anantchowdhary said:
A photon's energy is defined using the frequency of the EM wave.So if a photon is at rest,it will have no energy...and as a photon is a packet of energy...it ceases to exist at rest...therefore it CANNOT exist at rest
i'm one of the holdouts that do not like calling photons "massless" without at least a little qualification. photons have energy and photons have momentum. i don't think anyone disagrees with that. so, it depends on how one defines mass. if you say that the mass of a particle or object is the same m as in
E = m c^2
then, since E = h \nu, photons have an effective mass ("relativistic mass", whatever term they deprecate) of
m = \frac{E}{c^2} = \frac{h \nu}{c^2}
they travel at a speed of c relative to any observer, so their momentum is
p = m v = m c = \frac{h \nu}{c}
(this is the assumed definition of "effective mass" being the momentum of an object, relative to an observer, divided by the velocity of the object, relative to the same observer.)
for an object of velocity, v, (relative to some observer), the relationship this "relativistic mass", "effective mass", "inertial mass" m_0 (whatever term the "massless" folks want to see go away) and the rest mass (or invariant mass) is:
m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}
you will notice that if m_0 > 0, then if v approaches c, this mass goes to infinity. rearranging this:
m_0 = m \sqrt{1 - \frac{v^2}{c^2}}
so if m is finite (in fact it's m = (h \nu)/c^2), then when v=c, you can see that m_0 = 0 no matter what the inertial mass is.
