On Jun 28, 5:50*pm, Keith Dysart wrote:
Cecil has not found any and would rather prattle on about
the difference between energy and power than actually understand.

I have listed a number of the laws of physics that you are violating.
Of course, energy and power are different. Consider an ideal LC
oscillator at the instant of time when the voltage on the capacitor is
maximum and the current is zero. The energy is certainly not zero and
can be calculated knowing the voltage and capacitance (assume 1000
volts and 1uf). Yet the instantaneous power is zero because the
instantaneous current is zero.

energy = (V^2*C)/2 = 0.5 joule

power = V(t)*I(t) = 0

Please prove a one-to-one correspondence between energy and power. How
can you possibly say that you are tracking all the energy when the
power equals zero and the energy does not? Good grief!

I am not sure where you think there is an error.

Again, you need to cut off your output and enable your inputs. There
are no waves during DC steady-state. Therefore, there are no forward
waves and no reflected waves. Therefore, you basic assumptions are
invalid. For waves to exist there must be acceleration and
deceleration of carrier electrons and such does not exist during DC
steady-state. Why don't you know that?
--
73, Cecil, w5dxp.com

"Cecil Moore" wrote
...

Again, you need to cut off your output and enable your inputs. There
are no waves during DC steady-state. Therefore, there are no forward
waves and no reflected waves.
Therefore, you basic assumptions are
invalid. For waves to exist there must be acceleration and
deceleration of carrier electrons and such does not exist during DC
steady-state. Why don't you know that?

Why don't you know that oscillating current goes into the displacement
current. In EM no reflections.

You should decide: EM or electrons. The mixture is fun.
S*

On Jun 29, 11:44*am, "Szczepan Bialek" wrote:
You should decide: EM or electrons. The mixture is fun.

That's an interesting idea to which I don't know the answer. Let me
rephrase the question:

Although it is known that electrons cannot flow through the dielectric
of an ideal capacitor, how about photons? Can RF photons flow directly
through the dielectric layer of a capacitor? If they can, it would
explain a lot of things.
--
73, Cecil, w5dxp.com

Cecil Moore wrote:
On Jun 29, 11:44 am, "Szczepan Bialek" wrote:
You should decide: EM or electrons. The mixture is fun.

That's an interesting idea to which I don't know the answer. Let me
rephrase the question:

Although it is known that electrons cannot flow through the dielectric
of an ideal capacitor, how about photons? Can RF photons flow directly
through the dielectric layer of a capacitor? If they can, it would
explain a lot of things.
--
73, Cecil, w5dxp.com

photons can flow through a dielectric.. isn't that what EM propagation
is, after all?

On Jun 29, 12:54*pm, Jim Lux wrote:
photons can flow through a dielectric.. isn't that what EM propagation
is, after all?

Yes, after I posted it, I realized that it was a rhetorical question.
--
73, Cecil, w5dxp.com

On 29 jun, 15:08, Cecil Moore wrote:
On Jun 29, 12:54*pm, Jim Lux wrote:

photons can flow through a dielectric.. isn't that what EM propagation
is, after all?

Yes, after I posted it, I realized that it was a rhetorical question.
--
73, Cecil, w5dxp.com

Dear friends, I follow with interest your interesting digressions,
however in various different posts about differents matters, I notice
discussion arises about what is "real" and what is not. IMO that
contributes to the solution goes away from us (I remember making this
comment in a previous post). In this sense respecto to energy I would
like to quote a great physics:

"...there is a certain quantity, which we call energy, that does not
change in the manifold changes which nature undergoes. That is a most
abstract idea, because it is a mathematical principle; it says that
there is a numerical quantity which does not change when something
happens. It is not a description of a mechanism, or anything concrete;
it is just a strange fact that we can calculate some number and when
we finish watching nature go through her tricks and calculate the
number again, it is the same."

"It is important to realize that in physics today, we have no
knowledge of what
energy is. We do not have a picture that energy comes in little blobs
of a definite
amount. It is not that way. However, there are formulas for
calculating some
numerical quantity, and when we add it all together it gives "28"'—
always the
same number. It is an abstract thing in that it does not tell us the
mechanism or
the reasons for the various formulas."

From: Richard Feynman. "Six easy pieces"

Miguel LU6ETJ

On 29 jun, 20:57, lu6etj wrote:
On 29 jun, 15:08, Cecil Moore wrote:

On Jun 29, 12:54*pm, Jim Lux wrote:

photons can flow through a dielectric.. isn't that what EM propagation
is, after all?

Yes, after I posted it, I realized that it was a rhetorical question.
--
73, Cecil, w5dxp.com

Dear friends, I follow with interest your interesting digressions,
however in various different posts about differents matters, I notice
discussion arises about what is "real" and what is not. IMO that
contributes to the solution goes away from us (I remember making this
comment in a previous post). In this sense respecto to energy I would
like to quote a great physics:

"...there is a certain quantity, which we call energy, that does not
change in the manifold changes which nature undergoes. That is a most
abstract idea, because it is a mathematical principle; it says that
there is a numerical quantity which does not change when something
happens. It is not a description of a mechanism, or anything concrete;
it is just a strange fact that we can calculate some number and when
we finish watching nature go through her tricks and calculate the
number again, it is the same."

"It is important to realize that in physics today, we have no
knowledge of what
energy is. We do not have a picture that energy comes in little blobs
of a definite
amount. It is not that way. However, there are formulas for
calculating some
numerical quantity, and when we add it all together it gives "28"'—
always the
same number. It is an abstract thing in that it does not tell us the
mechanism or
the reasons for the various formulas."

From: Richard Feynman. "Six easy pieces"

Miguel LU6ETJ

Sorry, I forget to made clear that my comment not reference Cecil
recent post mentioning "real power" in mathematical sense, referencing
complex numbers. :)

Miguel

On 29 jun, 15:08, Cecil Moore wrote:
On Jun 29, 12:54*pm, Jim Lux wrote:

photons can flow through a dielectric.. isn't that what EM propagation
is, after all?

Yes, after I posted it, I realized that it was a rhetorical question.
--
73, Cecil, w5dxp.com

I learnt displacement current inside a condenser it was = eo* d(phi E)/
dt no EM radiation inside the condenser to made that current possible,
in any case EM radiation in physical condenser will come out from
condenser to the rest of the universe :).
I also learnt photons was necessary to explain certain energy
interchange phenomena such as fotoelectric effect or subatomic
particle interactions, wave-particle duality for me means "duality",
not "wave kaput" :) to account for EM wave well explainable
phenomenom.
As it was taught to me (I am not physicist), quantum nature of a 80 m
wavelenght energy it is useless for calculations and invisible to our
instrument resolution because its immensely large quantic number. Is
it wrong?

Miguel LU6ETJ

On Jun 29, 8:41*pm, lu6etj wrote:
I learnt displacement current inside a condenser it was = eo* d(phi E)/
dt no EM radiation inside the condenser to made that current possible,
in any case EM radiation in physical condenser will come out from
condenser to the rest of the universe :).

It depends on your definition of "radiation". In an ideal transmission
line, energy is not lost to radiation but photons (EM fields and
waves) necessarily exist all up and down the line. In an ideal coaxial
transmission line, the photons (EM fields and waves) are confined to
the dielectric. Electrons cannot travel at the speed of light. EM
waves travel at the speed of light. Therefore EM waves are photons.
Given the physical nature of a capacitor, refraction would be the
primary mechanism for losing energy to radiation and there's probably
very, very little refraction in the capacitor dielectric. If electrons
are being acelerated and decelerated in the capacitor, photons will be
emitted. It seems obvious now that when electrons are decelerated on
one capacitor plate, photons are emitted that propogate across the
capacitor dielectric and are absorbed by electrons on the opposite
plate. When the concept of displacement current was invented, nobody
knew that RF fields were actually made up of particles (photons) but
now we do know. Displacement current seems only to be EM radiation
from one capacitor plate to the other.

As it was taught to me (I am not physicist), quantum nature of a 80 m
wavelenght energy it is useless for calculations and invisible to our
instrument resolution because its immensely large quantic number. Is
it wrong?

The quantized nature of a single RF photon is no longer open to
argument and the energy in that single photon can be calculated,
(h*f), where h is Planck's constant, 6.626 x 10^-34 J*s. Whether there
are any instruments sensitive enough to detect a single 80m photon is
a moot point that does not change the nature of RF fields and waves.
What is important is that it is impossible to radiate a signal level
less than (h*f). If anyone asserts that RF fields and waves can
violate the laws of physics regarding photons, that person is wrong
and delusional. (Light left over from the time when the universe
became transparent is today red-shifted down to RF microwave
frequencies and called background radiation.)
--
73, Cecil, w5dxp.com

On Tue, 29 Jun 2010 18:41:52 -0700 (PDT), lu6etj
wrote:

As it was taught to me (I am not physicist), quantum nature of a 80 m
wavelenght energy it is useless for calculations and invisible to our
instrument resolution because its immensely large quantic number. Is
it wrong?

Yes.

We experience 80M activity every day irrespective of it being
Newtonian or Quantum. All it reveals is that something with a very,
very, very low energy is still quite measurable.

However, you "can" deliberately choose the wrong instrument to measure
the energy. That instrument reveals more about the choice-maker than
the energy.

For instance, a 1KW 80M energy source presents a near 0 degree
absolute temperature. A fever thermometer is not going to register
that energy.

73's
Richard Clark, KB7QHC