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The streaming velocity is the local mean stellar velocity obtained by averaging with the galactic distribution function. In an axisymmetric disk it usually denotes $⟨v_{\varphi }⟩$. It differs from the circular speed, which is set by the force field, and from the velocity dispersion, which measures fluctuations about the mean.

Consider two widely separated rigid spheres of equal radius $a$, with separation $R\mathbf{n}$, in Stokes flow. Apply a force $6\pi \mu a{\mathbf{U}}_0$ to the first sphere and let the second be force-free and torque-free. The second sphere samples the rate-of-strain tensor of the first sphere's Stokeslet, $\mathbf{E}=3a({\mathbf{U}}_0\cdot \mathbf{n})(\mathbf{I}-3\mathbf{nn})/(4R^2)$. Its leading disturbance is the stresslet of a sphere in a uniform straining Stokes flow. Evaluating that disturbance back at the first sphere givesThe hydrodynamic mobility matrix decreases at fixed force, consistently with the increase of resistance caused by an additional rigid inclusion.

For an elastic filament with reference arclength $s$, curvature $c$, axial strain $ϵ$ and excess twist rate $\Omega$, a harmonic energy density is $(K_b{c}^2+S{ϵ}^2+K_t{\Omega }^2)/2$. Bending and torsional rigidities have units of energy times length; the stretch modulus has units of force. Factoring out $k_{B}T$ therefore makes $A,C$ lengths but $B$ an inverse length. Axial strain under tension is approximately $F/S$. Twist is controlled by torque and rotational boundary conditions, not force alone. A three-dimensional thermally fluctuating worm-like chain has persistence length $K_b/(k_{B}T)$.

A uniform string with fixed ends $x=\pm \ell$, tension $T$, linear mass density $\rho$, and attached central point mass $m>0$ obeys the wave equation away from zero, continuous displacement at zero, and $m{y}_{tt}(0,t)=T[y_x(0^{+},t)-y_x(0^{-},t)]$. The normal modes split into antisymmetric modes that leave the mass at rest and symmetric modes that move it.

To prove the spectral equations, use $X=A\sin (k(x+\ell ))$ on the left and $X=B\sin (k(\ell -x))$ on the right, with ${\omega }^2=(T/\rho )k^2$. Continuity gives $(A-B)\sin (k\ell )=0$ and the force law gives $Tk(A+B)\cos (k\ell )=m{\omega }^{2}A\sin (k\ell )$. If $\sin (k\ell )=0$, then $k\ell =n\pi$, $n\ge 1$, and $B=-A$. Otherwise $A=B$ and cancellation gives $\cot (k\ell )=mk/(2\rho )$. In dimensionless variables $z=k\ell$, $\alpha =m/(2\rho \ell )$, strict monotonicity gives one symmetric root in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$. The apparent zero root of the sine equation has only the zero profile and is excluded.

For $m/(\rho \ell )\gg 1$, the lowest root tends to zero. The Taylor series $\cot z=1/z-z/3+O(z^3)$ gives $z\sim {\alpha }^{-1/2}$ and hence $\omega \sim \sqrt{2T/(m\ell )}$. The two nearly straight half-strings supply a restoring force $-2TY/\ell$, explaining this harmonic oscillator limit.

For a grain of mass density ${\rho }_p$ and volume $\mathcal{V}$ in a fluid of mass density ${\rho }_f$, its effective downward force is $({\rho }_p-{\rho }_f)g\mathcal{V}$, after subtracting buoyancy from weight.

Let downward velocity $v$ obey $\dot{v}=a-b{v}^2$, with constants $a,b>0$, terminal velocity $v_t=\sqrt{a/b}$ and initial speed $u>v_t$. Gravity minus buoyancy can supply the constant downward force represented by $a$. Since $\dot{z}=v$, the equation for $v^2$ is linear in depth: $d(v^2)/dz=-2b(v^2-v_t^2)$, giving the displayed decay. Exact terminal velocity is approached only asymptotically. A practical tolerance $v\le (1+ϵ)v_t$ is attained at $z_{ϵ}=(2b{)}^{-1}\ln [(u^2/v_t^2-1)/(2ϵ+{ϵ}^2)]$, provided the initial excess exceeds that tolerance. Its time-domain form is $v=v_t\coth (b{v}_{t}t+\chi )$ with $\chi =\mathrm{arctanh}(v_t/u)$, in contrast to the hyperbolic tangent for descent from rest under quadratic drag.

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.

For Stokes flow without body forces, $\nabla \cdot \mathbf{\sigma }=0$. Symmetry of the Newtonian fluid stress tensor also gives $\nabla \cdot (\mathbf{x}\times \mathbf{\sigma })=0$, interpreted componentwise as the angular-momentum flux. The divergence theorem makes the resultant force and torque the same on any two homologous enclosing surfaces lying in a nonsingular fluid region. In a boundary perturbation of a nearly spherical particle, evaluating each perturbation field on a fixed enclosing surface avoids spurious geometric force terms: the geometric terms from changing the surface normal and area cancel those from evaluating the base stress at the displaced surface.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.

For an initially intact link with no rebinding, survival probability obeys $dp/dt=-k(F)p$. A force-tilted activation barrier with force $F=ft$, constant $f>0$ and $p(0)=1$ integrates to the displayed law. The rupture-force probability density is $k(F)p(F)/f$, not $p(F)$ itself. The no-rebinding approximation requires separated partners to leave the capture region before appreciable reassociation; it is not guaranteed by the rate equation alone.

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 $\kappa {\ell }^2=mgh+(A-C){\Omega }^2\cos \alpha$.

A tangential force integral around an instantaneous moving loop is not generally the force work along the actual particle trajectory. For an expanding circular guide, the particle has both radial and tangential velocity; the magnetic radial and tangential powers cancel exactly even when the tangential circulation is nonzero.

MIT's Spring 2004 laboratory assignment Properties of Heat Treated Steel gives a concrete tensile testing protocol. Record transducer channels and calibration factors; measure specimen diameter and length, and the extensometer gauge length with its locking pin inserted. Zero the load cell, pin the threaded grips into their adapters, and attach the extensometer centrally. Apply the specified seating load, record actuator position, then acquire readings every second while moving at about $0.05\mathrm{cm}\mathrm{mi}{\mathrm{n}}^{-1}$. The handout derives engineering stress from load cell force, engineering strain from extensometer deformation, and Young's modulus by linear regression below the proportional limit; it warns that actuator-derived strain includes apparatus deformation.^[ref]

For a modulus-focused measurement, select a load window after seating, hold temperature and loading rate consistent, and record force, local extension and crosshead displacement together. Fit $F=F_0+k_g\Delta L_g$ within that window and calculate $E=k_g{L}_g/A_0$. Inspect the regression residuals for curvature; repeat at a lower maximum load and include unloading to check elastic deformation. These are diagnostic additions to the cited protocol, whose full test continues through failure. The handout's seating load is specific to its specimens and apparatus, rather than a universal preload.

An extensometer measures the change between its contact points; crosshead displacement measures travel of the machine's moving member. The manufacturer Instron explicitly distinguishes local strain from travel that includes machine deformation and machine compliance.^[ref] The gauge length $L_g$ can be shorter than the uniform length $L$ between grips. Thus even a perfectly rigid machine generally has $\Delta L_g\ne x$: the local instrument sees only its own segment. Do not divide one channel's extension by the other channel's reference length.

Grips must transmit axial force without slip. Misalignment adds bending moment and makes opposite sides experience different axial strains; an instrument on just one side may then report a biased modulus. A changing extension offset, hysteresis between loading and unloading, or load-dependent fitted slope is evidence to investigate seating, slip or material response. A straight line over a short interval alone does not identify which is responsible.

Tensile testing measures the response of a specimen to an axial pulling force. A tensile modulus experiment uses the initial proportional response to infer Young's modulus.

Tension is a pulling force transmitted along a taut string, rope, or similar body. For an ideal uniform string with linear mass density $\mu$, constant tension $T$ sets the small transverse wave speed $c=\sqrt{T/\mu }$.

Terminal velocity is a constant limiting velocity approached when a velocity-dependent resistance balances the other applied forces. A falling body may keep moving while its net force and acceleration tend to zero. For downward motion in a stationary medium, its limiting speed is set by balancing gravitational weight against drag force, including buoyancy if appreciable. The balance, rather than a universal numerical value, is the defining idea. ^[ref]

The terminal velocity under linear drag gives a simple example with an explicit approach to the limit. Real falling objects need an appropriate resistance model: linear drag is useful for slow motion of a small sphere in Stokes flow, whereas many faster falling objects require quadratic drag. Changing mass also changes the question: the existing accreting raindrop with quadratic drag has a speed-radius attractor of an accreting raindrop, rather than the fixed-mass balance used below.

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]

A unit of mass per length, equal to $1g/km$. Force divided by a linear density stated in tex gives tensile tenacity in newtons per tex.

A bead in a liquid keeps undergoing Brownian motion even when an optical trap holds its average position fixed. For a stationary harmonic optical trap in thermal equilibrium at absolute temperature $T$, the equipartition theorem assigns mean potential energy $k_{B}T/2$ to this one quadratic coordinate. With $k_B$ the Boltzmann constant and $x=0$ at equilibrium,At the same temperature, quadrupling the spring constant divides the position variance by four and the standard deviation by two. The Boltzmann factor gives the complete equilibrium probability density function, a Gaussian distribution:The prefactor makes ${\int }_{-\infty }^{\infty }p(x)dx=1$; its second moment is ${\int }_{-\infty }^{\infty }{x}^{2}p(x)dx=k_{B}T/k$. The probability density function has units of inverse length, and its area gives probability. This density is exact for the ideal quadratic model. In a finite optical trap it approximates the occupied well when thermal excursions remain within the harmonic region and escape is negligible during observation. Driven motion, appreciable nonconservative forces or laser heating need not give this equilibrium density.

Here $\ell =\sqrt{k_{B}T/k_0}$, $q=x/\ell$ and $\kappa =k/k_0$. The plotted density per unit $q$ is $p_q(q)=\ell p(\ell q)=\sqrt{\kappa /(2\pi )}{e}^{-\kappa q^2/2}$. Each curve has total area one; their dimensionless standard deviations are $1$, $1/\sqrt{2}$ and $1/2$. The curves are exact model densities, not noisy experimental data.

Brownian motion becomes a measuring tool: a calibrated position measurement supplies ${\sigma }^2=⟨(x-⟨x⟩{)}^2⟩$, and the known temperature supplies $k=k_{B}T/{\sigma }^2$. Once $k$ is known, a slowly applied constant external force $f$ shifts the mean by $⟨x⟩=f/k$, so $f=k⟨x⟩$ in the same harmonic range. Fluctuations therefore calibrate the conversion from displacement to force. Svoboda and Block used this equipartition theorem method when calibrating their bead traps. ^[ref].

For the example stiffness $0.01pN/nm$ and $T=300\mathrm{K}$, $k_{B}T\approx 4.142\mathrm{pN}\mathrm{nm}$ and $\sigma \approx 20.35\mathrm{nm}$. Using a variance measured in uncalibrated detector units would not determine that physical stiffness. Measurement noise, finite exposure time, drift and insufficient sampling of the correlated Brownian motion can bias the estimate; record positions over many optical trap relaxation times and account for the detector response. The deterministic relaxation curve above describes relaxation without thermal forcing; an individual thermal trajectory continues to fluctuate.

For a bond free of applied tension, use a local potential energy $V(a+x)=V(a)+\kappa x^2/2+\gamma x^3/6+\cdots$ with $\kappa >0$. Classical equipartition gives $⟨x^2{⟩}_0=k_{B}T/\kappa$; vanishing mean force then yields $⟨x⟩=-\gamma k_{B}T/(2{\kappa }^2)$ to first order in anharmonicity. Thus the linear thermal expansion coefficient is $\alpha =-\gamma k_B/(2a{\kappa }^2)$. For a Morse potential, $\gamma =-3\kappa /\xi$, giving positive expansion $3k_B/(2a\kappa \xi )$. This is a local bound-state approximation; the isolated cubic truncation is not a globally stable equilibrium potential.