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