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The 1988 experiment loaded sodium optical molasses from a slowed atomic beam. After stopping loading, a shutter extinguished the six cooling beams. Released atoms crossed a lower probe; a photomultiplier recorded fluorescence versus arrival time. Calculations integrated the measured initial cloud distribution and an assumed three-dimensional Maxwell-Boltzmann distribution, including gravity and detection geometry. The minimum was $43\pm 20\mu \mathrm{K}$, versus the $240\mu \mathrm{K}$ Doppler cooling limit.^[ref]

Probe separation, probe intensity, cloud density and shutter timing were varied. Fountain-height, horizontal-spreading and release/recapture measurements also gave below-limit results. Geometry remained an uncertainty; agreement among complementary thermometers made a detector artifact less plausible. These measured released atoms, rather than an in-situ velocity distribution.^[ref]

Steven Chu's account connects the explanation to optical pumping, Stark effect shifts and spatially changing electromagnetic polarization; Claude Cohen-Tannoudji's lecture separates this mechanism from subrecoil schemes. Their lectures and William D. Phillips's describe complementary experimental and theoretical routes.^[ref][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.

Consider a symmetric top with fixed vertex $Q$, axial unit vector $\mathbf{e}$, principal moments of inertia $A,A,C$, and centre of mass at $h\mathbf{e}$. Let a force $\kappa \ell (\mathbf{k}-\mathbf{e})$ act at $\ell \mathbf{e}$, where $\mathbf{k}$ is the upward vertical unit vector. Together with gravity, it produces torque $(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ and potential energy $mgh\mathbf{k}\cdot \mathbf{e}+\kappa {\ell }^2(1-\mathbf{k}\cdot \mathbf{e})$.

The axial angular velocity of a symmetric top $n$, vertical angular momentum $\mathbf{k}\cdot \mathbf{L}$ and total mechanical energy are conserved. For steady precession, $\dot{\mathbf{e}}=\Omega \mathbf{k}\times \mathbf{e}$ and $\mathbf{k}\cdot \mathbf{e}=\cos \alpha$. Substituting $\mathbf{e}\times \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\times \mathbf{k}$ into $A\mathbf{e}\times \ddot{\mathbf{e}}+Cn\dot{\mathbf{e}}=(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ gives the condition in the title for nonvertical inclination. If the total angular velocity itself is $\Omega \mathbf{k}$, then $n=\Omega \cos \alpha$, and the condition becomes $\kappa {\ell }^2=mgh+(A-C){\Omega }^2\cos \alpha$.

Consider a point mass of fixed mass $M>0$ released from rest in a stationary medium. Choose downward as positive, assume constant gravitational acceleration $g>0$, neglect buoyancy, and impose linear drag $F_d=-bv$ with constant linear friction coefficient $b>0$. Here $v$ is the signed downward velocity component; it equals speed during this descent. Newton's second law becomesThe terminal velocity balances the two forces, and the time constant $\tau$ controls how quickly that balance is approached:For the illustrative parameters $M=1\mathrm{kg}$, $b=2kg/s$ and $g=9.8m/{\mathrm{s}}^2$, the time constant is $0.5\mathrm{s}$ and the terminal velocity is $4.9m/s$ downward. At $t=1\mathrm{s}=2\tau$, the downward velocity is $4.9(1-e^{-2})m/s\simeq 4.23686m/s$ and the downward acceleration is $9.8e^{-2}m/{\mathrm{s}}^2\simeq 1.32629m/{\mathrm{s}}^2$. This is a toy linear drag model, not a claim that every $1\mathrm{kg}$ falling object has this resistance law.

For a proof outline, subtract the equilibrium value: $u=v-v_T$ obeys $du/dt=-u/\tau$. The exponential decay solution and the initial condition give $u=-v_T{e}^{-t/\tau }$; one derivative gives the acceleration. Every time constant multiplies the remaining gap $v_T-v$ by $e^{-1}$. The exact limit is reached only as $t\to \infty$. This makes the model useful for estimating a settling duration without mistaking vanishing acceleration for vanishing velocity.

Newton already analyzed vertical motion under uniform gravity and resistance proportional to velocity in Book II, Section I, Proposition III of the Principia. His geometric construction includes a greatest downward velocity, illustrating that a limiting rate can emerge from balancing a driving force and resistance. ^[ref]

Journal of Geophysical Research 81, 1903–1915 (1976). Primary ship-track gravity/bathymetry spectral analysis. doi.org/10.1029/JB081i011p01903

For a nearly flat interface, the extra surface energy in the small-slope approximation is $(\gamma /2)\int |\nabla h{|}^{2}dA$. Each nonzero Fourier mode therefore has thermal variance $⟨|h_{\mathbf{q}}{|}^2⟩=k_{B}T/(A\gamma q^2)$ in a convention with height reconstructed as the sum of its mode amplitudes. Summing over a two-dimensional annulus of wavevectors givesThe smallest wavevector is set by the lateral size or pinning, and the largest by the breakdown of the continuum description, commonly a molecular length. A fixed mean height removes the zero mode. Gravity supplies an additional restoring term and changes the long-wavelength cutoff when it is retained. Roughness here means a height variance; its square root is the root mean square amplitude, even when the mean height fluctuation is zero.

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.

Tidal deformation is the change in a body's shape caused by differential gravity across it, that is, by a tidal force. The familiar two-bulge picture represents an elongated response along the companion direction, rather than a separate force pulling only the nearest surface. If spin and orbital motion differ, the forcing sweeps through the body's material. With tidal dissipation, the response is out of alignment with the instantaneous forcing, allowing a net torque. ^[ref].

A mass above a sufficiently wide estuary experiences additional downward gravity when extra water is present. Approximating the tide as a plane slab of density $\rho$ and added thickness $h$ gives $\Delta g=2\pi G\rho h$. A linear spring with initial extension $x_0=mg/k$ changes by $\Delta x=(m/k)\Delta g=x_0\Delta g/g$. This isolates direct water attraction; it omits finite shoreline geometry, buoyancy, structural motion and terrestrial tides.

Work in Minkowski spacetime, neglect gravity, and choose the stay-at-home clock's inertial frame. Let departure be $O=(t,x)=(0,0)$ and reunion be $R=(T,0)$. The home clock follows $x_H(t)=0$. A traveller moves outward with velocity $+v$ and returns with velocity $-v$, where $0<v<c$ and $c$ is the speed of light. The ideal world line has a sharp turnaround at $P=(T/2,vT/2)$:Each straight leg lies in an inertial frame, but no single traveller inertial frame covers both legs.

For any differentiable timelike curve in this home inertial frame, its clock accumulates proper time through the integralApply it separately to the traveller's two legs. With the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For $v=0.6c$ and $T=10\mathrm{yr}$, the Lorentz factor is $\gamma =1.25$. The turnaround is $3$ light-years away at home time $5\mathrm{yr}$. The traveller records $4\mathrm{yr}$ outward and $4\mathrm{yr}$ returning: the reunited clocks read $10\mathrm{yr}$ and $8\mathrm{yr}$. These are illustrative model values, not experimental measurements.

The same integral gives a short geometric explanation: $\sqrt{1-v(t{)}^2/c^2}\le 1$, so every such competing timelike curve between these fixed spacetime events has $\Delta \tau \le T$. Equality requires zero velocity throughout, apart from sets of zero duration. Thus the straight home world line maximizes proper time in this Minkowski spacetime problem. This statement concerns fixed endpoints and an inertial home clock; in a gravitational field, compare paths using the appropriate metric tensor.

For two inviscid incompressible half-spaces, lower density/velocity ${\rho }_1,U_1$, upper density/velocity ${\rho }_2,U_2$, downward gravity $g$ and surface tension $\sigma$, a mode $e^{i(kx-\omega t)}$ with $k>0$ obeysA negative right-hand side produces exponential growth. The gravitational term recovers Rayleigh-Taylor instability for heavy-over-light ordering; the shear term can destabilize otherwise stable stratification. Finite layer thickness, viscosity, compressibility and magnetic tension lie outside this interface model.

Inside a homogeneous spherical mass, gravity is proportional to position. Projection onto a fixed straight chord gives $\ddot{y}=-{\omega }_0^{2}y$, independently of its perpendicular offset from the sphere centre. Release from one surface endpoint at rest reaches the opposite endpoint after $\pi /{\omega }_0$. The transverse force is supplied by the tunnel constraint. This idealization neglects removed tunnel mass, rotation, density variation and dissipative forces.

A horizontal bar of mass $M$ hangs from a spring of incremental stiffness ${\kappa }_0>0$, and two equal masses $m$ hang separately from the bar through springs of stiffness ${\kappa }_1>0$. The bar remains horizontal and all motion is vertical. For physical displacements $q=(q_1,q_2,q_3)$ from the loaded equilibrium, the kinetic energy and quadratic potential energy areGravity determines the static extensions; its linear terms cancel at equilibrium. Exchange of the lower masses splits the normal modes into the antisymmetric mode $(0,1,-1)$ with ${\omega }^2={\kappa }_1/m$, and two symmetric modes $(A,B,B)$. The latter obey $({\kappa }_0+2{\kappa }_1-M{\omega }^2)A-2{\kappa }_{1}B=0$ and $-{\kappa }_{1}A+({\kappa }_1-m{\omega }^2)B=0$. Their determinant gives the title quadratic in ${\omega }^2$, and $B/A={\kappa }_1/({\kappa }_1-m{\omega }^2)$. For $M=2,m=1,{\kappa }_0=4,{\kappa }_1=1$, their squared angular frequencies are $2\pm \sqrt{2}$, with displacement ratios $(1,1\mp \sqrt{2},1\mp \sqrt{2})$ respectively. The lower symmetric mode moves all masses in phase; the higher one moves the lower pair against the bar.

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].