Relativistic Mass, Momentum and Energy

When classical mechanics breaks down

Particle physicists at CERN accelerate protons to $0.9999999896c$ — just $10^{-7}$ m/s short of $c$. Each proton has a rest mass energy of $938\,\text{MeV}$, but at that speed its total energy is $6.5\,\text{TeV}$ — about 7,000 times larger than its rest energy. Doubling the energy barely changes the speed. Classical kinetic energy $E_k = \frac{1}{2}mv^2$ predicts a finite maximum — but experiments show that accelerating particles to higher speeds requires ever-increasing energy without limit.

Einstein showed that the correct relativistic expressions are:

These equations have profound properties:

• At rest ($v = 0$, $\gamma = 1$): $E = mc^2$. Even a stationary particle has energy — its rest energy.
• As $v \to c$: $\gamma \to \infty$, so $E_k \to \infty$. Infinite energy would be required to reach $c$ — therefore, massive objects cannot reach the speed of light.
• For massless particles ($m = 0$): $E = pc$ and $v = c$ exactly. Photons are massless and always travel at $c$.

Three energy equations: rest energy $E_0 = mc^2$; total energy $E = \gamma mc^2$; kinetic energy $E_k = (\gamma-1)mc^2$. Relativistic momentum: $p = \gamma mv$. As $v \to c$, $E_k \to \infty$ — massive objects cannot reach $c$. Massless photons ($m = 0$): $E = pc$ and naturally travel at $c$.

Write all three energy equations and note the difference between rest energy and kinetic energy.

The world's most famous equation

We just saw that relativistic energy is $E = \gamma mc^2$ and that even a stationary particle has rest energy $mc^2$. That raises a question: what does mass-energy equivalence actually mean physically, and where does it show up in the real world? This card answers it → nuclear fission, fusion, annihilation and the energy-momentum relation.

$E = mc^2$ tells us that mass is a form of energy. The conversion factor is $c^2$, an enormous number: 1 kg of mass equals $9\times10^{16}$ J of energy — roughly the output of a large power station running for several years.

Nuclear fission: A heavy nucleus (like uranium-235) splits into smaller fragments. The total mass of the fragments is slightly less than the original nucleus. This "missing mass" (mass defect) is converted to kinetic energy of the fragments — about 200 MeV per fission.

Nuclear fusion: Light nuclei combine to form a heavier nucleus with less total mass. The Sun converts about 4 million tonnes of mass to energy every second through fusion. About 0.7% of the mass of hydrogen fused is converted to energy, carried away as photons and neutrinos.

Particle-antiparticle annihilation: When an electron meets a positron, they annihilate, converting their entire rest mass into two gamma-ray photons: $e^- + e^+ \to \gamma + \gamma$. Each photon carries energy $mc^2 = 511$ keV.

Particle creation: The reverse process also occurs. High-energy photons or collisions can create particle-antiparticle pairs, converting energy into mass.

Applications of $E = \Delta mc^2$: nuclear fission releases ~200 MeV per U-235 event (mass defect); nuclear fusion powers the Sun ($4\times10^6$ t/s converted); $e^- + e^+ \to \gamma + \gamma$ produces two 511 keV photons. Energy-momentum relation: $E^2 = (pc)^2 + (mc^2)^2$; for photons ($m=0$): $E = pc = hf$.

Record the three real-world applications and the energy-momentum relation.

From voltage to velocity

We just saw the real-world manifestations of mass-energy equivalence — fission, fusion and annihilation. That raises a question: if a proton is accelerated through a large voltage, how do you calculate its $\gamma$, speed and momentum in sequence? This card answers it → a five-step worked example at 500 MV.

Worked strategy (proton at 500 MV): step 1 — $E_k = qV = 500$ MeV. Step 2 — $\gamma = 1 + E_k/mc^2 = 1.533$. Step 3 — $v/c = \sqrt{1-1/\gamma^2} = 0.758$. Step 4 — $pc = \sqrt{E^2 - (mc^2)^2} = 1090$ MeV/$c$. Classical would give 269 MeV: underestimate by 46%.

Write the four-step energy-to-speed chain and note when classical fails.

Avoiding the traps that cost marks

We just saw the step-by-step strategy for particle accelerator energy problems. That raises a question: what conceptual confusions most often trip students up on relativistic energy in the HSC? This card answers it → three misconceptions and a five-step exam strategy.

Misconception 1: $E = mc^2$ gives kinetic energy.
Wrong. $E = mc^2$ is the rest energy — the energy of a stationary particle. Total energy is $E = \gamma mc^2$. Kinetic energy is $E_k = (\gamma - 1)mc^2 = E_{total} - mc^2$. A common exam trap is to confuse these three quantities.

Misconception 2: Classical $E_k = \frac{1}{2}mv^2$ is a good approximation at any speed.
Wrong. When $\gamma > 1.1$ (roughly $v > 0.4c$), the classical formula is already off by more than 5%. At $v = 0.758c$ (the worked example), classical underestimates by 46%. Always check whether the problem is relativistic before applying classical formulas.

Misconception 3: Mass increases with speed.
The older "relativistic mass" concept ($m_{rel} = \gamma m_0$) is now deprecated. It is better to say that a particle's inertia increases (requiring more force for the same acceleration) and its energy and momentum increase, while its rest mass $m_0$ remains constant.

Three traps: (1) $E = mc^2$ is rest energy, NOT kinetic — kinetic energy is $E_k = (\gamma-1)mc^2 = E_{total} - mc^2$. (2) Classical $\frac{1}{2}mv^2$ fails when $\gamma > 1.1$ ($v > 0.4c$). (3) "Relativistic mass" is deprecated — rest mass $m_0$ is constant; it is inertia and energy that increase. HSC strategy: total energy check: $E_{total} = E_k + mc^2$.

Write the three misconception corrections and the total-energy identity.