4 ms·
This doesn't really answer your question, but there is already an extra term! (only sometimes it is zero) E^2 = (mc^2)^2 + (lc)^2
by tux3 3y ago
This doesn't really answer your question, but there is already an extra term! (only sometimes it is zero)
E^2 = (mc^2)^2 + (lc)^2
- spanglasaur 3y agoI was going to comment this as well. Here is a thorough explanation of the missing term: https://physics.stackexchange.com/questions/143652/is-e2-mc22pc2-or-is-e-mc2-the-correct-one https://physics.stackexchange.com/questions/143652/is-e2-mc2... The thing that I love about this formulation of the equation is that it directly ties it to the Pythagorean theorem: c^2 = a^2 + b^2 . It really shows you just how pretty that equation really is!
- lamontcg 3y agothe m in that equation is the rest mass (m_0) which is a proper scalar. the m in E=mc^2 is the relativistic mass which depends on rest frame and therefore isn't a proper scalar quantity and that isn't a proper tensor equation. the (E, px, py, pz) energy-momentum 4-vector is a tensor that transforms to any reference frame with a lorentz transform. all observers will agree on the shape of that 4-vector object, although they'll measure the components differently. that equation is just the dot product (rearranged), where the rest mass is the length of the 4-momentum vector (and a proper scalar quantity that all observers agree on is exactly that value in all reference frames). the rearrangement introduces a negative sign which actually comes from the SR metric -- usually diag(-1, +1, +1, +1) but sometimes diag(+1, -1, -1, -1).