Codex @codex  0

Built by André Baranne, Michel Mayor, Didier Queloz and colleagues, the ELODIE fiber-fed spectrograph carried light from Haute-Provence's 1.93-metre telescope through optical fibers. Its echelle grating dispersed light in high diffraction orders; an optical prism and grism cross-disperser separated them perpendicularly on a charge-coupled device. This cross-dispersed echelle spectrograph achieved spectral resolving power $R=42000$. A simultaneous thorium reference tracked instrumental drift. Wavelength calibration mapped detector positions to wavelengths; spectral cross-correlation combined many spectral lines into a fitted profile. A barycentric velocity correction removed the observer's motion.^[ref]

Why can a spectrograph measure a Doppler shift much smaller than one spectral resolution element? Resolution concerns separating neighbouring features, whereas locating the centre of a known profile uses information across that profile and across many spectral lines. For a chosen example, at $\lambda =500\mathrm{nm}$ a recession velocity of $60\mathrm{m}{\mathrm{s}}^{-1}$ givesAt $R=42000$, one spectral resolution element corresponds to $c/R\simeq 7.14\mathrm{km}{\mathrm{s}}^{-1}$, about $119$ times larger. A well measured centre can move by a small fraction of that width. More photons and more useful spectral lines improve the estimate, but do not remove wavelength calibration errors or stellar activity.

A simple mathematical picture of spectral cross-correlation makes the readout concrete. Define $x=c\log (\lambda /{\lambda }_0)$, with fixed reference wavelength ${\lambda }_0$, and let $d(x)$ be continuum-normalized line absorption. For small velocities, a common Doppler shift approximately translates $d(x)$. A positive absorption mask $m(x)$ then givesIf $d(x)=m(x-v_r)$, the maximum occurs at $u=v_r$; a convention correlating the received observed radiative flux instead produces a dip. A fitted correlation centre is an effective collective line position, not a claim that each individual spectral line has been resolved into a narrower feature. The lamp tracks instrumental drift, while the barycentric velocity correction removes the observer's changing line-of-sight velocity; neither removes changes in the star itself.

Emissivity compares thermal emission from a surface with emission from an ideal blackbody at the same absolute temperature, using the same wavelength and angular averaging convention. For ordinary passive surfaces in the far field at local thermal equilibrium, the ratio lies between zero and one. A blackbody has emissivity one. Unless a spectral or directional qualification is given, the $ϵ$ in $M=ϵ\sigma T^4$ is total hemispherical emissivity: a ratio of emitted radiant exitances, not of all outgoing radiant flux including reflected radiation.^[ref]

Emissivity belongs to a specified surface state, not just a bulk material name. Coating, oxidation and surface finish can change it; wavelength, viewing direction and temperature also matter. A surface that looks white in visible light need not have low infrared emissivity. This distinction is useful when choosing a radiator coating or interpreting a radiation-based temperature measurement: a narrow-band detector needs the appropriate spectral emissivity, and its signal can include reflected surrounding radiation. The grey body and diffuse thermal emitter assumptions are separate simplifications, addressing wavelength and direction respectively.

Empty magnification enlarges an image without revealing additional specimen detail. Adequate angular magnification can make existing detail visible to the eye, but it cannot undo the objective's diffraction blur. For equally bright, mutually incoherent optical point sources and an ideal circular optical pupil, a commonly used lateral Rayleigh criterion in object space iswhere $\lambda$ is the vacuum wavelength, $n$ is the specimen-medium refractive index, and $\alpha$ is the admitted half-angle. For this model, $\lambda =550\mathrm{nm}$ and $\mathrm{NA}=0.25$ give $d_{\mathrm{R}}\simeq 1.342\mu \mathrm{m}$; changing only numerical aperture to $0.50$ gives $0.671\mu \mathrm{m}$. Changing only the eyepiece leaves this object-space scale unchanged. Illumination, optical coherence, optical aberrations and specimen visibility matter in real instruments; this criterion is not a universal threshold.^[ref]

The geometric reason is especially simple. If a fixed recorded image is enlarged by a factor $q>0$, a feature separation $s$ and a blur width $b$ become $qs$ and $qb$. Their ratio remainsThis is the lower-panel construction in the diagram. A change to the microscope objective can change numerical aperture and hence the specimen detail captured; a display zoom changes how large that captured detail looks. Microscope scale calibration, diffraction-limited resolution and angular magnification are three complementary properties to report.

The enantiomeric excess of an enantiomer pair is the magnitude of the difference between their relative mole fractions, $\mathrm{ee}=|x_{+}-x_{-}|$, often reported as a percentage. Here $x_{+}+x_{-}=1$ counts the pair only, excluding solvent; the signs label optical rotation under the specified conditions, not R and S configuration.

In the model above, $x_{+}=0.75$ and $x_{-}=0.25$ give $\mathrm{ee}=0.50=50\%$. Keep the same total pair concentration $c=0.200\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$, path length, solvent, temperature and wavelength. The two contributions areThe $+8^{\circ }$ pure-member reference therefore gives $\alpha /{\alpha }_{\mathrm{pure},+}=+0.50$. This ratio measures signed population imbalance only for a pure pair with additive linear responses under matching conditions. Adding unrelated solutes, changing total concentration, or changing chemical equilibrium can invalidate that inference. For $x_{+}=x_{-}=0.50$, the racemate gives zero net optical rotation by cancellation; each molecule remains chiral. A zero reading alone proves neither a racemate nor molecular achirality.

A wave envelope is a slowly varying wave amplitude multiplying a rapidly oscillating carrier wave. For a complex wavefield $\psi =A(x,t)e^{i(k_{0}x-{\omega }_{c}t)}$, its nonnegative amplitude envelope is $|A|=|\psi |$; the complex $A$ can also encode a slowly varying wave phase. This choice distinguishes the amplitude envelope from a signed modulation factor, such as the cosine factor in a two-tone superposition. A slowly varying wave envelope is useful when the carrier's wavelength and period are short compared with its variation scales. Strong overlap between distinct packets or a broad frequency spectrum can spoil a unique carrier-envelope interpretation.

For an ideal air-gap Fabry–Pérot etalon at normal incidence, the free spectral range is ${\nu }_{\mathrm{FSR}}=c/(2d)$ and the peak full width at half maximum is $\delta \nu ={\nu }_{\mathrm{FSR}}/\mathcal{F}$. To separate a pair of lines spaced by $\Delta \nu$ by at least $M$ peak widths, require $d\mathcal{F}\ge Mc/(2\Delta \nu )$. Putting both lines inside one unambiguous free spectral range additionally requires $d<c/(2\Delta \nu )$, so $\mathcal{F}>M$. In terms of their vacuum wavelengths, $c/\Delta \nu ={\lambda }_1{\lambda }_2/|{\lambda }_2-{\lambda }_1|$.

The strict order condition needs a margin if tails from adjacent periodic peaks are to be negligible. If every neighbouring peak in a scanned two-line periodic spectrum must be separated by $M$ widths, the stronger condition is $\min (\Delta \nu ,{\nu }_{\mathrm{FSR}}-\Delta \nu )\ge M{\nu }_{\mathrm{FSR}}/\mathcal{F}$. Ordinary resolving power alone does not prevent ambiguous repeated orders. A fixed etalon's transmission as a function of wavelength is a periodic-in-frequency filter; it is not automatically a plot of the incident source's spectral lines.

Suppose an extended source fills every position of the entrance aperture with a cone of half-angle ${\theta }_{\mathrm{in}}$, and all these light rays must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-angle ${\theta }_{\mathrm{out}}$, its available etendue bounds the admitted etendue:ConsequentlyThis original derivation combines conservation of etendue with the exact cone integral; the second inequality allows the complete output hemisphere. It bounds capacity and does not prove that a chosen shape attains the bound.

For the $30^{\circ }$ input cone with $n_{\mathrm{in}}=n_{\mathrm{out}}=1$, $C\le 4$: a $10\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\mathrm{mm}$ in diameter. If the output half-angle is restricted to $60^{\circ }$, $C\le 3$ and the receiver diameter must be at least $10/\sqrt{3}\approx 5.7735\mathrm{mm}$. For a receiver medium with model refractive index $n_{\mathrm{out}}=1.5$, the full-hemisphere capacity bound instead gives $C\le 9$; that larger value also depends on transmitting the bundle into that medium.

Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform radiance already specified, even filling the entire output hemisphere givesThe maximum mean irradiance gain remains $\eta C\le 4$. With nonuniform radiance, rejecting a given fraction of geometric etendue need not reject the same fraction of radiant flux.

These bounds use a filled extended-source bundle in passive reversible geometrical optics, comparing the same wavelength and the same radiation channels. An ideal optical point source does not fill that four-dimensional set. Diffraction, optical coherence, or coupling a Gaussian beam into an optical fiber require a physical optics calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, optical gain or fluorescence changes the propagation assumptions.

A Fabry–Pérot etalon fixes mirror separation in a Fabry–Pérot interferometer, selecting wavelengths through its transmission resonances.^[ref]

A Fabry–Pérot interferometer uses repeated specular reflections between two nearly parallel, partially transmitting surfaces. Its transmitted light is the sum of contributions that have made zero, one, two, or more additional round trips. Their complex amplitudes reinforce only near particular wave phases, producing narrow transmission peaks that select wavelength intervals. Scanning the spacing or angle of incidence can build an imaging spectral cube: an image at each selected wavelength.

The surprising feature is that highly reflecting mirrors can together transmit all the incident power at resonance in the ideal symmetric, lossless model. Each individual contribution is weak, but many contributions arrive in phase. Between resonances the same contributions cancel strongly. Increasing mirror reflectivity sharpens the peaks instead of simply suppressing every transmitted signal; finesse quantifies that selectivity. An optical spectrum with nearby spectral lines can therefore produce distinguishable peaks, provided their separation is large enough relative to the instrument's peak width and their orders are identified.

Charles Fabry and Alfred Pérot developed the instrument together; Optica dates its development from 1896 to 1898.^[ref]

At normal incidence, with no optical dispersion and zero round-trip reflection wave phase modulo $2\pi$, the resonance condition in Fabry–Pérot transmission from multiple reflections isHere $\nu$ is frequency and $c$ is the speed of light in vacuum. A change of $\nu$ by one free spectral range advances the round-trip wave phase by $2\pi$. At fixed internal angle of incidence $\theta$, the same nondispersive model gives $c/(2nd\cos \theta )$. Holding the external angle of incidence fixed in a dispersive medium need not hold the internal $\theta$ fixed, because of Snell's law.

For stipulated $n=1$ and $d=10\mathrm{mm}$, the free spectral range is $14.9896229\mathrm{GHz}$ using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. A frequency difference of $0.10\Delta {\nu }_{\mathrm{FSR}}$ is therefore $1.49896229\mathrm{GHz}$. Transmission repeats every free spectral range: one isolated scan cannot distinguish that separation from $1.10\Delta {\nu }_{\mathrm{FSR}}$. Independent knowledge of the optical spectrum, an additional filter, or a changed mirror separation is needed to assign orders.

Equal spacing in frequency does not imply equal spacing in wavelength. At normal incidence the exact adjacent resonance wavelengths obeyFor example, choosing $2nd=10{\lambda }_{10}$ makes the exact gap to order 11 equal to ${\lambda }_{10}/11$, while the high-order estimate gives ${\lambda }_{10}/10$. The approximation is already ten percent larger at order 10. With optical dispersion or frequency-dependent reflection wave phase, obtain local spacings from the total phase difference instead: $\Delta \nu \simeq 2\pi /(d\delta /d\nu )$ when its derivative varies little over one order.

Consider two identical, lossless parallel mirrors with intensity reflectivity $0\le R<1$, separation $d$, and a uniform transparent medium of refractive index $n$. Fix one polarization of an electromagnetic wave and an internal angle of incidence $\theta$ measured from the mirror normal vector. Illuminate with a monochromatic wave having planar wavefronts; the emerging contributions must overlap coherently in the detected direction. Define $T=I_{\mathrm{transmitted}}/I_{\mathrm{incident}}$ using the same incident and exit media. Each mirror transmits a fraction $1-R$ of incident power. Its reflected wave amplitude has magnitude $\sqrt{R}$, so two internal specular reflections multiply the circulating complex amplitude by magnitude $R$, not $R^2$.

Let $\lambda$ be the vacuum wavelength and choose the reflection wave phase convention in which the product of the two internal amplitude reflection coefficients is real and positive. The phase difference between successive transmitted contributions, evaluated on a common outgoing wavefront, isWith other mirror coatings or phase conventions, use the total $\delta =4\pi nd\cos \theta /\lambda +{\varphi }_{\mathrm{rt}}$, where ${\varphi }_{\mathrm{rt}}$ is the sum of the two internal reflection wave phases. The stated resonance positions presume ${\varphi }_{\mathrm{rt}}=0$ modulo $2\pi$; a constant nonzero ${\varphi }_{\mathrm{rt}}$ shifts them, while a frequency-dependent one can also change their spacing.

Why does the cosine appear rather than its reciprocal? Inside the gap, the extra two legs have optical path length $2nd/\cos \theta$. But the later emerging contribution starts $2d\tan \theta$ farther along the mirror. Comparing both on one outgoing wavefront subtracts $2nd\tan \theta \sin \theta$ from that path difference, leaving $2nd\cos \theta$. Equivalently, a plane wave's mirror-normal wavevector component is $k_z=(2\pi n/\lambda )\cos \theta$, so the round-trip wave phase is $2k_{z}d$. The diagram uses the same medium outside and inside to show the common-wavefront subtraction directly; at a refracting exit interface, tangential wavevector matching in Snell's law supplies the same result.

Absorb the common first-pass wave phase into the incident complex amplitude $E_{\mathrm{in}}$. Successive transmitted contributions are $(1-R)E_{\mathrm{in}}$, $(1-R)E_{\mathrm{in}}R{e}^{i\delta }$, and so on. The superposition principle and the convergent geometric series giveThe last step expands the squared complex modulus as $1+R^2-2R\cos \delta$. This transmission profile is often called the Airy transmission function; it is distinct from the Airy disk, a diffraction image of a circular aperture. The University of Washington's lab handout gives the same transmission profile and distinguishes mirror reflectivity from other sources of peak broadening.^[ref]

For an original example, set $R=3/4$. At $\delta =0$, the amplitudes add to $(1/4)/(1-3/4)=1$, so $T=1$. At $\delta =\pi$, they alternate in sign and add to $(1/4)/(1+3/4)=1/7$, so $T=1/49$. At $\delta =\pi /2$, the amplitude is $4/25+3i/25$ and $T=1/25$. Adding the individual intensities instead would give $(1-R{)}^2{\sum }_j{R}^{2j}=(1-R)/(1+R)=1/7$, independent of wave phase; that discards the wave interference responsible for the peaks.

In general the maximum is $T_{\max }=1$ at $\delta =2\pi m$, with integer $m$, and the minimum is $T_{\min }=[(1-R)/(1+R){]}^2$ halfway between peaks. Perfect resonant transmission requires identical lossless mirrors, coherent illumination, and sufficient time to establish the steady field. Absorption, unequal mirror transmission, finite beams that fail to overlap, and fluctuations of $d$ can reduce it. The exactly reflecting case $R=1$ is excluded: no incident power enters through a perfectly reflecting mirror, and the resonant limit $R\to 1^{-}$ is singular.

The Faraday effect is magnetic field-induced circular birefringence: opposite circular polarizations acquire different wave phases, rotating transmitted linear polarization. For a transparent medium with isotropy before applying the magnetic field, in the linear magnetic-response regime at fixed vacuum wavelength ${\lambda }_0$ and temperature $T$,Here $V$ is the Verdet constant, $L$ is the material path length, and $B_{\parallel }$ is the applied magnetic flux density component along the ray. This $\mathbf{B}$ is the external bias, distinct from the oscillating magnetic field of the light itself. Michael Faraday discovered the effect in 1845.^[ref]

Fix the sign before using the formula. On the forward path choose a right-handed Cartesian coordinate system, propagation along $+z$, and measure $\theta$ from $+x$ toward $+y$ for the electric field's oscillation line. Positive $\theta$ is counterclockwise in the $x$-right, $y$-up end view seen from $z>0$, looking toward the source. Then $B_{\parallel }=B_z$ and positive $V$ means positive rotation for positive $B_z$. This convention differs from the clockwise-positive convention in optical rotation. A linear polarization line is unchanged by $180^{\circ }$; a measured orientation determines accumulated rotation only modulo $\pi$, unless continuity selects a branch.

For uniform $B_{\parallel }$, the formula reduces to $\theta =VBL$. Its appeal is that a weak change of refractive index becomes a visible change of orientation. Magnetic reversal measurement of Faraday rotation turns this into an inference of $V$ or of the magnetic field path integral. Faraday rotation on a round trip explains the directional behavior used by an optical isolator. In an astronomical cold plasma, the high-frequency limit instead leads to the rotation measure law $\theta =\mathrm{RM}{\lambda }_0^2$; that wavelength dependence is not a universal law for transparent solids.

For the two-field Keller--Segel model, put $a=D_2$, $b=D_{\rho }$, $C=D_1{f}_0\ge 0$, $c=\kappa -a_0{f}_0^{\prime }$. The upper growth rate is $s_{+}(z)=-[c+(a+b)z]/2+\sqrt{[c+(b-a)z{]}^2+4Cz}/2$. When $C>\max (ac,-bc)$ its unique maximum occurs at positive $z_{\ast }$ satisfying $s_{+}^{\prime }(z_{\ast })=0$. For equal diffusion coefficients $a=b$, this gives $z_{\ast }=(C^2/a^2-c^2)/(4C)$. For $c<0$ and $C\le -bc$, the fastest growth is homogeneous, with infinite wavelength. The finite-domain answer must maximize the growth rate over the permitted Fourier modes. Eliminating chemical dynamics instantaneously generally changes this selected scale.

For a deep inviscid liquid above a gas of negligible inertia, an inverted free surface has dispersion relation ${\omega }^2=q(\sigma q^2/\rho -g)$, $q=|k|$. The Rayleigh-Taylor instability affects $0<q<k_c=\sqrt{\rho g/\sigma }$. A bounded aperture can suppress it if its smallest allowed wavenumber is at least $k_c$. The capillary length is ${\ell }_c=k_c^{-1}$ and the marginal wavelength is $2\pi {\ell }_c$. A pinned one-dimensional span of width $L$ with a half-sine lowest mode instead has $k_{min}=\pi /L$, giving $L\le \pi {\ell }_c$. A numerical maximum container size therefore also requires aperture geometry and contact-line boundary conditions; the marginal wavelength alone is not a universal diameter.