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In a reflection-symmetric astrophysical disk, the vertical mass coordinate measures mass per unit area above the midplane. It satisfies $dm=\rho dz$, so ${\partial }_m={\rho }^{-1}{\partial }_z$ inside the material. If the total surface density is $\Sigma$, the upper face is $m=\Sigma /2$; the normalized coordinate $\zeta =2m/\Sigma$ ranges from zero to one. Integrals weighted by mass density become ordinary integrals in $m$, which is especially useful for radiative diffusion and the density-weighted mean kinematic viscosity.

A horizontal bar of mass $M$ hangs from a spring of incremental stiffness ${\kappa }_0>0$, and two equal masses $m$ hang separately from the bar through springs of stiffness ${\kappa }_1>0$. The bar remains horizontal and all motion is vertical. For physical displacements $q=(q_1,q_2,q_3)$ from the loaded equilibrium, the kinetic energy and quadratic potential energy areGravity determines the static extensions; its linear terms cancel at equilibrium. Exchange of the lower masses splits the normal modes into the antisymmetric mode $(0,1,-1)$ with ${\omega }^2={\kappa }_1/m$, and two symmetric modes $(A,B,B)$. The latter obey $({\kappa }_0+2{\kappa }_1-M{\omega }^2)A-2{\kappa }_{1}B=0$ and $-{\kappa }_{1}A+({\kappa }_1-m{\omega }^2)B=0$. Their determinant gives the title quadratic in ${\omega }^2$, and $B/A={\kappa }_1/({\kappa }_1-m{\omega }^2)$. For $M=2,m=1,{\kappa }_0=4,{\kappa }_1=1$, their squared angular frequencies are $2\pm \sqrt{2}$, with displacement ratios $(1,1\mp \sqrt{2},1\mp \sqrt{2})$ respectively. The lower symmetric mode moves all masses in phase; the higher one moves the lower pair against the bar.

A physical sheet under uniform isotropic tension per unit length T and with mass per unit area $\rho$ has small transverse displacement satisfying ${\partial }_t^2\psi =c^2{\nabla }^2\psi$, where $c^2=T/\rho$. Fixed edges impose zero displacement. This mechanical membrane model neglects bending stiffness, large slopes and dissipation.

A dynamical mass estimate inferred from an equilibrium size and velocity scale. For a pressure-supported spheroid, $M\sim C{R}_e{\sigma }_{\mathrm{los}}^2/G$ depends on the structural coefficient, orbital anisotropy and observational aperture.

Motion can weigh a system whose mass is otherwise inaccessible. For a self-gravitating isolated system in virial equilibrium, let $K$ and $U$ denote equilibrium averages and use its gravitational radius $R_g=-G_N{M}^2/U$ and a mass-weighted mean squared speed relative to its centre of mass. Then $K=M⟨v^2⟩/2$ gives $M=R_g⟨v^2⟩/G_N$. If random velocities are isotropic and mean streaming has been removed, $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$, where ${\sigma }_{\mathrm{los}}$ is the one-dimensional line-of-sight velocity dispersion. Observations sample many objects at one epoch; replacing a time average by that sample requires a representative, approximately stationary population, beyond the trajectory identity itself.^[ref]

A practical measurement route uses an optical telescope to locate members and a spectrograph to measure Doppler shifts of identified spectral lines. Their shifts relative to the system's mean give radial velocities; angular separations plus a distance estimate constrain its size. Subtracting instrumental broadening and measurement variance, testing membership, and looking for coherent rotation or subgroups help control the inference. The projected luminous size need not equal $R_g$, and luminous tracers need not have the same distribution as all gravitating mass. Unknown velocity-anisotropy parameters, interlopers, mergers, external tidal forces and appreciable boundary stresses can bias a naive estimate. A large velocity dispersion alone does not establish virial equilibrium.

For fixed cloud mass $M$ and isothermal sound speed $c_s$, approximate the binding energy by $-3G{M}^2/(5r)$. The boundary-pressure virial theorem gives $A=3M{c}_s^2/(4\pi )$ and $B=3G{M}^2/(20\pi )$. The required external pressure peaks at $r_c=4B/(3A)$. A homologous radial displacement is restoring on the branch $d{p}_{\mathrm{ext}}/dr<0$, or $r>r_c$; the smaller-radius positive-pressure branch is unstable. This is a one-radius virial model, not the exact hydrostatic isothermal-sphere sequence.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

For a solid spherical cap of height $0\le h\le 2R$ cut from a sphere of radius $R$,Centre the sphere at the origin with the cap on the side $z\ge R-h$. Its cross-section at height $z$ is a circle of squared radius $R^2-z^2$. A volume integral in cylindrical coordinates givesThe formula includes caps larger than a hemisphere, when $h>R$, and the whole solid sphere, when $h=2R$. For the retained region $z\ge -d$ with $0<d<R$, set $h=R+d$ to obtain $V=\pi (2R^3/3+R^{2}d-d^3/3)$. Multiplying by uniform mass density gives its mass.

The volume divided by the positive number of particles $N$. The Van der Waals equation uses this convention for $v$, whereas specific volume divides by mass and molar volume divides by the amount of substance. With the Avogadro constant $N_A$, the molar volume is $v_m=N_{A}v$.

Buoyancy changes the effective gravitational mass; a nearby no-slip wall changes the hydrodynamic resistance matrix and therefore the diffusion coefficient. These are distinct corrections. Write the horizontal resistance as ${\zeta }_{\parallel }=6\pi \eta a/f$, with $0<f<1$ representing wall hindrance at a specified height. Applying the Stokes–Einstein relation with that resistance givesif one mistakenly uses the bulk resistance in the inversion. A chosen $f=0.8$ would overestimate $N_A$ by $25\%$; this is a sensitivity example, not a correction measured in Jean Baptiste Perrin's apparatus. Normal and parallel coefficients of the hydrodynamic resistance matrix need not coincide. Stokes settling velocity near a wall can likewise bias a radius inferred using an unbounded-fluid formula.

For a passive dilute particle suspension at uniform temperature, a change of mobility alone does not change the Boltzmann distribution. In the absence of additional wall forces the correct local vertical flux is $j_z=-D_z(z)[dn/dz+m_{b}gn/(k_{B}T)]$, so zero flux still gives the same sedimentation scale height. Walls can nevertheless invalidate that simple inference through adsorption, additional interaction energies, or sampling of inaccessible grain-centre positions. Optical position errors and exposure averaging also affect displacement readout; varying distance from the wall, observation lag and optical depth tests different possible errors.

Electrically driven conversion of water into hydrogen and oxygen. Reversible thermodynamics sets a minimum voltage, while activation, resistance and mass transport increase the operating requirement.

Choose approximate atomic weights $A_{\mathrm{r}}(\mathrm{H})=1.008$ and $A_{\mathrm{r}}(\mathrm{O})=15.999$, and use $M_{\mathrm{u}}\approx 1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The chemical formula ${\mathrm{H}}_2\mathrm{O}$ then gives water molar massThese are deliberately chosen rounded inputs, not exact atomic weights or a universal value for every water sample. Different isotopic abundances, notably those in heavy water, change the mean molar mass; chemical formula alone does not specify them. Adding atomic weights also neglects the tiny mass change associated with chemical bond energies, immaterial at this rounding precision.

For an illustrative pure water sample of mass $4.50375\mathrm{g}$, this convention givesThe count refers to water molecules. Their hydrogen chemical atom count is approximately $2N=3.01107038\times 10^{23}$, and their oxygen chemical atom count is approximately $N$. The mean mass per water molecule is $M/N_{\mathrm{A}}\approx 2.99146\times 10^{-26}\mathrm{kg}$. Extra displayed digits expose the arithmetic; they do not imply that the sample composition or mass was measured that precisely.

The same conversion makes stoichiometry a mass calculation. For the chemical equation $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$, complete conversion of $0.250000\mathrm{mol}$ of molecular hydrogen and $0.125000\mathrm{mol}$ of molecular oxygen gives $0.250000\mathrm{mol}$ of water. With the chosen approximate atomic weights, the reactant masses are $0.50400\mathrm{g}$ and $3.99975\mathrm{g}$, summing to $4.50375\mathrm{g}$. The $2:1:2$ ratio belongs to molecule counts and amount of substance, not to masses; this bookkeeping neglects the much smaller reaction energy contribution to mass.

For a finite jump in a one-dimensional potential energy, with constant mass and no delta potential, both the wavefunction and its first spatial derivative are continuous. Integrating the time-independent Schrödinger equation across a vanishing interval makes the jump in ${\psi }^{\prime }$ vanish, since the integral of $(V-E)\psi$ vanishes. A jump in $\psi$ itself would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no matching term in the equation. The second derivative can jump. This matching allows transmission and quantum tunnelling; a finite step is not a hard wall.

Use the leading nonrelativistic magnetic coupling for an electronic level with total angular momentum operator $\mathbf{J}$, and choose the $z$ axis along a uniform external magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$, $B\ge 0$. Write its magnetic quantum number as $m_J$, so $J_z|J,m_J⟩=\hslash m_J|J,m_J⟩$. There are $2J+1$ possibilities from $-J$ to $J$. With Landé g-factor $g_J$, the first-order shift isHere $e>0$ is the magnitude of the electron electric charge, $m_e$ its mass, $h$ the Planck constant, and ${\mu }_B$ the Bohr magneton. Both $g_J$ and $m_J$ are dimensionless; with $B$ measured in tesla and ${\mu }_B$ in joules per tesla, $\Delta E$ is in joules. The photon frequency shift $\Delta {\nu }_{ul}$ is in hertz, not angular frequency; indices $u,l$ denote upper and lower levels. NIST states this weak-field convention and its orbital/spin assumptions explicitly.^[ref]

The sign follows directly from magnetic dipole potential energy. Within this electronic $J$ multiplet, the projected magnetic dipole moment is $\mathbf{\mu }=-g_J{\mu }_B\mathbf{J}/\hslash$. Therefore $H_Z=-\mathbf{\mu }\cdot \mathbf{B}$ has the displayed positive coefficient. Subtracting the shifts of the two levels and using photon energy $h{\nu }_{ul}=E_u-E_l$ gives the transition formula. For the orbital case, ${\mu }_B/h=e/(4\pi m_e)$ gives the same leading frequency spacing as the classical charged oscillator. This explains the historical sensitivity to charge divided by mass; the quantum state derivation here is modern.

“Weak” compares interaction energies: mixing with neighbouring fine structure levels must be small, so $J$ remains a useful quantum number. The formula keeps the term linear in $B$ and neglects quadratic shifts. It also omits hyperfine structure; if that is resolved and controls the coupling, the appropriate total electronic-plus-nuclear angular momentum and splitting factor must replace $J,g_J$. Weak compared with fine structure does not automatically mean weak compared with hyperfine structure.

An instrument for determining mass or weight; a comparison balance matches the effect of an unknown load against known loads under comparable conditions.

The weight-average molar mass weights each population by its contribution to total mass:Here $n_i$ counts molecules with molar mass $M_i$. SEC light-scattering detection can infer this average for an eluting slice under its optical and concentration assumptions; overlapping species need not have the same individual molar mass.

Here gravitational weight means the force on a body due to gravity. In the Newtonian model, a body of mass $m$ in a uniform gravitational acceleration $\mathbf{g}$ has gravitational weight $\mathbf{W}=m\mathbf{g}$, with magnitude $W=mg$. Its direction follows $\mathbf{g}$; a buoyant force is a separate contribution to the body's net force. ^[ref].

Band position gives an apparent mass, not unique protein identity. Intensity depends on retention, epitope recognition and detection response; it does not establish folded state, enzyme activity or successful protein targeting. Missing bands can reflect missing targets or failed Western blot transfer or Western blot antibody detection. ^[ref]

The yeast cross-species complementation experiment consequently combines two logically different readouts: Western blot evidence about production of an antibody-recognized protein, and growth evidence about functional complementation. A detectable protein need not perform the biological task required for growth.

In the common SDS-PAGE version, protein denaturation and binding of an anionic surfactant reduce differences due to folded shape and intrinsic electric charge. An electric field drives the resulting protein complexes through the gel; smaller chains generally migrate farther. A lane of reference proteins supplies positions for estimating apparent mass. This estimate depends on the separation conditions: it is not a direct weighing of a molecule, and different proteins can migrate together.