Sue..., the web page xerox queen, waxes irrelevant:
Bilge wrote:
Sue...:
Even if you think you are getting the right numerical
answers you won't be able to model near and far-field
wave impedance for EM coupling structures.
I'd say don't be an idiot, but it's too late for that.
Don't you ever get tired of misunderstanding the irrelevant
web pages you reference?
The fact you find them irrelevant is some clue why your
posts have to be limited to insults and malicious pranks.
Oddly enough, that is a very good guess. Now all you have to do is
finish making the logical connection between posting irrelevent drivel
and having the irrelevance pointed out.
Don't expect to read this at your preferred hang-out, alt.morons.
Why would I expect to read your reply in alt.morons? Since alt.morons
is one of the few newsgroups in which your posts could be relevant, I
don't expect you post there.
Learn some physics:
http://farside.ph.utexas.edu/teaching/em/lectures/node28.html
Feel free to compare what you know about physics to what I know
anytime and we'll find some physical phenomenon and a set of criteria
which may be used to determine which one of us is a windbag from
our respective analyses and calculations. Choose any branch of
physics you seriously think you know something about.
Sue...
If you
don't have the 1/r^2 and 1/r^3 appropriatly represented
in the E and H planes you'll have to invent magnetic
monopoles, funny clocks and FTL wave funcion collapse.
None of those are required for time dependent Maxwell's
equations.
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
http://en.wikipedia.org/wiki/Multiple_integral
http://arxiv.org/abs/physics/0204034
Sue...
Coulomb generators sum in a volume of space,
not through an imaginary line that you have called a vector.
Gobbledegook.
What about the force from a single charge on another
charge? I can't use F = k q1 q2 /r^2 to express that force
because it's "an imaginary line that I call a vector"?
Now I can factor out |r|
E = k*q*|r| * (|r| - 2*|R|*cos(theta)) / (|R-r|^2 * |R|^2)
Here's where the "dipole moment" comes in. The field
is proportional to this product q*|r| of the individual
charge sizes and their separation. So we define this
quantity as a new thing, the "dipole moment" p = q*|r|.
Charge sizes? Do they have UK, US and EU scales
like tennis shoes do?
They are measured in Coulombs in the mks system, in
which a proton carries a charge of 1.6 * 10^-19 Coulomb.
Other units have been used.
http://www.cactus2000.de/uk/unit/masschr.shtml
I think you'll find that one size (0.511MeV) fits all if you
do it this way.
No, I really won't, since that is an energy. It is the mass
of the electron in energy units. It is no particular number
associated with a proton.
- Randy
.
