"Waves of Average Power"
 

Jim Kelley wrote:
Cecil Moore wrote:
OK, the graphic is at http://www.w5dxp.com/thinfilm.gif

It's a decent start, however you must remember that power doesn't
reflect or propagate.

There's no power shown propagating. The quantities
in watts are the irradiance values in joules/sec/unit-area
existing within a unit-area on the surface of the thin-film.

You can't add power algebraically, so you won't
be able to take phase into consideration (sort of crucial if you want to
show cancellation).

Any physicist should be able to solve this problem using
the data given. The reason you cannot solve it is that you
have apparently never understood the irradiance equation.
It is true that one cannot add powers algebraically but
powers (irradiances) are added all the time in optical
physics using the irradiance equation. (Powers do not
superpose. There is special equation governing the
addition of powers.)

You have to use a vector quantity.

That's where you are wrong. One doesn't need to use
vectors. The only thing needed is the phase angle
between the external reflection and the 1st internal
reflection. You can figure out that phase angle by
knowing that the thin film is 1/4 wavelength thick.
If you cannot, I'll just tell you that the phase angle
between the external reflection and the internal
reflections is 180 degrees.

You have all you need to know to answer the question:
What happens to the total reflected power (irradiance)
in the air back toward the source when the first internal
reflection arrives assuming the reflections are coherent
and collinear?

I hope you will not go silent on this thread just because
you don't like the emerging technical facts. If you pursue
the discussion to its logical end, either you or I will
learn something new. I am eager to discover any flaws in
my technical logic.
--
73, Cecil http://www.w5dxp.com

"Waves of Average Power"
 

Cecil Moore wrote:
I am eager to discover any flaws in
my technical logic.

Then, instead of flapping your lips and waving your hands, consider
doing what I suggested.

ac6xg

"Waves of Average Power"
 

Jim Kelley wrote:

Cecil Moore wrote:
I am eager to discover any flaws in
my technical logic.

Then, instead of flapping your lips and waving your hands, consider
doing what I suggested.

I will save us some time and concede that a vector
analysis, which I already know how to do, would work
and yield valid results. You and I would not discover
anything new in the process. We would agree on the
results and not discover any points of disagreement.
If you want to divert the issue in that manner, I
will agree with any vector analysis that you perform
that doesn't contain errors.

Now it's your turn to do a power-density analysis
using the data given. Please use the irradiance
equation to determine the reflected power toward
the source after the external reflection and 1st
internal reflection encounter each other. All the
data that you need for such an analysis is contained
in the example which is the only data that optical
physicists had to work with 100 years ago when they
couldn't directly measure any phase angles.

It is in the area of the irradiance equation that
I might discover any flaws in my technical logic.
If you refuse to use the irradiance equation, I
will understand completely. Everyone else is afraid
of what they will discover.
--
73, Cecil http://www.w5dxp.com

"Waves of Average Power"
 

Cecil Moore wrote:
I will save us some time and concede that a vector
analysis, which I already know how to do, would work
and yield valid results. You and I would not discover
anything new in the process.

I think you might - if you worked it all the way from first transient
through to steady state. On the other hand the irradiance equation
can apparently lead some people to believe things about the nature of
the phenomenon which are not precisely correct.

73, ac6xg

"Waves of Average Power"
 

Jim Kelley wrote:
Cecil Moore wrote:
I will save us some time and concede that a vector
analysis, which I already know how to do, would work
and yield valid results. You and I would not discover
anything new in the process.

I think you might ...

I will agree with everything you say about a vector
analysis (unless you make a mistake).

Please let's proceed to the crux of our argument.
Please apply the same irradiance equation that
optical physicists were using before you were born.
Any physics professor should be able to accomplish
that simple task. If optical physicists were in
error in using that equation 100 years ago, please
explain their error.
--
73, Cecil http://www.w5dxp.com

"Waves of Average Power"
 

In article , Jim Kelley
wrote:

Cecil Moore wrote:

OK, the graphic is at http://www.w5dxp.com/thinfilm.gif

It's a decent start, however you must remember that power doesn't
reflect or propagate.

Hello, Jim , and power (energy per unit time) doesn't propagate? Where do
we get all that radiant heat from 93 million miles away? Sincerely, and
73s from N4GGO,

John Wood (Code 5550) e-mail:
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

"Waves of Average Power"
 

J. B. Wood wrote:
Where do
we get all that radiant heat from 93 million miles away?

When the infrared EM energy encounters the earth,
it is transformed from EM energy to heat energy.
Before that transformation, it is just another
set of EM waves differing only in frequency from
any other EM wave, e.g. RF and light.
--
73, Cecil http://www.w5dxp.com

"Waves of Average Power"
 

On Wed, 07 Nov 2007 06:30:41 -0500, (J. B. Wood)
wrote:

Hello, Jim , and power (energy per unit time) doesn't propagate? Where do
we get all that radiant heat from 93 million miles away? Sincerely, and

Hi John,

Is the volume of space immediately to the left of planet Earth as warm
as planet Earth? A balmy 13° to 17° C?

Same source, the sun.
Same distance, 93 million miles away;
Same power (sic), energy per unit time;
Same propagation, speed of light without obstruction;

Same heat? Not even close - even when accounting for and removing the
core temperature of Earth. A rather absolute chilly of 3° K.

Substitute the Moon by placing it to the left of Earth. Is it as warm
as planet Earth? A balmy 13° to 17° C?

Same source, the sun.
Same distance, 93 million miles away;
Same power (sic), energy per unit time;
Same propagation, speed of light without obstruction;

Same heat? Again, not even close. A rather toasty 107° C!

Two examples in proximity to Earth, and yet both seem to exhibit that
the Sun does not propagate power linearly. What could possibly
account for this error when the path is unobstructed and the Sun's
radiation is uniform? Even more quixotic is that the smaller load of
the Moon exhibits a higher dissipation of radiation (and has virtually
no core temperature effect to elevate that result).

Well, albedo has something to do with all three; but that is load
dependant and has nothing to do with energy per unit time from the
source. Insulation has something to do with all three; and that too
has nothing to do with energy per unit time from the source.

So, is the Sun selectively powerful (like a god)? Is the Sun like
some really, really huge radiant lamp (the kind with a toaster wire
wound like a slinky in the middle of a radar reflector)? Point the
radiant lamp at you, and you get toasty, step aside, and the air that
replaces your volume does not - hmmm. Put a pane of glass in the same
region, same chilling result.

Something going on here and there seems to be only one commonality to
power - the function of the load in the presence of energy.

Energy propagates much as we expect it does; power - well, not always
(hardly ever).

73's
Richard Clark, KB7QHC

"Waves of Average Power"
 

In article , Richard Clark
wrote:

Energy propagates much as we expect it does; power - well, not always
(hardly ever).

73's
Richard Clark, KB7QHC

Hello, and the above statement is simply not true as any undergraduate
textbook in electromagnetics will point out. Transmission lines, for
example, be they 60 Hz or at RF due in fact transmit power from source to
load. At any point along the line the average power is given by 1/2 the
real part of the product of the voltage and complex conjugate of the
current. Voltage and current contain their respective components of
travelling waves in both directions (source-to-load and load-to-source).
Of course the transport medium doesn't have to be a transmission line - it
can be free space, say from a transmitting antenna to a receiving antenna.

I have no idea how many of those in our ham hobby have taken any courses
in electromagnetics (traditionally called "fields" by undergrad EE
students). Such courses are part of an undergrad EE program and if you
major in electrophysics (as I did) at the grad level you delve much deeper
into the subject. Sincerely,

John Wood (Code 5550) e-mail:
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

"Waves of Average Power"
 

On Wed, 07 Nov 2007 11:41:29 -0500, (J. B. Wood)
wrote:

I have no idea how many of those in our ham hobby have taken any courses
in electromagnetics (traditionally called "fields" by undergrad EE
students).

Hi John,

I majored in English. Yet and all that intellectually crippled, I
found you couldn't respond to heat (the expression of power by your
own choice) being the dissipation of energy in a load, and that energy
is the entity that propagates - not power.

As the topic has migrated away from your original thesis of heat and
the sun towards fields; I, as an English major, am at a loss to
discover the correlation that resolves how fields can resurrect the
missing heat in an Earth equivalent volume of space, or provide the
surplus heat found in a smaller volume of the moon - all from the same
source of those fields, the sun.

Are fields selectively powerful (like a god)? Do the fields of
some really, really huge radiant lamp (the kind with a toaster wire
wound like a slinky in the middle of a radar reflector) propagate
differently? Point the radiant lamp at you, and you get toasty, step
aside, and the air that replaces your volume does not - hmmm. Put a
pane of glass in the same region, same chilling result.

The closest mechanism of the transfer of power (heat as you originally
premised) is found in Phonons, not Photons (aka sunlight, again, the
source of your choice). Phonons exist only in physical materials
(loads). Phonons are what is responsible for the flow of heat (does
this metaphor of power movement sound familiar?), In fact, Phonons
are specifically and explicitly power moving in a material. Again,
the commonality of this discussion is found in load specific energy
(photon-phonon) interaction as power and uniquely heat (your terms).

Phonons also exhibit field-like characteristics (there being optical
phonons and acoustic phonons that provide for both transverse and
longitudinal waves). In addition, once a phonon hits the margin of
the material it is migrating through it reflects or transforms.
However, in the transformation heat is left behind and energy moves on
(that is how your thermos works in the face of fields too).

I can provide you with pdf material from "Electron phonon interactions
: a novel semiclassical approach," Rose, Albert, to aid you in this
avenue of power transfer if you care to examine it. Other references
(sans pdf copy in the short term) can be made available too. I can
probably point you to resources in your own lab that amplify along
these lines of phononic interactions (I think I heard one presenter
from down your way talk about aerogels).

73's
Richard Clark, KB7QHC