13.4: Relativistic Energy
( \newcommand{\kernel}{\mathrm{null}\,}\)
The last three (or ‘spatial’) components of the momentum four-vector give us the regular components of the momentum, times the factor γ(v). What about the zeroth (or ‘temporal’) component? To interpret it, we expand γ(v), and find:
cp0=γ(v)mc2=[1+12(vc)2+38(vc)4+…]mc2=mc2+12mv2+…
The second term in this expansion should be familiar: it’s the kinetic energy of the particle. The third and higher terms are corrections to the classical kinetic energy - just like the higher-order terms in the spatial components are corrections to the classical momenta. The first term, however, is new: an extra energy contribution due to the mass of the particle. The whole term can now be interpreted as the relativistic energy of the particle:
E=γ(v)mc2=mc2+K
We can now write the zeroth component of the momentum four-vector as p0=E/c. Based on this interpretation, the four-vector is sometimes referred to as the energy-momentum four-vector.
A very useful relation can now easily be derived by calculating the length of the energy-momentum four-vector in two ways. On the one hand, it’s given by (leaving out the square root for convenience)
¯p⋅¯p=m2¯v⋅¯v=m2c2
while on the other hand, we could also simply expand in the components of ¯p itself to get:
¯p⋅¯p=(Ec)2−p⋅p
where p is again the spatial part of ¯p. Combining Equations ??? and ???, we get:
where p2=p⋅p. Equation ??? is the general form of Einstein's famous formula E=mc2, to which it reduces for stationary particles (i.e. when v=p=0).