Relativistic Energy and Momentum

We know classically that kinetic energy and momentum are related to each other, since

\({\text{KE}}_{\text{class}}=\cfrac{{p}^{2}}{2m}=\cfrac{(\mathrm{mv}{)}^{2}}{2m}=\cfrac{1}{2}{\mathrm{mv}}^{2}.\)

Relativistically, we can obtain a relationship between energy and momentum by algebraically manipulating their definitions. This produces

\({E}^{2}=(\text{pc}{)}^{2}+({\text{mc}}^{2}{)}^{2},\)

where \(E\) is the relativistic total energy and \(p\) is the relativistic momentum. This relationship between relativistic energy and relativistic momentum is more complicated than the classical, 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 \({\mathrm{mc}}^{2}\) (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 \(({\mathrm{mc}}^{2}{)}^{2}\) becomes negligible compared with the momentum term \((\mathrm{pc}{)}^{2}\); thus, \(E=\mathrm{pc}\) at extremely relativistic velocities.

If we consider momentum \(p\) to be distinct from mass, we can determine the implications of the equation \({E}^{2}=(\text{pc}{)}^{2}+({\text{mc}}^{2}{)}^{2},\) for a particle that has no mass. If we take \(m\) to be zero in this equation, then \(E=\text{pc}\), or \(p=E/c\). Massless particles have this momentum. There are several massless particles found in nature, including photons (these are quanta of electromagnetic radiation). Another implication is that a massless particle must travel at speed \(c\) and only at speed \(c\). While it is beyond the scope of this text to examine the relationship in the equation \({E}^{2}=(\text{pc}{)}^{2}+({\text{mc}}^{2}{)}^{2},\) in detail, we can see that the relationship has important implications in special relativity.

Problem-Solving Strategies for Relativity

Examine the situation to determine that it is necessary to use relativity. Relativistic effects are related to \(\gamma =\cfrac{1}{\sqrt{1-\cfrac{{v}^{2}}{{c}^{2}}}}\), the quantitative relativistic factor. If \(\gamma \) is very close to 1, then relativistic effects are small and differ very little from the usually easier classical calculations.
Identify exactly what needs to be determined in the problem (identify the unknowns).
Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Look in particular for information on relative velocity \(v\).
Make certain you understand the conceptual aspects of the problem before making any calculations. Decide, for example, which observer sees time dilated or length contracted before plugging into equations. If you have thought about who sees what, who is moving with the event being observed, who sees proper time, and so on, you will find it much easier to determine if your calculation is reasonable.
Determine the primary type of calculation to be done to find the unknowns identified above. You will find the section summary helpful in determining whether a length contraction, relativistic kinetic energy, or some other concept is involved.
Do not round off during the calculation. As noted in the text, you must often perform your calculations to many digits to see the desired effect. You may round off at the very end of the problem, but do not use a rounded number in a subsequent calculation.
Check the answer to see if it is reasonable: Does it make sense? This may be more difficult for relativity, since we do not encounter it directly. But you can look for velocities greater than \(c\) or relativistic effects that are in the wrong direction (such as a time contraction where a dilation was expected).

Check Your Understanding

A photon decays into an electron-positron pair. What is the kinetic energy of the electron if its speed is \(0.992c\)?

Solution

\(\begin{array}{lll}{\text{KE}}_{\text{rel}}& =& \left(\gamma -1\right){\mathrm{mc}}^{2}=\left(\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-1\right){\mathrm{mc}}^{2}\\ & =& \left(\frac{1}{\sqrt{1-\frac{\left(\text{0.992}c\right)^{2}}{{c}^{2}}}}-1\right)\left(\text{9.11}\times{\text{10}}^{-\text{31}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(\text{3.00}\times{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)^{2}=\text{5.67}\times{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}\end{array}\)

Relativistic Energy and Momentum