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Murray's law balances viscous pumping power against the metabolic cost of maintaining blood volume. For cylindrical vessels it gives $Q\propto R^3$ and hence $R_0^3=R_1^3+R_2^3$ at an ideal bifurcation.

For a sphere of radius $(1-ϵ)a$ displaced by $ϵ\lambda a$ from the axis of a tube of radius $a$, the narrow gap isIts thickness is $O(ϵa)$ and its axial length is $O(a\sqrt{ϵ})$, so lubrication pressure dominates shear stress by $O({ϵ}^{-1/2})$.

For a wall shear $u\simeq sz$, heating over streamwise length $L$ gives an outer temperature rise proportional to $L/(sz)$. Its transverse thermal conduction grows as $z^{-3}$ relative to the local source. If $\chi$ is thermal diffusivity, comparing conduction with heating gives $\delta \sim (\chi L/s{)}^{1/3}$, up to a numerical constant depending on the precise comparison. The same scale follows by comparing transverse diffusion time $z^2/\chi$ with residence time $L/(sz)$.

Nested measurements are responses grouped inside larger units, with each measured object belonging to just one such unit. Three leaves within each pot give nested measurements; repeatedly reading one leaf gives another level of repetition. Either can yield clustered data. A common random effect can describe the shared influence of the containing experimental unit.

For the constructed pot example, label treatments by $g\in \{A,B\}$, pots within each group by $i=1,\dots ,n$, and leaves within each pot by $j=1,\dots ,m$. Consider the statistical modelHere $Y_{gij}$ is a leaf length, $\mu +{\alpha }_g$ is its group expected value, $U_{gi}$ is a pot random effect, and ${\epsilon }_{gij}$ describes remaining variation among leaves and measurement error. Assume that all pot random effects are mutually independent, all remaining errors are mutually independent, and the two collections are independent of one another and of the randomization. These are explicit modelling assumptions, conditional on the assigned treatments; no normal distribution assumption is needed for the following variance calculation.

Two distinct leaf measurements in one pot share $U_{gi}$. Their covariance is ${\tau }^2$, their individual variance is ${\tau }^2+{\sigma }^2$, and their intraclass correlation coefficient isMeasurements from different pots have zero covariance and are independent under the stated statistical model. Statistical independence therefore concerns the whole response structure, not just the fact that the measured leaves are different objects.

The coefficient in $\mathbf{J}=-D\nabla \varphi$ for the scalar neutron flux convention. It has dimensions of length, unlike a particle-density diffusion coefficient with dimensions length squared per time. For a single fixed speed $v$, the corresponding density diffusion coefficient is $vD$.

Treat the incident and outgoing neutrons as free nonrelativistic particles outside the specimen. Their wave vectors, de Broglie wavelengths, momenta and kinetic energies obeyHere $k_i=|{\mathbf{k}}_i|$, $k_f=|{\mathbf{k}}_f|$, $m_n$ is the neutron mass, and $\hslash =h/(2\pi )$ uses the Planck constant. Subtracting the outgoing projectile quantities from the incoming ones gives the momentum transfer and energy delivered to the specimen:Thus $\mathbf{Q}$ has units of inverse length, rather than momentum. A positive $\omega$ means the specimen gains energy; negative $\omega$ means it supplies energy to the neutron. If the angle between incident and outgoing propagation directions is $2\theta$, the law of cosines givesKnowing the detector angle alone therefore leaves $Q$ undetermined unless the outgoing energy is known or an elastic scattering assumption is justified.

For a reproducible unit conversion choose $m_n=1.67492750056\times 10^{-27}\mathrm{kg}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. The first is a recommended central value; the latter two conversions are exact.^[ref] With $1$ ångström equal to $10^{-10}\mathrm{m}$ and $1\mathrm{meV}=10^{-3}$ electronvolts, substitution into $E=h^2/(2m_n{\lambda }^2)$ givesSquare brackets here mean the numerical value in the indicated unit. For an original chosen example, ${\lambda }_i=2$ ångströms and ${\lambda }_f=2.5$ ångströms give $E_i\simeq 20.451\mathrm{meV}$, $E_f\simeq 13.089\mathrm{meV}$ and $\hslash \omega \simeq 7.362\mathrm{meV}$. At $2\theta =60^{\circ }$, $Q\simeq 2.879{\mathring{\mathrm{A}}}^{-1}$. These values stipulate an allowed projectile geometry; they do not assert that any particular specimen has an excitation there.

The surface flux of a uniformly heated cylinder is $Ha/2$. Equating it to the Newton-cooling flux $\alpha [T(a)-T_{\mathrm{air}}]$ gives the displayed surface temperature. The full temperature profile adds the internal parabolic excess to this surface value. Electrical power per length equals the surface heat loss per length in steady state.

Albert Abraham Michelson received the award for precise optical instruments and their use in spectroscopy and metrology.^[ref] The Michelson interferometer converts relative wave phase into a readout for precise length measurements.

For axial stretch $\lambda$ and twist $\alpha$ per reference length, an incompressible neo-Hookean solid has current radius $a=R/\sqrt{\lambda }$. In local cylindrical orthonormal bases the deformation gradient is $A=\left(\begin{array}{ccc}{\lambda }^{-1/2}& 0& 0\\ 0& {\lambda }^{-1/2}& \alpha r\\ 0& 0& \lambda \end{array}\right)$. The Cauchy stress tensor is $\mu A{A}^T+qI$. Radial equilibrium gives $q^{\prime }=\mu {\alpha }^{2}r$, and a traction-free curved surface gives $q=-\mu /\lambda +\mu {\alpha }^2(r^2-a^2)/2$. The nominal end traction is the displayed vector, measured per reference area. Its moment about the axis is $\int (x_1{t}_2-x_2{t}_1)d{S}_0=\mu \alpha \int r^{2}d{S}_0=\mu \pi R^4\alpha /(2\lambda )$. This calculation uses current moment arms with reference-area tractions.

The distance for an order-one self-phase modulation at peak power $P$ in a lossless Kerr envelope model. With attenuation, the phase accumulates over an effective interaction length instead.

The distance over which a constant phase mismatch accumulates phase difference $\pi$. In unseeded, undepleted sum-frequency generation, the generated intensity first reaches a maximum at this length and returns to zero at $2{\ell }_c$. This length concerns relative phases of interacting waves, rather than the temporal optical coherence length of one source.

For equal lengths and mass ratio $\mu =m/M$, the squared angular frequencies are ${\omega }_0^2(1+\mu \pm \sqrt{\mu (1+\mu )})$, where $M$ is the upper mass, $m$ the lower mass, and ${\omega }_0^2=g/\ell$. The equal-mass case $\mu =1$ admits particularly simple mode shapes.

Use the ideal model of the double pendulum, with both angles measured from the downward vertical. Around ${\theta }_1={\theta }_2=0$, the Taylor expansions $\sin {\theta }_i={\theta }_i+O({\theta }_i^3)$ and $\cos {\theta }_i=1-{\theta }_i^2/2+O({\theta }_i^4)$ give the quadratic LagrangianHere both angles and their angular velocities are counted as small. Consequently the correction to $\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2$ is fourth order and does not enter $L_2$. The Euler-Lagrange equations give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$. The mass matrix is not diagonal: an angular motion of the upper rod carries the lower mass too.^[ref]

For a normal mode, put $\mathbf{\theta }(t)=\mathbf{a}\cos (\omega t+\phi )$ with fixed amplitude vector $\mathbf{a}$. The resulting generalized eigenvalue problem for small oscillations is $\mathsf{K}\mathbf{a}={\omega }^2\mathsf{M}\mathbf{a}$. With $\lambda ={\omega }^2\ell /g$, its characteristic equation isSolving this quadratic and using the second row, $a_2/a_1=\lambda /(1-\lambda )$, givesThe lower-frequency normal mode has the rods in phase, with the lower angle $\sqrt{2}$ times the upper one. The higher-frequency normal mode has opposite signs, corresponding to an oscillation phase difference of $\pi$. These are ratios of absolute angles. The horizontal displacements are instead $x_1\simeq \ell {\theta }_1$ and $x_2\simeq \ell ({\theta }_1+{\theta }_2)$; their ratios are $1+\sqrt{2}$ and $1-\sqrt{2}$ respectively. Thus the lower mass moves horizontally opposite to the upper mass in the faster normal mode.

Both squared angular frequencies are positive. Their ratio satisfies ${\omega }_{+}/{\omega }_{-}=1+\sqrt{2}$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell =1\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the angular frequencies are approximately $2.3972$ and $5.7874{\mathrm{s}}^{-1}$, and the oscillation periods $2\pi /{\omega }_{\pm }$ are $2.6211$ and $1.0857\mathrm{s}$. Doubling both masses leaves these unchanged; multiplying both rod lengths by four doubles both oscillation periods.

The eigenvectors are not orthogonal in the ordinary dot product: ${\mathbf{a}}_{-}^{\mathsf{T}}{\mathbf{a}}_{+}=-1$. They satisfy mass-matrix orthogonality of normal modes, sinceWriting $\mathbf{\theta }={\mathbf{a}}_{-}{q}_{-}+{\mathbf{a}}_{+}{q}_{+}$ therefore removes the cross terms from both quadratic energies. The two normal coordinates satisfy ${\ddot{q}}_{\pm }+{\omega }_{\pm }^2{q}_{\pm }=0$. This is why the normal modes are useful: one coupled system becomes two independently evolving harmonic oscillators, and any small release can be decomposed into them. At larger angles, the discarded terms in the double pendulum equations of motion matter and this exact linear superposition no longer applies.

The fractional change in the length of a material line in a specified direction. For infinitesimal deformation with unit direction $\mathbf{n}$, it is $\mathbf{n}\cdot \mathbf{\epsilon }\mathbf{n}$, where $\mathbf{\epsilon }$ is the small strain tensor.

The nuclear neutron scattering length is the effective length coefficient for low-energy nuclear scattering amplitude. For the elementary coherent calculation below, use a bound coherent coefficient $b_j$ for each site, and neglect absorption so that $b_j$ is real. This coefficient describes the atomic nucleus, including its isotope and nuclear spin dependence; it is not proportional to atomic number. Replacing nuclear coefficients by electronic magnetic moments would describe a different scattering channel, with additional direction and spatial-distribution factors, as explained in magnetic neutron diffraction.

A ruled scale viewed together with the intermediate microscope image. Its intervals acquire specimen length units through microscope scale calibration.

With fields proportional to $e^{i(kz-\omega t)}$, a homogeneous Ohmic medium obeys the displayed relation by combining the two curl Maxwell equations. The physical passive branch has $\mathrm{Im}k\ge 0$. For real positive conductivity much larger than $\omega ϵ$, it reduces to $k=(1+i)/\delta$ with skin depth $\delta =\sqrt{2/(\mu \sigma \omega )}$. Field amplitude and intensity then decay with respective lengths $\delta$ and $\delta /2$. A frequency-dependent conductivity must be retained when charge-carrier inertia matters.

Ohm's law states that an Ohmic conductor has electric current proportional to its applied electric potential difference, with constant electrical resistance under fixed material conditions:Here $V=\varphi (A)-\varphi (B)$ is the terminal electric potential difference, and positive $I$ enters at $A$ and exits at $B$. This sign convention makes $VI$ the electrical power absorbed. The straight-line relation through the origin lets one constant predict the response at many applied electric potential differences. Merely defining $R=V/I$ at one nonzero operating point does not establish Ohm's law: a component may have a ratio that varies with applied electric potential difference. Material and temperature must be specified and held fixed; many components have a nonlinear response. ^[ref].

For the ideal resistor model $R=2\Omega$ at electric potential difference $V=5\mathrm{V}$,This power is converted to internal energy by Joule heating; a watt is a joule per second. If the model holds for time $10\mathrm{s}$, the transferred energy is $125\mathrm{J}$. Doubling electric potential difference doubles electric current and quadruples power for the same fixed $R$. The model keeps material temperature fixed even while calculating Joule heating; appreciable changes in temperature may require a changing electrical resistance. ^[ref].

In 1827, Georg Ohm published his mathematical account of electrical transport in Die galvanische Kette, mathematisch bearbeitet. On printed page 36, his relation $S=A/L$ expresses electric current in terms of total driving tension and a reduced length incorporating electrical conductivity and geometry. This is historical notation: his $A$ is not the wire cross-sectional area used here, and his $L$ is not simply the wire's physical length. The striking idea is that material response and geometry can be combined into one quantity governing an electrical circuit. ^[ref].

A circular loop of radius $R$ carrying electric current $I$ has, on its axis,where the sign of $\hat{z}$ is fixed by the right-hand rule. At its centre, $B(0)={\mu }_{0}I/(2R)$.

Put the loop in the $xy$ plane, centered at the origin, with positive electric current counterclockwise when viewed from $+z$. The right-hand rule then gives the central magnetic field along $+z$. Parameterize source positions by the angle $\varphi$ and observe at $(0,0,z)$:Here $R$ without bold type is the loop radius, whereas $\mathbf{R}$ is the source-to-observer displacement. Their cross product isThe denominator $|\mathbf{R}{|}^3=(R^2+z^2{)}^{3/2}$ is constant around the loop. The integrals of sine and cosine over a full turn vanish, while the axial contribution adds to $2\pi R^2$. Substitution into the Biot-Savart law proves the on-axis formula. Symmetry cancels transverse components; it does not cancel the axial components.

For an original one-turn example in the same vacuum approximation, choose $R=0.04\mathrm{m}$ and $I=2\mathrm{A}$. With ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$,At $z=0.03\mathrm{m}$, the source-to-observer distance is $0.05\mathrm{m}$, so $B(z)/B(0)=64/125=0.512$ and $B(z)\approx 16.08\mu \mathrm{T}$. At $z=-0.03\mathrm{m}$, the magnetic field has the same magnitude and still points along $+z$. The point has crossed the loop plane, but the loop's electric current has not reversed. Reversing the electric current reverses the magnetic field on both sides.

For $|z|\ll R$, a Taylor expansion gives $B(z)/B(0)=1-3z^2/(2R^2)+O(z^4/R^4)$: the center is a smooth maximum, with zero first derivative. For $|z|\gg R$,This is the axial magnetic dipole field of magnetic dipole moment $\mathbf{m}=I\pi R^2\hat{\mathbf{z}}$.^[ref] A compact closed loop has an inverse-cube axial far field, in contrast to the inverse-distance field of the infinite straight wire. The SI units of the exact loop formula are ${\mu }_{0}I{R}^2/(R^2+z^2{)}^{3/2}\sim {\mu }_{0}I/\text{length}$, hence tesla.

For $N$ coincident turns carrying the same electric current, the superposition principle multiplies the ideal result by $N$. A real coil has radial width and axial length: if those are not negligible, integrate or sum over its individual turns. This is why a single measured coil diameter and a turn count may not determine the full spatial magnetic field precisely. A collection of loops becomes a solenoid; integrating their axial contributions gives the on-axis magnetic field of a finite solenoid.

Take a separated Boolean decomposition $f(x,y)={⨁}_{\ell }{u}_{\ell }(x)v_{\ell }(y)$. Feed $(u_{\ell },v_{\ell })$ to one independent PR box, obtaining outputs with $a_{\ell }\oplus b_{\ell }=u_{\ell }{v}_{\ell }$. Alice forms $A={⨁}_{\ell }{a}_{\ell }$, Bob forms $B={⨁}_{\ell }{b}_{\ell }$, and their parities satisfy $A\oplus B=f$. Sending the single bit $A$ lets Bob compute the answer with certainty. The truth-table decomposition needs at most $2^{\min (m,n)}$ boxes for input lengths $m,n$. Each box's local output is uniform, so Bob's uncommunicated output string carries no information about Alice's input; the final parity bit is essential for a general function depending on that input.

A single acoustic branch $\omega =v|k|$ has thermal energy per length $U_T/\ell =\int dk\hslash v|k|/[2\pi (e^{\beta \hslash v|k|}-1)]=\pi (k_{B}T{)}^2/(6\hslash v)$. Differentiating gives the displayed linear heat capacity. The regime requires temperatures below the microscopic phonon cutoff but above the finite-size level spacing, with the translational zero mode treated separately.