Part D · unit 27 · problems

Relativistic dynamics: four-vectors, and why colliders exist

the slides · the curriculum

Checkpoint. You compute a reaction threshold by evaluating the invariant \(s\) in whichever frame is easiest and compare it with a published accelerator design energy, and you derive \(\vec p = \gamma m\vec v\) from the action \(-mc^2\int\mathrm d\tau\) by the Euler–Lagrange equations of unit 11.

Work each problem on paper first. Under it, Hints opens a ladder one rung at a time — each rung appears only once the one before it is open, and the last gives the first line of the working — and Worked solution has the whole thing: the idea, the working, then a check. Read the solution after your attempt, including when you got it right. F folds everything back; D switches dark and light.

Twelve problems, about five hours. Conventions as in the deck: signature \((-,+,+,+)\), \(x^0 = ct\), four-momentum \(p = (E/c,\ \vec p)\) with \(\eta(p,p) = -m^2c^2\), and \(s = -c^2\eta(P,P)\) for a total four-momentum \(P\), so that \(\sqrt s\) is an energy. "Squaring" a four-vector \(w\) means forming \(\eta(w,w)\).

Constants: \(c = 299\,792\,458\ \mathrm{m\,s^{-1}}\), \(h = 6.62607015\times10^{-34}\) J s, \(hc = 1239.842\) MeV fm; rest energies \(m_ec^2 = 0.51099895\), \(m_pc^2 = 938.27209\), \(m_{\pi^\pm}c^2 = 139.57039\), \(m_{\pi^0}c^2 = 134.9768\), \(m_\mu c^2 = 105.65838\), \(m_\Lambda c^2 = 1115.683\) MeV; \(g = 9.80665\ \mathrm{m\,s^{-2}}\); one year is \(365.25\) days, so \(c/g = 0.968715\) yr and \(c^2/g = 0.968715\) ly; \(GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3\,s^{-2}}\).

Four-vectors and single particles

P1 Circular motion, and an accelerometer on a muon routine

exercises: four-velocity and four-acceleration in a frame, \(\eta(a,u) = 0\), proper acceleration, a number

A particle moves on a circle of radius \(r\) at constant speed \(v\): its worldline in an inertial frame is \(\big(ct,\ r\cos\omega t,\ r\sin\omega t,\ 0\big)\) with \(\omega = v/r\).

  1. Compute the four-velocity \(u\) and the four-acceleration \(a\), all four components of each, as functions of \(t\).
  2. Verify \(\eta(u,u) = -c^2\) and \(\eta(a,u) = 0\) directly from your components.
  3. Show that the proper acceleration is \(\alpha = \gamma^2v^2/r\), and say where each factor of \(\gamma\) comes from.
  4. Muons in the Fermilab \(g-2\) storage ring have \(\gamma = 29.3\) on a circle of radius \(7.112\) m. Evaluate the centripetal acceleration measured in the lab and the proper acceleration.
Hints · 3 rungs
Rung 1

Two four-component vectors, two identities to check, one scalar, and two numbers. The only tool is the chain rule: \(\mathrm d/\mathrm d\tau = \gamma\,\mathrm d/\mathrm dt\).

Rung 2

The speed is constant, so \(\gamma\) is constant here, and differentiating \(u\) does not touch it. That is why circular motion is the easy case.

Rung 3

First line: \(u = \gamma\,\dfrac{\mathrm d}{\mathrm dt}\big(ct, r\cos\omega t, r\sin\omega t, 0\big) = \gamma\big(c,\ -v\sin\omega t,\ v\cos\omega t,\ 0\big)\).

Worked solution

Idea. Differentiate the worldline twice with respect to proper time; \(\gamma\) is constant, so each differentiation brings in one factor of \(\gamma\).

1. \(u = \gamma(c,\ -v\sin\omega t,\ v\cos\omega t,\ 0)\), and \[ a = \gamma\frac{\mathrm du}{\mathrm dt} = \gamma^2\big(0,\ -v\omega\cos\omega t,\ -v\omega\sin\omega t,\ 0\big). \]

2. \(\eta(u,u) = \gamma^2(-c^2 + v^2\sin^2\omega t + v^2\cos^2\omega t) = -\gamma^2(c^2 - v^2) = -c^2\) ✓. \(\eta(a,u) = -0\cdot\gamma c + \gamma^3v^2\omega(\sin\omega t\cos\omega t - \cos\omega t\sin\omega t) = 0\) ✓.

3. \(\eta(a,a) = \gamma^4v^2\omega^2 = \gamma^4v^4/r^2\), so \(\alpha = \gamma^2v^2/r\). One \(\gamma\) comes from \(u = \gamma\,\mathrm dx/\mathrm dt\) (the clock runs slow, so the velocity per unit proper time is larger), the other from \(a = \gamma\,\mathrm du/\mathrm dt\) (the same effect again). In the particle's instantaneous rest frame this is the acceleration its accelerometer reads.

4. \(v/c = \sqrt{1 - 1/29.3^2} = 0.999417\). Lab: \(v^2/r = (0.999417c)^2/7.112 = 1.262\times10^{16}\ \mathrm{m\,s^{-2}}\). Proper: \(\gamma^2 = 858.49\) times that, \(\alpha = 1.084\times10^{19}\ \mathrm{m\,s^{-2}}\), about \(10^{18}\,g\).

Check. Dimensions: \(v^2/r\) is \(\mathrm{m\,s^{-2}}\) ✓. Limit \(v\ll c\): \(\alpha\to v^2/r\), the Newtonian centripetal acceleration ✓. The deck's general formula with \(\vec v\cdot\dot{\vec v} = 0\) gives \(a = (0,\gamma^2\dot{\vec v})\), the same ✓.

P2 Mass is not additive routine

exercises: invariant mass, null four-momenta, squaring a sum, a threshold angle

Two photons have energies \(E_1\), \(E_2\) and travel in directions making an angle \(\theta\).

  1. Show that the pair's invariant mass satisfies \(M^2c^4 = 2E_1E_2(1 - \cos\theta)\).
  2. For \(E_1 = E_2 = E\) travelling in opposite directions, show \(M = 2E/c^2\). Each photon is massless. Explain in one sentence where the mass came from, and describe what happens to \(M\) as \(\theta\to0\).
  3. Two photons of \(1\) MeV each can annihilate into an electron–positron pair only if \(\theta\) is large enough. Find the least angle.
Hints · 3 rungs
Rung 1

The answer to 1 is one scalar computed from one sum of two null vectors. Only the cross term survives.

Rung 2

Write each photon's four-momentum as \(\frac{E_i}{c}(1,\hat n_i)\) with \(\hat n_i\) a unit vector, and use \(\hat n_1\cdot\hat n_2 = \cos\theta\).

Rung 3

First line: \(M^2c^2 = -\eta(p_1 + p_2, p_1 + p_2) = -2\eta(p_1,p_2)\), because \(\eta(p_i,p_i) = 0\).

Worked solution

Idea. Square the total four-momentum. The squares of the parts vanish; the invariant mass lives entirely in the cross term, which depends on the angle.

1. \(\eta(p_1,p_2) = \frac{E_1E_2}{c^2}(-1 + \hat n_1\cdot\hat n_2) = -\frac{E_1E_2}{c^2}(1-\cos\theta)\). Hence \(M^2c^2 = -2\eta(p_1,p_2) = 2E_1E_2(1 - \cos\theta)/c^2\), that is \(M^2c^4 = 2E_1E_2(1-\cos\theta)\).

2. \(\theta = \pi\): \(M^2c^4 = 4E^2\), \(M = 2E/c^2\). The mass is the energy of relative motion: in the frame where the total momentum vanishes, the whole energy \(2E\) is "at rest" as a system, and invariant mass is exactly that energy over \(c^2\). As \(\theta\to0\), \(M\to0\) continuously: two photons moving together are, as a system, massless. Then \(P\) is null and there is no centre-of-momentum frame at all.

3. Pair creation needs \(M \ge 2m_e\) (the threshold theorem), so \(2E^2(1-\cos\theta) \ge 4m_e^2c^4\), i.e. \(\cos\theta \le 1 - 2(m_ec^2/E)^2 = 1 - 2(0.511)^2 = 0.4778\), so \(\theta \ge 61.46^\circ\). Head-on 1 MeV photons have \(M c^2 = 2\) MeV, comfortably above \(1.022\) MeV.

Check. Dimensions: \(E_1E_2\) is \(\mathrm{J^2}\), as is \(M^2c^4\) ✓. Limit \(\theta = 0\): \(M = 0\) ✓. The mass is not additive: \(0 + 0 \ne 2E/c^2\). This process, \(\gamma\gamma\to e^+e^-\), is Breit–Wheeler pair production; it limits how far very-high-energy gamma rays travel through the photon background between galaxies.

P3 An MeV electron, and Newton's answer routine

exercises: \(E = \gamma mc^2\), the mass shell, limits in both directions, a number

  1. An electron has kinetic energy \(T = 1.000\) MeV. Find \(\gamma\), \(v/c\) and \(pc\).
  2. Find the speed Newtonian mechanics predicts from \(T = \frac12mv^2\), and say in one line what is wrong with it.
  3. Show that in general \(pc = \sqrt{T(T + 2mc^2)}\), and that it reduces to Newton's \(p = \sqrt{2mT}\) when \(T\ll mc^2\) and to \(pc\approx T\) when \(T \gg mc^2\).
Hints · 3 rungs
Rung 1

Three numbers, one comparison, and one identity with two limits. Everything follows from \(E = T + mc^2\) and the mass shell.

Rung 2

\(\gamma = E/mc^2\); \(v/c = \sqrt{1 - \gamma^{-2}}\); \(p^2c^2 = E^2 - m^2c^4\).

Rung 3

First line: \(\gamma = 1 + T/mc^2 = 1 + 1/0.511 = 2.957\).

Worked solution

Idea. Kinetic energy fixes \(E\); \(E\) fixes \(\gamma\) and, through the mass shell, \(p\).

1. \(\gamma = 1 + 1.000/0.511 = 2.957\); \(v/c = \sqrt{1 - 1/2.957^2} = 0.9411\); \(pc = \sqrt{T(T + 2mc^2)} = \sqrt{1.000\times2.022} = 1.422\) MeV.

2. \(v = c\sqrt{2T/mc^2} = c\sqrt{2/0.511} = 1.978c\). It is faster than light, which is not the only thing wrong: \(\frac12mv^2\) is the first term of \((\gamma-1)mc^2\) and is a good approximation only while \(T\ll mc^2\). At \(T \approx 2mc^2\) the discarded terms are larger than the kept one.

3. \(p^2c^2 = E^2 - m^2c^4 = (T + mc^2)^2 - m^2c^4 = T^2 + 2Tmc^2\). If \(T\ll mc^2\), \(p^2c^2 \approx 2Tmc^2\), so \(p = \sqrt{2mT}\), Newton's relation ✓. If \(T\gg mc^2\), \(pc\approx T\): the particle behaves as if massless.

Check. Dimensions: \(T(T + 2mc^2)\) is \(\mathrm{J^2}\) ✓. Consistency: \(v/c = pc/E = 1.422/1.511 = 0.9411\) ✓, agreeing with part 1 by a different route.

Invariants, collisions and thresholds

P4 Compton, in four lines medium

exercises: four-momentum conservation, isolating the unknown and squaring, null vectors, a number

A photon of wavelength \(\lambda\) scatters off an electron at rest. The photon leaves at angle \(\theta\) to its original direction with wavelength \(\lambda'\). Use \(E = hc/\lambda\) for a photon (cited: the Planck–Einstein relation).

  1. Show, using only four-momentum conservation and \(\eta(k,k) = 0\) for a photon, that \[ \lambda' - \lambda = \frac{h}{m_ec}\,(1 - \cos\theta). \]
  2. Evaluate the Compton wavelength \(h/m_ec\).
  3. Molybdenum \(K\alpha\) X-rays have \(E = 17.479\) keV. Find \(\lambda\), the shift and the fractional shift at \(\theta = 90^\circ\), and the scattered photon's energy.
  4. Why did nobody see this shift with visible light?
Hints · 3 rungs
Rung 1

The answer is one scalar equation relating \(E\), \(E'\) and \(\theta\). You never need the recoiling electron's direction — so arrange for it to disappear.

Rung 2

Write \(k + p = k' + p'\) and isolate \(p'\), the electron after the collision, whose direction you do not know. Squaring it gives only its mass.

Rung 3

First line: \(\eta(p',p') = \eta(k + p - k',\ k + p - k')\), and the left side is \(-m_e^2c^2\) while \(\eta(k,k) = \eta(k',k') = 0\).

Worked solution

Idea. Isolate the particle you know least about, square, and evaluate the surviving cross terms in the electron's initial rest frame.

1. The four lines:

(i) \(-m^2c^2 = \eta(p',p') = \eta(k+p-k',k+p-k') = -m^2c^2 + 2\eta(k,p) - 2\eta(k,k') - 2\eta(p,k')\).

(ii) So \(\eta(k,p) = \eta(k,k') + \eta(p,k')\).

(iii) In the electron's rest frame \(p = (mc,\vec 0)\), \(k = \frac Ec(1,\hat n)\), \(k' = \frac{E'}c(1,\hat n')\): \(-Em = -\frac{EE'}{c^2}(1-\cos\theta) - E'm\).

(iv) Divide by \(EE'm\): \(\dfrac1{E'} - \dfrac1E = \dfrac{1-\cos\theta}{mc^2}\), and with \(1/E = \lambda/hc\): \(\lambda' - \lambda = \dfrac h{mc}(1-\cos\theta)\).

2. \(h/m_ec = 6.62607\times10^{-34}/(9.10938\times10^{-31}\times299\,792\,458) = 2.4263\times10^{-12}\) m \(= 2.4263\) pm.

3. \(\lambda = hc/E = 1239.842\ \text{MeV fm}/0.017479\ \text{MeV} = 70.93\) pm. At \(90^\circ\) the shift is \(\lambda_C = 2.426\) pm, a fractional shift of \(3.42\%\). \(E' = E/(1 + E/m_ec^2) = 17.479/(1 + 0.034206) = 16.90\) keV. Compton measured exactly this with molybdenum radiation in 1923.

4. At \(\lambda = 500\) nm the fractional shift is at most \(2\lambda_C/\lambda \approx 10^{-5}\), and in any case the electrons in matter are bound by a few eV — comparable to a visible photon's energy — so they are not free particles at rest and the kinematics above does not apply.

Check. Dimensions: \(h/mc\) is \(\mathrm{J\,s}/(\mathrm{kg\,m\,s^{-1}}) = \mathrm m\) ✓. \(\theta = 0\): no shift, as it must be when nothing happens ✓. \(m\to\infty\): no shift — classical (Thomson) scattering off an immovable charge ✓.

Another way

Components, as Compton did it. Energy: \(E + mc^2 = E' + E_e\). Momentum: \(\vec p_e = \vec k - \vec k'\), so by the law of cosines \(p_e^2c^2 = E^2 + E'^2 - 2EE'\cos\theta\). The mass shell \(E_e^2 = p_e^2c^2 + m^2c^4\) then gives \((E - E' + mc^2)^2 = E^2 + E'^2 - 2EE'\cos\theta + m^2c^4\), which expands to \(mc^2(E - E') = EE'(1 - \cos\theta)\) — line (iv) again. It is the same calculation with the frame chosen first: the law of cosines is the square of \(\vec p_e\), and the mass shell is the square of \(p'\). Putting the unknown alone before squaring (rung 2) is what lets the invariant route skip the components.

P5 Proof: what a lone photon cannot do medium

exercises: a proof about the sign of \(\eta\) on the future cone, impossibility of reactions, a threshold

  1. Prove: if \(v, w\in V\) are nonzero, future-pointing, and each timelike or null, then \(\eta(v,w)\le0\), with equality if and only if both are null and parallel.
  2. Deduce that a single free photon can never become an electron–positron pair, whatever its energy.
  3. Deduce that a free electron cannot absorb a photon: \(e + \gamma\to e\) is impossible.
  4. Pair creation does happen near a nucleus of mass \(M\): \(\gamma + N\to N + e^+ + e^-\). Find the threshold photon energy with the nucleus initially at rest, and show it tends to \(2m_ec^2\) as \(M\to\infty\). What is it when the target is an electron?
Hints · 3 rungs
Rung 1

Part 1 is a sign statement proved in a well-chosen frame; parts 2 and 3 are one-line contradictions using it and the invariance of \(s\); part 4 is the fixed-target threshold with unequal masses.

Rung 2

If one of the vectors is timelike, go to the frame where it is \((v^0,\vec 0)\). If both are null, write each as \(v^0(1,\hat n)\). "Future-pointing" means the time component is positive.

Rung 3

For part 3, compute \(s\) before and after: before, \(s = m^2c^4 - 2c^2\eta(p,k)\); after, \(s = m^2c^4\).

Worked solution

Idea. Everything reduces to one fact: two future-pointing causal vectors have negative \(\eta\)-product unless they are parallel light rays. Then an initial state's \(s\) is too small or too large for the proposed final state.

1. Case A: \(v\) timelike. By transitivity choose the frame with \(v = (v^0,\vec 0)\), \(v^0 \gt 0\). Then \(\eta(v,w) = -v^0w^0\), and \(w^0 \gt 0\) because \(w\) is nonzero, future-pointing and causal (\(|\vec w|\le w^0\) with \(w\ne0\) forces \(w^0 \gt 0\)). So \(\eta(v,w) \lt 0\). The case \(w\) timelike is symmetric. Case B: both null, \(v = v^0(1,\hat n)\), \(w = w^0(1,\hat m)\): \(\eta(v,w) = v^0w^0(-1 + \hat n\cdot\hat m)\le 0\) by Cauchy–Schwarz, with equality iff \(\hat n = \hat m\), i.e. \(v\parallel w\). \(\blacksquare\)

2. The initial state is one photon, \(s = -c^2\eta(k,k) = 0\). The final state needs \(s\ge(2m_ec^2)^2 \gt 0\) by the threshold theorem. Since \(s\) is conserved, impossible — in every frame at once, because \(s\) is invariant.

3. Before: \(s = -c^2\eta(p+k,p+k) = m^2c^4 - 2c^2\eta(p,k)\). By part 1 with \(p\) timelike, \(\eta(p,k) \lt 0\), so \(s \gt m^2c^4\). After: one electron, \(s = m^2c^4\). Contradiction. (Physically: in the electron's final rest frame there is no momentum, so the absorbed photon's momentum would have to be cancelled by the initial electron's, and then the energies cannot balance.)

4. Target at rest, photon energy \(E\): \(s = M^2c^4 + 2EMc^2\ge(M + 2m_e)^2c^4\), so \[ E \ge \frac{(M+2m_e)^2 - M^2}{2M}c^2 = 2m_ec^2\Big(1 + \frac{m_e}{M}\Big). \] As \(M\to\infty\) this tends to \(2m_ec^2 = 1.022\) MeV: a heavy nucleus absorbs momentum without taking energy. For a lead nucleus (\(Mc^2\approx 1.93\times10^5\) MeV) the correction is \(3\times10^{-6}\). For an electron target, \(M = m_e\): \(E\ge4m_ec^2 = 2.044\) MeV ("triplet production", three leptons in the final state).

Check. Dimensions: \(m_e/M\) dimensionless ✓. Part 4's formula at \(M = m_e\) reproduces the direct calculation \(s = m^2c^4 + 2Emc^2 \ge 9m^2c^4\) ✓. The nucleus's role is the one it plays in the proof of part 2: without a second body, \(s = 0\).

P6 Two thresholds, twelve orders of magnitude apart medium

exercises: threshold computation in the easiest frame, head-on kinematics, approximations justified, numbers

  1. Photons hit protons at rest: \(\gamma + p\to p + \pi^0\). Find the threshold photon energy.
  2. An ultra-high-energy cosmic-ray proton meets, head on, a photon of the cosmic microwave background with the background's mean energy \(\varepsilon = 6.34\times10^{-4}\) eV. Find the proton energy above which \(p + \gamma\to p + \pi^0\) is possible. Say which approximation you make and why it is harmless.
  3. Explain, with the formula for \(s\) in each case, why the two answers differ by about twelve orders of magnitude.
Hints · 3 rungs
Rung 1

Two numbers from the same inequality, \(s\ge(m_p + m_\pi)^2c^4\). Only the initial state's \(s\) differs.

Rung 2

In part 2 the proton has \(E\gg m_pc^2\), so \(|\vec p|c\) is \(E\) to excellent accuracy. Head on, the photon's momentum points opposite to the proton's, so the \(\vec p\cdot\vec k\) term adds rather than subtracts.

Rung 3

First line of part 2: \(s = m_p^2c^4 - 2c^2\eta(p,k) = m_p^2c^4 + 2\varepsilon(E + |\vec p|c)\approx m_p^2c^4 + 4E\varepsilon\).

Worked solution

Idea. Same reaction, same threshold \(s\); compute the initial \(s\) where it is easiest, which is the lab in both cases.

1. \(s = m_p^2c^4 + 2Em_pc^2\ge(m_p + m_\pi)^2c^4\) gives \[ E\ge\frac{m_\pi c^2\,(2m_p + m_\pi)}{2m_p} = \frac{134.977\times2011.521}{2\times938.272}\ \text{MeV} = 144.69\ \text{MeV}. \]

2. Proton \(p = (E/c,\ \vec p)\), photon \(k = \frac\varepsilon c(1,-\hat p)\). Then \(\eta(p,k) = -\frac{E\varepsilon}{c^2} - \frac{|\vec p|\varepsilon}{c}\), so \(s = m_p^2c^4 + 2\varepsilon(E + |\vec p|c)\). With \(|\vec p|c = E\sqrt{1 - (m_pc^2/E)^2}\) and \(E\sim10^{20}\) eV, the relative error in replacing \(|\vec p|c\) by \(E\) is about \(\tfrac12(m_pc^2/E)^2\sim10^{-22}\): harmless. So \(s\approx m_p^2c^4 + 4E\varepsilon\) and \[ E\ge\frac{m_\pi c^2\,(2m_p + m_\pi)c^2}{4\varepsilon} = \frac{2.7151\times10^{5}\ \text{MeV}^2}{4\times6.34\times10^{-10}\ \text{MeV}} = 1.07\times10^{14}\ \text{MeV} = 1.07\times10^{20}\ \text{eV}. \]

3. In both cases \(s - m_p^2c^4\) is linear in the product of the two energies: \(2Em_pc^2\) in part 1 and \(4E\varepsilon\) in part 2. The threshold needs that product to be about \(m_\pi c^2\cdot m_pc^2\sim 10^{5}\ \text{MeV}^2\). With a target of \(938\) MeV the beam needs \(\sim10^2\) MeV; with a target of \(6\times10^{-10}\) MeV it needs \(\sim10^{14}\) MeV. The targets differ by a factor \(938/(6.34\times10^{-10}) = 1.5\times10^{12}\), and so, inversely, do the thresholds: \(1.07\times10^{20}\) eV against \(1.45\times10^{8}\) eV, twelve orders of magnitude.

Check. Part 1 as \(m_p\to\infty\): \(E\to m_\pi c^2\), no recoil energy ✓. Part 2 dimensions: \(\text{MeV}^2/\text{MeV}\) ✓. This is the Greisen–Zatsepin–Kuzmin (GZK) cutoff (1966): protons above it lose energy on the microwave background within about \(10^8\) light years, so few arrive from further. The observed suppression sets in near \(4\text{–}5\times10^{19}\) eV rather than at this estimate, because the reaction is strongest at the \(\Delta(1232)\) resonance and the background photons have a spread of energies — refinements of the same kinematics (cited: Pierre Auger Collaboration, Phys. Rev. Lett. 101, 061101 (2008)).

P7 A two-body decay, and finding the Λ medium

exercises: two-body decay kinematics, factorising, invariant-mass reconstruction, sensitivity, numbers

  1. A particle of mass \(M\) at rest decays to masses \(m_1\), \(m_2\). Show that \[ E_1 = \frac{(M^2 + m_1^2 - m_2^2)\,c^2}{2M},\qquad |\vec p|c = \frac{c^2}{2M}\sqrt{\big[M^2 - (m_1+m_2)^2\big]\big[M^2 - (m_1-m_2)^2\big]}. \]
  2. Apply it to \(\Lambda\to p\,\pi^-\). Find \(|\vec p|\), \(E_p\) and \(E_\pi\).
  3. In a bubble chamber, two tracks leave a common vertex: a proton of momentum \(1000\) MeV/\(c\) and a \(\pi^-\) of momentum \(200\) MeV/\(c\), with opening angle \(35.9^\circ\). Compute the pair's invariant mass. How much does it change if the angle is \(36.9^\circ\) instead?
Hints · 3 rungs
Rung 1

Part 1 is the deck's pion-decay computation with \(m_2\ne0\), followed by a factorisation. Part 3 is a squared sum of two four-momenta, evaluated in the lab where they were measured.

Rung 2

Isolate \(p_2 = P - p_1\) and square. For the factorisation, \(a^2 - b^2 = (a-b)(a+b)\) twice.

Rung 3

Part 3: \(M^2c^4 = m_p^2c^4 + m_\pi^2c^4 + 2\big(E_pE_\pi - |\vec p_p||\vec p_\pi|c^2\cos\theta\big)\).

Worked solution

Idea. Squaring gives the energies; the mass shell gives the momentum; and the same squaring, run backwards on measured tracks, tells you what decayed.

1. \(p_2 = P - p_1\), so \(-m_2^2c^2 = -M^2c^2 - m_1^2c^2 - 2\eta(P,p_1)\). At rest \(\eta(P,p_1) = -ME_1\), giving \(E_1 = (M^2 + m_1^2 - m_2^2)c^2/2M\). Then \[ p^2c^2 = E_1^2 - m_1^2c^4 = \frac{c^4}{4M^2}\Big[(M^2 + m_1^2 - m_2^2)^2 - 4M^2m_1^2\Big]. \] The bracket is \((M^2 + m_1^2 - m_2^2 - 2Mm_1)(M^2 + m_1^2 - m_2^2 + 2Mm_1) = \big[(M-m_1)^2 - m_2^2\big]\big[(M+m_1)^2 - m_2^2\big]\) \(= (M - m_1 - m_2)(M - m_1 + m_2)(M + m_1 - m_2)(M + m_1 + m_2)\), which regroups as \(\big[M^2 - (m_1 + m_2)^2\big]\big[M^2 - (m_1 - m_2)^2\big]\).

2. \(M = 1115.683\), \(m_1 = 938.272\), \(m_2 = 139.570\) (MeV/\(c^2\)): \(|\vec p|c = 100.58\) MeV, \(E_p = 943.65\) MeV, \(E_\pi = 172.04\) MeV. The decay releases \(Q = (M - m_1 - m_2)c^2 = 37.84\) MeV of kinetic energy, and the light pion takes \(32.5\) MeV of it.

3. \(E_p = \sqrt{1000^2 + 938.272^2} = 1371.26\) MeV, \(E_\pi = \sqrt{200^2 + 139.570^2} = 243.89\) MeV. Then \[ M^2c^4 = 938.272^2 + 139.570^2 + 2\big(1371.26\times243.89 - 1000\times200\cos35.9^\circ\big)\ \text{MeV}^2, \] \(Mc^2 = 1115.65\) MeV — a \(\Lambda\), to within the precision of the angle. At \(36.9^\circ\): \(1117.51\) MeV, about \(1.9\) MeV per degree. A mass peak's width in such an experiment is set mostly by angle and momentum resolution.

Check. Energy in part 2: \(943.65 + 172.04 = 1115.69\approx M\) ✓ (rounding). Part 1 limits: \(m_2 = 0\), \(m_1 = m_\mu\) gives the deck's \(|\vec p|c = (M^2 - m_\mu^2)c^2/2M\) ✓; \(M = m_1 + m_2\) gives \(|\vec p| = 0\) ✓. Dimensions: every term in part 3 is \(\text{MeV}^2\) ✓.

The action and the Hamiltonian

P8 Proof: the Legendre transform, three ways medium

exercises: the Legendre transform as a supremum, the massless limit, Euler's homogeneous-function theorem, a proof

  1. Starting from \(H(\vec p) = \sqrt{p^2c^2 + m^2c^4}\), compute \(H^*(\vec v) = \sup_{\vec p}\big(\vec p\cdot\vec v - H(\vec p)\big)\) for every \(\vec v\in\mathbb R^3\), including \(|\vec v|\ge c\). Confirm that it returns \(-mc^2\sqrt{1 - v^2/c^2}\) inside the ball, and interpret what it does outside.
  2. Now \(m = 0\): \(H = |\vec p|c\). Compute \(H^*\). Explain why no Lagrangian \(L(\vec v)\) describes a massless particle usefully.
  3. Prove: if \(L(w)\) is homogeneous of degree one on a cone in \(\mathbb R^n\) and differentiable, then \(w^\mu\,\partial L/\partial w^\mu = L\). Deduce that the reparametrisation-invariant free particle has canonical Hamiltonian identically zero.
Hints · 3 rungs
Rung 1

Part 1 is a function on all of \(\mathbb R^3\) taking value \(+\infty\) somewhere; part 2 is unit 14's second column in three dimensions; part 3 is a one-line differentiation.

Rung 2

For \(|\vec v|\ge c\), take \(\vec p = t\hat v\) and let \(t\to\infty\); compare \(tv\) with \(\sqrt{t^2c^2 + m^2c^4}\). For \(|\vec v| \lt c\), the supremum is attained where \(\vec v = \partial H/\partial\vec p\).

Rung 3

Part 3: differentiate \(L(kw) = kL(w)\) with respect to \(k\) and set \(k = 1\).

Worked solution

Idea. The supremum definition handles the boundary of the speed limit automatically: it assigns infinite Lagrangian — infinite cost — to forbidden velocities.

1. Inside, \(|\vec v| \lt c\): the function \(\vec p\mapsto\vec p\cdot\vec v - H\) is strictly concave (its Hessian is minus \(H\)'s, and \(H\) is strictly convex), so its critical point is the maximum: \(\vec v = \vec pc^2/H\), solved by \(\vec p = \gamma m\vec v\), \(H = \gamma mc^2\), and \(H^* = \gamma mv^2 - \gamma mc^2 = -mc^2/\gamma\) ✓. On the sphere \(|\vec v| = c\): \(\vec p\cdot\vec v - H\le|\vec p|c - \sqrt{p^2c^2 + m^2c^4} \lt 0\), and along \(\vec p = t\hat v\) this tends to \(0\) as \(t\to\infty\); so \(H^* = 0\), not attained. Outside, \(|\vec v| \gt c\): along \(\vec p = t\hat v\), \(tv - \sqrt{t^2c^2 + m^2c^4}\ge t(v - c) - mc^2\to\infty\), so \(H^* = +\infty\). A Lagrangian that is \(+\infty\) outside the closed ball assigns infinite action to any path faster than light: the speed limit is built into the convex dual.

2. \(H^*(\vec v) = \sup_{\vec p}(\vec p\cdot\vec v - |\vec p|c)\). If \(|\vec v|\le c\), \(\vec p\cdot\vec v\le|\vec p|c\), so the bracket is \(\le 0\), with \(0\) at \(\vec p = 0\): \(H^* = 0\). If \(|\vec v| \gt c\), \(\vec p = t\hat v\) gives \(t(|\vec v| - c)\to\infty\). So \(L = 0\) on the closed ball and \(+\infty\) outside. Every admissible path then has action zero, every one is stationary, and the Euler–Lagrange equations say nothing: the Lagrangian carries only the constraint. A massless particle is described by its Hamiltonian \(|\vec p|c\), or by a reparametrisation-invariant action with an auxiliary variable (not needed in this course).

3. From \(L(kw) = kL(w)\) for \(k \gt 0\), differentiate in \(k\) by the chain rule: \(w^\mu\,(\partial L/\partial w^\mu)(kw) = L(w)\); at \(k = 1\), \(w^\mu\partial_\mu L(w) = L(w)\). \(\blacksquare\) For \(L(\gamma') = -mc\sqrt{-\eta(\gamma',\gamma')}\), homogeneous of degree one, the canonical Hamiltonian \(p_\mu\gamma'^\mu - L = \gamma'^\mu\partial_\mu L - L = 0\) identically.

Check. Dimensions in part 1: \(\vec p\cdot\vec v\) and \(H\) are both energies ✓. Involutivity: transforming \(H^*\) again returns \(H\), which is unit 14's theorem for a convex lower-semicontinuous function; the value \(0\) on the sphere is exactly what lower semicontinuity requires ✓. Limit \(m\to0\) of part 1: \(-mc^2/\gamma\to0\) inside the ball, which is part 2 ✓.

Rockets, acceleration and rigidity

P9 The photon rocket medium

exercises: four-momentum conservation in an instantaneous rest frame, rapidity additivity, numbers

  1. Rederive the rocket equation \(\theta = (w/c)\ln(M_i/M_f)\) for exhaust ejected at speed \(w\) in the rocket's rest frame. Say exactly where rapidity additivity is used.
  2. For a photon rocket (\(w = c\)) show that the final speed is \(v/c = (R^2 - 1)/(R^2 + 1)\) with \(R = M_i/M_f\).
  3. A photon rocket must reach \(0.9c\). Find \(R\); then \(R\) for reaching \(0.9c\) and stopping again at the destination; then for a round trip that stops at both ends. Repeat the first for an exhaust speed of \(0.1c\).
Hints · 3 rungs
Rung 1

Part 1 is a first-order separable ODE for \(\theta\) as a function of \(M\) (unit 00). Parts 2 and 3 are exponentials of rapidities.

Rung 2

Write four-momentum conservation for one instant in the rocket's rest frame, to first order in small quantities. Then \(\mathrm dv\) in that frame equals \(c\,\mathrm d\theta\).

Rung 3

Part 2: \(\tanh\theta = (e^{2\theta} - 1)/(e^{2\theta} + 1)\) and \(e^\theta = R\).

Worked solution

Idea. In the instantaneous rest frame the problem is Newtonian to first order; rapidity is the quantity whose small increments in successive rest frames simply add.

1. Rest frame at one instant: before, \((Mc,\vec 0)\). After: rocket \((M + \mathrm dM)(c,\ \mathrm dv)\) to first order, exhaust \(\mu\gamma_w(c, -w)\). Time components: \(\mu\gamma_w = -\mathrm dM\). Space: \(0 = M\,\mathrm dv - \mu\gamma_ww\), so \(M\,\mathrm dv = -w\,\mathrm dM\). A velocity increment \(\mathrm dv\) from rest is a rapidity increment \(\mathrm d\theta = \mathrm dv/c\). Here additivity is used: the rapidity in the launch frame is the sum of these increments because rapidities of collinear boosts add (unit 26). So \(\mathrm d\theta/\mathrm dM = -w/(cM)\), separable, and \(\theta = (w/c)\ln(M_i/M_f)\).

2. \(w = c\): \(e^\theta = R\), so \(v/c = \tanh\theta = (R^2 - 1)/(R^2 + 1)\).

3. \(\theta = \operatorname{artanh}0.9 = \tfrac12\ln19 = 1.4722\), so \(R = \sqrt{19} = 4.36\). Stopping doubles \(\theta\): \(R = 19\). A round trip stopping at both ends: \(R = 19^2 = 361\). With \(w = 0.1c\): \(R = e^{10\theta} = 19^5 = 2\,476\,099\).

Check. Part 2 with \(R = \sqrt{19}\): \((19 - 1)/(19 + 1) = 0.9\) ✓. Limit \(w\ll c\), small \(\theta\): \(v = w\ln R\), Tsiolkovsky ✓. Note \(\mu \lt -\mathrm dM\): with \(\gamma_w \gt 1\) some rest mass becomes exhaust kinetic energy; for photons all of it does (\(\mu = 0\), \(\mu\gamma_w\) read as \(E_{\text{exhaust}}/c^2\)).

P10 To Proxima at one \(g\) medium

exercises: hyperbolic motion, rapidity, the Rindler horizon, numbers

  1. A ship accelerates from rest at proper acceleration \(g\) for the first half of the \(4.2465\) ly to Proxima Centauri, then decelerates at \(g\) for the second half. Find the proper time of the trip, the Earth-frame time, and the top speed.
  2. During the first half, mission control at the launch point sends a message at Earth time \(t = c/2g\). Show that the ship receives it at proper time \(\tau = (c/g)\ln 2\), and evaluate. What happens to messages sent at \(t\ge c/g\) if the ship keeps accelerating for ever?
Hints · 3 rungs
Rung 1

Everything is the hyperbola \(x = (c^2/g)\cosh\theta\), \(ct = (c^2/g)\sinh\theta\), \(\theta = g\tau/c\), with the launch point at \(x = c^2/g\). The trip is two mirror-image halves.

Rung 2

Half-distance: \(\frac{D}{2} = \frac{c^2}{g}(\cosh\theta - 1)\) fixes \(\theta\) at the midpoint. For part 2, intersect the light line \(x = c^2/g + c(t - t_e)\) with the hyperbola.

Rung 3

Part 2, writing \(A = c^2/g\) and \(d = ct_e = A/2\): the intersection has \(ct = \dfrac{d(2A - d)}{2(A - d)} = \tfrac34A\), so \(\sinh(g\tau/c) = \tfrac34\).

Worked solution

Idea. Rapidity grows linearly in proper time; distance and Earth time are \(\cosh\) and \(\sinh\) of it.

1. \(\cosh\theta = 1 + \dfrac{gD}{2c^2} = 1 + \dfrac{2.12325}{0.968715} = 3.19182\), so \(\theta = 1.82824\). Each half takes proper time \((c/g)\theta = 1.7710\) yr and Earth time \((c/g)\sinh\theta = 0.968715\times3.03113 = 2.9363\) yr. Totals: \(\tau = 3.542\) yr, \(t = 5.873\) yr. Top speed \(\tanh\theta = 0.9497c\). (A photon rocket would need \(R = e^{2\theta} = 38.7\).)

2. Set \(A = c^2/g\). The signal \(x = A + (ct - d)\) meets \(x = \sqrt{A^2 + c^2t^2}\) where \((A + ct - d)^2 = A^2 + c^2t^2\), i.e. \(2ct(A - d) = d(2A - d)\). With \(d = A/2\): \(ct = \tfrac34A\). Then \(\sinh(g\tau/c) = ct/A = \tfrac34\), and \(\operatorname{arsinh}\tfrac34 = \ln\big(\tfrac34 + \sqrt{1 + \tfrac9{16}}\big) = \ln 2\). So \(\tau = (c/g)\ln2 = 0.6715\) yr — within the first half (\(1.771\) yr), so the ship is still accelerating ✓. If \(d\ge A\), that is \(t_e\ge c/g = 0.9687\) yr, the equation \(2ct(A - d) = d(2A - d)\) has no solution with \(ct \gt d\), that is after the emission (for \(d \gt 2A\) it has a positive root, but an earlier one, an artefact of squaring): the message never arrives. For \(d \lt A\) the root does exceed \(d\), since \(d(2A - d) \gt 2d(A - d)\). The launch point has crossed the ship's Rindler horizon.

Check. \(t = 5.873\) yr exceeds the light time \(4.2465\) yr ✓, and \(\tau \lt t\) ✓. Small-\(D\) limit: \(\cosh\theta - 1\approx\theta^2/2 = g^2\tau^2/2c^2\) gives \(D/2\approx\tfrac12g\tau^2\), Galileo's kinematics for each half ✓. Dimensions: \(gD/c^2\) dimensionless ✓.

P11 A Born-rigid train, and Bell's thread hard

exercises: Born rigidity, simultaneity on hyperbolae, proper distance, a paradox resolved

  1. A rod of proper length \(\ell\) is accelerated Born-rigidly, its front having proper acceleration \(\alpha_f\). Show that its back has \(\alpha_b = \alpha_f/(1 - \alpha_f\ell/c^2)\). For a \(100\) m train whose front accelerates at \(g\), find \(\alpha_b/\alpha_f - 1\). What is the longest rod whose front can be pulled Born-rigidly at \(g\)?
  2. Bell's spaceships. Two identical ships at rest a distance \(d\) apart in frame \(S\) start at \(t = 0\) and follow identical programmes of constant proper acceleration \(\alpha\). A thread is tied between them. Show that their separation measured in \(S\) stays \(d\). Then show that, when the rear ship has rapidity \(\theta\), the distance to the front ship measured in the rear ship's instantaneous rest frame is \[ d\cosh\theta + \sqrt{A^2 + d^2\sinh^2\theta} - A,\qquad A = c^2/\alpha, \] which is at least \(\gamma d\). Does the thread break?
Hints · 3 rungs
Rung 1

Part 1 uses the Born-rigid family \(x^2 - c^2t^2 = X^2\), proper acceleration \(c^2/X\). Part 2 is about two hyperbolae that are translates, not members of that family.

Rung 2

Put the rear ship's hyperbola centred at the origin, \(x^2 - c^2t^2 = A^2\), and the front's centred at \(x = d\). The rear ship's simultaneity line at rapidity \(\theta\) is the line through the origin with direction \((\sinh\theta,\cosh\theta)\).

Rung 3

Points on that line are \(r(\sinh\theta,\cosh\theta)\) in \((ct, x)\); the front ship is there when \((r\cosh\theta - d)^2 - r^2\sinh^2\theta = A^2\).

Worked solution

Idea. Born rigidity requires the members of a family of hyperbolae with a common centre; identical acceleration programmes give translated hyperbolae instead, and those drift apart in their own rest frames.

1. In the Born-rigid family, point \(X\) has proper acceleration \(c^2/X\). The front is at \(X_f = c^2/\alpha_f\) and the back at \(X_b = X_f - \ell\), so \(\alpha_b = c^2/(X_f - \ell) = \alpha_f/(1 - \alpha_f\ell/c^2)\). For \(\ell = 100\) m and \(\alpha_f = g\): \(g\ell/c^2 = 1.091\times10^{-14}\), and \(\alpha_b/\alpha_f - 1 = 1.091\times10^{-14}\) to that accuracy. The back's acceleration diverges as \(\ell\to c^2/\alpha_f\), so the longest rod is \(c^2/g = 0.9687\) ly — its back would sit on the front's Rindler horizon.

2. Both ships follow \(x_i(t) = x_i(0) + \sqrt{A^2 + c^2t^2} - A\), the same function plus a constant, so \(x_2(t) - x_1(t) = d\) for all \(t\) in \(S\). Now centre the rear hyperbola at the origin, \(x^2 - c^2t^2 = A^2\), so the front's is \((x - d)^2 - c^2t^2 = A^2\). The rear ship at rapidity \(\theta\) is at \(A(\sinh\theta,\cosh\theta)\) and its simultaneity line is \(\{r(\sinh\theta,\cosh\theta)\}\) (the deck's Born-rigidity slide). The front ship meets it where \((r\cosh\theta - d)^2 - r^2\sinh^2\theta = A^2\), i.e. \(r^2 - 2rd\cosh\theta + d^2 - A^2 = 0\), whose positive root is \(r = d\cosh\theta + \sqrt{d^2\cosh^2\theta - d^2 + A^2} = d\cosh\theta + \sqrt{A^2 + d^2\sinh^2\theta}\). Proper distance along the line is \(r - A\), which is the formula. Since \(\sqrt{A^2 + d^2\sinh^2\theta}\ge A\), it is at least \(d\cosh\theta = \gamma d\). The thread, which in its own rest frame wants length \(d\), is stretched by a factor that grows without bound: it breaks.

Check. \(\theta = 0\): \(d + A - A = d\) ✓. Dimensions: every term a length ✓. The paradox — "in \(S\) nothing changes, so why does the thread break?" — dissolves as in part 1: to keep the thread's proper length, the rear ship would need to accelerate harder than the front. John Bell reported that most of CERN's theory division initially said the thread would not break (J. S. Bell, Speakable and Unspeakable in Quantum Mechanics, ch. 9).

The instructive wrong answer

P12 Relativistic Kepler precesses — by the wrong amount hard

exercises: a non-\(T-V\) Lagrangian, Noether charges, the Binet substitution, a linear ODE, a number

Take the model \(L = -mc^2\sqrt{1 - v^2/c^2} + k/r\) in the plane, \(k \gt 0\): relativistic kinematics, Newtonian attraction.

  1. Show that \(E = \gamma mc^2 - k/r\) and \(\ell = \gamma mr^2\dot\phi\) are conserved.
  2. With \(u = 1/r\) and \(u' = \mathrm du/\mathrm d\phi\), show that \(\ell^2c^2(u'^2 + u^2) = (E + ku)^2 - m^2c^4\), and hence \[ u'' + \Big(1 - \frac{k^2}{\ell^2c^2}\Big)u = \frac{kE}{\ell^2c^2}. \]
  3. Solve it, and show that the perihelion advances by \(\delta = 2\pi\big[(1 - k^2/\ell^2c^2)^{-1/2} - 1\big]\approx\pi k^2/\ell^2c^2\) per revolution.
  4. For gravity, \(k = GM_\odot m\) and, to leading order, \(\ell^2 = m^2GM_\odot a(1 - e^2)\). Evaluate the advance for Mercury (\(a = 5.7909\times10^{10}\) m, \(e = 0.20563\), period \(87.969\) days) in arcseconds per century, and compare with general relativity's \(42.98''\) per century.
  5. What happens to the orbit if \(\ell c \lt k\)?
Hints · 3 rungs
Rung 1

Two Noether charges (unit 13), one first-order equation for the orbit, differentiated once to give a linear equation with constant coefficients (unit 06), whose solution is a cosine with the "wrong" period in \(\phi\).

Rung 2

Combine \(E + ku = \gamma mc^2\) with the mass shell \((\gamma mc^2)^2 = p^2c^2 + m^2c^4\), and split \(p^2 = (\gamma m\dot r)^2 + \ell^2/r^2\). The Binet trick: \(\gamma m\dot r = -\ell u'\).

Rung 3

\(\dot r = \dfrac{\mathrm dr}{\mathrm d\phi}\dot\phi = -\dfrac{u'}{u^2}\cdot\dfrac{\ell u^2}{\gamma m}\), so \(\gamma m\dot r = -\ell u'\). Then differentiate \(\ell^2c^2(u'^2 + u^2) = (E + ku)^2 - m^2c^4\) with respect to \(\phi\) and divide by \(2u'\).

Worked solution

Idea. Relativistic kinematics turns unit 08's \(u'' + u = \text{const}\) into \(u'' + \omega^2u = \text{const}\) with \(\omega\) slightly less than one, so the radial oscillation takes slightly more than \(2\pi\) of angle to repeat.

1. \(L\) has no explicit \(t\), so unit 13's energy function is conserved: \(h = \vec v\cdot\partial L/\partial\vec v - L = \gamma mv^2 + mc^2/\gamma - k/r = \gamma mc^2 - k/r\) (the deck's computation, plus \(-k/r\)). \(L\) depends on \(\phi\) only through \(v^2 = \dot r^2 + r^2\dot\phi^2\), so \(\phi\) is cyclic and \(\partial L/\partial\dot\phi = \gamma m\cdot r^2\dot\phi = \ell\) is conserved.

2. \(\gamma mc^2 = E + ku\). The momentum \(\vec p = \gamma m\vec v\) has radial part \(\gamma m\dot r\) and transverse part \(\gamma mr\dot\phi = \ell/r = \ell u\). By the rung, \(\gamma m\dot r = -\ell u'\), so \(p^2 = \ell^2(u'^2 + u^2)\). The mass shell \((\gamma mc^2)^2 = p^2c^2 + m^2c^4\) gives \(\ell^2c^2(u'^2 + u^2) = (E + ku)^2 - m^2c^4\). Differentiate in \(\phi\): \(2\ell^2c^2u'(u'' + u) = 2ku'(E + ku)\). On any non-circular orbit \(u'\ne0\) except at isolated points, so, by continuity everywhere, \(u'' + u = k(E + ku)/\ell^2c^2\), which rearranges to the stated equation.

3. Write \(\omega^2 = 1 - k^2/\ell^2c^2\), assumed positive. Unit 06: the general solution is \(u = \dfrac{kE}{\ell^2c^2\omega^2}\big[1 + e\cos(\omega(\phi - \phi_0))\big]\). Perihelion is a maximum of \(u\), recurring when \(\omega\,\Delta\phi = 2\pi\), so \(\Delta\phi = 2\pi/\omega\) and \(\delta = 2\pi(1/\omega - 1)\). With \(x = k^2/\ell^2c^2\) small, \((1 - x)^{-1/2} - 1 = x/2 + O(x^2)\), so \(\delta\approx\pi k^2/\ell^2c^2\).

4. \(k^2/\ell^2c^2 = G^2M^2m^2/(m^2GMa(1-e^2)c^2) = GM_\odot/(a(1-e^2)c^2) = 2.6625\times10^{-8}\). So \(\delta = 8.364\times10^{-8}\) rad per orbit. A century holds \(36\,525/87.969 = 415.20\) orbits, and one radian is \(206\,265''\): \(\delta = 7.16''\) per century. General relativity's \(6\pi GM_\odot/(a(1-e^2)c^2)\) gives \(42.98''\): the model gets exactly one sixth, and the prograde sign.

5. \(\omega^2 \lt 0\): the solutions are \(\cosh\) and \(\sinh\) in \(\phi\), and on an infalling orbit \(u\) grows exponentially in \(\phi\). It is finite at every finite \(\phi\), so the particle winds round infinitely often; but \(\mathrm dt = \gamma m\,\mathrm d\phi/(\ell u^2) = (E + ku)\,\mathrm d\phi/(\ell c^2u^2)\), which behaves like \(k\,\mathrm d\phi/(\ell c^2u)\) and so is integrable against an exponentially growing \(u\): it spirals into \(r = 0\) in a finite time. The relativistic centrifugal barrier \(\ell^2/r^2\) is under the square root in \(\sqrt{p^2c^2 + m^2c^4}\) and grows only like \(1/r\), no faster than the attraction. (In Sommerfeld's 1916 relativistic hydrogen atom, where this model is physically sound, the same condition becomes a bound on the nuclear charge.)

Check. Dimensions: \(k/\ell c\) is \((\mathrm{J\,m})/(\mathrm{J\,s}\cdot\mathrm{m\,s^{-1}})\), dimensionless ✓. Limit \(c\to\infty\): \(\omega\to1\), unit 08's closed ellipse with \(u = (mk/\ell^2)(1 + e\cos\phi)\), since \(E\to mc^2\) makes \(kE/\ell^2c^2\to km/\ell^2\) ✓. Why this is the most instructive wrong answer: the model has time dilation of the planet's clock and nothing else, and it misses by a factor of six. The other five sixths come from gravity slowing clocks deeper in the potential and curving space — things a force in flat spacetime cannot represent. Unit 37 derives the \(6\pi\) from \(u'' + u = GM/L^2 + 3GMu^2/c^2\).