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For an illustrative $n=2\to 1s$ emission calculation, retain only the first-order shifts above and neglect the quadratic shift of the hydrogen atom ground state. The unperturbed Lyman-alpha line frequency is ${\nu }_0=(E_2-E_1)/h$. The possible offsets areThe outer pair originates in the $p_0$ parts of $|{\psi }_{\pm }⟩$ and has transition electric dipole moments along the applied electric field, conventionally called $\pi$ components. The central pair originates in $p_{\pm 1}$ and has transverse rotating transition electric dipole moments, conventionally called ${\sigma }^{\pm }$ components. Because electromagnetic radiation is transverse, the $\pi$ components disappear when viewed along the electric field. Viewed from the side, the $\pi$ and $\sigma$ components have polarization of an electromagnetic wave respectively parallel and perpendicular to the applied electric field. Line intensities additionally require the initial quantum state populations and the collection geometry.

An Onsager reciprocal relation connects cross-couplings between near-equilibrium transport processes. Choose thermodynamic forces $X_j$ and fluxes $J_i$ so that their products give the entropy production, and write $J_i={\sum }_j{L}_{ij}{X}_j$. With microscopic reversibility, no imposed time-reversal-breaking field and variables with the appropriate matching time-reversal parity, the coefficients obey $L_{ij}=L_{ji}$. The choice of conjugate force and flux matters: arbitrary rescaling of just one coefficient need not preserve its numerical symmetry.

For coupled thermal and electrical transport, a thermal driving force can induce charge flow while an electrical driving force induces heat flow. Reciprocity relates those cross effects, reducing the number of independent transport measurements. Lars Onsager's 1931 paper develops this reasoning from microscopic reversibility and distinguishes it from thermodynamic positivity alone.^[ref] External magnetic fields and odd time-reversal variables require a generalized parity-sensitive relation rather than an unconditional symmetric matrix.

Causality constrains the real and imaginary parts of a frequency-dependent linear response. For a dielectric with ${ϵ}_r(\omega )\to 1$ at large frequency, the $e^{-i\omega t}$ convention givesThe integral is a Cauchy principal value. Absorption and optical dispersion therefore cannot be specified independently.

For an integrable linear response kernel $K$ and input $V$, the convolution theorem gives $\hat{I}(\omega )=\hat{K}(\omega )\hat{V}(\omega )$. The integrated output is therefore zero for every integrable input if and only if $\hat{K}(0)=\int K(t)dt=0$. For a sum of causal damped sines, $K(t)=\Theta (t){\sum }_j{a}_j{e}^{-{\gamma }_{j}t}\sin (p_{j}t)$, the condition is ${\sum }_j{a}_j{p}_j/({\gamma }_j^2+p_j^2)=0$. If all $a_j$ are positive, nonzero terms of both signs are needed; a term with $p_j=0$ vanishes identically.

A Kundt's tube turns an invisible standing wave into a measurable spatial pattern. A driven air column supports a longitudinal wave; measuring its frequency and the spacing of repeated nodes of a standing wave gives the speed of sound. The same drive can compare the wavelengths in a gas and a vibrating solid. August Kundt's 1866 paper introduced acoustic dust figures; its footnote explicitly counts successive nodes of a standing wave as half a wavelength.^[ref]

The attractive feature is the separation of measurements: a clock or frequency counter measures repetition in time, while a ruler measures repetition in space. Their product supplies a propagation speed. The pressure pattern and the fluid-motion pattern are different, however; identifying which one a detector responds to prevents a quarter-wavelength mistake.

The local fluid velocity perturbation in an acoustic wave. It describes the oscillating material, whereas the speed of sound describes propagation of the disturbance. A fluid parcel need not travel one wavelength during a wave period.

Sound pressure is the deviation of pressure from the undisturbed background. In linear acoustics, it is small compared with the background pressure; a microphone measures a response to this varying pressure, rather than directly recording fluid displacement.

How can the relative abundance of tracks and showers distinguish hypotheses? Consider an original teaching model, with chosen inputs throughout. Condition on a sample of $N$ selected events, each assigned exactly one of two labels: track or shower. Multitrack events and unclassified events lie outside this model. A shower label need not identify an electron; background can imitate it. Conversely, a true electron interaction need not receive a shower label.

Let $b$ be the background fraction among selected events. Among the selected signal events, let $f$ be the fraction whose associated charged lepton is an electron, with the remainder associated with a muon. These are fractions after event selection, rather than fractions of the incident neutrino particle flux. Define the conditional probabilitiesHere $a$ and $r$ describe classification within the selected sample. They are not measured historical detection efficiencies. The law of total probability gives the overall shower probabilityThis formula is the bridge between an interaction hypothesis and a visible label. It keeps beam contamination, background and mistaken classification conceptually distinct: genuine electron neutrino interactions belong in $f$, while non-signal events belong in $b$.

Assume independent random variables for the event labels, all with this same $q$. If $K$ counts shower labels, every specified pattern with $k$ showers has probability $q^k(1-q{)}^{N-k}$. Counting those patterns with a binomial coefficient yields the binomial distributionThe track count is $N-K$, so its expected value is $N(1-q)$. At fixed $N$, the two counts cannot fluctuate independently. The variance above measures sampling fluctuations under the stipulated model, not uncertainty in $a$, $r$, $b$ or $q_b$.

Choose $N=40$, $a=0.90$, $r=0.05$, $b=0.10$ and $q_b=0.40$. Compare a signal with only muon-associated interactions ($f=0$) against one with equal selected electron- and muon-associated interactions ($f=1/2$). These give $q=0.085$ and $q=0.4675$, respectively. Their expected shower counts are $3.4$ and $18.7$; their expected track counts are $36.6$ and $21.3$. A background-only model with the specified $q_b$ gives $16$ showers and $24$ tracks on average. These are expectations, so they need not be integers.

For a hypothetical observation $K=4$, the likelihood ratio of the two signal models isThe binomial coefficient cancels because the same hypothetical sample is used in both likelihoods. This ratio is not the probability that either hypothesis is true. A separate original check gives $\mathrm{Pr}(K\le 5)\simeq 0.880$, $4.57\times 10^{-6}$ and $1.44\times 10^{-4}$ for the muon-associated, equal-mixture and specified background-only models, respectively.

Why are the controls essential? If the background-only model may choose any $q_b$, taking $q_b=K/N$ can fit the observed shower fraction. Counts alone then fail to distinguish it from a signal model. Independent background measurements and calibrated particle identification constrain that freedom. Changing the decay path additionally tests where the candidate-producing particles originate. The power of the experiment comes from combining apparatus interventions with event classification; the small binomial distribution model isolates only the latter part of that reasoning.

Three Miller indices $(hkl)$ specify a lattice plane orientation in chosen unit cell axes. Choose a parallel plane away from the origin, express its intercepts in axis units, take reciprocals, and clear fractions. An infinite intercept gives index zero; a negative intercept gives a negative index, written with an overbar.^[ref]

For example, intercepts $(2a,3b,\infty )$ have reciprocal ratios $(1/2,1/3,0)$, so the orientation is $(320)$ after multiplying by six. Relative to the same axes, intercepts $(a,-b,\infty )$ give $(1\bar{1}0)$. Choosing a different parallel member changes the common intercept scale, not the orientation.

Primitive axes require relatively prime Miller indices for complete families. Centered conventional unit cells need not, as the FCC lattice $(200)$ example shows. Reflection labels can also retain higher diffraction orders. Use the reciprocal lattice for the normal vector with oblique axes; the direct vector $h{a}_1+k{a}_2+l{a}_3$ need not be perpendicular.

The interplanar spacing $d$ is the perpendicular distance between parallel planes in a specified family. Measuring along an oblique line gives a larger distance, so the direction of measurement matters.

There is a short reciprocal lattice derivation. With the angular convention $a_i\cdot b_j=2\pi {\delta }_{ij}$, put $G=h{b}_1+k{b}_2+l{b}_3$. Then $G\cdot r=2\pi m$ is the same plane equation as $hx+ky+lz=m$. Thus $G$ is a normal vector. Moving between consecutive sheets along this normal changes the dot product by $2\pi$, givingFor perpendicular cubic axes of length $a$, this becomesFor example, choose a primitive cubic lattice with $a=0.600\mathrm{nm}$. Its $(210)$ lattice planes have $d_{210}=0.600/\sqrt{5}\simeq 0.268328\mathrm{nm}$. This is a chosen geometry, not an experimental value. The distance from a point to a plane gives the same result: the planes $2X+Y=0$ and $2X+Y=a$ in ordinary Cartesian length coordinates are separated by $a/\sqrt{5}$.

Specify whether $d_{hkl}$ denotes spacing between successive occupied lattice planes or the formal spacing assigned to a reflection label. In a primitive cubic lattice, $(200)$ has formal spacing $a/2$ but every other sheet contains no lattice points; its reflection is the second diffraction order of $(100)$. In a face-centered cubic lattice with conventional cubic axes, the $(200)$ sheets at half-integer $X/a$ do contain lattice points. Here $a/2$ is the actual spacing of the complete family perpendicular to that cube edge. The Primitive cubic Bragg peak positions example explains how to avoid counting a reflection twice.

A north-east lattice path is a lattice path whose allowed steps are $E=(1,0)$ and $N=(0,1)$. From $(0,0)$ to $(a,b)$, with nonnegative integers $a,b$, it has $a$ east steps and $b$ north steps. The number isThere is one zero-step north-east lattice path when $a=b=0$.^[ref]

Seven positions, thirty-five paths. To reach $(4,3)$, choose the three positions occupied by north steps in a seven-symbol word. For example, the subset $\{2,4,7\}$ of $\{1,2,3,4,5,6,7\}$ gives ENENEEN: putting $N$ exactly in those positions and $E$ elsewhere reconstructs the north-east lattice path. Reading the north-step positions recovers the subset, so these two operations give a bijection. HenceThe numerator counts ordered choices of three different positions; each subset occurs in $3!=6$ orders. Dividing by this factorial counts each north-east lattice path once. A binary string recording $N$ as $1$ and $E$ as $0$ gives the same example as 0101001.

This interpretation is useful when arrangements have fixed totals but flexible order: a subset, a binary string, and a north-east lattice path become different views of one selection. It also separates endpoint restrictions from restrictions along the journey. Requiring the lattice path to avoid specified points changes the count; the unrestricted binomial coefficient no longer suffices, but the local north-east lattice path recurrence still suggests how to count allowed arrivals.

Doppler-free saturation spectroscopy pairs a strong pump with an opposed weak probe. Shared populations produce a narrow Lamb dip within the Doppler broadening of a spectral line.

Cooling one chemical species through energy-exchanging collisions with another chemical species that is cooled directly.