Superposition
 

On Wed, 21 Nov 2007 16:50:49 -0600, 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.

Hmmm, basic math demands that momentum be in units of rather more
prosaic terms, namely
kg·m/s

If we carry out the math of your own answer, it renders units in your
terms to:
(watts/m²)/(m/s)²
or
((kg·m²/s³)/m²)/(m/s)²
or
kg/(s·m²)

Perhaps I fumbled the product of powers, but it looks like you could
have as easily just divided by pi and still come out to the same
effect of "proving" a balance to conserve your dignity. ;-)

With momentum like this, Lyndon LaRouche would have been elected
president in 1991.

Superposition
 

On Thu, 22 Nov 2007 00:48:02 -0500, "AI4QJ" wrote:


"Richard Clark" wrote in message
.. .
On Wed, 21 Nov 2007 16:50:49 -0600, 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.

Hmmm, basic math demands that momentum be in units of rather more
prosaic terms, namely
kg·m/s

If we carry out the math of your own answer, it renders units in your
terms to:
(watts/m²)/(m/s)²
or
((kg·m²/s³)/m²)/(m/s)²
or
kg/(s·m²)

Perhaps I fumbled the product of powers, but it looks like you could
have as easily just divided by pi and still come out to the same
effect of "proving" a balance to conserve your dignity. ;-)

With momentum like this, Lyndon LaRouche would have been elected
president in 1991.

I think you never got a chance to take the basic courses in quantum
mechanics Richard. If you did, you would understand this. As it is, you seem
close to understanding it without having such a background which is really
curious. I think you would learn it quite easily.

Hi Dan,

You might be right, you might be wrong, but you don't really know
yourself, do you?

I did take QM, and I did learn it quite easily. However, this has
nothing to do with Momentum beyond what was already revealed above. If
you've found no math errors (and that could go either way, favoring
either me or Cecil - or we are both wrong), then what's your point?

Momentum is also "Radiation Pressure," a topic I've brought to this
forum in the past. Radiation Pressure has a very Newtonian result
that easily manifests itself in exactly the prosaic terms I posed
above. The computation may be tedious (and bordering well beyond
trivial), but it certainly isn't difficult. Do you care to offer a
solution? :-)

73's
Richard Clark, KB7QHC

Superposition
 

Tom Donaly wrote:
Actually, you're writing about momentum density. Momentum is
conserved, but momentum density isn't, ...

The momentum density may certainly change with area just
as the energy density may change with area. But in either
case, the total energy and total momentum are conserved.

As for any finite number being an infinite percentage above
zero, I think you should take that up with the next mathematician
you meet.

The equation for any percentage change from zero is
100(X-0)/0 Plug any value of X into that equation and
see what you get.
--
73, Cecil http://www.w5dxp.com

Superposition
 

Richard Clark wrote:
kg/(s·m²)

One needs to multiply by the area under consideration
to obtain the total momentum which is conserved.

It's the same as multiplying the Poynting vector
by the area under consideration to obtain the
total energy.
--
73, Cecil http://www.w5dxp.com

Superposition
 

AI4QJ wrote:
If you are considering adding structural supports for your transmission line
it is zero. If you want to understand the physics it is not zero. Photons
have mass (not rest mass but mass), light has pressure that has been
measured and it has momentum.

Quoting Maxwell in 1873: "In a medium in which waves are
propagated, there is a pressure in the direction normal
to the waves, and numerically equal to the energy in a
unit of volume."
--
73, Cecil http://www.w5dxp.com

Superposition
 

Gene Fuller wrote:
Let's see some real numbers.

The numbers are trivial. What is important is the concept.
In the experiment, the Pref1 wave disappears between
steady-state #1 and steady-state #2. Here's the question
that you and others have refused to answer.

When an EM wave disappears in its original direction of
travel, what happens to its energy?
--
73, Cecil http://www.w5dxp.com

Superposition
 

On Thu, 22 Nov 2007 05:20:10 -0600, Cecil Moore
were Cecil's dignity was not conserved, wrote:
If we carry out the math of your own answer, it renders units in your
terms to:
(watts/m²)/(m/s)²
or
((kg·m²/s³)/m²)/(m/s)²
or
kg/(s·m²)

One needs to multiply by the area under consideration
to obtain the total momentum which is conserved.

Which, of course, yields:
kg/s
which is not momentum, but oddly enough the units one might use in
describing how fast his bath tub was filling with photons. :-)

Oh well, third time's a charm! Keep fumbling with the conservation of
dignity, and eventually you might bumble into a job with Anderson
Consulting doing Enron's books.

Superposition
 

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?

That depends entirely on the units you're using.

If you measure the velocity of C in typical human-scale units (e.g.
metres per second) then C^2 is, indeed, a numerically-large value, and
50.95/C^2 is numerically small (and will be in units of
watts seconds-squared over metres-squared).

If, on the other hand, you measure velocities on another scale, things
look very different. Specifically, let's use a scale which represents
each velocity as a fraction of the maximum possible velocity... let's
call this unit of velocity "skedaddles". Measured this way, C is
precisely 1.0 skedaddle, C^2 is precisely 1.0 skedaddle squared, and
50.95/C^2 equals 50.95 (watts per square skedaddle).

Same result... only the units of measurement are different. Neither
result is zero. "Numerically small" is not equivalent to "zero for
all practical purposes".

(obLinguistic: "skedaddle" is a somewhat quirky American term of
uncertain heritage, which means "leave in a hurry, scram, escape", and
seems a reasonable term for a scale of zerch up to as-fast-as-
possible. I believe that the equivalent British term was defined as
"runawayrunaway" by King Arthur, as cited in "Monty Python and the
Holy Grail").

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

Superposition
 

Cecil Moore wrote:
Gene Fuller wrote:
Let's see some real numbers.

The numbers are trivial. What is important is the concept.
In the experiment, the Pref1 wave disappears between
steady-state #1 and steady-state #2. Here's the question
that you and others have refused to answer.

When an EM wave disappears in its original direction of
travel, what happens to its energy?

OK, but first let's set the ground rules.

The ONLY thing under discussion here is our disagreement about the
canceling waves heading back toward the source from the match point. You
claim those waves must exist and then cancel over a short distance (I
believe you reduced the distance to 'dx' or something similar.) I claim
those waves never exist at all and therefore don't need to be canceled.

If you have anything else in mind, then enjoy your solo activity,
whatever that might be.

So just what "trivial" numbers are required?

First, it is not clear how one make an instant transition from a 50 ohm
environment to a 300 ohm environment. Do you just connect a 50 ohm coax
to a 300 ohm coax? Or do you prefer to connect a 50 ohm twin-lead to a
300 ohm twin-lead? (Good luck with either of these.) If you want to
connect a 50 ohm coax to a 300 ohm twin-lead then you are going to need
some sort of transition device.

Oops! Where is the match point now? Ordinarily we would not really care
very much about such things, but you have stated that important things
are happening within the "dx" zone. It is a safe bet that the Z0
transition is not abrupt either. The "trivial" numbers just got a bit
more complicated.

Let's look at the conservation of energy part. You like to use the
Poynting vector, so we can stick with that. The first thing to note is
that the Poynting vector is E x H, not V x I. Perhaps only a minor bump
in the road, but the transition from E to V and H to I is not quite so
trivial at discontinuities such as the "match point".

But let's muddle ahead in any case. The integral form of the Poynting
theorem goes like the following.

* Define a test volume with a closed surface. There is no particular
size required, although infinite and zero don't work well for practical
reasons.

* Calculate the Poynting vector at all points on the closed surface.

* Integrate the 'normal' component of the Poynting vector over the
entire surface. This integral then represents the net electromagnetic
energy flowing into (or out of) the test volume. I believe this is what
you would consider the energy carried by the waves of interest. Note
that all waves are combined together; they are not treated separately.

* The Poynting theorem says that the net energy flow must be balanced by
the change in electromagnetic energy content within the test volume and
the work done by the fields on any charges within the test volume.

* Note that a change in field strength within the test volume is tied to
the change in electromagnetic energy content. Any charges within the
test volume can be accelerated. Remember, this sort of match point
cannot exist in free space, so there are charges in the region of interest.


This sort of description and the associated derivations can be found in
any ordinary E&M textbook.


You might notice that the Poynting theorem, i.e. conservation of energy
law for EM, says nothing about the sanctity of waves or about the
conservation of energy in waves. It does not say that the integral of
the Poynting vector over the test volume surface must be zero.

Even more importantly for this discussion, the Poynting theorem does not
help at all with your assertion that important things are happening in
the 'dx' zone. If you make the test volume size smaller than your 'dx'
then you run into trouble with the finite size of the transition region
described above. If you make the test volume large enough to contain the
'dx', then all of the purported interesting stuff happens inside. Again,
the Poynting theorem tells nothing.


If you want to believe in the conservation of waves, go right ahead.
Just don't expect conservation of energy to support your case.
Mathematically it cannot.


73,
Gene
W4SZ

Superposition
 

On Thu, 22 Nov 2007 07:34:14 -0800, Richard Clark
wrote:

On Thu, 22 Nov 2007 05:20:10 -0600, Cecil Moore
were Cecil's dignity was not conserved, wrote:
If we carry out the math of your own answer, it renders units in your
terms to:
(watts/m²)/(m/s)²
or
((kg·m²/s³)/m²)/(m/s)²
or
kg/(s·m²)

One needs to multiply by the area under consideration
to obtain the total momentum which is conserved.

Which, of course, yields:
kg/s
which is not momentum, but oddly enough the units one might use in
describing how fast his bath tub was filling with photons. :-)

Oh well, third time's a charm! Keep fumbling with the conservation of
dignity, and eventually you might bumble into a job with Anderson
Consulting doing Enron's books.

Hi All,

Well, no point in waiting for another rationalization when I am
perfectly capable of filling that in for Cec' (and more entertaining
than him when I do).

Let's see if I can pull together a good old-boy drawl and a scrub of
the boot toe in the dirt:
"One needs to multiply by the volume under consideration
to obtain the total momentum which is conserved!"

Umm, yes, if your Xeroxed authors need that much help in you
describing what they must have meant, but didn't say, then throwing in
previously undisclosed terms might do the trick.

However, looking aside from this obvious self-serving manipulation of
the books (Anderson would be proud, in a perverted sense) it then
gives us a third dimension of meters (the one you will have suckered
into the equation) which is also a time specification (and this would
then be called not Momentum, but Impulse, which does carry the same
units but is the Integration over time).

"Yes! Of course! This is called the Conservation of Impulse!
It is EXACTLY what my references meant to say...."

And then we return to an observation I made earlier about:
The short journey was described by the term "dt".

Ah, suffering the dt's.

As Ed McMahon would prompt Johnny:
"Just how short was that journey?"

My guess it will either be too short to do the job, or much too large
to be true.
and we had been left holding the bag once again with asking how big dt
is? Which, of course, also flummoxed Cecil (his having not yet had
the epiphany of what his references "meant to say" but left unsaid).
Some might begin to wonder how they earned a salary in the career of
teaching.

And to whip a dead horse, I also said:
This thread should be called:
"Supposition"
or
"Imposition"
or
"Superstition"

To the group,

Sorry for having unleashed yet another law of conservation that will
undoubtedly yield 100s of postings of "proofs" that in and of
themselves will actually contain no intellectual nourishment.

Next week:
"The Conservation of Radiation Buoyancy"

73's
Richard Clark, KB7QHC