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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 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.

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.

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 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]

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.

In the separated-impact approximation, an upper ball of mass $m_1$ falling at speed $u$ meets a lower ball of mass $m_2$ after the latter rebounds elastically from a fixed floor. Their outgoing upward velocities are $(3m_2-m_1)u/(m_1+m_2)$ and $(m_2-3m_1)u/(m_1+m_2)$. These follow from conservation of momentum and reversal of relative velocity. The upper ball receives a fraction $r(3-r{)}^2/(1+r{)}^3$ of the initial kinetic energy, where $r=m_1/m_2$; subtracting this fraction from unity gives $(1-3r{)}^2/(1+r{)}^3$. Hence complete transfer occurs at $r=1/3$, whereas a very light upper ball rebounds to nine times the common drop height.

Here $x=m_W^2/m_t^2$, the bottom quark mass is neglected, and the W boson is on shell. The weak charged current gives the spin-averaged squared scattering amplitude $g_W^2|V_{tb}{|}^2(m_t^2-m_W^2)(m_t^2+2m_W^2)/(4m_W^2)$. The two-body Lorentz-invariant phase space has $p^{\ast }=(m_t^2-m_W^2)/(2m_t)$; multiplying by $p^{\ast }/(8\pi m_t^2)$ gives the displayed decay width. Its longitudinal-vector contribution explains the large-$m_t$ cubic growth, and its threshold factor makes the width vanish as $m_t\downarrow m_W$. The formula requires $m_t>m_W$ and receives finite-mass and radiative corrections.

Oberlin College's documented demonstration uses a track, two collision carts, two force probes set to the $50\mathrm{N}$ range, added $500\mathrm{g}$ bars, and a laptop with a Vernier interface. Preparation consists of entering the measurement calibration, securing the force probes to the carts, and zeroing them. Rubber stoppers serve for elastic collisions, clay for inelastic collisions, and hooks with string for pulling. Comparing the two force records tests the mutual interaction; adding mass to one cart tests whether unequal masses change the equality. ^[ref]

Vernier's force probe uses a bending beam with strain gauges; changing electrical resistance produces a voltage readout calibrated as force. Its manual recommends zeroing in the working orientation. If recalibration is needed, use an unloaded reading and a known suspended gravitational weight; for the $50\mathrm{N}$ range its example is $1\mathrm{kg}$ giving $9.8\mathrm{N}$. Re-zero after changing orientation. The manual quotes calibrated accuracy $\pm 0.2\mathrm{N}$, distinct from the $0.05\mathrm{N}$ resolution of that range, and warns that the interface's data-collection rate can limit collision measurements. ^[ref]

The following sampling and control choices are proposed additions to that documented experiment. Level the track and check unloaded motion in both directions for slope and friction. Use a soft bumper and one interface for simultaneous records; start before contact and stop after separation. A trial sampling rate of $1000{\mathrm{s}}^{-1}$, if supported, would put about 100 samples across a $0.10\mathrm{s}$ interaction. Increase the sampling rate and repeat until the peak and integrated impulse comparison are stable; a large sample count alone does not establish adequate response. Measure each total moving mass, including the force probe and added bars.

Convert both channels to the same right-positive force convention. Oppositely facing force probes can have matching raw signs because their local positive directions differ; check each sign by a gentle push of known direction before the run. Overlay $F_{B\mathrm{on}A}(t)$ and $-F_{A\mathrm{on}B}(t)$, and also examine their signed sum. Integrate both over the same contact interval to compare impulses. Repeat with added mass on one cart, exchange the probes between carts, and check unloaded zeros before and after. These controls help distinguish a physical mass change from a channel offset or gain error.

Agreement within measurement error supports the pair-interaction model over the resolved interval; it does not establish exact equality at arbitrarily short time scales. Misalignment, a loose mounting, sensor-beam dynamics, saturation, unequal timing, and zero drift can spoil the comparison. Track friction and cable pulls are additional external forces; a probe measuring bumper contact need not measure the entire net force on its cart. If a massive connector carries appreciable changing momentum, compare the full cart–connector collection rather than assuming its endpoint forces are an instantaneous cart-to-cart pair. The constructed pulse diagram above illustrates what to compare, and supplies no measurements from this apparatus.

Consider one chosen model at a fixed total energy, with two open scattering channels, no absorption and no omitted outgoing modes. Channel 1 leaves the target unchanged; channel 2 excites it. Let $v_1=v>0$ and $v_2=v/2$. For the same projectile mass, conservation of energy then requires a target excitation $\Delta =3m{v}^2/8$. These are hypothetical parameters, not a fit to a particular atom. Each incoming or outgoing mode is normalized to unit probability current; the outgoing modes will have the angular profiles specified below.

Choose the S-matrixThis phenomenological unitary matrix specifies the channel conversion; it is not derived here from an interaction potential energy. For incidence in channel 1, its first column gives the outgoing flux amplitudesTo recover ordinary wave coefficients relative to an incident coefficient one, undo the velocity normalization:Thus $|c_2{|}^2=32/25>1$ is allowed: channel 2 travels more slowly. Its outgoing probability current fraction is $16/25$, not $32/25$. Renormalizing the unweighted squares would give $9/41$ and $32/41$, the wrong probabilities. More generally the ordinary coefficient map is $A=V^{-1/2}S{V}^{1/2}$, with $V=\mathrm{diag}(v,v/2)$; it obeys $A^{†}VA=V$. The conserved quantity is the squared coefficient norm weighted by velocity, which represents probability current.

Now specify an original angular realization of these two outgoing modes, with scattering angle $\theta$ and no azimuth dependence:The first normalization follows from $2\pi {\int }_{-1}^1(1+u^2)du=16\pi /3$, where $u=\cos \theta$. These angular profiles define the modes of this model; they are not asserted to follow from Born's approximation or an atomic potential energy. The joint directional probability densities are

Suppose a detector accepts the cone $D:0\le \theta \le \pi /3$, with full azimuth and unit efficiency. Its angular fractions within the channels areMultiplying the angular fractions by the outgoing flux weights givesFor $1600$ independent incident events, the expected values are therefore $171$ and $256$ accepted detections, compared with $576$ and $1024$ over all directions. The remaining expected $1173$ events fall outside the cone. The chosen incident number makes the expectations integers; actual counts fluctuate. Conditional on a detection inside the cone, the channel probabilities are $171/427$ and $256/427$, rather than $9/25$ and $16/25$. Angular acceptance changes the observed channel proportions even with equal detector efficiencies.

The final atomic states are orthogonal, so ignoring the atomic readout adds the two channel intensities; it does not produce $|b_1{g}_1+b_2{g}_2{|}^2$. Coherent paths that reach the same final state can instead interfere, as the existing probability amplitude and quantum superposition examples demonstrate. The distinction is physical distinguishability of final states, not whether the detector operator chooses to display their labels. Flux normalization and angular integration make the scattering interpretation useful: they turn complex outgoing waves into conserved event weights and then into the statistics of a specified detector.

For interaction density $g{\varphi }_1{\varphi }_2$, the combinations ${\varphi }_{\pm }={\varphi }_1\pm {\varphi }_2$ diagonalize the equations into independent Klein-Gordon equations with the displayed mass squares. When $0<g<m^2$, all Fourier frequencies are real and nonzero. Initial data in one original field generally excite both normal combinations.

For a deliberately simplified neutrino oscillation calculation, take two neutrino flavors related to two mass eigenstates by a real rotation matrix with angle $\theta$. A quantum superposition acquires different phases during propagation; projecting back onto the initial neutrino flavor gives the probabilityHere $L$ is path length and $E$ is neutrino energy; the numerical coefficient incorporates unit conversion.

The paper states the corresponding conversion formula.^[ref]

The path dependence has a geometrical origin. In an original spherical-Earth model, let $R$ be the radius and $h$ a fixed production height, neglecting detector depth. Let $\Theta$ be the zenith distance of the direction in the sky from which the neutrino arrives, so $c=\cos \Theta =+1$ means downward propagation. The backward ray from detector to production obeys $R^2+L^2+2RLc=(R+h{)}^2$. Its positive solution isThus $L(+1)=h$ and $L(-1)=2R+h$. The calculation explains why opposite directions test very different baselines, rather than attributing every observed direction to the same travel distance. Actual production heights and the angle between the outgoing charged lepton and incoming neutrino smear that relationship.

For an original chosen example, set $\theta =\pi /4$ and $\Delta m^2=2.5\times 10^{-3}{\mathrm{eV}}^2$. At $E=1\mathrm{GeV}$, $L=20\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.9960$, while $L=500\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.000279$. The first minimum is at $L/E\simeq 494.7km/GeV$. These are stipulated inputs, not a fit to the 1998 data. A broad distribution of true $L/E$, together with energy and directional uncertainty, smooths the sharp minima.

A direct amplitude check makes the mechanism explicit. DefineThen $S_{11}={\cos }^2\theta +{\sin }^2\theta e^{-2i\phi }$. Taking its squared absolute value yields the displayed survival formula because $1-\cos (2\phi )=2{\sin }^2\phi$. Unitarity gives $|S_{11}{|}^2+|S_{21}{|}^2=1$: disappearance from one channel becomes conversion to the other within this closed model.

This illustration does not describe all solar neutrino propagation. Coherent forward scattering in matter changes the effective propagation operator, and three active neutrino flavors require a larger mixing description. Even in the simplified model, neutrino oscillation measures a squared-mass difference; it does not determine the absolute neutrino mass.

The two-fluid model of superfluidity gives the normal component of a superfluid and the superfluid independent velocities.

An elementary uniform-flow example makes the distinction operational. Write the mass flux per unit area and the ideal entropy flux asHere $s$ is specific entropy, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and ${\rho }_s,{\rho }_n$ are response mass densities. The mass density ${\rho }_s$ used here gives the three-dimensional phase stiffness coefficient ${\rho }_s(\hslash /m{)}^2$ when ${\mathbf{v}}_s=(\hslash /m)\nabla \theta$ is substituted into the kinetic energy density ${\rho }_s{v}_s^2/2$. For a stipulated ${\rho }_s=3\rho /4$, ${\rho }_n=\rho /4$ and $v_n=3u$ along a channel, zero net mass flux requires $v_s=-u$. Yet $j_S=3\rho su$ is positive. Opposing velocities can therefore transport entropy while transporting no net mass.

This example explains why excellent apparent heat transport need not be ordinary diffusion of heat through a stationary liquid. In the ideal zero-net-mass-flux limit, the heat flux associated with this entropy transport is $T{j}_S$. Viscous losses, quantum vortices and external heat exchange require additional terms. The model also supports second sound.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

A two-track invariant mass and a direction test answer different questions. In units $c=1$, assign each track the pion mass $m_{\pi }$ and reconstruct its momentum before magnetic deflection. With $p_{\pm }=|{\mathbf{p}}_{\pm }|$ and opening angle $\alpha$,This follows by expanding the squared sum of the two four-momenta. The angle $\theta$ between the reconstructed parent momentum $\mathbf{P}={\mathbf{p}}_{+}+{\mathbf{p}}_{-}$ and the neutral particle beam direction $\hat{\mathbf{n}}$ satisfiesA fully reconstructed two-body decay channel gives the parent mass and $\theta \simeq 0$, within experimental resolution. A three-body decay channel with an unseen neutrino generally misses both constraints; an accidental correct invariant mass alone is insufficient. Notice that the track opening angle $\alpha$ need not be small even when the parent direction angle $\theta$ vanishes.

The 1964 analysis selected $490<M_{\pi \pi }<510$ MeV and remeasured events with $\cos \theta >0.9995$, comparing adjacent mass bands.^[ref]

For an original dimensionless example, take $m_{\pi }=1$, $\hat{\mathbf{n}}=(0,0,1)$ and ${\mathbf{p}}_{\pm }=(\pm \sqrt{3},0,3)$. Each energy is $\sqrt{13}$; the opening angle is $60$ degrees since $\cos \alpha =(9-3)/(9+3)=1/2$. Yet $\mathbf{P}=(0,0,6)$ points exactly along the beam, and $M_{\pi \pi }^2=4(13)-36=16$. Thus a parent of mass $4$ passes both ideal tests. These chosen numbers illustrate four-momentum conservation, not detector performance.

Suppose a synthetic polymer population contains three chains of molar mass $10000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ for every one chain of molar mass $40000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Multiplying both chain counts by any common factor leaves the averages unchanged. Substitution into the existing number-average molar mass and weight-average molar mass definitions givesThe larger chains are only one quarter of the chain count but supply $40000/70000=4/7$ of the mass. Thus a weighted mean that emphasizes mass gives a larger result than one that weights each molecule equally. Total mass divided by total chain amount of substance gives $M_n$, while $M_w/N_{\mathrm{A}}$ is not the mean mass of a randomly chosen chain. Inferring chain counts from size-exclusion chromatography still requires the existing SEC detector response and SEC molar mass calibration assumptions.

For zero initial velocity, body masses $kM,(1-k)M$ and fuel masses $k{m}_0,(1-k)m_0$, the rocket equation gives the displayed final second-stage speed after burning the first-stage fuel, discarding its body without impulse, and burning the second-stage fuel. It assumes $M>0$ and $0\le k<1$. The speed exceeds the single-stage value when $0<k<1$ and fuel is positive. Its limit as $k\to 1^{-}$ concerns a surviving second stage of vanishing mass; at $k=1$ no such stage exists.