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For $A,ϵ>0$, consider $\tan \theta =-ϵ\theta /\sqrt{A^2-{\theta }^2}$ on $0<\theta <A$. On each negative branch of the tangent, the left side increases strictly, while the right side decreases strictly. A unique intersection exists precisely when the branch starts at $p_n=(n-1/2)\pi <A$, since the difference increases from negative infinity to a positive value or positive infinity. Consequently the number of trapped normal modes is $\lceil A/\pi -1/2\rceil$. The floor function expression $\lfloor A/\pi +1/2\rfloor$ agrees except at $A=p_n$, where it counts an inadmissible cutoff state with zero exterior exponential decay rate.

This deliberately synthetic differential equation model illustrates response direction and lag. Every plotted number is chosen for teaching; none is a biological measurement or a fitted human parameter. Let externally prescribed, dimensionless $g>0$ represent a glucose stimulus relative to a reference. Let $I$ and $H$ be normalized insulin and glucagon response variables, with independently chosen time constants ${\tau }_I=4$ and ${\tau }_H=6$ in arbitrary time units:Here $I_{\infty }$ rises and $H_{\infty }$ falls with $g$. Their targets add to one by construction; the evolving $I+H$ need not equal one because their time constants differ. Start with $g=1$ and $I=H=1/2$. Set $g=1.6$ from $t=10$ to $40$, $g=0.65$ from $t=40$ to $70$, and $g=1$ otherwise. Within a constant-input interval beginning at $t_0$, the exponential decay solution isEach response is a continuous function of time even when the imposed stimulus jumps. The chosen starting values stay between zero and one, since each solution is a weighted arithmetic mean of its interval's starting value and target.

For an illustrative downstream comparison, define release tendency $R=0.3+0.6H/(1+I)$ and uptake tendency $U=(0.25+0.5I)g$, in the same arbitrary flux units. At the reference target, $R=U=0.5$. At the high-stimulus targets, $R\approx 0.343$ and $U\approx 1.094$; at the low-stimulus targets, $R\approx 0.742$ and $U\approx 0.212$. These chosen functions encode greater uptake with insulin and greater release with glucagon, moderated by insulin; they are not established enzyme kinetics. The $g$ trace is imposed externally, so the figure does not demonstrate self-correction of blood glucose. A closed model would also need a conservation of mass equation, nutrient input, finite glycogen stores, and additional tissue interactions. The qualitative restoring loops are shown separately in the insulin circuit.

Consider a point mass of fixed mass $M>0$ released from rest in a stationary medium. Choose downward as positive, assume constant gravitational acceleration $g>0$, neglect buoyancy, and impose linear drag $F_d=-bv$ with constant linear friction coefficient $b>0$. Here $v$ is the signed downward velocity component; it equals speed during this descent. Newton's second law becomesThe terminal velocity balances the two forces, and the time constant $\tau$ controls how quickly that balance is approached:For the illustrative parameters $M=1\mathrm{kg}$, $b=2kg/s$ and $g=9.8m/{\mathrm{s}}^2$, the time constant is $0.5\mathrm{s}$ and the terminal velocity is $4.9m/s$ downward. At $t=1\mathrm{s}=2\tau$, the downward velocity is $4.9(1-e^{-2})m/s\simeq 4.23686m/s$ and the downward acceleration is $9.8e^{-2}m/{\mathrm{s}}^2\simeq 1.32629m/{\mathrm{s}}^2$. This is a toy linear drag model, not a claim that every $1\mathrm{kg}$ falling object has this resistance law.

For a proof outline, subtract the equilibrium value: $u=v-v_T$ obeys $du/dt=-u/\tau$. The exponential decay solution and the initial condition give $u=-v_T{e}^{-t/\tau }$; one derivative gives the acceleration. Every time constant multiplies the remaining gap $v_T-v$ by $e^{-1}$. The exact limit is reached only as $t\to \infty$. This makes the model useful for estimating a settling duration without mistaking vanishing acceleration for vanishing velocity.

Newton already analyzed vertical motion under uniform gravity and resistance proportional to velocity in Book II, Section I, Proposition III of the Principia. His geometric construction includes a greatest downward velocity, illustrating that a limiting rate can emerge from balancing a driving force and resistance. ^[ref]

For a negative Coulomb energy, remove the exponential decay and regular radial power to obtain a confluent-hypergeometric series. Its recurrence terminates only when $n=q+\ell +1$, with $q$ a nonnegative integer. Regularity at the origin and normalizability at infinity then give the hydrogenic energy levels. Without those physical boundary conditions the radial differential equation has other solutions.

With thermal diffusivity $\xi$, a Fourier mode of angular wavenumber $k$ decays at rate $\beta =\xi k^2$. To avoid confusing this $k$ with thermal conductivity, use $\alpha =\xi$ and angular wavenumber $K$ below: the same rate is $\beta =\alpha K^2$, with units ${\mathrm{s}}^{-1}$.

Consider an insulated bar $0\le x\le L$ with constant thermal diffusivity, no sources and no heat flux through either end or the sides. An original initial temperature profile and its exact evolution areThe cosine has zero temperature gradient at both ends, satisfying the insulated Neumann boundary conditions. Its spatial mean is zero, so conservation of energy leaves the mean temperature $\overline{T}$ unchanged while the nonuniform part disappears. A short derivation substitutes $T=\overline{T}+A(t)\cos (K_{n}x)$ into the heat equation: the second spatial derivative supplies $-K_n^2$, hence $A^{\prime }=-\alpha K_n^{2}A$. The exponential decay time is ${\tau }_n=1/(\alpha K_n^2)$; doubling angular wavenumber quarters this time. If cycles-per-length spatial frequency $f_x$ is used instead, $K=2\pi f_x$ and the rate is $4{\pi }^2\alpha f_x^2$.

For the two model materials in thermal diffusivity and $L=0.010\mathrm{m}$, ${\tau }_{1,A}\approx 5.07\mathrm{s}$ and ${\tau }_{1,B}\approx 20.3\mathrm{s}$. After $10\mathrm{s}$, the remaining amplitude fractions are approximately $0.139$ and $0.610$. A narrow hot-and-cold pattern disappears faster than a broad one because its temperature curvature, and hence the imbalance in conductive heat flux, is larger.

Consider two rigid bodies forming an isolated system, with fixed composition and constant whole-body heat capacity at constant volume values $C_1=1\mathrm{kJ}{\mathrm{K}}^{-1}$ and $C_2=3\mathrm{kJ}{\mathrm{K}}^{-1}$. Their initial temperatures are $T_{1,0}=300\mathrm{K}$ and $T_{2,0}=400\mathrm{K}$. Assume each body remains internally uniform, their contact stores negligible energy, no phase transition occurs and no work is done or extracted. These are stipulated model values, not material measurements.

Because $d{U}_i=C_{i}d{T}_i$, write $U_i=a_i+C_i{T}_i$ with arbitrary fixed reference constants $a_i$. The conserved relative internal energy isThis is an energy reference convention, not a claim about actual heat capacities down to zero temperature. At thermal equilibrium the two temperatures equal $T_f$, so the first law of thermodynamics givesThe larger heat capacity pulls the common temperature closer to its initial value: body 1 warms by $75\mathrm{K}$, while body 2 cools by $25\mathrm{K}$.

To predict a timescale, stipulate a constant contact conductance $G=1/R_{\mathrm{th}}=0.1\mathrm{kJ}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, the reciprocal of the contact's thermal resistance. The rate of heat transfer into body 1 is $G(T_2-T_1)$, givingAdding proves conservation of $E$. Subtracting gives exponential decay of the temperature difference, with time constantand thereforeExact equality is approached as $t\to \infty$ in this ideal model. Changing $G$ changes the rate but not $T_f$: equilibrium and the speed of reaching it are separate predictions. Real bodies can have internal temperature gradients, changing heat capacities and contact losses, so the plotted $7.5\mathrm{s}$ is an assumed model timescale.

A thermometer measures temperature through a physical response whose relation to temperature is known. Examples include the length change from thermal expansion, a change in electrical resistance, and the pressure of a gas thermometer. If the response function $X=f(T)$ is an injective function over the working range, its measured value determines $T=f^{-1}(X)$ through the inverse function. Thermometer calibration establishes that relation for the particular instrument and measurement conditions. Two matching readings are meaningful only after calibration to a common scale and with the probes sampling comparable equilibrium states. ^[ref]

For a contact thermometer, the sensing part must reach thermal equilibrium with the region being measured. That region may differ from the rest of the object: an immersed probe near a heater can read a local temperature while the surrounding water remains cooler. A stable reading can also be a steady state set by competing heat transfer paths to the sample and the room. Good contact, adequate immersion, mixing where appropriate, and checks for spatial differences help establish which temperature the reading represents. These are measurement controls inferred from the thermal equilibrium requirement, not a guarantee that every flat trace is equilibrium.

A thermometer also changes the sample it touches. Consider an original toy calculation with an initially uniform sample at $T_o=350\mathrm{K}$, a probe at $T_p=300\mathrm{K}$, and constant whole-body heat capacities $C_o=100\mathrm{J}{\mathrm{K}}^{-1}$ and $C_p=1\mathrm{J}{\mathrm{K}}^{-1}$. Suppose together they form an isolated system, with no work or phase transition and negligible contact energy. The first law of thermodynamics and thermal equilibrium giveThe final thermometer reading underestimates the sample's original temperature by about $0.495\mathrm{K}$, although it correctly measures their new common temperature. The disturbance becomes small when the participating probe heat capacity is small compared with the sample's. A large controlled heat bath is a different limit: it supplies or removes heat while approximately maintaining its own temperature.

Waiting time is a separate issue from disturbance. Under a toy Newton's law of cooling model for a uniform probe in a maintained heat bath, a constant contact conductance $G$ givesThe time constant $\tau$ controls this exponential decay. For stipulated values $T_b=350\mathrm{K}$, $T_p(0)=300\mathrm{K}$ and $\tau =5\mathrm{s}$, the probe reads about $331.61\mathrm{K}$ after one time constant and $347.51\mathrm{K}$ after three. These are calculated values, not an instrument specification or measured data. Real probes may have multiple relaxation times and additional heat transfer paths; choose an equilibration criterion from the required accuracy, rather than assuming that a fixed waiting time works universally.

A useful original model treats the thermopile absorber as one approximately uniform temperature $T_h(t)$ with heat capacity $C_{\mathrm{th}}$, connected to a fixed reference temperature $T_c$ by thermal conductance $G_{\mathrm{th}}$. Both parameters are positive constants here. Let $P_{\mathrm{abs}}(t)$ be an added absorbed power above the initial background. Linearize the outgoing heat transfer about that background. With $\Delta T=T_h-T_c$ measured from an initial state having $\Delta T=0$, conservation of energy gives the first-order linear differential equationThe left side is the rate of stored energy; the two terms on the right are added power and outgoing heat transfer. The effective thermal conductance can include several paths, rather than just heat conduction along the wires. The model assumes a stable reference, negligible electrical loading, a common absorber temperature, and changes small enough for constant parameters.

For a step from zero to constant $P_0$ at $t=0$, subtract the steady solution $\Delta T_{\infty }=P_0/G_{\mathrm{th}}$. The difference obeys exponential decay, so the initial condition fixesThe time constant has units $(\mathrm{J}{\mathrm{K}}^{-1})/(\mathrm{W}{\mathrm{K}}^{-1})=\mathrm{s}$. With the same constructed $N=20$ and $\alpha =40\mu \mathrm{V}{\mathrm{K}}^{-1}$, choose $C_{\mathrm{th}}=0.02\mathrm{J}{\mathrm{K}}^{-1}$, $G_{\mathrm{th}}=0.01\mathrm{W}{\mathrm{K}}^{-1}$ and $P_0=0.03\mathrm{W}$. ThenAt one time constant, the response is $1-e^{-1}\approx 0.632$ of its final value: $V_{\mathrm{oc}}(2\mathrm{s})\approx 1.52\mathrm{mV}$. Reaching $95\%$ requires $t=-\tau \ln (0.05)\approx 5.99\mathrm{s}$. A voltmeter read too soon therefore underestimates the steady signal even when its electrical calibration is perfect.

The steady ratio $V_{\infty }/P_0=N\alpha /G_{\mathrm{th}}$ is $0.080\mathrm{V}{\mathrm{W}}^{-1}$ in this constructed model. Halving thermal conductance doubles this ratio but also doubles the time constant if heat capacity stays fixed. Reducing heat capacity alone speeds the response without changing its steady value. Increasing $N$ gives a larger signal only at fixed individual temperature differences: adding legs can also change thermal conductance, heat capacity and electrical resistance. The algebra identifies coupled design choices, rather than predicting a particular manufactured detector.

A finite electrical resistance produces Thermal electrical noise. For nearly uniform absolute temperature $T$ and effective measurement signal bandwidth $B$, its root mean square open-circuit voltage isHere $k_{\mathrm{B}}$ is the Boltzmann constant. Thermal electrical noise in the sensing element is an important contribution, and the signal bandwidth enters its voltage variance.^[ref] If identical pairs have fixed electrical resistance $r$ and fixed $\Delta T$, then $R_{\mathrm{pile}}=Nr$: the signal grows as $N$ while this noise grows as $\sqrt{N}$. The resulting signal-to-noise ratio grows only as $\sqrt{N}$ under those assumptions. Reference temperature drift, environmental electromagnetic radiation, and meter noise remain separate limits; reducing signal bandwidth trades fast information for less integrated electrical noise.

The time constant $\tau >0$ in exponential decay is the time over which the decaying quantity is multiplied by $e^{-1}$. A response approaching equilibrium as $q(t)=q_{\infty }+(q(0)-q_{\infty })e^{-t/\tau }$ has the same time constant. For a resistor-inductor circuit it is $L/R$.

The tunnelling decay constant is the inverse length controlling the exponential decay of an evanescent wave in a constant potential barrier with $E<V_0$. The wavefunction amplitude decay length is $1/\kappa$; the squared-amplitude decay length is $1/(2\kappa )$.

For $t$ in a compact positive interval $[a,b]$, split the integration region at one. Near zero the smallest exponent gives the displayed bound, whose integral is $1/a^2$. Above one the largest exponent gives the bound, integrable by exponential decay. A single integrable function therefore dominates all parameter derivatives locally. Dominated differentiation under the integral sign gives ${\Gamma }^{\prime }(t)={\int }_0^{\infty }{e}^{-x}{x}^{t-1}\log xdx$. The same argument with higher powers of $|\log x|$ gives all higher derivatives.

An underdamped RLC response is a free series RLC circuit response with $0<R<2\sqrt{L/C}$, equivalently $0<\alpha <{\omega }_0$. Its characteristic roots are $-\alpha \pm \mathrm{i}{\omega }_d$, and its general electric charge isFor initial conditions $q(0)=q_0$, $i(0)=0$,In the dimensionless example ${\omega }_0=1$, $\alpha =0.5$, the ringing angular frequency is ${\omega }_d=\sqrt{3}/2\approx 0.866025$, below the driven electric current resonance at ${\omega }_0=1$. The electric charge first crosses zero at $t=4\pi /(3\sqrt{3})\approx 2.41840$ and reaches its first negative extremum at $t=2\pi /\sqrt{3}\approx 3.62760$, where $q/q_0=-e^{-\pi /\sqrt{3}}\approx -0.163034$.

The sinusoidal factor changes sign while its envelope decays as $e^{-\alpha t}$. The total energy follows the exact balance $E^{\prime }=-R{i}^2$, rather than a pure exponential decay at every instant; in particular, its rate of loss is zero whenever $i=0$. As $\alpha \to 0$, the displayed solution tends to the lossless harmonic oscillator $q=q_0\cos ({\omega }_{0}t)$. As $\alpha \to {\omega }_0$ from below, $\sin ({\omega }_{d}t)/{\omega }_d\to t$, recovering the critically damped RLC response continuously. The overdamped RLC response reaches the same limit through $\sinh (\beta t)/\beta \to t$.

Viscous damping means a force proportional to, and opposite to, velocity: $F_{\mathrm{d}}=-b\dot{x}$ with $b\ge 0$. Its coefficient has units $\mathrm{kg}{\mathrm{s}}^{-1}$, equivalently $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. The name describes a force law; eddy-current damping can approximate the same law without a viscous fluid. Constant $m$, $b$ and $k$ produce a linear ordinary differential equation, allowing superposition of responses. Kinetic friction with approximately constant magnitude and quadratic drag obey different laws, so their decay need not have the exponential decay or critical threshold derived here.

The waveguide cutoff frequency separates oscillatory axial propagation from exponential decay for a particular normal mode. For the vacuum-filled rectangular electromagnetic waveguide, the dispersion relation isFor a transverse electric mode of a rectangular waveguide, $m,n$ are nonnegative integers but not both zero. For a mode with $B_z=0$ and nonzero $E_z$, both indices must be positive. The lowest cutoff for $a>b$ is consequently ${\omega }_{10}=\pi c/a$, or ordinary frequency $f_c=c/(2a)$.

A square cross-section instead gives equal ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$ cutoffs. The absence of a transverse electromagnetic mode is specific to this hollow pipe, rather than all electromagnetic waveguides.^[ref]

For $\omega >{\omega }_{mn}$, choose real $\beta >0$ for propagation toward $+z$. For $0<\omega <{\omega }_{mn}$, a field decaying into a semi-infinite guide hasA finite section can contain both growing and decaying axial solutions set by its two ends; the single decaying solution is not a complete finite-section transmission calculation. At exact cutoff $\beta =0$, the two axial solutions become a constant and a linear function of $z$: there is no finite axial wavelength for a traveling sinusoid.

In the chosen $a=30\mathrm{mm}$ guide at $f=0.8f_c=3.99723277\mathrm{GHz}$, the ${\mathrm{TE}}_{10}$ branch has $\alpha =20\pi {\mathrm{m}}^{-1}$. Its amplitude decay length is $1/\alpha \simeq 15.9155\mathrm{mm}$; after $50\mathrm{mm}$, its normalized wave amplitude is $e^{-\pi }\simeq 0.0432139$. This is geometric evanescence even with ideal walls, not Joule heating. A single decaying branch has zero cycle-averaged axial Poynting vector: substituting $\beta =i\alpha$ makes $B_x$ a quarter-cycle out of phase with $E_y$. Coupling across a finite section requires the complete boundary-value solution.