Richard Clark wrote:
On Wed, 05 Sep 2007 00:04:23 -0000, Jim Kelley
wrote:


Richard Clark wrote:


The description you offer requires a porous plate which is absent in
every radiometer that has come down the pike

On the other hand, the absence of porous plates in operating
radiometers tends to cast some doubt on your claim that the plates
must be porous.


Not "my claim," my report.

So be it. The absence of porous plates in operating radiometers tends
to cast doubt on your report that the plates must be porous.

The claim they must be porous arrives
through the math necessary to balance the kinetic forces.

But a balance of forces would result in the absence of an observable
effect. An imbalance in forces is required in order to produce
movement.

Now, if we simply move to another radiometer (Nichols, Tear, Hull, and
Webb already recited) without that partial vacuum, the vanes still
move, and expressely by Radiation Pressure.

By a different mechanism and in the opposite direction, yes.

In essence, these instruments indicate,
not measure.

A description which applies beautifully to power meters as well, don't
you agree? ;-)

The coy context of the thread was measuring the mass of a Photon.
Absolutely no SI Units have been named or any quantitative values
offered (the rather standard omission from claims made here).
However, feel free to introduce your own side thread's goal or even offer a
guess (your own quatitative value for the mass).

I'd like to offer m = E/c^2 as a guess.

73, ac6xg

On Wed, 05 Sep 2007 17:02:38 -0700, Jim Kelley
wrote:

Not "my claim," my report.

So be it. The absence of porous plates in operating radiometers tends
to cast doubt on your report that the plates must be porous.

Hi Jim,

You should distinguish between reporting, claims, and what you see.
The point about being porous is to substantiate the expectation of
explaining the full energy budget (and specifically for the Crookes
radiometer). I don't know how many times I have to emphasize this,
but NO METHOD achieves that balance.

The claim they must be porous arrives
through the math necessary to balance the kinetic forces.

But a balance of forces would result in the absence of an observable
effect. An imbalance in forces is required in order to produce
movement.

The balance is in the energy applied and the energy expended. You put
an HP into a car, and it will accelerate 550 foot-pounds/sec. You put
x photons into ANY radiometer, and the change in inertia WILL NOT
balance.

[This is why I expressed my question in Newtonian terms for the
benefit of the twins who are so devoted to the master (that they are
wholly lost in a simple 2 variable computation). The difference
between that computation and performance is extreme. What is more
compelling, is that it is quite a departure from what Quantum
Mechanics would predict. NO METHOD achieves that balance.]

Now, if we simply move to another radiometer (Nichols, Tear, Hull, and
Webb already recited) without that partial vacuum, the vanes still
move, and expressely by Radiation Pressure.

By a different mechanism and in the opposite direction, yes.

In essence, these instruments indicate,
not measure.

A description which applies beautifully to power meters as well, don't
you agree? ;-)

No. Power meters to even uncommonly high accuracy still conform to
Newtonian mechanics.

The coy context of the thread was measuring the mass of a Photon.
Absolutely no SI Units have been named or any quantitative values
offered (the rather standard omission from claims made here).
However, feel free to introduce your own side thread's goal or even offer a
guess (your own quatitative value for the mass).

I'd like to offer m = E/c^2 as a guess.

The photo-electron appears to even depart from that. More to follow.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

In essence, these instruments indicate,
not measure.

A description which applies beautifully to power meters as well, don't
you agree? ;-)


No. Power meters to even uncommonly high accuracy still conform to
Newtonian mechanics.

So it is because of Newtonian mechanics that an RF power meter is
actually measuring power rather than indicating power. What is the
value gained by this strain on credulity?

The coy context of the thread was measuring the mass of a Photon.
Absolutely no SI Units have been named or any quantitative values
offered (the rather standard omission from claims made here).
However, feel free to introduce your own side thread's goal or even offer a
guess (your own quatitative value for the mass).

I'd like to offer m = E/c^2 as a guess.


The photo-electron appears to even depart from that.

I've often wondered how one might go about recognizing a
photo-electron out of a group of other, less prominent electrons? :-)

73, ac6xg

On Thu, 06 Sep 2007 09:54:43 -0700, Jim Kelley
wrote:

So it is because of Newtonian mechanics that an RF power meter is
actually measuring power rather than indicating power. What is the
value gained by this strain on credulity?

Hi Jim,

Sounds like you should talk to your Chaplain about these issues.

I've often wondered how one might go about recognizing a
photo-electron out of a group of other, less prominent electrons? :-)

It is like complaining the Nobel winners are indistinguishable from
the crowd in the ceremonial hall.

Prominent is key, certainly. How many electrons can you motivate to
leap over the barrier of the work function of a metal? In Physics, a
simple population count would reveal the prominence. Tubes usually
have to boil them off incandescent filaments, or rip them out of their
matrix with 10's of kilovolts of nearby potential.

Hi All,

Let's examine those last two motivators. Photons hardly raise the
temperature of a metal vane to, what, 1000 degrees? And as for
kilovolts of excitation, how much potential is there in a photon?

Well, too often this group starves for information in response when I
toss these questions out - too technical for this forum of light
nappers I suppose. Too often, these threads turn into strings of
slaps at the snooze button (and "ether" has been the biggest snooze of
them all - self-fulfilling if one were to enlarge on the term's
rhetorical baggage).

Place a vane coated with sodium into an evacuated quartz tube.
Illuminate the sodium coated plate such that it absorbs one microwatt
per square meter. This is sufficient power to evoke the
photo-electric response (hopefully this is not too arcane a term). The
bulk of absorption will occur within a layer depth of 10 atoms.
Sodium, one atom thick, measures out to 10^19 atoms per square meter,
so we are absorbing the power throughout 10^20 atoms. Hence each atom
is illuminated with 10^-26 Watts OR 10^-7 eV/sec. (eV: electron Volt,
perhaps another prominence hard to embrace.)

The conundrum (sorry for hard words - but even those who use English
as a second language manage to cope) here is that to build a potential
to at least 1eV (and usually 3 to 5 times that for many metals) would
take nearly a year for a single electron to leap the Work Function
barrier. In reality, it occurs in less than a nanosecond.

It would be interesting to see Arthur's Newtonian math achieving a
10,000,000,000,000,000:1 leap of faith. False idolatry in place of
work is like putting a lottery ticket into the collection plate.

Strip away my stylistic excess and the facts fill maybe three
sentences. Still, I am four sentences ahead of the rest. ;-)

73's
Richard Clark, KB7QHC

Richard Clark wrote:

Still, I am four sentences ahead of the rest. ;-)

We now learn of the value in strained credulity. :-)

jk

On Sep 5, 5:02 pm, Jim Kelley wrote:
....

I'd like to offer m = E/c^2 as a guess.

73, ac6xg

A link is worth a thousand words (perhaps 10k-100k of Richard's...):
http://www.newton.dep.anl.gov/askasc...0/phy00332.htm
(in particular the first paragraph of the second response).

On Thu, 06 Sep 2007 12:47:46 -0700, K7ITM wrote:

On Sep 5, 5:02 pm, Jim Kelley wrote:
...

I'd like to offer m = E/c^2 as a guess.

73, ac6xg

A link is worth a thousand words (perhaps 10k-100k of Richard's...):
http://www.newton.dep.anl.gov/askasc...0/phy00332.htm
(in particular the first paragraph of the second response).

Hi Tom,

Your link is over valued (there is no second response), but it
maintains the standard of excellence here in the tradition of 10k-100k
more words than quantifiables - and someone else doing the work.

Care to walk us through your proffered math?

Well, I doubt it. Others may be interested in the curious form of
argument offered to a 15 year old however.

"For a particle with no mass, the relation reduces to E=pc."

The long and short of it is that there is no discussion of mass for a
photon (it is simply defined not to exist) and instead there is a
shuffle of math that youngster must imagine this bozo is pulling the
wool over his eyes through substituting p for Planck's constant h, and
c for Planck's energy formula variable v. This wool pulling is
another favorite past time here too.

Of course, there may be other meanings behind
"E=pc."
but in the model of thorough work, the description of terms is sadly
poor.

The typical legacy of offering links. It has all the appeal of a
Physicist's joke:
"How many milliseconds does it take to do a 5 minute car wash?"

73's
Richard Clark, KB7QHC

On Sep 6, 2:53 pm, Richard Clark wrote:
....(there is no second response)...

Pity you have so much trouble reading...

Richard Clark wrote:

On Thu, 06 Sep 2007 12:47:46 -0700, K7ITM wrote:


On Sep 5, 5:02 pm, Jim Kelley wrote:
...

I'd like to offer m = E/c^2 as a guess.

73, ac6xg

"E=pc."

Yes, and p=mv, so when v=c as is true for photons, and we substitute
mc for p in the equation above and then solve for m (the mass of a
photon was the original question), we're back at the equation offered
previously.

But we usually relate more directly to the frequency (or wavelength)
of the photon rather than its energy or momentum, so in such a case
E=h*nu would provide a more direct route to its mass equivalent.

ac6xg

On Thu, 06 Sep 2007 16:07:09 -0700, Jim Kelley
wrote:

Richard Clark wrote:

On Thu, 06 Sep 2007 12:47:46 -0700, K7ITM wrote:


On Sep 5, 5:02 pm, Jim Kelley wrote:
...

I'd like to offer m = E/c^2 as a guess.

73, ac6xg

"E=pc."

Yes, and p=mv,

Hi Jim,

Tom opines about my reading, but it is about the writing from the good
doctor that we find (in regard to your snippet above):
"we find that the momentum relation p=mv is
only an approximation. It is only correct when speed (v) is much
smaller than the speed of light (c).
which distinctly contradicts your tie-in:
so when v=c as is true for photons, and we substitute
mc for p in the equation above and then solve for m (the mass of a
photon was the original question), we're back at the equation offered
previously.

The circularity of Dr. Ken Mellendorf's foggy writing might suggest
it, if it weren't otherwise nipped in the bud by the bald statement.
"For a particle with no mass, the relation reduces to E=pc.
This works for a photon."

Hence the proximity of this to p=mv is textual, not factual.

What is the term p? Could it be (p)hoton? I've speculated about
Planck's constant (which you comment upon, below), but I find it very
sloppy writing for Dr. Mellendorf to wander into his own naming
conventions. Migrating through
E = mc²
something all can agree is a fair basis to begin with, we then have
expressly for a (p)hoton:
E = pc
Substituting for the previous E
pc = mc²
divide both sides by c
p = mc
which to me is new territory. What is mass times the speed of light
for a particle that has no mass?

Perhaps Tom's special reading skills can rescue this p term from the
oblivion of E = 0 for a (p)hoton.

But we usually relate more directly to the frequency (or wavelength)
of the photon rather than its energy or momentum, so in such a case
E=h*nu would provide a more direct route to its mass equivalent.

Yes, and it seems your daughter trumped me on Planck once before. ;-)

It is exceedingly obvious that the link offered amounts to
considerable wool gathering. Or maybe its the late hour....

73's
Richard Clark, KB7QHC