What a Convection Roll Costs: Entropy Export in Bénard Cells and Chemical Waves — Epoche C1
Heat a shallow layer of oil from below, gently and evenly, and at some point it stops being featureless: rolls or hexagonal cells appear and the fluid circulates in a regular way that was not there a moment before. Something comparable happens in chemistry. Put malonic acid, bromate and a ferroin catalyst in a shallow dish and concentric rings of colour spread outwards from scattered points — the Belousov–Zhabotinsky reaction, organising itself in space (Zaikin and Zhabotinsky 1970). Both look like trouble for the second law of thermodynamics, which says that entropy never decreases in an isolated system. Order appeared out of uniformity. Where did the entropy go? The answer is not merely that the surroundings pay. It is that the order is bought, that the price is computable, and that when computed the arithmetic turns out to be lopsided by a margin large enough to settle the question — which is what this essay does, once for the fluid and once for the reaction. Two entropies, not one The bookkeeping needed for a system that is not isolated splits the change in its entropy $S$ into what crosses the boundary and what is manufactured inside: $$\frac{dS}{dt}=\frac{d_{e}S}{dt}+\frac{d_{i}S}{dt},\qquad \frac{d_{i}S}{dt}\ge 0 .$$ Here $d_{e}S/dt$ is the rate at which entropy is carried in or out with heat and matter, and $d_{i}S/dt$ the rate at which it is produced irreversibly within. The split is a definition; the inequality is the second law in local form, and note that it constrains only the internally produced term (de Groot and Mazur 1962). Nothing forbids $d_{e}S/dt$ from being negative, and one consequence follows immediately. A pattern that persists unchanged has $dS/dt=0$, hence $d_{e}S/dt=-d_{i}S/dt\le 0$: a steady structure must export entropy continuously, at exactly the rate it produces it. The question "where did the entropy go" therefore has an answer with a number attached, and the rest is arithmetic. The production is written locally as a sum of fluxes multiplied by their conjugate forces, $\sigma=\sum_{k}J_{k}X_{k}$, and for heat conduction alone the conjugate pair can be derived rather than asserted. Local entropy obeys a balance $\partial s/\partial t+\nabla\!\cdot\!\mathbf{J}_{s}=\sigma$, with entropy flux $\mathbf{J}_{s}=\mathbf{J}_{q}/T$ since heat $\delta q$ carries entropy $\delta q/T$. Energy conservation with no sources gives $\partial u/\partial t=-\nabla\!\cdot\!\mathbf{J}_{q}$, and at fixed composition and volume $\partial s/\partial t=(1/T)\,\partial u/\partial t$. Substituting and using $\nabla\!\cdot\!(\mathbf{J}_{q}/T)=(1/T)\nabla\!\cdot\!\mathbf{J}_{q}+\mathbf{J}_{q}\!\cdot\!\nabla(1/T)$, the divergence terms cancel and $$\sigma=\mathbf{J}_{q}\cdot\nabla\!\left(\frac{1}{T}\right)=-\frac{\mathbf{J}_{q}\cdot\nabla T}{T^{2}} .$$ The form is not a convention: it vanishes when $T$ is uniform, and its non-negativity is precisely the statement that heat flows down a temperature gradient. Writing a local $T$ at all presumes local equilibrium — that each small parcel has relaxed internally on a timescale short compared with the pattern's. For water here that is safe with room to spare: the pattern scale is $5\times10^{-3}\,\mathrm{m}$ and the molecular scale about $3\times10^{-10}\,\mathrm{m}$, seven orders of magnitude apart. A threshold, not a drift Before pricing the pattern one needs to know when it appears, and the answer is that it does not creep in as the heating is raised — it switches on at a threshold. The control parameter for a layer of depth $d$ with a temperature difference $\Delta T$ across it is the Rayleigh number, and it is worth assembling from timescales because that is what explains its shape: $$\mathrm{Ra}=\frac{g\,\alpha\,\Delta T\,d^{3}}{\nu\,\kappa} ,$$ with $g$ the gravitational acceleration, $\alpha$ the thermal expansion coefficient, $\nu$ the kinematic viscosity and $\kappa$ the thermal diffusivity. A parcel displaced upward carries its temperature with it and so becomes buoyant, with acceleration of order $g\alpha\Delta T$. Viscosity limits the resulting speed to $u\sim g\alpha\Delta T\,d^{2}/\nu$, so the time to cross the layer is $t_{\mathrm{rise}}\sim d/u=\nu/(g\alpha\Delta T\,d)$. Against this stands the time for conduction to erase the parcel's temperature excess, $t_{\mathrm{diff}}=d^{2}/\kappa$. The Rayleigh number is their ratio, $t_{\mathrm{diff}}/t_{\mathrm{rise}}$, and the exponent on $d$ is three because two powers come from the diffusion time and one from the rise time. Convection wins when the parcel arrives before its buoyancy has been diffused away. Linear stability analysis converts "wins" into a number by asking at what $\mathrm{Ra}$ an infinitesimal perturbation stops decaying. The critical value depends on the boundaries: $\mathrm{Ra}_{c}\approx1708$ for rigid, perfectly conducting plates, and $657.5$ for stress-free ones, the case Rayleigh (1916) treated because it is analytically tractable (Chandrasekhar 1961). The same calculation fixes the size of the pattern, since one horizontal wavenumber goes unstable first: for rigid plates it corresponds to a wavelength $\lambda_{c}\approx2.0\,d$, so each roll is about half a layer-depth wide. Take water at $298\,\mathrm{K}$ in SI units: $\alpha=2.1\times10^{-4}\,\mathrm{K^{-1}}$, $\nu=1.0\times10^{-6}\,\mathrm{m^{2}\,s^{-1}}$, $\kappa=1.4\times10^{-7}\,\mathrm{m^{2}\,s^{-1}}$, thermal conductivity $k=0.60\,\mathrm{W\,m^{-1}\,K^{-1}}$ and $g=9.8\,\mathrm{m\,s^{-2}}$. For $d=5.0\,\mathrm{mm}$ the formula gives $\mathrm{Ra}=1.84\times10^{3}\,(\Delta T/\mathrm{K})$, so the threshold sits at $\Delta T_{c}=1708/1.84\times10^{3}=0.93\,\mathrm{K}$ — a temperature difference one could impose by mistake. Because $\mathrm{Ra}$ scales as $d^{3}$, halving the depth to $2.5\,\mathrm{mm}$ multiplies the required difference by eight, to $7.4\,\mathrm{K}$. The budget for that layer Now the two sides of the ledger, for one square metre of that water layer held just at threshold. At threshold