Illustrated Physics
physics, drawn
Concept

Reversibility — where it appears

The property of a process that can be run backwards leaving nothing changed, which no process at a finite rate has. It is a limit rather than an achievable state, and its usefulness is that it bounds everything real: no engine can beat a reversible one working between the same reservoirs.

Named by 14 essays across 4 fields — each of them below, with the objects they name alongside it.

A Carnot cycle on pressure–volume axes. Two isothermal steps joined by two adiabatic ones, forming a closed loop. The gas expands 9.6-fold in reaching the cold reservoir at 0.50 of the hot one, and the area enclosed is the net work done over one cycle.

The ceiling on every engine, set before it was designed

There is a maximum efficiency no heat engine can exceed, and it depends on nothing but two temperatures. Not the fuel, not the working substance, not the cleverness of the engineer.

thermodynamics · Heat engines The ceiling, inverted. How many joules of heat a perfect machine can move per joule of work, against the outside temperature, with the inside held at 21 °C. The upper curve is heating — T_h/(T_h − T_c), which is what the Carnot argument becomes when the cycle is run backwards — and the lower one is cooling the outside, T_c/(T_h − T_c). They differ by exactly one everywhere, to 1.8e-15 across the whole range as drawn, because the work put in is delivered as heat along with whatever was moved. The dashed line at one is a resistive heater, which is 100% efficient and is the worst option on the figure. At 7 °C and −7 °C the ideal coefficients are 21.0 and 10.5; a real machine reaching 25% of the ideal gets 5.3 and 2.6, which is still several times what burning the same energy would give.

The engine that pays back more than it takes

Carnot's argument puts a ceiling on how much work a flow of heat can be made to do. Run the same cycle backwards and the ceiling inverts into a floor that is greater than one — so a machine can deliver three or four joules of heat for every joule it consumes, and a perfectly efficient electric heater is the worst way to warm a room.

thermodynamics · Heat engines The work one molecule and one bit are worth. The pressure of a gas of one molecule against its volume, at 300 K, in units of the volume it starts in. The shaded area is the work the molecule does pushing a partition out isothermally, and it is measured here by integrating the drawn curve rather than written down: expanding by 1.5× yields 0.4055 kT against ln 1.5 = 0.4055; expanding by 2× yields 0.6931 kT against ln 2 = 0.6931; expanding by 4× yields 1.3863 kT against ln 4 = 1.3863; expanding by 8× yields 2.0794 kT against ln 8 = 2.0794, agreeing to 2.0e-10. The doubling is the one that matters, because a partition inserted in the middle leaves the molecule on one side or the other, and knowing which is what lets the load be attached to the right face. That single expansion delivers kT·ln2 = 2.87 zeptojoules at 300 K. It looks like work extracted from one temperature, and it is — until the engine is asked to run again, which requires forgetting which side the molecule was on.

The bit that has to be paid for

One molecule in a box, a partition, and the knowledge of which side it went — enough, between them, to extract work from a single reservoir, which the second law forbids. The engine is real and the arithmetic is right. What closes the loophole is that the cycle does not finish until the knowledge has been thrown away, and throwing away one bit costs exactly what the expansion delivered.

thermodynamics · Entropy What a thermal camera reads off a shiny surface. The temperature a camera calibrated for a black body reports, against the emissivity of the surface it is pointed at, for surfaces truly at 323 K, 373 K, 473 K in a room at 293 K. Every curve begins at the true temperature when the emissivity is one and ends at the room's temperature when it is zero, because a surface that emits nothing reflects everything and the camera is then looking at the room. polished aluminium at 0.05 reads 312 K; stainless steel, oxidised at 0.8 reads 451 K; matt black paint at 0.95 reads 468 K; human skin at 0.98 reads 471 K for a surface truly at 473 K. That is not a fault in the instrument. It follows from Kirchhoff's law — a poor emitter is a poor absorber and therefore a good reflector — so the shortfall in a shiny surface's own glow is made up almost exactly by whatever is reflected in it, and no measurement of the light leaving a surface can separate the two without knowing the emissivity in advance.

The glow that says nothing about the surface

A thermal camera pointed at a saucepan of boiling water reads a hundred degrees. Pointed at a polished aluminium block at the same temperature it reads about twenty-five, and the instrument is working perfectly. What it is measuring is emissivity as much as temperature — and a surface's emissivity is forced to equal its absorptivity, at every wavelength and every angle, by an argument with no physics of matter in it at all.

thermodynamics · Blackbody Each stage takes 25.0 per cent of what is left. Entropy against temperature for a spin-½ paramagnet at 0.25 T and 1 T, with the cooling cycle drawn between them: a vertical drop is isothermal magnetisation, a horizontal move is adiabatic demagnetisation. Starting from 1 K the treads are at 1.000 K, 0.250 K, 0.062 K, 0.016 K, 3.91 mK. Each is 0.2500 of the one before — a ratio read back off the drawn treads rather than written into them, and equal to the field ratio 0.25/1 because this refrigerant's entropy depends on the field and the temperature only through their quotient. The steps therefore shrink in proportion to what is left, and no finite number of them arrives.

The staircase that never reaches the floor

Absolute zero is unreachable, and the reason is not that the apparatus is not good enough. Every stage of cooling removes a fixed fraction of what is left rather than a fixed amount, so the steps shrink in proportion to the distance remaining — and the fixed fraction cannot be made one, because the entropy curves at two field strengths are required to meet where the axis is.

thermodynamics · Third law The least energy that fresh water can cost. The work needed to take fresh water out of a feed of 1150 mol/m³ of dissolved particles — seawater — against how much of the feed is taken, in kilowatt-hours per cubic metre of product. The lower curve is the reversible minimum, in which the pressure is raised continuously as the remaining feed gets saltier; the upper one is a single stage held throughout at the pressure the final, saltiest concentrate needs. At vanishing recovery both tend to the feed's own osmotic pressure, 28.5 bar or 0.79 kWh/m³, which is the floor for the first drop and is checked here against the limit of the formula. Taking more of the feed costs more per unit taken, because what is left behind is saltier: at 10% recovery 0.83 kWh/m³ reversibly and 0.88 in one stage, 30% recovery 0.94 kWh/m³ reversibly and 1.13 in one stage, 50% recovery 1.10 kWh/m³ reversibly and 1.58 in one stage, 60% recovery 1.21 kWh/m³ reversibly and 1.98 in one stage, 75% recovery 1.46 kWh/m³ reversibly and 3.17 in one stage. The horizontal line is what a good seawater plant actually uses, 3 kWh/m³, so the second-law efficiency of the industry is about 37 per cent. None of this is about membranes. It is the free energy of mixing salt into water, read backwards, and no technology of any kind can go below the lower curve.

What it costs to take the salt out

Salt dissolves in water because mixing is overwhelmingly the more probable arrangement, and separating the two again means paying back what the mixing gave away. The bill can be computed before any apparatus is chosen — 0.79 kilowatt-hours for the first cubic metre from seawater — and it rises with every further cubic metre taken, because what is left behind is saltier than what was started with.

fluids · Osmosis A wave that dies with nothing to rub against. The electric field of a plasma wave at kλ = 0.5, against time in plasma periods, on a logarithmic scale, obtained by integrating the collisionless kinetic equation as an initial-value problem. There are no collisions in the equation, no viscosity and no resistance; the only operator acting on the distribution is a rotation of phase whose rate depends on the particle's speed. The field nevertheless falls exponentially, at 0.1534 per plasma period, against the published root of the kinetic dispersion relation at this wavenumber, 0.1534, and Landau's asymptotic formula's 0.1514. Meanwhile the free energy of the perturbation — the weighted norm of the distribution plus the field energy, which the equation conserves exactly — moves by 2.4e-10. So nothing has been dissipated: every joule the field loses is still in the distribution, and the accounting closes to a part in ten thousand million. The energy has gone into the particles' ordered motion, and the information about the wave is wound into structure at finer and finer scales in velocity.

The wave that dies with nothing to rub against

Every damping in this collection so far removes energy from a wave and puts it somewhere warmer. This one removes it and produces no heat at all: there are no collisions in the equation, the entropy is unchanged, the whole thing runs backwards perfectly, and the wave still dies exponentially. What it dies into is structure in velocity too fine for a field to see.

astrophysics · Plasma oscillation Runs that break the second law, and how often. The work done in a process repeated many times, and the same for the process run in reverse with its work reflected, for a free-energy change of 4 kT and a dissipation of 3 kT. The average work exceeds the free-energy change, which is the second law, and individual runs do not have to: the shaded tail is the fraction of runs that do less work than the free energy — trajectories in which the entropy of the universe went down — and it is 11.03% here. The two curves cross exactly at the free-energy change, whatever the dissipation, which is what makes an irreversible measurement able to report an equilibrium quantity.

The second law, with a probability attached

Entropy increases, on average. For a small system pulled quickly, individual runs go the other way — and how often is not a matter of taste but an exact number, fixed by a relation with no adjustable constant in it and no requirement that anything be near equilibrium.

thermodynamics · Entropy The rectangle every cycle is equal to. An ideal Otto cycle for air on a temperature–entropy diagram: compression ratio 9, intake at 300 K, 1400 kJ/kg added at constant volume. Compression takes the charge to 722 K, combustion to 2672 K, expansion back to 1110 K, and the exhaust cools at constant volume. The shaded loop is the work; the region under the lower curve is the heat rejected. Heat enters over a range of temperatures, and its entropy-weighted mean — heat divided by the entropy it brings — is 1491 K; the heat leaves at a mean of 619 K. The dashed rectangle between those two temperatures has exactly the loop's area, and 1 − 619/1491 = 58.5%, which is the Otto efficiency, checked to rounding. A Carnot engine between the coldest and hottest points of the same cycle would reach 88.8%.

The temperature an engine really takes its heat at

Carnot's ceiling is set by two temperatures, and no engine that burns fuel takes its heat in at one temperature or gives it out at another. It takes heat over a range, from the moment combustion starts to the moment it ends. For any reversible cycle there is an exact replacement for Carnot's two numbers: the average temperature at which heat arrives and the average at which it leaves, each weighted by the entropy the heat carries. The gap between a real cycle and Carnot is a gap between those averages and the extremes.

thermodynamics · Heat engines Two bodies an engine draws together. Two equal bodies of 4.186 kJ/K — a kilogram of water each — one at 90.0 °C and one at 10.0 °C, against the heat drawn from the hot one. The solid curves are the best possible engine running between them, a reversible one, which leaves the product of the two temperatures unchanged and brings both to the geometric mean, 320.7 K (47.5 °C). It draws 177.8 kJ from the hot body and delivers 20.8 kJ of work, C(√T₁ − √T₂)², checked against the heat balance. The dashed lines are the same bodies simply touching: they meet at the arithmetic mean, 323.1 K, having exchanged 167.4 kJ and delivered nothing. The 2.5 K between the two endpoints is the work, left behind as heat.

The work left in two buckets of water

Carnot's ceiling assumes reservoirs so large that taking heat from one and giving it to the other changes neither temperature. Two buckets of water are not reservoirs. Run the best possible engine between a hot one and a cold one and both temperatures move, the efficiency available shrinks as they do, and the engine stops when they meet — at the geometric mean of the starting temperatures, not the ordinary one. The work it delivered is exactly the difference between those two meeting points, and it is far less than the starting temperatures promise.

thermodynamics · Heat engines A fridge with no work going into it, and its ceiling. How much heat a three-reservoir machine can lift out of a cold space per unit of heat supplied to drive it, against the temperature of the driving heat, for 3 cold temperatures and an ambient of 300 kelvin. No work enters or leaves: the machine takes heat in at the top, takes heat in at the bottom, and rejects the sum at ambient. That it can do anything at all is the surprise — the second law allows heat to be moved up a gradient provided a larger flow is moved down one, and the accounting is a single inequality in the three entropy flows. The ceiling is the product of two familiar expressions, and at 450 kelvin driving a 253-kelvin space it is 1.79. The marks are what real machines achieve, which is a fifth to a third of it — absorption refrigeration is not efficient and is chosen when the heat is free and the silence and the absence of moving parts are worth something.

A fridge with no work going into it

Every engine here so far turns heat into work or work into a heat flow. A machine exchanging heat with three reservoirs and doing no work at all can still move heat from cold to hot, and the ceiling on how much is the product of two Carnot expressions — an engine's efficiency times a fridge's coefficient of performance. A gas flame makes ice, and the accounting is one inequality in three entropy flows.

thermodynamics · Heat engines A rod in a shear flow turns in fits and starts. The angle between a rigid rod's axis and the flow direction, unwrapped, against the strain of a steady shear flow (shear rate times time), for spheroids of aspect ratio 1, 5, 20 lying in the plane of the shear, each starting along the flow. A sphere (r = 1) turns steadily at half the shear rate. A rod spends most of its time nearly aligned with the flow, turning slowly, and then flips through half a turn quickly; at r = 5 it is within 10° of the flow direction 46 per cent of the time, at r = 20, 82 per cent. Each flip is half of Jeffery's period, π(r + 1/r) units of strain.

The rod that tumbles for ever in a stream

Drop a short rod into honey being sheared between two plates and it does not settle into line with the flow. It lies nearly along the stream for a long time, then flips end over end in a moment, then lies along it again, and it keeps doing that for as long as the shear lasts. George Jeffery worked out the motion in 1922: every orientation belongs to a closed orbit, every orbit takes the same time, and the flow has no way to choose between them — so a suspension of rods never reaches a steady state, and its thickness rises and falls in time with the tumbling.

fluids · Viscosity A ring of balls that forgets, and then remembers. Kac's ring: 400 balls on 400 sites round a circle, all white at the start, with 52 of the edges between sites marked; each tick every ball moves one site and changes colour if it crosses a marked edge. The solid line is the excess of white over black, as a fraction of all the balls, against ticks; the dashed line is Boltzmann's prediction (1 − 2μ)^t with μ = 0.1, the fraction of edges marked. For the first few dozen ticks the ring follows Boltzmann's decay exactly as an irreversible process would: the colours mix and the excess falls to zero, then wanders by a few hundredths. At tick 400 every ball has gone once round, through all 52 markers — an even number of colour changes — and the ring is all white again. A rule with no randomness in it, run forward, looked irreversible for 400 ticks and then undid itself.

The ring that forgets and then remembers

Put a few hundred balls on a ring, all white, and mark a handful of the gaps between them. Each tick, every ball moves one place round and changes colour if it crosses a mark. Nothing is random and everything can be run backwards. Yet the colours mix, half and half, exactly as Boltzmann's equation for a gas says they should, and the entropy rises to its maximum and stays there — until, after one full turn, every ball is white again. Mark Kac built the model in 1956 to show where the arrow of time in Boltzmann's equation comes from, and what it quietly assumes.

thermodynamics · Entropy A signal that dies, and comes back when the spins are turned over. The size of the signal from an ensemble of proton spins whose precession rates are spread by 20 Hz (root mean square) about the mean, with a true relaxation time T₂ of 80 ms, against time after they are tipped. The free signal falls to 1/e in about 10 ms and to 0.03 by 20 ms: not because any spin has relaxed but because they have fanned out. A π pulse at 30 ms (marked) turns every spin over, and at 60 ms the signal returns to 0.472 — exactly e^(−2τ/T₂), the dashed envelope, since every static difference in rate has cancelled. Without the pulse the signal at 60 ms is below one part in a million.

The echo that gives back everything but the wandering

Tip a sample of protons in a slightly uneven magnetic field and their signal dies within milliseconds. None of them has relaxed: they have fanned out, each precessing at the rate its own patch of field sets. Turn every spin over halfway through and the fan closes again, and the signal comes back as an echo, smaller only by the decay that was real. What the echo cannot return is a phase gained while a spin moved, and that one loss, which grows as the cube of the waiting time, is how a scanner finds a stroke within minutes.

quantum · Spin

Named alongside it

The objects these essays reach for when they reach for this one.

EntropyEfficiencyThe second lawFree energyThe Carnot cycleIrreversibilityTemperatureHeat capacityDetailed balanceExergyHeat engineThe p–V diagram

All concepts