relativistic energy momentum and rest energy

In special relativity, the four-momentum of a free particle is $p^\mu = (E/c, \mathbf p)$, and it transforms between inertial frames by the same Lorentz transformation that preserves the space-time interval $s^2 = c^2t^2 - x^2$.

The time component is the total energy $E$ divided by $c$, and the spatial part is relativistic momentum

$$\mathbf p = \gamma m \mathbf v, \qquad \gamma = (1 - v^2/c^2)^{-1/2}.$$

Because $p^\mu p_\mu$ is a Lorentz scalar, the combination

$$E^2 - (pc)^2 = (mc^2)^2$$

is the same in every inertial frame; the quantity $mc^2$ is called the rest energy. Therefore the total energy is $E = \gamma mc^2$, and the kinetic energy is $K = E - mc^2 = (\gamma - 1)mc^2$. For $v \ll c$, expansion gives $K \approx \frac12 mv^2$ and $E \approx mc^2 + p^2/(2m)$, recovering the Newtonian results.

An electron (rest mass $m_e = 9.11\times 10^{-31}\,\mathrm{kg}$, rest energy $m_ec^2 = 0.511\,\mathrm{MeV}$) moves at $v = 0.80c$. Find $\gamma$, the total energy $E$, the momentum $p$, and the kinetic energy $K$.

A proton (rest energy $m_pc^2 = 938\,\mathrm{MeV}$) has kinetic energy $K = 500\,\mathrm{MeV}$. Find its total energy, momentum (in $\mathrm{MeV}/c$), and speed as a fraction of $c$.

A particle of mass $M$, initially at rest, decays into two identical particles of mass $m$, with $m < M/2$. Using conservation of relativistic energy and momentum, find the energy and momentum of each daughter particle, and its speed.

Self-study with a spreadsheet or a browser relativistic-dynamics simulation (e.g., PhET or an equivalent open simulation). Choose a particle mass (electron or proton) and tabulate $\beta$ from 0.01 to 0.99 in steps (e.g., 0.05). For each row compute $\gamma = 1/\sqrt{1-\beta^2}$, $E = \gamma mc^2$, $pc = \gamma \beta mc^2$, $K = E - mc^2$. Check that $E^2-(pc)^2$ is constant and equals $(mc^2)^2$. Then compute the fractional difference $(K - \frac12 mv^2)/K$ and find the smallest tabulated $\beta$ where it exceeds 1%. If a simulation is available, repeat for one momentum value and confirm that the invariant is the same in a moving frame (apply a Lorentz boost to $E/c$ and $p$). Do not use radioactive sources or accelerators; all work is numerical or simulated.

To show that allowed states of a free particle lie on the mass-shell hyperbola and to make visible the invariant $E^2-(pc)^2=(mc^2)^2$, the rest-energy intercept, and the photon limit.

A labelled $E$–$pc$ graph: the hyperbola $E^2-(pc)^2=(mc^2)^2$ in the first quadrant, with intercept $E=mc^2$ on the vertical axis; the asymptote $E=pc$ (photon line); points marked for a slow particle ($pc \ll mc^2$, $E\approx mc^2$), a relativistic particle, and a photon ($m=0$). Dashed arrows show the Lorentz-invariant distance from the origin.

Usain Bolt covered $100\ \mathrm{m}$ in $9.58\ \mathrm{s}$ and reached a top speed of about $12.4\ \mathrm{m/s}$; his mass is $94\ \mathrm{kg}$. His Newtonian kinetic energy is $K \approx \frac{1}{2}mv^2 \approx 7.2\ \mathrm{kJ}$, while his rest energy is $E_0 = mc^2 = 94\ \mathrm{kg} \times (3.00\times 10^{8}\ \mathrm{m/s})^2 \approx 8.5\times 10^{18}\ \mathrm{J}$ — the output of a $1\ \mathrm{GW}$ power station running for about 270 years.

This exercises $E = \gamma mc^2$ together with $K = (\gamma-1)mc^2$. With $\beta = 12.4/(3.0\times 10^{8}) = 4.1\times 10^{-8}$ we get $\gamma - 1 \approx \frac{1}{2}\beta^2 = 8.5\times 10^{-16}$, so $K = (\gamma-1)mc^2 = 7.2\ \mathrm{kJ}$ — exactly what $\frac{1}{2}mv^2$ gives. The ratio $K/E_0 \approx 8.5\times 10^{-16}$ shows that at everyday speeds the relativistic expression is indistinguishable from Newton's, while $E_0 = mc^2$ remains the dominant energy store of the body.

A $1500\ \mathrm{kg}$ electric car drives at $30\ \mathrm{m/s}$ ($108\ \mathrm{km/h}$). Its kinetic energy is $\frac{1}{2}mv^2 = 6.75\times 10^5\ \mathrm{J} = 675\ \mathrm{kJ}$, and a $60\ \mathrm{kW\cdot h}$ battery holds $2.16\times 10^8\ \mathrm{J}$ — 320 times as much. The rest energy of the same mass is $E_0 = mc^2 = 1500\ \mathrm{kg} \times 9\times 10^{16}\ \mathrm{m^2/s^2} = 1.35\times 10^{20}\ \mathrm{J}$.

The same relations $E = \gamma mc^2$ and $K = (\gamma-1)mc^2$ apply: at $v = 30\ \mathrm{m/s}$, $\gamma - 1 \approx \frac{1}{2}(10^{-7})^2 = 5.0\times 10^{-15}$, so $(\gamma-1)mc^2 = 5.0\times 10^{-15} \times 1.35\times 10^{20} = 6.75\times 10^5\ \mathrm{J}$, matching $\frac{1}{2}mv^2$. The relation $E_0 = mc^2$ also runs backwards: a change of energy $\Delta E$ changes the mass by $\Delta m = \Delta E/c^2$, so a fully charged battery is heavier than a discharged one by $2.16\times 10^8/(9\times 10^{16}) = 2.4\times 10^{-9}\ \mathrm{kg}$, i.e. by $2.4\ \mathrm{\mu g}$.

A single lightning stroke transfers about $10^9\ \mathrm{J}$ ($1\ \mathrm{GJ}$) — for instance a charge of $15\ \mathrm{C}$ across a potential difference of $7\times 10^7\ \mathrm{V}$. The equivalent mass is $\Delta m = E/c^2 = 10^9/(9\times 10^{16}) \approx 1.1\times 10^{-8}\ \mathrm{kg} \approx 11\ \mathrm{\mu g}$ — less than a grain of salt. The rest energy of $1\ \mathrm{kg}$ of matter is $9\times 10^{16}\ \mathrm{J}$, i.e. 90 million such strokes.

This is a direct application of $E_0 = mc^2$: the energy carried away by light, thunder and heated air corresponds to a mass decrease $\Delta m = \Delta E/c^2$, and the same link explains why this decrease is negligible in chemical reactions but appreciable in nuclear ones. The general relation $E^2 = (pc)^2 + (mc^2)^2$ additionally shows that a massless particle ($m = 0$) carries energy only through momentum, $E = pc$.