15.24: Kinetic Energy

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.