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The superconducting isotope effect is the dependence of the superconducting transition temperature on isotopic mass $M$ at fixed chemical composition. Its discovery in 1950 was a crucial clue that phonons participate in conventional superconductivity.^[ref]

The model prediction can be derived from two assumptions. For harmonic motion with a fixed restoring-force constant $K$, a phonon frequency scales as $\omega \propto \sqrt{K/M}$. If the BCS dimensionless coupling is independent of $M$, the weak coupling result $k_B{T}_c\simeq 1.134\hslash {\omega }_D{e}^{-1/\lambda }$ then gives $T_c\propto M^{-1/2}$ and $\alpha =1/2$. Changes in the effective interaction can change this exponent; it is not a universal law for every material exhibiting superconductivity. For a stipulated mass ratio $M_2/M_1=1.04$, the ideal model predicts $T_{c,2}/T_{c,1}=1/\sqrt{1.04}\simeq 0.98058$. This is an original scaling example, not a pair of measured transition temperatures.

For a sphere with constant mass density $\rho$ collecting mass at rate $\beta$ per unit surface area, $\dot{m}=4\pi r^2\beta$ and $m=4\pi \rho r^3/3$ give $\dot{r}=\beta /\rho$. If it falls through stationary material under constant gravitational acceleration $g$ without drag, its velocity satisfies $\dot{v}+3\dot{r}v/r=g$. With $v(0)=0$, an integrating factor $r^3$ gives $v=g[r-r_0^4/r^3]/(4\dot{r})$.

At radius $R$ outside a spherical mass $M$, the Newtonian relative-acceleration tensor has eigenvalues $-g/R,-g/R,2g/R$ with $g=GM/R^2$. It compresses free fall separations tangentially and stretches them radially. The displayed time scale is the inverse square root of the curvature scale controlling this tidal evolution. An ideal circular orbit at that radius has period $2\pi t_{\mathrm{tide}}$, providing a second physical interpretation. For Earth's surface it is about $806$ seconds, while the corresponding ideal low-orbit period is about $84$ minutes. A uniform gravitational acceleration does not determine this relative motion; spatial variation of the field does.

A transition density with absorbing boundaries counts the joint event of survival up to time $t$ and a specified final position. Its integrated mass is survival probability, not generally one. Dividing by that mass gives the endpoint density conditional on survival, a different object.

Suspended sediment is particulate material carried within a fluid rather than solely along the bed. Its concentration balance includes advection, erosion into suspension and deposition from suspension. For water depth $h$ and suspended mass concentration $c$, the conservative balance $(hc{)}_t+(huc{)}_x={\rho }_s(v_E-v_D)$ reduces using water continuity to $h(c_t+u{c}_x)={\rho }_s(v_E-v_D)$.

A tabletop mass $m$ joined through a hole by a massless inextensible string of length $l$ to a hanging mass $M$ has radial inertia $M+m$ and conserved angular momentum $J=m{r}^2\dot{\theta }$. The radial energy is $\frac{1}{2}(M+m){\dot{r}}^2+U_J(r)$, with $0<r<l$. Its effective potential diverges at the hole for $J\ne 0$ and has a unique minimum at $r_0^3=J^2/(mMg)$. Since $U_J^{\prime \prime }(r_0)=3Mg/r_0$, small radial oscillations at fixed $J$ have squared angular frequency $3Mg/[(M+m)r_0]$.

A negative squared mass in a relativistic field spectrum signals instability of the background about which the field is expanded. The bosonic string mass spectrum contains such an unexcited state.

For an axisymmetric planar lamina of total mass $M$, let $I_z$ be its moment of inertia about the central axis normal to its plane. The perpendicular axis theorem and rotational symmetry give $I_x=I_y=I_z/2$ about central diameters. An in-plane tangent line a perpendicular distance $R$ from a parallel central diameter therefore has moment of inertia $I_{\tan }=I_z/2+M{R}^2$ by the parallel axis theorem. An axis normal to the lamina through the same rim point instead has $I_z+M{R}^2$; specifying the point without the axis direction is insufficient.

The visible angular distribution and laboratory energy in ${\tau }^{-}\to h^{-}{\nu }_{\tau }$ carry information about the parent's spin. For fixed hadronic invariant mass, $E_h=\gamma (E_h^{\ast }+\beta p_h^{\ast }\cos \vartheta )$. A spin-zero recoil has analyzing coefficient ${\alpha }_h=1$. For a spin-one recoil coupled through the V-A interaction, longitudinal and transverse partial rates have weights $m_{\tau }^2:2m_h^2$ and opposite slopes, giving ${\alpha }_h=(m_{\tau }^2-2m_h^2)/(m_{\tau }^2+2m_h^2)$. Thus the harder-helicity ordering can reverse when the recoil mass exceeds $m_{\tau }/\sqrt{2}$. A mean invariant mass without spin-dependent spectral weights cannot fix the sign for a mixed sample.

For a nearly massless incident tau neutrino and stationary target of mass $M_T$, the initial Mandelstam variables give $s=M_T^2+2M_T{E}_{\nu }$. At threshold the final tau lepton and the lightest allowed recoil system have no relative momentum, so $s=(m_{\tau }+M_X{)}^2$. This proves the formula. A free neutron recoil channel gives approximately $3.45\mathrm{GeV}$; an electron target has a different recoil and a threshold near $3.09\mathrm{TeV}$.

Temperature labels thermal equilibrium: two internally equilibrated bodies have the same temperature when contact permitting only heat exchange produces no net heat transfer. The zeroth law of thermodynamics makes the comparison transitive. If bodies A and B separately reach thermal equilibrium with the same thermometer without appreciably disturbing either body, they agree with each other. This is what lets a gas thermometer compare a metal block with water, despite their different internal energies and heat capacities. Insulation alone makes no such comparison: it can prevent heat transfer between bodies at different temperatures. ^[ref]

The zeroth law of thermodynamics establishes equality comparisons, but does not choose a numerical scale. Thermodynamic temperature supplies an absolute scale consistent with Carnot's theorem and entropy. For a simple compressible system at thermodynamic equilibrium with fixed particle number $N$ and composition,Here $U$ is internal energy, $S$ is entropy, and $V$ is volume; the second equality assumes the derivative can be inverted. Temperature is a thermodynamic conjugate variable to entropy and belongs to the intensive and extensive thermodynamic quantities as an intensive quantity. Its partial derivative tells us how much internal energy must change per unit change of entropy under the stated constraints. More general systems require their other mechanical and material variables to be held fixed too. ^[ref]

For an infinitesimal quantity $\delta Q>0$ of heat transferred from A to B, with no work or matter exchange, the equilibrium entropy changes giveFor ordinary positive temperatures, spontaneous heat transfer from $T_A>T_B$ increases total entropy; thermal equilibrium has $T_A=T_B$. This comparison uses changes in entropy, rather than the total internal energy contained in either body. A small hot object can have less internal energy than a large cool one. See entropy maximization under thermal contact for the equilibrium criterion and negative temperature for the special case of bounded energy spectra.

Express absolute thermodynamic temperature $T$ in kelvin, and a Celsius reading $t_C$ in degree Celsius:The offset is exact: $300\mathrm{K}$ corresponds to $26.85{}^{\circ }\mathrm{C}$, and $400\mathrm{K}$ to $126.85{}^{\circ }\mathrm{C}$. Their absolute temperature ratio is $4/3$; dividing their Celsius readings gives a different number. Thus gas thermometer pressure ratios and Carnot's theorem require thermodynamic temperature. The modern SI definition of the kelvin fixes the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$; a reference bath realizes or transfers the scale rather than defining that constant.^[ref]

Temperature is not universally the mean kinetic energy of a particle. In a classical, nonrelativistic ideal gas at thermal equilibrium, the equipartition theorem gives the mean translational kinetic energy relative to the gas's bulk motion:Here $m$ is particle mass, $\mathbf{v}$ its velocity, and $\mathbf{u}$ the mean flow velocity. In this regime, different particle masses at the same temperature have equal mean translational kinetic energies but different characteristic speeds. A moving gas can also carry bulk kinetic energy without being hotter in its rest frame. Internal molecular modes and quantum statistics require further care; the equipartition theorem has classical assumptions, whereas the equilibrium entropy definition applies more widely. ^[ref]

A sheet of mass per area $m$ and elastic-sheet tension $T$ backed by an identical outgoing fluid half-space presents the stated surface acoustic impedance to the other half-space, using upward sheet velocity $i\omega \eta =-V$. The fluid term is its normal acoustic impedance; the remaining term is structural. Infinite elastic-sheet tension suppresses every fixed nonzero spatial wavenumber, but not $k=0$, whereas infinite mass suppresses even spatially uniform motion. A massless untensioned sheet between identical fluids is transparent.

For steady upward bubbles in a stationary liquid, assume fixed bubble mass, quadratic drag force, negligible gas weight and inertia, and negligible wall force. Then $a\propto {\rho }_g^{-1/3}$. Hydrostatic mixture support and terminal drag force balance give $v^2\propto a(1-\alpha )$. The gas mass flux $J=\alpha {\rho }_{g}v$ is constant, hence $J^2\propto {\alpha }^2(1-\alpha ){\rho }_g^{5/3}$. Comparing at a reference point proves the displayed relation. The approximation requires a small gas/liquid mass density ratio and macroscopic variation scales; it need not hold near a gas-dominated endpoint or rapid acceleration region.

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]

Mass per unit length of a filament or yarn. The tex unit equals one gram per kilometre, or $10^{-6}\mathrm{kg}{\mathrm{m}}^{-1}$.

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.

The original study combines optical light curves, companion-star radial velocities, rotational broadening and a distance constraint to infer the compact object's mass. The result $14.8\pm 1.0$ solar masses is the estimate from this 2011 analysis, rather than a claim about the latest mass determination.

A geometrically thin, optically thick disk has a radially varying effective temperature. Relativistic transport and an atmosphere model convert its thermal spectrum into an inferred emitting area and inner radius. Black-hole mass, distance, inclination and spectral hardening are important external constraints. Selecting thermal-state observations with moderate luminosity limits contamination and the breakdown of the thin-disk approximation. Relating the fitted inner radius to the innermost stable circular orbit remains a physical modeling assumption.

The thermomechanical effect couples a temperature difference to a pressure difference across a superleak. For the closed-chamber model in the apparatus figure, consider neighboring equilibrium states with small velocities, negligible height difference and pure helium-4. Let ${\mu }_m$ be chemical potential per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The Gibbs free energy differential per unit mass givesThe superleak allows the superfluid to redistribute until the two sides have equal ${\mu }_m$. Setting $d{\mu }_m=0$ therefore givesThe sign matters: heating one side at initially equal pressure lowers its ${\mu }_m$; inflow raises its pressure until chemical balance is restored. The hotter side has the higher pressure. The integral follows the constant-${\mu }_m$ path; it need not be evaluated at fixed pressure when material properties vary appreciably.

For an original local example, stipulate constant $\rho =150\mathrm{kg}{\mathrm{m}}^{-3}$, $s=100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $T_h-T_c=0.020\mathrm{K}$. The predicted pressure difference is $300\mathrm{Pa}$. These are teaching assumptions, not helium measurements. This static thermodynamics relation predicts neither the required heater power nor the filling time: those require heat leakage, channel conductance and chamber heat capacity. An outlet would introduce kinetic energy and a flowing balance, so the static $300\mathrm{Pa}$ is not a predicted fountain height.