hello friends, i have heard that E=mc2 is incomplete.. That is e2=(mc2)2+ (pc)2. But i haven't found the proof... Can you tell me the proof??

In order to get the units to match (ie both sides have units of energy) it is necessary to multiply mass by speed squared and for momentum just a single factor of speed. The reason it is c and not some other speed is that \(E^{2} = (mc^{2})^{2} + (pc)^{2}\) is a Lorentz invariant expression, which means that when you apply a Lorentz transform to the vector (E,p), mapping it to new energy and momentum (E',p') it is necessary that \(-E^{2} + (pc)^{2} = (mc^{2})^{2} = -(E')^{2} + (p'c)^{2}\). Lorentz transforms leave the speed of light unchanged and that is why c arises in this expression.

Have you seen Einstein's paper, where he has derived his equation e=mc*c? In that paper Einstein considered a 'rest mass' in a rest frame and another moving frame with a relative velocity. Here he considered that the rest mass is being converted into 'Light Energy'. That is 'rest energy'. If the mass is not at rest in the rest frame, it will be having some additional 'mechanical energy'. So, when the mass in motion is converted into Light-Energy, it will be having some additional energy than 'rest-energy'. Thats why additional term 'pc' is incorporated in the modified energy equation to account for the additional energy.

Basically... E2=(mc2)2+(pc)2 means... Energy sqaured is equals to Mass (m) x Speed of light (c) squared Squared plus Momentum (p) x Speed of light (c) squared... E=mc2 is only applicable for objects that has mass but not moving but if there's motion.. meaning there's momentum created by the kinetic energy which stands for P and another speed f light in which you see the object moving.. Please Register or Log in to view the hidden image!