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Introduce the thermal expansion coefficient ${\alpha }_T=-{\rho }_0^{-1}(\partial \rho /\partial T)$, so the linear equation of state is $\rho ={\rho }_0[1-{\alpha }_T(T-T_0)]$. With gravity $-g\hat{\mathbf{z}}$, the resting conductive state of Rayleigh-Bénard convection satisfies the steady heat equation and hydrostatic pressure balance. Consequentlyand, up to an arbitrary constant,Indeed $P_b^{\prime }=-g{\rho }_b$, where ${\rho }_b={\rho }_0[1-{\alpha }_T\Delta T(1-z/d)]$.

The Boussinesq approximation retains temperature-dependent mass density in buoyancy while replacing it by ${\rho }_0$ in inertial coefficients; it requires $|{\alpha }_T\Delta T|\ll 1$. Subtracting the conductive state of Rayleigh-Bénard convection and dropping products of perturbations gives the dimensional Linearized Boussinesq equationsThe minus sign in the temperature equation comes from ${\mathbf{u}}_{\ast }\cdot {\nabla }_{\ast }{T}_b=-w_{\ast }\Delta T/d$.

Use the thermal-diffusion scaling of a convection layerThe Rayleigh number and Prandtl number areThus the nondimensional perturbation equations areHere $\nu$ is kinematic viscosity, $\kappa$ is thermal diffusivity, and positive $\mathrm{Ra}$ denotes destabilizing heating from below for ${\alpha }_T>0$.

Take the layer depth as the length unit, its thermal diffusion time as the time unit, and its imposed temperature difference as the temperature unit. Then $\mathbf{u}=(u,0,w)$ is the velocity field, $P$ the pressure after the conductive hydrostatic pressure has been subtracted, $\theta$ the departure from the conductive temperature $T_0=1-z$, ${\mathbf{e}}_z$ the upward unit vector, and $R=g\alpha \Delta T{d}^3/(\nu {\kappa }_T)$ the Rayleigh number. Here $d$ is the dimensional depth, $\nu$ the kinematic viscosity, ${\kappa }_T$ the thermal diffusivity and $\alpha$ the thermal expansion coefficient. The gradient, divergence and Laplacian act on the dimensionless coordinates. The term $w$ in the heat equation is advection of the background gradient, since $-\mathbf{u}\cdot \nabla T_0=w$. Mass density variations enter only the buoyancy term under the Boussinesq approximation; the incompressible flow constraint is $\nabla \cdot \mathbf{u}=0$.

For fixed, impermeable, stress-free plates, the no-penetration boundary condition, stress-free boundary condition and perfectly conducting thermal boundary condition areFor a nonzero horizontal Fourier mode, incompressibility gives $u=i{w}_z/k$, so the velocity conditions are equivalently $w=w_{zz}=0$ on both plates. A uniform horizontal velocity is not fixed by these boundary conditions; take its mean to be zero when discussing the stationary forced pattern.

The printed small-Prandtl-number label does not generally justify these equations. In thermal-time units the full momentum equation contains ${\mathrm{Pr}}^{-1}({\partial }_t\mathbf{u}+\mathbf{u}\cdot \nabla \mathbf{u})$, where the Prandtl number is $\mathrm{Pr}=\nu /{\kappa }_T$. Dropping that term gives infinite-Prandtl-number convection or another explicitly justified Stokes flow approximation. It is not the general $\mathrm{Pr}\to 0$ limit. For example the horizontal shear $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$, satisfies the full linear momentum equation and the plate conditions, and persists at fixed $t$ as $\mathrm{Pr}\to 0$. The printed inertia-free equation would instead require this shear to vanish. The following parts solve the equations actually displayed; applying them at small Prandtl number needs an additional slaving or creeping-flow ordering.