Work and Energy

Suppose that the force ${\bf f} = f\,{\bf e}_x$, that acts on the particle discussed in the previous section, causes the particle to displace a distance $d{\bf r}=dx\,{\bf e}_x$. The net work done on the particle is clearly

$\displaystyle dW = {\bf f}\cdot d{\bf r} = f\,dx= \frac{d(m\,v)}{dt}\,dx= v\,d(m\,v),$ (3.164)

because, by definition, $v=dx/dt$. (See Section 1.3.2.) Here, use has been made of Equations (3.160) and (3.161). However,

$\displaystyle m = \gamma\,m_0 = \left(1-\frac{v^2}{c^2}\right)^{-1/2}\,m_0,$ (3.165)

so

$\displaystyle v = c\left(1-\frac{m_0^{\,2}}{m^2}\right)^{1/2}.$ (3.166)

The previous equation can be combined with Equation (3.164) to give

$\displaystyle dW$ $\displaystyle = c^2\left(1-\frac{m_0^{\,2}}{m^2}\right)^{1/2}d\!\left[m\left(1-\frac{m_0^{\,2}}{m^2}\right)^{1/2}\right]$    
  $\displaystyle = c^2\left(1-\frac{m_0^{\,2}}{m^2}\right)^{1/2}d\!\left(\!\!\sqrt{m^2-m_0^{\,2}}\right)$    
  $\displaystyle =c^2\left(1-\frac{m_0^{\,2}}{m^2}\right)^{1/2}\frac{m\,dm}{\sqrt{m^2-m_0^{\,2}}}$    
  $\displaystyle = c^2\,dm.$ (3.167)

Suppose that the particle is initially at rest, so that its initial relativistic mass is $m_0$. Let the force perform net work $W$ on the particle, in the process causing its relativistic mass to increase to $m$. It is clear from the previous equation that

$\displaystyle W= (m-m_0)\,c^2.$ (3.168)

However, we know that the net work that a force does on a particle causes the particle's kinetic energy, $K$, to increase by a corresponding amount. (See Section 1.3.2.) Thus, given that the particle's initial kinetic energy is zero, we deduce that its kinetic energy is

$\displaystyle K = (m-m_0)\,c^2$ (3.169)

when its relativistic mass is $m$.

Equation (3.169) can be combined with Equation (3.165) to give

$\displaystyle K = m_0\,c^2\left[\left(1-\frac{v^2}{c^2}\right)^{-1/2}-1\right].$ (3.170)

In the limit that the particle is moving at a non-relativistic speed, such that $v/c\ll 1$, the previous equation reduces to

$\displaystyle K \simeq m_0\,c^2\left[\left(1+ \frac{1}{2}\,\frac{v^2}{c^2}+\cdots\right)-1\right],$ (3.171)

or

$\displaystyle K= \frac{1}{2}\,m_0\,v^2.$ (3.172)

This is consistent with the Newtonian definition of kinetic energy, as long as we identify the rest mass of the particle with its mass in Newtonian dynamics. (See Section 1.3.2.)