Illustrated Physics
physics, drawn
The collection

Every essay — page 10

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence.

Mechanics

Motion, force, and the quantities that refuse to change.

Rough contact grows as W^1.000, and one smooth bump as W^0.667. Real contact area against load, both logarithmic, three ways. Plastic flow of the asperities gives A = W/H, a slope of exactly one, which is the account Bowden and Tabor's argument uses. A single elastic sphere gives Hertz's answer, A ∝ W^(2/3), and a friction coefficient falling as W^(−1/3) — Amontons' law would be false. A rough elastic surface, with many asperities spread exponentially in height and integrated here rather than quoted, gives a slope of 1.0000, because pressing harder mostly recruits new contacts rather than enlarging existing ones. The dotted curve repeats that integration with a Gaussian spread of heights and gives 0.9695 — the upper tail of a Gaussian is nearly exponential, so the answer is nearly and not exactly linear. Amontons' law does not need plasticity. It needs roughness, it is exact for one distribution and good to three per cent for another, and it survives whichever way the individual junctions deform.

The grip that is not a coefficient

A friction coefficient is independent of load and of area, and the reason is usually given as plastic flow of the asperities. It is not — a rough elastic surface gives the same law, integrated here to a slope of 1.0000, because pressing harder recruits new contacts rather than enlarging old ones. What breaks the law is a contact that is smooth, or a material that dissipates in its bulk — where μ passes one and stops being a coefficient at all.

8 figures · part 3 on Friction Indistinguishable for 10.2 seconds, then not. The path of the lower bob for two double pendulums released 1e-8° apart, over 11 seconds, with the arms drawn at the final instant. The two traces lie on top of each other for the first 10.2 seconds — the point at which they are two pixels apart on this canvas — and after that they have nothing to do with one another. Neither is more correct: both are exact solutions of the same equations, differing only in a release angle that no apparatus could set apart. The separation is growing at 2.05 per second the whole time, including during the stretch where the picture shows one curve.

The error that doubles on a schedule

Two double pendulums released a hundred-millionth of a degree apart follow one curve for ten seconds and then have nothing to do with each other. The separation grows exponentially the whole time, including while the picture shows a single trace — which turns unpredictability into a rate, and makes the length of a forecast the logarithm of the precision rather than anything proportional to it.

8 figures · part 1 on Chaos Phase portraits of the standard map at 2 couplings. The standard map p → p + K sin θ, θ → θ + p, iterated 220 times from 12 starting points, at couplings of 0.6 and 1.3. Both coordinates run from 0 to 2π. An orbit that lies on a curve spanning the picture from left to right is an invariant circle, and nothing can cross it; an orbit that fills an area is chaotic; an orbit that circulates round a centre is trapped in a resonance island. At K = 0.6, orbits launched on p = 0 get no further than 1.55 in p, so a spanning curve is still there. At K = 1.3, orbits launched on p = 0 get no further than 10.78 in p, so a spanning curve is gone and transport is global. The point of the pair is that the change between them is not a change of character in any single orbit — chaotic orbits and regular ones coexist on both sides — but the loss of the barriers that kept the chaotic ones local.

The last curve to go

Chaos does not arrive all at once. Order is destroyed a resonance at a time, and there is a coupling — 0.971635, known to six figures — at which the final barrier separating one part of the phase space from another gives way. Below it a chaotic orbit is still trapped; above it nothing stops it.

6 figures · part 2 on Chaos What a scale reads while a chain falls onto it. The reading of a scale, in units of the whole chain's weight, against the length of chain that has already landed, for two ways of putting the same chain down. Lowered gently, the scale reads the weight of what is resting on it and nothing else, so the reading climbs along the diagonal to one and stops. Dropped from rest with its lower end just touching, the scale reads three times that at every instant of the fall: one part is the pile's weight and two parts is the force needed to stop the links that are arriving, which is λv² with v² = 2gx and is therefore exactly twice λgx however far the fall has got. The peak, read off the drawn curve, is 3.00 chain weights. It is reached at the instant the last link lands, and the reading then falls discontinuously to one, because the momentum flux stops all at once. The discontinuity is the part a real experiment does not show — a real chain has links of a finite size and a scale has a response time — and it is the reason a chain dropped into a bucket on a kitchen scale reads high and then settles.

The pile that lands heavier than it weighs

Drop a chain onto a scale and the reading is three times the weight of the part that has landed — not approximately, exactly, all the way through the fall. The extra two parts are the force needed to stop links that are still arriving, and the same arithmetic run backwards says that picking a chain up wastes exactly half the energy it takes to get it moving.

4 figures · part 4 on Momentum

Waves

Oscillation, and everything that turns out to be an oscillation.

An f² law that is right in shape and out by 30× in size. Two absorption curves for air against frequency, both logarithmic, in decibels per kilometre. The lower one is the classical Stokes–Kirchhoff result computed from air's viscosity and thermal conductivity alone, and it goes as f^2.000 — exactly two, because the loss per cycle is fixed and the number of cycles per metre is proportional to the frequency. The upper one is the measured atmospheric absorption at 20 °C and 50 per cent humidity, which fits f^1.42 and is 30 times larger at 1 kHz and 211 times at 125 Hz. The excess is not a correction to viscosity: it is nitrogen and oxygen storing energy in vibration and giving it back late, at a rate the water vapour sets, and it is the mechanism that actually removes the treble from a distant sound.

The distance that takes the treble out

Spreading treats every frequency alike; absorption does not. The loss per cycle is roughly fixed and the number of cycles per metre goes as the frequency, so absorption climbs as f² and a sound gets duller with distance as well as quieter — which is the whole account of why a nearby thunderclap cracks and a distant one rumbles.

5 figures · part 1 on Attenuation A source moving faster than its own waves. Circles showing where each crest has reached, centred on where the source was when it emitted them. Ahead of the source the crests are closer together and the frequency heard is higher; behind, they are spread out and it is lower.

The cone the source leaves behind

Take the Doppler construction past the speed of the wave and the wavefronts acquire an envelope. Its half-angle obeys sin θ = 1/M, an expression with no pressure, no density and no shape of the object in it — so a photograph of the cone is a speedometer. And the bang is not an event at the moment of crossing — it is a signature dragged along the ground for the whole of the flight.

8 figures · part 4 on Doppler The same number read off a decay and off a linewidth. A lightly damped oscillator released and left alone, above, and the power spectrum of exactly those samples, below. The decay falls to 1/e of its starting amplitude after 8.0 cycles, which makes the quality factor π times that, or 25.0. The spectrum peaks at 1.0000 radians per second and falls to half its power 0.04001 radians per second wide, which makes the quality factor the peak divided by the width, or 25.0. The two disagree by 0.02 per cent, which is the resolution of the frequency grid rather than a difference in the physics. They cannot disagree by more, because they are the same statement: a resonance is narrow because its ringing is long, and the transform that turns one into the other is not an approximation but an identity. A measurement of either is a measurement of both — which is why a bell can be characterised by hitting it and listening, or by driving it and sweeping, and why the two instruments never argue.

The width that is a lifetime

Hit a bell and time how long it rings; drive it and measure how narrow its response is. The two numbers are the same number, and they cannot disagree — not because the physics conspires but because a decay and a linewidth are one function seen in two coordinate systems.

4 figures · part 4 on Resonance A graded junction against a quarter-wave layer. Reflectance against wavelength for four ways of joining a medium of index 1 to one of 1.52. The bare interface reflects 4.26 per cent at every wavelength. A single quarter-wave layer of index √(n₀n_s) takes that to zero at 550 nm exactly and rises symmetrically either side, which is the shape of every single-layer coating and every single-section transformer. The remaining curves are graded junctions of 150 nm and 400 nm, built as 120 thin layers whose index climbs geometrically and put through the same matrix product. They do not have a design wavelength at all. Below a cutoff they are flat and negligible; above it they climb steeply toward the bare value, and the cutoff is set by the length: 398 nm for the 150 nm taper, 1062 nm for the 400 nm taper. Roughly, a taper works for every wavelength shorter than about twice its own optical length, which is the statement that a reflection needs a partner a quarter of a wavelength further in to cancel against. The engineering versions of this are everywhere: the horn on a loudspeaker, the moth's eye, the flared transition between two waveguides, and the graded layer that lets an ultrasound probe reach tissue across a hundredfold impedance step.

The taper that matches every note

A quarter-wave layer cancels a reflection at one wavelength and only near it. Spread the same change of impedance over a distance instead, and the reflection vanishes for every wavelength shorter than about twice that distance — not by cancelling one echo against another, but by leaving no step anywhere for an echo to come from.

7 figures · part 3 on Impedance

Optics

Light, and the small number of rules it obeys.

Abbe's ratio, measured on the traced rays. The quantity h divided by the sine of the angle at which the ray converges on the focus, in units of the paraxial focal length, against how far up the aperture the ray entered. Abbe's sine condition says that a system already free of spherical aberration images a small region round the axis faithfully only if this ratio is the same for every ray. A horizontal line means the condition is met. The parabola departs by 12.96 per cent across the aperture; The sphere departs by 7.18 per cent across the aperture. The paraboloid is the interesting case, because it is exactly stigmatic on axis — every ray from infinity crosses at one point, which is the definition of the shape — and it still fails this test. Perfection at one point buys nothing at the next one along. What the departure predicts is coma, a blur that grows linearly with the distance off axis and quadratically with the aperture, and the offaxis figure measures exactly that blur on the same surfaces. The condition is not a design rule invented for telescopes: it follows from requiring that the same optical path length join object and image for every route, and any instrument that images a field rather than a point has to meet it.

The condition a lens must meet

A paraboloid brings every parallel ray to exactly one point. Move the source a fifth of a degree off axis and the image is a fan rather than a point, and the reason is a condition Abbe wrote down that has nothing to do with the axis — perfection at one point buys nothing at the next one along.

4 figures · part 4 on Imaging Fringe contrast against baseline, for four stellar diameters. The visibility of the fringes an interferometer would obtain at 575 nm, against the separation of its two apertures, for uniform discs of angular diameter 10, 20, 47, 100 milliarcseconds. Each curve is 2J₁(πθB/λ)/(πθB/λ), the transform of a uniform disc, and each first reaches zero at 14.47 m for 10 mas, 7.23 m for 20 mas, 3.08 m for 47 mas, 1.45 m for 100 mas. Dividing each of those by λ/θ returns the same number, 1.2197, which is the 1.22 in every textbook and is the first zero of J₁ divided by π — recovered here from the four curves rather than written into them. The practical content is that a smaller star needs a longer baseline, in exact inverse proportion, and that the measurement is of a contrast rather than of a picture. Michelson and Pease found the fringes from Betelgeuse vanishing at a 3.07 m separation in 1920 at a wavelength of 575 nm, which by the same arithmetic is a disc 47.1 milliarcseconds across — and no telescope resolved that star for another seventy years.

The fringe that measures a star

Set two apertures 3.07 metres apart in 1920 and the fringes from Betelgeuse vanish. That single fact gives the star's angular diameter to two significant figures, without ever forming an image of it — because the contrast of a fringe pattern is a Fourier component of the source's own shape.

4 figures · part 3 on Coherence The wavelength at which a fibre stops smearing a pulse. Group-delay dispersion against wavelength for fused silica, computed by differencing Malitson's Sellmeier fit twice rather than read off a table, in picoseconds of spread per nanometre of source width per kilometre of fibre. The material curve crosses zero at 1273 nm — a property of the glass, fixed by where the ultraviolet and infrared absorptions balance, and not adjustable. Below it a fibre is normally dispersive and above it anomalously so, and at 1550 nm, where silica is most transparent, it is 21.9. The second curve adds a waveguide term, which is negative because a mode confined by a core spreads into the cladding differently at different wavelengths, and which depends on the fibre's geometry rather than on the glass; it is drawn here as a constant offset chosen to put the total zero at 1310 nm, which is what standard single-mode fibre is built to do. The consequence in a system is the thing worth carrying away: a source one nanometre wide sent 100 km through fibre at 1550 nm arrives 1833 picoseconds broader than it left, against 0.0 at the zero. That is the whole reason a network runs at one wavelength rather than another, and the reason the two quantities an engineer wants — lowest loss and zero dispersion — sit at different wavelengths and have to be reconciled by design rather than chosen.

The wavelength a fibre does not smear

Fused silica's index has a second derivative that passes through zero at 1,273 nanometres, and that is not a design choice — it is where the ultraviolet and infrared absorptions balance. A pulse sent at that wavelength arrives the shape it left. A pulse at the wavelength of least loss arrives 1,800 picoseconds wider after a hundred kilometres.

5 figures · part 3 on Dispersion

Electromagnetism

Charge, field, and the lines drawn between them.

Three vector potentials for one magnetic field. Three different vector potentials, drawn as arrow fields over the same square, every one of which describes the same uniform magnetic field of 1 tesla out of the page. The symmetric gauge circulates about the origin; the two Landau gauges are unidirectional and point in perpendicular directions, and neither has any circulation about anything a reader can see. Underneath each is the field recovered from its own arrows, by adding up the potential round a small square and dividing by the area enclosed — which is what the curl is. The three numbers 1.000000, 1.000000, 1.000000 agree to the last digit computed. The picture is the argument: a vector potential has no physical direction, no physical magnitude and no physical circulation of its own, because a change of gauge alters all three and alters nothing that can be measured. What survives is the curl, and the curl is the field.

The potentials that are not unique

Nobody solves Maxwell's equations for the fields. They are solved for potentials instead, and the potentials are not unique — three completely different vector potentials describe the same uniform magnetic field, and one of them changes everywhere the instant a charge moves, at any distance, without anything having outrun light.

4 figures · part 2 on Maxwell equations Skin depth against frequency, over eleven decades. The skin depth of copper, stainless steel, seawater, mu-metal against frequency, both axes logarithmic. Every line has the same slope, −½, because the depth goes as the inverse square root of the frequency for all of them; what separates them is conductivity and permeability, which enter the same way. Three points are marked and each is a practical fact. Copper at mains frequency has a skin depth of 9.22 mm, so a busbar of that thickness carries current throughout and a thicker one does not. Copper at a gigahertz has 2.1 µm, so the current in a microwave component runs in a layer thinner than the plating and the surface finish becomes the conductor. Seawater at the frequency navies use to signal submerged submarines has 28.9 m, which is why that link is measured in tens of hertz and carries a few characters a minute. Mu-metal is the outlier and the reason it exists: its conductivity is forty times worse than copper's and its permeability twenty thousand times better, so at low frequency it is the one material on the chart that is thin.

How far a field gets into metal

The first three rungs of this ladder all say the field inside a conductor is zero, and all three assume the electrons have had time to move. Give them less time and the field gets in — 9.2 millimetres into copper at mains frequency, 2.1 microns at a gigahertz — and the metal box that silences a radio does almost nothing about the cable running past it.

4 figures · part 4 on Conductors How much of the answer a finite wire gives back. The field beside a straight segment of wire, divided by what the infinite-wire formula would give, against the length of the segment in units of the distance to the field point, on a logarithmic horizontal axis. The field is integrated element by element along the segment rather than evaluated from a closed form. A wire ten times as long as the distance already gives 92.8 per cent of the infinite answer, and one as long as the distance gives 45 — which is the practical content, and the reason the infinite-wire result is used for laboratory wires without apology. The second column is the part that matters for the law rather than for the number. The circulation of this field round a circle of radius d is not μ₀I; it falls short by exactly the fraction the segment fails to subtend. What makes up the difference is the displacement current of the charge piling up at the segment's two ends, and the two terms, both integrated here, sum to μ₀I to within 2.2e-8 per cent at every length. So Ampère's law is not approximately true for an open circuit and exactly true for a closed one — it is exactly true always, and it is a computation only when symmetry supplies the direction and the magnitude along the loop. Symmetry is doing the work; the law is doing the bookkeeping.

The law that is always true and rarely useful

Ampère's law holds for every loop and every current. Apply it to a wire of finite length and it gives an answer that is wrong by half — until the displacement current of the charge piling up at the wire's two ends is put back in, whereupon the two terms sum to μ₀I exactly at every length. The law was never approximate. It was never a computation either.

4 figures · part 3 on Ampere law

Thermodynamics

Heat, disorder, and the one law with a direction in it.

The fraction of walks that have come home, against how long they have walked. The proportion of 1,600 lattice random walks that have returned to their starting point at least once, against the number of steps taken, on a logarithmic horizontal axis, in one, two and three dimensions. In one dimension almost every walk is home almost at once and the fraction climbs toward one. In two it climbs more slowly — the return is still certain, but only logarithmically, so a two-dimensional walk that has not come back after a thousand steps is unremarkable. In three the curve flattens: it reaches 0.3481 and stops, against the exact value 0.3405, drawn as a line. That number is Watson's integral, and it is the probability that a three-dimensional walk ever comes home at all. The difference between the cases is not one of degree. In one and two dimensions the expected number of returns is infinite and a diffusing particle visits every site eventually; in three it is finite, and a molecule released in a room will, with probability two-thirds, never pass through its starting point again. The same statement runs the other way round: a reaction that needs two diffusing partners to meet is a very different problem on a membrane from what it is in a cell.

The walk that comes home

A particle wandering at random on a line returns to where it started, with certainty. On a plane it returns, with certainty. In three dimensions the probability is 0.3405 — so two out of three molecules released in a room never pass through their starting point again, and the difference between the cases is not a matter of degree.

4 figures · part 3 on Diffusion Entropy that is still there at absolute zero, counted and measured. Four substances whose entropy does not go to zero when they are cooled as far as anybody can cool them, with the entropy counted from the arrangements they froze into beside the entropy measured by integrating their heat capacities. Ice is the famous one: every oxygen has four hydrogen bonds and the rule is that two hydrogens sit near it and two far, which leaves six of the sixteen placements legal, and Pauling's count of the whole crystal collapses to R ln(3/2) = 3.371 J per mole per kelvin against a measured 3.41 — an agreement to one per cent from an argument on one line. Carbon monoxide and nitrous oxide are the easy cases, molecules that can lie either way round in the lattice and have too little to gain by choosing. The worst of the four is off by 25 per cent, which is the honest state of this subject: the counts are crude, they ignore the correlations between neighbouring choices, and they still land within sight of a calorimeter. What the figure is really about is that the third law has an escape clause and the escape clause is measurable. A perfect crystal has one arrangement and zero entropy; a crystal that ran out of time while it had many has the logarithm of however many it stopped at, permanently.

The entropy that is still there at zero

The third law says a perfect crystal has no entropy at absolute zero. Ice has 3.41 joules per kelvin per mole left over, and the number can be recovered from one line of counting — two hydrogens near each oxygen and two far, six legal arrangements out of sixteen, R ln(3/2). The law has an escape clause and the escape clause is measurable.

4 figures · part 2 on Third law

Relativity

Space and time, drawn on the same axes.

One expression over 3 decades of area times rate. Sagnac time difference against the product of enclosed area and rotation rate, both logarithmic. The relation Δt = 4AΩ/c² is linear in that product — the fitted slope of the drawn points is 1.0000 — and it contains no refractive index, no shape of the loop and no position of the axis inside it. a 1 km fibre gyroscope on a 10 cm coil: 1.62e-19 s, 3.14e-5 fringes; Sagnac's own ring, 1913: 4.84e-17 s, 0.0666 fringes against a reported 0.07; a laboratory turntable at one revolution a second: 6.99e-17 s, 0.0331 fringes; Michelson and Gale, 1925: 4.50e-16 s, 0.2364 fringes against a reported 0.23. Michelson and Gale's rectangle in Illinois is the one that carries the check: 0.236 fringes predicted from its own dimensions and the vertical component of the Earth's rotation at its latitude, and 0.230 reported.

The ring where the two beams disagree

Send light both ways round a closed loop on a turntable and the two beams come back at different times, by 4AΩ/c² — an expression with no refractive index in it, no shape of the loop and no position of the axis. The same number is what a set of clocks round the rim fails to close by, which is the sharper statement — on a rotating platform there is no global simultaneity to be had.

4 figures · part 3 on Simultaneity A horizon 0.97 light years behind, made by nothing but the motion. Position across and time up, in units where light travels at 45°, for a rocket holding a constant proper acceleration of 1 gravity. The worldline is the hyperbola x² − c²t² = (c²/a)², asymptotic to the light line it never crosses. Three light signals are drawn: one released at x = 0.55 catches up at t = 0.63, one released at x = 0 never arrives, one released at x = -0.6 never arrives. The dividing line is the asymptote itself. Everything at or behind it is permanently out of reach, and for one gravity that boundary sits 0.97 light years behind the rocket's starting point. Nothing is there — no mass, no field, no surface. The horizon is a consequence of never stopping.

The wall of silence behind a rocket that never stops

Hold a constant acceleration and the worldline is a hyperbola asymptotic to a light ray — so there is a light ray that never catches it. An observer who never stops accelerating has a horizon a distance c²/a behind, made by nothing but the motion, and at one gravity it sits 0.97 light years back. Nothing is there. No mass, no surface, no field.

4 figures · part 1 on Accelerated frames

Fluids

Matter that will not hold a shape, and the forces that act in it anyway.

A pore that lifts 100 m has to be 0.1 µm or finer. Capillary rise against pore radius, both logarithmic, for a liquid of surface tension 72.8 mN/m at a contact angle of 20°. The relation is a straight line of slope −1 — halve the pore and double the rise — and the two horizontal marks are the height in question, 100 m, and the 10.3 m that one atmosphere supports. 0.01 µm lifts 1394.7 m; 0.1 µm lifts 139.5 m; 1 µm lifts 13.9 m; 5 µm lifts 2.8 m; 20 µm lifts 69.7 cm; 50 µm lifts 27.9 cm. The conducting vessels of a tree are tens of microns across and lift under a metre; the pores in the membranes between them are tens of nanometres and would lift kilometres. Those are the same expression at two scales, and only one of them is a pipe.

The column that is pulled, not pushed

A capillary fine enough to lift a hundred metres is far too fine to carry any flow, and one wide enough to carry the flow lifts under a metre. Neither is how the water gets up a tree. The column is under tension — an absolute pressure of −0.88 MPa at the top, which a gas cannot have — held together by cohesion and prevented from tearing by pores a few tens of nanometres across.

4 figures · part 3 on Capillarity Everything is decided against one line at 9.76 K per kilometre. Temperature against height for five environments, with the dry adiabat drawn heavy. A parcel lifted from the ground cools along the adiabat, at g/c_p = 9.76 K/km — a number with no meteorology in it, only gravity and the heat capacity of air. If the environment cools faster than that, a lifted parcel finds itself warmer than its surroundings and keeps going; if it cools more slowly, the parcel finds itself colder and sinks back. -5 K/km gives N² = 5.02e-4 s⁻², a period of 4.7 min; 0 K/km gives N² = 3.32e-4 s⁻², a period of 5.7 min; 6.5 K/km gives N² = 1.11e-4 s⁻², a period of 9.9 min; 9.8 K/km gives N² = -1.43e-6 s⁻², an e-folding time of 835 s; 12 K/km gives N² = -7.63e-5 s⁻², an e-folding time of 114 s. The classification is a comparison of two slopes and nothing else: no density appears in it, and the same cold air is stable under one profile and unstable under another.

The layer a parcel cannot leave

Whether a column of air overturns is not decided by its density but by a difference of two gradients — the rate the environment cools with height, and the rate a lifted parcel cools on its own. Subtract one from the other and what is left is a restoring force per unit displacement, so a stable atmosphere rings at a period of minutes and an unstable one has no period at all.

7 figures · part 1 on Stratification The thickness a paste can hold on a slope. The greatest thickness a yield-stress fluid can rest at without flowing, against the angle of the surface it is resting on, on a logarithmic vertical axis, for four materials. A layer of thickness h puts a shear stress ρgh sin α on its own base; it stays put while that is below the yield stress and flows when it is not, so the critical thickness is τ_y divided by ρg sin α and it depends on nothing else — not on the viscosity, not on how long it is left, not on how it was put there. On a vertical wall the numbers are 1.4 mm of ketchup, 9.2 mm of mayonnaise, 15.7 mm of toothpaste, 15.1 mm of basaltic lava, which is why toothpaste stays on a brush and ketchup does not stay on a plate held up. Read the other way it is a measurement: a lava flow that came to rest 3 centimetres thick on a 30° slope had a yield stress of about 400 pascals, and that is how the rheology of a flow nobody was standing next to is recovered from its shape a thousand years later. The model stops where the layer is thin enough for surface tension to matter and where the material's yield stress depends on how long it has been left alone, which for most of these it does.

The paste that holds up its own hill

Toothpaste stands on a brush and ketchup does not stand on a plate, and the difference is a single number with the units of a pressure. Below it a material does not flow slowly — it does not flow. That threshold turns a rheological property into a length, and the length is why a lava flow's thickness says what the lava was made of.

5 figures · part 3 on Rheology The cross a shaken cylinder leaves in a stratified fluid. The beams radiated by a small body oscillating in a fluid of buoyancy frequency 1.053e-2 per second — a period of 9.9 minutes — at 0.3, 0.6, 0.9 times that frequency. The disturbance does not spread in circles. It leaves along four rays, and the angle of those rays to the horizontal is fixed entirely by the ratio of the driving frequency to the buoyancy frequency: 17.5° at 0.3N, 36.9° at 0.6N, 64.2° at 0.9N. Nothing about the size of the body, the amplitude of the shaking or the wavelength enters. Drive it faster and the beams stand up; drive it slower and they lie down; drive it above N and there are no beams at all, because the dispersion relation ω = N cos φ has no solution. The short arrows across each beam are the wavevector, which is perpendicular to the beam — the dot products drawn here are 6e-17 — so the crests travel sideways across the ray while the energy travels along it, and a fluid doing this looks, in a photograph, as though its waves are moving at right angles to where they are going.

The wave that picks an angle

Shake a rod slowly in a tank of salty water layered by density and the disturbance leaves along four straight beams, at an angle fixed entirely by how fast the shaking is. Change the wavelength and the angle does not move. Change the frequency and it does. Above the buoyancy frequency there is no wave at all.

4 figures · part 2 on Stratification

Astrophysics

Gravity read as geometry, and the laws carried where no laboratory can follow.

The self-force matters at 6.27 × 10⁻²⁴ s, and nowhere a charge has ever been. The ratio of the radiation reaction to the applied force, which is τ divided by the time the force takes to change, on a logarithmic axis. τ = μ₀q²/6πmc = 6.266e-24 s for an electron, and light crosses 1.879 femtometres in that time — two thirds of the classical electron radius. a 3 GHz accelerating cavity: 1.9e-14; a 500 nm optical field: 3.8e-9; an electron orbiting a proton, ground state: 4.1e-8; an X-ray at 0.1 nm: 1.9e-5; over a classical electron radius: 6.7e-1. The largest of them, "over a classical electron radius", is still 1.5e+0 times too slow. So the correction is never large for any force anybody can apply, and the only regime where it would be is one in which the charge's own structure has already made the whole description meaningless.

The force a charge exerts on itself

Larmor's formula says how much an accelerating charge radiates and says nothing about who pays. Conservation says the charge does, so there is a force on it — and the equation that force produces has a free particle accelerating for ever with nothing pushing it, or else beginning to move before it is pushed. Both solutions are absurd, and the interval over which they are absurd is smaller than the electron the equation was written for.

9 figures · part 2 on Radiating charge One line, 21 decades of density, and every mirror on it. Plasma frequency against electron density, both logarithmic, with the horizontal rules at frequencies a reader already has a feel for. A wave is reflected by everything to the right of where its own rule meets the line and passes through everything to the left. a fluorescent tube at 1.0e+17 m⁻³ cuts off at 2.84 GHz; ionosphere, D layer at night at 1.0e+8 m⁻³ cuts off at 90 kHz; ionosphere, E layer at 1.0e+11 m⁻³ cuts off at 2.84 MHz; ionosphere, F2 layer at 1.0e+12 m⁻³ cuts off at 8.98 MHz; aluminium's conduction electrons at 1.8e+29 m⁻³ cuts off at 3819.9 THz. So the F2 layer turns back a 1 MHz broadcast and lets a 100 MHz one straight out, which is why one of them is heard across an ocean at night and the other stops at the horizon; and aluminium's cutoff sits in the far ultraviolet, which is why it is a mirror for everything visible and a window above 78 nm. The dependence is a square root, so the line has slope one half: a hundredfold denser plasma reflects only ten times the frequency. Arriving at an angle helps by a factor of sec θ — 1.00 at 0°, 1.15 at 30°, 2.00 at 60° — because only the component of the motion along the density gradient has to be turned round.

The frequency below which nothing gets in

Free charges give a medium a permittivity that is negative, and a negative permittivity is not an absorbing medium — it is one in which no wave exists at all. Below that frequency the reflection is total, exactly rather than nearly, because there is no transmitted wave and nothing to absorb. The same expression puts the number at 9 MHz for the ionosphere and 3.8 PHz for aluminium.

8 figures · part 1 on Plasma oscillation

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