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.