Illustrated Physics
physics, drawn
Concept

Free energy — where it appears

The part of a system's energy available to do work at a given temperature, which is what a system in contact with a reservoir actually minimises. It is the internal energy less the temperature times the entropy, so a state can be favoured either by being low in energy or by being numerous.

Named by 13 essays across 4 fields — each of them below, with the objects they name alongside it.

Ways to arrange 12 coins. The number of distinct arrangements giving each number of heads, for 12 coins. Every individual arrangement is equally likely; the middle wins because there are more ways to reach it.

What a system actually minimises

A ball falls to the bottom of a bowl and a gas fills a room, and neither of those is the rule. A system in contact with a large reservoir minimises U − TS, and the minus sign is the reservoir's own entropy written in the system's variables — which is why a rubber band pulls harder when it is heated.

thermodynamics · Entropy 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 column a dissolved thing holds up. Two arms of one vessel, joined below by a membrane that passes water and not solute. On the right is 10 mol/m³ of dissolved particles at 25 °C; on the left, pure water. Water crosses into the solution until the extra weight of the right-hand column has raised its pressure by the osmotic pressure — 24.8 kPa, which is 2.53 m of water, drawn here to scale. Nothing is pulling. The solvent is at a lower chemical potential where it is mixed, so it moves that way, and it stops when mechanical pressure has made up the difference. A solute a thousand times more dilute than seawater lifts a column taller than a person.

The pressure that comes from counting

Dissolve a teaspoon of anything in a litre of water, put a membrane between it and pure water, and the solution will hold up a column of water two and a half metres tall. Nothing is pulling. The pressure does not depend on what was dissolved, only on how many particles it made — which is the ideal gas law, with the solute in place of the gas.

fluids · Osmosis 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 The field a classical partition function cannot see. The momentum plane of one classical charge, in units of the root-mean-square thermal momentum. With no field the Boltzmann weight is a set of circles about the origin; with a field the same circles sit about p = qA, displaced and otherwise unaltered, because the energy depends on p only through p − qA. Integrating over the whole plane therefore cannot notice the displacement, and the numbers beside the drawing are that integral evaluated at 6 displacements: the largest departure from the zero-field value is 3.3e-16, which is the precision of the arithmetic and not a physical effect. The classical free energy has no B in it, so the classical magnetisation is exactly zero at every field and every temperature — no paramagnetism, no diamagnetism, no ferromagnetism. Every magnetic material is therefore evidence of something classical mechanics does not contain.

The magnetism classical physics forbids

Write down the partition function of any collection of classical charges in a magnetic field, and the field cancels out. Not approximately, not to leading order — the integral is over all of momentum space and the field only shifts where the middle of it is. So classical statistical mechanics predicts no paramagnetism, no diamagnetism and no ferromagnetism, and a compass needle is a quantum instrument.

electromagnetism · Magnetisation 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.

fluids · Capillarity 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 The barrier a new phase has to climb. The free energy of a droplet against its radius, at four supersaturations. Two terms compete: the volume term is a gain and goes as r³, the surface term is a cost and goes as r². At small radius the surface wins, so a droplet that forms by chance is more expensive than the vapour it came from and evaporates again; past a critical radius the volume wins and the droplet grows without limit. The maximum between them is the barrier. At S = 1.5 the critical radius is 2.66 nm and the barrier 533.8 kT, S = 2 the critical radius is 1.56 nm and the barrier 182.6 kT, S = 3 the critical radius is 0.98 nm and the barrier 72.7 kT, S = 5 the critical radius is 0.67 nm and the barrier 33.9 kT. The critical radius contains a few hundred molecules at low supersaturation and a handful at high, which is the first sign that a theory built on a surface tension and a bulk free energy is being applied outside its comfort. Both the critical radius and the barrier are located here by searching the drawn curve and checked against the closed forms 2γ/|Δg| and 16πγ³/3Δg², which agree to a part in a thousand.

The barrier a new phase has to climb

Water vapour three times supersaturated is thermodynamically desperate to condense and will sit there indefinitely if it is clean enough. The obstacle is that a droplet has to start small, and a small droplet is nearly all surface — so the first nanometre of every phase transition costs energy rather than releasing it, and what decides whether anything happens is the height of that cost divided by kT.

thermodynamics · Phase change 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 The only part of the field that can pull. A dielectric slab part-way into a parallel-plate capacitor. Everywhere except at the slab's edge the field is perpendicular to the plates and therefore perpendicular to the direction the slab can move, so it exerts no force along that direction at all: inside the parallel-plate model, which is uniform between the plates and zero outside them, nothing pulls the slab anywhere. The force lives in the bowed lines drawn at the edge, where the field leaks past the end of the dielectric and acquires a component along the plates. Those lines are what the model throws away as a small correction near the boundary, and they are the entire mechanism. The energy method sidesteps the drawing altogether: differentiate the total energy with respect to the insertion and the answer is 1.328e-3 newtons, inward, without ever asking where on the slab the force is applied.

The force that lives where the model is not

A slab of glass held at the mouth of a charged capacitor is pulled in. Inside the parallel-plate model there is no force at all — the field is perpendicular to the slab's motion everywhere — and the same model's energy nevertheless gives the pull exactly right. The mechanism is entirely in the part of the field the model throws away.

electromagnetism · Dielectrics Melting curves, and the one that leans the wrong way. Melting temperature against pressure for water, benzene, naphthalene, each measured from its own melting point at one atmosphere, with pressure in bars. The slope of every coexistence line is the latent heat divided by the temperature and the change in volume, and the latent heat of melting is positive for everything — so the sign of the slope is the sign of the volume change, and nothing else. Almost everything expands on melting and its line leans forwards. Water's solid is less dense than its liquid, so its line leans backwards at 135 bars a kelvin: pressing on ice at just below zero melts it, and it takes 135 atmospheres to gain a single degree. The anomaly is not in the thermodynamics; it is in the fact that ice floats.

The melting curve that leans the wrong way

The slope of any coexistence line is the latent heat divided by the temperature and the change in volume. Latent heat is always positive, so the sign of the slope is the sign of the volume change — and for water the volume change is negative, which is the whole of why ice floats and why the melting curve leans backwards.

thermodynamics · Phase change 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 Every reaction's free energy has its lowest point inside. An ideal reaction A ⇌ B at 298 K. Across: how far it has gone, from pure A towards pure B. Up: the Gibbs energy of the mixture per mole, relative to pure A. Left, for standard reaction Gibbs energies ΔG° of −4, 0, +4 kJ/mol: the dashed straight lines are what the energy would be if A and B did not mix, and the solid curves add the entropy of mixing them. For ΔG° = −4 kJ/mol the lowest point is at 83.4 per cent B; for ΔG° = 0 kJ/mol the lowest point is at 50.0 per cent B; for ΔG° = +4 kJ/mol the lowest point is at 16.6 per cent B. Right, magnified near pure A, a reaction with ΔG° = +10 kJ/mol, whose straight line climbs from the start and which looks as if it should not proceed at all: its curve first falls, to a minimum of −43 J/mol at 1.74 per cent B, because the mixing term falls infinitely steeply away from a pure end. Each minimum was found by search and sits where the ratio of B to A equals exp(−ΔG°/RT).

The reaction that cannot go all the way

Chemistry speaks of reactions that go to completion and reactions that do not happen, and at equilibrium there are neither. The reason is a logarithm. The free energy of a half-finished reaction contains the entropy of mixing, whose slope is infinite at both pure ends, so every reaction's lowest point lies strictly inside — and the slope of that free energy, the chemical potential, is to particles what temperature is to heat.

thermodynamics · Chemical potential

Named alongside it

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

ReversibilityEntropyThe second lawThe Boltzmann factorEntropy of mixingEquilibriumIrreversibilityLatent heatMetastabilityPhase transitionSurface tensionChemical potential

All concepts