# Kinetic Energy and Momentum of a Relativistic Particle

1. Dec 4, 2016

### Mister T

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.

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!

Last edited: Dec 12, 2017
2. Dec 5, 2016

### robphy

Riggs' article [thanks @Mister T for making this known]
is available at
A Comparison of Kinetic Energy and Momentum in Special Relativity and Classical Mechanics
Peter J. Riggs
Phys. Teach. 54, 80 (2016)
http://scitation.aip.org/content/aapt/journal/tpt/54/2/10.1119/1.4940169 [appears to be open access, for now]

Along these lines, there is an older article [not referenced by the above]:
Parallels between relativistic and classical dynamics for introductory courses
Donald E. Fahnline
Am. J. Phys. 43, 492 (1975)
http://scitation.aip.org/content/aapt/journal/ajp/43/6/10.1119/1.9775

I think these are useful, as you say, to more easily show the classical limit of the relativistic expressions.. without calculus or a series expansion.
Unfortunately, these relativistic expressions, as derived in these works, don't "fall out" naturally from first principles... but are arranged to resemble the classical limit. So, the physical intuition is limited to being an algebraic expression relating the two expressions.
To be clear... it's very useful and should be better known, but it is limited.

(By the way, you forgot the square-root in the expression for gamma.)