Gaussian statics law
 

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

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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