Suppose that the force
, that acts on the particle discussed in the previous section, causes
the particle to displace a distance
. The net work done on the particle is clearly
![$\displaystyle dW = {\bf f}\cdot d{\bf r} = f\,dx= \frac{d(m\,v)}{dt}\,dx= v\,d(m\,v),$](img2600.png) |
(3.164) |
because, by definition,
. (See Section 1.3.2.)
Here, use has been made of Equations (3.160) and (3.161).
However,
![$\displaystyle m = \gamma\,m_0 = \left(1-\frac{v^2}{c^2}\right)^{-1/2}\,m_0,$](img2601.png) |
(3.165) |
so
![$\displaystyle v = c\left(1-\frac{m_0^{\,2}}{m^2}\right)^{1/2}.$](img2602.png) |
(3.166) |
The previous equation can be combined with Equation (3.164) to give
Suppose that the particle is initially at rest, so that its initial relativistic mass is
. Let the
force perform net work
on the particle, in the process causing its relativistic mass to increase to
.
It is clear from the previous equation that
![$\displaystyle W= (m-m_0)\,c^2.$](img2609.png) |
(3.168) |
However, we know that the net work that a force does on a particle causes the particle's kinetic energy,
, 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
![$\displaystyle K = (m-m_0)\,c^2$](img2610.png) |
(3.169) |
when its relativistic mass is
.
Equation (3.169) can be combined with Equation (3.165) to give
![$\displaystyle K = m_0\,c^2\left[\left(1-\frac{v^2}{c^2}\right)^{-1/2}-1\right].$](img2611.png) |
(3.170) |
In the limit that the particle is moving at a non-relativistic speed, such that
, the previous equation
reduces to
![$\displaystyle K \simeq m_0\,c^2\left[\left(1+ \frac{1}{2}\,\frac{v^2}{c^2}+\cdots\right)-1\right],$](img2612.png) |
(3.171) |
or
![$\displaystyle K= \frac{1}{2}\,m_0\,v^2.$](img2613.png) |
(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.)