Gene Fuller wrote:
It is easy to give examples where the waves survive the superposition,
because they always do. It is rather strange that you are making this
argument after all the back and forth about traveling waves and standing
waves. Do we now have multiple flavors of EM waves? Some that obey
superposition and some that don't?

They all obey superposition which can occur with or
without interference. And you are wrong about all
waves surviving superposition. Canceled waves do
not survive wave cancellation in the direction that
they are traveling. Access this web page and set
the two waves to equal frequencies, equal magnitudes,
and opposite phases, i.e. 0 and 180 degrees.

http://micro.magnet.fsu.edu/primer/j...ons/index.html

When you do that, the waves are canceled in their
original direction of travel. The energy in those
canceled waves certainly survives, but those two
original waves cease to exist never to be seen
again.

I must have missed class the day they went over the theory of
"cancellation".

You must have. Please run the above java application
and alleviate your ignorance about what you missed. Why
do the waves disappear when they are of equal magnitude
and opposite phase?

I stand 100% behind my two messages to Walt. If you actually read them
you would note that I said for most cases it makes no difference whether
the waves interfere forever or whether they interact and "cancel".

Of course it makes all the difference in the world. That's
what the entire argument is all about. You simply cannot
sweep the truth under the "does not matter" rug. And until
you can say "all cases" instead of "most cases" your
argument is irrelevant. If it doesn't work for all, it
doesn't work at all.

The bottom line is that EM waves do not interact in free space.

It is indeed difficult to get two beams of light collinear
in space space. But it is not difficult at all to get two
RF waves collinear in a transmission line. It happens every
time someone adjusts his antenna tuner for a Z0-match.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
Do Born and Wolf offer crisp definitions of the boundaries
between coherent, partially coherent, and mutually incoherent?
Or is it a continuum arbitrarily divided into 3 regions for
the purposes of discussion?

The IEEE Dictionary has an interesting definition for
"degree of coherence". Imax means intensity maximum
and Imin means intensity minimum.

Visibility = (Imax-Imin)/(Imax+Imin)

Light is considered "highly coherent" when Visibility
exceeds 0.88 and "partially coherent" when Visibility
is less than 0.88. Incoherent for "very small values"
of Visibility.
--
73, Cecil http://www.w5dxp.com

On Apr 8, 7:09 pm, Cecil Moore wrote:
Keith Dysart wrote:
Do Born and Wolf offer crisp definitions of the boundaries
between coherent, partially coherent, and mutually incoherent?
Or is it a continuum arbitrarily divided into 3 regions for
the purposes of discussion?

The IEEE Dictionary has an interesting definition for
"degree of coherence". Imax means intensity maximum
and Imin means intensity minimum.

Visibility = (Imax-Imin)/(Imax+Imin)

Light is considered "highly coherent" when Visibility
exceeds 0.88 and "partially coherent" when Visibility
is less than 0.88. Incoherent for "very small values"
of Visibility.

This is good; a continuum with high coherence at one end, low
coherence at the other and medium in the middle, and, of course,
since the ends are infinitely small, no such thing as perfect
coherence or "NO" coherence (at least in the real world).

This then takes us back to the original point; there is always
some interference, though it may be small enough that an
engineer does not find it of interest.

....Keith

Cecil Moore wrote:
Gene Fuller wrote:
It is easy to give examples where the waves survive the superposition,
because they always do. It is rather strange that you are making this
argument after all the back and forth about traveling waves and
standing waves. Do we now have multiple flavors of EM waves? Some that
obey superposition and some that don't?

They all obey superposition which can occur with or
without interference. And you are wrong about all
waves surviving superposition. Canceled waves do
not survive wave cancellation in the direction that
they are traveling. Access this web page and set
the two waves to equal frequencies, equal magnitudes,
and opposite phases, i.e. 0 and 180 degrees.

http://micro.magnet.fsu.edu/primer/j...ons/index.html


When you do that, the waves are canceled in their
original direction of travel. The energy in those
canceled waves certainly survives, but those two
original waves cease to exist never to be seen
again.

I must have missed class the day they went over the theory of
"cancellation".

You must have. Please run the above java application
and alleviate your ignorance about what you missed. Why
do the waves disappear when they are of equal magnitude
and opposite phase?


[snip]


Cecil,

That's really funny. A grad student and a programmer put together a
simply java applet to try to illustrate the concept of interference, and
you treat it as a new bible. I bet the authors would be appalled by your
interpretation.

By the way, did you look beyond the pretty pictures and read the section
where the authors said,

"All of the wave examples presented in Figure 1 portray waves
propagating in the same direction, but in many cases, light waves
traveling in different directions can briefly meet and undergo
interference. After the waves have passed each other, however, they will
resume their original course, having the same amplitude, wavelength, and
phase that they had before meeting."


Hmmm, I think that is exactly what I said in this thread on RRAA.



73,
Gene
W4SZ

Keith Dysart wrote:
This is good; a continuum with high coherence at one end, low
coherence at the other and medium in the middle, and, of course,
since the ends are infinitely small, no such thing as perfect
coherence or "NO" coherence (at least in the real world).

But remember that definition is for fiber optics
sources, not amateur radio sources. Coherency
in amateur radio systems can get as close as
a zero reading on a reflected power meter.

Still there are those nagging assertions of Born
and Wolf that for two equal magnitude signals,
the total intensity possible for incoherent
signals is double the intensity of one signal.
The total intensity possible for coherent
signals is four times the intensity of one
signal.
--
73, Cecil http://www.w5dxp.com

Gene Fuller wrote:
That's really funny. A grad student and a programmer put together a
simply java applet to try to illustrate the concept of interference, and
you treat it as a new bible. I bet the authors would be appalled by your
interpretation.

One more example of an ignorant person making fun of something
he doesn't understand. One of those signals is s11(a1). The
other is s12(a2). Added together they equal zero. That's the
S-Parameter equation for reflections toward the source.

b1 = s11(a1) + s12(a2) = 0

If s11, a1, s12, and a2 are all not zero, the above equation
describes wave cancellation, something you say never happens.

By the way, did you look beyond the pretty pictures and read the section
where the authors said,

"All of the wave examples presented in Figure 1 portray waves
propagating in the same direction, but in many cases, light waves
traveling in different directions can briefly meet and undergo
interference. After the waves have passed each other, however, they will
resume their original course, having the same amplitude, wavelength, and
phase that they had before meeting."

Yes, that happens "in many cases" but NOT IN ALL CASES. You
apparently missed the point which is the part where they said:

"... when two waves of equal amplitude and wavelength that are
180-degrees ... out of phase with each other meet, they are not
actually annihilated, ... All of the photon energy present in
these waves must somehow be recovered or redistributed in a
new direction, according to the law of energy conservation ...
Instead, upon meeting, the photons are redistributed to regions
that permit constructive interference, so the effect should be
considered as a redistribution of light waves and photon energy
rather than the spontaneous construction or destruction of light."

Hmmm, I think that is exactly what I said in this thread on RRAA.

No, what you have said on RRAA is that wave cancellation never
happens because wave cancellation doesn't occur in many
cases. That is obviously faulty logic and all it takes to prove
you wrong is one case of wave cancellation. That case happens
every time a ham adjusts his antenna tuner for zero reflected
power. If we consider the java ap as the reflected waves flowing
toward the source, setting them to 0 and 180 degrees is exactly
what happens at the antenna tuner.
--
73, Cecil http://www.w5dxp.com

On Mon, 09 Apr 2007 02:45:52 GMT, Cecil Moore
wrote:

All of the photon energy present in
these waves must somehow be recovered or redistributed in a
new direction, according to the law of energy conservation

New heights of sheer stupidity. This sounds like bingo night in the
church basement where no one actually loses any money, it just gets
shuffled around.

"Recovered OR redistributed" ... this certainly qualifies for next
year's Oscar for chuckles. Luckily the Nobel committee doesn't follow
Hollywood or they would have awarded Mighty Mouse the Physics award
for antigravity (certainly 50 million children's admiration couldn't
lead them astray on this choice!).

I'd vowed that I wouldn't hit this tarbaby yet again. But here I go.

Among the junk science being bandied about here is the following
supposition:

Suppose you have beams from two identical coherent lasers which, by a
system of (presumably partially reflective and partially transmissive)
mirrors, are made to shine in exactly the same direction from the same
point (which I'll call the "summing point"). Further, suppose that the
paths from the two lasers to this summing point differ by an odd number
of half wavelengths. So beyond the summing point, where the laser beams
exactly overlie each other, there is no beam because the two exactly
cancel. Or, in other words, the sum of the two superposed fields is
zero. The recurring argument is that because each laser is producing
energy and yet there is no net field and therefore no energy in the
summed beams, something strange has happened at the summing point (or
"virtual short circuit"), and creative explanations are necessary to
account for the "missing energy". One such proposed explanation is that
the mere meeting of the two beams is the cause of some kind of a
reflection of energy, and that each wave somehow detects and interacts
with the other.

Well, here's what I think. I think that no one will be able to draw a
diagram of such a summing system which doesn't also produce, due solely
to the reflection and transmission of the mirrors, a beam or beams
containing exactly the amount of energy "missing" from the summed beam.
No interaction(*) of the two beams at or beyond the summing point is
necessary to account for the "missing" energy -- you'll find it all at
other places in the system. Just as you do in a phased antenna array,
where the regions of cancelled field are always accompanied by
complementary regions of reinforced field. Somewhere, in some bounce
from a mirror or pass through it, the beams will end up reinforcing each
other is some other direction. My challenge is this: Sketch a system
which will produce this summation of out-of-phase beams, showing the
reflectivity and transmissivity of each mirror, and showing the beams
and their phases going in all directions from the interactions from each
mirror. Then show that simple interaction of the beams with the mirrors
is insufficient to account for the final distribution of energy.

Next, do the same for a transmission line. Show how two coherent
traveling waves can be produced which will propagate together in the
same direction but out of phase with each other, resulting in a net zero
field at all points beyond some summing point. But also calculate the
field from waves reflected at the summing point and elsewhere in the
system due to simple impedance changes. Show that this simple analysis,
assuming no interaction between the traveling waves, is insufficient to
account for all the energy. A single case will do.

Until someone is able to do this, I'll stand firm with the unanimous
findings of countless mathematical and practical analyses which show
superposition of and no interaction between waves or fields in a linear
medium.

(*) By "interaction" I mean that one beam or wave has an effect on the
other, altering it in some way -- for example, causing it to change
amplitude, phase, orientation, or direction. I'm not including
superposition, that is the fact that the net field of the two waves is
the sum of the two, in the meaning of "interaction".

Roy Lewallen, W7EL

Richard Clark wrote:
Cecil Moore wrote:
All of the photon energy present in
these waves must somehow be recovered or redistributed in a
new direction, according to the law of energy conservation

New heights of sheer stupidity. This sounds like bingo night in the
church basement where no one actually loses any money, it just gets
shuffled around.

One more example of an ignorant person making fun of
things he doesn't understand. The principle of the
conservation of energy indeed states that energy is
not gained or lost - it just gets shuffled around.
When EM wave cancellation occurs, the energy remains
in EM wave form and simply gets redistributed.

As we tune our antenna tuners while watching the
reflected power indication, we are varying the
magnitude and phase of the two component reflected
waves at the tuner input. When those two component
waves are adjusted to equal magnitudes and opposite
phase, reflections toward the source are obviously
canceled since they can be measured as going to zero
in the direction of the source. The following S-Parameter
equation describes the final outcome toward the source.

b1 = s11(a1) + s12(a2) = 0, total destructive interference

Since energy cannot be destroyed by an antenna tuner
and since there are only two directions available, in
a lossless system, the energy in waves canceled toward
the source must be redistributed (re-reflected) to waves
traveling toward the load as constructive interference.
That's the other S-Parameter equation.

b2 = s21(a1) + s22(a2) = Vfor/SQRT(Z0)

Square those equations and you get the power equations.
|b1|^2 = net reflected power = 0
|b2|^2 = net forward power = all the available power
--
73, Cecil http://www.w5dxp.com

"Cecil Moore"
Still there are those nagging assertions of Born
and Wolf that for two equal magnitude signals,
the total intensity possible for incoherent
signals is double the intensity of one signal.
The total intensity possible for coherent
signals is four times the intensity of one signal.
________

It is a fairly common practice in broadcast designs to combine the outputs
of two r-f amplifiers of equal power rating, using a 4-port, 3 dB coaxial
hybrid. The two amplifiers are driven by a single exciter through a
suitable splitter. The antenna connects to one output port of the hybrid,
and the other output port is connected to a dummy load.

When the relative r-f phases of the two txs are suitably set, the antenna
connection of the hybrid receives the total output power of the two txs, and
the dummy load port receives zero. When the relative r-f phases of the txs
are changed by 90 degrees from that setting, then the conditions at the
output ports are reversed.

The total average power available at the hybrid output for both of these
conditions is twice that of a single tx without the hybrid.

Does the quote from Born and Wolf support this?

RF