The_Thinker said:
how come light does not have mass, we know e=mc^2, so what gives; why is it that light doesn't have mass or rather what is the explanation for light not having mass?
The answer to your question lies in the definition of the "m" in that expression. The definition of "m" is a much discussed topic in most physics forums, this one being no diffferent.
One can define the "m" of a
particle as the "m" in
p =
mv where
p is the particle's 3-momentum and
v is the particle's 3-velocity. The
inertial energy, E, of a particle is defined as the sum of the particle's rest energy, E
0, and the particle's kinetic energy, K. I.e. E = K + E
0. It can then be shown that E = mc
2. However, from the definition above, i.e.
p =
mv, it also follows that anything with momentum has a non-zero "m." If the particle is a tardyon (i.e. a particle that always travels at a speed less than the speed of light) then it can be shown that m is a function of speed, i.e. m = m(v). Let m
0 = m(0). Then it can be shown that
m = \gamma m_0 = \frac{m_0}{\sqrt{1-v^2/c^2}}
and thus
E = mc^2 = \gamma m_0c^2
It can easily be shown that
E^2 - (pc)^2 = m_0^2c^2
Since the magnitude of a photon's momentum is related to its energy as E = pc it follows that m
0 = 0.
Another way to define "m" is to assume the particle is a tardyon and assume that it's momentum can be expressed as
p = M(v)
v. Then we define "m" as m = M(0). Now we have to write
E = \gamma mc^2
and
E^2 - (pc)^2 = m^2c^2
In the particle's rest frame
E = mc^2
As for me? I don't like this whole "photon mass = zero" thingy. The relationship between E, p and for m
0 is derived on the basis that the particle is a tardyon. One then extrapolates for a luxon (particle that always travels at the speed of light) by letting E = pc, which holds for all luxons, in the expression
E^2 - (pc)^2 = m_0^2c^2
it follows that m
0 = 0.
Pete
?, Well, because they exhibit the U(1)-symmetry which is the local symmetry of the EM-interactions and which is the general symmetry of the QED (or QCD)-vacuum. This has nothing to do with gauge-invariance. It is quite the opposite, as a matter of fact, gauge-symmetries are broken in the spontaneous breaking of symmetry.
