Relativistic Momentum and Mass-Energy Equivalence
Relativistic Momentum and E = mc²
In classical Newtonian mechanics, momentum is simply mass times velocity: p equals m times v. This works perfectly well at everyday speeds, but it breaks down catastrophically as speeds approach the speed of light. If momentum were just m-v, then a constant force applied to an object could accelerate it past the speed of light — violating Einstein's second postulate.
The resolution is relativistic momentum. In special relativity, momentum is redefined as p equals gamma times m times v, where gamma is the Lorentz factor and m is the rest mass of the object. At low speeds, gamma is essentially one and this reduces to the familiar Newtonian formula. But as v approaches c, gamma grows without bound, causing the momentum to increase far more rapidly than in classical mechanics. This means that the faster an object moves, the harder it becomes to accelerate — effectively requiring more and more force to achieve smaller and smaller increases in speed. Reaching the speed of light would require infinite momentum, which would require infinite energy, which is physically impossible for any object with mass.
This leads directly to Einstein's most famous equation: E equals m-c-squared. The total energy of a particle is E equals gamma times m times c-squared. When the particle is at rest, gamma equals one and this reduces to E-zero equals m-c-squared — the rest energy. This equation tells us that mass is a form of energy. A kilogram of matter contains c-squared joules of energy — about 90 quadrillion joules, or 90 petajoules. This is the energy released in nuclear reactions and the energy that powers the sun.
The kinetic energy of a relativistic particle is the total energy minus the rest energy: K equals the quantity gamma minus one, times m-c-squared. At low speeds this reduces to the familiar one-half m-v-squared. The relationship between energy, momentum, and mass is captured by the relativistic energy-momentum relation: E-squared equals p-c-quantity-squared plus m-c-squared-quantity-squared. For a photon, which has zero rest mass, this gives E equals p-c — a photon's energy is entirely due to its momentum.
$$p = \gamma m v = \frac{mv}{\sqrt{1 - v^2/c^2}}$$
As $v \to c$, $p \to \infty$. No massive object can reach $c$.
Rest energy: $$E_0 = mc^2$$
Total energy: $$E = \gamma mc^2$$
Relativistic kinetic energy: $$K = (\gamma - 1)mc^2$$
$$E^2 = (pc)^2 + (mc^2)^2$$
For a photon ($m = 0$): $E = pc$.
Find the rest energy of a 1 kg object.
$$E_0 = mc^2 = 1 \times (3 \times 10^8)^2 = 9 \times 10^{16} \text{ J} = 90 \text{ PJ}$$
That's equivalent to about 21.5 megatons of TNT — more than the largest nuclear weapon ever tested (Tsar Bomba at ~50 MT was about 2.3 kg of mass converted to energy).
A proton ($m = 1.67 \times 10^{-27}$ kg) moves at $0.9c$. Find its relativistic momentum and kinetic energy.
Step 1: $\gamma = 1/\sqrt{1-0.81} = 1/\sqrt{0.19} \approx 2.294$
Step 2: Momentum:
$$p = \gamma mv = 2.294 \times 1.67 \times 10^{-27} \times 0.9 \times 3 \times 10^8$$
$$p \approx 2.294 \times 4.509 \times 10^{-19} \approx 1.034 \times 10^{-18} \text{ kg·m/s}$$
Step 3: Kinetic energy:
$$K = (\gamma - 1)mc^2 = (2.294 - 1) \times 1.67 \times 10^{-27} \times 9 \times 10^{16}$$
$$K \approx 1.294 \times 1.503 \times 10^{-10} \approx 1.945 \times 10^{-10} \text{ J}$$
In nuclear fission, a uranium-235 nucleus splits and releases about 200 MeV of energy. How much mass is converted?
Step 1: Convert MeV to joules: $200 \text{ MeV} = 200 \times 1.602 \times 10^{-13} = 3.204 \times 10^{-11}$ J
Step 2: $\Delta m = E/c^2 = 3.204 \times 10^{-11} / (9 \times 10^{16}) = 3.56 \times 10^{-28}$ kg
That's about 0.09% of the uranium nucleus's mass — yet it produces an enormous amount of energy.
Show that $K = (\gamma - 1)mc^2$ reduces to $\frac{1}{2}mv^2$ at low speeds.
Step 1: For $v \ll c$, use the binomial approximation:
$$\gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2}$$
Step 2: $\gamma - 1 \approx \frac{v^2}{2c^2}$
Step 3: $K \approx \frac{v^2}{2c^2} \cdot mc^2 = \frac{1}{2}mv^2$ ✓
Practice Problems
Show Answer Key
1. $E_0 = 9.11 \times 10^{-31} \times 9 \times 10^{16} = 8.20 \times 10^{-14}$ J $= 0.511$ MeV.
2. $\gamma = 1/\sqrt{0.75} \approx 1.155$. $p = 1.155 \times 2 \times 1.5 \times 10^8 \approx 3.46 \times 10^8$ kg·m/s.
3. $K = (\gamma - 1)mc^2 = (5/3 - 1)mc^2 = \frac{2}{3}mc^2$. Kinetic energy is $2/3$ of rest energy.
4. $E = 4 \times 10^9 \times 9 \times 10^{16} = 3.6 \times 10^{26}$ J/s $= 3.6 \times 10^{26}$ W.
5. $\gamma \approx 7.089$. $K = (7.089 - 1) \times 0.511 \approx 6.089 \times 0.511 \approx 3.11$ MeV.
6. $E = pc = 10^{-27} \times 3 \times 10^8 = 3 \times 10^{-19}$ J $\approx 1.87$ eV.
7. $\gamma = 2 \implies v = c\sqrt{3}/2 \approx 0.866c$.
8. $\Delta m = E/c^2 = 10^{18}/(9 \times 10^{16}) \approx 11.1$ kg.
9. Since $E = \gamma mc^2$ and $p = \gamma mv$, both transform with $\gamma$. The combination $E^2 - (pc)^2$ cancels the $\gamma^2$ dependence, leaving $(mc^2)^2$, which is a constant independent of frame.
10. $K = mc^2 \implies \gamma - 1 = 1 \implies \gamma = 2 \implies v = c\sqrt{3}/2 \approx 0.866c$.