Equivalence of mass and energy
The above observation led Einstein to the suggestion that the mass of a body can be expressed more simply than by the formula (15.1), if we say that the mass is equal to the total energy content divided by $c^2$. If Eq. (15.11) is multiplied by $c^2$ the result is \begin{equation} \label{Eq:I:15:12} mc^2=m_0c^2+\tfrac{1}{2}m_0v^2+\dotsb \end{equation} Here, the term on the left expresses the total energy of a body, and we recognize the last term as the ordinary kinetic energy. Einstein interpreted the large constant term, $m_0c^2$, to be part of the total energy of the body, an intrinsic energy known as the “rest energy.” Let us follow out the consequences of assuming, with Einstein, that the energy of a body always equals $mc^2$. As an interesting result, we shall find the formula (15.1) for the variation of mass with speed, which we have merely assumed up to now. We start with the body at rest, when its energy is $m_0c^2$. Then we apply a force to the body, which starts it moving and gives it kinetic energy; therefore, since the energy has increased, the mass has increased—this is implicit in the original assumption. So long as the force continues, the energy and the mass both continue to increase. We have already seen (Chapter 13) that the rate of change of energy with time equals the force times the velocity, or \begin{equation} \label{Eq:I:15:13} \ddt{E}{t}=\FLPF\cdot \FLPv. \end{equation} We also have (Chapter 9, Eq. 9.1) that $F = d(mv)/dt$. When these relations are put together with the definition of $E$, Eq. (15.13) becomes \begin{equation} \label{Eq:I:15:14} \ddt{(mc^2)}{t}=\FLPv\cdot\ddt{(m\FLPv)}{t}. \end{equation} We wish to solve this equation for $m$. To do this we first use the mathematical trick of multiplying both sides by $2m$, which changes the equation to \begin{equation} \label{Eq:I:15:15} c^2(2m)\,\ddt{m}{t}=2m\FLPv\cdot\ddt{(m\FLPv)}{t}. \end{equation} We need to get rid of the derivatives, which can be accomplished by integrating both sides. The quantity $(2m)\,dm/dt$ can be recognized as the time derivative of $m^2$, and $(2m\FLPv)\cdot d(m\FLPv)/dt$ is the time derivative of $(mv)^2$. So, Eq. (15.15) is the same as \begin{equation} \label{Eq:I:15:16} c^2\,\ddt{(m^2)}{t}=\ddt{(m^2v^2)}{t}. \end{equation} If the derivatives of two quantities are equal, the quantities themselves differ at most by a constant, say $C$. This permits us to write \begin{equation} \label{Eq:I:15:17} m^2c^2=m^2v^2+C. \end{equation} We need to define the constant $C$ more explicitly. Since Eq. (15.17) must be true for all velocities, we can choose a special case where $v= 0$, and say that in this case the mass is $m_0$. Substituting these values into Eq. (15.17) gives \begin{equation*} m_0^2c^2=0+C. \end{equation*} We can now use this value of $C$ in Eq. (15.17), which becomes \begin{equation} \label{Eq:I:15:18} m^2c^2=m^2v^2+m_0^2c^2. \end{equation} Dividing by $c^2$ and rearranging terms gives \begin{equation} m^2(1-v^2/c^2)=m_0^2,\notag \end{equation} from which we get \begin{equation} \label{Eq:I:15:19} m=m_0/\sqrt{1-v^2/c^2}. \end{equation} This is the formula (15.1), and is exactly what is necessary for the agreement between mass and energy in Eq. (15.12).  Ordinarily these energy changes represent extremely slight changes in mass, because most of the time we cannot generate much energy from a given amount of material; but in an atomic bomb of explosive energy equivalent to $20$ kilotons of TNT, for example, it can be shown that the dirt after the explosion is lighter by $1$ gram than the initial mass of the reacting material, because of the energy that was released, i.e., the released energy had a mass of $1$ gram, according to the relationship $\Delta E=\Delta(mc^2)$. This theory of equivalence of mass and energy has been beautifully verified by experiments in which matter is annihilated—converted totally to energy: An electron and a positron come together at rest, each with a rest mass $m_0$. When they come together they disintegrate and two gamma rays emerge, each with the measured energy of $m_0c^2$. This experiment furnishes a direct determination of the energy associated with the existence of the rest mass of a particle.