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Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

Irradiance measures incident radiant flux per actual receiving area. Its units $\mathrm{W}{\mathrm{m}}^{-2}$ match radiant exitance, but it counts the incoming side of the radiation exchange. Measure incoming angles from the unit normal pointing toward the incident hemisphere. ThenFor $2\mathrm{W}$ distributed uniformly over $0.010{\mathrm{m}}^2$, $E=200\mathrm{W}{\mathrm{m}}^{-2}$. If the receiving plane is tilted $60^{\circ }$ in the same broad uniform parallel beam, the incident power per actual area is halved to $100\mathrm{W}{\mathrm{m}}^{-2}$. This receiving-surface cosine factor is geometric and does not assume Lambertian reflectance. Together, irradiance and radiant exitance describe how radiation supplies and removes energy at a surface, while radiance preserves the directional information.

For an ideal diffuse emitting or reflecting planar surface with uniform direction-independent radiance, the radiant intensity in a direction at angle $\theta$ to its unit normal is $I(\theta )=I(0)\cos \theta$. This follows immediately from $I=LA\cos \theta$ and the projected area formula. The worked radiance example and diagram show the $0^{\circ }$ and $60^{\circ }$ cases. Lambertian reflectance supplies one physical idealization realizing the law; an ideal blackbody provides an emitting example.

The historical setting is Johann Heinrich Lambert's Photometria sive de mensura et gradibus luminis, colorum et umbrae, published in Augsburg in 1760. Its Part I, §139, explicitly compares normal and oblique illumination using the sine of the source's elevation, equivalently the cosine of its angle from the receiving normal vector. This is an early receiving-surface projection calculation; the modern outgoing-patch statement above keeps radiance, radiant intensity and irradiance distinct. ^{[ref] [ref]}.

A Lie group is a group that is also a finite-dimensional smooth manifold, with multiplication $G\times G\to G$ and inversion $G\to G$ both smooth maps. Its tangent space at the identity element carries a natural Lie algebra structure. Smoothness lets us differentiate the group operation; being a group means that every transformation can be composed and undone. The dimension of a manifold counts independent local parameters, rather than the number of elements of the group. See MIT 18.745, Lecture 2, Definition 2.1 and Example 2.5.

For example, the planar special orthogonal group $SO(2)$ consists of rotation matricesThe sine and cosine entries vary smoothly. Four entries appear, but they are constrained by $R_{\theta }^T{R}_{\theta }=I$ and $\det R_{\theta }=1$: locally, one angle specifies the whole matrix. Multiplication adds angles, with angles differing by $2\pi$ representing the same rotation. Thus $SO(2)$ is an infinite group of dimension one, geometrically a circle group.

The positive-dimensional examples motivate the language of continuous symmetry, but the definition also admits finite groups with the discrete topology as zero-dimensional Lie groups. Conversely, being an infinite group alone says nothing about having a continuous angle parameter. The compatibility between smooth maps and the group operation is what makes differentiation useful.

A lunar phase is the appearance of the sunlit portion of the Moon as seen from Earth. Outside a lunar eclipse, the distant Sun illuminates approximately one hemisphere of the Moon, but an observer sees a changing overlap of that hemisphere with the hemisphere facing Earth. The lunar phase therefore changes even though synchronous rotation keeps approximately the same near side of the Moon facing Earth. ^[ref].

The cycle runs through new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, last quarter and waning crescent. “Waxing” means that the lunar illuminated fraction is increasing; “waning” means it is decreasing. A crescent has less than half its apparent disk sunlit, and a gibbous lunar phase has more than half. First and last quarter each show half a disk: “quarter” refers to progress around the cycle, rather than a quarter of the disk being lit. ^[ref].

An instructive circular orbit model puts Earth at the origin, the Sun far away along $+x$, and the Moon at $R(\cos \theta ,\sin \theta )$. The sunward position $x>0$, $\theta =0$, gives new Moon; the opposite position $x<0$, $\theta =\pi$, gives full Moon. Counterclockwise motion gives first quarter at $\theta =\pi /2$ and last quarter at $\theta =3\pi /2$. The Sun's direction is fixed during this snapshot of the cycle. Every small globe in the orbital view is illuminated on its $+x$ half; the adjacent apparent disks are separate views from Earth, rather than globes viewed from above.

The disk views choose north up, with the upward direction perpendicular to the orbital plane. In this convention waxing lunar phases are bright on the right and waning lunar phases on the left. Local horizon orientation and viewing from different locations on Earth can rotate the apparent disk; its lit edge always points toward the projected direction of the Sun. The lunar illuminated fraction is unaffected by this rotation.

The useful angle is the phase angle $\alpha =\angle SME$, with $S$ the Sun, $M$ the Moon and $E$ the observer at Earth's center: its vertex is at the Moon. Full Moon has $\alpha =0$, quarter has $\alpha =\pi /2$, and new Moon has $\alpha =\pi$ in the ideal model. By contrast, the elongation $ϵ=\angle SEM$ has its vertex at Earth. For a distant Sun, its rays are parallel, soHere $k$ is the lunar illuminated fraction: a fraction of projected disk area, not a fraction of the Moon's total surface area or its received radiant flux relative to full Moon. The phase angle and unsigned elongation each lie between $0$ and $\pi$, so either alone gives the illuminated amount without distinguishing waxing from waning. The orbital direction resolves that ambiguity.

The lunar illuminated fraction has an elegant geometric derivation. Model the Moon as a sphere of unit radius and view it along the $z$ direction. Take the Sun's unit direction to be $(\sin \alpha ,0,\cos \alpha )$. A visible surface point above apparent disk coordinates $(x,y)$ has $z=\sqrt{1-x^2-y^2}$. Direct illumination requires the outward normal's dot product with the sunward direction to be positive. The day-night terminator therefore obeys $x\sin \alpha +z\cos \alpha =0$. At a fixed $y$, set $q=\sqrt{1-y^2}$; the visible day-night terminator is $x_t=-q\cos \alpha$, and the illuminated strip extends from $x_t$ to the right edge $q$. Integrating its width givesThe integral is the area of half a unit circle. Thus a three-dimensional visibility problem reduces to a circle's area and a cosine. The derivation assumes parallel sunlight, a distant observer and a sphere, and ignores a lunar eclipse. Real observed brightness additionally depends on surface scattering of light and observing geometry, so $k=1/2$ does not promise half the full-Moon radiant flux.

For a worked example, elongation $ϵ=60^{\circ }$ gives phase angle $\alpha =120^{\circ }$ and $k=(1+\cos 120^{\circ })/2=1/4$: a crescent with $25\%$ of its apparent disk sunlit. At elongation $120^{\circ }$, the phase angle is $60^{\circ }$ and $k=3/4$: a gibbous lunar phase. At $90^{\circ }$, $k=1/2$. In each case the same $k$ occurs once while waxing and once while waning in the ideal cycle.

Ordinary lunar phases are not caused by Earth's shadow. A lunar eclipse occurs only when the Moon actually enters that shadow near full Moon. The orbital inclination of the Moon is about $5^{\circ }$ relative to Earth's orbital plane around the Sun; most full Moons pass above or below the shadow. This is why full Moon does not imply a lunar eclipse every month. ^[ref].

The phase cycle is a synodic month, about $29.5$ days, whereas the Moon's orbital period relative to distant stars is a sidereal month, about $27.3$ days. During one sidereal month, Earth advances around the Sun, so the Moon must move farther to repeat the sunward alignment. For coplanar circular orbits traversed in the same direction, subtract their mean motions: $n_{\mathrm{syn}}=n_{\mathrm{sid}}-n_{\mathrm{year}}$, with $n=2\pi /P$. HenceUsing $P_{\mathrm{sid}}=27.321661\mathrm{days}$ and $P_{\mathrm{year}}=365.256\mathrm{days}$ gives $P_{\mathrm{syn}}=29.530591\mathrm{days}$, consistent with the mean synodic month of $29.53059$ days. After one sidereal month, the sunward direction has advanced about $26.93^{\circ }$; catching up takes about $2.21$ more days. These are mean-period estimates: orbital eccentricity and changing geometry make individual synodic months vary. ^[ref]. ^[ref].

The lunar phase is also a practical clock for observing. Full Moon is opposite the Sun and rises roughly at sunset; first quarter is about a right angle from the Sun and is convenient in the evening, while last quarter is convenient before dawn. These are rough guides, not exact rise-time predictions for every place and season. ^[ref]. Historically, repeated lunar phases provided the month in lunar calendars, while half-lit disks gave Aristarchus of Samos a way to infer astronomical distances. ^[ref].

Mathematics studies patterns in quantity, structure, space, and change by giving objects precise mathematical definitions and examining what follows from them. Arithmetic starts with operations on numbers; algebra organizes relations and transformations; geometry and topology studies shapes and the structures preserved by maps; analysis studies limits and continuous change. These fields connect because the same structure can appear in different forms.

A small example shows the value of changing viewpoint. A point moving around the unit circle has coordinates $x=\cos \theta$, $y=\sin \theta$. Regard the plane as complex numbers and the same point is $z=e^{i\theta }$, by Euler's formula. Adding two angles now corresponds to multiplying their complex exponentials: $e^{i\alpha }{e}^{i\beta }=e^{i(\alpha +\beta )}$. A geometric rotation has become an algebraic operation.

The connection reaches calculus too. Differentiating twice gives $x^{\prime \prime }+x=0$ and $y^{\prime \prime }+y=0$, so the two coordinates solve the harmonic oscillator equation. At angle $\theta =\pi /3$, $z=1/2+i\sqrt{3}/2$; one point supplies an exact cosine, an exact sine, and the state of an oscillation. The Pythagorean trigonometric identity records that the modulus stays one. Seeing these as descriptions of one object makes each formula easier to remember and reuse.

This is why a useful mathematical explanation provides both a reliable mathematical proof and a way to see what the result is doing. In his 1994 essay, William Thurston discusses mathematical progress in terms of advancing understanding, and uses the many interpretations of a derivative to illustrate this aim. ^[ref] A diagram, a computed example, and a concise argument can make different parts of the same idea accessible.

Applications ask us to select a mathematical model of a physical or other system. Its assumptions determine which conclusions transfer to that system. Within mathematics, the same practice of making assumptions explicit lets a result about groups, vector spaces, or probability distributions apply to many concrete examples. The beauty is the combination of precision and reuse: an argument can reveal more than the particular case that first suggested it.

In elliptic coordinates $x=\cosh \mu \cos \nu$, $y=\sinh \mu \sin \nu$, a harmonic function in $a<\mu <b$, with $0<a<b$ and period $2\pi$ in $\nu$, satisfies $u_{\mu \mu }+u_{\nu \nu }=0$. Prescribe $u(b,\nu )=f(\nu )$ and $u_{\mu }(a,\nu )=g(\nu )$. Expand both boundary functions in Fourier series, writing their constant terms as $f_0,g_0$. For either the sine or cosine coefficient of order $n\ge 1$, separation of variables givesThe constant angular mode is $u_0(\mu )=f_0+g_0(\mu -b)$. These expressions follow by solving $u_n^{\prime \prime }-n^2{u}_n=0$ and checking $u_n(b)=f_n$, $u_n^{\prime }(a)=g_n$. Their Fourier series gives the solution for sufficiently regular boundary data.

The phrase normal derivative needs care: the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$, so the outward unit normal derivative on the inner ellipse is ${\partial }_{n}u=-h^{-1}{u}_{\mu }$. Prescribing a physical Neumann boundary condition therefore differs from prescribing $u_{\mu }$.

For uniqueness, subtract two solutions to obtain $v$. Multiply $v_{\mu \mu }+v_{\nu \nu }=0$ by $v$ and use integration by parts over the periodic coordinate rectangle. The outer Dirichlet boundary condition, inner homogeneous derivative condition and periodic side cancellation give ${\int }_a^b{\int }_0^{2\pi }(v_{\mu }^2+v_{\nu }^2)d\nu d\mu =0$. Thus $v$ is constant, and the outer condition makes it zero.

When the FM modulation index satisfies $|\beta |\ll 1$, a Taylor expansion gives a useful first-order approximation. With ${\omega }_c=2\pi f_c$ and ${\omega }_m=2\pi f_m$,The product-to-sum formula for sines exposes a carrier and one pair of sidebands, with opposite signs for this wave phase convention. Higher sidebands are neglected, not identically absent. For $\beta =0.1$, the predicted normalized first-sideband magnitude $0.05$ is close to the original exact evaluation $J_1(0.1)\approx 0.0499375$. The neglected waveform term is bounded in absolute value by ${\beta }^2/2=0.005$, using the second-derivative bound for the cosine. Small frequency deviation relative to $f_c$ alone does not justify this approximation: it is $\Delta f/f_m$ that must be small.

Choose one driving electric field component and one responding electric polarization component. A schematic local, instantaneous expansion isHere ${ϵ}_0$ is the vacuum permittivity, ${\chi }^{(1)}$ is the linear electric susceptibility, and the higher coefficients describe nonlinear response. In SI, $P$ has units $\mathrm{C}{\mathrm{m}}^{-2}$, $E$ has units $\mathrm{V}{\mathrm{m}}^{-1}$, and ${\chi }^{(2)}$ has units $\mathrm{m}{\mathrm{V}}^{-1}$. The selected scalar coefficient depends on polarization of waves and crystal orientation; it is not a universal material number independent of geometry.

For a chosen monochromatic wave $E(t)=E_0\cos \omega t$, the quadratic contribution is exactlyThe cosine identity separates a constant contribution, associated with optical rectification, from an oscillation at twice the angular frequency, the source for second-harmonic generation. A term oscillating at $2\omega$ is a local driving source; efficient outgoing light additionally requires propagation and phase matching.

For a material with optical dispersion, the scalar instantaneous expansion is only a teaching model. The actual quadratic response is a tensor with output and input frequency arguments: ${\chi }_{ijk}^{(2)}(2\omega ;\omega ,\omega )$ describes the harmonic, whereas ${\chi }_{ijk}^{(2)}(0;\omega ,-\omega )$ describes the constant response. Their coefficients need not coincide. The cosine calculation establishes which frequencies are available, without assigning measured coefficients to any crystal.

The Nyquist–Shannon sampling theorem says that a bandlimited function can be recovered exactly from equally spaced samples, provided the sampling rate is high enough. It is the bridge between a continuous waveform and a sequence of numbers: the missing values between samples are determined by the restriction on its frequencies.

Here is a precise finite-energy version. Let $f\in L^2(\mathbb{R})$, meaning ${\int }_{\mathbb{R}}|f(t){|}^{2}dt<\infty$, and suppose its Fourier transform, defined through the Plancherel theorem, is zero outside $[-2\pi B,2\pi B]$. Thus $B>0$ is the largest allowed ordinary frequency, in cycles per second, rather than an angular frequency. Set $T=1/f_s$, where $f_s$ is the number of samples per second. If $f_s\ge 2B$, the continuous representative of $f$ is recovered byThe quotient is the normalized sinc function, with value one when its argument is zero. Each shifted sinc function is one at its own sampling time and zero at every other sampling time. Its coefficient is therefore exactly the required sample value. For this L2 space formulation, the reconstruction converges in the L2 space norm and uniformly in $t$. The critical rate $f_s=2B$ is allowed for this class; choosing $f_s>2B$ leaves a gap between the occupied frequencies and the sampling boundary.

For a numerical example, a bandlimited function with no frequencies above $3$ cycles per second can be reconstructed from its samples every $0.1$ seconds: $f_s=10>6$. Exact reconstruction uses the whole sequence indexed by $\mathbb{Z}$. A finite recording and a finite sum of sinc functions introduce truncation error; joining adjacent samples by straight lines is generally a different reconstruction.

The obstruction is aliasing. At sampling rate $f_s=10$, the two distinct waveformshave identical samples at $t=n/10$, becauseThe samples alone cannot tell whether the waveform oscillates twice or eight times per second. If we know in advance that all frequencies are below $5$, the $8$-cycle waveform is excluded and the ambiguity disappears. These pure cosines are an illustrative aliasing example, rather than members of the finite-energy L2 space class in the stated theorem.

Exactly at a sampling boundary, the function class matters. At rate $f_s$, the sine $\sin (2\pi (f_s/2)t)$ has zero values at every $n/f_s$. It is a nonzero waveform indistinguishable from zero on that grid, but has infinite energy and is excluded by the finite-energy hypothesis above. This explains why a rule intended to cover arbitrary pure sinusoids should demand a strict inequality. For spatial data, replace seconds by distance: the Nyquist spatial frequency is $1/(2p)$ for sample pitch $p$.

The proof idea is short: the samples are Fourier coefficients of the Fourier transform restricted to a sampling-sized frequency interval. Because a bandlimited function has no contribution outside that interval, its Fourier series determines the whole Fourier transform; Fourier inversion then gives the sinc function expansion. See sampling expansion by periodic Fourier projection for this connection.

Claude E. Shannon's Communication in the Presence of Noise (1949), Section II, Theorem 1 and equation (7), gives the sampling statement and reconstruction using shifted sinc functions in a communication setting. Shannon explicitly notes earlier mathematical forms and credits Nyquist with emphasizing the sampling interval in telegraphy.^[ref]

A circular loop of radius $R$ carrying electric current $I$ has, on its axis,where the sign of $\hat{z}$ is fixed by the right-hand rule. At its centre, $B(0)={\mu }_{0}I/(2R)$.

Put the loop in the $xy$ plane, centered at the origin, with positive electric current counterclockwise when viewed from $+z$. The right-hand rule then gives the central magnetic field along $+z$. Parameterize source positions by the angle $\varphi$ and observe at $(0,0,z)$:Here $R$ without bold type is the loop radius, whereas $\mathbf{R}$ is the source-to-observer displacement. Their cross product isThe denominator $|\mathbf{R}{|}^3=(R^2+z^2{)}^{3/2}$ is constant around the loop. The integrals of sine and cosine over a full turn vanish, while the axial contribution adds to $2\pi R^2$. Substitution into the Biot-Savart law proves the on-axis formula. Symmetry cancels transverse components; it does not cancel the axial components.

For an original one-turn example in the same vacuum approximation, choose $R=0.04\mathrm{m}$ and $I=2\mathrm{A}$. With ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$,At $z=0.03\mathrm{m}$, the source-to-observer distance is $0.05\mathrm{m}$, so $B(z)/B(0)=64/125=0.512$ and $B(z)\approx 16.08\mu \mathrm{T}$. At $z=-0.03\mathrm{m}$, the magnetic field has the same magnitude and still points along $+z$. The point has crossed the loop plane, but the loop's electric current has not reversed. Reversing the electric current reverses the magnetic field on both sides.

For $|z|\ll R$, a Taylor expansion gives $B(z)/B(0)=1-3z^2/(2R^2)+O(z^4/R^4)$: the center is a smooth maximum, with zero first derivative. For $|z|\gg R$,This is the axial magnetic dipole field of magnetic dipole moment $\mathbf{m}=I\pi R^2\hat{\mathbf{z}}$.^[ref] A compact closed loop has an inverse-cube axial far field, in contrast to the inverse-distance field of the infinite straight wire. The SI units of the exact loop formula are ${\mu }_{0}I{R}^2/(R^2+z^2{)}^{3/2}\sim {\mu }_{0}I/\text{length}$, hence tesla.

For $N$ coincident turns carrying the same electric current, the superposition principle multiplies the ideal result by $N$. A real coil has radial width and axial length: if those are not negligible, integrate or sum over its individual turns. This is why a single measured coil diameter and a turn count may not determine the full spatial magnetic field precisely. A collection of loops becomes a solenoid; integrating their axial contributions gives the on-axis magnetic field of a finite solenoid.

Orientation selectivity is a systematic dependence of a visual neuron's response on the axis of a bar, edge or stripe pattern presented in its receptive field, under specified stimulus conditions. Here the angle $\theta$ measures the stripe axis from horizontal: $0^{\circ }$ is horizontal and $90^{\circ }$ vertical. An axis has no arrow, so $\theta$ and $\theta +180^{\circ }$ describe the same orientation. A preference for one direction of movement across that axis is an additional property; the 1962 recordings included both equal and unequal responses to opposite movements. ^[ref]

A particularly clear experiment using extracellular single-unit recording appears in the 1959 paper, Figures 2–3. A bright $1^{\circ }\times 8^{\circ }$ slit was centered on the same receptive field and rotated through successive angles; each presentation lasted one second. The vertical slit strongly increased the extracellularly recorded spike train, whereas the horizontal slit was ineffective. Small-spot tests had identified ON and OFF receptive-field subregions, with a central ON region and opposing flanks: rotation illuminated less of the center and more of the flanks. Geometry and location therefore mattered together. These dimensions belong to that recorded unit, not to every neuron. ^[ref]

An orientation tuning curve summarizes a controlled angle sweep. Keep location, size and visual contrast fixed; for a stripe pattern also specify spatial frequency and phase. The same signed spatial filter model of a simple cell can produce strong orientation selectivity and strong dependence on phase, as the following original toy illustrates. Its smooth curves are calculated predictions, not digitized or invented biological measurements.

For the sinusoidal density-of-states model, divide the oscillatory amplitude of $\Phi (T,\mu )-\Phi (0,\mu )$ by its smooth magnitude $g_0{\pi }^2(k_{B}T{)}^2/6$. This gives the title ratio. Its limit as $T\to 0$ is $|\alpha |$, although both changes vanish at exactly zero temperature. The expansion is $|\alpha |[1-7X^2/60+O(X^4)]$, and its large-$X$ envelope is $6|\alpha |/X^2$. A fixed chemical potential adds the actual cosine phase, which can make the oscillatory contribution zero even when the envelope is nonzero. Exponential suppression of the full oscillatory potential therefore must not be confused with this algebraic thermal-change ratio.

Normalize both leading coefficients to one. For the indicial root $a=1/2$, the recurrence relation from part (b) givesRecognizing the Taylor series of the sine gives the first Frobenius solutionFor the smaller indicial root $-1/2$, $n(n-1)d_n=-d_{n-2}$. The equation at $n=1$ is $0=0$; choose the free coefficient $d_1=0$ to avoid adding the first Frobenius solution. The remaining coefficients areThus the Taylor series of the cosine givesTheir leading terms are $x^{1/2}$ and $x^{-1/2}$, as required for Case IIIb. A direct check uses $y=x^{-1/2}u$: the differential equation becomes $x^{3/2}(u^{\prime \prime }+u)=0$, so these two functions solve it on $x>0$. Their Wronskian is $-1/x\ne 0$, proving linear independence. The general solution is $(C_1\sin x+C_2\cos x)/\sqrt{x}$.

Use the Poincaré-Lindstedt method. Introduce $\theta =\Omega t$ and $x=x_0(\theta )+ϵx_1(\theta )+\cdots$. The leading equation gives ${\omega }_0=3$ and $x_0=A_0\cos \theta$, after fixing the arbitrary phase. At order $ϵ$,Here $4{\cos }^2\theta \sin \theta =\sin \theta +\sin 3\theta$. Periodicity requires the two resonant coefficients to vanish. With $A_0>0$, this givesTherefore the leading limit cycle is $x(t)=\cos (3t)+O(ϵ)$, with frequency $3+O({ϵ}^2)$. For example the nonresonant first correction contains $-ϵ\sin (9t)/24$; its homogeneous sine/cosine pieces depend on phase and amplitude conventions. The absence of a first-order frequency shift is a solvability condition, not an assumption.

Let $\mathcal{R}=|r{|}^2$ be the single-mirror power reflectivity and ${\theta }_g$ the propagation angle inside the gas. Define $\lambda$ as the vacuum wavelength. Successive transmitted complex amplitudes differ by a round-trip factor $\mathcal{R}{e}^{i\delta }$, where, for real dielectric reflection coefficients whose product has positive sign,The cosine arises by comparing successive emerging rays on the same outgoing wavefront; it is not the literal zigzag distance $2d/\cos {\theta }_g$. Equivalently, the round-trip phase is $2k_{z}d$. If the source's $\theta$ is the angle in external air, Snell's law gives $n\sin {\theta }_g=\sin \theta$ andIf $\theta$ denotes the internal angle, simply set ${\theta }_g=\theta$.

For identical lossless mirrors, their power transmission is $1-\mathcal{R}$. In an energy-normalized amplitude convention this is $|t{|}^2=1-\mathcal{R}$; more general interface amplitudes require the corresponding wave impedance factors. The geometric series of transmitted amplitudes givesHenceThis is Fabry–Pérot transmission from multiple reflections.^[ref]

At normal incidence, for nondispersive $n$ and zero net reflection phase modulo $2\pi$, the resonance condition is $\delta =2\pi m$. The resonance frequencies and free spectral range areFor finite $\mathcal{R}<1$ the resonances have finite width, and transmission away from them is nonzero. Thus “only discrete frequencies” is an ideal high-reflectivity description of the selected resonances, not an exact statement about the support of the transmitted spectrum. For example, $\mathcal{R}=0.9$ at $\delta =\pi$ still gives $I_t/I_0=(0.1/1.9{)}^2\simeq 0.00277$.

At half the unit peak, $Q{\sin }^2(\delta /2)=1$. Measuring the phase displacement from a resonance yieldsThe last expression is the high-reflectivity approximation, with finesse $\pi \sqrt{\mathcal{R}}/(1-\mathcal{R})$. At $\mathcal{R}=0.9$, the full phase width is about $0.2109$ radians and $\Delta {\nu }_{\mathrm{FWHM}}/\Delta {\nu }_{\mathrm{FSR}}\simeq 0.03357$. At $\mathcal{R}=0.1$, the minimum transmission is $0.6694$, so the trace never reaches half the peak and this full width at half maximum is not defined. The exact half-maximum condition requires $\mathcal{R}\ge 3-2\sqrt{2}$.

For the absorbing Fabry–Pérot mirror phase and transmission, write $r=\sqrt{\mathcal{R}}{e}^{i{\varphi }_r}$ and let $\mathcal{T}$ be the power transmission of one mirror. Optical absorption makes $\mathcal{R}+\mathcal{T}<1$. The transmitted intensity becomesHere the phase formula is at normal incidence. The peak is reduced below unity, while a change in $\mathcal{R}$ changes the width. The complex reflection phase also moves the resonances: if ${\varphi }_r$ is approximately constant,If either $n$ or ${\varphi }_r$ depends on frequency, this is an implicit condition and the free spectral range need not be constant. No numerical shift is determined without the mirrors' optical response.

Let $u,w\in M$ both solve the variational inequality. Test the inequality for $u$ with $w$ and the one for $w$ with $u$, then add. Writing $d=u-w$ givessince cosine is Lipschitz with constant one. Since $d(0)=0$, the fundamental theorem of calculus and Cauchy-Schwarz inequality give $|d(x)|\le \sqrt{l}\Vert d^{\prime }{\Vert }_2$; integrating gives $\Vert d{\Vert }_2^2\le l^2\Vert d^{\prime }{\Vert }_2^2$. ThereforeFor $\boxed{0<\gamma <2c_0/l^2}$, the coefficient is positive, so $d^{\prime }=0$ and the zero boundary value gives $d=0$. This proves uniqueness with an explicit sufficient smallness condition. More generally any Poincaré inequality constant $C_P$ gives the condition $\gamma C_P^2<2c_0$; optimizing that constant is unnecessary for the requested conclusion. The underlying principle is uniqueness for a strongly monotone variational inequality.

Take eastward $x$, upward $z$, positive wavenumbers $k,K$, and a real instantaneous amplitude $A$. A common growing factor does not affect the phase geometry. Put $C=\cosh (Kz)$ and $S=\sinh (Kz)$, using the hyperbolic cosine and hyperbolic sine. The prescribed streamfunction iswhereFollowing a fixed phase givesTherefore the streamfunction pattern tilts westward with height. The lower boundary leads the upper boundary eastward.

In the conventional quasi-geostrophic streamfunction normalization, hydrostatic balance gives $b^{\prime }=f_0{\psi }_z^{\prime }$. ThereforeFor the phase calculation, multiplication by the constant $f_0$ is immaterial. Define a continuous phase ${\delta }_b=atan2(\alpha C,S)$; this avoids a spurious discontinuity at $z=0$. ThenThe buoyancy perturbation pattern tilts eastward with height, opposite to the streamfunction pattern. At $z=0$, ${\psi }^{\prime }=A\cos (kx)$ while $b^{\prime }=-f_{0}AK\alpha \sin (kx)$, so the two fields are a quarter wavelength apart there.

Both perturbation amplitudes are largest at the boundaries. Indeed,which increase with $|z|$. The physical reason is particularly clear through stratified quasi-geostrophic inversion: the Eady model has no interior background potential-vorticity gradient, and a growing normal mode has zero interior potential vorticity anomaly. The independent thermal anomalies reside at the rigid upper and lower boundaries, where meridional motion advects the basic horizontal buoyancy gradient. Each Eady edge wave induces an interior balanced flow that decays away from its boundary on scale $K^{-1}$. Their interacting, phase-locked combination produces the growing baroclinic instability. Boundary confinement of the thermal sources, rather than a rigid-wall requirement ${\psi }^{\prime }=0$, explains the amplitude pattern.

For a sketch with a quarter-wavelength difference between the boundary streamfunction patterns, chooseAn upper positive streamfunction crest is then ${\lambda }_x/4$ west of its lower counterpart, where ${\lambda }_x=2\pi /k$. The buoyancy perturbation crests slope the other way. The quarter-wavelength boundary assumption is used for the streamfunction; differentiating in $z$ fixes the buoyancy perturbation and does not generally give it the same boundary phase separation.

The connection to poleward heat transport can be checked without relying on the sketch. The meridional geostrophic flow and buoyancy perturbation areAverage over one zonal wavelength, using equal averages $1/2$ for squared sine and cosine and zero for their product:In the Northern Hemisphere this is positive: warm anomalies preferentially move northward and cold anomalies southward. For the same eastward-increasing basic thermal wind in the Southern Hemisphere, $f_0<0$ and the basic buoyancy gradient reverses; the negative northward eddy buoyancy flux is again poleward. For thermal expansion coefficient ${\alpha }_T>0$ and specific heat capacity at constant pressure $c_p$, the corresponding sensible heat flux in a thermal Boussinesq approximation is ${\rho }_0{c}_p\overline{v^{\prime }{T}^{\prime }}={\rho }_0{c}_p\overline{v^{\prime }{b}^{\prime }}/(g{\alpha }_T)$. The growing amplitude makes this flux grow as $A^2$. The tilt supplies the correlation absent in an untilted disturbance and reduces the background available potential energy.

Take $m,k,\gamma ,F_0>0$ and $\omega >0$, and write $D_{\omega }=k-m{\omega }^2$ and $B_{\omega }=m\gamma \omega$. Substitution of $z=A\cos \omega t$ into the forced harmonic oscillator equation givesEquating the independent sine and cosine coefficients yieldsThus the required phase matching for a displacement referenced driven oscillator isFor a positive amplitude $A$, one must also choose the correct quadrant: $\cos {\varphi }_0=D_{\omega }/\sqrt{D_{\omega }^2+B_{\omega }^2}$ and $\sin {\varphi }_0=B_{\omega }/\sqrt{D_{\omega }^2+B_{\omega }^2}$. Squaring and adding the coefficient equations givesFor arbitrary forcing phase, the steady response is instead $z=A\cos (\omega t+{\varphi }_0-\delta )$, where $\delta$ has the sine and cosine just specified. A change of time origin can put this displacement into $A\cos \omega t$, but it also changes the stated forcing phase.

The source's requested parameter list for $A$ omits $\gamma$; in general it cannot be omitted. As a concrete counterexample, take $m=k=F_0=\omega =1$ and ${\varphi }_0=\pi /2$. The prescribed cosine response exists for every $\gamma >0$, but $A=1/\gamma$, so fixing all four listed parameters does not fix the amplitude. Likewise, with fixed ${\varphi }_0=0$, positive damping and nonzero $\omega$, the second coefficient equation prevents a nonzero pure-cosine response. The forcing phase is an essential part of the condition.

To invert the complex cosine, let $z=\cos w$ and put $u=e^{iw}$. Since the exponential function is nonzero, $u\ne 0$, and Euler's formula gives $2z=u+u^{-1}$. Multiplying by $u$ yields the quadratic equation $u^2-2zu+1=0$. ThusChoosing either square root of $1-z^2$ merely exchanges the two signs. The two values have product one, so neither vanishes; taking a complex logarithm is legitimate.

For each choice of sign, all solutions of $e^{iw}=u$ satisfy $iw=\mathrm{Log}u+2\pi in$. Therefore the full inverse relation isConversely, exponentiating any displayed value recovers one root of the quadratic equation, and hence $(e^{iw}+e^{-iw})/2=z$, proving completeness. With a multivalued complex logarithm, the $2\pi n$ term is already included and this is the source's compact formula for inverse cosine.

The principal inverse cosine requires compatible choices of logarithm branch, square root and sign. Both signs cannot represent that same principal function indiscriminately: at $z=0$, the two principal-log values are $\pi /2$ and $-\pi /2$, whereas the principal $\arccos 0$ is $\pi /2$. Thus the printed $\pm$ is correct for the inverse relation, not a pair of interchangeable principal-value formulas.