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