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# 15.24: Kinetic Energy

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

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.