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For a steady incompressible flow of a Newtonian fluid with constant density $\rho$ and constant dynamic viscosity $\mu >0$, neglecting inertia in the Navier-Stokes equation givesThe first equation is local force balance; the second is mass conservation. The term $\mu {\nabla }^2\mathbf{u}$ is the divergence of the viscous part of the Newtonian fluid stress tensor. The body force $\mathbf{f}$ has units of force per unit volume, $\mathrm{N}{\mathrm{m}}^{-3}$, and equals $\rho \mathbf{g}$ for uniform gravity. Appropriate boundary data for velocity or traction, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the no-slip boundary condition.

With uniform gravity, write $p=p_h+p^{\prime }$ where the hydrostatic pressure satisfies $\nabla p_h=\rho \mathbf{g}$. Then the disturbance equation is $-\nabla p^{\prime }+\mu {\nabla }^2\mathbf{u}=0$. This removes gravity from the fluid disturbance equation, while the hydrostatic pressure still supplies buoyancy on an immersed body. Omitting it from the total force balance would give the wrong Stokes settling velocity.

Take downward as positive and let gravity have magnitude $g$. For a rigid sphere of density ${\rho }_p$ in otherwise stationary fluid of density ${\rho }_f$, its weight minus buoyancy is $\frac{4}{3}\pi a^3({\rho }_p-{\rho }_f)g$. At terminal velocity, this balances the Stokes drag law:This signed downward settling velocity is negative for a buoyantly rising sphere and zero for equal densities. For fixed density contrast and dynamic viscosity, the submerged weight grows as $a^3$ while linear drag at fixed speed grows as $a$; their balance gives the settling-velocity radius scaling $U_s\propto a^2$. Meanwhile,Thus doubling the radius multiplies the predicted Stokes settling speed by four and the particle Reynolds number by eight. This makes the formula useful for separating particles by size, while also providing its own check against extrapolation to large particles.

For a computed water example, take ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and $a=10.0\mu \mathrm{m}$. These are specified rounded inputs rather than a claim of exact water properties at every temperature. They giveThe diameter is $20.0\mu \mathrm{m}$, and both the net downward weight and the upward drag force are $6.16\times 10^{-11}\mathrm{N}$. At this terminal velocity, falling $1.00\mathrm{mm}$ takes $3.06\mathrm{s}$. At $a=20.0\mu \mathrm{m}$ the prediction is $1.308\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.05232$; at $a=50.0\mu \mathrm{m}$ it is $8.175\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.8175$, which is no longer a convincing small-Reynolds number case. The figure labels these as predictions of the Stokes settling velocity formula and shows where its inertia check fails.

The calculation describes the eventual terminal velocity, not the release transient governed by particle acceleration and possibly the Unsteady Stokes equation. Walls, hydrodynamic interactions, departures from a rigid sphere, or a different surface condition require a different resistance. For sufficiently small particles, Brownian motion can obscure individual settling trajectories; the formula still specifies the gravitational drift within its assumptions. The balancing argument already appears in article 43 of ^[ref].

For a liquid interface with constant surface tension $\gamma$, the mechanical surface energy is the energy associated with its surface area $A$:The additive constant $C$ does not affect the work needed to change surface area. For several interfaces with the same surface tension, $A$ means their total surface area; the two faces of a thin soap film must both be counted. ^[ref]

With assumed $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, creating $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ of one interface requires $7.2\mu \mathrm{J}$ of reversible work. In the sliding-rim example illustrated under surface tension, moving a $20\mathrm{mm}$ rim outwards by $1\mathrm{mm}$ creates $4.0\times 10^{-5}{\mathrm{m}}^2$ across the two faces and requires $2.88\mu \mathrm{J}$. Independently, multiplying its $2.88\mathrm{mN}$ holding force by the displacement gives the same work.

This makes the preference for a round isolated droplet precise. At fixed volume $V$, the three-dimensional isoperimetric inequality giveswith equality for a sphere. Thus a sphere minimizes surface energy among sufficiently regular closed shapes enclosing the same volume, when surface tension is uniform and other energy contributions are negligible. This is an application of the isoperimetric inequality, rather than a proof of it. Walls, wetting, and gravity can change the shape by adding constraints or other energy contributions. ^[ref]

The differential rule $d{E}_{\mathrm{s}}=\gamma dA$ integrates directly to $E_{\mathrm{s}}=\gamma A+C$ because surface tension is constant. At fixed liquid volume, decreasing surface area can drive viscous fluid flow; the released surface energy can supply viscous dissipation. This mechanical bookkeeping is useful for predicting which deformations can release energy. If surface tension depends on temperature or composition, distinguish this mechanical area-work relation from the complete interfacial thermodynamic free energy and internal energy.

A thermodynamic system is in thermodynamic equilibrium when its macroscopic state has no spontaneous tendency to change through any exchange allowed by its constraints. For an isolated system, stable thermodynamic equilibrium maximizes entropy over the accessible states at fixed total energy and conserved material inventories. A wall can forbid an exchange, so always specify what can cross or move. Thermal, mechanical and chemical conditions answer different questions about that same constrained state. ^[ref]

For example, two compartments separated by a rigid, impermeable wall that transmits heat can reach thermal equilibrium while retaining different pressures and compositions. Allowing an unloaded piston to move adds a mechanical equilibrium condition. Allowing one chemical species to cross adds chemical potential equality for particle exchange; allowing a chemical reaction adds its chemical equilibrium condition. A fixed barrier can therefore support a constrained thermodynamic equilibrium that ceases to be an equilibrium when the barrier is removed.

Uniformity is not the definition of thermodynamic equilibrium. An isothermal atmosphere in hydrostatic equilibrium can have a common temperature while its pressure and mass density vary with height: the pressure gradient balances gravity. In an external field, the exchange condition must include the relevant potential energy; for charged matter this leads to the electrochemical potential. Relativistic gravitational corrections to thermal equilibrium are described separately by the Tolman temperature law.

A steady state only says that the macroscopic description is constant in time. A bar maintained between hot and cold thermal reservoirs has steady heat flux and produces entropy, so it is not in thermodynamic equilibrium. Even matching temperatures at one perfect-contact interface can coexist with heat transfer through the surrounding material; matching those two local values does not establish equilibrium of the whole apparatus. See thermal conduction for that spatial transport problem.

The useful simplification is that equilibrium conditions predict an endpoint without computing every collision or flow during relaxation. The first law of thermodynamics supplies the conserved quantities, and the Second law of thermodynamics supplies the direction of change. For a system held at fixed temperature and volume, the corresponding criterion is minimization of Helmholtz free energy; at fixed temperature and pressure, it is minimization of Gibbs free energy, with the material and exchange constraints still imposed.

Here gravitational weight means the force on a body due to gravity. In the Newtonian model, a body of mass $m$ in a uniform gravitational acceleration $\mathbf{g}$ has gravitational weight $\mathbf{W}=m\mathbf{g}$, with magnitude $W=mg$. Its direction follows $\mathbf{g}$; a buoyant force is a separate contribution to the body's net force. ^[ref].