Antonio Vernucci wrote:
But if both waves are sumultaneouly present, the power carried by each
wave when alone is no longer a meaningful number.

Why is the ExB Poynting vector of each wave no longer
proportional to the energy content? Why does the energy
content of the component waves have to change when they
superpose? Where does that energy change go? Do the
necessary joules disappear and/or appear from thin air?

As a matter of fact
when superposing two coherent waves (same frequency, fixed phase
relationship), one MUST first sum voltages (or currents) and then
calculate power.

That's what I did and the result was 171 joules/sec.
The Poynting vector for each of the two source waves
is 50 joules/sec. Why is the energy content of the
component waves not a meaningful number?

Summing wave powers could only be done in case of incoherent waves.

No, there is a special equation to be used for summing coherent
waves, i.e. the irradiance equation from optical physics. For
power density:

Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)

where 'A' is the angle between the two E-fields.

In conclusion, the answer to your question is that the apparent extra 71
joules/s come front the fact that 100 joules/s taken as reference is a
number having no physical meaning.

For every second that passes, 50 + 50 = 100 joules has no
physical meaning? Are you saying that an EM wave is not
associated with ExB joules/sec?
--
73, Cecil http://www.w5dxp.com

Why is the ExB Poynting vector of each wave no longer
proportional to the energy content? Why does the energy
content of the component waves have to change when they
superpose? Where does that energy change go? Do the
necessary joules disappear and/or appear from thin air?

Joules do not disappear, they just get distributed over the free space in a
non-uniform manner.

In certain regions of the space the two waves add up (apparently creating extra
power), in other regions they cancel out (apparently destroying power). The
integral of total radiated power does not change.

In your example you considered a location where the two waves have a 45 deg.
shift. At another location, where the two waves have a zero deg. shift, you
would observe an even higher apparent power creation. Conversely, at locations
where the two waves have a 180 deg. shift you would observe absence of power.

The principle causing the apparent power creation at your location is the same
principle by which an antenna formed by two stacked dipoles features a gain of
up to 3 dB with respect to a single dipole, and can then deliver up to twice the
power to a receiver placed at the maximum radiation heading (and zero power at a
receiver placed at 90 degrees from that heading). .

That's what I did and the result was 171 joules/sec.
The Poynting vector for each of the two source waves
is 50 joules/sec. Why is the energy content of the
component waves not a meaningful number?

Each wave produces 50 joules/s when alone. When the two waves are superimposed,
each wave produces not only its 50 joules/s but also 35.5 more joules that it
"robs" from other regions of the space. If you would plainly sum the power of
two components (i.e. 50 + 50), you would neglect the fact that coherent waves
necessarily interfere with each other in the space, in constructive or
destructive manner depending on the receiver location.

Summing wave powers could only be done in case of incoherent waves.

No, there is a special equation to be used for summing coherent
waves, i.e. the irradiance equation from optical physics. For
power density:

Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)

where 'A' is the angle between the two E-fields.

Please re-read my sentence more carefully. My statement was that summing powers
(that is. Ptotal = P1 + P2) would only be correct for incoherent waves.

For coherent waves, plainly summing powers would generally be incorrect (apart
from one particular phase angle), and one must nstead use the equation you have
shown.

For every second that passes, 50 + 50 = 100 joules has no
physical meaning? Are you saying that an EM wave is not
associated with ExB joules/sec?

see previous remarks.

73

Tony I0JX

Antonio Vernucci wrote:
Each wave produces 50 joules/s when alone. When the two waves are
superimposed, each wave produces not only its 50 joules/s but also 35.5
more joules that it "robs" from other regions of the space.

My point exactly! The same thing is true when it happens
in a transmission line at a Z0-match point. The region of
constructive interference toward the load (forward energy
wave) "robs" energy from the region of destructive
interference toward the source (reflected energy waves).
That's how antenna tuners work.
--
73, Cecil http://www.w5dxp.com

"Cecil Moore" wrote in message
news
Antonio Vernucci wrote:
Each wave produces 50 joules/s when alone. When the two waves are
superimposed, each wave produces not only its 50 joules/s but also 35.5
more joules that it "robs" from other regions of the space.

My point exactly! The same thing is true when it happens
in a transmission line at a Z0-match point. The region of
constructive interference toward the load (forward energy
wave) "robs" energy from the region of destructive
interference toward the source (reflected energy waves).
That's how antenna tuners work.
--
73, Cecil http://www.w5dxp.com

You can come up with a lot simpler example that at first might look like a
paradox. Consider two DC current sources of 1 amp each. Each current source
will deliver 50 W to a 50 Ohm resistor. Now connect the two current sources
in parallel, and the resultant 2 amps will deliver 200W to the same 50 Ohm
resistor. There is nothing wrong here.

Tam

Tam/WB2TT wrote:
You can come up with a lot simpler example that at first might look like a
paradox. Consider two DC current sources of 1 amp each. Each current source
will deliver 50 W to a 50 Ohm resistor. Now connect the two current sources
in parallel, and the resultant 2 amps will deliver 200W to the same 50 Ohm
resistor. There is nothing wrong here.

Of course, the sources simply deliver the extra power.
But in our previous examples, the source is nowhere around
the point of interference and therefore cannot be the source
of the extra power. Away from any source, the energy required
by constructive interference *always* comes from destructive
interference somewhere else.

In a Z0-matched transmission line with reflections, the
constructive interference is toward the load at the Z0-match
point and the destructive interference is toward the source
at the Z0-match point.

An understanding of the constructive and destructive
interference at a Z0-match point is a necessary and
sufficient condition for understanding where the
reflected energy goes which is the whole purpose
of this thread.
--
73, Cecil http://www.w5dxp.com

"Tam/WB2TT" wrote in message
. ..

"Cecil Moore" wrote in message
news
Antonio Vernucci wrote:
Each wave produces 50 joules/s when alone. When the two waves are
superimposed, each wave produces not only its 50 joules/s but also 35.5
more joules that it "robs" from other regions of the space.

My point exactly! The same thing is true when it happens
in a transmission line at a Z0-match point. The region of
constructive interference toward the load (forward energy
wave) "robs" energy from the region of destructive
interference toward the source (reflected energy waves).
That's how antenna tuners work.
--
73, Cecil http://www.w5dxp.com

You can come up with a lot simpler example that at first might look like a
paradox. Consider two DC current sources of 1 amp each. Each current
source will deliver 50 W to a 50 Ohm resistor. Now connect the two current
sources in parallel, and the resultant 2 amps will deliver 200W to the
same 50 Ohm resistor. There is nothing wrong here.

Tam

Wouldn't you have to double the voltage to get the 1 amp from each source to
flow ?

On Sat, 17 Nov 2007 12:02:26 -0500, "Ralph Mowery"
wrote:

You can come up with a lot simpler example that at first might look like a
paradox. Consider two DC current sources of 1 amp each. Each current
source will deliver 50 W to a 50 Ohm resistor. Now connect the two current
sources in parallel, and the resultant 2 amps will deliver 200W to the
same 50 Ohm resistor. There is nothing wrong here.

Tam

Wouldn't you have to double the voltage to get the 1 amp from each source to
flow ?

Hi Ralph,

That is one of the usual properties of a current source.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Sat, 17 Nov 2007 12:02:26 -0500, "Ralph Mowery"
wrote:

You can come up with a lot simpler example that at first might look like
a
paradox. Consider two DC current sources of 1 amp each. Each current
source will deliver 50 W to a 50 Ohm resistor. Now connect the two
current
sources in parallel, and the resultant 2 amps will deliver 200W to the
same 50 Ohm resistor. There is nothing wrong here.

Tam

Wouldn't you have to double the voltage to get the 1 amp from each source
to
flow ?

Hi Ralph,

That is one of the usual properties of a current source.

73's
Richard Clark, KB7QHC

If you want to drive 1 amp into a 100K resistor, make sure the current
source can develop 100,000 Volts!

Tam

Tam/WB2TT wrote:

You can come up with a lot simpler example that at first might look like a
paradox. Consider two DC current sources of 1 amp each. Each current source
will deliver 50 W to a 50 Ohm resistor. Now connect the two current sources
in parallel, and the resultant 2 amps will deliver 200W to the same 50 Ohm
resistor. There is nothing wrong here.

I see Cecil is still superposing his waves of average power.

I have an example that's more fun yet.

Take a 10 volt source and connect it through a 10 ohm resistor to
another 10 volt source. The positive terminals of the sources are
connected to the ends of the resistor, and the negative terminals are
connected together -- "grounded", if you prefer. Turn on one source,
leaving the other off. (An "off" voltage source is a short circuit.)
Result: 10 watts of dissipation in the resistor. Turn off the first
source and turn on the second. Result: 10 watts of dissipation in the
resistor. Now turn both sources on. Result: An exercise for the reader.

This is a linear circuit for which superposition holds.

Roy Lewallen, W7EL

Roy Lewallen wrote:
I see Cecil is still superposing his waves of average power.

That's at least unfair and at most unethical, Roy. I am not
superposing power. I am using the accepted irradiance equations
from optical physics to predict the energy result of superposing
EM waves, something that was being done by physicists before you
were born.

I have an example that's more fun yet.
Take a 10 volt source ...

It may be more fun but irrelevant. Please explain how the sun
can adjust its energy output depending upon what might or might
not be happening on earth. If you can do that, I will retract
everything I have said about this subject.
--
73, Cecil http://www.w5dxp.com