Question: Write out the total energy of a collection of interacting massive objects, break each term into internal and kinetic parts, and show that equation (2-27) that is $\mathit{\Delta }\mathit{K}\mathit{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathit{\Delta }\mathit{m}{\mathit{c}}^{\mathbf{2}}$ follows.

The total energy of the system is the sum of kinetic energy and internal energy of the system.

It is given by the formula

$T=KE+U$ ….. (1)

Here, is the total energy, is the kinetic energy, and is the potential energy.

The relativistic kinetic energy is

$KE=\left({\gamma }_u-1\right)m_o{c}^2$ ….. (2)

Here, is the relativistic factor, is the rest mass, and is the speed of light,

$U=m_o{c}^2$

And internal energy is just the rest mass of the particle that is

role="math" localid="1659089805123" $U=m_o{c}^2$ ….. (3)

Therefore, the system's total energy can be given by substituting equation (2) and (3) into (1).

$T=\left({\gamma }_u-1\right)m_o{c}^2+m_o{c}^2$

Consider a collection of massive objects interacting with each other. The system's total energy is conserved. Thus, you can write,

$$\begin{gathered} T_f-T_i=0 \\ \sum \left[\left({\gamma }_{u_f}-1\right)m_f{c}^2+m_f{c}^2\right]-\sum \left[\left({\gamma }_{u_i}-1\right)m_i{c}^2+m_i{c}^2\right]=0 \\ \sum \left[\left({\gamma }_{u_f}-1\right)m_f-\left({\gamma }_{u_i}-1\right)m_i\right]c^2+\sum \left(m_f-m_i\right)c^2=0 \\ \Delta KE+\Delta m{c}^2=0 \end{gathered}$$

Hence, we get the expression for the change in Kinetic energy of the system

$\Delta KE=-\Delta m{c}^2$

Thus, the equation holds for a collection of massive objects interacting with each other. Any change in kinetic energy results from the change in mass.