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SEC molar mass calibration measures standards under the same column, solvent, and temperature conditions as the unknown. Over a suitable fractionation interval, a common empirical approximation iswhere $M_{\mathrm{ref}}$ specifies the molar mass unit and $b$ has inverse-volume units. Interpolation is conditional on the standards' size–molar mass relationship. SEC measures access associated with hydrodynamic size, so a branched polymer, an extended chain, and a compact protein need not share that relationship.

In this model the $80,000$, $40,000$, and $20,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ standards have hydrodynamic radii $2\sqrt{2}$, $2$, and $\sqrt{2}\mathrm{nm}$ and elute at $7$, $9$, and $11\mathrm{mL}$. An unknown at $8\mathrm{mL}$ would be assigned $M_{\mathrm{app}}=40,000\sqrt{2}\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ by this calibration. But a $40,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ species that expands from $2$ to $2\sqrt{2}\mathrm{nm}$ elutes at $7\mathrm{mL}$ and is assigned $80,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ without gaining any mass. The numbers demonstrate ambiguity, rather than predict a real unfolding transition.

Changes in conformational isomerism, protein folding, protein denaturation, and protein aggregation can therefore shift peaks. Association usually increases size; compaction can reduce size at unchanged molar mass. Adsorption creates a different ambiguity: delayed elution need not indicate a smaller species. A different solvent can change both polymer coil dimensions and interactions with the packing. The SEC protein conformation experiment tests these alternatives with additional detectors.

During continued compression of a first hydrostatic core, molecular hydrogen dissociation absorbs energy and softens the effective equation of state. This can trigger a second rapid contraction before a compact stellar core becomes pressure-supported. It is distinct from the initial nearly isothermal collapse and from later Kelvin-Helmholtz contraction at roughly fixed mass.

Second sound is an entropy and temperature wave. At negligible thermal expansion, it has almost zero net mass flux. Lev Davidovich Landau predicted two sound modes in 1941.^[ref]

The sedimentation coefficient $s$ is radial drift velocity divided by local centrifugal acceleration. It has units of time and combines particle properties with those of the surrounding liquid. In a dilute, deterministic mathematical model with negligible particle inertia, an effective displaced volume $V$, particle mass $m$, local liquid density ${\rho }_f$ and linear friction coefficient $f>0$ giveThe subtraction represents buoyancy. Equal mass does not imply equal $s$: changing shape changes drag force, and changing the medium changes both buoyancy and viscosity.^[ref]

For an isolated sphere of radius $b$ and uniform density ${\rho }_p$, Stokes's law supplies $f=6\pi \eta b$, where $\eta$ is dynamic viscosity, and $m=(4\pi /3)b^3{\rho }_p$. ThereforeThis original force-balance check predicts fourfold $s$ on doubling $b$ with everything else fixed, and zero drift when ${\rho }_p={\rho }_f$. It applies to dilute isolated spheres in Stokes flow, not arbitrary folded proteins or crowded suspensions. If $s$ and $\omega$ remain constant in a uniform medium, the ordinary differential equation $dr/dt=s{\omega }^{2}r$ has the exponential function solution $r(t)=r(0)e^{s{\omega }^{2}t}$. Arrival at an outer boundary therefore depends on starting position as well as $s$; a separation has no universal size cutoff.

Take identical rigid spheres of radius $a$, particle density ${\rho }_p$ and displaced liquid density ${\rho }_f$, at uniform absolute temperature $T$. Let $z$ increase upwards from a reference layer and write $\Delta \rho ={\rho }_p-{\rho }_f>0$. The buoyancy correction gives the effective gravitational massThe actual grain mass is ${\rho }_{p}V$; the displaced liquid subtracts ${\rho }_{f}V$ from the effective gravitational mass. In a dilute, noninteracting particle suspension, the Boltzmann distribution therefore givesHere $n$ is number density and $k_B$ is the Boltzmann constant. Equivalently, the dilute osmotic pressure $\Pi =n{k}_{B}T$ supports the grains above it: $d\Pi /dz=-n{m}_{b}g$. Dividing by $n$ and integrating yields the same exponential function. No dynamic viscosity appears in this equilibrium argument. If ${\rho }_p<{\rho }_f$, the population instead increases upwards; at equal densities this gravitational model predicts uniform $n$.

A microscope counts a finite optical slab, not an infinitesimal plane. For equal viewed area $A$ and slab thickness $\delta$, an ideal uniform detection window centred at $z$ givesThus equal optical volumes preserve the exponential function count ratio exactly, even for finite $\delta$, provided the windows lie wholly inside the accessible chamber and have the same detection response. Unequal depth response or overlapping a boundary can corrupt this inference. Repeated observations estimate the mean count; closely spaced visits need not be independent because grains remain in view.

For an original chosen model, set $a=0.250\mu \mathrm{m}$, ${\rho }_p=1200\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $T=298\mathrm{K}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. These are illustrative inputs, not Jean Baptiste Perrin's measurements. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesIf equal observation effort gives expected count $400$ in the reference layer, the expected counts $20$, $40$ and $60\mu \mathrm{m}$ higher are approximately $214.3$, $114.8$ and $61.49$. Fractions describe expected values over observations, not fractional grains. Using the actual grain mass instead of its buoyancy-corrected value would make the inferred Avogadro constant six times too small for these chosen densities.

Sediment resuspension transfers deposited bed particles into a particle suspension. It can supply particle buoyancy flux to a particle-laden gravity current if erosion or upward mixing exceeds settling losses. Ambient fluid without particles dilutes concentration without supplying particle mass.

In the type-I regime $M_L\simeq 0$, $M_R\gg m_D$, the light mass is $m_D^2/M_R$. Specifying only the right-handed scale does not fix a numerical light mass. A choice of Dirac mass is an extra assumption, not a result of matrix diagonalization.

The type-I seesaw mechanism combines a Dirac neutrino mass $m_D$ with a much larger right-handed-neutrino Majorana mass term $M$. Diagonalizing the mass matrix gives a light eigenvalue of order $m_D^2/M$ and a heavy eigenvalue of order $M$.

In four-dimensional $\mathcal{N}=2$ supersymmetric gauge theory, Seiberg–Witten theory determines exact low-energy quantities from the geometry of the vacuum parameter space. For rank one, the central charge in supersymmetry is proportional to $n_{e}a+n_m{a}_D$, with integral electric and magnetic charges and $a_D$ the dual scalar period. The BPS bound in supersymmetry fixes the mass of a short state by this charge combination. Its vanishing identifies loci where a magnetic monopole or dyon can become massless, even when the original electric description is strongly coupled.

A seismometer compares ground motion with the motion of an inertial test mass. A mechanical seismograph records their relative displacement. Its frequency response depends on the suspension and on which relative or absolute velocity enters the damping law.

An enzyme accelerates particular chemical reactions, with catalytic selectivity arising from recognition and from the relative ease of the subsequent chemistry. Strong binding alone does not guarantee fast conversion. Nor does catalysis change the chemical equilibrium for an unchanged net chemical reaction. These distinctions explain how enzymes can direct the flow of material without creating an extra source of free energy.^[ref]

For an original model of competing routes, let a substrate $S$ become either a desired reaction product $P$ or another reaction product $Q$:The labels represent hypothetical chemical species with the same elemental composition, not molecular structures. Assume fixed volume, irreversible first-order reactions, constant effective rate constants, and initially $100\mu \mathrm{mol}$ of $S$ with neither reaction product present. The model approximates a limited operating regime; real enzyme kinetics can include saturation and reverse reactions.

The reaction-rate equations give $d{n}_S/dt=-(k_P+k_Q)n_S$, $d{n}_P/dt=k_P{n}_S$ and $d{n}_Q/dt=k_Q{n}_S$, where $n$ denotes amount of substance. Consequently,With $k_P=0.08$ and $k_Q=0.02\mathrm{mi}{\mathrm{n}}^{-1}$, the eventual split is $80\mu \mathrm{mol}$ of $P$ and $20\mu \mathrm{mol}$ of $Q$. At ten minutes, the calculated amounts are approximately $36.8$, $50.6$ and $12.6\mu \mathrm{mol}$ for $S$, $P$ and $Q$. Selectively increasing $k_P$ fourfold changes the eventual desired fraction to $0.32/0.34=16/17\approx 94.1\%$. Increasing both rate constants fourfold instead preserves the $80\%$ split and only shortens the conversion time. Speed and selectivity answer different questions.

The bookkeeping matters independently of that model. In ideal alcoholic fermentation, complete conversion of $3\mathrm{mmol}$ of glucose yields $6\mathrm{mmol}$ each of ethanol and carbon dioxide. Of the initial $18\mathrm{mmol}$ of carbon chemical atoms, $12$ remain in ethanol and $6$ enter carbon dioxide. This is a chosen material-balance example using the net stoichiometry explained under Saccharomyces cerevisiae; it gives neither an observed yield nor the mechanism of the individual steps. Equal amounts of the two reaction products also do not mean equal masses.

A shell of fixed mass $M_s$ and radius $R$ has self-energy $-G{M}_s^2/(2R)$. Its inward generalized force opposing an increase of radius is $G{M}_s^2/(2R^2)$. Including this in a swept-gas shell threshold changes an interior-only factor $f(1-f)$ to $f(1-f/2)$ when the original total mass is held fixed.

Three identical like point charges, initially at rest at an equilateral triangle's vertices, remain equilateral as they move away from the fixed centroid. Each resultant force is radial by symmetry. If the centroid radius is $r$, its magnitude is $k_e{q}^2/(\sqrt{3}{r}^2)$. In a nonrelativistic model neglecting radiation, energy conservation gives the final speed $v=\sqrt{2k_e{q}^2/(ma)}$ for initial side $a$ and charge mass $m$.

In units $\hslash =1$, periodic-angle quantization gives canonical charge $Q_{\theta }=\ell \in \mathbb{Z}$. The mechanical charge is $Q_0=\ell +\theta T/(2\pi )$ and the rotating-kink mass gives the displayed leading semiclassical spectrum. The $T=1$ and $T=-1$ families are related by reversing the kink. Theta periodicity relabels the integer; fluctuation corrections are not excluded by this leading calculation.

Semiclassical quantization uses classical trajectories and actions to approximate quantum states in a regime where relevant actions are large compared with $\hslash$. Phase consistency on a periodic orbit leads to Bohr-Sommerfeld quantization. Soliton collective coordinates can be quantized this way, while fluctuations supply additional corrections to the leading classical mass.

Beta decay Q values depend on small differences between much larger atomic masses. A smooth semi-empirical mass formula can therefore approximate absolute masses reasonably while predicting the wrong sign for a MeV-scale decay threshold. Nuclear pairing explains offset even-even and odd-odd mass branches, but does not guarantee that every specific even-even neighbor lies below the intervening odd-odd nucleus. Measured masses or additional nuclear shell corrections are needed for quantitative threshold tests near a model's residual error scale.

Sensible heat is transferred heat associated with a temperature change within a single thermodynamic phase. By contrast, latent heat can change the phase fraction while the temperature stays fixed during equilibrium phase coexistence. Neither name denotes an additional stored substance: both describe how transferred heat participates in a process.

For fixed mass $m$ and composition, use the heat capacity appropriate to the heating constraint. Under the quasistatic process assumptions in heat capacity, with only pressure-volume work and unchanged bulk kinetic energy and potential energy, fixed-pressure heating givesHere $c_p$ is specific heat capacity at constant pressure; fixed-volume heating instead uses specific heat capacity at constant volume $c_V$. A cooling step has negative $Q_{\mathrm{sens}}$ with the sign convention in heat. A temperature difference has the same numerical value in units of the kelvin or degree Celsius. The integral follows by adding the small transfers $\delta Q=m{c}_p(T)dT$, rather than assuming a constant heat capacity.

As a stylized ice-to-water example for a fixed-composition closed system at fixed pressure $1\mathrm{atm}$, take $m=0.5\mathrm{kg}$, an initial temperature of $-10{}^{\circ }\mathrm{C}$, melting at $0{}^{\circ }\mathrm{C}$, and a final temperature of $20{}^{\circ }\mathrm{C}$. Assume constant specific heat capacity at constant pressure values $c_{\mathrm{ice}}=2.1\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $c_{\mathrm{water}}=4.18\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and latent heat of fusion $L_f=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. These are stipulated, rounded model values; ignore the container and heat losses, retain all the material, and assume only pressure-volume work under the quasistatic process conditions above. The three steps areOnly the first and third terms are sensible heat; the middle term is latent heat. The plateau begins after $10.5\mathrm{kJ}$ and ends after $177.5\mathrm{kJ}$ of total supplied heat. At $94\mathrm{kJ}$ total input, half the mass is liquid water and the temperature remains $0{}^{\circ }\mathrm{C}$. The equilibrium phase transition temperature depends on pressure; this example stipulates a single fixed pressure. ^[ref]

This split makes heating budgets easy to interpret: melting consumes $167/219.3\approx 76.2\%$ of the supplied heat in this model despite causing no temperature rise. Within either single thermodynamic phase, the heating curve slope is $dT/dQ=1/(m{c}_p)$; a larger heat capacity makes the slope shallower. A flat interval represents changing phase fraction, so it cannot be modeled by applying $Q=m{c}_p\Delta T$ across the phase transition. The complementary constructed example in latent heat tracks that phase fraction explicitly.

Identical linear springs of stiffness $k$ have equivalent stiffness $2k$ in parallel and $k/2$ in series. Parallel springs have equal displacement and add forces; series springs carry equal force and add extensions. A mass $m$ attached to an equivalent massless spring has angular frequency $\sqrt{k_{\mathrm{eff}}/m}$.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

A physicist.^[ref]