John E. Davis wrote:
On Wed, 14 Mar 2007 05:53:43 GMT, Tom Donaly
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

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

On Wed, 14 Mar 2007 09:09:27 -0800, Tom Donaly
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

John E. Davis wrote:
On Wed, 14 Mar 2007 09:09:27 -0800, Tom Donaly
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

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

On 14 Mar, 12:40, "Tom Donaly" wrote:
John E. Davis wrote:
On Wed, 14 Mar 2007 09:09:27 -0800, Tom Donaly
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

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- Hide quoted text -

- Show quoted text -

Tom,
don,t worry about it! A poster has stated it was invented years ago.
I haven,t found it in any of my books so perhaps he will tell you
where
you can look at it. I assume my patent will now be turned down when
it is pointed out where the Gaussian antennas can be seen. When he
describes it
it to you then it should be much easier for you to understand the
logic behind it
and to determine whether it is all a todo about nothing. As for me I
think
the subject can be said as proven, albiet over 100 years or more ago
and the whole subject
started by that blithering idiot, sycopath and all those other phrases
can now
seen as closed. I will hang around a bit to see what I could have
been if
only I had been born 100 or 200 years ago where somebody said I will
have all
the manufacturers knocking on my door. Maybe that inventor of the
gaussian array left me a morsel on the cutting room floor which I can
exploit and which I can reveal after the existing Gaussian presence is
revealed by the poster. But that still leaves the question why haven't
ham
radio users not picked up the slack and tried them? Maybe it is the
'not invented in my town' thinking and where their heads still rest in
the sand. I also have this other invention
that I want.............no, I have learned my lesson I will take it to
my grave that would be so much easier.
Art The Englishman

On 14 Mar 2007 13:26:46 -0700, "art" wrote:

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.

73's
Richard Clark, KB7QHC

On 14 Mar, 12:40, "Tom Donaly" wrote:
John E. Davis wrote:
On Wed, 14 Mar 2007 09:09:27 -0800, Tom Donaly
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

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- Hide quoted text -

- Show quoted text -

Tom
I was rereading this thread as to why people have a hard time in
understanding Gauss's law
with respect to conservative fields and a transition to a non
conservative field where with the addition of time one can consider
what is outside the enclosed surface. Since you pursued the
mathematical side of the subject to a minor conclusion ( you stated
you didn't understand what I was proposing) with John E Davis of
M.I.T. I wish to share with you some notes on the Internet by David J
Raymond called "a radically modern aproach"
which to me is the best I have seen on Radiation in it's entirety.
Obviously there is a lot written that as hams it is not essential
reading for hams but what it does do is explain in a very clear way
the mechanics of radiation with specific applications with respect to
the transition from conservative fields ala Gaussian law of statics to
non conservative fields where at the cessation of time one can
reconcile what is outside the enclosed surface with that which is
inside the surface where what is inside the enclosure is in
equilibrium and the enclosing surface is frictionless. As can
obviously seen a Yagi inside the enclosed border cannot be considered
since at the cessation of time interaction between elements is still
taking place after the cessation of time. The notes are so well
written that one not conversant
with upper math can still follow the implications of the discussion at
hand and thus can be considered as recommended reading for all hams
interested in antennas as a subject. It also gives a very clear
mathematical progression from Gaussian law to the subject of non
conservative fields can be formed with the activation of curl during a
moment in time.
It is this progression that leads designers to design around cluster
arrays that are in equilibrium regardless of orientation ie without
continuing coupling effects after the cessation of time and is very
well chronicalled in the above stated notes.
Best regards
Art

On 18 Apr 2007 13:07:43 -0700, art wrote:

I was rereading this thread as to why people have a hard time in
understanding Gauss's law
with respect to conservative fields and a transition to a non
conservative field where with the addition of time one can consider
what is outside the enclosed surface.

Hi Art,

That single, obscure, and ponderously long sentence is a clue as to
why...

since at the cessation of time interaction between elements is still
taking place after the cessation of time.

One has to wonder what new meaning you have for the word "cessation."

How can there be any time (for anything to take place) when there is
no more time (for anything to take place).

73's
Richard Clark, KB7QHC

On 18 Apr, 13:36, Richard Clark wrote:
On 18 Apr 2007 13:07:43 -0700, art wrote:

I was rereading this thread as to why people have a hard time in
understanding Gauss's law
with respect to conservative fields and a transition to a non
conservative field where with the addition of time one can consider
what is outside the enclosed surface.

Hi Art,

That single, obscure, and ponderously long sentence is a clue as to
why...

since at the cessation of time interaction between elements is still
taking place after the cessation of time.

One has to wonder what new meaning you have for the word "cessation."

How can there be any time (for anything to take place) when there is
no more time (for anything to take place).

73's
Richard Clark, KB7QHC

When power supply is stopped to an array inside an enclosed surface
kinetic energy is in evidence by radiation between elements PRIOR to
emerging from the enclosed surface.
You cannot have an equation with reference to time if equality is not
obtained at the cessation of the time under consideration.
You should read up on conservative and non conservative fields before
you succumb to temptation by replying while you still have your foot
in your mouth. I have responded to you this one time only to show
others how stupid you can be when you allow animosity to over ride
cfommon sense.

art wrote:


When power supply is stopped to an array inside an enclosed surface
kinetic energy is in evidence by radiation between elements PRIOR to
emerging from the enclosed surface.
You cannot have an equation with reference to time if equality is not
obtained at the cessation of the time under consideration.
You should read up on conservative and non conservative fields before
you succumb to temptation by replying while you still have your foot
in your mouth. I have responded to you this one time only to show
others how stupid you can be when you allow animosity to over ride
cfommon sense.


Yes, and well-done

On 18 Apr 2007 13:54:19 -0700, art wrote:

When power supply is stopped to an array inside an enclosed surface
kinetic energy is in evidence by radiation between elements PRIOR to
emerging from the enclosed surface.
You cannot have an equation with reference to time if equality is not
obtained at the cessation of the time under consideration.

Hi Art,

Unless the enclosed surface is immense, all times (even for HF)
considered are on the scale of nanoseconds. At common excitations
considered for Amateur application would reveal power issues in the
microwatts. The ratio of scales (energy/time-volume) would be 9
orders of magnitude and well outside the accuracy of any modeler, and
vastly beyond the cares of useful theory.

73's
Richard Clark, KB7QHC