# 13.4: Relativistic Energy

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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 $$v = p = 0$$).

This page titled 13.4: Relativistic Energy is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Timon Idema (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.