Where Does the Power Go?
 

Cecil Moore wrote:
Richard Clark wrote:

Gene Fuller wrote:

reversing momentum


Imagine the G forces of that for an infinitesmal 1/10¹² second or so.


The pressure of such a momentum change can be calculated.
From "Optics", by Hecht, 4th edition, page 57:
"When the surface under illumination is *perfectly
reflecting*, the beam that entered with a velocity
of +c will emerge with a velocity of -c. This
corresponds to twice the change in momentum that
occurs on absorption, and hence,"

Pressure of a absorbed wave = S(t)/c

Pressure of a reflected wave = 2*S(t)/c

I'll bet the reason no one can measure the radiation pressure
resulting from your "4th mechanism of reflection" is because it
cancels out with the radiation pressure from the cancelled reflection
in the other direction. Right, Cecil? :-)

73, Jim, AC6XG

Where Does the Power Go?
 

On Tue, 03 Oct 2006 10:47:58 -0700, Jim Kelley
wrote:

Pressure of a reflected wave = 2*S(t)/c

I'll bet the reason no one can measure the radiation pressure
resulting from your "4th mechanism of reflection" is because it
cancels out with the radiation pressure from the cancelled reflection
in the other direction. Right, Cecil? :-)

Hi Jim,

Who needs reasoning when a Xeroxed formula, like a "4th mechanism of
reflection," is simpler to cut and paste than actually offer a
practical answer to? This is akin to his sophomoric allusion to
anti-glare coating yet again (yawn), he couldn't work the math on that
one anywhere closer than pi = 22/7.

Soon we will be down the path of 66% allowable error to prove a
concept.

73's
Richard Clark, KB7QHC

Where Does the Power Go?
 

Jim Kelley wrote:
I'll bet the reason no one can measure the radiation pressure resulting
from your "4th mechanism of reflection" is because it cancels out with
the radiation pressure from the cancelled reflection in the other
direction. Right, Cecil? :-)

There is certainly radiation pressure pushing outward
from a 1/4WL thin-film non-reflecting surface even
though the reflected waves cancel each other.
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

On 2 Oct 2006 11:11:40 -0700, "Denny" wrote:

Assume that we put a 1 second squirt at the rate of 100 Joules per
second, into the input end of the line, it is an open line at the far
end, and yank the input end of the line out of the transmitter at
exactly 1 second... Now we have a lossless line with a wave going
forward and a reflected wave coming back... Do you still claim 200
joules?

Hi Denny,

A question of my own:
Did you really expect your question above wouldn't be dodged?

I suppose not. Anyway, your observations revealed Cecil's usual lack
of rigor. You certainly know how to step back while he juggles
un-pinned hand grenades.

73's
Richard Clark, KB7QHC

Where Does the Power Go?
 

Richard Clark wrote:
Who needs reasoning when a Xeroxed formula, like a "4th mechanism of
reflection," is simpler to cut and paste than actually offer a
practical answer to? This is akin to his sophomoric allusion to
anti-glare coating yet again (yawn), he couldn't work the math on that
one anywhere closer than pi = 22/7.

Richard, I wasn't the one who, through superposition of powers,
came up with an irradiance brighter than the surface of the sun
at the non-reflective surface interface. Did you ever figure
out that superposing powers is a no-no?
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

Cecil Moore wrote:

There is certainly radiation pressure pushing outward
from a 1/4WL thin-film non-reflecting surface even
though the reflected waves cancel each other.

Cecil,

Radiation pressure without waves?

Isn't that the sort of thing that started the Boston Tea Party?

8-)

73,
Gene
W4SZ

Where Does the Power Go?
 

OK, so we've established you don't know the G forces for changed
momentum, only how to sniff toner at the Xerox.

On Tue, 03 Oct 2006 18:34:17 GMT, Cecil Moore
wrote:

Richard, I wasn't the one who, through superposition of powers,
Yes, you did have a problem translating power to energy and back. I
could offer any number of common scenarios that would have you gasping
for air:
There is a common bare light bulb 1 meter away;
it illuminates a cm² target with 3µW @ 55nM of POWER;

what is:
the number of candela per steradians,
at the target,
from total bandwidth radiation?
or:
How much power is being supplied to the bulb?

came up with an irradiance brighter than the surface of the sun
at the non-reflective surface interface.

No, true to your form, you rounded errors and fudged numbers to prove
light was black.

In that regard I will offer you a third choice question from above:
Can you see this amount of light on the target?
(choose this one, you might guess it right - it doesn't demand any
math skill.)

:-0

Where Does the Power Go?
 

Gene Fuller wrote:
Cecil Moore wrote:
There is certainly radiation pressure pushing outward
from a 1/4WL thin-film non-reflecting surface even
though the reflected waves cancel each other.

Radiation pressure without waves?

Please don't deliberately obfuscate things, Gene.

There are no reflected waves on the outside of the
thin-film and therefore no reflected wave pressure
from the outside.

All the reflected waves are on the inside of the
thin-film outer plane. That's why the reflected wave
pressure is pushing outwards.

The pressure is where the waves are. There's no
wave pressure without waves.
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

wrote:
Do you mean the mashed-potatoes energy equations where
the wave energy components are completely ignored?

Calling the steady state solution "mashed-potatoes" is ridiculous.

I'm not calling the solutions ridiculous. I am calling some
of the conceptual conclusions ridiculous. For instance, since
the *net* Poynting vector equals delivered source watts, there
are zero watts in the reflected wave even though there are
joules in the reflected wave moving at the speed of light.

If you want to know what happens during the transient, you MUST start
from the original differential equation that describes how the fields
were set up in the line in the first place, but it doesn't matter in
the steady state. The answer is the same.

Not according to some of the gurus posting here. The joules
supplied during the initial transient state are being swept
under the steady-state rug. Witness "Food For Thought #1"
on www.eznec.com

I don't want to have to do it but one of these days I think it might be
necessary.

What you will find is that there is exactly the energy in the
transmission line required to support the real speed-of-light
forward traveling wave and the real speed-of-light reflected
traveling wave - no more and no less. The argument that there
is no more energy in a transmission line with reflections than
is being supplied by the source is simply false.

The Poynting vector in a mismatched line has been worked out by me,
here, and I'd appreciate comments on the MATH.

http://en.wikipedia.org/wiki/User:Dan_Zimmerman/Sandbox

What about the forward Poynting vector, Pz+, and the reflected
Poynting vector, Pz-. Reference page 291, "Fields and Waves ...",
Ramo and Whinnery, 2nd edition?

My claim is that that's it. That's all. No handwaving, no setting up
steady state startup and claiming that that energy remains in the line.
The real part of the Poynting vector represents real power flux and
the imaginary part represents reactive power flux. To get the energy
that exists in the line you just add it all up...

The forward wave's average Poynting vector is real. The reflected
wave's average Poynting vector is real. Adding two real power flow
vectors is a lot easier than integrating a power flow vector with
real and imaginary parts. And, bottom line, one obtains exactly
the same results with 10% of the work. Subtract the reverse
Poynting vector from the forward Poynting vector to determine
the net power. Add the two Poynting vectors and multiply by the
length in seconds of the feedline to determine the total energy.
What could be simpler?

Am I wrong, Cecil et. al?

No, you are correct but you are going around the world to get
from California to New York. My method, supported by Ramo and
Whinnery, can be done in much less time and can be more easily
understood by people not familiar with your method. By the time
you get out your calculator, I can have your answer waiting for
you and it will be the same answer you get after wasting a lot
of time.

When the SWR is infinite, there is no energy in the line associated
with flowing power. None at all. The reflection coefficient is 1, and
all energy in the line is associated with circulating power (reactance
only, as you would expect from a stub)

You are, of course, talking about *net* "flowing" power. But your
answer is exactly the same as component forward power in real
joules/sec plus component reflected power in real joules/sec.
And thinking in terms of those real component powers is a lot
easier than your imaginary stuff.

Pz+ - Pz- = Pz-tot sourced and delivered

(Pz+ + Pz-)* feedline length in seconds = total feedline energy

There is also a physics problem with the imaginary concept. EM
wave energy must necessarily move at the speed of light. A small
amount of TV modulation will prove that those forward and reflected
waves are still moving end to end at the speed of light and still
transferring information.
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

Cecil Moore wrote:
Gene Fuller wrote:
Cecil Moore wrote:
There is certainly radiation pressure pushing outward
from a 1/4WL thin-film non-reflecting surface even
though the reflected waves cancel each other.

Radiation pressure without waves?

Please don't deliberately obfuscate things, Gene.

There are no reflected waves on the outside of the
thin-film and therefore no reflected wave pressure
from the outside.

All the reflected waves are on the inside of the
thin-film outer plane. That's why the reflected wave
pressure is pushing outwards.

The pressure is where the waves are. There's no
wave pressure without waves.

Cecil,

Sorry, I guess my closing was not clear. Try this.

8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-)
8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-)

73,
Gene
W4SZ