13.4: Relativistic Energy

The last three (or ‘spatial’) components of the momentum four-vector give us the regular components of the momentum, times the factor $\gamma(v)$. What about the zeroth (or ‘temporal’) component? To interpret it, we expand $\gamma(v)$, and find:

$$\begin{align} c p_{0} &= \gamma(v) m c^{2} \\[4pt] &=\left[1+\frac{1}{2}\left(\frac{v}{c}\right)^{2}+\frac{3}{8}\left(\frac{v}{c}\right)^{4}+\ldots\right] m c^{2} \\[4pt] &=m c^{2}+\frac{1}{2} m v^{2}+\ldots \end{align}$$

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:

$$\begin{align} E&=\gamma(v) m c^{2} \\[4pt] &=m c^{2}+K \label{13.4.1} \end{align}$$

We can now write the zeroth component of the momentum four-vector as $p_{0}=E / c$. 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)

$$\overline{\boldsymbol{p}} \cdot \overline{\boldsymbol{p}}=m^{2} \overline{\boldsymbol{v}} \cdot \overline{\boldsymbol{v}}=m^{2} c^{2} \label{13.4.3}$$

while on the other hand, we could also simply expand in the components of $\overline{\boldsymbol{p}}$ itself to get:

$$\overline{\boldsymbol{p}} \cdot \overline{\boldsymbol{p}}=\left(\frac{E}{c}\right)^{2}-\boldsymbol{p} \cdot \boldsymbol{p} \label{13.4.4}$$

where $\boldsymbol{p}$ is again the spatial part of $\overline{ \boldsymbol{p}}$. Combining Equations $\ref{13.4.3}$ and $\ref{13.4.4}$, we get:

$$E^{2}=m^{2} c^{4}+p^{2} c^{2} \label{13.4.5}$$

where $p^{2}=\boldsymbol{p} \cdot \boldsymbol{p}$. Equation $\ref{13.4.5}$ is the general form of Einstein's famous formula $E = m c^{2}$, to which it reduces for stationary particles (i.e. when $v = p = 0$).