Superposition
 

Jim Kelley wrote:
Yes. The short journey was described by the term "dt". According to
Cecil, that is the amount of time after energy is reflected and before
it 'turns around and goes the other way as it is required to do by the
law of conservation of energy'. You may recall that it is forced to go
the other way 'because there are only two directions in a transmission
line'.

So you don't even accept differential calculus? :-)
Jim, you have never answered the tough questions so I
will keep asking. I just posted an example with one
question that should be easy for you to answer.

In the example, what happens to the energy and momentum
in Pref1 when the load is switched from 300 ohms to 50
ohms? It's a simple question. Please be specific in your
answer.
--
73, Cecil http://www.w5dxp.com

Superposition
 

On Wed, 21 Nov 2007 14:00:39 GMT, Cecil Moore
wrote:

What is the momentum of 50.95 W?

momentum? Please be specific.
ditto. :-)

Superposition
 

Cecil Moore wrote:
Gene Fuller wrote:
Waves are useful. However, they are not living objects. They have no
will to survive. There is nothing in the standard E&M science based on
Maxwell's laws that requires waves to be "canceled" if they no longer
exist. There is no conservation law of wave-ality.

All EM waves must obey the conservation of energy and
conservation of momentum principles. It is not a will
to survive - it is simply the laws of physics.

Here is an example for you to explain. The source is
a signal generator equipped with an ideal circulator
and a load resistor:

Steady-state #1: Rho at '+' equals 0.7143. Load equals
300 ohms.

100w SGCL--50 ohm feedline--+--1/2WL 300 ohm feedline--300 ohm load
Pfor1=100w-- Pfor2=49w--
--Pref1=51w --Pref2=0w

Pref1 is an 51w EM wave whose energy and momentum must be
conserved.

Steady-state #2: Rho at '+' equals 0.7143. Load is switched
to 50 ohms.

100w SGCL--50 ohm feedline--+--1/2WL 300 ohm feedline--50 ohm load
Pfor1=100w-- Pfor2=204W--
--Pref1=0w --Pref2=104w

*Note that Rho has NOT changed!*

The only question that you need to answer is during the
process that changes Pref1 from 51 joules/sec in the direction
of the source to 0 joules/sec (canceled), *exactly* what happens
to the energy and momentum? Please be specific.


Cecil,

Nice try.

Combining "steady state" with "switched" and "during the process that
changes" makes a very messy problem for an analytical solution.

You first.

And you won't get any closer to the correct solution through all of your
handwaving arguments either.

73,
Gene
W4SZ

Superposition
 

Richard Clark wrote:
On Wed, 21 Nov 2007 14:00:39 GMT, Cecil Moore
wrote:

What is the momentum of 50.95 W?

momentum? Please be specific.
ditto. :-)

If 50.95 watts is the Poynting vector, actually
watts/unit-area, then the momentum is 50.95/c^2.
Please reference pages 56,57 of "Optics", by Hecht,
4th edition.
--
73, Cecil http://www.w5dxp.com

Superposition
 

Gene Fuller wrote:
Cecil Moore wrote:
Gene Fuller wrote:
Waves are useful. However, they are not living objects. They have no
will to survive. There is nothing in the standard E&M science based
on Maxwell's laws that requires waves to be "canceled" if they no
longer exist. There is no conservation law of wave-ality.

All EM waves must obey the conservation of energy and
conservation of momentum principles. It is not a will
to survive - it is simply the laws of physics.

Here is an example for you to explain. The source is
a signal generator equipped with an ideal circulator
and a load resistor:

Steady-state #1: Rho at '+' equals 0.7143. Load equals
300 ohms.

100w SGCL--50 ohm feedline--+--1/2WL 300 ohm feedline--300 ohm load
Pfor1=100w-- Pfor2=49w--
--Pref1=51w --Pref2=0w

Pref1 is an 51w EM wave whose energy and momentum must be
conserved.

Steady-state #2: Rho at '+' equals 0.7143. Load is switched
to 50 ohms.

100w SGCL--50 ohm feedline--+--1/2WL 300 ohm feedline--50 ohm load
Pfor1=100w-- Pfor2=204W--
--Pref1=0w --Pref2=104w

*Note that Rho has NOT changed!*

The only question that you need to answer is during the
process that changes Pref1 from 51 joules/sec in the direction
of the source to 0 joules/sec (canceled), *exactly* what happens
to the energy and momentum? Please be specific.

You first.

Cop out. Why am I not surprised that you, yet once again, refuse
to answer the question? Could it be because you would immediately
be proven wrong? Do you really believe that diversions are a tool
of technical knowledge?

The original 51 joule/sec reflected wave toward the source interacts
with the newly reflected wave from the load and is partially
canceled which through constructive interference, delivers more
forward power toward the load, which results in an increase in the
energy in the reflected wave from the load, which results in more
wave cancellation at '+', etc. until steady-state #2 is reached.
--
73, Cecil http://www.w5dxp.com

Superposition
 

Cecil Moore wrote:
Richard Clark wrote:
On Wed, 21 Nov 2007 14:00:39 GMT, Cecil Moore
wrote:

What is the momentum of 50.95 W?

momentum? Please be specific.
ditto. :-)

If 50.95 watts is the Poynting vector, actually
watts/unit-area, then the momentum is 50.95/c^2.
Please reference pages 56,57 of "Optics", by Hecht,
4th edition.

50.95 divided by the speed of light squared? So, for all
practical purposes - if that's right - it's zero. Why not
just say so?
73,
Tom Donaly, KA6RUH

Superposition
 

Cecil Moore wrote:


Cop out. Why am I not surprised that you, yet once again, refuse
to answer the question? Could it be because you would immediately
be proven wrong? Do you really believe that diversions are a tool
of technical knowledge?

The original 51 joule/sec reflected wave toward the source interacts
with the newly reflected wave from the load and is partially
canceled which through constructive interference, delivers more
forward power toward the load, which results in an increase in the
energy in the reflected wave from the load, which results in more
wave cancellation at '+', etc. until steady-state #2 is reached.

Blah, Blah, Blah. Totally useless drivel. Let's see some real numbers.
Then we can discuss cop out.

You might also check your favorite reference to try to figure out what
conservation of energy really means. Then you would realize that even a
full solution to your idealized problem would demonstrate absolutely
nothing with respect to conservation of energy.

73,
Gene
W4SZ

Superposition
 

Tom Donaly wrote:
Cecil Moore wrote:
Richard Clark wrote:
On Wed, 21 Nov 2007 14:00:39 GMT, Cecil Moore
wrote:

What is the momentum of 50.95 W?

momentum? Please be specific.
ditto. :-)

If 50.95 watts is the Poynting vector, actually
watts/unit-area, then the momentum is 50.95/c^2.
Please reference pages 56,57 of "Optics", by Hecht,
4th edition.

50.95 divided by the speed of light squared? So, for all
practical purposes - if that's right - it's zero. Why not
just say so?

The percentage difference between zero and that momentum
is infinite. And whatever value it is must be conserved.
Sweeping it under the rug in violation of the laws of
physics is just not acceptable.
--
73, Cecil http://www.w5dxp.com

Superposition
 

Gene Fuller wrote:
Cecil Moore wrote:
Cop out. Why am I not surprised that you, yet once again, refuse
to answer the question? Could it be because you would immediately
be proven wrong? Do you really believe that diversions are a tool
of technical knowledge?

The original 51 joule/sec reflected wave toward the source interacts
with the newly reflected wave from the load and is partially
canceled which through constructive interference, delivers more
forward power toward the load, which results in an increase in the
energy in the reflected wave from the load, which results in more
wave cancellation at '+', etc. until steady-state #2 is reached.

Blah, Blah, Blah. Totally useless drivel. Let's see some real numbers.
Then we can discuss cop out.

In the words of the biggest cop out artist I know, "You first".
I've already posted the numbers for the graphic at:
http://www.w5dxp.com/thinfilm.gif

Instead of responding, you tucked tail and ran.

You might also check your favorite reference to try to figure out what
conservation of energy really means. Then you would realize that even a
full solution to your idealized problem would demonstrate absolutely
nothing with respect to conservation of energy.

That's exactly the problem. Lots of people pay lip service to
the conservation of energy principle without realizing they
advocate violation of it.
--
73, Cecil http://www.w5dxp.com

Superposition
 

Cecil Moore wrote:
Tom Donaly wrote:
Cecil Moore wrote:
Richard Clark wrote:
On Wed, 21 Nov 2007 14:00:39 GMT, Cecil Moore
wrote:

What is the momentum of 50.95 W?

momentum? Please be specific.
ditto. :-)

If 50.95 watts is the Poynting vector, actually
watts/unit-area, then the momentum is 50.95/c^2.
Please reference pages 56,57 of "Optics", by Hecht,
4th edition.

50.95 divided by the speed of light squared? So, for all
practical purposes - if that's right - it's zero. Why not
just say so?

The percentage difference between zero and that momentum
is infinite. And whatever value it is must be conserved.
Sweeping it under the rug in violation of the laws of
physics is just not acceptable.

Actually, you're writing about momentum density. Momentum is
conserved, but momentum density isn't, any more than energy
density, or any other kind of density, with the possible
exception of the bone density in the heads of some people.
As for any finite number being an infinite percentage above
zero, I think you should take that up with the next mathematician
you meet. Mathematicians need to laugh once in a while, too.
73,
Tom Donaly, KA6RUH