Exercises — Time dilation — derivation, twin paradox

WHAT: Plug $\beta =0.8$ into the formula. WHY: $\gamma$ is the single number that tells us how much time stretches; we always need it first. $\gamma =\frac{1}{\sqrt{1-0.8^2}}=\frac{1}{\sqrt{1-0.64}}=\frac{1}{\sqrt{0.36}}=\frac{1}{0.6}\approx \mathrm{1.667.}$ Answer: $\gamma \approx 1.667$.

WHAT: Identify where the two events (start of trip, end of trip) happen at the same place. WHY: Proper time is defined as the interval measured by a single clock present at both events — here, the wristwatch travels with the ship, so both "start" and "end" occur right at that watch. The wristwatch reading, $5$ years, is the proper time $\Delta t_0$. Earth's $8$ years is the dilated $\Delta t$. Check: proper time is the smaller one, and $5<8$. ✓ Answer: $\Delta t_0=5$ years.

WHAT: Find $\gamma$, then multiply. $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{0.64}}=\mathrm{1.25.}$ WHY multiply? The lab sees the moving particle's clock run slow, so its lifetime looks longer. Longer = multiply by $\gamma \ge 1$. $\Delta t=\gamma \Delta t_0=1.25\times 2.6\times 10^{-8}\text{s}=3.25\times 10^{-8}\text{s}.$ Answer: $\Delta t\approx 3.25\times 10^{-8}\text{s}$.

WHAT: The events (leave Earth, arrive) are both on the ship, so the ship measures proper time $\Delta t_0$; Earth measures the dilated $\Delta t=10$ yr. $\gamma =\frac{1}{\sqrt{1-0.36}}=\mathrm{1.25.}$ WHY divide? We know the dilated time and want the proper time, so invert $\Delta t=\gamma \Delta t_0$: $\Delta t_0=\frac{\Delta t}{\gamma }=\frac{10}{1.25}=8\text{years}.$ Answer: $8$ years on the ship.

WHAT: Solve $\gamma =2$ for $v$. WHY invert this way? Here the answer we want is the speed, but the speed sits buried inside a square root in the denominator of $\gamma$. So we peel the formula apart one operation at a time — reciprocal, then square, then isolate ${\beta }^2$ — undoing each step that wraps around $\beta$. Physically we are asking: "for the clock to tick at exactly half rate, how fast must it fly?" — the speed is the unknown, $\gamma$ is the given. $2=\frac{1}{\sqrt{1-{\beta }^2}}\Rightarrow \sqrt{1-{\beta }^2}=\frac{1}{2}\Rightarrow 1-{\beta }^2=\frac{1}{4}\Rightarrow {\beta }^2=\frac{3}{4}.$ $\beta =\frac{\sqrt{3}}{2}\approx \mathrm{0.866.}$ Answer: $v\approx 0.866c$. (Notice: to merely double the tick-time you already need $87\%$ of light speed — dilation is a high-speed effect.)

WHAT (a): Naive distance $=v\Delta t_0$. WHY (a): We compute this deliberately wrong first — assuming (as Newton would) that the muon's own $2.2\mu s$ lifetime is the time available in the ground frame too. This relies on the false assumption that time is universal. The point is to expose the contradiction: this baseline is what relativity must fix. $d_{\text{naive}}=0.98\times (3\times 10^8)\times (2.2\times 10^{-6})\approx 6.47\times 10^2\text{m}\approx 0.65\text{km}.$ That is far short of $10$ km — so classically the muon should never reach us. WHAT (b): Find $\gamma$, dilate the lifetime, recompute distance. WHY (b): The muon's clock runs slow in the ground frame, so from the ground the muon survives $\gamma$ times longer, giving it far more time to cover distance. $\gamma =\frac{1}{\sqrt{1-0.98^2}}=\frac{1}{\sqrt{1-0.9604}}=\frac{1}{\sqrt{0.0396}}\approx \mathrm{5.025.}$ $\Delta t=\gamma \Delta t_0=5.025\times 2.2\mu s\approx 11.05\mu s.$ $d=v\Delta t=0.98\times (3\times 10^8)\times (11.05\times 10^{-6})\approx 3.25\times 10^3\text{m}\approx 3.25\text{km}.$ WHY it matters: With dilation the muon covers several km, so a big fraction actually reaches the ground — the classic experimental proof (see Muon Decay Experiment). Answer: (a) $\approx 0.65$ km; (b) $\Delta t\approx 11.05\mu s$, $d\approx 3.25$ km.

WHAT: In the muon's frame it is at rest and the $10$ km atmosphere rushes past, contracted by $\gamma$ (see Length Contraction). $L=\frac{L_0}{\gamma }=\frac{10\text{km}}{5.025}\approx 1.99\text{km}.$ WHY: The muon experiences proper time $2.2\mu s$; in that time the contracted distance flies past at $0.98c$: $d_{\text{muon}}=v\Delta t_0=0.98\times (3\times 10^8)\times (2.2\times 10^{-6})\approx 0.647\text{km}.$ Compared to the contracted atmosphere thickness $1.99$ km, the muon covers a fraction $0.647/1.99\approx 0.325$ — the same fraction of the trip as the ground observer computed ($3.25$ km of $10$ km $=0.325$). Both frames agree on the physical outcome. ✓ Answer: contracted distance $\approx 1.99$ km; the muon crosses $\approx 0.65$ km of it — consistent with the ground calculation.

WHAT (a): Round-trip distance $=12\text{ly}$ at $0.6c$: $\Delta t_A=\frac{12\text{ly}}{0.6\text{ly/yr}}=20\text{years}.$ WHY (a): Time = distance ÷ speed. This is Twin A's own frame, where A is at rest, so A just watches the ship cover the fixed $12\text{ly}$ path at $0.6\text{ly/yr}$ — an ordinary "distance over speed" the same as any everyday travel-time, because A never changes frames. WHAT (b): $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{0.64}}=1.25$. Then $\Delta t_B=\frac{\Delta t_A}{\gamma }=\frac{20}{1.25}=16\text{years}.$ WHY (b): B's clock rides along with B, so both trip-events (leave Earth, return to Earth) happen right at that clock — it measures proper time, the smaller one. We know the dilated Earth time $\Delta t_A$ and want the proper time, so we divide by $\gamma$ (as in L2·Q2). WHAT (c): $\Delta t_A-\Delta t_B=20-16=4\text{years}.$ WHY (c): Both twins meet again at the same place (Earth) at the end, so we may directly subtract their two logged times to get the real, permanent age gap. Only such a reunion makes an age comparison meaningful. WHY asymmetric: Only B turns around at the star (changes inertial frame, feels a force). See deepdives/dd-physics-2.3.29-d4-s02.png — B's world-path is bent, A's is straight, and the straight path through spacetime is the one that clocks the most time. Answer: (a) $20$ yr; (b) $16$ yr; (c) B is $4$ years younger.

WHAT: In B's frame the $6\text{ly}$ gap is contracted: $L=\frac{6\text{ly}}{1.25}=4.8\text{ly}.$ WHY: B sits still while the star rushes toward it at $0.6c$ across the contracted $4.8\text{ly}$: $t_{\text{out}}=\frac{4.8\text{ly}}{0.6\text{ly/yr}}=8\text{years}.$ This is exactly half of $\Delta t_B=16$ yr. ✓ Length contraction and time dilation are two faces of the same Lorentz Transformations. Answer: $8$ years outbound, matching $16/2$.

WHAT (a): Here we combine two speeds: let $v_1=0.8c$ (ship P past Earth) and $v_2=0.5c$ (probe past ship P). We want $u$, the probe's speed past Earth. Newtonian addition would give $v_1+v_2=0.8c+0.5c=1.3c>c$ — impossible. Use the relativistic rule (see Relativistic Velocity Addition): $u=\frac{v_1+v_2}{1+\frac{v_1{v}_2}{c^2}}=\frac{0.8c+0.5c}{1+(0.8)(0.5)}=\frac{1.3c}{1.4}\approx 0.9286c.$ WHY: The denominator $1+v_1{v}_2/c^2$ keeps the result below $c$ — nothing outruns light. WHAT (b): With the probe's Earth-frame speed known, $\beta =0.9286$: $\gamma =\frac{1}{\sqrt{1-0.9286^2}}=\frac{1}{\sqrt{1-0.8623}}=\frac{1}{\sqrt{0.1377}}\approx \mathrm{2.695.}$ WHY (b): $\gamma$ always uses the speed relative to the observer whose clock we ask about. We want how slow the probe's clock runs as seen from Earth, so we must feed in the probe's Earth-frame speed $u$ — not $v_1$ or $v_2$, which are speeds relative to other frames. Only $u$ answers "how much does Earth see the probe's time stretch?" Answer: (a) $\approx 0.929c$; (b) $\gamma \approx 2.70$.

WHAT (a): $v=0\Rightarrow \beta =0\Rightarrow \gamma =1/\sqrt{1-0}=1$, so $\Delta t=\Delta t_0$. WHY (a): A clock that isn't moving relative to you has no diagonal light-path to lengthen — the photon just goes straight up and down as in the clock's own frame. With no extra path there is no extra time, so dilated time collapses back to proper time. This is the sanity check that relativity contains ordinary Newtonian time as its slow-speed limit. WHAT (b): As $v\to c$, $\beta \to 1$, so $1-{\beta }^2\to 0^{+}$ and $\gamma \to \infty$. Then $\Delta t=\gamma \Delta t_0\to \infty$. WHY (b): Physically, one tick of the moving clock would take infinite observer-time — the clock (and every process on it) appears frozen. This is exactly why a massive object can never reach $c$: to do so would demand infinite time-stretch, which in turn requires infinite energy. Light itself experiences no proper time at all. Look again at the steep right edge of deepdives/dd-physics-2.3.29-d4-s01.png. Answer: (a) $\gamma =1$, no dilation; (b) $\gamma \to \infty$, clock appears frozen.

WHAT: For tiny $\beta$, expand $\gamma =(1-{\beta }^2{)}^{-1/2}\approx 1+\frac{1}{2}{\beta }^2$. WHY the expansion? At everyday speeds ${\beta }^2$ is minuscule; a linear (Taylor) approximation is exact enough and avoids catastrophic rounding. $\frac{1}{2}{\beta }^2=10^{-12}\Rightarrow {\beta }^2=2\times 10^{-12}\Rightarrow \beta =\sqrt{2\times 10^{-12}}\approx 1.414\times 10^{-6}.$ $v=\beta c\approx 1.414\times 10^{-6}\times 3\times 10^8\approx 424\text{m/s}.$ WHY it matters: This is a jet-airliner speed — real flying-clock experiments confirm dilation at exactly this level. Answer: $v\approx 424\text{m/s}$.