Problemata: Physics

Most of this notebook is derivation: long chains of steps written out in full so the reasoning stays recoverable. This page is the opposite exercise. Each entry is a short question about the physical world — statable in a few sentences, answerable rightly or wrongly before any machinery is switched on. The solution sits folded beneath the question.

The intended use is strict. Read the question, commit to an answer — say it out loud if you have to — and only then open the fold. A question you answer before checking teaches something; a question you read through teaches almost nothing.

The genre is old. The Peripatetic Problemata collected hundreds of short physical questions — why does…, whether… — and the format has survived because it exercises something derivations do not: saying what happens before calculating why. Where the machinery is genuinely needed, each solution links back down into the notebook where it lives.

Questions of this kind are folklore. Most have been asked in one form or another for a century, some for much longer; the statements and solutions here are my own wording. The collection opens with relativity; mechanics and quantum questions will join it, and the numbering simply continues.

Relativity

Ten questions, roughly in order of increasing depth. The first five need nothing beyond special relativity; the middle three run on the equivalence principle; the last two need a black hole and an expanding universe.

I. The lever that outruns light

Problema I

A rod one light-year long, made as stiff as physics allows, floats at rest in front of you. You shove your end forward by one meter. When does the far end move — and if the answer were “immediately,” could you use the rod to send messages faster than light?

Solution

Tens of thousands of years later — and no.

A perfectly rigid body is a Newtonian fiction. Your shove propagates down the rod as a compression wave at the material’s sound speed, $c_s=\sqrt{\partial p/\partial \rho }$, and relativistic causality caps every signal — elastic waves included — at $c$. Stiffness is not a loophole: the elastic moduli enter the stress-energy tensor, and a material with $c_s>c$ would have to violate the causal structure of the field equations that govern it. For diamond, $c_s\approx 12km/s$, so the far end of a light-year rod answers your shove roughly $c/c_s\approx 25,000$ years later.

The deeper point is that relativity does not merely forbid building a rigid rod; it forbids defining one that can be commanded from one end. The strongest notion of rigidity the theory admits — Born rigidity, constant proper distance between neighboring pieces — is so restrictive (this is the Herglotz–Noether theorem) that a Born-rigid body has essentially no independent degrees of freedom left to push.

Deeper in the notebook: 01. Timelike, Null and Spacelike Separation; the Causal Order · 03. Tachyons and the Limits of Causality (signpost) · 03. Bell’s Spaceship Paradox and Born Rigidity

II. The photograph of a passing sphere

Problema II

A sphere flies past you at $0.9c$ and you photograph it side-on. Length contraction flattens it along its direction of motion by a factor of $\gamma \approx 2.3$. Does the photograph show a flattened ellipsoid?

Solution

No — the outline in the photograph is still a perfect circle.

A photograph does not record where the parts of the sphere are at one instant; it records photons arriving at one instant, which left different parts of the sphere at different times. The light from the trailing edge left earlier, when the sphere was further back, and this light-travel delay exactly compensates the Lorentz contraction of the outline. What survives is an apparent rotation of the surface pattern — the Terrell–Penrose effect: a fast sphere looks like a sphere turned, never a sphere squashed.

The trap is conflating two operations. Measuring length means locating both endpoints simultaneously in your frame — that gives the $\gamma$-contraction. Seeing folds light-travel time into the picture, and aberration happens to map circles to circles. Contraction is real; it is just not what a camera measures.

Deeper in the notebook: 07. Relativistic Doppler Effect, Aberration, and Beaming · 01. Minkowski Spacetime and the Lorentz Group

III. Two rockets and a thread

Problema III

Two identical rockets float at rest, one ahead of the other, joined by a taut, fragile thread. At $t=0$ in your frame both fire identical engines and follow identical velocity profiles, so in your frame their separation never changes. Does the thread break?

Solution

Yes.

In your frame the separation is constant by construction — but the thread is now moving, so its unstressed length is Lorentz-contracted. A thread forced to span an uncontracted gap while its equilibrium length shrinks is a thread under growing strain, and it snaps. In the rockets’ instantaneous rest frames the same story reads differently: there the separation between the ships grows (the leading ship, by relativity of simultaneity, “started earlier”), and the thread is stretched outright.

To keep the thread relaxed, the ships would have to maintain constant proper separation — Born-rigid motion — and that requires the trailing ship to accelerate harder than the leader, the two following different Rindler hyperbolae with $a\propto 1/x$. Identical accelerations and rigid formation are incompatible demands; Bell’s spaceships are the cleanest demonstration that length contraction has dynamical teeth.

Deeper in the notebook: 03. Bell’s Spaceship Paradox and Born Rigidity · 01. Uniform Acceleration; the Rindler Wedge and Horizon

IV. The pole and the barn

Problema IV

A 20-meter pole is carried at $\gamma =2$ toward a 10-meter barn with doors at both ends. The farmer says: contracted to 10 meters, the pole fits — slam both doors simultaneously, and for an instant it is entirely inside. The runner says: the barn is contracted to 5 meters, and the pole never fit at any moment. Both cannot be right about the doors — can they?

Solution

Both are right, and the doors never contradict either of them.

“The pole fits” secretly means “both ends are inside at the same time” — it is a simultaneity claim, not an invariant statement. The two door-slams are simultaneous in the barn frame only. In the runner’s frame the rear door slams and reopens before the front end of the pole reaches the far door; the slams happen in sequence, and at no moment is the pole enclosed. The two events are spacelike separated, so their time-order is frame-dependent, and no door ever touches the pole in either account.

The paradox only acquires teeth if you demand the pole stop inside the barn — and then Problema I collapses it: the pole cannot stop rigidly. The front stops first, a compression wave runs back, and what ends up enclosed in the barn is a genuinely shortened, crumpled pole in every frame.

Deeper in the notebook: 02. The Ladder (Pole-and-Barn) Paradox · 01. Timelike, Null and Spacelike Separation; the Causal Order

V. A box full of light

Problema V

A perfectly mirrored box sits on a scale. You fill it with photons of total energy $E$ — particles with zero mass. Does the scale read more?

Solution

Yes: heavier by exactly $E/c^2$.

Mass is not the sum of the masses of the parts; it is the norm of the total four-momentum. Already two photons of energy $E$ flying in opposite directions form a system with ${\mathbf{p}}_{\mathrm{tot}}=0$ and $E_{\mathrm{tot}}=2E$, hence invariant mass $2E/c^2$ — massless constituents, massive whole. The box of light is a composite at rest with rest energy increased by $E$, and it gravitates, and weighs, accordingly.

The scale learns this through radiation pressure. Photons bouncing off the floor have fallen through the box’s height and arrive slightly blueshifted; photons hitting the ceiling arrive slightly redshifted. The floor is pushed harder than the ceiling, and integrating the imbalance over the photon gas gives precisely $(E/c^2)g$. The bookkeeping device that makes all such accounts come out consistent is the stress-energy tensor: energy density, momentum flux, and pressure all gravitate together.

Deeper in the notebook: 06. Relativistic Mechanics and the Stress-Energy Tensor · 01. Gravitational Redshift

VI. The thrown clock

Problema VI

You must send a clock away from your lab bench and have it back in exactly $T$ seconds of bench time — but you want the clock itself to have ticked off as much time as possible when it returns. Carry it, drive it, fly it, throw it: which round trip wins?

Solution

Throw it straight up, so that it is in free fall the whole time, and gravity brings it back at $T$. (The winning toss peaks at height $g{T}^2/8$ — about 20 meters for $T=4\mathrm{s}$.)

In the weak-field limit a clock’s rate is $d\tau \approx \left(1+\frac{gh}{c^2}-\frac{v^2}{2c^2}\right)dt$: it ticks faster the higher it sits and slower the faster it moves. Maximizing $\int d\tau$ therefore means buying as much altitude as possible while spending as little speed as possible — and the optimal compromise is exactly the free-fall parabola. This is no coincidence: maximizing $\int \left(gh-\frac{v^2}{2}\right)dt$ is the same variational problem as extremizing $\int (T_{\mathrm{kin}}-U)dt$. Hamilton’s principle of least action, for a projectile, is proper-time maximization read in Newtonian light.

That is the honest content of the geodesic hypothesis: free fall is not a force-driven motion but the worldline of greatest aging. Things fall because falling is how you age the most.

Deeper in the notebook: 03. Local Inertial Frames; the Geodesic Hypothesis · 04. Gravity as Geometry - the Heuristic Argument · 01. The Twin Paradox

VII. Satellite clocks: fast or slow?

Problema VII

Compared to a clock on the ground, does an astronaut’s clock on the ISS (400 km up) run fast or slow? Same question for a GPS satellite clock (20,200 km up). Is the sign even the same?

Solution

Opposite signs: the ISS clock runs slow by about $25\mu \mathrm{s}$ per day; a GPS clock runs fast by about $38\mu \mathrm{s}$ per day.

Two effects compete. Sitting higher in the potential makes a clock run faster (gravitational blueshift, about $+46\mu \mathrm{s}/\mathrm{day}$ for GPS, only $+4$ for the ISS); orbiting fast makes it run slower (about $-7\mu \mathrm{s}/\mathrm{day}$ for GPS, $-28$ for the low, fast ISS). For a circular orbit both effects pack into one clean formula, $d\tau /dt=\sqrt{1-3GM/r{c}^2}$, to be compared against a ground clock’s $\sqrt{1-2GM/R{c}^2}$ (ignoring Earth’s spin, a fraction-of-a-microsecond refinement).

Setting them equal gives the crossover at $r=\frac{3}{2}R$ — about 3,200 km of altitude. Below that, speed wins and orbiting clocks lag; above it, potential wins and they lead. GPS engineering does not get to debate whether relativity is real: uncorrected, the system’s position error would grow by roughly ten kilometers per day.

Deeper in the notebook: 02. Time Dilation and Clocks (GPS) · 01. Gravitational Redshift

VIII. The marbles in the falling elevator

Problema VIII

Einstein’s elevator: sealed in a freely falling lab, you are supposed to be unable to distinguish it from a lab floating in deep space. You have two marbles and as much patience as you like. Can you nevertheless prove there is a planet outside?

Solution

Yes — the equivalence principle is local, and two marbles are enough to probe “non-local.”

Release the marbles a horizontal distance $d$ apart: each falls toward the center of the Earth, so their paths converge, and they drift together at tidal acceleration $\sim GMd/r^3$. Release them one above the other instead: the lower one sits in a stronger field and pulls ahead, so they separate at $\sim 2GMd/r^3$. A lab in empty space shows neither. Squeeze in the horizontal plane, stretch along the vertical — that residue is what no choice of freely falling frame can erase.

The formal statement: coordinates can flatten the metric and kill its first derivatives at a point (that is the freely falling frame), but the second derivatives — the Riemann tensor — stay. Gravity’s irreducible signature is not the pull, which you can transform away, but the squeeze, which you cannot. Relative acceleration of nearby free-fallers is geodesic deviation, and it is the curvature tensor read off with rulers and marbles.

Deeper in the notebook: 01. Tidal Forces and Geodesic Deviation · 02. Weak, Einstein, and Strong Equivalence Principles · 01. Jacobi Fields

IX. Hover or orbit?

Problema IX

At the same Schwarzschild radius $r$ outside a black hole, one observer hovers on rockets while another passes by in a circular orbit, meeting the hoverer once per revolution. Problema VI said free fall maximizes aging — so the orbiter, who is in free fall, ages more between meetings. Correct?

Solution

No: the hoverer ages more — and Problema VI is not contradicted, only read too generously.

The rates are unambiguous: $d\tau /dt=\sqrt{1-2GM/r{c}^2}$ for the hoverer versus $\sqrt{1-3GM/r{c}^2}$ for the circular orbit — the orbiter pays the full gravitational toll plus a velocity toll, with nothing in compensation. As $r$ approaches the photon sphere at $3GM/c^2$, the orbiter’s clock rate slides toward zero while the hoverer’s stays finite.

The resolution is that geodesics are local extrema of proper time: maximal only against nearby competing worldlines, and only up to the first conjugate point. A full revolution is not a small variation away from hovering, and it contains a conjugate point (neighboring orbits refocus within an epicyclic period), so the circular orbit between two meetings is a saddle of the proper-time functional, not a maximum. The true champion between those same two events is a third worldline: the radial toss of Problema VI, thrown upward hard enough to fall back exactly one revolution later, which out-ages hoverer and orbiter alike.

Deeper in the notebook: 02. Geodesics, Orbits, the Effective Potential · 02. Conjugate Points and the Second Variation of Arc Length · 01. The Twin Paradox

X. The missing energy of an old photon

Problema X

A CMB photon has been traveling since recombination, and the expansion has stretched its wavelength about 1,100-fold: it has lost 99.9% of the energy it started with. Where did that energy go?

Solution

Nowhere — the question assumes a ledger the universe does not keep.

Conservation laws are bought with symmetries: energy conservation, in particular, with time-translation invariance — in geometric language, a timelike Killing field. An expanding FLRW universe has none; there is no global “total energy of the universe” whose books must balance, so no account the photon’s loss must be debited from. (In a static spacetime the ledger exists: Schwarzschild redshift really is bookkept, via the Killing field, between photon and gravitating source.)

What survives everywhere is the local law ${\nabla }_{\mu }{T}^{\mu \nu }=0$ — a continuity equation, not a global conservation law. For the cosmic fluid it reads $\dot{\rho }=-3H(\rho +p)$: dilution for dust ($\rho \propto a^{-3}$), dilution plus one extra factor for radiation ($\rho \propto a^{-4}$) — and that extra $a^{-1}$ is exactly the redshift. Kinematically, each comoving observer the photon passes is receding from the last one who measured it: energy is observer-dependent, $E=-p_{\mu }{u}^{\mu }$, and a chain of mutually receding observers simply records ever-smaller values. Nothing is lost to anywhere; there was never one number to conserve.

Deeper in the notebook: 02. Redshift · 02. Noether’s Theorem in GR · 07. Killing Vectors and Isometries · 01. The Friedmann Equations and Cosmic Dynamics