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Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.

The torsion constant, or torsional spring constant, is the positive restoring torque per unit twist angle of a complete suspension in its linear elasticity range:Its SI unit is $\mathrm{N}m/rad$. The minus sign describes restoration toward the untwisted orientation. The derivative specifies a local stiffness if the response is not globally linear; using one constant over a finite range additionally requires a straight calibration curve. Return to the same zero after unloading tests recoverable elastic deformation; repeat loading in both directions to check whether the response depends on its previous loading. Staying below permanent deformation is necessary, but does not alone prove proportionality between torque and twist.

For an ideal uniform circular wire of length $\ell$, radius $r$, and shear modulus $G_s$, the filament torsional stiffness is $C=G_s{J}_p$, with polar second moment of area $J_p=\pi r^4/2$. Uniform twist gives $\theta /\ell$ as the twist per unit length, soHere $U$ is stored elastic energy. Doubling the wire's radius multiplies the ideal torsion constant by $16$; doubling its length halves it. This explains why a long thin fiber is useful for weak torque measurement. The formula assumes linear elasticity with uniform material and circular section; it does not calibrate arbitrary ribbons, clamps, or a compound suspension. Within this ideal model, the equipartition theorem predicts an equilibrium thermal mean-square angular displacement $⟨{\theta }^2⟩=k_B\mathcal{T}/\kappa$, where $k_B$ is the Boltzmann constant and $\mathcal{T}$ the uniform absolute temperature. A softer suspension magnifies both the signal and its thermal fluctuations.

Static calibration supplies known torques without assuming the unknown interaction being measured. Apply a known perpendicular force $F$ at lever arm $a$, record the change from the unloaded twist, and useSeveral positive and negative settings test proportionality; fit the slope of applied torque against twist and retain a possible intercept. A nonzero intercept must not be silently absorbed into the slope. Convert degrees to radians before reporting $\mathrm{N}m/rad$. PASCO's procedure turns the Coulomb balance on its side and adds small known masses to generate a gravitational weight at the sphere's lever arm, returning the rod to its index mark by turning the dial. Its tabulated coefficient is force per degree: multiply by the lever arm and $180/\pi$ to obtain our torsion constant. ^[ref]

The figure's ideal static calibration uses $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$: an applied torque of $2.0\times 10^{-10}\mathrm{N}\mathrm{m}$ gives $0.020\mathrm{rad}$. Both signs lie on the same straight line. The plotted points are calculated examples, not experimental observations; they cannot demonstrate a real fiber's linearity.

Dynamic calibration instead measures the oscillation period $T_0$ with known total suspended moment of inertia $I$. For negligible damping, small free oscillations about a fixed reference obey simple harmonic motion:Include the rod, mirrors, and any attachments in $I$ unless their contributions are demonstrably negligible. Keep the applied torque absent or independent of angle. If an external torque has derivative $g_{\tau }=d{\tau }_{\mathrm{ext}}/d\theta$ at equilibrium, linearization gives effective stiffness $\kappa -g_{\tau }$: timing alone then measures that combination, and stability requires $\kappa -g_{\tau }>0$. A constant applied torque merely shifts equilibrium. Non-negligible damping changes the measured period; damped torsional oscillation gives a correction under a specified damping model. Agreement between static and dynamic calibrations is a useful check on the assumed moment of inertia, reference stability, and restoring law, rather than a reason to ignore their assumptions.

A harmonic oscillator in which a restoring torque is proportional to angular displacement; its moment of inertia and torsion constant set the natural angular frequency.

In a steady Keplerian accretion disk with $\Sigma ={\Sigma }_0{R}^{-2}$, $\nu ={\nu }_0{R}^2$, and tracer diffusivity $k=\zeta \nu$, the trace passive scalar concentration obeys the displayed constant-coefficient advection-diffusion equation in logarithmic radius. Contaminant mass is proportional to $CdY$, because $2\pi R\sigma dR=2\pi {\Sigma }_{0}CdY$. A radial impulse $C_0\delta (R-R_0)$ therefore has logarithmic amplitude $C_0/R_0$ and evolves by the Gaussian heat kernel centered at $\ln R_0-vt$. The full-line solution idealizes radii far from the disk boundaries.

A low-mean molecular weight hydrogen-helium envelope can add a large radius while adding a small fraction of the mass. The exoplanet transit photometry radius is the altitude where tangential optical depth becomes order unity, so temperature, opacity, and exoplanet clouds can move the measured radius above the deep material radius. Atmospheric loss and cooling change this contribution over time.

A square wire loop partly in a uniform field has linear drag $B^2{a}^{2}v/R$, giving velocity-relaxation time ${\tau }_t=mR/(B^2{a}^2)$. For uniform wire of mass density ${\rho }_m$ and resistivity ${\rho }_e$, ${\tau }_t=16{\rho }_m{\rho }_e/B^2$, independent of wire dimensions.

For a free molecule of mass $m$ in a large three-dimensional volume $V$, replacing the translational state sum by an integral gives the displayed partition function. For identical classical molecules the many-molecule canonical partition function is $q_{\mathrm{trans}}^N/N!$, not simply $q_{\mathrm{trans}}^N$. Differentiating its Helmholtz free energy at fixed volume gives the ideal-gas translational entropy.

JEOL's minimum-dose method focuses and corrects astigmatism away from the recording position before exposing it, reducing irradiation of the chosen field.^[ref] A downstream objective aperture cannot undo irradiation that has already reached the specimen: rejecting more outgoing electrons does not itself reduce specimen exposure.

For an original dose budget, define $D$ as incident electron count per specimen area, rather than absorbed energy per mass. A uniform beam of electric current magnitude $I$, illuminating area $A$ for time $T$, giveswhere $e$ is the elementary charge. Choosing $I=1\mathrm{pA}$, $T=1\mathrm{s}$ and $A=1\mu {\mathrm{m}}^2$ gives approximately $6.24\times 10^6$ incident electrons, or $6.24\mathrm{electrons}\mathrm{n}{\mathrm{m}}^{-2}$. This is an invented exposure calculation, not a safe-dose recommendation. Focusing and searching exposures must also enter the accumulated budget.

If detected counts follow a Poisson distribution, a mean count $\mu$ has standard deviation $\sqrt{\mu }$. Ignoring background, $100$ accepted counts therefore have $10\%$ relative counting uncertainty, while $25$ have $20\%$. Reducing accepted counts can reveal useful diffraction contrast while increasing counting uncertainty. It is useful to specify both contrast and count budget when discussing electron microscope resolution.

A proposed measurement calibration sequence records a beam-blocked detector background, an illuminated empty support region, and a specimen image with fixed voltage, exposure and detector gain. Test detector linearity with exposure time before comparing corrected intensities. Calibrate the image scale against a known spacing, as in microscope scale calibration, and independently calibrate electron diffraction spacing at its own lens settings; one calibration does not supply the other. Register compared regions, record objective aperture selection, specimen tilt and accumulated exposure, then repeat the first setting to test drift and irradiation changes. These are proposed controls, not undocumented steps attributed to JEOL's silicon example.

A transport equation carries a quantity along the flow of a velocity field $b$. For a smooth solution of $u_t+b\cdot \nabla u=0$, the value of $u$ is constant along each characteristic curve. With constant speed $c$, the chain rule gives $\frac{d}{dt}u(x_0+ct,t)=u_t+c{u}_x=0$, hence $u(x,t)=g(x-ct)$. This is exact translation: it preserves the shape rather than smoothing it, as the periodic heat and transport comparison shows.

For a two-body W boson decay into a charged lepton $\ell$ and a neutrino, neglect both daughter masses and use natural units $c=1$. Let $\Delta \varphi$ be the angle between the daughter transverse momenta. Identifying the missing transverse momentum with the neutrino's givesThis observable uses no $p_{z\nu }$. The 1983 UA1 experiment paper already used transverse mass as a lower bound on parent mass, before making additional decay-model assumptions for a mass estimate.^[ref]

The bound follows directly from the energy–momentum relation. For each massless daughter write its four-momentum, using rapidity $y$ and azimuth $\varphi$, asUsing the Minkowski inner product with positive time sign, $p^2=E^2-|\mathbf{p}{|}^2$, the two daughter four-momenta give the true parent invariant mass $M$:Since $\cosh u\ge 1$, $m_T\le M$, with equality at equal daughter rapidities when both transverse momenta are nonzero. Thus the unmeasured longitudinal information can enlarge the true invariant mass, but cannot lower it below this ideal transverse mass.

For an original example, stipulate $p_{T\ell }=p_{T\nu }=40\mathrm{GeV}$ and $\Delta \varphi =\pi$. Then $m_T=80\mathrm{GeV}$. Equal rapidities give $M=80\mathrm{GeV}$; choosing $y_{\ell }-y_{\nu }=1$ with exactly the same transverse readout gives $M=80\cosh (1/2)\simeq 90.210\mathrm{GeV}$. One measured transverse configuration therefore permits different true parent masses.

For two outgoing massless leptons, ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$ is transverse to $q=q_1+q_2$. This follows from the massless Dirac equation and anticommutation of the chirality matrix with the gamma matrices. Thus the $q^{\mu }{f}_{-}(q^2)$ term in a pseudoscalar-to-pseudoscalar form factor does not contribute to this scattering amplitude. A nonzero charged-lepton mass spoils exact transversality and restores a mass-suppressed scalar-form-factor contribution.

For a uniform solid cylinder of length $l$, radius $r$ and mass $m$, a thin circular slice at axial coordinate $z$ has transverse moment of inertia $(z^2+r^2/4)dm$ about a diametral axis through the cylinder's midpoint. Integrating from $-l/2$ to $l/2$ gives $I_{\perp ,G}=m{l}^2/12+m{r}^2/4$. The parallel axis theorem gives $I_{\perp ,\mathrm{end}}=m{l}^2/3+m{r}^2/4$ about a parallel axis through an end-face centre. The transverse and symmetry-axis moments of inertia coincide exactly when $l=\sqrt{3}r$.

A concrete example separates a trapped surface from a failure of coordinates. Use geometrized units $G=c=1$, metric signature $(-,+,+,+)$, and the regular ingoing Eddington-Finkelstein coordinates of a Schwarzschild black hole with mass $M>0$:For a round sphere, choose future null vector normalsThe spherical null expansion in ingoing coordinates follows directly from $A=4\pi r^2$: ${\theta }_k=2k(r)/r$. At $r=3M$, $({\theta }_{+},{\theta }_{-})=(1/(9M),-2/(3M))$; at $r=3M/2$, it is $(-2/(9M),-4/(3M))$. Both future light fronts shrink in the second case. At $r=2M$, ${\theta }_{+}=0$, so this sphere is marginally outer trapped surface, not a strict trapped surface. In the spherical model this boundary is an event horizon; that identification does not extend to arbitrary geometries.

Although $k_{-}$ is affine as written, $k_{+}$ generally obeys ${\nabla }_{k_{+}}{k}_{+}=(M/r^2)k_{+}$. These are valid initial normals for the sign test; use affine rescaling of a null normal before applying the affine Null Raychaudhuri equation. Regularity of the displayed metric tensor at $r=2M$ shows why a divergence in another chart would not establish null geodesic incompleteness.

On a compact trapped surface, both null expansions have a uniform negative margin. Sufficiently small smooth perturbations controlling the relevant metric tensor and surface derivatives preserve trapping. This conditional robustness does not prove that every nonspherical gravitational collapse creates a trapped surface.^[ref]

For the Higgs doublet vacuum $(0,v{)}^T/\sqrt{2}$, the neutral gauge-boson mass matrix in $(W^3,B)$ is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. Its zero eigenvector is proportional to $(g^{\prime },g)$, the photon; the perpendicular vector is the Z boson. The charged fields $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$ have the stated W boson mass. Thus $m_W=m_Z\cos {\theta }_W$, where ${\theta }_W$ is the Weinberg angle. The unbroken charge generator annihilates the vacuum and explains the zero eigenvalue.

The maximum primordial occupation of a collisionless fermion distribution limits the coarse-grained phase-space density attainable in a halo. Combining that bound with an observed velocity dispersion and mass density gives a lower candidate mass. This tests light-fermion halo models independently of whether the candidate is optically visible.^[ref]

A two-sided line plume whose axial velocity and density anomaly both have triangular transverse profiles. For half-width $b$, ${\int }_{-b}^{b}f(x/b{)}^{n}dx=2b/(n+1)$. These profile moments set the different coefficients of stored mass, momentum, and their fluxes. It differs from a wall plume because both edges entrain ambient fluid.

A penetrating instrument and associated tube used historically to enter and drain a cavity. Its historical use does not establish safety for modern ovarian-mass management.

An abnormal tissue mass. The term includes benign as well as malignant growths, so tumour and cancer are not always interchangeable.

Cloudiness measured through attenuation or scattering of light. actin filaments, bundles and aggregates can contribute differently, so the signal is not automatically proportional to polymer mass.