College-level introductory physics textbooks usually devote a chapter to special relativity. Peter J. Riggs in his article appearing in the February 2016 issue of The Physics Teacher (pp 80-82) derives a couple of expressions for the kinetic energy of a massive (as opposed to massless) particle that I find very useful. I don't recall having seen them in those textbooks, and Riggs claims that they rarely do. They are, I think, time-savers for those of us trying to teach relativity in this course.(adsbygoogle = window.adsbygoogle || []).push({});

The first is ##T=\frac{p^2}{(\gamma+1)m}## and the second is ##T=(\frac{\gamma^2}{\gamma+1})mv^2##.

In these expressions ##T## is the kinetic energy, ##p## is the magnitude of the 3-momentum, ##m## is the mass, ##v## is the speed, and ##\gamma## is the relativistic factor defined by $$\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}.$$

It's easy for students to see that in the low speed limit ##\gamma \approx 1## and that these expression then reduce to the familiar newtonian versions: ##T=\frac{p^2}{2m}## and ##T=\frac{1}{2}mv^2## respectively. There is therefore no need to carry out the series expansion that is usually done to demonstrate the latter.

To derive these expressions start with the familiar expression ##E^2=(pc)^2+(mc^2)^2##.

Therefore ##p^2c^2=E^2-m^2c^4=(E-mc^2)(E+mc^2)=T(E+mc^2)=T(\gamma mc^2+mc^2)=Tmc^2(\gamma+1)##.

Dividing both sides by ##c^2## gives ##p^2=Tm(\gamma+1)##. Solve for ##T## and you have the first expression. Replace ##p## with ##\gamma mv## and you get the second expression.

Edit: Fixed the missing square root in the expression for ##\gamma##. Thanks!

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# Kinetic Energy and Momentum of a Relativistic Particle

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