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A bubble-chamber track is a line of growing vapor bubbles seeded by ionization along a charged particle path in a superheated liquid. The bubbles scatter illumination and can be photographed from multiple directions to reconstruct the path in three dimensions.

In a known magnetic field, curvature constrains the particle's momentum and the sign of its electric charge; increasing ionization toward a stopping point can add information about energy loss. Bubble size and photographic blur set the visible trail width, so that width is not the particle's physical diameter. Neutral particles can remain invisible until an interaction produces charged secondaries.

For an entire function $f$, let $F(z)=f(z)z^{-\alpha }$ with noninteger real $\alpha$, and choose the branch of the complex logarithm with $\varphi <\arg z<\varphi +2\pi$. On the cut ray $z=r{e}^{i\varphi }$, the bank whose argument approaches $\varphi$ has value $F_{+}=f(r{e}^{i\varphi })r^{-\alpha }{e}^{-i\alpha \varphi }$, while the other bank has argument approaching $\varphi +2\pi$ and value $F_{-}=e^{-2\pi i\alpha }{F}_{+}$. The signed jump is therefore $(1-e^{-2\pi i\alpha })F_{+}$ when defined as $F_{+}-F_{-}$. Stating the branch and order of subtraction removes the sign ambiguity. A keyhole contour uses this phase relation to convert its two bank integrals into a constant multiple of one real or radial integral.

The effective electric charge associated with electric polarization is bound charge. For polarization density $\mathbf{P}$, its volume charge density is ${\rho }_{\mathrm{b}}=-\nabla \cdot \mathbf{P}$. At a boundary with unpolarized surroundings, its surface charge density is ${\sigma }_{\mathrm{b}}=\mathbf{P}\cdot \mathbf{n}$, where $\mathbf{n}$ is the outward unit normal vector.

For a real shell potential $V(r)=V_0\delta (r-R)$, define $g=2m{V}_{0}R/{\hslash }^2$ and $\mathrm{sinc}u=\sin u/u$, with value one at zero. Angular integration of its three-dimensional Fourier transform gives $4\pi V_0{R}^2\mathrm{sinc}(qR)$, where the momentum transfer has wave-number magnitude $q=2k\sin (\theta /2)$. Substitution in the first Born approximation proves the displayed scattering amplitude. It is real. Its squared modulus is a nonzero order-$V_0^2$ differential scattering cross-section, whose optical theorem counterpart requires the order-$V_0^2$ imaginary part of the forward amplitude from the next Born term. Thus squaring the first-order amplitude does not supply a consistently unitary second-order amplitude.

For $V(r)=A{e}^{-\kappa r}/r$, with real $A$ and $\kappa >0$, the first Born approximation gives the scattering amplitudeIndeed, orienting the polar axis along the momentum transfer gives the angular integral $4\pi \sin (qr)/(qr)$, and the remaining integral is ${\int }_0^{\infty }{e}^{-\kappa r}\sin (qr)dr=q/(q^2+{\kappa }^2)$. The differential scattering cross-section is $|f_B(q){|}^2$.

The limit $\kappa \to 0^{+}$ at fixed nonzero $q$ produces the Rutherford differential cross-section for a Coulomb potential. The unscreened interaction has long range, so this limiting calculation by itself does not establish the validity of a naive Born approximation to the full Rutherford scattering state. The exact Rutherford scattering amplitude has an additional phase, while its modulus agrees with the Rutherford differential cross-section.

In On the Constitution of Atoms and Molecules, Part I (1913), pp. 8–9, Niels Bohr interprets existing hydrogen discharge and astronomical spectra. Electrical discharge in a vacuum tube removes electrons; recombination supplies spectral lines. This is a theoretical interpretation, without a reported spectrograph of his own.^[ref]

An actual optical obstacle appears in Theodore Lyman's measurements: hydrogen discharge light entered an evacuated spectrograph directly through a slit, avoiding a transmitting window. A concave reflection diffraction grating dispersed the light onto photographic film. Pumping preserved the vacuum; a second slit supplied a known spectral displacement for wavelength calibration. Gas contamination and vacuum leaks limited the measurement. This apparatus resolved ultraviolet spectral lines; it did not image orbits.^[ref]

For a separate original calibration exercise, suppose an idealized spectrograph returns synthetic vacuum line positions $x_n={\lambda }_{n\to 2}+b$ for $n=3,4,5,6$, with chosen additive offset $b=0.200\mathrm{nm}$. Assume its dispersion scale is already known and only this offset is uncertain. Compare the finite-mass prediction with the fixed-nucleus prediction ${\lambda }_n^{(\infty )}=(\mu /m_e){\lambda }_{n\to 2}$. Calibrate each model using only the $3\to 2$ line: $\hat{b}=x_3-{\lambda }_3^{\mathrm{model}}$. Then hold that offset fixed and test the other lines with residualsFor the finite-mass model every $d_n=0$. The fixed-nucleus model has $\hat{b}=0.55733\mathrm{nm}$ and residuals $d_4=-0.09264\mathrm{nm}$, $d_5=-0.12100\mathrm{nm}$, $d_6=-0.13400\mathrm{nm}$. Algebraically $d_n=(1-\mu /m_e)({\lambda }_n-{\lambda }_3)$: one additive calibration cannot absorb a multiplicative change of the wavelengths. Allowing an unknown multiplicative dispersion factor as well would make the two models degenerate in this exercise, so independent scale calibration matters.

For emission take integers $n_i>n_f\ge 1$. Applying conservation of energy to the model's internal energy difference, while neglecting the much smaller whole-atom photon recoil, givesThus Johannes Robert Rydberg's Rydberg formula is a difference of two energy levels expressed as a spectroscopic wavenumber. The same constants determine every line; there is no new parameter for each electronic transition. With the chosen inputs, $R_H=1.096775834\times 10^7{\mathrm{m}}^{-1}$. All wavelengths below are calculated in vacuum, not air, and are rounded model predictions rather than historical measurements.

The conventional names commemorate Theodore Lyman and Johann Jakob Balmer. Fixing $n_f=1$ produces the ultraviolet Lyman series, beginning with the Lyman-alpha line; fixing $n_f=2$ produces the Balmer series. As $n_i\to \infty$, the wavelength approaches $n_f^2/R_H$: $91.1763\mathrm{nm}$ at the Lyman limit and $364.705\mathrm{nm}$ for the Balmer series. Crowding follows directly from $E_{n+1}-E_n\sim 2B_H/n^3$. Discreteness therefore coexists with a series limit.

Correct line positions do not give wave intensities, lifetimes or electric-dipole selection rules for hydrogen. Those require a theory of the wavefunctions and their coupling to electromagnetic radiation. The circular model also omits spin, relativistic corrections and the Lamb shift. Its successful gross energy levels do not make its assumed orbits modern atomic orbitals.

Consider a single electron of mass $m_e$ and a point atomic nucleus of mass $M$ and positive electric charge $Ze$, with $e>0$ the elementary charge. Use nonrelativistic classical mechanics, omit external fields and interactions with other electrons, and writeHere $\mu$ is the Coulomb two-body reduced mass, ${ϵ}_0$ the vacuum permittivity, and $h$ the Planck constant. The separation $r$ and relative speed $v$ describe the two-particle relative coordinate; neither body is silently pinned in place. The electron's own radius about the centre of mass is $Mr/(M+m_e)$.

For the assumed circular orbit, Coulomb's law provides the required centripetal acceleration, and the imposed orbital angular momentum condition isSubstitute $v=n\hslash /(\mu r)$ into the first equation and cancel $r^{-2}$ to obtainThe kinetic energy is $K=\mu v^2/2=\kappa /(2r)$ and the Coulomb potential energy is $U=-\kappa /r$, with zero energy at infinite separation. HenceThis derives the energy sequence from the stated circular model; the modern Schrödinger equation derivation is the distinct series-termination quantization of the Coulomb radial equation. The radius above scales as $n^2/Z$, whereas the binding energy scales as $\mu Z^2/n^2$. The Bohr radius article supplies the corresponding Coulomb potential length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the speed of light.

For an original numerical example choose $Z=1$, $M=1.67262192369\times 10^{-27}\mathrm{kg}$, $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, ${ϵ}_0=8.8541878128\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. Treat these as stipulated constants for the calculation. They give $\mu /m_e=0.99945568$, $r_1=0.0529465\mathrm{nm}$, $E_1=-13.598287\mathrm{eV}$ and $v_1/c\simeq 0.00730$, where $\mathrm{eV}$ denotes an electronvolt. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state ionization energy is $-E_1$.

Taking $M\to \infty$ gives $\mu \to m_e$. A finite nuclear mass reduces every binding energy by $\mu /m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual spectral lines. Replacing the atomic nucleus by a heavier isotope changes the same factor, providing another check of the model.

The Bloch Hamiltonian acts on the cell-periodic part of a Bloch state at fixed wavevector. If ${\psi }_{n\mathbf{k}}(\mathbf{r})=e^{i\mathbf{k}\cdot \mathbf{r}}{u}_{n\mathbf{k}}(\mathbf{r})$, defineFor a scalar periodic potential, its kinetic term is $(-i\hslash \nabla +\hslash \mathbf{k}{)}^2/(2m)$ on cell-periodic functions.

Solving this family of cell problems produces electronic band structure. In a tight-binding description, the Bloch Hamiltonian is a matrix on the orbitals within a primitive cell, not the full infinite real-space matrix. Basis choices can change its entries without changing the physical band energies. The distinction between the full wavefunction and its periodic factor also matters when computing matrix elements and geometric phases.

Invariance under spatial inversion about a specified origin.

A crystal with $N$ primitive unit cells has $N$ distinct wavevectors in one Brillouin zone. For a rectangular periodic sample, $k_i=2\pi l_i/L_i$, and the product of the numbers of allowed values along the three directions equals the number of cells. Each orbital-derived energy band therefore holds $N$ states per spin orientation, or $2N$ with twofold spin degeneracy. This means two states per primitive cell per spin-degenerate band, independently of how many atoms the cell contains.

William Shockley's proposal used transport through the bulk of a semiconductor, rather than coupling between two surface points. The base must be thin enough for injected minority carriers to reach the collector. Two separately wired diodes do not supply that shared transport region. The proposal was a design principle; producing a working device required further control of material properties.^[ref]

Take a wire segment on the $y$ axis, with source positions ${\mathbf{r}}^{\prime }=(0,s,0)$ for $s_1\le s\le s_2$, and positive electric current along $+y$. Observe at $\mathbf{r}=(a,0,0)$, where $a>0$ is the perpendicular distance to the wire's supporting line. Choose $x$ rightward and $y$ upward, so $+z$ points out of the page. ThenThus the magnetic field points into the page by the right-hand rule. The Biot-Savart law and an elementary antiderivative giveFor an original worked example, choose $s_1=-0.12\mathrm{m}$, $s_2=0.12\mathrm{m}$, $a=0.05\mathrm{m}$, and $I=3\mathrm{A}$. Use ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$ for these rounded calculations. Each endpoint contributes $0.12/0.13=12/13$, soThese are calculated inputs and output, not laboratory data. The two endpoint angles measured from the perpendicular each have sine $12/13$; this recovers the equivalent endpoint-angle formula without depending on a diagram's angle-sign convention.

For a segment centered on the perpendicular foot, $s_1=-L$, $s_2=L$. Its magnetic field magnitude isThe limit $L/a\to \infty$ gives $B_{\infty }={\mu }_{0}I/(2\pi a)$, consistent with Ampère's circuital law for an infinite straight wire. In the example, $B_{\infty }=12\mu \mathrm{T}$ and $B/B_{\infty }=12/13$: the infinite-wire value overestimates the segment value by $1/12$, or about $8.33\%$ of the segment value. For $a\gg L$, the isolated segment contribution instead behaves as $B\sim {\mu }_{0}I(2L)/(4\pi a^2)$. The complete localized electrical circuit has a different far-field behavior: return-path contributions cancel this leading segment term, leaving a magnetic dipole field when its magnetic dipole moment is nonzero. Segment and whole-circuit limits must be distinguished.

The SI units check directly: ${\mu }_{0}I/a$ has units $(\mathrm{T}m/A)\mathrm{A}/\mathrm{m}=\mathrm{T}$, the tesla. The apparent divergence as $a\to 0$ is a failure of the thin-wire model near the conductor, not a prediction of infinite magnetic field inside a real wire. This calculation gives only this segment's contribution; include the return wires before comparing with an apparatus.

A current element is the oriented infinitesimal wire contribution $Id{\mathbf{\ell }}^{\prime }$ inside the Biot-Savart law. For uniform axial current density across wire cross-sectional area $A$, $I=JA$ and $d^3{r}^{\prime }=Ad{\ell }^{\prime }$ imply $\mathbf{J}{d}^3{r}^{\prime }=Id{\mathbf{\ell }}^{\prime }$. This reduces a volume integral to a line integral in the thin-wire approximation. It is an integration contribution, not an isolated steady source. Conventional current runs opposite the drift of negatively charged electrons.

For a circular arc of radius $R$ carrying clockwise electric current $I$ and subtending angle $\theta$, with two tangent semi-infinite leads extending on the crossing side of the arc, the axial magnetic field at height $h$ above the centre isThe crossing is insulated. The positive term is from the two leads and the negative term from the arc, with $+z$ out of the clockwise-current plane. The Biot-Savart law yields the lead integral ${\int }_0^{\infty }R(R^2+h^2+s^2{)}^{-3/2}ds=R/(R^2+h^2)$. Cancellation of the axial component gives the title formula, expanding to $\theta =2+h^2/R^2+O(h^4/R^4)$ in radians. This is cancellation of one component away from the plane, not necessarily the entire magnetic field.