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. :-)

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.
--
73, Cecil http://www.w5dxp.com

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

Tom Donaly wrote:
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?

The percentage difference between zero and that momentum
is infinite. And whatever value it is must be conserved.
Sweeping it under the rug in violation of the laws of
physics is just not acceptable.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Tom Donaly wrote:
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?

The percentage difference between zero and that momentum
is infinite. And whatever value it is must be conserved.
Sweeping it under the rug in violation of the laws of
physics is just not acceptable.

Actually, you're writing about momentum density. Momentum is
conserved, but momentum density isn't, any more than energy
density, or any other kind of density, with the possible
exception of the bone density in the heads of some people.
As for any finite number being an infinite percentage above
zero, I think you should take that up with the next mathematician
you meet. Mathematicians need to laugh once in a while, too.
73,
Tom Donaly, KA6RUH

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

Cecil Moore wrote:
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.

Division by zero is not infinity, Cecil, it's undefined.
It's good to see you agree that there's no conservation of
______ (fill in the blank)density, any more than there's a
law of the conservation of power. Have a good thanksgiving.
73,
Tom Donaly, KA6RUH

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.

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

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.