ARCHETYPE: Relativity Numericals (γ, Dilation, E=mc²)

The formulas you just drilled are engineering constraints. Particle accelerators at CERN push protons to $\gamma$ of several thousand, every magnet setting computed from $p=\gamma mv$ and $E=\gamma mc^2$; treat the beam as Newtonian and it hits the wall of the ring. The Global Positioning System corrects its atomic clocks for the special-relativistic slowing of the fast-orbiting satellites (about $7\ \mu$s per day) — ignore it and navigation drifts kilometres off within hours. Nuclear power and the Sun both run on $E=mc^2$, converting a sliver of rest energy into light and heat. And the muon of example 2 is now a working tool: cosmic-ray muons, reaching the ground only because of time dilation, are used to "X-ray" the interiors of volcanoes and pyramids. A student's exercise in $\gamma$ underlies the satnav in your pocket and the scan in a hospital.

Bookkeeping first, because in a numerical gym the accounting is the physics. Two observers move at a steady speed $v$ relative to each other, each carrying identical rods and clocks. The whole of special relativity, for exam purposes, is that they will disagree about times, lengths, momenta and energies in a precise, calculable way — and every disagreement is governed by one dimensionless number. Define the speed ratio $\beta = v/c$ (always between 0 and 1, since nothing with mass reaches $c$) and the Lorentz factor

where: $v$ — the relative speed of the two frames; $c = 3.0\times10^{8}$ m/s — the speed of light, the same in every inertial frame (Einstein's second postulate); $\beta$ — that speed as a fraction of $c$; and $\gamma$ — the factor by which times stretch and lengths shrink. Read $\gamma$ as a "relativity dial": at rest $\beta=0$ and $\gamma=1$ (nothing happens, Newton is exact); as $v$ climbs toward $c$, $\beta\to1$ and $\gamma\to\infty$. The dial is lazy at first and then explosive — you have to get remarkably close to $c$ before $\gamma$ departs from 1 at all, and then it runs away. That laziness decides your whole method. At $\beta=0.1$ (a tenth of light speed, already 30 000 km/s), $\gamma = 1.005$ — a half-percent correction you would ignore in any exam. Memorise the anchor pairs — $0.6c \to 1.25$, $0.8c \to 1.667$, $0.98c \to 5.03$, $0.99c \to 7.09$ — because examiners reuse them relentlessly, and recognising "$0.6c$" as "$\gamma = 1.25$" on sight saves you a square root every time.

With $\gamma$ in hand, the entire subject is two short families of formulas. The kinematic family describes what happens to space and time. A clock moving past you ticks slower than one at rest beside you, and a ruler moving past you is shorter along its direction of motion:

where: $\Delta t_0$ — the proper time, the interval read by a single clock present at both events (the shortest possible reading); $\Delta t$ — the longer interval a moving observer assigns; $L_0$ — the proper length, measured in the object's own rest frame (the longest reading); $L$ — the contracted length seen by an observer it flies past. The two effects pull in opposite directions on purpose: times stretch (multiply by $\gamma\ge1$), lengths shrink (divide by $\gamma$). The classic memory hook is that the proper quantities — proper time and proper length — are always the "extreme" ones: proper time is the smallest elapsed time, proper length the largest length. The dynamic family describes what happens to momentum and energy. Newton's $p = mv$ and $KE = \tfrac12 mv^2$ are only the low-speed corners of deeper relations:

where: $m$ — the rest mass (also called invariant mass), a fixed property of the particle that every observer agrees on; $p$ — the relativistic momentum, Newton's $mv$ boosted by $\gamma$; $E$ — the total energy; $E_0 = mc^2$ — the rest energy, the energy locked up in mass even when the particle sits still; and $KE = E - E_0 = (\gamma-1)mc^2$ — the kinetic energy, everything above rest. Read the kinetic-energy form carefully: at low speed $\gamma\to1$ so $KE\to0$ as it must, but it grows without limit as $v\to c$, which is exactly why no massive particle ever reaches light speed — you would have to pour in infinite energy. Tying the two dynamic quantities together is the single most useful relation in the whole chapter, the energy–momentum invariant:

where: the same $E$, $p$ and $m$ as above. This is a relativistic Pythagoras: total energy is the hypotenuse of a right triangle whose legs are $pc$ (the motion part) and $mc^2$ (the rest part). Two limits fall straight out of it. For a particle at rest, $p=0$ and $E = mc^2$ — pure rest energy. For a massless particle such as a photon, $m=0$ and the relation collapses to $E = pc$ — pure momentum energy, no rest energy at all. That is why light, which has no mass, still carries momentum.

Six moves cover the whole family. Compute $\gamma$ from equation (1) as line one of every problem (and invert with $\beta = \sqrt{1 - 1/\gamma^2}$ if handed $\gamma$ — for example $\gamma = 2$ means $\beta = 0.866$). Dilation and contraction: decide which quantity is proper, then multiply time by $\gamma$ or divide length by $\gamma$ (equation 2). The muon proof: a cosmic-ray muon with a $2.2\ \mu$s rest-frame lifetime reaches sea level because its clock runs slow from the ground (or the atmosphere is contracted from the muon) — the signature numerical. Energies: rest $mc^2$, total $\gamma mc^2$, kinetic $(\gamma-1)mc^2$, in MeV for particles ($m_ec^2=0.511$ MeV, proton $938.3$ MeV) or joules for masses ($1$ kg $\Rightarrow 9\times10^{16}$ J). Momentum and the invariant: $p=\gamma mv$, with $E^2=(pc)^2+(mc^2)^2$ giving any of $E$, $p$, $m$ from the others — essential for photons ($E=pc$) and for annihilation, where an electron and positron convert into two $0.511$ MeV photons. The twin paradox: the stay-at-home ages by the Earth time, the traveller by that time divided by $\gamma$; the turnaround breaks the symmetry, so there is no contradiction — Langevin's 1911 traveller who returns younger than everyone he knew.

1 · Board level — the Lorentz factor and a slow-running clock (IB HL pattern)

A spacecraft passes Earth at $0.6c$. (a) Find its Lorentz factor. (b) The crew measure a $1.0$-s interval on their own clock between two events on board. How long does that interval last as measured from Earth? (c) The ship is $90$ m long in its own frame; how long does it appear from Earth?

Step 1. Lorentz factor, from equation (1): $\gamma = 1/\sqrt{1 - 0.6^2} = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.8 = \boxed{1.25}$. Recognise "$0.6c$" as "$\gamma = 1.25$" on sight — it is the single most reused anchor in the subject.

Step 2. Time dilation, equation (2). The $1.0$ s is proper time (one clock on the ship reads both events), so $\Delta t_0 = 1.0$ s and Earth measures the longer $\Delta t = \gamma\,\Delta t_0 = 1.25 \times 1.0 = \boxed{1.25\ \text{s}}$. The moving clock runs slow: what the crew call one second, Earth clocks as a second and a quarter.

Step 3. Length contraction, equation (2). The $90$ m is proper length (measured on the ship), so $L_0 = 90$ m and Earth sees the shorter $L = L_0/\gamma = 90/1.25 = \boxed{72\ \text{m}}$. Note the opposite operations: time was multiplied by $1.25$, length was divided by it.

2 · Board level — the muon that should not reach the ground (A-Level pattern)

A muon is created $3.0$ km up in the atmosphere and hurtles down at $0.98c$. Its lifetime at rest is $2.2\ \mu$s. (a) Without relativity, how far would it travel before decaying? (b) With time dilation, what is its lifetime as seen from the ground, and how far does it actually get? Does it reach sea level?

Step 1. Lorentz factor: $\gamma = 1/\sqrt{1 - 0.98^2} = 1/\sqrt{1 - 0.9604} = 1/\sqrt{0.0396} = \boxed{5.03}$. At $98\%$ of light speed the relativity dial is fully open.

Step 2. Newtonian distance, ignoring relativity. In its own lifetime the muon would cover $d_0 = v\,\Delta t_0 = 0.98 \times 3.0\times10^8 \times 2.2\times10^{-6} = \boxed{647\ \text{m}}$ — barely a fifth of the way down. On this reasoning almost no muons should survive to the ground.

Step 3. Dilated lifetime, equation (2). From the ground the muon's clock runs slow, so its lifetime is $\Delta t = \gamma\,\Delta t_0 = 5.03 \times 2.2 = \boxed{11\ \mu\text{s}}$ (precisely $11.1\ \mu$s), five times longer.

Step 4. Actual distance and verdict. Travelling at $0.98c$ for this longer lifetime, it covers $d = v\,\Delta t = 5.03 \times 647 = \boxed{3.25\ \text{km}}$ — which exceeds the $3.0$ km it had to fall, so it reaches sea level with room to spare. (From the muon's own frame the story is length contraction: the $3.0$ km atmosphere is squeezed to $3.0/5.03 = 0.60$ km, crossed easily in $2.2\ \mu$s. Same landing, either way.) The steady rain of muons at sea level is a daily, direct confirmation of time dilation.

3 · Competitive level — rest, kinetic and total energy, and the invariant (A-Level and IB HL pattern)

An electron ($m_e c^2 = 0.511$ MeV) is accelerated to $0.6c$. Find (a) its kinetic energy, (b) its total energy, (c) the magnitude of its momentum via the invariant. (d) Separately, how much energy is locked in $1.0$ kg of any matter?

Step 1. With $\gamma = 1.25$ at $0.6c$, kinetic energy from equation (3): $KE = (\gamma-1)m_ec^2 = (1.25 - 1)\times0.511 = 0.25 \times 0.511 = \boxed{0.128\ \text{MeV}}$. (Compare the Newtonian $\tfrac12 m_e v^2 = \tfrac12(0.6)^2 m_e c^2 = 0.18 \times 0.511 = 0.092$ MeV — already a $28\%$ undercount at only $0.6c$.)

Step 2. Total energy: $E = \gamma m_e c^2 = 1.25 \times 0.511 = \boxed{0.639\ \text{MeV}}$, which is of course rest plus kinetic, $0.511 + 0.128 = 0.639$ MeV.

Step 3. Momentum via the invariant, equation (4): rearrange to $pc = \sqrt{E^2 - (m_ec^2)^2} = \sqrt{0.639^2 - 0.511^2} = \sqrt{0.408 - 0.261} = \sqrt{0.147} = \boxed{0.383\ \text{MeV}}$, so $p = 0.383$ MeV$/c$. As a check, the direct route $p = \gamma m_e v \Rightarrow pc = \gamma\beta\, m_ec^2 = 1.25\times0.6\times0.511 = 0.383$ MeV$/c$ — the two roads agree exactly.

Step 4. Rest energy of a kilogram: $E_0 = mc^2 = 1.0 \times (3.0\times10^8)^2 = \boxed{9\times10^{16}\ \text{J}}$ — about the yield of a twenty-megatonne bomb, sleeping inside every kilogram of ordinary matter. This is why nuclear reactions, which liberate a whisker of it, release so much. Timing note: about two minutes. The whole problem is "get $\gamma$, then plug." Keep every energy in MeV and $m_ec^2 = 0.511$ MeV on the page. The one habitual slip is reaching for $\tfrac12 mv^2$ in part (a) — at any speed where they bother to ask, it is wrong; use $(\gamma-1)mc^2$.

4 · Competitive level — Langevin's twin, resolved two ways (IB HL and A-Level pattern)

A twin travels to a star $4.0$ light-years away at $0.8c$ and immediately returns at $0.8c$. (a) By Earth's clocks, how long is the round trip, so how much does the stay-at-home twin age? (b) How much does the travelling twin age? (c) Confirm the traveller's answer by length contraction. (d) What is the age gap?

Step 1. Lorentz factor: $\gamma = 1/\sqrt{1 - 0.8^2} = 1/\sqrt{0.36} = 1/0.6 = 1.667$.

Step 2. Earth-frame round-trip time. Each leg is $4.0$ ly at $0.8c$, taking $4.0/0.8 = 5.0$ yr, so the round trip is $10.0$ yr. The stay-at-home twin ages the full $\boxed{10.0\ \text{years}}$.

Step 3. Traveller's own ageing. The traveller carries the single clock that is present at both departure and return, so his elapsed time is proper time, the Earth interval divided by $\gamma$: $\Delta t_0 = \Delta t/\gamma = 10.0/1.667 = \boxed{6.0\ \text{years}}$. He returns having aged only six years.

Step 4. Cross-check by length contraction. In the traveller's frame the distance contracts: the $4.0$ ly becomes $L = L_0/\gamma = 4.0/1.667 = 2.4$ ly each way, so the round trip $(2\times2.4)/0.8 = 6.0$ yr — exactly the proper time of Step 3, as it must. The age gap is $10.0 - 6.0 = 4.0$ years, the traveller the younger. There is no paradox: only the traveller fires his engines to turn around, so the situation is not symmetric, and it is unambiguously his clock that logs the shorter interval — the essence of Langevin's 1911 resolution. Timing note: about two minutes. Do it entirely in the Earth frame — round-trip time $= 2 \times \text{distance}/v$, then divide by $\gamma$ for the traveller. Reserve the length-contraction route as a cross-check, never as the main line, and never try to "average" the two twins' frames — the asymmetry of the turnaround is the whole point.

Not computing $\gamma$ first. Almost every mistake downstream traces to a wrong or skipped $\gamma$. Write $\gamma = 1/\sqrt{1-\beta^2}$ as line one, plug the anchors you know ($0.6c\to1.25$, $0.8c\to1.667$, $0.98c\to5.03$), and only then choose a formula. If $\gamma$ rounds to $1.00$ at your precision, the problem is secretly Newtonian and you should say so.

Multiplying length by $\gamma$. Time dilation multiplies ($\Delta t=\gamma\Delta t_0$); length contraction divides ($L=L_0/\gamma$). Mixing them up is the most common single error. Anchor it on "proper length is the longest reading," so the moving observer must see something shorter — divide.

Choosing the wrong proper quantity. Proper time is read by one clock present at both events; proper length is measured in the object's own rest frame. In the muon problem the lifetime is proper in the muon's frame; in the ship problem the ship's length is proper in the ship's frame. Identify the proper one, then multiply or divide once.

Using $\tfrac12 mv^2$ at relativistic speed. It undercounts, badly. The true kinetic energy is $(\gamma-1)mc^2$; at $0.8c$ it is more than double the Newtonian value. Only in the $\gamma\approx1$ corner are the two the same.

Talking about "relativistic mass." Keep the rest mass $m$ fixed and invariant; put the speed dependence in $\gamma$ ($p=\gamma mv$, $E=\gamma mc^2$). This is the modern convention both HL and A-Level mark, and it avoids a swamp of confusion about "which mass" to use in $E^2=(pc)^2+(mc^2)^2$ — always the rest mass.

Board scope — this whole topic is enrichment. Special relativity is not in the CBSE, ICSE, JEE or NEET core, nor in AP Physics 2. It is examined on IB Physics HL (the Relativity option) and on some A-Level specifications (for example AQA's "Turning Points in Physics" option). If you are sitting an Indian board or AP, this material is background enrichment that sharpens your feel for energy and momentum — valuable, but not directly on your paper. Where it is examined, the numbers above ($\gamma$, dilation, contraction, $E=\gamma mc^2$, the muon, the twin) are exactly the exam currency.

Texts. IB Physics Option A: Relativity (Lorentz factor, dilation, contraction, relativistic energy and momentum) · AQA A-Level Turning Points in Physics, the special-relativity section (muon experiment, time dilation, $E=\gamma mc^2$) · A. P. French, Special Relativity (the classic light-clock and muon treatments) · for the nuclear cross-link, NCERT Physics Class 12 Ch. 13.

Provenance. Standard exam-style relativity numericals; every value is recomputed in verify_hs2-relativity-numericals.py. Recompute: $\gamma(0.6c)=1.25$, $\gamma(0.8c)=1.667$, $\gamma(0.98c)=5.03$, $\gamma(0.99c)=7.09$; $0.6c$ dilation turns $1.0$ s into $1.25$ s and $90$ m into $72$ m; muon at $0.98c$: reach $647$ m, dilated lifetime $\approx11\ \mu$s, reach $3.25$ km through a $3.0$ km (contracted $0.60$ km) atmosphere; electron at $0.6c$: $KE=0.128$, total $0.639$, $pc=0.383$ MeV ($m_ec^2=0.511$ MeV); at $0.8c$: $KE=0.341$, total $0.852$ MeV; $1$ kg $\Rightarrow 9\times10^{16}$ J, $1$ g $\Rightarrow 9\times10^{13}$ J; annihilation $\Rightarrow$ two $0.511$ MeV photons ($1.022$ MeV); $100$ m at $0.8c \to 60$ m; twin at $0.8c$ to a $4.0$ ly star: Earth $10.0$ yr, traveller $6.0$ yr, gap $4.0$ yr; $\gamma=2$ at $\beta=0.866$.

Prerequisites. Time Dilation, Length Contraction and the Muon Proof (`hs2-time-dilation-length-contraction`) — the concept article that builds the kinematic pair and the muon experiment — and Mass–Energy Equivalence and Relativistic Energy (`hs2-relativistic-energy`), which develops $E=\gamma mc^2$, $KE=(\gamma-1)mc^2$ and the invariant. This gym assumes both and drills using them at speed.

Leads to. The university seams Relativistic Collision Kinematics (`aq-sr-relativistic-collision-kinematics`), where the invariant $E^2=(pc)^2+(mc^2)^2$ is turned into the four-momentum bookkeeping of particle collisions and decays, and Relativistic Velocity Addition (`aq-sr-relativistic-velocity-addition`), which replaces the naive $u+v$ with the exact formula $u' = (u+v)/(1+uv/c^2)$ that keeps every speed below $c$.

Go deeper (UG). The light-clock derives time dilation, but the full Lorentz transformation — the pair $x'=\gamma(x-vt)$, $t'=\gamma(t-vx/c^2)$ that maps any event between frames — is only quoted here as a linear form; its axiomatic derivation, with the relativistic Doppler and velocity-addition formulas, belongs to the university seams named above. All of it is enrichment: every numerical in this gym stands on $\gamma$, the kinematic pair and the dynamic set alone, none of which needs anything beyond algebra and Pythagoras.