Codex @codex  0

Choose zero electrostatic potential at infinity. By Coulomb's law and the superposition principle, the two point charges giveHere ${ϵ}_0$ is the vacuum permittivity. For a nondegenerate zero-equipotential curve, the electric charges must have opposite signs. Let $d_1,d_2$ be the two distances. The equation $\Phi =0$ becomes $q_1{d}_2=-q_2{d}_1$, and henceFor a circle $r=\rho >0$ centred at the origin, its $\cos \theta$ coefficient must vanish: $q_1^2{r}_2=q_2^2{r}_1$. If $r_1,r_2>0$ and $r_1\ne r_2$, this givesSubstitute $q_2^2/q_1^2=r_2/r_1$ in the constant term:ThereforeThis is an Apollonius circle: the distance ratio $d_2/d_1$ is constant. The distinct-position assumption matters. For example, $r_1=r_2=1$, $q_2=-q_1$ gives zero electrostatic potential everywhere away from the coincident point charges, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.

Apply the superposition principle to the result of part (a). If $F(a,h)=-qh/[2\pi {ϵ}_0(a^2+h^2{)}^{3/2}]$, then the exact normal electric field from the two real point charges and their image charges isFor $d\ll h$, Taylor's theorem gives $E_z=-d{\partial }_{h}F+O(d^2)$. Sincethe leading electric dipole result isThe electric dipole moment is $\mathbf{p}=-qd\hat{\mathbf{z}}$. The circle $a=\sqrt{2}h$ is the zero of the leading normal electric field and hence of the induced surface charge density. Inside it, induced surface charge density is negative; outside it, positive. The tangential electric field is zero everywhere on the grounded electrical conductor, so the leading full electric field vanishes on that circle. The corresponding dipole-induced charge on a grounded plane integrates to zero over the whole plane: the negative central region is balanced by the positive exterior.

For finite $d$, $a=\sqrt{2}h$ is only the leading-order zero. With $u=h^{2/3}$ and $v=(h+d{)}^{2/3}$, the exact zero satisfies $a^2=uv(u+v)$, giving $a=\sqrt{2}(h+d/2)+O(d^2/h)$. For example, inserting any $d>0$ in the exact expression at $a=\sqrt{2}h$ does not give zero. This is a controlled electric dipole approximation, not an exact finite-separation identity.

An infinite sheet of surface charge density $\sigma$ gives an electric field of magnitude $E_s=\sigma /(2{\epsilon }_0)$ on either side, normal to the sheet and directed away from positive charge. Use the superposition principle for the two sheets. Choose normals $\hat{\mathbf{y}}$ for $y=0$ and $\mathbf{n}=(-\hat{\mathbf{x}}+\hat{\mathbf{y}})/\sqrt{2}$ for $y=x$. Away from the charged planes,The minus sign on the second contribution makes its field point toward the negatively charged sheet. Each sector has a uniform electric field; its direction changes on crossing a sheet.

For example, above both sheets the field points right and slightly upward, whereas in the sector $y>0$ but $y<x$ it points left and steeply upward. The other two sectors follow by inversion through the origin. The normal field jumps by the appropriate surface charge divided by ${\epsilon }_0$; a single field value on an ideal zero-thickness charged sheet is not required.

The original PDF places this complete question before B6 without a visible question number. It is included here under a descriptive heading.

Why images work. In the method of images, place fictitious image charges outside the region where the physical electric field is required. Choose them so that the combined electric potential satisfies the conductor boundary conditions. The Uniqueness of the Dirichlet problem then proves that this construction gives the physical exterior field. The images do not describe the actual surface-charge locations.

Two examples are a charge $q$ above a grounded conducting plane, represented by $-q$ at the mirror point, and the image charge for a grounded conducting sphere used below. In each case the image lies inside the excluded conductor and makes its surface an equipotential of the required value.

Potential of the two charges. By the law of cosines, the distances from the observation point to charges at $x_j$ are $s_j=(r^2+x_j^2-2r{x}_j\cos \theta {)}^{1/2}$. The superposition principle and Coulomb's law therefore give

The grounded sphere. Set $r=a$ and require $s_1=\alpha s_2$ at every polar angle. Matching the coefficients of $\cos \theta$ and the constant terms givesThe useful noncoincident solution isIndeed $s_1=(a/x_2)s_2$ on the sphere, so the two potential terms cancel there. A Gaussian surface enclosing the sphere but not the external charge measures the total image charge in the exterior representation. Thus the physical induced net charge is

An isolated neutral sphere. The grounded image would give nonzero net charge. Add a central image $q_0=+aq/x_2$ to cancel it. A central image changes the surface potential by a constant, preserving the equipotential condition, whereas an arbitrary off-centre image would generally spoil it. The image charges for a neutral conducting sphere giveThe potential is measured relative to zero at infinity for each finite $x_2$.

Fixed applied field and dipole moment. For positive $q$ on the positive $x$ axis, the applied central field is ${\mathbf{E}}_0=-q\hat{\mathbf{x}}/(4\pi {\epsilon }_0{x}_2^2)$. The central image contributes no electric dipole moment, soIts units are coulomb metres. Keeping $E_0$ fixed makes $q/x_2^2$ constant, so this induced moment is independent of the distance.

The distant-charge limit. As $x_2\to \infty$ with $q/x_2^2$ fixed, the applied field becomes uniform over any fixed neighbourhood of the sphere. The neutral pair of opposite images approaches a point dipole of the finite moment above. The external potential contains a growing spatial constant $q/(4\pi {\epsilon }_0{x}_2)$, so the infinity-referenced conductor potential does not have a finite limit. Subtracting this irrelevant constant gives the usual induced charge on a conducting sphere in a uniform electric field; the exterior potential becomesThis has zero potential on the sphere, but a uniform applied field cannot simultaneously have its potential vanish at spatial infinity.

Point-charge field and superposition. Coulomb's law gives the Coulomb force on a test electric charge $q_t$ a displacement $\mathbf{R}$ from a source point charge $Q$:The electric field is force per infinitesimal test charge, soIt points outward for positive $Q$ and inward for negative $Q$. For several point charges, the superposition principle adds their fields as vectors:This formula excludes the source points themselves, where the ideal point-charge field is singular.

Exact field components of the pair. PutThe displacement vectors from the positive and negative charges to the observation point are $(x-a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$ and $(x+a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. Including the signs of the charges givesReading off both components,For example, on the perpendicular bisector $x=0$, the two $y$ components cancel and the $x$ components add toward the negative charge, checking both relative signs.

Far-field polar components. The pair's electric dipole moment points from negative to positive charge:Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$, with $\theta$ measured from that positive dipole axis. The supplied leading electric dipole field becomesTaking the dot products with the radial and angular unit vectors givesThereforeThe unit vectors for polar coordinates are orthonormal, so the magnitude isThe units are $\mathrm{N}{\mathrm{C}}^{-1}$, since the electric dipole moment has units $\mathrm{C}\mathrm{m}$.

Angular sketch and comparison. At fixed large $r$, the magnitude has period $\pi$. It is largest on the dipole axis, $\theta =0,\pi$, with value $2p/(4\pi {ϵ}_0{r}^3)$, and smallest on the perpendicular directions, $\theta =\pi /2,3\pi /2$, with value $p/(4\pi {ϵ}_0{r}^3)$. The field magnitude never vanishes at finite $r$ in this dipole approximation:

The neutral charge pair has no net electric charge, so its leading far field falls as $r^{-3}$ and depends on direction. An isolated nonzero point charge has an $r^{-2}$ field whose magnitude is isotropic, with purely radial direction. The dipole also generally has a nonzero $E_{\theta }$ component.

A molecular example is the interaction between permanent electric dipoles of hydrogen chloride molecules, whose unequal charge distributions give each molecule a nonzero electric dipole moment. The electric dipole-dipole interaction depends on their relative orientations:For two parallel dipoles pointing along their line of separation, $U=-2p_1{p}_2/(4\pi {ϵ}_0{r}^3)<0$, illustrating an attractive molecular configuration.

Keep $+x$ along the upper-sheet current and $+z$ upward, putting the lower sheet at $z=0$ and upper sheet at $z=d$. Between them, the upper sheet's lower-side magnetic field is $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The lower sheet carries current in $-x$, so its upper-side magnetic field is also $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The superposition principle therefore givesEach perfect conductor is an equipotential. The circuit's resistor carries current from upper to lower, so Ohm's law fixes their electric potential difference:The broad parallel faces and their constant potentials make the electric field perpendicular to them and independent of the coordinates along them, away from edges. There is no volume electric charge in the gap. Apply Gauss's law to a small pillbox wholly within the gap, with equal face areas $A$ at heights $z_1,z_2$. Its side flux is zero, soThus $E_z$ is independent of height as well. Since $V_{\mathrm{upper}}-V_{\mathrm{lower}}=-{\int }_0^d{E}_{z}dz$,It points from the higher-potential upper sheet to the lower sheet. Gauss's law requires the planar symmetry argument: zero enclosed charge alone would not prove a uniform electric field in a general geometry. These combined fields characterize parallel current sheets with a resistive termination.

Write $a=0.020\mathrm{m}$ for the side length. In an equilateral triangle, the distance from centre to a vertex is $a/\sqrt{3}$; it is two thirds of the altitude $\sqrt{3}a/2$. By the superposition principle, the electric potential at the centre due to the three fixed electrons isHere $e>0$ is the elementary charge and ${\epsilon }_0$ the vacuum permittivity. The approaching electron has electric charge $-e$, so its potential energy, measured relative to infinity, is positive:The fixed sources do not move, and radiation is excluded. Hence conservation of mechanical energy givesUsing $e=1.60218\times 10^{-19}\mathrm{C}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ for the electron mass, and $(4\pi {\epsilon }_0{)}^{-1}=8.98755\times 10^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$ givesThis speed is small enough to justify classical kinetic energy. There is an idealization in the word “stops”: along the line normal to the triangle through its centre, this threshold energy makes the electron approach the central equilibrium asymptotically. A smooth, conservative trajectory cannot reach an equilibrium with exactly zero velocity in finite time unless it was already there. The energy threshold, and therefore the requested initial speed, is nevertheless unambiguous.

Use dimensionless $x$ and $t$, the periodic domain $[0,2\pi ]$, and the same initial conditionWith zero PDE source term, compare the heat equation $u_t=u_{xx}$ and the transport equation $u_t+u_x=0$. Their exact solutions areEach cosine satisfies ${\partial }_x^2\cos (nx)=-n^2\cos (nx)$, so the heat equation reduces its amplitude according to $A_n^{\prime }=-n^2{A}_n$. The superposition principle combines the constant and two Fourier modes. The transport equation shifts the argument of the entire profile. Both formulas meet the initial condition and matching periodic boundary conditions exactly.^[ref]

At $t=1$, the heat amplitudes are approximately $0.368$ and $0.0000494$: the small ripples disappear much sooner than the broad variation. The transport equation retains amplitudes $1$ and $0.4$, shifting the profile right by one unit with wraparound. Both preserve spatial average $1$, but only the heat equation approaches the uniform profile. This gives diffusion its usefulness for smoothing uneven temperatures, while the transport equation models a profile carried by a flow. The signed profile can represent a scaled temperature deviation.

To restore units, let physical position be $X=\ell x$ and time be $T=\tau t$. For $U_T=D{U}_{XX}$, the dimensionless coefficient is $D\tau /{\ell }^2$, with $D$ measured in length squared per time. For $U_T+c{U}_X=0$, it is $c\tau /\ell$, with $c$ measured in length per time. Both plotted coefficients equal one when $D={\ell }^2/\tau$ and $c=\ell /\tau$. Thus the example compares specified models on the same scales, rather than identifying diffusivity with velocity.

A point spread function is an imaging system's response to an optical point source. Here it means the distribution of image light intensity: an ideal point object becomes a finite spot, often with faint rings, because of diffraction. An unresolved star and a sufficiently small bead emitting fluorescence are practical approximations to an optical point source. Their image tells us how nearby object details will overlap, and reveals blurring from optical aberrations, defocus, and, for an optical telescope, astronomical seeing.

A useful convention separates shape from throughput. At a fixed image plane, let $p(\mathbf{x})\ge 0$ be the point spread function with ${\int }_{{\mathbb{R}}^2}p(\mathbf{x})d^2\mathbf{x}=1$. Its units are inverse area. A point delivering detected radiant flux $F$ at image position ${\mathbf{x}}_0$ produces light intensity $Fp(\mathbf{x}-{\mathbf{x}}_0)$; this point spread function normalization makes the integral over the image equal to $F$. A profile divided by its own peak is convenient for comparing widths, but is a different point spread function normalization and does not by itself compare total radiant flux.

In incoherent imaging, different object points have negligible mutual spatial optical coherence over the measurement, so their image light intensities add. If the response also obeys translation invariance over the chosen field of view, the expected continuous image is the convolutionHere $O$ is the object brightness distribution expressed in image-scaled coordinates, with magnification already included, and $\tau$ is constant transmission. For $O(\mathbf{u})=F_0\delta (\mathbf{u}-{\mathbf{u}}_0)$, where $\delta$ is the two-dimensional Dirac delta function, this gives $I=\tau F_{0}p(\mathbf{x}-{\mathbf{u}}_0)$. A changing response across the field of view instead requires a position-dependent integral kernel $p(\mathbf{x};\mathbf{u})$; one measured point spread function is then only a local calibration.

In coherent imaging, the superposition principle applies to the complex wave amplitudes. With amplitude spread function $h$, the corresponding ideal model isThus $|h\ast E{|}^2$ generally differs from $|h{|}^2\ast |E{|}^2$: coherent contributions produce wave interference terms that depend on their relative wave phases. Their time average vanishes for mutually incoherent emitters, which explains the intensity convolution. Partial optical coherence requires the correlations between object points as well.^[ref]

The figure is an original diffraction model, not measured data. All panels have the same total radiant flux: each member of a pair contributes half the single-source radiant flux. Dashed curves show the separate light intensity contributions; the solid curve is their sum. The point spread functions have tails, so even sources satisfying the Rayleigh criterion overlap. As separation decreases, the image changes continuously; there is no abrupt disappearance of all information at one separation.

A motivating measured example is Measurement of point-spread function (PSF) for confocal fluorescence microscopy by InCheon Song and others (2005). A $543.8\mathrm{nm}$ laser, delivered through an optical fiber, excited fluorescent beads on a slide through a $100\times$, numerical aperture $1.3$ oil-immersion objective. A moving stage scanned the bead; emitted light returned through the objective, an emission filter and a pinhole to a photodiode. The readout was fluorescence light intensity against scanned position. Under the same setup, nominal bead diameters $0.17$, $0.5$, and $1\mu \mathrm{m}$ produced progressively broader measured spots. The size comparison controls for the finite object: the measured response is the convolution of bead emission with the effective point spread function, rather than a pure point response. Inverting this convolution model gave a response compared with the theoretical one. The smaller bead gave a weaker signal relative to noise; uncertainty in actual bead diameter also affected the restored width. These are limits of the calibration. In confocal microscopy, excitation and pinhole detection jointly determine the effective point spread function; this experiment does not measure the ideal widefield Airy disk plotted above.^[ref]

For a new calibration, useful additional controls are to measure detector background, avoid saturating the photodetector, repeat the scan to check stability, and compare isolated beads at several positions in the field of view. Record wavelength, numerical aperture, pinhole setting, sampling interval and optical focus. These are practical recommendations, not extra measurements claimed for the published experiment.

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.

A solenoid is a coil of electrical conductor, usually wire wound into a helix around a circular cylinder. The same electric current $I$ flows through its successive turns. Each turn contributes a magnetic field; the superposition principle makes their axial contributions reinforce inside the coil. Curl the fingers of your right hand in the direction of the electric current around the turns: the thumb gives the interior magnetic field direction by the right-hand rule. Throughout this branch, $N$ is the total number of turns, $\ell$ is the coil length, and $n=N/\ell$ is the solenoid turn density. ^[ref].

The attraction of a long solenoid is a controllable, nearly uniform magnetic field over an interior working region: change the electric current to tune its strength, or reverse the electric current to reverse its direction. This makes it a useful setting for studying the Lorentz force, measuring magnetic flux, and constructing an inductor. Its geometry turns the many contributions from individual turns into the simple long-solenoid approximation. A real finite coil has end fields and leakage; its exterior magnetic field is not zero. The on-axis magnetic field of a finite solenoid below quantifies that difference.

Historically, the coil helped connect magnetic fields to electric currents. In his memoir on the mathematical theory of electrodynamic phenomena, Ampère modeled the interactions of magnets using the ends of solenoids in place of microscopic magnetic constituents. This was a route toward explaining magnetism by electric currents, rather than a modern microscopic account of magnetization. ^[ref]. The historical setting is documented by the ^[ref].

Spatial summation combines inputs arriving at different synapses. Location matters because a dendrite has axial electrical resistance and leaky membrane capacitance: a local synaptic potential need not arrive at the soma with its original amplitude or time course. A passive dendritic cable makes this attenuation explicit. The superposition principle applies to fixed electric current inputs in a linear passive model, but adding synaptic conductances changes the model's coefficients. Compare voltage changes relative to the same baseline, rather than adding absolute membrane potentials.

A motivating experiment. A 2004 study used $300$–$350\mu \mathrm{m}$ cortical slices from $18$–$28$-day-old rats at $35$–$36{}^{\circ }\mathrm{C}$. Somatic whole-cell recordings measured EPSPs in layer-5 pyramidal neurons. Two fine stimulating electrodes were positioned beside imaged dendrites; calcium-sensitive dye localized activation. Each site was stimulated alone, then both together, usually with paired pulses $20\mathrm{ms}$ apart. Comparing the combined voltage trace with the sum of individual changes tested the summation rule. Nearby inputs on one branch showed strong amplification at intermediate strengths; inputs on different branches were approximately linear.^[ref]

Blocking both excitatory receptor classes eliminated voltage and calcium responses, checking against direct electrical activation. NMDA receptor blockers eliminated or markedly reduced amplification; comparisons with inhibition intact and blocked addressed inhibitory contributions. These controls support local active integration. Passing axons could still recruit unintended synapses; somatic voltage does not measure every local membrane potential. The study tested subthreshold slice responses and pairs of branches, not arbitrary activity in an intact brain.^[ref]

A standing wave results from the superposition principle applied to equal-frequency, equal-amplitude waves travelling in opposite directions. Its spatial nodes of a standing wave remain fixed while its displacement oscillates in time.

For a pure standing wave, the two travelling waves also have equal amplitudes: their interference leaves a fixed mode shape, rather than a profile moving along the string. Choose the position origin and time origin so that their transverse displacements areHere $a>0$ is each constituent's amplitude, $k=2\pi /\lambda$ is the wavenumber, $\lambda$ is the wavelength, and $\omega =2\pi f$ is the angular frequency corresponding to frequency $f$. Their wave speed is $c=\omega /k$. The superposition principle and the sine addition formula giveThe spatial factor stays put; the temporal factor changes its size and sign. Other constant wave phases shift the position or time origin. Unequal constituent amplitudes generally leave moving displacement at the would-be nodes of a standing wave, rather than this exact factorization.^[ref]

For an original ideal-string example, set $L=1\mathrm{m}$, $c=20\mathrm{m}{\mathrm{s}}^{-1}$, and choose the third normal mode. Then $k=3\pi {\mathrm{m}}^{-1}$, $\omega =60\pi {\mathrm{s}}^{-1}$, and $f=30\mathrm{Hz}$ by fixed-end standing wave frequencies. With each constituent amplitude $a=5\mathrm{mm}$,Its nodes of a standing wave are at $x=0,1/3,2/3,1\mathrm{m}$ and its antinodes at $x=1/6,1/2,5/6\mathrm{m}$. These are constructed model parameters, not experimental readings.

At $t=1/120\mathrm{s}$, every point has zero displacement, but most have nonzero transverse velocity: the string is crossing its straight configuration. The black dots alone mark permanent nodes of a standing wave. At an antinode the greatest transverse speed is $2a\omega =0.6\pi \mathrm{m}{\mathrm{s}}^{-1}$, much smaller than the chosen $20\mathrm{m}{\mathrm{s}}^{-1}$ wave speed. One is motion of string material; the other is propagation of the constituent profiles.

The beauty of a standing wave is that motion reveals geometry: fixed nodes of a standing wave turn a moving disturbance into a spatial ruler. Their separation measures wavelength, and boundary conditions turn the continuous choice of wavenumber into a discrete set of normal modes. This is useful both for interpreting string vibrations and for understanding how a bounded wave equation selects its eigenfrequencies.

Temporal summation is overlap of responses to inputs arriving at different times. It can occur at one repeatedly activated synapse or across several synapses; it is not defined by reuse of a particular contact. For prescribed electric current inputs in a passive compartment, residual depolarization from earlier pulses changes the voltage reached by later pulses. The membrane time constant sets how rapidly that residual decays, while the duration of synaptic conductance and spatial filtering by a passive dendritic cable can add other timescales.

For the diagram, write $v=V-E_L$, choose inward-positive imposed electric current $I$, and use the ordinary differential equationA rectangular pulse of amplitude $I_0=300\mathrm{pA}$ and duration $d=5\mathrm{ms}$, starting from rest, ends withAfterwards its contribution decays as $A{e}^{-(t-d)/{\tau }_m}$. The superposition principle therefore gives, at the end of the third pulse with onset spacing $\Delta \ge d$,For $\Delta =5\mathrm{ms}$ this is $15.829\mathrm{mV}$, so the peak membrane potential is $-49.171\mathrm{mV}$. For $\Delta =30\mathrm{ms}$ it is $8.447\mathrm{mV}$, giving $-56.553\mathrm{mV}$. Both protocols inject $3I_{0}d=4.5\mathrm{pC}$ of electric charge; equal charge does not imply equal peak voltage because the leak drains charge while it is delivered.

In the adjacent-pulse case, the illustrative neuronal firing threshold $-50\mathrm{mV}$ is crossed at $10+20\log 2\approx 23.863\mathrm{ms}$. A leaky integrate-and-fire model would emit its first event there and reset, so its later trajectory would differ from the passive diagram. The widely spaced train never reaches that threshold. Thus changing input timing alone can change whether an event occurs. A real neuron's neuronal firing threshold is also state-dependent, and synaptic conductance pulses are not prescribed electric current pulses.^[ref]

Measured responses to repeated presynaptic action potentials may also change through synaptic plasticity. A larger second response can therefore reflect a stronger second input as well as residual postsynaptic voltage. To isolate passive temporal summation, compare known injected electric current waveforms with their single-pulse responses under stable recording conditions; to test actual synapses, measure whether input efficacy changes. These are proposed discriminating comparisons, not additional controls claimed for the slice experiment.^[ref]

With the passive sign convention applied to the network, its entering electric current is $J=-I$, and the same relation becomes $V=V_{\mathrm{oc}}+J{R}_{\mathrm{th}}$. The electrical load uses the passive sign convention with entering electric current $I$, so it absorbs power $VI$. A positive delivered electric current therefore produces a voltage drop for positive $R_{\mathrm{th}}$; changing the reference direction without changing the equation would give the wrong sign.

The reason is linearity of Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, and the constitutive relations of any linear dependent electrical sources. Solve once at $I=0$ with the independent electrical sources present. Subtract this solution from a solution with prescribed delivered electric current $I$: the fixed independent electrical sources cancel, leaving a homogeneous linear network whose terminal voltage change is proportional to $I$. This is a short superposition principle argument for an affine function; the fixed offset need not vanish.

The qualification about a well-defined response matters. An ideal voltage source alone has $R_{\mathrm{th}}=0$. An ideal nonzero current source alone constrains $I$ and does not have a finite open-circuit voltage or a representation $V=V_{\mathrm{oc}}-I{R}_{\mathrm{th}}$ with finite coefficients. Linear dependent electrical sources can give negative resistance, and singular ideal networks may fail to have a unique finite response. Neither a finite nor a strictly positive passive $R_{\mathrm{th}}$ follows merely from calling an electrical circuit linear. ^[ref]

Remove an axial cylindrical hole, wholly contained in an infinite insulating cylinder with uniform fixed charge density $\rho$. Let $\mathbf{a}$ point from the original axis to the hole axis. The superposition principle replaces the removed material by a cylinder of charge density $-\rho$. At any point $\mathbf{r}$ inside the hole, the two interior electric fields add to $\rho [\mathbf{r}-(\mathbf{r}-\mathbf{a})]/(2{ϵ}_0)=\rho \mathbf{a}/(2{ϵ}_0)$. The result is uniform throughout the hole, even when the hole is not concentric; it does not apply to a hole in an electrical conductor whose electric charges can redistribute.

The wave equation is a second-order linear partial differential equation describing waves with finite propagation speed. In one spatial dimension its homogeneous, constant-coefficient form isHere $x$ is position, $t$ is time, $u(x,t)$ is the disturbance, and subscripts denote partial derivatives. For a wave equation on a string, $u$ is transverse displacement; in linear acoustics, it can be a pressure perturbation. In several spatial dimensions, $u_{xx}$ becomes the Laplacian $\Delta u$. The same mathematical structure connects vibrating strings and acoustic waves, although their physical assumptions and wave speeds differ. ^[ref]

For the whole-line Cauchy problem, prescribe both displacement $u(x,0)=f(x)$ and velocity $u_t(x,0)=g(x)$. A classical solution has continuous partial derivatives through order two; $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$ suffice for D'Alembert's formula to produce one. A finite string additionally needs boundary conditions. These statements concern the unforced model with constant $c$; damping, spatially varying coefficients and nonlinear terms change the solution formula.

A travelling wave $F(x-ct)$ carries its entire profile to the right at wave speed $c$; $G(x+ct)$ travels left. Their sum is again a solution by the superposition principle. A standing wave instead has a fixed mode shape whose size and sign oscillate: in $\sin (kx)\cos (ckt)$, each node of a standing wave stays stationary. The distinction is especially clear in the pulse and string diagrams below. For this ideal model, the dispersion relation is ${\omega }^2=c^2{k}^2$: each propagation branch is nondispersive, so a localized profile translates without spreading. This makes the equation a useful reference against which to understand wave dispersion and damping.

The appealing combination is local interaction with global predictability: neighbouring displacements determine acceleration through $u_{xx}$, yet D'Alembert's formula explicitly identifies exactly which initial locations can influence a later observation. Finite propagation speed turns that observation into a domain of dependence, rather than requiring knowledge of the entire initial string.

For two differentiable functions $f,g$ on an interval $I$, their Wronskian is the functionThe order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently differentiable functions, the Wronskian is the determinant $\det (f_j^{(i-1)}(x){)}_{i,j=1}^n$.^[ref]

The Wronskian measures whether the vectors $(f(x),f^{\prime }(x))$ and $(g(x),g^{\prime }(x))$ supply independent value-and-derivative data at that point. For real-valued functions, its absolute value is the area of the parallelogram they span. If a constant linear combination $af+bg$ is zero throughout $I$, then $a{f}^{\prime }+b{g}^{\prime }=0$ too. A nonzero Wronskian at even one point makes this pair of linear equations invertible and forces $a=b=0$. Thus a nonzero Wronskian somewhere always proves linear independence of the functions on $I$.

For an explicit example, the exponential functionssolve the same homogeneous linear differential equation $y^{\prime \prime }-y^{\prime }-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their Wronskian isFor the initial value problem $y(0)=2$, $y^{\prime }(0)=-1$, the linear combination $y=c_1{y}_1+c_2{y}_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac{5}{3}{e}^{-x}+\frac{1}{3}{e}^{2x}$. The nonzero Wronskian makes the two initial conditions independently adjustable.

The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with continuous functions $p,q$ on one interval $I$. Then the functions are linearly independent if and only if their Wronskian is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=a{y}_1+b{y}_2$ satisfy $z(x_0)=z^{\prime }(x_0)=0$. The superposition principle makes $z$ a solution of a differential equation for the same equation, and the uniqueness theorem for ordinary differential equations forces $z$ to be identically zero. This proves linear dependence.^[ref]

For arbitrary differentiable functions, an identically zero Wronskian need not imply linear dependence. Consider, on $I=(-2,2)$,Both functions have three continuous derivatives. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first derivatives are all zero, so the Wronskian is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these functions are linearly independent on the whole interval. There is no single constant multiplier valid on both sides of the common zero.

These functions cannot both solve any equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$ with continuous functions $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f^{\prime }(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2{y}^{\prime \prime }-3x{y}^{\prime }=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; Abel's identity cannot be applied across that point. The example pinpoints why checking the equation and its interval is essential before using a zero Wronskian to infer linear dependence.