13.4: Relativistic Energy

Last updated
Save as PDF

Page ID: 17448

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

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 \(v = p = 0 \)).