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.