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Thermal convection is fluid motion driven by temperature-dependent buoyancy. It transports heat by bulk motion in addition to thermal conduction.

For a small steady lower-boundary height $b(x,y)$, impermeability gives $w={\mathbf{u}}_g\cdot \nabla b$ at the reference plane. Substitution in the buoyancy equation $D_g(f_0{\psi }_z)+N^{2}w=0$ yields the displayed boundary condition. For nonzero Fourier modes in a uniform crossing current, this fixes ${\psi }_z(0)=-N^{2}b/f_0$. An upper radiation condition or decay condition is still needed to select the physical response.

Let positive transverse displacement point along a uniform load per unit length $w$. The filament bending modulus $B$ gives energy ${\int }_0^L[B(h_{xx}{)}^2/2-wh]dx$. The variational derivative is $B{h}_{xxxx}-w$, so resistive-force theory gives ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}+w$. Clamping imposes $h(0)=h_x(0)=0$, while a free tip imposes $h_{xx}(L)=h_{xxx}(L)=0$. The steady solution isIts tip displacement is $w{L}^4/(8B)$. For a submerged cylindrical filament under Newtonian gravity, buoyancy gives $w=({\rho }_f-\rho )\pi a^{2}g$. This linear approximation requires small slope and negligible inertia.

For total plume flux $P(h)$, fixed ceiling exhaust $F$, zero ambient buoyancy, and uniform upper-layer reduced gravity $b$, the ideal two-layer balances areAt $P(h_{\ast })=F$ and $b_{\ast }=B/F$, linearization yields the decay times ${\tau }_h=A/P^{\prime }(h_{\ast })$ and ${\tau }_b=A(H-h_{\ast })/F$. For $P(h)\propto h^{5/3}$, ${\tau }_h=3A{h}_{\ast }/(5F)$. Upper-layer buoyancy relaxation usually dominates when $h_{\ast }\ll H$. Exact equilibrium is approached asymptotically; a fixed small relative tolerance takes order $\max ({\tau }_h,{\tau }_b)\log (1/\epsilon )$ after the initial transient.

Buoyancy changes the effective gravitational mass; a nearby no-slip wall changes the hydrodynamic resistance matrix and therefore the diffusion coefficient. These are distinct corrections. Write the horizontal resistance as ${\zeta }_{\parallel }=6\pi \eta a/f$, with $0<f<1$ representing wall hindrance at a specified height. Applying the Stokes–Einstein relation with that resistance givesif one mistakenly uses the bulk resistance in the inversion. A chosen $f=0.8$ would overestimate $N_A$ by $25\%$; this is a sensitivity example, not a correction measured in Jean Baptiste Perrin's apparatus. Normal and parallel coefficients of the hydrodynamic resistance matrix need not coincide. Stokes settling velocity near a wall can likewise bias a radius inferred using an unbounded-fluid formula.

For a passive dilute particle suspension at uniform temperature, a change of mobility alone does not change the Boltzmann distribution. In the absence of additional wall forces the correct local vertical flux is $j_z=-D_z(z)[dn/dz+m_{b}gn/(k_{B}T)]$, so zero flux still gives the same sedimentation scale height. Walls can nevertheless invalidate that simple inference through adsorption, additional interaction energies, or sampling of inaccessible grain-centre positions. Optical position errors and exposure averaging also affect displacement readout; varying distance from the wall, observation lag and optical depth tests different possible errors.