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We know classically that kinetic energy and momentum are related to each other, because:
Relativistically, we can obtain a relationship between energy and momentum by algebraically manipulating their defining equations. This yields:
where E is the relativistic total energy, $E=m{c}^{2}\text{/}\sqrt{1-{u}^{2}\text{/}{c}^{2}},$ and p is the relativistic momentum. This relationship between relativistic energy and relativistic momentum is more complicated than the classical version, but we can gain some interesting new insights by examining it. First, total energy is related to momentum and rest mass. At rest, momentum is zero, and the equation gives the total energy to be the rest energy $m{c}^{2}$ (so this equation is consistent with the discussion of rest energy above). However, as the mass is accelerated, its momentum p increases, thus increasing the total energy. At sufficiently high velocities, the rest energy term ${(m{c}^{2})}^{2}$ becomes negligible compared with the momentum term ${(pc)}^{2};$ thus, $E=pc$ at extremely relativistic velocities.
If we consider momentum p to be distinct from mass, we can determine the implications of the equation ${E}^{2}={(pc)}^{2}+{(m{c}^{2})}^{2},$ for a particle that has no mass. If we take m to be zero in this equation, then $E=pc,\phantom{\rule{0.2em}{0ex}}\text{or}p=E\text{/}c.$ Massless particles have this momentum. There are several massless particles found in nature, including photons (which are packets of electromagnetic radiation). Another implication is that a massless particle must travel at speed c and only at speed c . It is beyond the scope of this text to examine the relationship in the equation ${E}^{2}={(pc)}^{2}+{(m{c}^{2})}^{2}$ in detail, but you can see that the relationship has important implications in special relativity.
Check Your Understanding What is the kinetic energy of an electron if its speed is 0.992 c ?
$\begin{array}{cc}\hfill {K}_{\text{rel}}& =\left(\gamma -1\right)m{c}^{2}=\left(\frac{1}{\sqrt{1-\frac{{u}^{2}}{{c}^{2}}}}-1\right)m{c}^{2}\hfill \\ & =\left(\frac{1}{\sqrt{1-\frac{{(0.992c)}^{2}}{{c}^{2}}}}-1\right)(9.11\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{-31}\phantom{\rule{0.2em}{0ex}}\text{kg}){(3.00\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s})}^{2}=5.67\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{-13}\phantom{\rule{0.2em}{0ex}}\text{J}\hfill \end{array}$
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