Recall that work done by a force gives change in the kinetic energy, \(K_2 - K_1\text{.}\)
\begin{align*}
K_2 - K_1 \amp = W_{12} = \int_1^2 \vec F \cdot d\vec r\\
\amp = \int_1^2 \dfrac{d\vec p}{dt} \cdot d\vec r = \int_1^2 \vec v \cdot d\vec p,
\end{align*}
where I have used \(d\vec r /dt = \vec v\text{,}\) the velocity. When we use relativistic momentum \(\vec p = m_v \vec v\text{,}\) where \(m_v\) is velocity-dependent mass, the differential element will become
\begin{equation*}
d\vec p = d( m_v\vec v ) = \vec v dm_v + m_V d\vec V.
\end{equation*}
Therefore,
\begin{equation}
K_2 - K_1 = \int_1^2 v^2 dm_v + \frac{1}{2} \int_1^2 m_v d v^2,\tag{49.95}
\end{equation}
where I used \(vdv = \frac{1}{2}dv^2\text{.}\) Now from the definition of speed-dependent mass we have
\begin{equation}
m_v = \dfrac{m_0}{\sqrt{1-v^2/c^2}}\ \ \Longrightarrow\ \ m_v^2 c^2 - m_v^2 v^2 = m_0^2 c^2.\tag{49.96}
\end{equation}
Working out the differential of this equation gives
\begin{equation}
d(m_v^2) c^2 - \left[d(m_v^2) u^2 + m_v^2 d(v^2)\right]= 0.\tag{49.97}
\end{equation}
Simplifying this expression yields
\begin{equation}
v^2\: dm_v + m_v\: v\: d v = c^2 dm_v.\tag{49.98}
\end{equation}
Putting this back in Eq.
(49.95) gives rise to a simple integral for the relativistic kinetic energy. We use the limits of integration as
\(m_0\) when
\(u=0\) and
\(m=m\) when
\(u=u\text{.}\) Therefore energy of the particle due to motion, or kinetic energy when it has a speed
\(u\) will be
\begin{equation}
K = c^2 \int_{m_0}^{m_v}dm_v = m_v c^2 - m_0 c^2.\tag{49.99}
\end{equation}
Therefore, the relativistic kinetic energy has the following definition.
\begin{equation}
K = m_v c^2 - m_0 c^2 = m_0 c^2 \left[ \dfrac{1}{\sqrt{1-v^2/c^2}} - 1\right]. \tag{49.100}
\end{equation}
This unfamiliar formula for kinetic energy actually reduces to \(K= \dfrac{1}{2} m_0 v^2\) when \(v \lt c\text{,}\) i.e. in the non-relativistic regime when we expect Newton’s formulas to be applicable. To show this we first replace \(m_v\) by its relativistic expression in this equation.
\begin{align*}
K \amp = \dfrac{m_0}{\sqrt{1-v^2/c^2}} c^2 - m_0 c^2\\
\amp \approx m_0c^2\left[ 1 + \dfrac{1}{2}\: \dfrac{v^2}{c^2} \right] - m_0 c^2 = \dfrac{1}{2} m_0 v^2.
\end{align*}
We notice that when
\(v=0\text{,}\) kinetic energy
\(K\) is zero. That is,
\(K\) is equal to the energy for increasing the speed of particle from zero to
\(v\text{.}\) Rewriting Eq.
(49.100) as
\begin{equation*}
m_vc^2 = K + m_0c^2
\end{equation*}
Einstein claimed that \(m_vc^2\text{,}\) often written as \(mc^2\text{,}\) was the total energy \(E\) of the moving particle with \(m_0 c^2\) being its energy at rest (rest energy) and \(K\) energy added to speed up the particle to speed \(v\text{.}\)
\begin{equation}
E = m_vc^2 = K + m_0c^2,\tag{49.101}
\end{equation}
with
\begin{equation}
\textrm{Rest Energy } = m_0 c^2.\tag{49.102}
\end{equation}
You might say that rest energy represents an innate energy possesed by a particle, thus illustrating that mass and energy are somewhat equivalent. Equation
\(E=mc^2\) is perhaps the most famous formula of physics; it is attributed to its dicovery in Einstein in 1905 although others found similar formulas relating energy and inertia around 1900-1905. See
Scientific American.
