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