36.10.1 Rest Energy and Total Energy

With that as a possibility, subject to experimental verification, let’s define the total energy E of a particle to be

E = γ_{p} m c^{2} = E_{0} + K = rest energy + kinetic energy

(36.42)

This total energy consists of a rest energy

E_{0} = m c^{2}

(36.43)

and a relativistic expression for the kinetic energy

K = (γ_{p} - 1) m c^{2} = (γ_{p} - 1) E_{0}

(36.44)

This expression for the kinetic energy is very nearly $\frac{1}{2} m u^{2}$ when $u ≪ c$ but, as FIGURE 36.36 shows, differs significantly from the classical value for very high velocities.

The figure illustrates the concept of relativistic and Newtonian kinetic energies graphically.

FIGURE 36.36 Full Alternative Text

Equation 36.43 is, of course, Einstein’s famous $E = m c^{2}$ , perhaps the most famous equation in all of physics. Before discussing its significance, we need to tie up some loose ends. First, we can use Equations 36.42 and 36.43 to rewrite Equation 36.39 as $E^{2} - {(p c)}^{2} = E_{0}^{2}$ . This is easily rearranged to give the useful result

E = \sqrt{E_{0}^{2} + {(p c)}^{2}}

(36.45)

The quantity $E_{0}$ is an invariant with the same value $m c^{2}$ in all inertial reference frames.

Second, notice that we can write

p c = (γ_{p} m u) c = \frac{u}{c} (γ_{p} m c^{2})

But $γ_{p} m c^{2}$ is the total energy E and $u / c = β_{p}$ , where the subscript p, as on $γ_{p}$ , indicates that we’re referring to the motion of a particle within a reference frame, not the motion of two reference frames relative to each other. Thus

p c = β_{p} E

(36.46)

FIGURE 36.37 shows the “velocity-energy-momentum triangle,” a convenient way to remember the relationships among the three quantities.

The figure illustrates the velocity-energy-momentum triangle.

FIGURE 36.37 Full Alternative Text

EXAMPLE 36.11 Kinetic energy and total energy

Calculate the rest energy and the kinetic energy of (a) a 100 g ball moving with a speed of 100 m/s and (b) an electron with a speed of 0.999c.

MODEL The ball, with $u ≪ c$ , is a classical particle. We don’t need to use the relativistic expression for its kinetic energy. The electron is highly relativistic.

SOLVE

For the ball, with $m = 0.10 kg,$

$\begin{array}{l} E_{0} & = & m c^{2} = 9.0 \times 10^{15} J \\ K & = & \frac{1}{2} m u^{2} = 500 J \end{array}$
For the electron, we start by calculating

$γ_{p} = \frac{1}{{(1 - u^{2} / c^{2})}^{1 / 2}} = 22.4$

Then, using $m_{e} = 9.11 \times 10^{- 31} kg,$ we find

$\begin{array}{l} E_{0} & = & m c^{2} = 8.2 \times 10^{- 14} J \\ K & = & (γ_{p} - 1) E_{0} = 170 \times 10^{- 14} J \end{array}$

REVIEW The ball’s kinetic energy is a typical kinetic energy. Its rest energy, by contrast, is a staggeringly large number. For a relativistic electron, on the other hand, the kinetic energy is more important than the rest energy.

36.10.1 Rest Energy and Total Energy

FIGURE 36.36 The relativistic kinetic energy.

FIGURE 36.37 The velocity-energy-momentum triangle.

EXAMPLE 36.11 Kinetic energy and total energy