what happens to reflected energy ?
 

On Jun 28, 7:20 pm, K1TTT wrote:
On Jun 28, 10:50 pm, Keith Dysart wrote:
On Jun 28, 5:11 pm, K1TTT wrote:

On Jun 27, 11:59 pm, Keith Dysart wrote:
I suppose, but then you have to give up on P(t)=V(t)*I(t), generally
considered to be a rather fundamental equation.

very fundamental, and very restricted. only good for one point in
space at one time, and for one pair of voltage and current
measurements... can not be applied to separate waves that are
superimposed, only to the final total voltage and current at the
measurement point at that instant.

True, but others reject it completely.

In your example, the RF energy does seem to disappear and re-appear,
when tracked on a moment by moment basis.

when doing conservation of energy you must include the WHOLE system!
it doesn't work on one section of a transmission line any more than it
works for the infamous undergraduate teaser:

take a refrigerator, put it in a perfectly insulated room, and then
open the doors... what happens to the temperature in the room?

The teaser is amusing, but hardly relevant. In my example, all of
the energy is tracked. Or, I invite you to point out that which was
overlooked. Cecil has not found any and would rather prattle on about
the difference between energy and power than actually understand.

of course its relevant... so what happens to the temperature?

Any energy that might raise the temperature is accounted for in the
source and load resistors, so I still suggest that no energy sources
or sinks have been missed.

Well, it would help if you could actually find and articulate a flaw
inhttp://sites.google.com/site/keithdysart/radio6.

...Keith

that site is rather worthless... you say Vs can be used to get the
time reference for the other signals, but time is a variable, as is
space. you seem to have a snapshot of a bunch of sine waves on an
angular scale, but is that scale time or distance?

Time, of course. I agree, though, it is not as clear as it could
have been. It helps a bit if you look at Cecil’s schematic.

Still, it is complicated and will probably take some effort to
understand.

It would probably be better to start with the step wave example
offered previously in another post and copied below for
convenience:

example
I am not sure where you think there is an error. Perhaps you can
point them out in the following example:
Generator:
- 100V step in to an open circuit
- 50 ohm source impedance
Line:
- 50 ohm
- open circuit
Generator is commanded to produce a step.
This will produce 50 V and 1 A at the line input which will
propagate down the line.
The open end of the line has a reflection co-efficient of 1.0.
Just before the 50 V step reaches the end of the line, the
whole line will be at 50 V and 1 A will be flowing everywhere.
The 50 V step hits the end and is reflected, producing a 50 V
step (on top of the 50V already there) which propagates back
to the generator. In front of the 50 V step, the current is
still 1 A (which provides the charge necessary to produce
the reverse propagating 50 V step. Behind the step, the
current is 0.
When the reverse 50 V step (which is actually a step from
50V to 100V) reaches the generator, the source impedance
matches the line impedance so there is no further reflection.
The line state is now 100V and 0A all along its length.
The settling time was one round-trip.
The generator is still producing the step, so the forward
step voltage wave is still 'flowing' and being reflected so
there is still a reflected step voltage wave, each of 50 V.
Since the generator open circuit voltage is 100 V and the
line voltage is now 100 V, current is no longer flowing
from the generator to the line.
Does this agree with your understanding?
/example

...Keith

this is different than what you claimed before. on june 17th you
claimed:
If(t) = 50/50 = 1a
Ir(t) = 50/50 = 1a
now you say current is no longer flowing.

i will object to you saying the voltage 'wave' is still flowing. The
line is at a constant 100v, there is no current, there can be no em
wave without current AND voltage, therefore voltage can not be
flowing.

I think I detect a bit of a problem originating with definitions.

Start by considering that old stalwart from the text books: the
infinite transmission line. For convenience, the characteristic
impedance is 50 ohms and at the left end is attached a generator
with a 50 ohm source impedance. The transmission line extends
to the right forever, and this time the generator produces a
sinusoid.

Turn on the generator and it starts to produce a sinusoidal wave.
For greater certainty in the description let us say that the
function which describes the wave is:
V(t) = 100 cos(wt)
where w is the ascii excuse for omega, the angular frequency.

Because the line is accepting this signal and transporting it
away from the generator, the voltage distribution on the line
is also a sinusoid, but spatially distributed along the line.

We can visualise this spatially distributed sine wave as
moving along the line and thus we have the term travelling wave.
If we run fast enough along the line we can get to a point
where the signal has not yet arrived and the line is at 0V,
and if we wait, the sine wave will eventually arrive and
we will observe the same sinusoidal spatially distributed
voltage pattern moving along the line.

We can run the experiment with the generator producing other
functions of time and we would observe the same function
spatially distributed along the line and moving.

A generator producing a square wave will produce a square
wave voltage distribution moving along the line.

A generator producing pulses will result in pulses moving
along the line.

A voice signal will result in a complicated voltage pattern
moving along the line.

This holds for any wave, V(t), produced by the generator.

But most lines are not infinite. What happens when the line
is terminated? Most texts will explain that if a section
of line is terminated in its characteristic impedance, that
section will behave as if it was connected to a line that
continued forever. That is, the observations on that
section will be no different than if the line went on and
on and on.

Thus in that section we would observe the moving sine,
square, pulse, or whatever wave, but, we can no longer
run far down the line to find a place where the wave
has not yet arrived, since the line has been truncated.

Now what happens if our section of line is open-circuited?
The wave reflection model allows us to compute that the
reflection coefficient (rc) will be 1. This means that if
a forward travelling voltage wave, Vf(t), encounters the
open circuit, it will be reflected as a reverse wave,
Vr(t).

Mathematically
Vr(t) = rc * Vf(t)

Going back to our waveforms, this is most readily
visualised with pulses where the repetition rate is low
enough that only one pulse is on the line at a time.
The pulse can be seen to travel down the line, hit the
end and travel in the reverse direction back to the
generator. What happens when the pulse gets back to
the generator? Well the generator was specified to
have a source impedance that was the same as the line
characteristic impedance, so the behaviour on the
section of line will be exactly the same as if the
line continued forever. There is no reflection, and
we can forget about the pulse.

The wave reflection model also tells us that the voltage
at any point on the line will be
V(t) = Vf(t) + Vr(t)
where Vf(t) is the voltage from the forward wave at that
point on the line and Vr(t) is the voltage from the
reverse wave at that point on the line.

For pulses, where there is only one pulse on the line
at a time, this is easy to visualise, but for continuous
signals, the voltage distribution on the line is now
the sum of the forward and reverse travelling waves and
for the most part, the forward and reverse travelling
waves are no longer easily visible. For example, with
a sine wave, the resulting voltage distribution on the
line is the time varying, so called, standing wave.

It is important to note that even though the original
travelling waves are no longer visible in the spatial
distribution of the voltage on the line, the travelling
wave continues to be provided by the generator, it
travels to the end of the line where it is reflected and
this reverse wave travels back to the generator where it
behaves as if the line continued forever because the
line is terminated by its characteristic impedance.

Recall that all of this holds for any Vf(t) produced by
the generator. So what happens with that original
sticking point, the step function?

Consider the infinite line again. The step is moving
along the line. If we run far enough ahead, we can
watch the step pass, but behind the step the voltage
is constant. This makes it hard to visualise as a
moving wave because there are no peaks and valleys
to observe, but it is still moving. If in doubt,
just run further down the line to observe the step
pass again.

Now if we terminate the line, we can no longer run
on ahead to find the step, but the wave is still
moving; we just can’t see it because of the constant
nature of the wave after the step.

With the open circuited line, the step function
is reflected and travels as a reverse wave back to
the generator. Once the step reaches the generator,
just as happens for any other V(t) wave, the wave
continues to move; it just is not visible. And the
voltage at any point on the line can be correctly
computed from
V(t) = Vf(t) + Vr(t)

Another way to think of this is to produce the step
wave in a different manner. Start with a pulse wave.
Then increase the pulse width until it approaches
the pulse period. The variation in voltage can still
be observed on the line. Then slowly increase the
width until it is equal to the period. There is now
no longer any variation in the wave to be observed,
but the wave is still moving just as it was when
the width was infinitesimally shorter than the
period.

Another view is to start with a square wave, then
increase the period until it is much longer than
the duration of the experiment. Over the duration
of the experiment, no change in voltage will be
observed, but it is a square wave. One could
perform a Fourier transform on the square wave,
treat all the constituent sine waves as travelling
waves, work out the reflections for each, and sum
them back. Exactly the same results will be
obtained as will be for treating the step as a
moving wave, even though there are no peaks and
valleys to aid the visualisation.

So I hope that I have persuaded you that it is
completely valid to view a step function as a
wave that continues to move (though it is hard
to see) even after the step has passed.

If not, re-read the above to locate the flaw in
my argument.

If you have been convinced, then we can get back
to
If(t) = 50/50 = 1a
Ir(t) = 50/50 = 1a

Consider a generator, 50 ohms, that produces a
50 V step in to 50 ohms. For this
Vf(t) = 50V
The line has a 50 ohm impedance, so
If(t) = 50/50 = 1A

This is the voltage and current that would be
observed on the infinite or terminated line
after the step passes.

With the open circuit line, where the reflection
coefficient is 1, the voltage will be reflected
so that
Vr(t) = 50V
after the reverse step passes and
Ir(t) = 50/50 = 1A

Recalling that at any point on the line
V(t) = Vf(t) + Vr(t)
and
I(t) = If(t) – Ir(t)

we have, after the reverse step has returned to the
generator, everywhere on the line:
V(t) = 50 + 50 = 100V
I(t) = 1 – 1 = 0A

This is exactly as expected for a generator with
an open circuit voltage of 100V and an open circuit
line that has no current flowing.

So the wave reflection model, with continuously
forward and reverse waves, accurately describes
the resulting observations on the line, even for
a step wave.

....Keith

what happens to reflected energy ?
 

On Jun 29, 9:46*am, Cecil Moore wrote:
On Jun 28, 5:27*pm, Keith Dysart wrote:

And yet, you have not located any errors in the math or the models.

Superposition of power *IS* an error!

You have really latched on to that haven't you. I remember the good
ol'
days when you were superposing powers and the very same people with
whom you are arguing today were trying to convince you that
superposing
power was an invalid operation.

Now you have bought in to the rule and apply it even where it does
not apply.

Just because a sum is being computed does not mean that superposition
is at work.

When you add up the energies to validate that energy is not created
or destroyed, you are not superposing energy.

When I add up the energy flows for the same purpose, that does not
mean that superposition is at work.

From a basic calculus, if the energy balances, so must the energy
flow:

f(t), g(t), etc. are functions that describes the energy in
individual entities of a system with respect to time. Then

f(t) + g(t) + h(t) = k

is the expression that says the sum of all the energies must be
a constant, for energy is neither created nor destroyed.

From basic calculus

d(f(t) + g(t) + h(t))/dt = d(k)/dt
d(f(t) + g(t) + h(t))/dt = 0
d(f(t))/dt + d(g(t))/dt + d(h(t))/dt = 0

d(f(t))/dt is the derivitive with respect to time of the energy
(i.e. energy flow, or power) for the entity.

So it is completely valid to sum these energy flows to
track the energy within the system; as my example does.

You add and subtract powers
willy-nilly as if that mathematical step were valid which it is not.
Two coherent 50w waves do not add up to a 100w wave, even using
average powers, except for the special case of zero interference where
the waves are 90 degrees out of phase with each other.

True. You must always use the actual energy flows in the entity and
not the powers that are not real.

In order to use power as an energy tracking tool, we must be very
careful to ensure that there is a one-to-one correspondence between
energy and power, i.e. every joule passing a point in one second must
result in one watt of power (no VARs allowed).

Also true. And as is done in my example.

If that one-to-one
correspondence doesn't exist, no valid conclusion can be drawn from
tracking the power and any valid conclusion must be based on tracking
the energy which is no small task. The key is that there is no such
thing as imaginary energy. All energy is real. Some "power" is not
real.

Agreed. Especially the power in the waves that are being superposed.
That is why you need a fudge factor when you sum your partial powers
from your waves. I carefully do not sum partial powers from
superposed waves.

A one-to-one correspondence does not exist in a standing wave.
Therefore, tracking power as if it were equivalent to energy in
standing waves is invalid. You have made that error for years.

Please provide an example of such an error, preferably from one
of my expositions.

One-to-one correspondence also does not exist over a fraction of a
wave.

Of course it does. Energy must always balance; not just at the end
of the cycle.

Therefore, instantaneous power is irrevelent in tracking the
energy. That's your latest error which is the same conceptual error as
before. In general, average power in the traveling waves over at least
one complete cycle (or over many cycles) has a one-to-one
correspondence to the average energy in the traveling waves. But that
one-to-one correspondence is more often than not violated within a
fraction of each cycle.

Now you are allowing us to create and destroy energy as long as it
balances at the end of the cycle? Are you sure you want to do this?
It does not align with the well understood principle of conservation
of energy.

If I follow that rule, I just make the cycle long enough, say, my
lifetime, and I am golden.

But I do not hold out any hope for this.

Here's a quote from "Optics", by Hecht, concerning power density
(irradiance).

"If however, the 'T' is now divided out, a highly practical quantity
results, one that corresponds to the average energy per unit area per
unit time, namely 'I'." - where 'I' is the irradiance (*AVERAGE* power
density).

There you go again, assuming that the 'practical' challenges of optics
also apply to transmission lines. On a transmission line, we can
measure the voltage and the current. We do not suffer from the same
constraints as optics.

Use the tools that are available for transmission lines. Don't tie
two hands behind your back just because those tools do not work in
the optical domain.

....Keith

what happens to reflected energy ?
 

On Jun 29, 10:13*am, Cecil Moore wrote:
On Jun 28, 5:33*pm, Keith Dysart wrote:

Do you reject *P(t)=V(t)*I(t) ?

I certainly reject as a moronic method for attempting to track
instantaneous energy.

Dodging the question again! It is not that hard.

I agree completely. And my analysis does successfully track all the
energy at all times.

That is obviously false. You have tracked the *power* after assuming a
one-to-one correspondence between power (watts) and energy (joules).
Your assumption is most likely false. You usually cannot use
instantaneous watts to track joules within a fraction of a cycle.

Of course you can. It is just like water flowing from multiple pipes
in to a tank. The sum of the instantaneous flows is exactly equal
to the rate at which the volume of water is increasing in the tank.

Not hard at all.

If
the voltage and current are out of phase, some of the joules are
occupied as reactive power, and not available as watts of real power.

Mostly true. But this is accurately accounted for by using the
instantaneous voltage and current and not the RMS. This is an
important difference expressed in

P(t) = V(t) * I(t)

and why I do not use

Pavg = Vrms * Irms

which has all the problems you mention.

Why do you think the power companies spend so much money trying to
balance the power factor?

Not for this reason. The energy meter at my house accurately
measures the power (using P(t) = V(t) * I(t)) and integrates
so that they bill me for the energy I have used. And so they do.

Poor power factor does not affect this, but increases the losses
experienced by the power company when delivering the energy to me.
They want the customer to pay for these losses so they monitor
VAR.

....Keith

what happens to reflected energy ?
 

On Jun 29, 11:02*am, Cecil Moore wrote:
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

Indeed. Power and energy are related in that power is the derivitive
with
respect to time of energy. They are very closely related. See my
other
post for the derivation.

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!

'cause that is how calculus works!? If one tracks all the flows, then
one knows where the energy is because it can only move by flowing.
Think water pipes and tanks for an analogy.

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.

Please see my other post where this is covered.

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?

You, yourself, explained once how you would use the Fourier transform
to analyze such an example. After you have transformed the signal
to the frequency domain, there are enough sinusoidal waves to
keep anyone happy.

Or are you saying that a step function does not have a Fourier
transform?

....Keith

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Jun 30, 1:30*am, "Szczepan Bialek" wrote:
Photons are in the real light. The natural light is not coherent. It is
emitted in the portions (packets).
Radio waves are emitted continously. Radar waves are in the portions.

Laser light can usually be considered to be coherent light as are
waves generated by an RF transmitter. But of course, nothing is
perfect.

Radio waves and the light photons flow directly through the dielectric layer
of a capacitor.

Energy flow, just not as "current".

Electrons are compressed in the ends of the open circuit (condenser and
antenna).

"Compressed" may be overstating the phenomenon. Have you ever
calculated exactly how far an electron moves at 10 MHz?

Electrons are simple.

Yes, too simple to move at the speed of light.

what happens to reflected energy ?
 

On Jun 30, 6:43*am, Keith Dysart wrote:
You have really latched on to that haven't you. I remember the good
ol' days when you were superposing powers and the very same people with
whom you are arguing today were trying to convince you that
superposing power was an invalid operation.

And I remember the good ol' days when you were still messing your
diapers. Do you still do that? I probably did try to superpose power
back when I was young and foolish but I learned the error of my ways
and corrected it.

When I add up the energy flows for the same purpose, that does not
mean that superposition is at work.

But you are NOT adding up the energy flows - you are adding up the
power. That's superposition of power and is a no-no. Power does not
obey any conservation of power principle. The instantaneous maximum
charge on a capacitor contains any amount of energy while power equals
zero. What is it about that concept that you don't understand?

True. You must always use the actual energy flows in the entity and
not the powers that are not real.

What happens when energy = 1 joule, and de/dt = 0 watts. This happens
all the time during an RF cycle so you are not using actual energy
flows. You are using power which goes to zero even when maximum energy
is still present. There is obviously NOT a one-to-one correspondence
between power and energy.

YOU CANNOT USE WATTS TO TRACK ENERGY UNLESS THERE IS A ONE-TO-ONE
CORRESPONDENCE BETWEEN WATTS AND JOULES.

Also true. And as is done in my example.

Actually false, as I have proved above. Please plot the joules, not
the watts.

Please provide an example of such an error, preferably from one
of my expositions.

At a current node in a standing wave, you claim there is zero power
which is true. You claim that since there is zero power, there is also
zero energy which is absolutely false since that is a voltage maximum
point and you will fry yourself if you grab it. The energy in the
voltage maximum is unrelated to the power being zero - just one more
proof that energy does not correspond to power. At the zero power
points, there is a maximum of energy in one of the fields which you
are ignoring.

Of course it does. Energy must always balance; not just at the end
of the cycle.

Energy must balance but power doesn't have to balance and usually does
not balance. Power only balances for special cases. You are using
power in the general case as an example so anything you say is
invalid.

Now you are allowing us to create and destroy energy as long as it
balances at the end of the cycle?

No, energy cannot be created or destroyed. But power is created and
destroyed all the time because there is no conservation of power
principle. Your glaring error is that you are using power as if it
were energy and it is not. I have given numerous examples. In an LC
oscillator, when all of the non-destructable energy is stored in the
capacitor, there is ZERO power, i.e. power has been destroyed. Why do
you refuse to discuss that fact of physics? 90 degrees later in the
cycle, power has been created.

Every time one of your instantaneous power curves crosses the zero
axis, power has been destroyed. Every time one of your instantaneous
power curves reaches a peak, power has been created.

We do not suffer from the same constraints as optics.

You suffer from the delusion that power obeys the same laws of physics
that energy does. You willy-nilly interchange power and energy. Until
you admit the error of your ways, there is little that can be done to
alleviate your ignorance.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 30, 7:24*am, Keith Dysart wrote:
Or are you saying that a step function does not have a Fourier
transform?

Your example has the step function disappearing during steady-state.
DC steady-state has no Fourier transform since omega equals zero. A
Fourier analysis requires a recurring function of the sum of sinusoids
involving integer multiples of omega*time. Your single step does not
recur so your conclusions are invalid.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 30, 10:11*am, Cecil Moore wrote:
On Jun 30, 7:24*am, Keith Dysart wrote:

Or are you saying that a step function does not have a Fourier
transform?

Your example has the step function disappearing during steady-state.
DC steady-state has no Fourier transform since omega equals zero. A
Fourier analysis requires a recurring function of the sum of sinusoids
involving integer multiples of omega*time. Your single step does not
recur so your conclusions are invalid.

Check the a0 coefficient in the Fourier transform. This represents
the DC component of the signal. It is perfectly valid for it to be
the only non-zero coefficient.

Without this, how would you deal with a signal such as
V(t) = 10 + 2 cos(3t)


Alternatively, one can use the standard trick for dealing with
non-repetitive waveforms: choose an arbitrary period. 24 hours
would probably be suitable for these examples and transform from
there. Still, you will have zero frequency component to deal
with, but there will be some at higher frequencies (if you
choose your function to make it so).

....Keith

what happens to reflected energy ?
 

On Jun 30, 10:03*am, Cecil Moore wrote:
On Jun 30, 6:43*am, Keith Dysart wrote:

You have really latched on to that haven't you. I remember the good
ol' days when you were superposing powers and the very same people with
whom you are arguing today were trying to convince you that
superposing power was an invalid operation.

And I remember the good ol' days when you were still messing your
diapers. Do you still do that?

Warning! Warning! Descent in to scatology.

I probably did try to superpose power
back when I was young and foolish but I learned the error of my ways
and corrected it.

Learning is good. But now you want to over apply the rules.

When I add up the energy flows for the same purpose, that does not
mean that superposition is at work.

But you are NOT adding up the energy flows - you are adding up the
power.

Ummm. Energy flow is power. Joules/s!

If it helps, any place I have written 'power', please replace with
'energy flow'.

That's superposition of power and is a no-no. Power does not
obey any conservation of power principle.

Try the multiple pipe, water container, water flow example to
understand how flows must also be conserved if matter (or
energy) is not to be created or destroyed.

The instantaneous maximum
charge on a capacitor contains any amount of energy while power equals
zero. What is it about that concept that you don't understand?

Excellent concept. There is no conflict with conserving flows, or
energy
for that matter.

True. You must always use the actual energy flows in the entity and
not the powers that are not real.

What happens when energy = 1 joule, and de/dt = 0 watts. This happens
all the time during an RF cycle so you are not using actual energy
flows. You are using power which goes to zero even when maximum energy
is still present.

Yes, indeed. That is a fundamental possibility and occurs on
transmission lines with infinite VSWR.

There is obviously NOT a one-to-one correspondence
between power and energy.

Correct. Power is the time derivitive of energy. They are related
but definitely not one-to-one.

YOU CANNOT USE WATTS TO TRACK ENERGY UNLESS THERE IS A ONE-TO-ONE
CORRESPONDENCE BETWEEN WATTS AND JOULES.

This is quite incorrect. Energy flows must balance, otherwise energy
is being created or destroyed to sustain a difference in flow.

Also true. And as is done in my example.

Actually false, as I have proved above. Please plot the joules, not
the watts.

If it makes you happier, since it is a discrete simulation, you can
simply substitute joules per degree wherever there is a curve labeled
Watts. You need to scale, of course, between degrees and seconds, but
the shape and sums will be identical.

Please provide an example of such an error, preferably from one
of my expositions.

At a current node in a standing wave, you claim there is zero power
which is true.

Excellent. I am glad you have come to accept that.

You claim that since there is zero power, there is also
zero energy

I have never claimed that. In fact, the minima and maxima are
where the energy peaks will be found. I have claimed there is
no energy flow across (i.e. power) these points.

which is absolutely false since that is a voltage maximum
point and you will fry yourself if you grab it. The energy in the
voltage maximum is unrelated to the power being zero

Not quite. It follows from the functions for each. Since power is
the derivitive of energy, it should come as no surprise that
maximum energy occurs at the point of minimum power. In basic
calculus looking for the zero in the derivitive is how you
locate the maximum value of the curve.

- just one more
proof that energy does not correspond to power. At the zero power
points, there is a maximum of energy in one of the fields which you
are ignoring.

Of course it does. Energy must always balance; not just at the end
of the cycle.

Energy must balance but power doesn't have to balance and usually does
not balance. Power only balances for special cases. You are using
power in the general case as an example so anything you say is
invalid.

Unfortunately wrong. Energy flows must balance as well. Otherwise,
energy is coming from nowhere to sustain the flow.

Now you are allowing us to create and destroy energy as long as it
balances at the end of the cycle?

No, energy cannot be created or destroyed. But power is created and
destroyed all the time because there is no conservation of power
principle. Your glaring error is that you are using power as if it
were energy and it is not. I have given numerous examples. In an LC
oscillator, when all of the non-destructable energy is stored in the
capacitor, there is ZERO power, i.e. power has been destroyed.

Yes, indeed. At that instant, zero energy is flowing from the inductor
to the capacitor. But very soon, energy will be flowing from the
capacitor to the inductor. The balance is that the energy flowing
out of the capacitor is always and exactly equal to the energy flowing
in to the inductor. That is the energy flow balance. The only way for
this not to be true is for energy to be created or destroyed.

Why do
you refuse to discuss that fact of physics? 90 degrees later in the
cycle, power has been created.

One way of thinking about it, I suppose. But not too useful.

Instead, think that at every instant, the energy flow between the
entities in the experiment must balance.

Every time one of your instantaneous power curves crosses the zero
axis, power has been destroyed. Every time one of your instantaneous
power curves reaches a peak, power has been created.

I think you may be confused because you are only looking at the
flow in and out of a single entity. This is clearly not conserved.
Nor for that matter is the energy within that entity. It is the
total energy within the system that is conserved, just as it is
the total of the flows of energy between the entities within the
system that must be conserved.

Put more strictly: The sum of all the energy flows in to all of
the entities within the system must equal the energy flow in to
the system.

We do not suffer from the same constraints as optics.

You suffer from the delusion that power obeys the same laws of physics
that energy does. You willy-nilly interchange power and energy. Until
you admit the error of your ways, there is little that can be done to
alleviate your ignorance.

An intriguing accusation, but you do need to provide a concrete
example of where my analysis has gone wrong.

....Keith