Take a static uniform fluid of mass density $\rho >0$, constant gas pressure, and background magnetic field ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$ with $B_0>0$. Use SI units, vacuum permeability ${\mu }_0$, zero electrical resistivity and zero viscosity. Write $\mathbf{B}={\mathbf{B}}_0+{\mathbf{b}}_{\perp }(z,t)$ and $\mathbf{u}={\mathbf{u}}_{\perp }(z,t)$, with both disturbances perpendicular to ${\mathbf{e}}_z$. Retain only linear order in their amplitudes. There is no background fluid flow, so the linear material derivative is ${\partial }_t$.

The continuity equation has $\nabla \cdot {\mathbf{u}}_{\perp }=0$, hence ${\partial }_t\delta \rho =0$; choosing no initial mass density fluctuation gives $\delta \rho =0$. For this shear mode the gas pressure fluctuation is also zero. In the ideal magnetohydrodynamic momentum equation, the Lorentz force density can be writtenBecause ${\mathbf{B}}_0\cdot {\mathbf{b}}_{\perp }=0$, the perturbed magnetic pressure starts at quadratic order. The surviving transverse magnetic tension gives the first equation below. Expanding the ideal magnetohydrodynamic induction equation gives the second:Differentiate the first in time and substitute the second. Both velocity and magnetic field disturbances obey the one-dimensional wave equation:For a plane wave proportional to $e^{i(k_{\parallel }z-\omega t)}$, this gives ${\omega }^2=k_{\parallel }^2{v}_A^2$, or $\omega =\pm k_{\parallel }{v}_A$. The magnetic and velocity amplitudes satisfy ${\mathbf{b}}_{\perp }=-(B_0{k}_{\parallel }/\omega ){\mathbf{u}}_{\perp }$: they have opposite signs for propagation along $+{\mathbf{e}}_z$ and equal signs for propagation along $-{\mathbf{e}}_z$. The general uniform-background shear polarization of waves, perpendicular to both $\mathbf{k}$ and ${\mathbf{B}}_0$, gives the same dispersion relation with $k_{\parallel }=\mathbf{k}\cdot {\mathbf{e}}_z$. Its group velocity is $\pm v_A{\mathbf{e}}_z$, whereas its phase velocity normal to an oblique wavefront has magnitude $v_A|k_{\parallel }|/|\mathbf{k}|$.

U. Essmann and H. Träuble reported triangular vortex arrays, including defects, using electron microscopy on lead-indium and niobium samples in the remanent state.^[ref] Such a prepared state need not display the equilibrium array at a uniform applied field. Images of regular positions and a proof of unrestricted vortex motion would answer different questions.

A measure of spatial nonuniformity in the order parameter, with averages taken over one lattice cell. Set $X=|\psi {|}^2$. ThenEquality requires uniform $|\psi {|}^2$, so a nonconstant core pattern costs more in the quartic term than a uniform amplitude with the same mean squared amplitude. This identity alone does not compare two different vortex arrays: their admissible profiles must also satisfy the field equations.

Assume a triangular array of straight, singly quantized Abrikosov vortices with nearest-neighbor distance $a$. Here $\overline{B}>0$ is the magnitude of the spatially averaged internal magnetic field, and ${\Phi }_0=h/(2e)$ is the superconducting flux quantum; $e>0$ is the elementary charge. Two lattice vectors of length $a$ meeting at $60^{\circ }$ span area $A_c=\sqrt{3}{a}^2/2$. One flux quantum per cell givesThus multiplying $\overline{B}$ by five decreases $a$ by $\sqrt{5}$; it does not multiply the flux of each vortex by five. Using the exact SI Planck constant and elementary charge, ${\Phi }_0\simeq 2.06783\times 10^{-15}\mathrm{Wb}$.^[ref]

For chosen internal fields $0.20\mathrm{T}$ and $1.0\mathrm{T}$, the spacings are approximately $109\mathrm{nm}$ and $48.9\mathrm{nm}$. These are geometric model calculations, not measurements from a particular sample. The local magnetic field varies within a cell; replacing $\overline{B}$ by the field at a core would not give this counting relation.

At leading order, for final fermions much lighter than the Z boson, the partial decay width is proportional to $N_c(|g_L{|}^2+|g_R{|}^2)$. Here $N_c$ is the final-state color multiplicity in a decay width, and $g_L=T_{3L}-Q{\sin }^2{\theta }_W$, $g_R=T_{3R}-Q{\sin }^2{\theta }_W$ are the dimensionless left- and right-chirality couplings. The common normalization cancels in ratios.

For an electron, $g_L=-1/2+{\sin }^2{\theta }_W$, $g_R={\sin }^2{\theta }_W$. For one active neutrino flavour, $g_L=1/2$, $g_R=0$. ThereforeThere is no extra flavour factor for a single specified neutrino species. Summing the three active species produces three times that invisible partial width, within the same massless approximation.

Max von Laue proposed crystal X-ray diffraction and wrote the theoretical section; Walther Friedrich and Paul Knipping performed the experiments.^[ref]

An X-ray tube, lead apertures and shielding supplied a narrow beam through a crystal on a goniometer. Photographic plates recorded the direct beam and separate spots. Copper(II) sulfate trials preceded oriented zinc blende. Optical alignment was checked with X-rays. Removing the crystal removed the extra spots; grinding destroyed their ordered arrangement; translation preserved it, while tilt changed it. Rotation about the beam rotated the pattern.^[ref]

Spots and orientation dependence supported wave interference and three-dimensional crystal periodicity. They determined neither all atomic positions nor an absolute wavelength scale. The authors left open whether the selected radiation originated as crystal fluorescence or was already in the incident beam. This differs from the modern Laue method.^[ref]

The 1914 Nobel Prize in Physics went to Max von Laue; the discovery's experimental account remains jointly credited to Walther Friedrich and Paul Knipping.^[ref] The subsequent inference of particular crystal structures is developed under X-ray crystallography.

In the Laue method, a fixed single crystal structure receives a collimated beam containing many wavelengths. Each reciprocal lattice vector can select its own wavelength; waves with different wavelengths need not interfere with one another. For $\mathbf{G}\ne 0$, the elastic scattering constraint gives the particularly useful testA candidate reflection requires ${\lambda }_{\mathbf{G}}>0$ within the source spectrum and an outgoing ray that reaches the detector. Varying wavelength changes the radius and center of the Ewald sphere while leaving the reciprocal lattice fixed. A monochromatic wave instead fixes the radius: rotating the crystal structure or varying the incident wavelength in a scan finds reflections sequentially. A white-beam photograph integrates many spectral selections in one exposure; it is not an angular scan at one known wavelength.

The Laue method is useful for orienting a single crystal structure: tilting it changes which reciprocal lattice vectors meet the spectral and detector conditions. Absolute scale has an ambiguity if the wavelengths are unknown. Scaling all direct lattice translations by $c>0$ and all selected wavelengths by $c$ leaves every outgoing direction unchanged. Independently known spectral information or length calibration is needed to remove this ambiguity.

For incident and outgoing unit propagation vectors ${\mathbf{s}}_{\mathrm{in}}$ and ${\mathbf{s}}_{\mathrm{out}}$, wavelength $\lambda >0$, and primitive lattice translation vectors $a_1,a_2,a_3$, the Laue equations areAll three must hold for the same wavelength and outgoing direction. The integers label a reciprocal lattice vector; they need not be coprime. For wavevectors ${\mathbf{k}}_{\mathrm{in},\mathrm{out}}=(2\pi /\lambda ){\mathbf{s}}_{\mathrm{in},\mathrm{out}}$, equivalentlyThis uses angular wavenumbers; the crystallographic convention without $2\pi$ instead uses $\mathbf{s}/\lambda$.^[ref]

For a short proof, translate a scatterer through $a_i$. Its incident wave phase changes by ${\mathbf{k}}_{\mathrm{in}}\cdot a_i$ and propagation to the same distant observation direction contributes $-{\mathbf{k}}_{\mathrm{out}}\cdot a_i$. The relative wave phase factor is therefore $\exp (-i\mathbf{G}\cdot a_i)$. It equals one for all lattice translations exactly when $\mathbf{G}$ belongs to the reciprocal lattice. A finite crystal lattice structure factor broadens the ideal maxima; the crystal basis structure factor can cancel them. Geometric phase matching is necessary for a sharp reflection, but does not guarantee a visible spot.

For elastic scattering, the additional constraint isIt is an Ewald sphere centered at $-{\mathbf{k}}_{\mathrm{in}}$ in the space of transfer vectors $\mathbf{G}$. The parent elastic Bragg scattering condition uses $q={\mathbf{k}}_{\mathrm{in}}-{\mathbf{k}}_{\mathrm{out}}=-\mathbf{G}$, so its displayed condition has the opposite sign. Both give Bragg's law for the same lattice planes.

Powder X-ray diffraction measures X-ray diffraction from many crystal grains with different orientations. For each allowed set of lattice spacings, suitable grains satisfy Bragg's law and produce a cone of diffracted directions, seen as a ring on an area detector or a peak in an angular scan. Several peak positions and systematic absences can identify a crystal structure even when a sample no longer retains one single-crystal orientation. Preferred orientation, strain and phase coexistence can complicate interpretation.

To isolate conversion losses, suppose a converter receives 100 blue photons at ${\lambda }_b=450\mathrm{nm}$, absorbs a fraction $a=0.60$, and has fluorescence quantum yield $\Phi =0.90$: 0.90 converted photons per absorbed blue photon. Stipulate that every converted photon has ${\lambda }_c=560\mathrm{nm}$, that unabsorbed blue is transmitted, and that both outputs escape without further losses. The output is then 40 blue and 54 converted photons. The total photon yield is $0.94$, while the retained optical energy fraction isThe lower energy of each converted photon matters even if $\Phi =1$. Combining this stipulated converter with the preceding chosen light-emitting diode gives $P_{\mathrm{after}}=16.5430\mathrm{mW}$ and an overall optical/electrical power ratio $0.258485$. These chosen narrow lines and photon counts illustrate conversion accounting; they do not establish a white appearance, color quality or luminous efficacy. Those require the full spectrum and a model of human visual sensitivity.

Adding a spatially constant Hermitian matrix $V=dI+\mathbf{b}\cdot \mathbf{\sigma }$ to an isotropic Weyl Hamiltonian preserves each momentum sector. By the Hermitian two-by-two matrix in the Pauli basis decomposition, the exact energies are $d\pm |\hslash v\mathbf{k}+\mathbf{b}|$. The touching point moves to ${\mathbf{k}}_{\ast }=-\mathbf{b}/(\hslash v)$ at energy $d$, without opening a gap in three-dimensional momentum space.

For a fixed $k_z=0$ slice, however,This slice has minimum branch separation $2|b_z|$ if $b_z\ne 0$, because it misses the shifted node. A one-dimensional cut can likewise look gapped while a crossing exists elsewhere. No nonzero two-by-two matrix anticommutes with all three Pauli matrices, which is another way to see why an independent constant mass matrix is unavailable for this isolated three-dimensional two-component model.

Intermodal dispersion is the temporal separation of components in different propagating normal modes because their group velocities differ. A pulse exciting several modes can therefore broaden even when its individual modal components have negligible bandwidth. It is distinct from group velocity dispersion within a single mode.

Neglect final-state masses and QCD corrections. Each of the three leptonic W boson decay channels contributes one common unit of width. Each open quark channel contributes $3|V_{ij}{|}^2$, with the factor three counting final colours. The top quark is kinematically inaccessible in an on-shell W decay because $m_t>m_W$. Thus $B(W\to \ell \nu )=1/D\simeq 1/9$ and $B(W^{-}\to b\bar{u},b\bar{c})=3(|V_{ub}{|}^2+|V_{cb}{|}^2)/D$. The large $V_{tb}$ does not open a forbidden top-quark channel.

An experiment of size $L$ stops the renormalization-group flow at a finite ${\ell }_{\max }$. In the leading scaled model take the original initial conditions $Y_0=0.06$ and $X_0=-0.10,-0.06,-0.03$. The invariant is respectively $0.0064,0,-0.0027$. The first approaches $X=-0.08$, hence $K=(2.08)/\pi$; the second follows the critical separatrix; the third eventually leaves the small-vortex fugacity regime. These are chosen model trajectories, with no fitted experimental parameters.

On the critical separatrix, $X=-Y$ implies $Y^{\prime }=-Y^2$, henceEven a critical finite sample therefore has a phase stiffness above its asymptotic jump value, with a slow logarithmic correction in $L$. Slightly above the transition, a sample with $L\ll \xi$ can resemble the algebraic phase before the flow reaches unbound vortices. A trap additionally varies the local density and phase stiffness; neither its whole-cloud coherence nor a finite observation window automatically measures the homogeneous thermodynamic limit.

A dimensionless statistical weight per microscopic phase vortex core. In a dilute convention it is $y_0\simeq \exp [-E_c/(k_{B}T)]$, with core energy $E_c$ and a specified microscopic integration measure. Its numerical normalization fixes coefficients in the vortex-pair renormalization flow.

Let a bounded planar region $D$ have positive area $A$ and centroid $\overline{\mathbf{x}}=A^{-1}{\int }_D\mathbf{x}dA$. If a solid has nonnegative thickness $h(\mathbf{x})=c+\mathbf{v}\cdot \mathbf{x}$ above $D$, its volume integral givesThus the average of an affine function over a region is its value at the centroid. The proof is the linearity of integration. For a disk the centroid is its centre, making the formula particularly useful for a cylindrical bore cut through a solid whose width varies linearly with height.

For a sufficiently smooth real scalar field on Euclidean space, set ${\Sigma }_{kj}={\partial }_k\psi {\partial }_j\psi -{\delta }_{kj}|\nabla \psi {|}^2/2$. The product rule and equality of mixed partial derivatives giveThe divergence theorem, applied twice, consequently gives the useful pure-boundary identityHere ${\partial }_n\psi =\hat{\mathbf{n}}\cdot \nabla \psi$ uses the outward unit normal. This is the Euclidean zero-potential counterpart of the scalar stress-energy divergence identity.

The molar volume $v_m=V/n_{\mathrm{mol}}$ is the volume per amount of substance, not the total container volume or the volume per particle. In the molar convention of the Van der Waals equation, take constant positive parametersThenor, for any fixed amount of substance,Both terms in the first equation are pressures. The total excluded volume is $n_{\mathrm{mol}}{b}_m$: using $V-b_m$ would subtract quantities with different units. Changing the sample size at fixed molar volume and temperature changes $V$ and $n_{\mathrm{mol}}$ proportionally, while leaving pressure unchanged. This is a useful unit and extensivity check before fitting a real gas.

For the molar Van der Waals critical point, differentiate with respect to molar volume at fixed temperature:Setting both to zero gives $R{T}_c=2a_m(v_{m,c}-b_m{)}^2/v_{m,c}^3$ and $R{T}_c=3a_m(v_{m,c}-b_m{)}^3/v_{m,c}^4$. Equating them and dividing by the nonzero factors yields $2v_{m,c}=3(v_{m,c}-b_m)$, soHere $v_{m,c}$ has units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T_c$ has units $\mathrm{K}$, and $p_c$ has units $\mathrm{Pa}$. This is the molar version of the particle derivation below: $v_{m,c}=N_A{v}_c$, with the same physical $T_c$ and $p_c$. If one writes $v_c=3b$, $T_c=8a/(27Rb)$ and $p_c=a/(27b^2)$ without subscripts, then all of $v_c,b,a$ in those formulas must be molar quantities.

Now define the dimensionless reduced variables explicitly as $x=v_m/v_{m,c}$, $t=T/T_c$ and $q=p/p_c$. Substituting the three critical values cancels $a_m,b_m,R$ and givesBecause $v_m/v_{m,c}=v/v_c$, these are exactly the reduced variables in the Van der Waals isotherm discussion. The Van der Waals law of corresponding states is therefore independent of whether one starts with particle or molar units. It predicts a critical ratio $p_c{v}_{m,c}/(R{T}_c)=3/8$, providing a third check on any proposed parameter fit.

For one mole with temperature-independent parameters $a,b$, the equation of state is $P=RT/(V-b)-a/V^2$. A Maxwell relation gives $(\partial S/\partial V{)}_T=R/(V-b)$, so the volume dependence is $R\ln (V-b)$. Matching the ideal gas entropy at infinite volume fixes the excess at the same $V,T$ to $R\ln (1-b/V)$. The attraction contributes $-a/V$ to the Helmholtz free energy but no excess entropy when $a$ is independent of temperature. A temperature-dependent attraction parameter would change this conclusion.