selfAdjoint said:
Two answers, which are consistent with one another:
1. If I look at an object which is moving with a velocity v with respect to me, and somehow I measure its energy, in my frame, then its energy will be [tex]\gamma E[/tex], where E is its energy in its own frame, and [tex]\gamma = \frac {1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]. So if I equate this to [tex]mc^2[/tex] then [tex]\gamma E = mc^2[/tex] or [tex]E = \frac{mc^2}{\gamma}[/tex].
the notation we used when i was in school 3 decades ago was
[tex]E = T + E_0 = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}} = m c^2[/tex]
where [itex]T = E - E_0[/itex] is the kinetic energy (which depends on whose frame of reference you are in), [itex]E_0 = m_0 c^2[/itex] is the rest energy (the "energy in its own frame") and [itex]m_0[/itex] is the rest mass or the apparent mass in the frame of the body (also called "invariant mass"). we also were presented with the out of vogue concept of "relativistic mass"
[tex]m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \ge m_0[/tex]
so [tex]E = m c^2[/tex]
can hold in all cases with [itex]E[/itex] and [itex]m[/itex] both having corresponding meaning in the same frame. if [itex]m[/itex] is rest mass, than [itex]E[/itex] is rest energy. we were able to equate this [itex]E[/itex] (
total energy, the sum of kinetic and rest energies) to the same [itex]E[/itex] that relates particle energy to wave function frequency:
[tex]E = h \nu = \hbar \omega[/tex]
resulting in an apparent inertial mass of
[tex]m = \frac{E}{c^2} = \frac{h \nu}{c^2}[/tex]
and given that relativistic mass we always defined momentum to be
[tex]p = mv[/tex]
and assuming that photons always move at a speed of [itex]c[/itex] for any reference frame, then the momentum was simply
[tex]p = mv = \frac{h \nu}{c^2} v = \frac{h \nu}{c}[/tex]
and we got the same expression (keeping notational consistency):
[tex]E^2 = p^2 c^2 + m_0^2 c^4[/tex]
but this notation has evidently been frowned on in current times. now [itex]m[/itex] always seems to mean only "rest mass" (what we used to call "[itex]m_0[/itex]") and there is no "relativistic mass" to talk about.
2. The general law of energy for particles in SR is
[tex]E^2 = p^2c^2 + m^2c^4[/tex] (p is the magnitude of the three momentum), so for a massless particle (m=0 in its own frame), the second term drops out and we would have [tex]E = pc[/tex]. On the other hand, for a massive particle in its own frame, the first term drops out (in its own frame its momentum is zero), and we have [tex]E = mc^2[/tex].
Both answers give you the same info: the famous equation refers to a massive particle in its own rest frame. In all other cases the law is subject to relativistic corrections. This is not at all academic, either. Particle physicsts have to use these relatiionships in calcualting the effects of the fast moving paricles in their experiments.
we didn't call photons "massless", we said only that their rest mass was 0.
