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Power describes how quickly energy is transferred. For mechanical work, the instantaneous power supplied by a force isThe SI unit is $\mathrm{W}=J/s=\mathrm{N}m/s$, distinct from the velocity unit $m/s$. The italic symbol $W$ for work and the upright unit $\mathrm{W}$ for power mean different things. Two processes can transfer the same energy while requiring different powers: doing $12\mathrm{J}$ of work in $2\mathrm{s}$ requires an average $6\mathrm{W}$, whereas doing it in $4\mathrm{s}$ requires an average $3\mathrm{W}$. ^[ref]

Pressure measures the compressive normal force per unit surface area. For a uniformly loaded surface of surface area $A$, the magnitude of the perpendicular force is $F_{\perp }=PA$; locally, $P=d{F}_{\perp }/dA$. Its SI unit is the pascal, $1\mathrm{Pa}=1\mathrm{N}{\mathrm{m}}^{-2}$. A force parallel to the surface contributes shear stress, rather than pressure. In a gas, the steady macroscopic force comes from many microscopic transfers of momentum; ideal gas pressure from wall collisions makes this connection explicit. ^[ref]

For a computed example, a uniform pressure of $100\mathrm{kPa}$ on a piston of surface area $10\mathrm{c}{\mathrm{m}}^2=10^{-3}{\mathrm{m}}^2$ produces a perpendicular force of $100\mathrm{N}$. If the opposite face is exposed to $80\mathrm{kPa}$, the net force due to the two pressures is $(100-80)\times 10^3\times 10^{-3}=20\mathrm{N}$, directed toward the lower-pressure side. The pressure on either face and their difference answer different questions; gauge pressure records a difference from a specified reference, often atmospheric pressure.

A scalar pressure describes a gas in local thermodynamic equilibrium with isotropic random particle velocities: every wall orientation has the same normal force per unit surface area. Without this isotropy, normal stresses can depend on direction and a pressure tensor is needed. Thus random motion alone is insufficient; a preferred direction matters.

In thermodynamics, pressure and volume are thermodynamic conjugate variables. For a simple compressible system, reversible expansion delivers work $\delta W=PdV$, and at fixed entropy $S$ and particle number $N$,where $U$ is internal energy. A large pressure therefore means both a large wall force per unit surface area and a large amount of expansion work per small increase in volume.

In Gaussian units, the gyrotropic pressure tensor adds directional stress to the magnetic tension. A transverse displacement of a uniform magnetic field obeys $\rho {\partial }_t^2{\mathbf{\xi }}_{\perp }=\mathcal{T}{\partial }_{\parallel }^2{\mathbf{\xi }}_{\perp }$. Positive $\mathcal{T}$ gives a restoring force; negative $\mathcal{T}$ gives the fluid firehose instability. This follows by combining $-\nabla \cdot \mathsf{P}$ with the divergence of the Maxwell stress tensor. In SI replace $B^2/(4\pi )$ by $B^2/{\mu }_0$.

A low-frequency, strongly oblique plasma perturbation in which perpendicular magnetic pressure equilibrates rapidly compared with the retained dynamics. In Gaussian units, its leading total transverse pressure perturbation vanishes as shown. Thus $\delta B/B_0=-(\beta /2)\delta p_{\perp }/p_0$, where $\beta$ is plasma beta. A small fractional magnetic field perturbation can still exert a leading force.

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.

The resultant of distributed surface pressure; in a fluid its force per unit volume is $-\nabla P$. A spherical shell in hydrostatic equilibrium balances this against its gravitational weight.

A spatially varying pressure exerts net force per unit volume $-\nabla p$ on a small fluid element. Its contribution to acceleration is $-\nabla p/\rho$. Pressure increases in the direction opposite to this force.

A Prony brake measures shaft power using a resisting band: its force difference gives torque $\tau =r(F_1-F_2)$, and rotation gives $P=\tau \omega$, where $\omega$ is angular velocity. The Stirling engine experiment above uses this independent load measurement to distinguish shaft output from gas-cycle work.

An ideal receding mirror reflecting a source beam of constant power $P$ receives force $(2P/c)(1-\beta )/(1+\beta )$ after including both Doppler-shifted reflected momentum and reduced impact rate. For fixed rest mass $M$, use $dt/d\tau =\gamma$ in the momentum equation. It reduces to $d\beta /d\tau =(1-\beta {)}^2/{\tau }_0$, where ${\tau }_0=M{c}^2/(2P)$. Starting from rest gives $\beta =\tau /(\tau +{\tau }_0)$. This continuum mirror model neglects individual-photon recoil and finite beam interception.

The tensile force magnitude needed to detach contacting surfaces during retraction. A force-probe measurement reads it from the force minimum relative to the noncontact baseline; roughness, capillary forces and contact geometry affect its interpretation.

A lossless intermediate section of positive real characteristic impedance $Z_M$ matches unequal real exterior characteristic impedances $Z_L,Z_R$ when $Z_M^2=Z_L{Z}_R$ and its propagation wave phase is an odd multiple of $\pi /2$. The length is an odd quarter of the internal wavelength. This principle also applies to the characteristic impedance of a stretched string.

For a uniform intermediate string of length $l$, put $\delta =k_{M}l$. Continuity of displacement and transverse force gives the displacement reflection coefficientAt real positive mechanical impedances, the denominator never vanishes. Its numerator vanishes only if $(Z_L-Z_R)\cos \delta =0$ and $(Z_M^2-Z_L{Z}_R)\sin \delta =0$. Thus unequal exterior strings require the quarter-wave condition. Equal exterior strings with a different middle string instead transmit perfectly at integer half-wavelengths; if all three mechanical impedances agree, every frequency is matched.

A weak tangential force can move a satellite through a slowly varying sequence of nearly circular Kepler orbits. For constant satellite mass, $E=-\mu m/(2r)$ and $v_c=\sqrt{\mu /r}$ give $\dot{E}=F{v}_c$, hence $F/m=\sqrt{\mu }\dot{r}/(2r^{3/2})$. Integrating the thrust magnitude for a monotone outward transfer proves the displayed formula. The integrated thrust per mass is not the vector difference of the endpoint velocities; gravity acts throughout the transfer. The approximation requires slow orbital evolution and negligible eccentricity.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

A point mass constrained to a straight guide through the axis of a uniformly rotating reference frame, with signed central displacement $x$ and Hooke's law force $-kx$, satisfies $m\ddot{x}=-kx+m{\Omega }^{2}x$. The transverse Coriolis force is balanced by the guide. The central equilibrium is stable for ${\Omega }^2<k/m$, neutral at equality and unstable above it. The angular frequency is the positive square root in the title.

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

A 2013 experiment by Paul A. Williams and collaborators mounted a 25-mm dielectric mirror on a commercial force-restoration balance. Removing its pan-weight compensation spring allowed horizontal force sensing with a horizontal laser path. They measured mirror optical reflectance and varied incidence from about $7^{\circ }$ to $45^{\circ }$. A 1071-nm laser, capable of 530 W, was switched at 0.1 Hz. Digitized mass-equivalent readings were converted to force using $g$; quadrature demodulation isolated the switched signal from drift. ^[ref]

Measurement calibration used a 20-g mass, with checks down to 1 mg. The optical forces lay below that checked range, limiting the accuracy claim. An angle scan tested the predicted cosine dependence at fixed intercepted power. Replacing the mirror with a carbon nanotube coated absorber increased absorbed power while reducing the predicted optical force; no modulation peak was detected. This bounded gas-heating artifacts under those conditions, rather than proving their universal absence. The fitted slope was about 7% below $2/c$. Inferring laser power requires calibrated force, optical reflectance, radiometric absorptance and angle of incidence, with measurement uncertainty for each. ^[ref]

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

The leading uniform stellar radiation pressure is a differential acceleration relative to a planet, of scale $\beta G{M}_{\star }/a_p^2$. Relative to planetocentric gravity at dust radius $a$, its magnitude isThus lying inside a Hill sphere alone does not guarantee small radiative perturbations. Approximately circular tightly bound dust requires this ratio sufficiently small, as well as weak tides; the uniform radiative force can otherwise excite orbital eccentricity or unbind grains.

Place two equal point masses of $1\mathrm{kg}$ each on opposite sides of a fixed origin, at equal radius $r$, and let them share a signed angular velocity $\omega =\dot{\theta }$ in a plane. Their total mass is $M=2\mathrm{kg}$ and their moment of inertia about the perpendicular axis is $I=M{r}^2$. Radial pulling forces have zero torque about that origin, soTake an initial radius $r_i=1\mathrm{m}$ and angular velocity ${\omega }_i=1rad/s$, and a final radius $r_f=0.5\mathrm{m}$. ThenandThese are endpoint states with ${\dot{r}}_i={\dot{r}}_f=0$. During the pull, the full kinetic energy also contains radial motion: $K=M{\dot{r}}^2/2+K_{\mathrm{rot}}$. Assume the pulling apparatus supplies all net work to the two point masses; the work-energy theorem then fixes its net work at $+3\mathrm{J}$, including any radial acceleration and braking. Each mass spirals inward during the pull, allowing its inward force to do positive work. Conservation of angular momentum therefore predicts the final angular velocity without requiring kinetic energy conservation. For an apparatus included in a larger system, its stored energy supplies the increase consistently with conservation of energy. OpenStax uses the same mechanism to explain a skater pulling her arms inward. ^[ref]

The geometrical optics account of an optical trap adds the momentum flux changes of refracted and reflected light rays. It applies when the bead is large compared with the surrounding medium's wavelength. When the bead is much smaller, an induced electric dipole gives a different approximation. For a homogeneous dielectric sphere of radius $a$, define $\mathbf{p}(t)=\mathrm{Re}[\alpha {\mathbf{E}}_0{e}^{-i\omega t}]$, with ${\mathbf{E}}_0$ the peak complex electric field and $\alpha$ its SI polarizability. In a transparent, nonmagnetic, nondispersive host,where $n_p,n_m$ are particle and host refractive indices, $I$ is cycle-averaged light intensity, ${ϵ}_0$ is the vacuum permittivity and $c$ is the speed of light. The quasistatic electric dipole approximation requires $2\pi n_{m}a/{\lambda }_0\ll 1$, with ${\lambda }_0$ the vacuum wavelength, and a field nearly uniform across the bead.

For real quasistatic $\alpha$, building an induced electric dipole gives instantaneous potential energy $-\alpha E^2/2$. Averaging over one optical cycle givesThus $n_p>n_m$ gives positive polarizability and attraction toward larger light intensity; $n_p<n_m$ reverses this contribution. The factor $1/4$ uses a peak field: substituting a root mean square field without changing that factor would halve the force incorrectly. Dimensionally $\alpha |E_0{|}^2$ is energy, so its gradient is force.

This leading conservative term does not remove the optical scattering force. Even a nonabsorbing induced electric dipole reradiates light: at fixed wavelength and index contrast, its Rayleigh scattering cross-section scales as $a^6$, while $\alpha$ scales as $a^3$. A smaller bead can therefore improve the ratio of gradient to scattering forces while weakening the well relative to $k_{B}T$, where $k_B$ is the Boltzmann constant and $T$ the temperature. Optical force balance and resistance to thermal escape are separate questions. Intermediate bead sizes require a wave-scattering calculation; neither a ray calculation nor this small-dipole expression is uniformly valid there.