Minkowski space: relativity as the geometry of a quadratic form

Idea. Boosts along one line are hyperbolic rotations; the rapidity plays the angle, and angles add. Velocity addition is then nothing but the addition formula for \(\tanh\).

1. Since \(\cosh^2 - \sinh^2 = 1\), \(\cosh\varphi = (1 - \tanh^2\varphi)^{-1/2} = \gamma\) and \(\sinh\varphi = \gamma\beta\). So \[ B(\varphi) = \begin{pmatrix}\cosh\varphi & -\sinh\varphi\\ -\sinh\varphi & \cosh\varphi\end{pmatrix}. \] Multiplying, \[ B(\varphi)B(\psi) = \begin{pmatrix}\cosh\varphi\cosh\psi + \sinh\varphi\sinh\psi & -(\cosh\varphi\sinh\psi + \sinh\varphi\cosh\psi)\\ \cdot & \cdot\end{pmatrix}, \] with the second row the mirror image. From \(\cosh a = (e^a + e^{-a})/2\), \(\sinh a = (e^a - e^{-a})/2\), expanding gives \(\cosh\varphi\cosh\psi + \sinh\varphi\sinh\psi = \cosh(\varphi + \psi)\) and \(\sinh\varphi\cosh\psi + \cosh\varphi\sinh\psi = \sinh(\varphi + \psi)\). So \(B(\varphi)B(\psi) = B(\varphi + \psi)\). The composite velocity is \(\tanh(\varphi + \psi)\); dividing numerator and denominator of \(\sinh(\varphi+\psi)/\cosh(\varphi+\psi)\) by \(\cosh\varphi\cosh\psi\), \[ \beta_w = \frac{\tanh\varphi + \tanh\psi}{1 + \tanh\varphi\tanh\psi} = \frac{\beta_u + \beta_v}{1 + \beta_u\beta_v}. \]

2. Each stage has \(\varphi_1 = \operatorname{artanh}0.5 = \tfrac12\ln 3 = 0.549306\). Ten stages: \(\varphi = 5\ln 3 = 5.493061\), and \(e^{2\varphi} = 3^{10} = 59049\). So \[ \beta = \tanh\varphi = \frac{3^{10} - 1}{3^{10} + 1} = \frac{59048}{59050} = 0.99996613, \] short of \(1\) by \(3.39\times10^{-5}\). And \(\gamma = \cosh\varphi = \tfrac12(3^5 + 3^{-5}) = \tfrac12(243 + \tfrac1{243}) = 121.502\).

3. Galileo adds velocities: \(10\times0.5c = 5c\). In the formula, small rapidities have \(\tanh\varphi \approx \varphi\), so for small boosts the rapidities and the velocities both add. The first stage alone agrees with Galileo exactly; already after two, rapidity \(\ln 3\) gives \(0.8c\), not \(1.0c\).

Check. Dimensions: all quantities are pure numbers ✓. Two stages: \((3^2 - 1)/(3^2 + 1) = 0.8\), agreeing with \((0.5 + 0.5)/(1 + 0.25)\) ✓ — the deck's twins. \(\gamma\) and \(\beta\) consistent: \(1/\sqrt{1 - 0.99996613^2} = 121.50\) ✓.

Another way

By Doppler factors: a boost of rapidity \(\varphi\) multiplies received periods by \(e^{\varphi}\), and a relay of observers multiplies the factors. Each stage has factor \(\sqrt{1.5/0.5} = \sqrt3\); ten stages give \(3^5 = 243 = \sqrt{(1+\beta)/(1-\beta)}\), so \((1+\beta)/(1-\beta) = 3^{10}\), the same equation.

Idea. A negative \(K\) makes the invariant form \(k^2t^2 + x^2\) positive definite, so the group is the rotation group of the plane — compact, with no invariant cone, and rotating far enough turns time around.

1. \(-Kv = v/k^2\), so \(t' = \gamma(t + vx/k^2)\) and \(kt' = \gamma(kt + (v/k)x)\). Put \(\tan\theta = v/k\), \(\theta\in(-\pi/2, \pi/2)\): then \(\cos\theta = (1 + v^2/k^2)^{-1/2} = \gamma\) and \(\sin\theta = \gamma v/k\). So \[ \begin{pmatrix}T'\\ x'\end{pmatrix} = \begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}T\\ x\end{pmatrix}, \qquad \theta(v) = \arctan(v/k), \] a rotation, which preserves \(T^2 + x^2 = k^2t^2 + x^2\).

2. Rotations compose by adding angles, and \(\theta(k) = \pi/4\). Two boosts at \(v = k\) give \(\theta = \pi/2\): \(T' = x\), \(x' = -T\). The worldline \(x = 0\) is sent to \(T' = 0\): every tick of the clock happens at the same instant \(t' = 0\) of the new frame, spread out along \(x'\). The clock's own history has become a line of simultaneity. The velocity-addition formula \(w = (u + v)/(1 - uv/k^2)\) has a zero denominator at \(u = v = k\): the composite "frame" moves at infinite speed. The velocities do not form a group; only the angles do.

3. Four boosts give \(\theta = \pi\): \(T' = -T\), \(x' = -x\). A clock at rest at the origin is seen, in an allowed inertial frame, with its ticks in reverse order. Its ticks are events on one worldline, each a cause of the next (the gear that moved moves the next gear), and the order of cause and effect along a single material worldline — timelike-related events — would be frame-dependent. No physics with causes survives that; so \(K \lt 0\) is excluded. In the Lorentz case \(K \gt 0\) section 4 of the deck proves exactly the opposite: timelike order is invariant.

4. No. Invariance of a speed \(u\) would need \((u + v)/(1 - uv/k^2) = u\) for all \(v\), i.e. \(v(1 + u^2/k^2) = 0\): impossible for \(v \ne 0\). A rotation has no invariant real direction (other than for angle \(0\) or \(\pi\)), so there is no light cone.

Check. Dimensions: \(v/k\) is a pure number, so \(\theta\) is an angle ✓. Limit \(k\to\infty\) (so \(K\to0^-\)): \(\theta \approx v/k \to 0\) and \(T' \approx T + (v/k)x\) gives \(t' = t + vx/k^2 \to t\), Galileo ✓.

Idea. Two continuous \(\{\pm1\}\)-valued invariants separate four pieces; the decomposition \(\Lambda = BR\) shows the identity piece is connected, and the others are translates of it.

1. \(\det(\Lambda^{\mathsf T}\eta\Lambda) = \det(\Lambda)^2\det\eta = \det\eta\), so \(\det(\Lambda)^2 = 1\). The \(00\) entry: \((\Lambda^{\mathsf T}\eta\Lambda)_{00} = \sum_{\mu}\eta_{\mu\mu}(\Lambda^\mu{}_0)^2 = -(\Lambda^0{}_0)^2 + \sum_{i}(\Lambda^i{}_0)^2 = -1\), so \((\Lambda^0{}_0)^2 = 1 + \sum_i(\Lambda^i{}_0)^2 \ge 1\).

2. Let \(C^+ = \{X : \eta(X,X) \lt 0,\ X^0 \gt 0\}\). It is convex: for \(X, Y \in C^+\) and \(s, t \ge 0\) not both zero, \(\eta(sX + tY, sX + tY) = s^2\eta(X,X) + 2st\,\eta(X,Y) + t^2\eta(Y,Y) \lt 0\) because \(\eta(X, Y) \lt 0\) (by \(\eta(X,Y) \le -X^0Y^0 + |\vec x||\vec y| \lt 0\)), and the time component is positive. So \(C^+\) is connected. \(\Lambda\) preserves \(\eta\), so maps \(C^+\) into the timelike vectors, which are the disjoint union of the open sets \(C^+\) and \(-C^+\). A connected image lies in one of them — the one containing \(\Lambda e_0\), whose time component is \(\Lambda^0{}_0\). So \(\Lambda^0{}_0 \ge 1 \iff \Lambda(C^+) \subseteq C^+\). Maps with \(\Lambda(C^+) \subseteq C^+\) are closed under composition; and for inverses, \(\Lambda^{-1} \in O(1,3)\) and if \(\Lambda^{-1}\) sent \(C^+\) into \(-C^+\) then \(\Lambda\) would send \(-C^+\) into \(C^+\), hence (by linearity) \(C^+\) into \(-C^+\), contradiction. So the set is a subgroup.

3. Let \(\Lambda \in SO^+(1,3)\) and \(u = \Lambda e_0 \in C^+\) with \(\eta(u,u) = -1\); write \(u = (\gamma, \gamma\vec\beta)\), \(|\vec\beta| \lt 1\), \(\gamma = (1 - |\vec\beta|^2)^{-1/2}\). Let \(B\) be the boost with velocity \(\vec\beta\): rotate \(\vec\beta\) to the \(x\)-axis, boost along \(x\), rotate back. Then \(Be_0 = u\). So \(R = B^{-1}\Lambda\) is in \(SO^+(1,3)\) and fixes \(e_0\). It preserves \(e_0^\perp = \operatorname{span}(e_1, e_2, e_3)\) (if \(\eta(e_0, Y) = 0\) then \(\eta(e_0, RY) = \eta(Re_0, RY) = \eta(e_0, Y) = 0\)), and there \(\eta\) is the Euclidean inner product, so \(R\) restricted to it is orthogonal; \(\det R = 1\) makes it a rotation. So \(\Lambda = BR\). Paths: a rotation has an axis (\(\det(R - I) = -\det(R - I)\) in odd dimension, so \(1\) is an eigenvalue) and turns the orthogonal plane by some \(\theta\); \(s\mapsto\) (the same axis, angle \(s\theta\)) joins \(I\) to \(R\). The boosts \(B_s\) with velocity \(s\vec\beta\) join \(I\) to \(B\). Then \(s\mapsto B_sR_s\) is a continuous path in \(SO^+(1,3)\) from \(I\) to \(\Lambda\).

4. \(\det\) and \(\operatorname{sign}\Lambda^0{}_0\) are continuous into \(\{\pm1\}\), so constant on each connected component: at least four components, one in each nonempty cell (\(I\), \(P\), \(T\), \(PT\) witness nonemptiness). Each cell is a single component: the cell of \(P\) is \(P\cdot SO^+(1,3)\) (if \(\det\Lambda = -1\) and \(\Lambda^0{}_0 \ge 1\) then \(P\Lambda \in SO^+\), since \(P\) does not change the \(00\) entry), the image of a path-connected set under the homeomorphism \(\Lambda\mapsto P\Lambda\); similarly for \(T\) and \(PT\). Exactly four.

Check. A special case: the boost at \(\beta = 0.8\) has \(\Lambda^0{}_0 = \gamma = 5/3 \ge 1\) and \(\det = \cosh^2 - \sinh^2 = 1\) ✓. \(T = \operatorname{diag}(-1,1,1,1)\) has \(\Lambda^0{}_0 = -1\), \(\det = -1\) ✓. Dimension count: \(O(1,3)\) is \(6\)-dimensional, three boost and three rotation parameters, matching \(\Lambda = BR\) ✓.

Idea. In two dimensions the cone has only two null directions, and nothing ties the scale along one to the scale along the other, so any pair of increasing reparametrisations preserves the order.

1. \(-t^2 + x^2 = -(t + x)(t - x) = -uw\). Causal: \(\Delta u\Delta w \ge 0\). Future: \(\Delta u + \Delta w \gt 0\). Together with \(q \ne p\): both \(\ge 0\), not both zero. Conversely both \(\ge 0\) and not both zero gives both conditions.

2. \(F(s) = s^3\) is a strictly increasing bijection of \(\mathbb R\), so \(\Delta u \ge 0 \iff \Delta(u^3) \ge 0\), with \(\Delta u = 0 \iff \Delta(u^3) = 0\); likewise for \(w\). By part 1 the map preserves \(\prec\) both ways. It is not affine: it sends the points \(u = 0, 1, 2\) on the line \(w = 0\) to \(u = 0, 1, 8\), not equally spaced. Back in \((t, x)\) it is a genuinely non-linear map of the plane preserving all causal relations.

3. A linear bijection \(A\) preserving \(\prec\) both ways maps the closed quadrant \(Q = \{u \ge 0, w \ge 0\}\) onto itself. The edge rays of \(Q\) are its extreme rays (not positive combinations of two independent vectors of \(Q\)), and a linear bijection of \(Q\) onto itself maps extreme rays to extreme rays. So either \(A(u, w) = (au, bw)\) or \(A(u, w) = (bw, au)\), with \(a, b \gt 0\). Write \(a = \lambda e^{\varphi}\), \(b = \lambda e^{-\varphi}\) with \(\lambda = \sqrt{ab}\). The map \((u, w)\mapsto(e^{\varphi}u, e^{-\varphi}w)\) is the boost of rapidity \(-\varphi\) (check: \(t + x \mapsto e^{\varphi}(t + x)\) is what \(\cosh\varphi\,t + \sinh\varphi\,x\), \(\sinh\varphi\,t + \cosh\varphi\,x\) do); the swap \(u \leftrightarrow w\) is \(x\mapsto -x\); and \(\lambda\) is a dilation. So the linear causal automorphisms are exactly orthochronous Lorentz transformations times dilations — Zeeman's conclusion, which in \(1+1\) holds for linear maps and fails without linearity.

Check. A special case: the boost at \(\beta = 0.8\) has \(e^{\varphi} = 3\), and acts as \((u, w)\mapsto(u/3, 3w)\) or its inverse; it preserves \(uw\), hence \(\eta\) ✓. The dilation by \(\lambda\) multiplies \(\eta\) by \(\lambda^2 \gt 0\), keeping every sign ✓.

Idea. The muon's proper time from production to sea level is a frame-independent number; each frame computes it in its own way.

1. \(\gamma = 10.0125\). The trip: \(t = 15000/(0.995\times299\,792\,458) = 50.286\ \mu\)s. The muon's proper time is \(t/\gamma = 5.0223\ \mu\)s. Survival \(e^{-5.0223/2.19698} = e^{-2.2860} = 0.1017\). About one muon in ten reaches sea level. (Equivalently, the lab lifetime \(\gamma\tau_\mu = 21.997\ \mu\)s against \(50.286\ \mu\)s of flight.)

2. In the muon's frame the \(15.0\) km column is contracted to \(15000/10.0125 = 1498.1\) m, and passes at \(0.995c\) in \(1498.1/(0.995c) = 5.0223\ \mu\)s of the muon's own time: survival \(0.1017\). Exactly equal, because \((L/\gamma)/v = (L/v)/\gamma\): both are the proper time \(\tau = \sqrt{t^2 - L^2/c^2}\) between production and arrival, the invariant interval divided by \(c\). Indeed \(\sqrt{50.286^2 - 50.035^2} = 5.022\ \mu\)s, with \(L/c = 50.035\ \mu\)s.

3. No dilation: survival \(e^{-50.286/2.19698} = e^{-22.889} = 1.15\times10^{-10}\). Almost no muons would reach the ground; in fact they are the dominant charged component of cosmic rays at sea level.

Check. Dimensions: \(\tau/\tau_\mu\) is a ratio of times ✓. Limit \(v\ll c\): \(\gamma\to1\) and the two predictions merge ✓. Order of magnitude: \(\gamma\approx10\) makes the exponent \(10\) times smaller, \(22.9\to2.29\) ✓.

Idea. "The pole is inside the barn" means "both ends are inside at one instant" — a statement about simultaneity. The frames disagree about which events are simultaneous, and about nothing else.

1. In the barn frame the pole has length \(20/\gamma = 10\) m. At \(t = 0\) its rear is at \(x = 0\), so its front is at \(x = 10\) m: exactly spanning the barn.

2. \(\beta = \sqrt{1 - 1/4} = \sqrt3/2 = 0.8660\). \(A = (ct, x) = (0, 0) \mapsto (0, 0)\). \(B = (0, 10\ \mathrm m) \mapsto ct'_B = -\gamma\beta\cdot10\ \mathrm m = -17.32\ \mathrm m\), \(x'_B = \gamma\cdot10 = 20\) m. So in the pole's frame the back door shuts first, by \(17.32\ \mathrm m/c = 57.8\) ns.

3. Pole frame: the pole is at rest with rear at \(x' = 0\), front at \(x' = 20\) m (the rear is at the origin because the rear passes the front door at \(A\), the common origin). \(B\) is at \(x' = 20\) m: the front end ✓. \(A\) is at \(x' = 0\): the rear end ✓. Separate argument: in the pole frame the barn has length \(10/\gamma = 5\) m and moves at \(-\beta c\). At \(B\) the back door is at the pole's front, \(x' = 20\), so the front door is at \(x' = 15\) m. It must travel \(15\) m to reach the rear at \(x' = 0\), taking \(15\ \mathrm m/(0.8660c) = 17.32\ \mathrm m/c = 57.8\) ns ✓. So in the pole frame: the back door shuts and reopens just as the front arrives, the barn slides on, and \(57.8\) ns later the front door shuts and reopens just behind the rear. Never both shut, never the pole inside, never a collision.

4. \(A\) and \(B\) are spacelike separated (interval \((10\ \mathrm m)^2 \gt 0\)), so their time order is frame-dependent and "both shut at once" is not a frame-independent statement; every frame-independent fact — which pole end is at which door at each closing, and that nothing collides — is the same in both frames.

Check. Invariance: \(-(ct'_B)^2 + (x'_B)^2 = -300 + 400 = 100\ \mathrm m^2 = (10\ \mathrm m)^2\) ✓. Dimensions: metres over \(c\) is seconds, \(17.32/299\,792\,458 = 5.78\times10^{-8}\) s ✓. Limit \(\beta\to0\): the delay \(\gamma\beta\cdot10\ \mathrm m/c\to0\), and the paradox disappears along with the contraction ✓.

Idea. Radar coordinates are the roots of the equation "\(E - \tau u\) is null", and the sum and product of the roots are read off the coefficients, which are values of \(\eta\).

1. Both \(\tau_1\) (outgoing: \(E - \tau_1u\) future null) and \(\tau_2\) (returning: \(\tau_2u - E\) future null) satisfy \(\eta(E - \tau u, E - \tau u) = 0\), which expands to \(\tau^2 + 2\eta(E,u)\tau - \eta(E,E) = 0\). For \(E\) not on the worldline this quadratic has two real roots (the discriminant \(4\eta(E,u)^2 + 4\eta(E,E) = 4d_E^2\) is positive, as follows), and they are \(\tau_1 \lt \tau_2\). Vieta: \(\tau_1 + \tau_2 = -2\eta(E,u)\), \(\tau_1\tau_2 = -\eta(E,E)\). So \(\tau_E = -\eta(E,u)\), and \(d_E^2 = \tfrac14\big((\tau_1 + \tau_2)^2 - 4\tau_1\tau_2\big) = \eta(E,u)^2 + \eta(E,E)\). Positivity: write \(E = aU + Z\) with \(U = u\), \(Z \perp u\); then \(\eta(E,u) = -a\) and \(\eta(E,E) = -a^2 + \eta(Z,Z)\), so \(d_E^2 = \eta(Z,Z) \ge 0\), positive unless \(Z = 0\), i.e. unless \(E\) is on the worldline (the deck's lemma: \(u^\perp\) is spacelike).

2. \(\tau_E = 0 \iff \eta(E, u) = 0 \iff E \in u^\perp\), a hyperplane since \(\eta\) is non-degenerate. On it, \(d_E^2 = \eta(E,E)\), so \(d_E = \sqrt{\eta(E,E)}\): the observer's radar distance to simultaneous events is the \(\eta\)-length, and \(u^\perp\) with \(\eta\) restricted is the observer's Euclidean space.

3. \(\eta(E, u) = -\gamma t + \gamma\beta x = 0 \iff t = \beta x\) (with \(c = 1\)): the \(x'\) axis, and the parallel hyperplanes \(\eta(E, u) = -\tau\) are \(\gamma(t - \beta x) = \tau\), i.e. \(t' = \tau\) — exactly the Lorentz transformation's lines of constant \(t'\).

Check. A special case: \(u = e_0\) (\(\beta = 0\)), \(E = (t, x)\): \(\tau_E = t\) and \(d_E^2 = -t^2 + x^2 + t^2 = x^2\) ✓. Dimensions (restoring \(c\)): \(\eta(E, u)\) with dimensionless \(u\) is a length, \(c\tau_E\) ✓. Number: for the deck's traveller on the outbound leg, \(u = (5/3, 4/3)\) and \(E = \) Earth at \(t = 1.8\): \(\tau_E = -(-\tfrac53\cdot1.8 + 0) = 3\), the traveller's clock at the turn ✓.

Idea. A light ray, direction and frequency together, is one null vector. A boost acts on it linearly; aberration is its direction changing, Doppler its time component changing.

1. \(k = (\omega'/c)\big(\gamma(1 + \beta\cos\theta'),\ \gamma(\cos\theta' + \beta),\ \sin\theta',\ 0\big)\). So \[ \omega = \gamma(1 + \beta\cos\theta')\,\omega', \qquad \cos\theta = \frac{\cos\theta' + \beta}{1 + \beta\cos\theta'}, \qquad \sin\theta = \frac{\sin\theta'}{\gamma(1 + \beta\cos\theta')}. \] Null check: \(\gamma^2(\cos\theta' + \beta)^2 + \sin^2\theta' - \gamma^2(1 + \beta\cos\theta')^2 = \gamma^2(1 - \beta^2)(\cos^2\theta' - 1) + \sin^2\theta' = 0\) ✓.

2. \(1 - \beta\cos\theta = \dfrac{1 + \beta\cos\theta' - \beta\cos\theta' - \beta^2}{1 + \beta\cos\theta'} = \dfrac{1 - \beta^2}{1 + \beta\cos\theta'}\). Hence \(\gamma(1 - \beta\cos\theta)\cdot\gamma(1 + \beta\cos\theta') = \gamma^2(1 - \beta^2) = 1\), and \(\omega'/\big(\gamma(1 - \beta\cos\theta)\big) = \gamma(1 + \beta\cos\theta')\omega' = \omega\) ✓.

3. Half the photons have \(\theta' \lt 90°\). \(\theta' = 90°\) maps to \(\cos\theta = \beta\), \(\sin\theta = 1/\gamma\), and the map \(\theta'\mapsto\theta\) is increasing, so half the photons lie within \(\theta_{1/2} = \arcsin(1/\gamma)\). At \(\gamma = 10\): \(\arcsin 0.1 = 0.10017\) rad \(= 5.739°\).

4. In the Sun's frame the starlight arrives perpendicular to the Earth's velocity; in the Earth's frame it arrives tilted forward by \(\alpha\) with \(\sin\alpha = \beta\) to first order (from part 1's \(\cos\theta = \beta\) at \(\theta' = 90°\): \(\theta = 90° - \alpha\)). \(\beta = 29\,780/299\,792\,458 = 9.934\times10^{-5}\), so \(\alpha = 9.934\times10^{-5}\) rad \(= 20.49''\). The star traces a small ellipse of semi-major axis \(20.5''\) over the year; Bradley announced it in 1729.

Check. \(\beta = 0\): \(\theta = \theta'\), \(\omega = \omega'\) ✓. \(\theta' = 0\): \(\cos\theta = 1\), \(\omega = \gamma(1+\beta)\omega' = e^{\varphi}\omega'\), the collinear approaching Doppler factor ✓. \(\gamma\to\infty\): the cone closes to zero angle ✓.

Idea. The boost changes \(\Delta t\) by an amount proportional to \(v\Delta x\). For spacelike separation that amount can exceed \(\Delta t\); for timelike separation it never can.

1. Rotate so \(\Delta\vec x = (\Delta x, 0, 0)\), \(\Delta x \gt 0\). With \(v_0 := c^2\Delta t/\Delta x\), spacelike gives \(|v_0| \lt c\). For a boost at \(v\): \(\Delta t' = \gamma(\Delta t - v\Delta x/c^2) = \gamma(\Delta x/c^2)(v_0 - v)\), negative exactly for \(v_0 \lt v \lt c\) — a nonempty range. (At \(v = v_0\) the events are simultaneous; below it, the order is kept.)

2. Timelike future: \(\Delta t \gt 0\) and \(|\Delta\vec x| \lt c\Delta t\). For any boost at \(|v| \lt c\) along any direction, the change is \(v\Delta x_\parallel/c^2\) with \(|v\Delta x_\parallel/c^2| \lt c\cdot c\Delta t/c^2 = \Delta t\), so \(\Delta t' = \gamma(\Delta t - v\Delta x_\parallel/c^2) \gt 0\).

3. \(\Delta t = 0.5\) yr, \(\Delta x = 1\) ly: \(v_0 = 0.5\). At \(v = 0.8\), \(\gamma = 5/3\): \(\Delta t' = \tfrac53(0.5 - 0.8) = -0.5\) yr. The signal arrives half a year before it left. It arrives before leaving for every boost with \(0.5c \lt v \lt c\).

4. A single signal whose reception precedes its emission in some frame is strange but contains no loop: the sender still sends before (in her frame) the receiver receives. A contradiction needs a round trip. By the principle of relativity, the receiver, moving at \(v\), may send a speed-\(2c\) signal backwards in his frame, and for \(v \gt 2u/(1 + u^2) = 0.8c\) (with \(u = 2c\)) it reaches the original sender before she sent the first signal — she could then decide not to send it. Relativity plus "no causal loops" forbids faster-than-light signals; relativity alone forbids only their being harmless.

Check. Invariance: \(-(0.5)^2 + 1^2 = 0.75\), and in the moving frame \(\Delta x' = \tfrac53(1 - 0.8\cdot0.5) = 1\), \(-(0.5)^2 + 1 = 0.75\) ✓. Limit: a signal at exactly \(c\) has \(v_0 = c\), and the range \((c, c)\) is empty — light signals keep their order ✓.

Idea. The deck's three computations, with new numbers; if they disagree, an arithmetic slip has been made.

1. Earth: \(40\) yr. Traveller: each leg \(20\sqrt{1 - 0.36} = 20\times0.8 = 16\) yr, total \(32\) yr.

2. \(\sqrt{1.6/0.4} = 2\). Traveller receives: outbound, \(16\) yr at one pulse per \(2\) yr, \(8\) pulses; inbound, \(16\) yr at one per \(\tfrac12\) yr, \(32\) pulses; total \(40\) = Earth's age ✓. The flip is at the turn, traveller's year \(16\) of \(32\). Earth receives: light from the turn \((20, 12)\) arrives at \(t = 32\); for \(32\) yr pulses arrive every \(2\) yr, \(16\) pulses; in the last \(8\) yr every \(\tfrac12\) yr, \(16\) pulses; total \(32\) = the traveller's age ✓. Earth sees the flip at year \(32\) of \(40\).

3. Outbound simultaneity through \((20, 12)\): \(t - 20 = 0.6(x - 12)\), at \(x = 0\): \(t = 12.8\). Inbound: \(t = 20 + 7.2 = 27.2\). Jump: \(14.4\) yr. Pieces: \(12.8 + 14.4 + 12.8 = 40\) ✓, and \(12.8 = 16/\gamma\) ✓.

Check. The reversed triangle inequality: \(\lVert(20, 12)\rVert = \sqrt{400 - 144} = 16\), and \(16 + 16 = 32 \le 40 = \lVert(40, 0)\rVert\) ✓. The Earth events spacelike to the turn have \(|t - 20| \lt 12\), i.e. \(8 \lt t \lt 32\): the jump \([12.8, 27.2]\) lies inside, and \(8\) and \(32\) are the flip-related times (the last pulse the traveller gets before turning was sent at \(8\); Earth sees the turn at \(32\)) ✓. Limit \(\beta\to0\): all three give equal ages ✓.

Idea. Proper time is the Lorentzian arc length; parametrised by the frame's \(t\) it is an ordinary integral of \(\sqrt{1 - \beta^2}\), which is at most \(1\), so every answer must come out less than the coordinate time.

1. With \(\beta = 0.8t/T\), \(\mathrm dt = (T/0.8)\,\mathrm d\beta\): \[ \tau = \frac{T}{0.8}\int_0^{0.8}\sqrt{1 - \beta^2}\,\mathrm d\beta = \frac{T}{0.8}\cdot\tfrac12\Big[\beta\sqrt{1-\beta^2} + \arcsin\beta\Big]_0^{0.8} = \frac{T}{1.6}\,(0.48 + 0.92730) = 0.880\,T. \] (The antiderivative by \(\beta = \sin s\): \(\int\cos^2 s\,\mathrm ds = \tfrac12(s + \sin s\cos s)\).)

2. \(\beta = 7660/299\,792\,458 = 2.555\times10^{-5}\). Loss per day \(= 86\,400\,(1 - \sqrt{1 - \beta^2})\ \mathrm s \approx 86\,400\,\beta^2/2 = 2.820\times10^{-5}\) s \(= 28.2\ \mu\)s. (The exact and first-order values agree to five figures.)

3. \(\dot x = c^2t/\sqrt{\ell^2 + c^2t^2}\), \(1 - \dot x^2/c^2 = (\ell^2 + c^2t^2 - c^2t^2)/(\ell^2 + c^2t^2) = \ell^2/(\ell^2 + c^2t^2)\). So \[ \tau = \int_0^{t_1}\frac{\ell\,\mathrm dt}{\sqrt{\ell^2 + c^2t^2}} = \frac{\ell}{c}\operatorname{arsinh}\frac{ct_1}{\ell}, \] by \(ct = \ell\sinh s\), \(c\,\mathrm dt = \ell\cosh s\,\mathrm ds\), \(\sqrt{\ell^2 + c^2t^2} = \ell\cosh s\). Numbers: \(\ell = c^2/g = 9.1648\times10^{15}\) m \(= 0.96872\) ly, so \(\ell/c = 0.96872\) yr, and \(ct_1/\ell = 10/0.96872 = 10.323\). \(\tau = 0.96872\operatorname{arsinh}(10.323) = 2.935\) yr. Distance \(\sqrt{0.96872^2 + 10^2} - 0.96872 = 9.078\) ly. Ten years pass on Earth, under three on board. (This worldline has constant proper acceleration \(g\), as unit 27 shows; it is the rocket that would feel like standing on Earth.)

Check. Every \(\tau\) is below its coordinate time: \(0.880T \lt T\), \(2.935 \lt 10\) ✓. Limits: in part 1, \(0.8\to0\) gives \(\tau\to T\) ✓; in part 3, \(ct_1 \ll \ell\) gives \(\operatorname{arsinh}z \approx z\) and \(\tau \approx t_1\) ✓, while for \(ct_1 \gg \ell\), \(\tau \approx (\ell/c)\ln(2ct_1/\ell)\), only logarithmic growth ✓. Dimensions: \(\ell/c\) is a time ✓.

Idea. Split any vector along and across a timelike \(X\); across is spacelike, and because \(\eta(X,X) \lt 0\) the spacelike part pushes the inequality the "wrong" way. For curves, go to the frame where the endpoints are at one place: there proper time is \(\int\sqrt{1 - |\dot{\vec x}|^2/c^2}\,\mathrm dt\) and the integrand is at most \(1\).

1. Suppose \(Z \ne 0\), \(\eta(X, Z) = 0\), \(\eta(Z, Z) \le 0\). \(Z\) is not a multiple of \(X\) (else \(\eta(X,Z) = a\eta(X,X) \ne 0\)), so \(W = \operatorname{span}(X, Z)\) is 2-dimensional, and \(\eta(aX + bZ, aX + bZ) = a^2\eta(X,X) + b^2\eta(Z,Z) \le 0\) on \(W\). The spatial span \(S = \operatorname{span}(e_1, e_2, e_3)\) has \(\dim(W \cap S) \ge 2 + 3 - 4 = 1\), so some nonzero \(w \in W \cap S\) has \(\eta(w,w) \gt 0\) — contradicting \(\eta \le 0\) on \(W\).

2. Let \(a = \eta(X,Y)/\eta(X,X)\), \(Z = Y - aX\); then \(\eta(X, Z) = 0\) and \(\eta(Y,Y) = a^2\eta(X,X) + \eta(Z,Z)\). So \[ \eta(X,Y)^2 - \eta(X,X)\eta(Y,Y) = a^2\eta(X,X)^2 - a^2\eta(X,X)^2 - \eta(X,X)\eta(Z,Z) = -\eta(X,X)\,\eta(Z,Z) \ge 0, \] since \(-\eta(X,X) \gt 0\) and, by part 1, \(\eta(Z,Z) \ge 0\) with equality iff \(Z = 0\), i.e. \(Y = aX\). For the triangle inequality: future timelike \(X, Y\) have \(\eta(X,Y) \le -X^0Y^0 + |\vec x||\vec y| \lt 0\), so \(-\eta(X,Y) = |\eta(X,Y)| \ge \lVert X\rVert\lVert Y\rVert\), and \[ \lVert X + Y\rVert^2 = \lVert X\rVert^2 + \lVert Y\rVert^2 - 2\eta(X,Y) \ge (\lVert X\rVert + \lVert Y\rVert)^2 . \] \(X + Y\) is future timelike (the cone is convex), so taking square roots is legitimate. Equality forces \(Y = aX\), \(a \gt 0\).

3. Translate \(p\) to \(0\). \(q = (c\Delta t, \Delta\vec x)\) with \(|\Delta\vec x| \lt c\Delta t\); rotate \(\Delta\vec x\) onto the \(x\)-axis and boost with \(\beta = |\Delta\vec x|/(c\Delta t)\): now \(q = (cT, \vec 0)\) with \(cT = \lVert q - p\rVert\). Proper time is invariant under these maps. A future timelike curve has \(\mathrm dt/\mathrm d\lambda \gt 0\), so \(t\) parametrises it on \([0, T]\) with \(\vec x(0) = \vec x(T) = \vec 0\), and \[ \tau = \int_0^T\sqrt{1 - |\dot{\vec x}(t)|^2/c^2}\,\mathrm dt \le \int_0^T 1\,\mathrm dt = T = \lVert q - p\rVert/c . \] If \(\tau = T\), the non-negative piecewise-continuous function \(1 - \sqrt{1 - |\dot{\vec x}|^2/c^2}\) has integral zero, so vanishes on each piece: \(\dot{\vec x} \equiv 0\), \(\vec x \equiv 0\), the straight segment. And the segment attains \(T\).

4. In the same frame, for \(0 \lt \beta \lt 1\) and \(n \ge 1\), the zig-zag moving at \(+\beta c\) for time \(T/2n\) then \(-\beta c\) for \(T/2n\), repeated \(n\) times, is future timelike, piecewise \(C^1\), from \(p\) to \(q\), with \(\tau = T\sqrt{1 - \beta^2}\). As \(\beta\to1\), \(\tau\to0\): the infimum is \(0\). It is not attained, because the integrand is strictly positive on a timelike curve, so \(\tau \gt 0\).

Check. On the deck's twins, \(X = (5, 4)\), \(Y = (5, -4)\): \(\eta(X,Y)^2 = 41^2 = 1681 \ge 81 = (-9)(-9)\) ✓, and \(\lVert X + Y\rVert = 10 \ge 3 + 3\) ✓. A number for part 4: at \(\beta = 0.99\) the zig-zag between the twins' meeting events has \(\tau = 10\sqrt{1 - 0.9801} = 1.41\) yr. Limit \(c\to\infty\): \(\tau\to T\) for every curve, absolute time ✓.