# 15.24: Kinetic Energy

- Page ID
- 8477

If a force \( \bf{F}\) acts on a particle moving with velocity \( \bf{u}\), the rate of doing work – i.e. the rate of increase of kinetic energy \( T\) is \( \dot{T}=\bf{F\cdot u}\). But \( \bf{F=\dot{p}}\) where \( \bf{p}=m\bf{u}=\gamma m_{0}u\).

(A point about notation may be in order here. I have been using the symbol \( \bf{v}\) and \( v\) for the velocity and speed of a frame \( \Sigma'\) relative to a frame \( \Sigma\), and my choice of axes without significant loss of generality has been such that \( \bf{v}\) has been directed parallel to the \( x\)-axis. I have been using the symbol \( \bf{u}\) for the velocity (speed = \( u\)) of a particle relative to the frame \( \Sigma\). Usually the symbol \( \gamma\) has meant \( \left(1-\frac{v^{2}}{c^{2}}\right)^{-\frac{1}{2}}\), but here I am using it to mean \( \left(1-\frac{u^{2}}{c^{2}}\right)^{-\frac{1}{2}}\). I hope that this does not cause too much confusion and that the context will make it clear. I toyed with the idea of using a different symbol, but I thought that this might make matters worse. Just be on your guard, anyway.)

We have, then

\[ \bf{F}=m_{0}(\dot{\gamma}u+\gamma\dot{u}) \label{15.24.1}\]

and therefore

\[ \dot{T}=m_{0}(\dot{\gamma}u^{2}+\gamma\bf{\dot{u}\cdot u}). \label{15.24.2}\]

Making use of Equations 15.23.5 and 15.23.6 we obtain

\[ \dot{T}=\dot{\gamma}m_{0}c^{2} \label{15.24.3}\]

Integrate with respect to time, with the condition that when \( \gamma\) = 1, \( T\) = 0, and we obtain the following expression for the kinetic energy:

\[ T=(\gamma-1)m_{0}c^{2}. \label{15.24.4}\]

*Exercise. *Expand \( \gamma\) by the binomial theorem as far as \( \frac{u^{2}}{c^{2}}\), and show that, to this order, \( T=\frac{1}{2}mu^{2}\).

I here introduce the dimensionless symbol

\[ K=\frac{T}{m_{0}c^{2}}=\gamma-1 \label{15.24.5}\]

to mean the kinetic energy in units of \( m_{0}c^{2}\). The second half of this was already given as Equation 15.3.5.