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A chi-squared test compares the sum of squared standardized regression residuals with the appropriate chi-squared distribution. With both line parameters fitted, the data give ${\chi }^2=4.6537$ on $6-2=4$ degrees of freedom, with upper-tail P-value about $0.325$: the fitted exponential decay model describes the data adequately.

First consider fixing the alternative lifetime at its central value ${\tau }_a=0.478\mathrm{ps}$. Its slope is $m_a=-1/{\tau }_a=-2.09205{\mathrm{ps}}^{-1}$. One must refit the normalization rather than retain the earlier intercept. Minimization over the intercept gives $c_a=\bar{y}-m_a\bar{x}=7.29773$, andon $6-1=5$ degrees of freedom, with upper-tail P-value about $0.00130$. Thus that fixed central lifetime gives a substantially worse fit than $\hat{\tau }=0.4216\mathrm{ps}$. Equivalently, the increase after profiling over the intercept isThis fixed-value comparison does not yet include the uncertainty of the alternative measurement.

For the compatibility of independent lifetime measurements, include both errors. Under the approximately Gaussian lifetime-error model, equality of the two true lifetimes is tested byThis has one degree of freedom and P-value approximately $0.032$, equivalent to a $2.15$-standard deviation discrepancy. The two measurements are therefore in mild tension: equality is rejected at a $5\%$ significance threshold, but not at $1\%$. There is no unique yes-or-no compatibility verdict without choosing such a threshold. Treating the alternative central value as exact would overstate the disagreement between the experiments.

The lifetime that provides the best fit to these six rates remains $\boxed{\tau =0.4216\mathrm{ps}}$, not $0.478\mathrm{ps}$. The compatibility calculation above assumes the two experiments have independent uncertainties; a shared systematic error would require its covariance.

Classically, a particle with $0<E<V_0$ is completely reflected and never enters the raised region. In quantum mechanics the wavefunction penetrates the classically forbidden side as an evanescent wave. Define the tunnelling decay constantBoundedness at positive infinity excludes $e^{+\kappa x}$, soMatching the wavefunction and its derivative givesThe reflected probability amplitude has unit modulus, hence quantum reflection probability one. The decaying tail carries zero probability current: this semi-infinite potential step has no transmitted beam at infinity, unlike a finite potential barrier. The sketch shows an exact matched example rather than attempting to assign a single real value to a complex wavefunction.

Because the probability density decays as $|C{|}^2{e}^{-2\kappa x}$, the requested $1/e$ penetration distance isFor a proton, take $m_p=1.67262\times 10^{-27}\mathrm{kg}$, $V_0-E=10\mathrm{MeV}=1.60218\times 10^{-12}\mathrm{J}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe wavefunction amplitude decay length $1/\kappa =1.44\mathrm{fm}$ is twice this value. The nonrelativistic equation supplied in the paper is appropriate for this estimate since $10\mathrm{MeV}$ is small compared with the proton rest energy, approximately $938\mathrm{MeV}$. If the original PDF's unbarred $h$ is instead read literally as the Planck constant, the displayed differential equation gives $d_{\mathrm{literal}}=2\pi d=4.53\mathrm{fm}$. This is a valid calculation for that literal coefficient, but $0.720\mathrm{fm}$ is the physical estimate using the standard Schrödinger equation.

Noninteracting spectral line. Treat $\mathbf{k}$ as a wavevector, so physical crystal momentum is $\hslash \mathbf{k}$. Write ${\Omega }_{\mathbf{k}}={\epsilon }_{\mathbf{k}}/\hslash$; an energy zero at the chemical potential can be chosen throughout. The temporal factor of a noninteracting state is $e^{-i{\Omega }_{\mathbf{k}}t}$. With the Fourier transform convention $\tilde{\psi }(\omega )=\int dt{e}^{i\omega t}\psi (t)$,Thus its normalized frequency spectral function isOne should not square the delta distribution as if it were an ordinary finite-time transition amplitude. More precisely, the fermionic retarded Green function is $G_0^R(\mathbf{k},t)=-i\Theta (t)e^{-i{\Omega }_{\mathbf{k}}t}$. Its Fourier transform is $(\omega -{\Omega }_{\mathbf{k}}+i{0}^{+}{)}^{-1}$, whose negative imaginary part divided by $\pi$ is the same delta peak.

Interactions and lifetime. Interactions change the dispersion through the real part of the self-energy and permit decay into other excitations through its imaginary part. In the simplest quasiparticle-pole model, exponential decay of the retarded amplitude givesPerforming the causal integral,The corresponding Lorentzian quasiparticle spectral peak isIts total frequency integral is one. The exponential amplitude lifetime is $1/{\Gamma }_{\mathbf{k}}$; the probability lifetime is $1/(2{\Gamma }_{\mathbf{k}})$. Describing ${\Gamma }_{\mathbf{k}}$ as a scattering rate therefore requires the lifetime convention to be specified. The energy half-width is $\hslash {\Gamma }_{\mathbf{k}}$.

A general interacting electron is more accurately described by $G^R=Z_{\mathbf{k}}/(\omega -{\Omega }_{\mathbf{k}}^{\ast }+i{\Gamma }_{\mathbf{k}})+G_{\mathrm{inc}}^R$ near the pole. The coherent peak then integrates to the quasiparticle residue $Z_{\mathbf{k}}$ and an incoherent continuum carries the remaining weight. In an energy-unit self-energy convention,Linearizing the real denominator gives $Z=[1-{\hslash }^{-1}{\partial }_{\omega }\mathrm{Re}{\Sigma }^R{]}^{-1}$ and $\Gamma =-Z\mathrm{Im}{\Sigma }^R/\hslash$ at the pole. The requested unit-weight Lorentzian is the normalized quasiparticle model, not the most general exact interacting spectrum.

Peak position and width. At fixed $\mathbf{k}$, put $\gamma =\gamma (\mathbf{k})>0$ and $\Omega =\hslash k^2/(2m^{\ast })$. Differentiation givesIt is positive for $\omega <\Omega$ and negative for $\omega >\Omega$, so the only maximum is at $\Omega$, with height $1/(\pi \gamma )$. Setting $A$ to half this height gives $(\omega -\Omega {)}^2={\gamma }^2$. ThereforeIn energy units the maximum is ${\hslash }^2{k}^2/(2m^{\ast })$ and the half-width is $\hslash \gamma (\mathbf{k})$. Frequency dependence of the self-energy would require reconsidering both the shape and peak position.

Fermi gas and Fermi liquid. A noninteracting Fermi gas has a delta-function spectral line at its bare dispersion for every momentum. A normal three-dimensional Fermi liquid has a shifted coherent peak plus incoherent spectral weight; close to the Fermi surface its width scales as $(\epsilon -\mu {)}^2+(\pi k_{B}T{)}^2$. At zero temperature the peak becomes arbitrarily sharp at the Fermi surface, while it broadens farther away. Spectral functions include addition and removal weight; occupied-only photoemission contains an extra Fermi factor.

Photoemission measurement. A photon of known energy $h\nu$ ejects an electron. Measure its kinetic energy $E_{\mathrm{kin}}$ and emission direction. For binding energy $E_B\ge 0$ below the Fermi level and the appropriate calibrated work function $\Phi$, conservation of energy givesConservation of momentum parallel to a periodic surface giveswhere ${\mathbf{G}}_{\parallel }$ is a surface reciprocal-lattice vector. At suitable photon energies the photon momentum can be neglected; at higher energies it must be retained. The surface breaks perpendicular translation symmetry, so the perpendicular crystal momentum generally needs a final-state model. A two-dimensional band can be reconstructed directly from the parallel components.

Within the sudden approximation and electric-dipole approximation, photoemission intensity and spectral function are related byHere $E$ is measured relative to the chemical potential. Instrumental resolution convolves the product and inelastic scattering adds background. Thus ARPES maps the occupied electron removal spectral function, weighted by a matrix element; it does not measure the full unoccupied spectrum or an unweighted spectral function automatically. Bright ridges give dispersions, and intrinsic widths give decay information after instrumental effects are removed.

Reading the supplied image. Its bright ridges form a graphene-like cone. The crossing near $k_y=0$ is the Dirac point, not in general the Fermi level. In the paper's plotted energy coordinates its energy is roughly $E_D\sim 1.4$–$1.5\mathrm{eV}$. A velocity follows from a branch slope independently of the energy zero:For example, the lower branch moves by about $0.3\mathrm{eV}$ between $|k_y|\sim 0.03$ and $0.11{\mathring{\mathrm{A}}}^{-1}$. This gives $dE/dk\sim 3.8\mathrm{eV}\mathring{\mathrm{A}}$ andReading other points changes the estimate by tens of percent. The original exam image has unequal energy increments on nearly equally spaced labelled ticks, so it does not justify a precise fitted velocity. The value is an estimate of the cone slope extrapolated to the Fermi crossing, rather than a measured derivative exactly at a visibly identified Fermi cutoff.

To determine a Fermi energy, first calibrate the Fermi edge, then subtract the chosen reference energy. Neither a Fermi-level marker nor an explicit energy-zero convention is supplied in this crop. If its upper termination near $1.9\mathrm{eV}$ is intended to represent the Fermi cutoff, the conditional graph answer is $E_F\simeq 1.9\mathrm{eV}$ on that axis, or $E_F-E_D\sim 0.4\mathrm{eV}$ relative to the Dirac crossing. Identifying the crossing itself as the Fermi energy would incorrectly assume undoped graphene.

The research article cited by the exam identifies this as linearly polarized graphene ARPES and independently states $E_F-E_D\simeq 0.4\mathrm{eV}$ for the substrate-doped sample. That resolves the relative energy when this external calibration is admitted.^[ref] The crop alone does not uniquely fix an absolute Fermi energy; the source-supported relative value and the conditional $1.9\mathrm{eV}$ axis reading should be distinguished.

The persistent current in a superconducting loop experiment prepares a closed cold path, changes the externally linked magnetic flux, and follows the resulting superconducting persistent current through its magnetic field. The essential measurement starts after the inducing magnetic field has stopped changing.

Heike Kamerlingh Onnes's 1914 Communication 140c describes a lead coil closed by fused lead connections. Cooling in a magnetic field and removing the inducing magnet prepared the circulation. A magnetic compass sensed it without contacts; a separate copper coil, cooled in liquid air, supplied a measured balancing magnetic field. Reversing orientation and compensation reduced offsets. ^[ref]

The liquid-helium cryostat maintained low temperature, but refilling temporarily warmed the apparatus and changed the inferred electric current. A later run used a smaller initial electric current. Apparatus deformation and needle-zero drift limited comparison; cutting the closure still left a smaller magnetic signal, warning against assigning every signal to circulation around the whole loop. The report consequently gave an upper bound under a residual Ohm's law hypothesis. ^[ref]

For interpreting such a measurement, write a locally calibrated magnetic signal asHere $k$ is the coupling and gain, while $b$ contains the background magnetic field and readout offset. Only after establishing stable $k$ and correcting $b$ does a ratio of signals determine a ratio of electric currents. A gradual change in geometry, a drifting background magnetic field, or a change of temperature can imitate or conceal a decay. A pickup coil connected only to a voltage meter obeys Faraday's law of induction and detects a changing linked magnetic flux; its zero steady output cannot demonstrate that the underlying static magnetic field has disappeared. A magnetic compass can respond to that static magnetic field.

Opening the path, reversing the preparation, and repeating with smaller initial electric current are useful controls for separating whole-loop circulation, magnetic background and a superconducting critical current. Keep the temperature and external magnetic field stable during the interval used to infer electrical resistance. A sudden reduction near a superconducting critical current is not evidence for the constant-$R$ exponential decay model at smaller electric current. Persistence also does not replace the Field-cooled Meissner experiment: transport around a hole and magnetic expulsion from the material test different properties.

A superconducting persistent current is an electric current circulating in a closed path with superconductivity after the external driving electromotive force has ceased. Its associated magnetic field makes the circulation observable without inserting a measuring instrument into the path. This article concerns macroscopic circulation in superconductivity. A finite experiment establishes persistence over its observation window with a stated measurement error, rather than directly observing an infinite lifetime.

The ideal circuit connection is flux conservation in a superconducting loop. In a fixed loop with self-inductance $L$, a single circulating electric current $I$, and externally imposed linked magnetic flux ${\Phi }_{\mathrm{ext}}$, the lumped circuit approximation givesWhen the electrical resistance $R$ vanishes, integrating gives $LI+{\Phi }_{\mathrm{ext}}=\text{constant}$. Removing the external linked magnetic flux therefore changes the circulating electric current; holding it fixed afterwards preserves that electric current. This is a circuit approximation with a specified self-inductance, not a derivation of the microscopic state of superconductivity. For thin paths where the carrier contribution matters, fluxoid and flux quantization in a superconductor supplies the more careful description.

For constant $L>0$, constant effective electrical resistance $R\ge 0$, and no changing external linked magnetic flux, the RL circuit instead predictsThis exponential decay follows by separating $dI/I=-(R/L)dt$. A long relaxation time tests extremely small electrical resistance without trying to resolve an equally tiny steady electric potential difference. The stored magnetic energy is $U=L{I}^2/2$, so the same model gives $dU/dt=-R{I}^2$, exactly the Joule heating loss. The finite energy stored in the loop supplies the magnetic readout; the circulating electric current is not a continuously powered source of useful work.

Choose the following original phenomenological mathematical model solely to calculate a paired-pulse readout. It is not experimental data, a fit to the teaching experiment, or a simulation of the Hodgkin-Huxley model. The conditioning action potential defines time zero. Test pulses have fixed width $0.2\mathrm{ms}$; $I$ denotes their inward electric current amplitude. Impose an absolute neuronal refractory period $t_{\mathrm{abs}}=2\mathrm{ms}$. No finite test amplitude generates an event for $0<\Delta <t_{\mathrm{abs}}$. After that hold, stipulate a required amplitudewith $I_{\infty }=0.2\mathrm{nA}$, dimensionless $A=2$ and recovery time constant ${\tau }_r=3\mathrm{ms}$. Emit a test event exactly when $I\ge I_{\mathrm{req}}(\Delta )$, including equality. For this ideal event readout, assign events to pulse onset and assume the conditioning event is successfully evoked. There is no calculated membrane potential, spike waveform, response latency or stochastic success probability.

The exponential decay represents a combined recovery deficit; it is neither the Hodgkin-Huxley model's $h$ nor a measured potassium electrical conductance. One such curve cannot separately identify sodium recovery and persistent potassium electrical conductance. A constant pulse-width amplitude criterion also does not establish an equivalent rule for other pulse widths or voltage trajectories.

At the hold boundary, $I_{\mathrm{req}}(2\mathrm{ms})=0.2(1+2)=0.6\mathrm{nA}$. At $\Delta =4\mathrm{ms}$,at $\Delta =6\mathrm{ms}$,Thus a $0.35\mathrm{nA}$ test fails at $1\mathrm{ms}$ because of the imposed hold, fails at $4\mathrm{ms}$ because recovery is insufficient for that amplitude, and succeeds at $6\mathrm{ms}$. At $4\mathrm{ms}$, a $0.5\mathrm{nA}$ pulse would instead succeed: failure of the weaker pulse was relative refractoriness.

For $I>I_{\infty }$, solve $I\ge I_{\mathrm{req}}(\Delta )$ using the logarithm. The earliest allowed delay isFor $I=0.35\mathrm{nA}$ this is $2+3\ln (8/3)\approx 4.942488\mathrm{ms}$; for $I=0.5\mathrm{nA}$ it is $2+3\ln (4/3)\approx 2.863046\mathrm{ms}$. If $I\ge 0.6\mathrm{nA}$, the imposed $2\mathrm{ms}$ hold is the only delay restriction. If $I\le I_{\infty }$, no finite delay passes the rule: the required amplitude stays strictly above its asymptotic baseline. A finite experimental “fully recovered” time therefore also needs a tolerance and a readout definition.

Photobleaching is an irreversible loss of fluorescent capacity caused by illumination. It changes the number of active fluorophores over experimental time; fluorescence lifetime concerns the delay after an individual excitation. Reversible dark states and ordinary return to the ground state are different processes.

In a stipulated population-loss model, let $T$ denote elapsed illumination time and $t$ denote delay after a pulse. If active population declines as $e^{-k_{\mathrm{b}}T}$ while surviving fluorophores retain the same decay rates, then $I(t;T)=A_0{e}^{-k_{\mathrm{b}}T}{e}^{-t/\tau }$. Brightness falls while $\tau$ stays fixed. This example does not assert that all bleaching follows exponential decay: chemical changes and selective loss from a mixture can also change the observed decay shape. The existing FRET photobleaching controls illustrate why deliberate bleaching requires careful interpretation.

The mean lifetime of an unstable particle measured in its rest frame, using elapsed proper time. In the exponential decay model, survival to proper time $\tau$ has probability $e^{-\tau /{\tau }_0}$. For constant Lorentz factor $\gamma$, time dilation gives laboratory mean lifetime $\gamma {\tau }_0$. A mean lifetime is a statistical mean, rather than a fixed decay time.

A charge $q$ undergoing simple harmonic motion of angular frequency $\omega$ and slowly varying amplitude $A$ radiates the averaged powerIts mechanical energy $E=m{A}^2{\omega }^2/2$ therefore has exponential decay rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$ and half-life

Radioactive decay is the spontaneous transformation of an unstable atomic nucleus or nuclear state. Its timing can be unpredictable for an individual atomic nucleus while a large population has a reproducible average behavior. A constant nuclear decay constant $\lambda >0$ gives the familiar exponential decay law. The half-life describes how quickly the parent population falls, while radioactive activity describes its rate of transformations.^[ref]

This connection between randomness and a regular population law makes radioactive decay useful for measuring elapsed time and for understanding sources of radiogenic heating. It also distinguishes two questions in an experiment: what the source emits, and what an instrument records. Becquerel's uranium-radiation experiment used a photographic plate; the Curie ionization-current measurement measured an electrical consequence of radiation. Their readouts reveal different aspects of the same investigation.

A radiocarbon age estimates elapsed time from a measured carbon-14 ratio $R$, an assumed initial ratio $R_0$ and a specified nuclear decay constant $\lambda$. The displayed exponential decay inversion gives the age on that model's timescale. A conventional radiocarbon age uses a prescribed reference and decay convention; it is not generally a calendar age. Radiocarbon calibration accounts for changes in the atmospheric isotope ratio and maps the measurement to a calendar-age estimate.

An idealized radiochemical neutrino counting model separates production, survival, recovery and readout. Suppose a freshly cleared target produces atoms at a constant rate $R$ during exposure $T$. Each atom has nuclear decay constant $\lambda =\log 2/h$, where $h$ is its half-life. The expected surviving population obeysThe proof is to multiply the differential equation by $e^{\lambda t}$ and apply integration. The expression also follows by summing the exponential decay survival probabilities of atoms produced at every earlier time. Production does not imply eventual recovery: atoms can decay during exposure.

Let $\eta$ be the fraction recovered instantaneously at extraction, $d$ the subsequent delay before counting, $ϵ$ the probability of recording a decay, and $W$ the counting duration. The expected signal pulses areWith constant background rate $b$, the expected total is $\mu ={\mu }_s+bW$. During counting, the expected instantaneous pulse rate is $s(t)+b$, with $s(t)=ϵ\eta \lambda N(T)e^{-\lambda (d+t)}$. Equal time windows therefore lose signal by the factor $e^{-\lambda \Delta t}$ while a constant background stays unchanged. Fitting the time dependence helps distinguish radioactive decay from that background; it does not recover individual capture times.

For an original worked example choose $Rh=20$, $T=W=h$, $d=0$, $\eta =0.8$, $ϵ=0.5$ and $bh=0.2$. These are illustrative assumptions. Then $N(T)=10/\log 2$, andDoubling the exposure does not double the surviving inventory: $N(2h)/N(h)=3/2$. A delay of one half-life halves the signal. At $\lambda T\ll 1$, $N(T)\simeq RT$; at long exposure it saturates at $R/\lambda$. These limits explain why a large target and repeated extraction can matter more than simply waiting longer. The model neglects time-dependent production, finite extraction duration, residual atoms between runs and time-dependent backgrounds.

Use the complex amplitude convention $v(t)=\mathrm{Re}(\hat{V}{e}^{\mathrm{i}\omega t})$, where $\omega =2\pi f$ is the angular frequency and $f$ is the frequency in cycles per second. Substitution into the RC low-pass filter equation replaces the time derivative with multiplication by $\mathrm{i}\omega$, giving its continuous-time frequency responseHere $\varphi$ is the output-minus-input phase difference; a negative value means the output lags. Equivalently, the voltage divider formed from the electrical impedances $R$ and $1/(\mathrm{i}\omega C)$ gives the same ratio. For an input $A\cos (\omega t)$, the sustained output is $A|H|\cos (\omega t+\varphi )$ after the initial exponential decay has died away.

Define the cutoff frequency by a voltage amplitude ratio $1/\sqrt{2}$ relative to the low-frequency value:On a Bode plot, the voltage gain in decibels isAt the cutoff frequency it is $-3.0103\mathrm{dB}$, and far above that frequency it approaches a decrease of $20\mathrm{dB}$ per factor of ten in $f$. There is a smooth transition rather than an abrupt wall. The squared voltage ratio is one half at the corner; converting that statement to a power ratio requires the same real electrical resistance for the compared voltages. It does not mean that an unloaded capacitor dissipates half the source power.

For $R=2.2\mathrm{k}\Omega$ and $C=47\mathrm{nF}$, the cutoff frequency is $f_c\approx 1.5392\mathrm{kHz}$. A $1\mathrm{V}$ peak input at $0.1f_c$, $f_c$, and $10f_c$ produces peak output amplitudes of approximately $0.9950\mathrm{V}$, $0.7071\mathrm{V}$, and $0.09950\mathrm{V}$, with phase differences $-5.711^{\circ }$, $-45^{\circ }$, and $-84.289^{\circ }$. Thus a wanted oscillation at $0.1f_c$ is almost preserved while an unwanted one at $10f_c$ is reduced by about a factor of ten in voltage. The superposition principle lets these two responses be added when both oscillations occur in the same signal.

The step response describes a sudden persistent input change. Suppose the input becomes a constant $V_s$ at $t=0$, and the initial capacitor voltage is $v_0$. Finite electric current leaves that voltage continuous. Subtract $V_s$ from the RC low-pass filter equation: the remaining error satisfies $\tau de/dt+e=0$. Separating variables givesFor an initially uncharged capacitor and a $0$ to $3\mathrm{V}$ step in the constructed example,At one time constant, $t=103.4\mu \mathrm{s}$, the voltage is $1.8964\mathrm{V}$, about $63.2\%$ of its final value. The voltage error then continues its exponential decay. This is different from a pure delay: the output starts changing immediately, yet takes time to approach the target.

Define settling time relative to the total voltage change: require $|v_{\mathrm{out}}-V_s|\le ϵ|v_0-V_s|$ thereafter, with $0<ϵ<1$. The exponential decay is monotone, soFor $1\%$ settling, $t_{0.01}=4.6052\tau \approx 476.17\mu \mathrm{s}$; the $10\%$ to $90\%$ rise takes $\tau \ln 9\approx 227.19\mu \mathrm{s}$. Reducing the cutoff frequency by doubling $C$ doubles both times. This reciprocal relation is the useful connection between signal bandwidth and response speed: stronger suppression of rapid noise comes with slower tracking of wanted changes.

A freely moving unstable particle of speed $v$ travels a laboratory distance $R$ in time $R/v$, accumulating proper time $R/(\gamma v)$. For rest-frame half-life $T$ and exponential decay, its survival probability is the displayed expression. At least half survive exactly when $\gamma vT\ge R$. The minimum speed is thereforeThe relevant flight length is $\gamma vT$, rather than $\gamma cT$ for a particle whose speed is appreciably less than $c$.

Consider an original ideal model with sample volume $V_s$, bath volume $V_b$, and initial sample solute concentration $c_0$. Assume each volume is fixed, a freely permeant, nonbinding solute, no chemical reactions, complete mixing on each side, and equal concentrations at equilibrium. Each fresh bath initially contains none of this solute. The protein is perfectly retained in this model.

The first equilibration redistributes the initial amount of substance $c_0{V}_s$ through both compartments. Conservation givesThis denominator includes the sample itself. For the chosen $V_s=1\mathrm{mL}$ and $V_b=9\mathrm{mL}$, $c_1/c_0=1/10$, rather than $1/9$. Nine-tenths of the initial solute is in the bath and one-tenth remains in the sample; nothing has vanished during equilibration.

Discarding the equilibrated bath removes its solute. A second fresh bath starts with only $c_1{V}_s$ of solute in the entire system, so the same argument applies again. Induction gives the geometric progressionHere $n$ counts completed equilibrations with fresh baths: $n=1$ is the first bath, $n=2$ follows one replacement, and $n=3$ follows two replacements. The normalized sample concentrations are respectively $0.1$, $0.01$ and $0.001$. Leaving the first bath in place for longer cannot lower the ideal equilibrium value below $0.1$.

The same conservation argument handles a fresh bath already containing a permeant solute at fixed concentration $b$:Thus the sample approaches $b$, not necessarily zero. For example, if a hypothetical new buffer component starts at zero in the sample and at $b$ in every bath, its sample concentration reaches $0.9b$, $0.99b$ and $0.999b$ in the three equilibrations. Outward removal and inward replacement follow the same mathematics.

Equilibrium fractions do not specify elapsed time. An illustrative well-mixed linear ordinary differential equation model within one bath isHere $K$ is an effective transfer coefficient with units of volume per time. Adding the two equations conserves $V_s{c}_s+V_b{c}_b$; subtracting their normalized forms givesThe difference undergoes exponential decay. A bath replacement resets $c_b$, renewing the driving difference without making $c_s$ jump instantly. The plotted points are equilibrium endpoints, not a measured time course or a timetable for an arbitrary protein.

Real sample volume can change through osmosis; proteins can be lost through leakage or adsorption to the semipermeable membrane. Consequently, a measured decrease in protein concentration alone cannot distinguish dilution from loss of protein. Measuring recovered volume as well as protein concentration permits an estimate of recovered protein mass.^[ref]

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.

An RL circuit combines electrical resistance $R$ and self-inductance $L$. In the series model, the same electric current $I$ flows through both components. An ideal battery maintains terminal electric potential difference $V$; the drops in the direction of positive electric current obey Ohm's law and the inductor relation:This is Kirchhoff's voltage law with the inductor's terminal drop, rather than its oppositely signed induced electromotive force. For constant $R>0$, $L>0$, and $V$, the first-order linear differential equation has solutionThe time constant $\tau$ measures exponential decay of the difference from the final electric current. ^[ref].

For a stipulated example, take electrical resistance $R=2\Omega$, self-inductance $L=0.5\mathrm{H}$, terminal electric potential difference $V=6\mathrm{V}$, and initial electric current $I(0)=0$. ThenWith $t$ measured numerically in seconds, this is $I(t)=3(1-e^{-4t})$ in amperes. At one time constant,and the terminal electric potential difference drops areTheir sum is exactly $6\mathrm{V}$. At $t=0^{+}$, the inductor takes the entire terminal electric potential difference while the electric current is still zero. At steady electric current, the inductor's terminal drop is zero and the entire drop is across the electrical resistance.

The final magnetic energy stored by the inductor is $U_{\infty }=L{I}_{\infty }^2/2=2.25\mathrm{J}$. At one time constant, the electric current has reached about $63.21\%$ of its final value, but its magnetic energy has reached only about $39.96\%$:This difference follows from the square of electric current in the inductor's magnetic energy.

For release, take $t=0$ at the start of a separate experiment with initial electric current $3\mathrm{A}$. Remove the battery from the model while maintaining a closed path through the same electrical resistance and inductor. ThenNow $v_L=-v_R$: the induced electromotive force supports the continuing positive electric current as its magnitude decreases. This closed-loop model describes exponential decay, not opening the sole path and imposing an instantaneous jump to zero electric current. The inductor's magnetic energy and the power of Joule heating areTheir time constant is $\tau /2=0.125\mathrm{s}$. The total energy dissipated by Joule heating is exactly the initial magnetic energy:The second diagram in inductor shows this energy transfer. A finite closed path lets us predict both how long the electric current persists and how much energy becomes heat. ^[ref].

A second-order reaction has overall order of reaction two: a rate law may depend on $[\mathrm{A}{]}^2$, or on $[\mathrm{A}][\mathrm{B}]$ with partial orders of reaction one and one. Doubling all relevant reactant concentrations multiplies the reaction rate by four at fixed conditions. This experimentally determined dependence does not establish molecularity or a unique reaction mechanism. For an elementary reaction obeying the law of mass action, molecularity and overall order of reaction agree; a net chemical equation alone cannot establish that the conversion is elementary.^[ref]

For the single-concentration model, explicitly choose the disappearance rate conventionHere $k_{\mathrm{obs}}$ is the coefficient for loss of A, held fixed at the specified temperature and other conditions. Its subscript labels the chosen observed disappearance rate; it does not imply a pseudo-first-order reaction. Assume a uniformly mixed closed system, fixed volume, and no appreciable reverse or competing chemical reactions. The integrated rate law isThe reciprocal-concentration plot is a straight line with slope $k_{\mathrm{obs}}$ and intercept $1/[\mathrm{A}{]}_0$. A brief derivation uses the chain rule: $d(1/[\mathrm{A}])/dt=-[\mathrm{A}{]}^{-2}d[\mathrm{A}]/dt=k_{\mathrm{obs}}$, then applies the initial condition. This is a hyperbolic decay, rather than the exponential decay of a first-order reaction.^[ref]

As an original prescribed example, choose $[\mathrm{A}{]}_0=0.020\mathrm{mol}{\mathrm{L}}^{-1}$ and $k_{\mathrm{obs}}=0.50\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. At $t=100\mathrm{s}$, the reciprocal-concentration calculation is $50+50=100\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, giving $[\mathrm{A}]=0.010\mathrm{mol}{\mathrm{L}}^{-1}$. The initial disappearance rate is $0.50(0.020{)}^2=2.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$; after $100\mathrm{s}$ it is $5.0\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Halving the concentration quarters the disappearance rate.

Dimensional analysis checks every formula: $k_{\mathrm{obs}}[\mathrm{A}{]}^2$ has units $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $k_{\mathrm{obs}}t$ has units $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, and $k_{\mathrm{obs}}[\mathrm{A}{]}_{0}t$ is a dimensionless quantity. Changing concentration units changes the numerical rate constant: the same example is $[\mathrm{A}{]}_0=20\mathrm{mol}{\mathrm{m}}^{-3}$, $k_{\mathrm{obs}}=5.0\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$, giving $[\mathrm{A}](100\mathrm{s})=10\mathrm{mol}{\mathrm{m}}^{-3}$.

For an elementary reaction $2\mathrm{A}⟶\mathrm{P}$, instead define the stoichiometrically normalized reaction rate $r=k[\mathrm{A}{]}^2$. Two A entities disappear per P formed, soConsequently the reciprocal-concentration slope is $2k$, not $k$. In the prescribed example, the normalized rate constant would be $k=0.25\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$; initially $r=1.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ while the disappearance rate is twice that. With $[\mathrm{P}](0)=0$, stoichiometry gives $[\mathrm{P}](100\mathrm{s})=([\mathrm{A}{]}_0-[\mathrm{A}])/2=0.0050\mathrm{mol}{\mathrm{L}}^{-1}$. Both rate constants have the same dimensions, so a units check alone cannot catch their factor-of-two difference. The symbol $r$ denotes the same normalization as $v$ in reaction rate.

The half-life, meaning the time to reach half a specified starting concentration, follows by setting $[\mathrm{A}]=[\mathrm{A}{]}_0/2$:For $r=k[\mathrm{A}{]}^2$ in $2\mathrm{A}⟶\mathrm{P}$, this becomes $t_{1/2}=1/(2k[\mathrm{A}{]}_0)$. Its units are seconds because $k_{\mathrm{obs}}[\mathrm{A}{]}_0$ has units ${\mathrm{s}}^{-1}$. Doubling the starting concentration halves the first half-life; a quoted half-life needs its starting concentration. This differs from a first-order reaction, whose half-life is independent of starting concentration.^[ref]

Within the prescribed run, $[\mathrm{A}]$ reaches $0.010$, $0.0050$, and $0.0025\mathrm{mol}{\mathrm{L}}^{-1}$ at $100$, $300$, and $700\mathrm{s}$. The successive halving intervals are $100$, $200$, and $400\mathrm{s}$: each begins at a smaller concentration. More generally, reaching fraction $f$ of the initial concentration, with $0<f<1$, requires $t=(f^{-1}-1)/(k_{\mathrm{obs}}[\mathrm{A}{]}_0)$, so $90\%$ consumption takes nine initial half-lives, rather than a fixed sequence of equal waiting times.

A straight reciprocal-concentration plot tests this particular single-concentration rate equation. It does not uniquely identify a reaction mechanism, nor does curvature rule out every second-order reaction: two independently changing reactant concentrations require their own integrated rate law. At low concentration, a small measurement error $\delta [\mathrm{A}]$ becomes approximately $-\delta [\mathrm{A}]/[\mathrm{A}{]}^2$ in the reciprocal, magnifying late-time errors. Use the line as a diagnostic; a nonlinear regression of the original concentrations can retain their measurement scale and explicit error model.

Use the rotating frame and positive gyromagnetic ratio convention of free induction decay. Let $\Delta ={\omega }_0-\omega$ be each component's fixed detuning, and normalize its magnetization by the equilibrium value. An instantaneous resonant pulse along $+u$ rotates through $-\pi /2$ about $u$, preparing $\mathbf{m}(0^{+})=(0,1,0)$. A second pulse along $+v$ rotates through $-\pi$ about $v$. These ideal pulses require a drive much stronger than the relevant detunings and negligible relaxation during either pulse. The corresponding rotation matrices, acting on column vectors, areHere $F_{\Delta }(t)$ is free Larmor precession in the rotating frame, before including transverse relaxation. Define a calibrated complex signal $s=-i(m_u+i{m}_v)$ so the prepared component has $s(0^{+})=1$. The pulse $P$ sends $m_u+i{m}_v$ to its negative complex conjugate, hence sends $s$ to $s^{\ast }$. Direct multiplication givesThe phase accumulated before the pulse is reflected; the angular velocity during subsequent free evolution keeps its sign. At $2\tau$ every fixed detuning has zero residual phase. Equivalently, $P{F}_{\Delta }(t)P=F_{\Delta }(-t)$, so $F_{\Delta }(\tau )P{F}_{\Delta }(\tau )=P$. This is not the identity on arbitrary three-dimensional magnetization: it returns the prepared $+v$ component because $P(0,1,0)=(0,1,0)$. The two transverse pulse axes specified here also fix the receiver sign of the echo.

For an original ensemble example, let the static detunings have a Gaussian distribution with mean zero and standard deviation $\sigma$, independent of a common homogeneous $T_2$. Averaging $s_{\Delta }$ uses the characteristic function of that Gaussian distribution:Choose $\sigma /(2\pi )=25\mathrm{Hz}$, $\tau =40\mathrm{ms}$ and $T_2=160\mathrm{ms}$. Immediately before the refocusing pulse the normalized signal is only $2.08\times 10^{-9}$, yet $S(80\mathrm{ms})=e^{-1/2}=0.60653$. A one-pulse free induction decay at 80 ms would instead be about $3.11\times 10^{-35}$. These chosen numbers describe a calculation, not a measured trace. Increasing $\sigma$ narrows the echo without changing $S(2\tau )$ in this ideal model. Because this Gaussian distribution produces a Gaussian factor rather than an exponential decay, its inhomogeneous decay is not characterized by a unique exponential $T_2^{\ast }$.