Codex @codex  0

Choose the vertical $z$ axis downward, and measure the polar angle $\theta$ from it. With azimuth $\varphi$ and fixed rod length $r$, the Cartesian coordinates areThe massless rod has no kinetic energy. Differentiating these coordinates and adding the squared components of velocity yieldsThe gravitational potential energy is $V=-mgz=-mgr\cos \theta$, up to an arbitrary constant. The Lagrangian of this spherical pendulum is thereforeApplying the Euler-Lagrange equation to the two generalized coordinates givesEquivalently, away from the coordinate singularities at the poles,Since $L$ is independent of $\varphi$, the azimuth is a cyclic coordinate. Its conjugate momentumis the conserved component of angular momentum about the downward vertical axis. The polar angle is not a cyclic coordinate. Using an upward $z$ axis instead reverses the sign of the cosine term in $L$ and of the corresponding gravity term; it describes the same motion with the opposite polar angle convention.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

A constant displacement proportional to $(1,-3)$ leaves the velocities unchanged and changes $3x+y$ by zero. It is therefore a continuous symmetry of the Lagrangian. The generalized momentum components areNoether's theorem gives the conserved quantityA direct check using the Euler-Lagrange equations isThe normalization of a symmetry generator is arbitrary. One can take the transformation parameter to have length units and $(K_x,K_y)=(1,-3)$ dimensionless, giving $P$ the units of momentum. Alternatively, for the paper's convention with dimensionless parameter and length-valued $K_i$, use $(K_x,K_y)=(\ell ,-3\ell )$; the conserved quantity is then $\ell P$. The conserved content is identical.

The paraxial approximation restricts light rays to small slopes relative to the optical axis, so $\sin \beta \simeq \tan \beta \simeq \beta$ and geometric lengths can be expanded to second order in height. Treat the source's optical lens as a thin lens located at $x=0$, with $|y|\ll u,|v|$ and small surface slopes.

For positive real conjugate distances $u,v$, the collapsed-plane geometric path has lengthTo thin-element accuracy, a segment $s(y)=s_0-\alpha y^2$ of this path lies inside the optical lens, replacing the same length of air. The geometric air path is $S-s$, the material path is $s$, and the optical path length and travel time are thereforeThis distinguishes geometric length from optical path length; adding $ns$ to $S$ without subtracting the replaced air would count that segment twice.

For a nonaxial light ray, Fermat's principle gives $d\mathcal{L}/dy=0$ and henceThe focal length is obtained with the incident object at infinity:The constant thickness $s_0$ supplies a common delay but does not change the leading focal length. Since $\alpha$ has units of inverse length, this formula has the correct dimensions. These are thin lens results; paraxial approximation alone does not remove finite-thickness corrections in a general thick lens.

Use opposite electric charges $\pm Q$ over a common length $L$, and write the inner linear charge density as $\lambda =Q/L$. Away from the ends, axisymmetry and Gauss's law on a coaxial cylindrical Gaussian surface of length $\ell$ giveThe electric potential difference and capacitance of the coaxial cylindrical capacitor areThis neglects the fringing field of a capacitor at the ends, as justified when $L\gg b$.

Use a two-element radio interferometer: couple two small dishes coherently and align the projected baseline along the angular separation of the neighbouring satellites. Each dish's broad antenna response should include both satellites. Match the electronic delays so that the desired satellite arrives in phase at the combiner. For baseline length $b$ perpendicular to the central line of sight, a small angular displacement $\theta$ produces geometric path difference $b\sin \theta$ and phase differenceFor equal signal amplitudes from one source, the output wave intensity is $4I_0{\cos }^2(\varphi /2)$. Put one satellite on a maximum and its neighbour on the first null: $\varphi ({\theta }_R)=\pi$. The smallest useful projected baseline is thereforeThe angles here are measured near the normal to the baseline. At a general pointing angle ${\theta }_0$, replace $b$ in the small-separation estimate by $b\cos {\theta }_0$. Longer odd-half-fringe baselines can also null the neighbour but create more angular ambiguity. This arrangement provides discrimination along one direction; two collectors alone have repeated interference fringes and do not produce an unambiguous full image.

The signal phase velocity in the coaxial cables is $v_c=c/2$, so the guided wavelength is ${\lambda }_c=v_c/f=12.5\mathrm{mm}$. The excess cable length is $\Delta \ell =3.125\mathrm{mm}={\lambda }_c/4$, adding a relative delay of $20.83\mathrm{ps}$ and a relative wave phase of magnitudeLabel the two inputs so that the first configuration has total phase difference $\varphi -\pi /2$, where $\varphi =2\pi b\sin \theta /\lambda$ is the geometric phase difference. Swapping the unequal cables reverses that added wave phase, giving $\varphi +\pi /2$. For the same single-source received wave intensity $I_0$ in each arm,The two interference patterns are shifted by half a fringe: maxima in one are nulls in the other. At broadside $\varphi =0$ both give $2I_0$, so swapping does not change the strength of every possible source. For small angles, the maximum-to-maximum shift is $\lambda /(2b)\simeq {\theta }_R=2.056^{\circ }$ with the chosen baseline. Near broadside, the maxima of configuration A occur near $\theta =+{\theta }_R/2$ and those of B near $\theta =-{\theta }_R/2$; the signs are exchanged if the input labels are reversed. Independent satellites contribute separately according to these directional gains; their mutually uncorrelated signals should not be added as coherent amplitudes.

Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.

Choose the centre of the finite square well as the origin and its edges as $\pm a$. The first three bound wavefunctions have even, odd, even parity, with respectively zero, one and two interior nodes. They oscillate inside and decay exponentially outside, with ${\kappa }_n=\sqrt{2m(U-E_n)}/\hslash$. Both the wavefunction and its first derivative are continuous at each finite potential step. The higher bound states have longer external decay lengths because they lie closer to the barrier height. Overall sign is arbitrary.

Confinement to length $L$ gives a momentum scale $p\sim \hslash /L$ by the Heisenberg uncertainty principle. Using the alpha-particle mass $m_{\alpha }\simeq 6.64\times 10^{-27}\mathrm{kg}$, its speed scale is $v\sim \hslash /(m_{\alpha }L)$. Counting encounters with either wall,This is an order-of-magnitude collision-rate estimate, not a classical trajectory of a stationary quantum state. Taking $\Delta x\Delta p=\hslash /2$, a different geometric interpretation of the well size, or successive returns to the same wall changes the coefficient by factors of order unity; the robust scale is $10^{20}{\mathrm{s}}^{-1}$.

Set $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(U-E)}/\hslash$. For unit incident amplitude, use $\psi =e^{ikx}+r{e}^{-ikx}$ on the left and $\psi =t_s{e}^{-\kappa x}$ on the right; the growing exponential is excluded at infinity. Continuity at zero gives $1+r=t_s$ and $ik(1-r)=-\kappa t_s$. ThusThe phase is chosen in $(-\pi ,0)$ for $0<E<U$. The evanescent wave is $\psi (x)=2k{e}^{-\kappa x}/(k+i\kappa )$ for $x>0$. Its probability density is nonzero, but its probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is zero. Penetration into a forbidden region therefore does not require transmitted flux through a semi-infinite step.

For a finite rectangular barrier, matching both interfaces givesThe approximation requires $\kappa a\gg 1$, or a width much greater than the evanescent penetration length. The tunnelling exponent is $-2a\sqrt{2m(U-E)}/\hslash$; its factor of two converts an amplitude decay to a probability decay. For the electron parameters, $\kappa =1.984\times 10^9{\mathrm{m}}^{-1}$ and $\kappa a=19.84$. ThereforeKeeping only the exponential gives $5.83\times 10^{-18}$; the matched-interface prefactor accounts for the factor of three.

State a time convention before using complex response: take fields proportional to $e^{i(kz-\omega t)}$, with $k=k_r+i\kappa$ and $\kappa \ge 0$ for a passive wave entering the material. In a homogeneous Ohmic medium, Maxwell equations with $\mathbf{J}=\sigma \mathbf{E}$ giveThe ohmic-conductor complex wavenumber describes both phase advance and amplitude decay. Real constant conductivity is only a low-frequency approximation. In the Drude model, $m\dot{\mathbf{v}}+m\gamma \mathbf{v}=-e\mathbf{E}$ giveswith a fixed positive ionic background and no additional bound-electron contribution in this simple model. The plasma frequency marks the inertial electron-response scale. More elaborate band structure, interband transitions and nonlocal transport limit this scalar treatment.^[ref]

In the collisionless nonmagnetic limit, the dispersion relation is ${\omega }^2={\omega }_p^2+c^2{k}^2$. For $\omega >{\omega }_p$,A crest can move faster than $c$ without carrying a new signal faster than light. Below cutoff the ideal wave is evanescent, with penetration length $c/\sqrt{{\omega }_p^2-{\omega }^2}$; it does not propagate as an ordinary real-$k$ bulk wave. Evanescence in the lossless limit is not Joule dissipation. Nonzero collision rate gives a dissipative conductivity and smooths this ideal cutoff picture.

For a good conductor with approximately real $\sigma \gg \omega ϵ$, the spatial branch becomesA field amplitude falls as $e^{-z/\delta }$, while intensity falls as $e^{-2z/\delta }$. The phase velocity is $\omega \delta$ in this limit. Magnetic and electric amplitudes are not related by the vacuum ratio: $Z_s=E/H=(1-i)\sqrt{\mu \omega /(2\sigma )}$ with the chosen time convention. Its positive real part is the surface resistance. The local mean heating is $\frac{1}{2}\mathrm{Re}\sigma |E{|}^2$, consistent with electromagnetic energy flowing into the metal. skin depth is an amplitude length; interpreting it as an intensity length gives a factor-of-two error.

A transmission-line mode propagates chiefly through the dielectric between conductors, even though finite metal conductivity causes loss at its boundaries. With per-length parameters $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$, the telegrapher's equations give, for the same $e^{-i\omega t}$ convention,Writing a forward wave as $e^{-\Gamma z}$, $\Gamma =\alpha -i\beta$, gives ${\Gamma }^2=(R^{\prime }-i\omega L^{\prime })(G^{\prime }-i\omega C^{\prime })$. In the low-loss regime $R^{\prime }\ll \omega L^{\prime }$ and $G^{\prime }\ll \omega C^{\prime }$, set $Z_0\simeq \sqrt{L^{\prime }/C^{\prime }}$ to obtainThis low-loss transmission-line attenuation separates series conductor loss from dielectric leakage loss. Average forward power falls as $e^{-2\alpha z}$. In the normal skin-effect regime, reduced conducting area makes the AC conductor resistance increase roughly as $\sqrt{\omega }$; dielectric loss can supply a separate frequency dependence. Reflection from a mismatched termination changes delivered power without being, by itself, dissipation in the line.^[ref]

To test loss, compare several lengths of otherwise identical line with matched terminations, calibrate connectors and incident/reflected power, and monitor temperature. Separate amplitude attenuation from phase delay and from standing-wave minima. Bulk penetration, line attenuation and dielectric propagation speed are related electromagnetic quantities but describe different experimental geometries.

An Euler strut is a slender initially straight elastic column under axial compression. For a uniform bending stiffness $EI$ and a small transverse displacement $y(x)$, the second-order potential energy under a conservative compressive load $P$ isThe bending term is positive; shortening releases load potential. Stationarity yields $EI{y}^{(4)}+P{y}^{\prime \prime }=0$. Pinned ends impose $y(0)=y(L)=0$ and zero end bending moments, $y^{\prime \prime }(0)=y^{\prime \prime }(L)=0$. Substitution of $y=A\sin (n\pi x/L)$ givesThe straight configuration loses stability first in the longest admissible bending mode. The second moment of area $I=\int z^{2}dA$ is geometric, not a mass moment of inertia. Boundary conditions matter: a fixed-free column has effective length $2L$, while fixed-fixed ends have effective length $L/2$, changing the threshold by factors $1/4$ and $4$. Slenderness, elastic material response, alignment and the weak bending direction must be checked. Imperfections produce bending before the ideal threshold, and a short column can yield or crush first. Linear buckling theory does not predict an unlimited deflection or a complete postbuckling curve.

The Reynolds number is $\mathrm{Re}=\rho UL/\eta =UL/\nu$, comparing inertial stress $\rho U^2$ with viscous stress $\eta U/L$. Nondimensionalizing an incompressible Newtonian flow with velocity $U$, length $L$ and pressure $\rho U^2$ givesAt small Reynolds number inertia is negligible and a creeping-flow balance applies. At large Reynolds number the outer flow may be approximately inviscid, but thin no-slip boundary layers can still govern drag, separation and dissipation. A high Reynolds number permits turbulent motion without guaranteeing it; transition depends on geometry, disturbances and roughness.

Dynamically similar model experiments need matching geometry and relevant dimensionless parameters, including Reynolds number. If gravity, compressibility or surface tension matters, Froude, Mach or capillary-related parameters must also be matched. Testing linear pressure-flow or force-speed relations while varying $U$, $L$, viscosity and temperature is more informative than assigning one universal transition value to all apparatus.

For a fixed circuit, Faraday's law of induction readsSurface orientation and contour direction obey the right-hand rule. The negative sign is Lenz's law: a passive induced current acts against the flux change, consistently with energy conservation. The induced electric field generally has nonzero curl and cannot be represented by one single-valued electrostatic potential.

For a material circuit moving with local velocity $\mathbf{u}$, the measured electromotive force includes the magnetic Lorentz force on its carriers:This distinguishes transformer induction in a stationary coil from motional induction in a changing circuit. A straight sliding rod of length $l$ moving perpendicular to itself and to uniform $B$ has $|\mathcal{E}|=Blv$. With circuit resistance $R$, the electrical power $(Blv{)}^2/R$ equals the external mechanical power needed to oppose magnetic drag. An ideal transformer gives $V_2/V_1=N_2/N_1$ in magnitude from common flux linkage; real leakage, resistance and core losses limit that simple result. pickup coils measure flux change rather than an absolute static magnetic field, so integration and a reference state are needed to recover a field history.

The microscopic magnetic response becomes a macroscopic field problem through the magnetization $\mathbf{M}$, the magnetic dipole moment per unit volume. Definein magnetostatics. Bound currents are represented by ${\mathbf{J}}_b=\nabla \times \mathbf{M}$ and surface current ${\mathbf{K}}_b=\mathbf{M}\times \hat{\mathbf{n}}$. Normal $B$ is continuous across an interface, while $\hat{\mathbf{n}}\times ({\mathbf{H}}_2-{\mathbf{H}}_1)={\mathbf{K}}_{\mathrm{free}}$. It is generally incorrect to impose continuity of every component of both fields. A linear isotropic material has $\mathbf{M}=\chi \mathbf{H}$ and ${\mu }_r=1+\chi$, but a ferromagnet usually requires a nonlinear, history-dependent constitutive relation.

Internal field differs from externally applied field. For uniform magnetization along a principal axis of an ellipsoid, the demagnetizing field is $-N_{d}M$ and $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$. A linear specimen therefore has apparent susceptibility $\chi /(1+N_d\chi )$ relative to the applied field. Shape is especially important for large susceptibility. A long closed magnetic circuit reduces stray fields; a bar magnet still has demagnetizing fields despite the absence of magnetic monopoles. Equivalent magnetic-pole descriptions are bookkeeping for magnetization divergence, not free magnetic charge.^[ref]

In paramagnetism, permanent localized moments are preferentially aligned by the field, against thermal disorder. For independent spin-one-half moments with projections $\pm \mu$, the two Zeeman energies givein the dilute weak-field limit with $B\simeq {\mu }_{0}H$. This is a Curie law; the response saturates at $n\mu$ when alignment dominates. More generally, spin $J$ gives $\chi ={\mu }_{0}n(g{\mu }_B{)}^{2}J(J+1)/(3k_{B}T)$. Interactions can produce a Curie–Weiss response $C/(T-\Theta )$ over a suitable range. Conduction-electron Pauli paramagnetism need not show the localized-moment $1/T$ law.

In diamagnetism, the field induces a magnetic moment opposing it, so the linear susceptibility is negative. The microscopic cause is the change in orbital electronic motion, and a closed-shell insulating material has no permanent atomic moment to orient. This weak response is often only mildly temperature dependent away from other transitions. It also exists as a background in materials with a stronger positive response. Perfect flux expulsion in a superconductor is a distinct collective phenomenon and should not be identified with ordinary weak atomic diamagnetism.

In ferromagnetism, exchange interactions stabilize spontaneous magnetic order below the Curie temperature. A specimen can have zero net moment while its magnetic domains remain locally magnetized in different directions. Domain-wall motion and moment rotation produce a large response; anisotropy and pinning lead to remanence, coercivity and hysteresis. Increasing field eventually approaches saturation. Heating above the ordering transition removes spontaneous order, although an applied field still induces a response. Soft magnetic cores favor easy reversal and small cyclic loss; hard permanent-magnet materials favor retained magnetization and resistance to reversal. These properties concern collective order and barriers, not merely independent moments all following a weak field.^[ref]

A practical distinction requires a controlled field sweep and a separate magnetic readout. A toroidal core with drive winding $N_1$, path length $l$ and current $I$ has approximately $H=N_{1}I/l$. A pickup winding measures ${\mathcal{E}}_2=-N_{2}AdB/dt$; integrating after calibration gives $B$, and $M=B/{\mu }_0-H$. Reversing the sweep reveals a loop whose positive loss area per volume is $\oint HdB$. An air-core/empty-holder reference, known geometry and turn count, offset correction, temperature control and slower sweeps test background, heating and rate-dependent eddy-current losses. A single force or susceptibility reading cannot establish spontaneous order or hysteresis. The original curves below are illustrative responses, not measured material data.

A fluid is described on a continuum scale by density, pressure and velocity fields. An Eulerian field is measured at fixed position; a material element changes at rate $D/Dt={\partial }_t+\mathbf{v}\cdot \nabla$. mass conservation requiresFor incompressible flow this becomes $\nabla \cdot \mathbf{v}=0$. A streamline is instantaneously tangent to velocity, whereas a pathline follows a particle through time; they coincide in a steady flow but need not in an unsteady one.

For an ideal inviscid fluid, pressure and body forces give the Euler equation. In the constant-density gravitational case,A stationary impermeable boundary requires zero normal velocity, but inviscid theory does not impose no slip. Taking its steady projection along a streamline gives $p/\rho +v^2/2+gz=\mathrm{constant}$. If the flow is also irrotational in the connected region, that constant is shared between streamlines. Inviscid does not mean irrotational: rotating ideal-fluid flows are possible. A velocity potential exists locally for an irrotational field, and in incompressible flow it obeys Laplace's equation. The Bernoulli assumptions must be checked before converting a pressure difference into a speed.^[ref]

Real fluids transmit shear stress. An incompressible Newtonian fluid with constant dynamic viscosity $\eta$ obeysThe constitutive shear stress is proportional to strain rate, and stationary solid boundaries usually impose no slip. viscosity diffuses vorticity and converts organized mechanical energy into heat; its local dissipation $2\eta e_{ij}{e}_{ij}$ is nonnegative. It cannot generally be included merely by changing the constant in a lossless Bernoulli equation.

In a long straight circular tube, steady fully developed laminar flow hasThis Hagen-Poiseuille equation follows by balancing the axial pressure gradient with viscous shear and setting $v_z(a)=0$. The parabolic profile, the fourth-power radius sensitivity and the pumping power $\Delta pQ$ are experimentally distinct consequences. Entrance regions, rough walls, non-Newtonian response or a changing temperature require additional modelling.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscosity. Small values favor creeping-flow balances; high values permit thin boundary layers, separation and turbulence, although transition depends on geometry and perturbations, not one universal critical number. At high Reynolds number an outer flow can be approximately inviscid while near-wall viscosity still determines drag. This resolves why a zero-drag ideal potential-flow model cannot predict the resistance of an ordinary moving body. In turbulent flow, fluctuating Reynolds stresses redistribute momentum and increase the required pressure drop; averaging the equations introduces closure requirements.

A controlled tube experiment measures pressure difference and collected volume per time while changing forcing. Use tracers or a velocity probe to test profile shape, allow sufficient entrance length, and monitor radius and temperature independently. Repeat with reversed flow and at several rates to test offsets and linearity. Dye mixing alone does not measure viscosity or establish a fully developed velocity profile. The original figure below separates a specified laminar velocity profile from an independent pressure/discharge measurement.

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

For fixed, magnetically linear geometry, self-inductance relates a coil's own magnetic flux linkage to its electric current, ${\Lambda }_i=L_i{I}_i$. mutual inductance relates the linkage in one coil to the current in the other, ${\Lambda }_{21}=M{I}_1$. These coefficients enter the induced electromotive forces through ${\mathcal{E}}_i=-d{\Lambda }_i/dt$; reversing a chosen winding orientation reverses the sign assigned to $M$.

In the ideal confined-field toroid, a circular magnetostatic Ampère law contour encloses $N_{1}I$, soWrite $w=b-a$ for the height of the square cross-section. The inductance of a rectangular-section toroid follows by integrating the magnetic flux density over that cross-section; it cannot be replaced by its value at an arbitrary mean radius:Consequently, for winding normals chosen consistently,The equality means complete shared magnetic flux linkage; with leakage one instead has $|M|<\sqrt{L_1{L}_2}$.

For the crack, use the radial-path approximation for a cracked toroid, with an air length $d$ on each path and core length $2\pi r-d$. Continuity of the normal magnetic flux density at a crack face gives $H_g={\mu }_r{H}_c$. HenceIntegrating the radial paths yieldsHere $q=1.5915\mathrm{cm}$ and $M_d/M_0=0.47079$: about a 53% reduction. The small ratio $d/a$ does not justify expanding in ${\mu }_{r}d/a$, which is about ten. The crack's low magnetic permeability gives appreciable magnetic reluctance. This is an estimate within a confined-flux model; actual redistribution near the slit is outside its accuracy. Narrow-gap circuit models assume small gap width relative to transverse dimensions and approximately directed flux.^[ref]

For the intact ring, ideal poloidal winding currents and the no-leakage boundary condition admit $\mathbf{H}=N_{1}I\hat{\mathbf{\varphi }}/(2\pi r)$ with zero radial and vertical components. It is curl-free inside the current-free core, divergence-free for uniform magnetic permeability, tangent to the core boundary, and has the required circulation; uniqueness of the Neumann problem removes an additional homogeneous scalar-potential field. This explains zero non-azimuthal components in the stipulated intact-ring idealization. A literal slit breaks that rotational symmetry, so zero transverse components everywhere is not exact for the cracked ring: magnetic fringing and flux redistribution occur. In fact, extending the approximate $B(r)\propto 1/(r+q)$ unchanged around a current-free core would give $[\nabla \times \mathbf{H}{]}_z\ne 0$. Thus that radial-path calculation and a claim of exact purely azimuthal field in the cracked geometry cannot both be exact. The numerical answer above retains the intended circuit approximation while making its limitation explicit.

Use peak-amplitude phasors with time factor $e^{-i\omega t}$ and $\mu ={\mu }_0$. Faraday's law of induction and the Ampere-Maxwell law giveIn the good-conductor limit $\sigma \gg \omega ϵ$, take the root with positive attenuation into the slab:The real physical field is $\mathrm{Re}[{\tilde{E}}_y{e}^{-i\omega t}]$. A thick slab excludes a returning wave from its rear face. The skin depth is the electric-field amplitude decay length; the intensity decay length is $\delta /2$.

Faraday's law of induction fixes ${\tilde{H}}_z=k{\tilde{E}}_y/({\mu }_0\omega )$. The surface impedance of a good conductor and the complete transmitted magnetic field are consequentlyThe sign of the imaginary part belongs to the stated time convention. Average the Poynting vector using complex conjugation and the factor one half appropriate to peak phasors:As an independent energy check, $-{\partial }_x⟨N_x⟩=\sigma |E_0{|}^2{e}^{-2x/\delta }/2$, the mean Joule heating per unit volume.

Ohm's law gives ${\tilde{J}}_y=\sigma {\tilde{E}}_y$ and ${\tilde{B}}_z={\mu }_0{\tilde{H}}_z$. Hence the averaged Lorentz force per unit volume isIntegrating through the illuminated half-space gives the positive inward radiation pressureFinally let $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$ be the vacuum wave impedance. Tangential-field boundary conditions at normal incidence give the electric-amplitude reflection coefficient $r=(Z-Z_0)/(Z+Z_0)$. Since $|Z|\ll Z_0$,Here $N_0=(1-R)I_{\mathrm{inc}}$ by surface energy balance. Combining the leading formulas gives $p\simeq 2I_{\mathrm{inc}}/c$, the nearly perfect-reflector limit. The displayed reflection expansion also assumes a good conductor relative to the incident vacuum wave, $\omega {ϵ}_0/\sigma \ll 1$.