6.P — Worked Problems: Relativity

Fourteen problems across Part 6, climbing from standard to genuinely awkward. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — Reading gamma off a decay

A particle created in an accelerator has a proper lifetime of 2.20 μs and is measured to travel 1.20 km before decaying. Find its speed and its Lorentz factor.

Solution

Let the speed be \beta c. In the lab frame the lifetime is dilated to $\gamma\times2.20\ \mu$s, and the distance covered is:

d = \beta c\gamma\tau_0

1200 = \beta(3\times10^{8})\gamma(2.20\times10^{-6}) = 660\,\beta\gamma

\beta\gamma = 1.818

Now solve for \beta. Since \gamma = 1/\sqrt{1-\beta^2}:

\frac{\beta}{\sqrt{1-\beta^2}} = 1.818 \quad\Longrightarrow\quad \frac{\beta^2}{1-\beta^2} = 3.305

\beta^2 = 3.305 - 3.305\beta^2 \quad\Longrightarrow\quad 4.305\beta^2 = 3.305

\beta^2 = 0.7677 \quad\Longrightarrow\quad \beta = 0.8762

\gamma = \frac{1}{\sqrt{1-0.7677}} = \frac{1}{\sqrt{0.2323}} = \frac{1}{0.4820} = 2.075

Check: \beta\gamma = 0.8762\times2.075 = 1.818. ✔

What to notice. The combination \beta\gamma is what a distance measurement gives you directly, and it is worth solving for it first rather than guessing \beta and iterating. This is also exactly how particle lifetimes are measured: you cannot start a stopwatch on a muon, so you measure how far it gets and work backwards. Without time dilation this particle would travel only \beta c\tau_0 = 578 m, so it has gone twice as far as classical physics allows.

Problem 2 — The pole and the barn, quantified

A 20 m pole is carried at 0.866c through a 10 m barn with doors at both ends. In the barn frame, find the time interval between the two door closings as measured in the pole's frame.

Solution

\gamma = \frac{1}{\sqrt{1-0.75}} = \frac{1}{0.5} = 2

In the barn frame the pole is contracted to 20/2 = 10 m, exactly fitting. Both doors close simultaneously, at \Delta t = 0, separated by \Delta x = 10 m.

Transform to the pole's frame:

\Delta t' = \gamma\left(\Delta t - \frac{v\Delta x}{c^2}\right) = 2\left(0 - \frac{(0.866)(3\times10^{8})(10)}{9\times10^{16}}\right)

= 2\times\left(-\frac{2.598\times10^{9}}{9\times10^{16}}\right) = 2\times(-2.887\times10^{-8}) = -5.77\times10^{-8}\ \text{s}

57.7 nanoseconds apart in the pole's frame.

Sanity check from the pole's own geometry. In the pole's frame the barn is contracted to 10/2 = 5 m, and the pole is 20 m. The front door must open before the pole's tip reaches it, and the back door must close after the tail is in. The barn passes the pole at 0.866c, so travelling the extra 20 - 5 = 15 m of pole takes 15/(0.866\times3\times10^{8}) = 5.77\times10^{-8} s. The same 57.7 ns. ✔

What to notice. The two calculations agree, and the second one shows why: the interval between door closings in the pole frame is exactly the time for the barn to traverse the difference in lengths. Nothing was crushed, nothing was contradictory — the entire "paradox" is the assumption that "simultaneously" means the same thing to both observers.

Problem 3 — Doppler shift of a receding galaxy

A galaxy's hydrogen line, emitted at 656.3 nm, is observed at 720.0 nm. Find its recession speed using the relativistic formula, and compare with the non-relativistic estimate.

Solution

\frac{\lambda_{\text{obs}}}{\lambda_0} = \frac{720.0}{656.3} = 1.0971

Since \lambda \propto 1/f, the relativistic receding formula inverts:

\frac{\lambda_{\text{obs}}}{\lambda_0} = \sqrt{\frac{1+\beta}{1-\beta}} = 1.0971

Square both sides:

\frac{1+\beta}{1-\beta} = 1.2036

1+\beta = 1.2036 - 1.2036\beta \quad\Longrightarrow\quad 2.2036\beta = 0.2036

\beta = 0.0924 \quad\Longrightarrow\quad v = 2.77\times10^{7}\ \text{m/s}

Non-relativistic estimate:

z = \frac{\Delta\lambda}{\lambda_0} = \frac{63.7}{656.3} = 0.0971 \quad\Longrightarrow\quad v \approx zc = 2.91\times10^{7}\ \text{m/s}

5 % higher.

What to notice. At z = 0.1 the simple v = zc is already 5 % wrong, and the error grows fast — at z = 1 it would give v = c, which is nonsense. There is a further and more important caveat for real cosmology: at large redshift the shift is not a Doppler effect at all but a stretching of the wavelength by the expansion of space itself, and neither formula above applies. Chapter 12.5 does it correctly with the scale factor.

Problem 4 — Energy to accelerate a spacecraft

Find the energy needed to accelerate a 1000 kg spacecraft to (a) 0.1c, (b) 0.9c, (c) 0.999c, and compare with world annual energy consumption of 6\times10^{20} J.

Solution

KE = (\gamma-1)mc^2, \qquad mc^2 = 1000\times9\times10^{16} = 9\times10^{19}\ \text{J}

(a) 0.1c: \gamma = 1/\sqrt{1-0.01} = 1.00504

KE = 0.00504\times9\times10^{19} = 4.5\times10^{17}\ \text{J}

Newtonian estimate: \frac{1}{2}(1000)(3\times10^{7})^2 = 4.5\times10^{17} J. Identical to three figures.

(b) 0.9c: \gamma = 1/\sqrt{1-0.81} = 2.294

KE = 1.294\times9\times10^{19} = 1.16\times10^{20}\ \text{J}

Newtonian would give 3.6\times10^{19} — a factor of 3.2 too low.

(c) 0.999c: \gamma = 22.37

KE = 21.37\times9\times10^{19} = 1.92\times10^{21}\ \text{J}

More than three years of total world energy consumption, for one tonne.

What to notice. The energy is not the only problem, and it is not even the worst one. At 0.999c the interstellar medium — about one hydrogen atom per cubic centimetre — arrives as a beam of 20 GeV protons, delivering roughly a kilowatt per square metre of hull as hard radiation. And there is no way to decelerate at the far end without carrying the same energy again. This arithmetic is why every serious interstellar propulsion study lands somewhere between 0.05c and 0.2c, and why Chapter 12.9 examines whether the shortcuts in fiction have any physics behind them.

Problem 5 — Colliding particles and available energy

Two protons collide. Compare the energy available for making new particles when (a) a 100 GeV proton hits a stationary proton, and (b) two 50 GeV protons collide head-on. Proton rest energy is 0.938 GeV.

Solution

The quantity that matters is the invariant mass of the system, from Chapter 6.5:

(Mc^2)^2 = \left(\sum E\right)^2 - \left(\sum \vec{p}c\right)^2

(a) Fixed target. The moving proton has E_1 = 100 GeV and:

p_1c = \sqrt{E_1^2-(m c^2)^2} = \sqrt{10000 - 0.880} = 99.9956\ \text{GeV}

The target has E_2 = 0.938 GeV, p_2 = 0.

(Mc^2)^2 = (100+0.938)^2 - (99.9956)^2 = 10188.5 - 9999.1 = 189.4

Mc^2 = 13.8\ \text{GeV}

(b) Head-on collider. Each has E = 50 GeV, and the momenta are equal and opposite so they cancel:

(Mc^2)^2 = (50+50)^2 - 0 = 10000

Mc^2 = 100\ \text{GeV}

What to notice. The same total beam energy — 100 GeV — gives 13.8 GeV of usable energy in a fixed-target arrangement and 100 GeV in a collider. A factor of seven wasted, because in the fixed-target case most of the energy goes into the centre-of-mass motion of the debris rather than into making anything.

Worse, the fixed-target available energy grows only as \sqrt{E}, so doubling the beam energy buys only 41 % more. This is the entire reason every high-energy machine since the 1970s is a collider despite the enormous difficulty of making two thin beams hit each other. To match the LHC's 13 TeV with a fixed target you would need a beam of about 90,000 TeV.

Problem 6 — Mass defect of an alpha particle

The helium-4 nucleus has mass 4.001506 u. Given the proton at 1.007276 u and the neutron at 1.008665 u, find the binding energy and the binding energy per nucleon.

Solution

Mass of separate parts:

2(1.007276)+2(1.008665) = 2.014552 + 2.017330 = 4.031882\ \text{u}

Mass defect:

\Delta m = 4.031882 - 4.001506 = 0.030376\ \text{u}

Energy, using 1\ \text{u}\cdot c^2 = 931.494 MeV:

E_B = 0.030376\times931.494 = 28.30\ \text{MeV}

Per nucleon:

\frac{28.30}{4} = 7.07\ \text{MeV}

What to notice. Helium-4 is exceptionally tightly bound for such a light nucleus — its 7.07 MeV/nucleon is far above its neighbours lithium (5.6) and beryllium-8 (7.06 but unstable). This is why alpha particles are emitted as intact units in radioactive decay rather than as separate nucleons, why helium is the second most abundant element in the universe, and why the helium-burning step in stars proceeds through the awkward three-body triple-alpha process (Chapter 12.1) — beryllium-8 falls apart in 10^{-16} s, so two alphas cannot simply stick.

Note also the fractional mass loss: 0.0304/4.032 = 0.75 %. Three quarters of one percent of the mass became binding energy, and that is the number behind fusion power.

Problem 7 — Relativistic rocket

A rocket accelerates at a constant 1 g as felt by its crew. Find the speed and distance after 1 year of ship time, and after 10 years.

Solution

For constant proper acceleration a, the relations (derived from integrating the four-velocity) are:

\beta = \tanh\left(\frac{a\tau}{c}\right), \qquad d = \frac{c^2}{a}\left[\cosh\left(\frac{a\tau}{c}\right)-1\right]

Useful number: a/c for g = 9.81 m/s² is 9.81/3\times10^{8} = 3.27\times10^{-8} s⁻¹. In a year (3.156\times10^{7} s):

\frac{a\tau}{c} = (3.27\times10^{-8})(3.156\times10^{7}) = 1.032

After 1 year of ship time:

\beta = \tanh(1.032) = 0.7748

d = \frac{(9\times10^{16})}{9.81}\left[\cosh(1.032)-1\right] = (9.17\times10^{15})(1.5808-1) = 5.33\times10^{15}\ \text{m}

which is 0.56 light years.

After 10 years of ship time: a\tau/c = 10.32

\beta = \tanh(10.32) = 0.99999998

\cosh(10.32) = \frac{e^{10.32}+e^{-10.32}}{2} \approx \frac{30{,}300}{2} = 15{,}150

d = (9.17\times10^{15})(15149) = 1.39\times10^{20}\ \text{m} = 14{,}700\ \text{light years}

What to notice. Ten years of ship time carries you almost 15,000 light years — while nearly 15,000 years pass on Earth. Continue to 12 years of ship time and you cross the observable universe. This is not a loophole in the speed limit. The traveller's own speed never reaches c in any frame, and Earth measures the journey taking the light-travel time or more. What compresses is the traveller's proper time, together with the contracted distance in their own frame.

The practical obstacle is fuel. Sustaining 1 g for a year requires, even with perfect matter–antimatter annihilation, a fuel-to-payload mass ratio of about 4:1. For 10 years it is about 10^{4}:1, and that assumes 100 % efficient conversion and perfect exhaust collimation, neither of which is remotely achievable.

Problem 8 — Gravitational redshift from a white dwarf

Sirius B has mass 1.02\,M_\odot and radius 5850 km. Find the gravitational redshift of light from its surface, and the equivalent Doppler velocity.

Solution

\frac{\Delta\lambda}{\lambda} = \frac{GM}{Rc^2}

GM = (6.674\times10^{-11})(1.02\times1.989\times10^{30}) = 1.354\times10^{20}

\frac{\Delta\lambda}{\lambda} = \frac{1.354\times10^{20}}{(5.85\times10^{6})(8.988\times10^{16})} = \frac{1.354\times10^{20}}{5.258\times10^{23}} = 2.575\times10^{-4}

Equivalent velocity:

v = c\frac{\Delta\lambda}{\lambda} = (3\times10^{8})(2.575\times10^{-4}) = 7.7\times10^{4}\ \text{m/s} = 77\ \text{km/s}

What to notice. 77 km/s is a large and easily measurable shift, and it was measured by Walter Adams in 1925 — one of the earliest confirmations of general relativity, though his value was contaminated by scattered light from Sirius A and modern measurements give about 80 km/s. Compare with the Sun's 0.636 km/s, which is buried under convective Doppler noise. White dwarfs are a good laboratory precisely because M/R is about 400 times the Sun's.

The measurement also does something else: combined with the mass from the binary orbit, it gives the radius, and that is how the extraordinary density of white dwarfs was first established. Sirius B has the Sun's mass in the Earth's volume.

Problem 9 — How close before you are spaghettified?

Find the distance from a 10-solar-mass black hole at which the tidal acceleration across a 1.8 m human reaches a lethal 10 g.

Solution

\Delta g = \frac{2GM\ell}{r^3}

M = 10\times1.989\times10^{30} = 1.989\times10^{31}\ \text{kg}

r^3 = \frac{2GM\ell}{\Delta g} = \frac{2(6.674\times10^{-11})(1.989\times10^{31})(1.8)}{98.1}

= \frac{4.779\times10^{21}}{98.1} = 4.872\times10^{19}

r = (4.872\times10^{19})^{1/3}

Take logs: \log_{10}(4.872\times10^{19}) = 19.688, divided by 3 is 6.563, so r = 3.66\times10^{6} m.

3660 km. The Schwarzschild radius is r_s = 2.95\times10 = 29.5 km, so this is 124 Schwarzschild radii out.

Now repeat for the supermassive hole in our galaxy, Sagittarius A* at 4.3\times10^{6}\,M_\odot:

r^3 = \frac{2(6.674\times10^{-11})(8.55\times10^{36})(1.8)}{98.1} = \frac{2.054\times10^{27}}{98.1} = 2.094\times10^{25}

r = 2.76\times10^{8}\ \text{m}

And r_s = 2.95\times4.3\times10^{6} km = 1.27\times10^{10} m. So r = 0.022\,r_s — well inside the horizon.

What to notice. For the stellar-mass hole you are destroyed 124 horizon radii out, long before you could reach it. For the supermassive one, the tidal force does not become lethal until you are deep inside the horizon, so you cross the event horizon alive and feeling nothing. Tidal acceleration at the horizon scales as M/r_s^3 \propto 1/M^2, so a billion-solar-mass hole is a billion-billion times gentler at its horizon than a solar-mass one. This is the fact behind the observation in Chapter 12.3 that stars are torn apart outside small black holes and swallowed whole by large ones.

Problem 10 — Orbit at the ISCO

Find the orbital period at the innermost stable circular orbit (r = 3r_s) of a 10-solar-mass black hole, as measured by a distant observer.

Solution

r_s = 2.95\times10 = 29.5\ \text{km}, \qquad r = 3r_s = 88.5\ \text{km}

For circular orbits in Schwarzschild spacetime, the coordinate angular velocity happens to take the same form as the Newtonian result:

\Omega = \sqrt{\frac{GM}{r^3}}

GM = (6.674\times10^{-11})(1.989\times10^{31}) = 1.3275\times10^{21}

r^3 = (8.85\times10^{4})^3 = 6.93\times10^{14}

\Omega = \sqrt{\frac{1.3275\times10^{21}}{6.93\times10^{14}}} = \sqrt{1.9156\times10^{6}} = 1384\ \text{rad/s}

T = \frac{2\pi}{\Omega} = \frac{6.283}{1384} = 4.54\times10^{-3}\ \text{s}

4.5 milliseconds per orbit — 220 orbits per second.

The orbital speed:

v = \Omega r = 1384\times8.85\times10^{4} = 1.22\times10^{8}\ \text{m/s} = 0.41c

What to notice. Matter at the inner edge of an accretion disc around a stellar-mass black hole orbits 220 times a second at 40 % of light speed. Hot spots in the disc therefore produce X-ray flickering at hundreds of hertz, and these quasi-periodic oscillations are actually observed — they are one of the few direct probes of the region just outside a black hole, and their frequencies are used to estimate the hole's mass and spin.

Note that the ISCO exists at all only because of general relativity. Newtonian gravity permits a stable circular orbit at any radius whatsoever; the extra term in the orbit equation of Chapter 6.9 is what destabilises everything inside 3r_s.

Problem 11 — GPS from scratch

Verify the 38 μs/day figure for a GPS satellite, then find at what orbital radius the two relativistic effects would exactly cancel.

Solution

Net rate difference between a satellite at radius r and a clock on the ground at R:

\frac{\Delta t}{t} = \underbrace{\frac{GM}{c^2}\left(\frac{1}{R}-\frac{1}{r}\right)}_{\text{gravitational, satellite gains}} - \underbrace{\frac{v^2}{2c^2}}_{\text{velocity, satellite loses}}

For a circular orbit, v^2 = GM/r, so:

\frac{\Delta t}{t} = \frac{GM}{c^2}\left(\frac{1}{R}-\frac{1}{r}\right)-\frac{GM}{2rc^2} = \frac{GM}{c^2}\left(\frac{1}{R}-\frac{3}{2r}\right)

With R = 6.371\times10^{6} m and r = 2.656\times10^{7} m:

\frac{1}{R} = 1.5696\times10^{-7}, \qquad \frac{3}{2r} = \frac{3}{5.312\times10^{7}} = 5.648\times10^{-8}

\frac{\Delta t}{t} = 4.435\times10^{-3}\times(1.5696\times10^{-7}-5.648\times10^{-8}) = 4.435\times10^{-3}\times1.0048\times10^{-7}

= 4.456\times10^{-10}

Per day:

4.456\times10^{-10}\times86400 = 3.85\times10^{-5}\ \text{s} = 38.5\ \mu\text{s}$$ ✔ **Where the effects cancel.** Set the bracket to zero: $$\frac{1}{R} = \frac{3}{2r} \quad\Longrightarrow\quad r = \frac{3R}{2} = 9.56\times10^{6}\ \text{m}

That is an altitude of 9560 - 6371 = 3190 km.

What to notice. Below 3190 km altitude a satellite clock runs slow overall, above it runs fast. The International Space Station at 420 km is well below, so ISS clocks lose about 28 μs/day and astronauts return very slightly younger — Sergei Krikalev, with 803 days in orbit, is about 20 milliseconds younger than he would otherwise be, and holds the record for time travel into the future.

Note also that ignoring the velocity effect entirely would give 45.7 μs/day instead of 38.5 — a 19 % error, which is 2 km of position error per day. Both relativities are needed and neither alone is enough.

Problem 12 — Chirp mass from a waveform

A gravitational wave signal is observed at 100 Hz sweeping upward at 5.0 Hz per second. Find the chirp mass.

Solution

\frac{df}{dt} = \frac{96}{5}\pi^{8/3}\left(\frac{G\mathcal{M}}{c^3}\right)^{5/3}f^{11/3}

Solve for \mathcal{M}:

\left(\frac{G\mathcal{M}}{c^3}\right)^{5/3} = \frac{5}{96}\pi^{-8/3}f^{-11/3}\frac{df}{dt}

Numbers. \pi^{8/3}: \ln\pi = 1.1447, times 8/3 is 3.0526, so \pi^{8/3} = 21.17.

f^{11/3} = 100^{11/3}: \ln 100 = 4.6052, times 11/3 is 16.886, so f^{11/3} = 2.15\times10^{7}.

\left(\frac{G\mathcal{M}}{c^3}\right)^{5/3} = \frac{5}{96}\times\frac{1}{21.17}\times\frac{1}{2.15\times10^{7}}\times5.0

= (0.05208)(0.04724)(4.651\times10^{-8})(5.0) = 5.72\times10^{-10}

Take the 3/5 power: \ln(5.72\times10^{-10}) = -21.28, times 0.6 is -12.77, so:

\frac{G\mathcal{M}}{c^3} = e^{-12.77} = 2.84\times10^{-6}\ \text{s}

\mathcal{M} = \frac{(2.84\times10^{-6})(2.7\times10^{25})}{6.674\times10^{-11}}

using c^3 = 2.7\times10^{25}:

\mathcal{M} = \frac{7.67\times10^{19}}{6.674\times10^{-11}} = 1.15\times10^{30}\ \text{kg} = 0.58\,M_\odot

What to notice. A chirp mass of 0.58 solar masses corresponds to two objects of roughly 0.67 solar masses each — too light for the black hole mergers LIGO usually sees, and in the range where a binary would be white dwarfs or something more exotic. In practice such a system would sweep through 100 Hz only in its final moments and would be a candidate for a sub-solar-mass primordial black hole, which is a live search topic.

Note also that the chirp mass is extracted from timing alone — the frequency and its rate of change — with no reference to distance or amplitude. That separation is what makes the amplitude a clean distance measurement afterwards, and it is why mergers work as standard sirens.

Problem 13 — Energy radiated in a merger

Two 30-solar-mass black holes merge to leave a 57-solar-mass hole. Find the energy radiated, the average power if it takes 0.2 s, and compare with the Sun's total lifetime output.

Solution

Mass converted:

\Delta M = 60 - 57 = 3\,M_\odot = 5.967\times10^{30}\ \text{kg}

Energy:

E = \Delta Mc^2 = (5.967\times10^{30})(8.988\times10^{16}) = 5.36\times10^{47}\ \text{J}

Average power:

P = \frac{5.36\times10^{47}}{0.2} = 2.68\times10^{48}\ \text{W}

The Sun's lifetime output. It burns for about 10^{10} years at 3.85\times10^{26} W:

E_\odot = (3.85\times10^{26})(10^{10}\times3.156\times10^{7}) = 1.2\times10^{44}\ \text{J}

\frac{5.36\times10^{47}}{1.2\times10^{44}} = 4470

What to notice. The merger releases in a fifth of a second more than four thousand times everything the Sun will emit in its entire ten-billion-year life. And the peak power exceeds the combined electromagnetic output of every star in the observable universe, roughly 10^{49} W.

The reason it is nonetheless so hard to detect is the coupling constant 2G/c^4 = 1.65\times10^{-44} from Chapter 6.10. Spacetime is extraordinarily stiff, so an enormous energy produces a minute strain. That same stiffness is why the waves pass through 1.3 billion light years of galaxies without measurable absorption.

Problem 14 — The twins with a realistic profile

Instead of an instant turnaround, a traveller accelerates at 1 g for 1 year of ship time, decelerates for 1 year, then repeats to return. Find the total ship time, the Earth time, and the distance reached.

Solution

Use the relations from Problem 7, with a\tau/c = 1.032 per year of ship time.

Leg 1, accelerating for 1 ship year. Reaches \beta = \tanh(1.032) = 0.7748 and covers 0.56 light years. Earth time elapsed:

t = \frac{c}{a}\sinh\left(\frac{a\tau}{c}\right) = \frac{3\times10^{8}}{9.81}\sinh(1.032) = (3.058\times10^{7})(1.2222) = 3.737\times10^{7}\ \text{s}

= 1.184\ \text{years}

Leg 2, decelerating for 1 ship year. By symmetry, it covers another 0.56 ly and takes another 1.184 Earth years, arriving at rest.

Turnaround point: 1.12 light years out, 2 ship years, 2.37 Earth years.

Legs 3 and 4 mirror the first two.

Totals:

\tau_{\text{ship}} = 4\ \text{years}, \qquad t_{\text{Earth}} = 4\times1.184 = 4.74\ \text{years}

Difference: 0.74 years — about 9 months younger.

What to notice. A four-year round trip at a comfortable 1 g reaches barely a light year and buys nine months of age difference. The effect is small because the traveller spends most of the trip well below 0.775c, and \gamma barely departs from 1 until then.

Push it further and the numbers explode. A 10-year ship-time trip on the same profile — 2.5 years per leg — reaches about 30 light years and takes 62 Earth years, so the traveller returns 52 years younger than their twin. The asymmetry lives entirely in the acceleration, and Problem 7's arithmetic shows why it is nonetheless not a shortcut to the stars: the fuel requirement grows exponentially with ship time.