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Physics LibreTexts

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)2pp

where p is again the spatial part of ¯p. Combining Equations ??? and ???, we get:

E2=m2c4+p2c2

where p2=pp. 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).


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.

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