Codex @codex  0

Uniform samples at $t_n=n/f_s$ cannot distinguish frequencies separated by integer multiples of $f_s$:For a real cosine, opposite frequencies also coincide. Thus an unknown high-frequency component can appear in the sampled record at its folded frequency $f_{\mathrm{alias}}\in [0,f_s/2]$. For example, 2 Hz and 8 Hz cosines give identical samples at 10 samples per second. This ambiguity is aliasing, rather than uniform quantization error.

The Nyquist–Shannon sampling theorem permits reconstruction of an ideally bandlimited function of bandwidth $B$ when $f_s>2B$. Use an analogue anti-aliasing filter before conversion and allow a transition band between the desired passband and the frequencies that fold into it. Sampling faster relaxes the filter requirement. Digital filtering after undersampling cannot separate already identical sample sequences. Finite observation time, clock jitter, noise and imperfect filter rejection impose further practical limits.

Solving the time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

For a plane wave propagating along $\hat{\mathbf{k}}$ in vacuum, Maxwell equations give $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$, with $\mathbf{E}\perp \hat{\mathbf{k}}$. HenceTaking $E=E_0\cos (\mathbf{k}\cdot \mathbf{r}-\omega t)$, the period average of the squared cosine is $1/2$. ThusThe fields are in phase, so the travelling-wave energy flux never reverses. The peak amplitude convention is essential; using a root-mean-square electric field would remove the factor $1/2$.

A Fourier series separates a periodic physical signal into harmonics. For example, a periodic voltage applied to a linear circuit can be treated one frequency at a time; the harmonic responses then superpose to give the output waveform.

Write $w=\Delta T$ for the pulse width and $\delta =w/T$ for its duty cycle, with $0<w<T$. The waveform is a train of positive rectangular pulses centred on $t=jT$, each with area one. It is even about $t=0$, so every sine coefficient vanishes. Cosine orthogonality givesand, for $n\ge 1$,This is the Fourier series of a periodic rectangular pulse:The actual constant term in the series is $a_0/2=1/T$, not $a_0$. As usual for a jump discontinuity, the series converges to the average of the one-sided values at each pulse edge, namely $1/(2w)$, irrespective of the value assigned at that isolated endpoint. Elsewhere it converges to the rectangular waveform.

Counting the constant as the first term, the first three terms areso their sum isIf “first three” instead means the first three nonconstant harmonics, add $a_3\cos (6\pi t/T)$; the original plot shows that alternative partial sum as well. The three-term approximation rounds the narrow peaks and ripples between them; many harmonics are needed to represent a sharp pulse.

The original sketches use the illustrative duty cycle $\delta =0.15$ and cover $-T\le t\le T$. The coefficient panel also displays the first spectral zero.

For $n\delta \ll 1$, expand the sinc function:Thus the low-index coefficients, including $a_0$, have nearly equal values, although their constant-series contribution still has the factor $1/2$. The first sinc zero occurs at $n\delta =1$, or $n=T/w$. If this is not an integer, the closest integer coefficient is near zero rather than exactly zero: writing $n\delta =1+\epsilon$ gives $a_n\simeq -(2/T)\epsilon$. This pulse-width scale of Fourier coefficients means that a narrower pulse requires significant harmonics extending to proportionally larger indices, the reciprocal time-width/frequency-width relation.

In a homogeneous, source-free, nonmagnetic dielectric the electric wave equation isSet $\theta =\omega t-kx$ and $E_y=f(z)\cos \theta$. Substitution gives $f^{\prime \prime }+({ϵ}_r{\omega }^2/c^2-k^2)f=0$. For the labelled field form (a), $f=E_0\cos Kz$, so $K^2={ϵ}_r{\omega }^2/c^2-k^2>0$. For form (b), $f=E_0^{\prime }{e}^{s{K}^{\prime }z}$ with $s=\pm 1$, so $K^{\prime 2}=k^2-{ϵ}_r{\omega }^2/c^2>0$. These give the two stated inequalities, with equality a zero-transverse-wavenumber limiting case.

Faraday's law gives the associated wave magnetic field, without an added static field,Thus in case (a), $H_x=-K{E}_0\sin (Kz)\sin \theta /({\mu }_0\omega )$ and $H_z=k{E}_0\cos (Kz)\cos \theta /({\mu }_0\omega )$. In case (b), $H_x=s{K}^{\prime }{E}_0^{\prime }{e}^{s{K}^{\prime }z}\sin \theta /({\mu }_0\omega )$ and $H_z=k{E}_0^{\prime }{e}^{s{K}^{\prime }z}\cos \theta /({\mu }_0\omega )$. In both cases $H_y=0$.

For the planar dielectric waveguide the core is dielectric but the exterior is vacuum. Consequently the exterior decay constant has $K^{\prime 2}=k^2-{\omega }^2/c^2$, while the interior oscillation has $K^2={ϵ}_r{\omega }^2/c^2-k^2$. Adding gives the relation labelled (c):For an even bound mode the complete profile can be writtenContinuity of tangential $E_y$ fixes the exterior amplitude; continuity of tangential $H_x$ makes $f^{\prime }$ continuous because both media have magnetic permeability ${\mu }_0$. At $z=d$,which is the relation labelled (d). The same even profile also matches at $z=-d$; continuity of $H_z$ follows from that of $E_y$.

In the requested $K,K^{\prime }$ plot, (c) is a first-quadrant circle and (d) rises from the origin to its first pole at $Kd=\pi /2$. Equivalently put $u=Kd$, $w=K^{\prime }d$, and $\mathcal{V}=(\omega d/c)\sqrt{{ϵ}_r-1}$. At $u=0$ the circle has $w=\mathcal{V}>0$ whereas $u\tan u=0$. At the other end of $0<u<\min (\mathcal{V},\pi /2)$, the rising tangent curve exceeds the falling circle. Monotonicity therefore gives one intersection on this first branch for every $\mathcal{V}>0$.

The fundamental even mode of a symmetric dielectric slab has no positive cutoff in index contrast or thickness. Its electric profile is a nodeless cosine maximum at the mid-plane, smoothly joined to decaying exterior tails. The qualifier ${ϵ}_r>1$ is essential: for the printed limiting value ${ϵ}_r=1$, the circle collapses to $K=K^{\prime }=0$ and there is no transversely confined bound mode. A uniform unconstrained wave is not a guided mode. This even-mode argument does not assert that the specified cosine describes every possible odd guide mode.

The original diagram uses the illustrative normalized radius $\mathcal{V}=1.4$; it is a scaled $K{K}^{\prime }$ plot with $u=Kd,w=K^{\prime }d$, followed by the corresponding field profile.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

Fixed endpoints require $k_n=n\pi /L$, with $n=1,2,\dots$. Writing $c_s=\sqrt{T/\rho }$,For equal amplitudes $A$, the two requested profiles areThe first is asymmetric, with an interior zero at $x=2L/3$; the second has its interior node at $L/2$. For two-mode string recurrence, orthogonality means the whole initial profile recurs only when both cosine coefficients equal one. Since ${\omega }_2=2{\omega }_1$,

The weak-stiffness string frequency shift follows by expanding the supplied dispersion relation:Higher normal modes acquire larger fractional frequency shifts. Let $\eta =\alpha k_1^2/c_s^2\ll 1$ and retain ${\omega }_1^{(0)}=c_s{k}_1$ for the original ideal frequency. At the originally defined time $t_1=\pi /(2{\omega }_1^{(0)})$,Thus the $t=0$ sketch is unchanged, whereas the original quarter-period sketch gains a small negative fundamental component. If $t_1$ is redefined using the new fundamental frequency, its fundamental component is exactly zero and the second-mode change there is only second order. Both conventions agree that generic later shapes change because of relative phase drift.

The new frequency ratio isAn exact recurrence now requires ${\omega }_{1}t=2\pi q$ and ${\omega }_{2}t=2\pi p$ for integers $p,q$. If $r=p/q$ in lowest terms, the first recurrence is $2\pi q/{\omega }_1$; for irrational $r$ there is no exact finite recurrence. Near-recurrences remain possible. Positive stiffness alone does not prove that the ratio is irrational.

The band effective mass is determined by the second derivative of the dispersion relation:With the stated sign convention and $t>0$, the band minimum is at $k=0$. Expanding the cosine, or differentiating twice, givesComparison with a free electron's kinetic energy ${\hslash }^2{k}^2/(2m_e)$ therefore requiresThe hopping integral has units of energy. A negative $t$ would move the minimum to the edge of the Brillouin zone and give the same positive bottom-of-band effective mass with $|t|$ in this expression.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

For an even function, the interior sine term must vanish, because sine is odd and cosine is even. Comparing reflected exterior tails at any $x>a$ gives $D{e}^{-kx}=A{e}^{-kx}$. Hence the parity condition alone isIt does not yet impose the matching condition on $B$. Continuity at $x=a$, when imposed in the next part, additionally gives $A=D=B{e}^{ka}\cos (\kappa a)$.

For a material volume, mass conservation is $d{\int }_{V(t)}\rho dV/dt=0$. The transport theorem and arbitrariness of that volume give $\boxed{{\rho }_t+\nabla \cdot (\rho u)=0}$. Balance of linear momentum includes the pressure traction and body force:Apply transport, the divergence theorem and the mass equation to obtainFor small perturbations of the uniform resting barotropic state, write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=p_0+c_0^2{\rho }^{\prime }+\cdots$. Linearized mass and momentum areTaking the curl shows that initially vanishing vorticity remains zero, so the acoustic velocity has a potential $u=\nabla \varphi$. Integrating momentum, using the additive freedom of a velocity potential to remove the spatially constant integration term, gives ${\rho }^{\prime }=-({\rho }_0/c_0^2){\varphi }_t$. Continuity now yields $\boxed{{\varphi }_{tt}=c_0^2{\nabla }^2\varphi }$.

For spherical symmetry, set $\varphi =e^{-i\omega t}F(r)$ and $k=\omega /c_0$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }+k^2{r}^{2}F=0$, or $(rF{)}^{\prime \prime }+k^2(rF)=0$. HenceRigid boundaries require radial velocity $F^{\prime }=0$ at both radii. A form automatically satisfying the condition at $a$ isAt $b$, the condition $b(rF{)}^{\prime }-(rF)=0$ becomesTherefore the nonzero oscillatory eigenfrequencies obeyThe undivided sine-cosine form avoids artificial issues from dividing by a cosine. The formal $k=0$ root is the spatially constant zero-frequency potential additive freedom of a velocity potential, not a propagating acoustic oscillation. This is the radial acoustic spectrum between rigid concentric spheres.

Measure both angles from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,Thus $|AC{|}^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. The requested expression implicitly uses a zero-rest-length spring: its energy is $\kappa |AC{|}^2/2$. The exact Lagrangian for equal masses isExpand the cosines and count angles and velocities as first order. The cosine correction multiplying $\dot{\theta }\dot{\varphi }$ is fourth order and is dropped. After removing constants,The positive spring term in $L_2$ means the spring lowers its energy when the straight assembly bends. If its natural length were $s_0$, this term would instead be $\kappa \ell (2\ell -s_0)(\theta -\varphi {)}^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.

For the cosine mode in the question, the Laplacian acts by multiplication by $-s$, whereThese modes obey the Neumann boundary conditions. In the chemotactic term, put $q=1+u$, $c=\eta +v$. Then $\nabla c=\nabla v$, and $q\nabla c=\nabla v+u\nabla v$; the second product is a quadratic polynomial in perturbation amplitude. Thus the linearized equations areSubstituting the modal time dependence givesSeparating this matrix into its transport part and the reaction Jacobian matrix gives the required form, with the same $J$ as above.

A periodic function repeats exactly after a nonzero shift of its input. For $f:\mathbb{R}\to Y$, a positive period of a function $T$ satisfiesOne interval of length $T$ therefore determines the whole function. For another domain of a function $D$, a period must also preserve the domain, $D+T=D$; equality only where two domains happen to overlap is weaker. Periodicity is an exact translation of a function symmetry, rather than merely two similar-looking pieces of its graph.

The sine $\sin (2\pi x)$ has fundamental period one. The function ${\sin }^2(2\pi x)$ repeats after half a unit: the sine changes sign under that shift, but squaring removes the sign. The mixed function $g(x)=2+\sin (2\pi x)+\frac{1}{2}\cos (4\pi x)$ again has fundamental period one. All positive integer multiples of a period are periods, so specifying one period need not identify the shortest repetition.

If the input is time and $T_0$ is the fundamental period, the frequency is $1/T_0$ and the angular frequency is $2\pi /T_0$. Repetition lets a Fourier series describe a cycle using sines and cosines; the period average separates the reference level from the oscillating terms. Strang develops this representation and its Fourier coefficients in §4.1 of Computational Science and Engineering by Gilbert Strang.^[ref]

This connection grew out of a physical problem. In his 1822 book Théorie analytique de la chaleur by Joseph Fourier, Joseph Fourier used periodic functions to extend profiles beyond their initial interval while treating heat conduction. Chapter III, §§233–234 combines even functions and odd functions into sine–cosine expansions and rescales their period.^[ref] A modern Fourier series interpretation still requires the appropriate convergence conditions.

A constant function has every real shift as a period and no fundamental period. Without continuity, even a nonconstant periodic function can lack a smallest positive period; periodic function without a least positive period gives a concrete example. The period group of a function organizes all these possibilities.

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.

Periodicity is invariance under translation by a nonzero period $P$: a periodic function satisfies $f(x+P)=f(x)$ at every point in its domain, with translation preserving the domain. For an angular coordinate on a full circle, single valuedness requires $2\pi$-periodicity. The sine and cosine factors $\sin (k\theta )$ and $\cos (k\theta )$ are $2\pi$-periodic for nonzero real $k$ precisely when $k$ is an integer. Their Fourier series supply the angular modes for the harmonic expansion on a disk.

A nonzero complex number has polar formwhere $r$ is its modulus and $\theta$ is an argument. The equality uses Euler's formula; sine and cosine recover the imaginary part and real part from polar coordinates. For $z=3+4i$, take $r=5$ and $\theta =\arctan (4/3)\approx 0.927295$ radians, so $\cos \theta =3/5$ and $\sin \theta =4/5$.

If $w=s{e}^{i\varphi }\ne 0$, thenThus multiplication multiplies moduli and adds arguments modulo $2\pi$; division divides moduli and subtracts arguments modulo $2\pi$. Since $1+i=\sqrt{2}{e}^{i\pi /4}$, multiplying $3+4i$ by $1+i$ gives $-1+7i$, of modulus $5\sqrt{2}$ and argument $\theta +\pi /4\approx 1.712693$ radians. One formula simultaneously describes its new size and direction.

Repeated multiplication now becomes especially simple: $(r{e}^{i\theta }{)}^n=r^n{e}^{in\theta }$ for a positive integer $n$. This explains the evenly spaced arguments of the roots of a complex number. For example, $(3+4i{)}^2=-7+24i$ has modulus $25$ and principal argument $2\theta \approx 1.854590$ radians. The Cartesian calculation and the polar form provide independent checks on both its location and its size. Choose polar form when the task concerns multiplication, division, or repeated turns; use real parts and imaginary parts for addition.

Polarimetry measures optical rotation with a polarimeter. The vertical Vernier Go Direct instrument uses a $589\mathrm{nm}$ light-emitting diode source, fixed polarizer, sample cell, motor-driven polarimeter analyzer, photodetector and optical rotary encoder. Calibrate using the same solvent at $10\mathrm{cm}$ fill height; rinse with sample, fill to $10.0\mathrm{cm}$, and read the meniscus to $0.1\mathrm{cm}$. The readout plots illumination against analyzer angle; a maximum near zero or a cosine-squared fit locates $\alpha$.^[ref]

An actual teaching experiment in the Truman State University-hosted Optical Activity Laboratory handout, revised July 2013, uses sucrose chemical solutions. Weigh the sucrose, dissolve completely without heating, and measure final solution volume for concentration. Use one polarimeter throughout; measure the water blank's extinction reading, then each sample's extinction reading and actual liquid path length near $10\mathrm{cm}$. Subtract the blank and plot corrected optical rotation against concentration. The attached apparatus description specifies a vertical light path, an incandescent or tungsten source with yellow filter, sheet polarizers and a rotating eyepiece polarimeter analyzer. Its coarse demonstration permits temperature control to the nearest $5^{\circ }\mathrm{C}$, path length to the nearest millimeter and approximately $1^{\circ }$ angular precision. These are design tolerances, not measured results.^[ref]

For a quantitative comparison, equilibrate blank, reference and sample at the same recorded temperature; use a thermostatted cell when needed. This is a recommended control, not a claim that the teaching instruments provide a thermostat. A yellow filter also passes a band of wavelengths, so its result need not match a narrow-source reference exactly. The laboratory's concentration series tests the linear model rather than establishing it in advance; a second series varying only path length tests the other factor.

For our original ideal model, Malus's law giveswhere $I_0$ is light intensity immediately before the ideal polarimeter analyzer, after sample transmission. It is not the intensity of the original unpolarized source. The angle difference is evaluated in consistent units; the plotted degree values are converted to radians before computing the cosine. Assume transmitted light retains linear polarization, with negligible scattering and no polarization-dependent absorption. The pure positive model, $75:25$ mixture and racemate shift the maxima to $8^{\circ }$, $4^{\circ }$ and $0^{\circ }$, respectively. Minima shift by the same amount, to $98^{\circ }$, $94^{\circ }$ and $90^{\circ }$. Thus an extinction measurement and a maximum measurement recover the same shift after their respective blanks are subtracted.

Repeat blanks and use a reference to check drift and scale. Bubbles, particles, stressed windows and sample absorption can spoil the ideal curve. Vernier specifies $\pm 1^{\circ }$ accuracy; the manual's $\pm 0.1^{\circ }$ fit reproducibility does not establish that accuracy. It warns against interpreting its illumination readout quantitatively and requires a transparent homogeneous sample.^[ref] Neither a zero reading nor a matching specific rotation uniquely identifies a molecule.

A Michelson interferometer turns displacement into a count of interference fringes: a moving mirror changes both the outward and return paths. This lets a wavelength act as a ruler whose divisions need no engraving. Assume normal incidence, a fixed reference arm, stable refractive index $n$ along the moving arm, and fixed vacuum wavelength ${\lambda }_0$. Write $\lambda ={\lambda }_0/n$ for the wavelength in that medium and take $\Delta L$ positive away from the beam splitter. The change in optical path difference, the accumulated phase difference, and the signed number of complete interference fringe cycles areThe factor two comes from propagation geometry: moving the mirror by $\Delta L$ adds $\Delta L$ before reflection and $\Delta L$ afterward. One bright-to-bright cycle means mirror travel $\lambda /2$; bright-to-dark means $\lambda /4$, half a cycle. All lengths must use compatible units, and $\Delta N$ is dimensionless. A constant reflection phase difference changes the initial fringe position but cancels from this change formula.

For an original hypothetical example, suppose $\lambda =600\mathrm{nm}$ in the actual surrounding medium and a continuous scan gives $40.25$ cycles, including the final fractional cycle. Then $\Delta L=(600\mathrm{nm})(40.25)/2=12075\mathrm{nm}=12.075\mu \mathrm{m}$. With the wavelength treated as exact, a fringe-reading measurement uncertainty of $0.02$ cycle contributes $6\mathrm{nm}$ to the displacement measurement uncertainty. An uncertainty of $0.01\mathrm{rad}$ in a local phase difference instead corresponds to $600(0.01)/(4\pi )\mathrm{nm}\approx 0.477\mathrm{nm}$. This illustrates measurement resolution, not a complete claim of accuracy: wavelength calibration, air changes, mirror tilt and missed cycles can dominate.

To separate the oscillating readout from its optical coherence envelope, use the local modelHere $I$ is detected light intensity, $\bar{I}$ is the mean level, ${\varphi }_b$ is a fixed bias wave phase, and $\mathcal{V}$ is interferometric visibility. A fine scan of $D$ measures the carrier phase difference; a coarse scan can map the temporal optical coherence envelope through $\mathcal{V}(D)$. The envelope maximum locates matched delays only under the source and apparatus assumptions; it is not a bright fringe identifying an arbitrary mirror position. Even with known $\mathcal{V}$, one intensity reading leaves the cosine sign and whole-cycle ambiguities. Continuous motion of known direction, or an independently obtained pair of signals in phase quadrature, is needed to recover signed travel through reversals.

The figure's original control example stipulates equal arm light intensities, matching linear polarizations, and a Lorentzian optical spectrum giving $\mathcal{V}(D)=\exp (-|D|/L_c)$ with $L_c=1\mathrm{mm}$. In vacuum, arm mismatches $|L_2-L_1|=0.60\mathrm{mm}$ and $0.10\mathrm{mm}$ give $\mathcal{V}=e^{-1.2}\approx 0.301$ and $e^{-0.2}\approx 0.819$. Matching the arms improves the phase difference readout without changing its ${\lambda }_0/2$ travel per cycle. A fine mirror scan of one cycle changes $D$ by only $600\mathrm{nm}$ for the stipulated $600\mathrm{nm}$ source, much less than $L_c$. These are calculated values, not historical data.