaussian statics law
 

On 11 Mar, 18:45, Cecil Moore wrote:
Richard Clark wrote:
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?
--
73, Cecil http://www.w5dxp.com

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
Art

Gaussian statics law
 

On 12 Mar 2007 08:51:57 -0700, "art" wrote:

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?

73's
Richard Clark, KB7QHC

Gaussian statics law
 

Richard Clark wrote:
On Sun, 11 Mar 2007 21:23:28 +0000, Ian White GM3SEK
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.

No, it's the simplest. A couple of days ago, I quoted and explained the
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.



--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Gaussian statics law
 

On Mar 12, 8:51 am, "art" wrote:
On 11 Mar, 18:45, Cecil Moore wrote:

Richard Clark wrote:
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?
--
73, Cecil http://www.w5dxp.com

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
Art

Gaussian statics law
 

So we have this group lining up with space Antenna engineering on the
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

Gaussian statics law
 

Richard Clark 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.

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.

Gaussian statics law
 

On 12 Mar, 17:45, "Derek" wrote:
So we have this group lining up with space Antenna engineering on the
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

NO,NO,NO.
Let us have Fox news on one side with the CB's and NASA (don't
stigmatize ham radio) and CNN with MIT on the other side. NASA starts
of with Maxwells theme and MIT east coast comes back with Maxwell has
nothing to do with it
it was Gauss that started the logic based on equilibrium
and Maxwell didn't evolve anything about equilibrium. Of course IEEE
could still be the science adjudicator only.
Then NASA could come back with La Place and then we go to a commercial
followed by a rebuttal from the CB ers who will then interject
photons.

Heh T.V. or public radio would have a lot of fun doing the background
with segments on Green Gauss, Heaviside, Newton e.t.c. to give the
debate more flavor tho
the verbalising is going to get so fierce that the West coast NASA and
the CB ers may demand that Maxwell be dug up to talk on his own behalf
and offcourse MIT will make similar
demands for Gauss so both Gauss and Maxwell can apply for a visa
beforehand. As for the USPTO they have got the money so the outcome of
the patent rerquest doesn't really matter!
People with colds must not be allowed in as both Gauss and
Maxwell won't want to hear anything to do with coffin!!
Maybe U tube would be a better stage to air the subject so the public
can see a punch up!!!!!
Somebody alert 60 minuits
Art

Gaussian statics law
 

On Mon, 12 Mar 2007 23:21:32 +0000, Ian White GM3SEK
wrote:

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.

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?

73's
Richard Clark, KB7QHC

Gaussian statics law
 

On 12 Mar, 18:16, "art" wrote:
On 12 Mar, 17:45, "Derek" wrote:

So we have this group lining up with space Antenna engineering on the
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

NO,NO,NO.
Let us have Fox news on one side with the CB's and NASA (don't
stigmatize ham radio) and CNN with MIT on the other side. NASA starts
of with Maxwells theme and MIT east coast comes back with Maxwell has
nothing to do with it
it was Gauss that started the logic based on equilibrium
and Maxwell didn't evolve anything about equilibrium. Of course IEEE
could still be the science adjudicator only.
Then NASA could come back with La Place and then we go to a commercial
followed by a rebuttal from the CB ers who will then interject
photons.

Heh T.V. or public radio would have a lot of fun doing the background
with segments on Green Gauss, Heaviside, Newton e.t.c. to give the
debate more flavor tho
the verbalising is going to get so fierce that the West coast NASA and
the CB ers may demand that Maxwell be dug up to talk on his own behalf
and offcourse MIT will make similar
demands for Gauss so both Gauss and Maxwell can apply for a visa
beforehand. As for the USPTO they have got the money so the outcome of
the patent rerquest doesn't really matter!
People with colds must not be allowed in as both Gauss and
Maxwell won't want to hear anything to do with coffin!!
Maybe U tube would be a better stage to air the subject so the public
can see a punch up!!!!!
Somebody alert 60 minuits
Art


Re
MIT vs Nasa and the CBersc

Oh forgot to mention, astronaut diapers will be made available for
hams ooooops no, CB ers incase things go against them
Art

Gaussian statics law
 

Now thats more like it..

Who's next?

D

Gaussian statics law
 

wrote:
Richard Clark 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.



--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Gaussian statics law
 

Ian White GM3SEK wrote:

[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