what is the actual equation of e=mc^2? this is only the simplified equation, and i have forgotten the actual one already...
It's a lowercase p, and it stands for (linear) momentum. I don't know what you mean by "homogenous." - Warren
It all depends on whether you are a massist or an energist. ------ A massist is willing to attribute mass values to anything, in any state of motion. For a massist, this m is really m_{0}, a mass attributed to something in its rest frame of reference. For a massist E^{2} = p^{2}c^{2} + m_{0}^{2}c^{4} p = mv are always true in any inertial frame. For light quanta, E = pc p = mc , because m_{0} = 0 for light quanta. But m = p/c = E/c^{2}, a mass value dependent upon total energy of a quantum. So E = mc^{2} is true for a light quantum as well as a particle with a non-zero rest mass. ------ An energist is willing to attribute energy values to anything, in any state of motion. For an energist, m can only be attributed to something in its rest frame, so the subscript 0 is never needed. For an energist, p^{2} = E^{2}/c^{2} - m^{2}c^{2} is always true in any inertial frame. The energy E must come from other physics. For light quanta, p = E/c is a given, so p^{2} = p^{2} - m^{2}c^{2} , so m^{2}c^{2} = 0 . Since c > 0, m = 0 for a light quantum. So, E = mc^{2}/(1 - v^{2}/c^{2})^{1/2} only in the case of a particle with non-zero rest mass. ------ Most modern day physicists, especially high-energy physicists, tend to be energists rather than massists.
I said: So, E = mc^{2}/(1 - v^{2}/c^{2})^{1/2} only in the case of a particle with non-zero rest mass. I should have said: So, E = mc^{2}/(1 - v^{2}/c^{2})^{1/2} only in the case of a particle with non-zero mass.
Re: e=mc^2 The equation E = mc^{2} is the mass-energy equation relating the mass m of a particle to the free-particle energy E. The proof can be found here www.geocities.com/physics_world/sr/mass_energy_equiv.htm If the particle is a tardyon (i.e. a particle which travels at speeds less than light) then m = m_{0}/sqrt[1-(v/c)^{2}] Multiply both sides by c^{2} mc^{2} = m_{0}c^{2}/sqrt[1-(v/c)^{2}] Substitute in E = mc^{2} to get E = m_{0}c^{2}/sqrt[1-(v/c)^{2}] This equation can be rewritten as E^{2} - (pc)^{2} = (m_{0}c^{2})^{2} Pete