pmb_phy said:
A very odd way of looking at it. Something seems wrong but I can't quite put my finger on it.
i was involved in an argument about this (at least about the pedagogy of this) in that other
photon mass thread. seems as if only myself and
bernhard.rothenstein look at it this way and i am convinced it is because of the time that we took our first physics courses (30 years ago for me) and something changed in what was the fashionable pedagogy since then.
it sure wasn't a very odd way of looking at it then. perhaps you can find a copy of Arthur Beiser
Concepts of Modern Physics 2
nd edition, © about 1974. this was precisely the approach taken in that book. they
never said that photons are massless particles, only that they have no rest mass (for the reason stated in my post). the expressions
E^2 = \left( m_0 c^2 \right)^2 + (p c)^2
nor
p = \frac{h \nu}{c}
were never stated as first principles but were derived from
m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}
E = m c^2
p = m v (where m is relativistic mass)
from these three equations, we can derive
E^2 = \left( m_0 c^2 \right)^2 + (p c)^2.
and for photons,
E = h \nu = \hbar \omega .
from this and the 3 previous "axiomatic" equations immediately above, and from the presumption that the particles that manifest the particle-like properties of light have the same velocity as of the E&M wave, that is v = c for photons, you get:
m_0 = 0
m = \frac{E}{c^2} = \frac{h \nu}{c^2} (
not rest mass)
and p = m v = m c = \frac{E}{c^2} c = \frac{h \nu}{c} .
i would still like to see a cogent explanation, from first principles, on how to derive
E^2 = \left( m_0 c^2 \right)^2 + (p c)^2
for the general body of invariant mass m_0 and, for photons,
p = \frac{h \nu}{c}
without a concept differentiating rest mass from relativistic mass.
Photons with energy greater than zero are not "massless" particles, but they have no rest mass (or "invariant mass") because their velocity is (practically by definition) that of light.