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Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

The image charge construction replaces conductor-induced charge by fictitious sources outside the physical domain. Match the conductor potentials and the far-field condition; uniqueness then fixes the potential in the physical region. For a grounded plane, reflect a charge and reverse its sign, giving attraction $q^2/(16\pi {ϵ}_0{h}^2)$. For a grounded sphere of radius $a$, a charge at distance $R>a$ has image $-qa/R$ at radial distance $a^2/R$. These are examples with prescribed zero potential, not arbitrary isolated conductors.^[ref]

Use horizontal $x$ and upward $y$ in the upper right-angle wedge. The sheets meet electrically, so take their common potential as zero and assume no externally imposed field. For a real charge at $(0,y)$, the image charges in a right-angle conducting wedge are $-q$ at $(y,0)$ and $(-y,0)$, and $+q$ at $(0,-y)$. The first two contributions have cancelling horizontal components. With $K_e=1/(4\pi {ϵ}_0)$,Thus the force magnitude at the original height is $\beta /y_0^2$, directed down toward the intersection. The sphere is treated as a point charge, with size much smaller than its distances from the sheets.

Since gravity is already balanced in the uncharged equilibrium, the charged vertical equilibrium satisfies $k(y_e-y_0)+\beta /y_e^2=0$. For $\beta /(k{y}_0^3)\ll 1$, $y_e\simeq y_0-\beta /(k{y}_0^2)$. Differentiating the force givesThe electrostatic softening above a conducting wedge reduces the restoring force; the small-charge vertical equilibrium remains stable.

For the lateral stability of a spring-supported charge in a wedge, the off-axis images of $(x,y)$ are $-q$ at $(y,x)$ and $(-y,-x)$ and $+q$ at $(-x,-y)$. Expansion near the bisector givesSo the electrostatic force is laterally destabilizing: moving toward either sheet strengthens its attraction. In the simplified model where the spring supplies only a vertical force, the overall equilibrium is a saddle and is unstable.

A three-dimensional spring suspension also has geometrical transverse stiffness. If its charged-equilibrium tension is $T$ and length is $\ell$, a small sideways displacement gives spring force $-(T/\ell )x$, with $T=mg+\beta /y_e^2$. The full stability conditions in this suspension model areMotion parallel to the intersection line has no electrostatic stiffness in the ideal infinite-wedge geometry and is stabilized by the positive spring tension. Thus lateral electrostatic destabilization alone does not prove instability of a real hanging spring. At fixed finite support geometry, a sufficiently small charge preserves stability because $D_x=O(q^2)$ while $T/\ell \to mg/\ell >0$. The given vertical-displacement approximation does not specify the attachment length or by itself establish the transverse inequality for an arbitrary charge. The general criterion above states what can be concluded; the saddle verdict applies to the distinct idealization with zero lateral support stiffness.

For distributed inductance $L$ and capacitance $C$, a small segment obeys $V_z=-L{I}_t$ and $I_z=-C{V}_t$. Differentiation gives $V_{zz}=LC{V}_{tt}$ and $I_{zz}=LC{I}_{tt}$. For a forward wave,A backward voltage wave has current $I^{-}=-V^{-}/Z$. This sign matters at an impedance discontinuity.

The source figure is the stacked-strip parallel-plate transmission line geometry, despite its use of the word coplanar. Gauss's law gives $C={ϵ}_0{ϵ}_{r}a/d$. Opposite surface currents give magnetic field $B={\mu }_{0}I/a$ between the strips by Ampère's law; the magnetic flux per unit length is $Bd$, hence $L=Bd/I={\mu }_{0}d/a$. ThereforeNo numerical $a/d$ is given; the impedance is appropriately left in terms of that ratio. The line is taken to be nonmagnetic and lossless.

Increasing the gap by $\alpha$ changes $L$ to $\alpha L$ and $C$ to $C/\alpha$, so $Z_2=\alpha Z_1$ while $v$ is unchanged. voltage and current continuity at the step giveConsequently the voltage and energy-flux transmission coefficients areHere amplitudes are peak voltage amplitudes, with mean power $|V{|}^2/(2Z)$. The reciprocal gaps have $r_V(1/\alpha )=-r_V(\alpha )$: equal reflected energy but opposite voltage phase. A voltage transmission coefficient above one does not imply energy gain.

For a finite impedance section of a transmission line, put $\ell =l_2-l_1$ and $\delta =\omega \ell /v$. Multiple reflections inside the altered section interfere. Summing their geometric series, relative to the first boundary,Transmission is unity at $\delta =n\pi$ and minimal at $\delta =(n+1/2)\pi$, with minima $4{\alpha }^2/(1+{\alpha }^2{)}^2$. Resonances are separated by $\Delta f=v/(2\ell )$; the long-wavelength limit also transmits fully. Matching the far termination removes end echoes, not the two internal reflections.

For steady thin beams a distance $d$ apart, let ${\lambda }_1,{\lambda }_2$ be their laboratory charge per length. Their current magnitudes are $|{\lambda }_j|v$, giving force magnitudes per lengthThe electric force is repulsive and the magnetic force attractive for the parallel electron velocities. The net force remains repulsive for $v<c$.

Assume uniform charge per unit length and the same angular speed $\Omega$. A charge element at distance $r$ has $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$, so the Biot-Savart law gives the centre field of a rigidly rotating charged loop as $dB={\mu }_0\Omega dq/(4\pi r)$. For one square side, with $b=a/2$,Using $dq=Qdx/(4a)$ on each side givesThus $B_{\square }/B_{\circ }=\ln (1+\sqrt{2})\simeq 0.8814$: the circular field is larger by $1/\ln (1+\sqrt{2})\simeq 1.135$. This is a rigidly rotating charge distribution, not a current constrained to the sides of a stationary square.

Applying Gauss's law for magnetism to a pillbox and Ampère's law to an interface-crossing rectangle givesHere $\hat{n}$ points from material 1 to 2 and ${\mathbf{K}}_f$ is free surface current. With no such current, $B_n$ and $H_t$ are continuous. For linear isotropic media, $B_{t2}/{\mu }_2=B_{t1}/{\mu }_1$ and ${\mu }_2{H}_{n2}={\mu }_1{H}_{n1}$. The magnetic fields need not have equal magnitudes on either side.

With negligible leakage, Gauss's law for magnetism makes each cross-section of a magnetic flux tube carry the same flux $\mathcal{F}$. High permeability guides it through the core; a short gap permits neglect of fringing. Flux conservation across these particular geometric sections is an approximation, not a consequence of zero divergence without controlling side leakage.

Let $L$ be the constant-radius arc marked in the figure, and ${\mu }_r=\mu /{\mu }_0$. For each straight taper, the radius changes linearly from $R$ to $r$, giving the reluctance of a conical magnetic poleSumming magnetic reluctances to find the gap field of a tapered electromagnet, and using $B_g=\mathcal{F}/(\pi r^2)$ givesIf $L$ is instead defined as the total magnetic-material length, replace the displayed uniform-arc length by $L-2s$. The comparison untapered core has uniform radius $R$ and material length $L+2s$, soThe original PDF specifies $s/L\ll r/R$; the TeX transcription reverses this inequality. In the printed limit the taper reluctances are small relative to that of the broad arc. The gain is approximately $[h+L/{\mu }_r]/[h+L{r}^2/({\mu }_r{R}^2)]$, approaching $(R/r{)}^2$ if the gap reluctance is also negligible. If the gap dominates, both fields instead approach $NI/h$ and the gain approaches one. magnetic saturation near the small poles, gap fringing, leakage and winding heating limit practical enhancement.

$Z$ is the terminating load impedance, $Z_0$ the line's characteristic impedance, and $k$ the propagation wave number; $l$ is its length. In the lossless convention printed, an open circuit gives $Z_{\mathrm{in}}=-i{Z}_0\cot kl$, a short circuit gives $i{Z}_0\tan kl$, and a matched load $Z=Z_0$ gives $Z_{\mathrm{in}}=Z_0$ for every length. The divergent or zero limits at isolated electrical lengths are understood continuously.

Put $\theta =kl$, $c_{\theta }=\cos \theta$, $s_{\theta }=\sin \theta$. The voltage/current matrix of one section isFor a pair beginning with $Z_1$, write $M_1{M}_2=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Its entries are $A=c_{\theta }^2-(Z_1/Z_2)s_{\theta }^2$, $D=c_{\theta }^2-(Z_2/Z_1)s_{\theta }^2$, $B=i(Z_1+Z_2)s_{\theta }{c}_{\theta }$, and $C=i(1/Z_1+1/Z_2)s_{\theta }{c}_{\theta }$. The alternating-impedance transmission line has unchanged input $X$, its Bloch impedance, obeying $X=(AX+B)/(CX+D)$. Hence, away from the removable dividing zeros,For real positive $Z_1,Z_2$, select the branch carrying energy into the line or decaying into it. A positive real part selects the pass-band root; infinitesimal passive loss resolves the stop-band branch. At $\theta =m\pi$ the cell matrix is the identity, so the exact fixed-point condition alone does not determine $X$; at odd quarter-wave lengths the appropriate limiting impedance can be zero or infinite.

In a transmission-line stop band, a lossless semi-infinite ladder totally reflects because its physical input is reactive. Equivalently its cell hasThus, for nonzero contrast, the stop-band interiors satisfyWith ${\theta }_c=\arcsin [2\sqrt{Z_1{Z}_2}/(Z_1+Z_2)]$, these are $m\pi +{\theta }_c<kl<(m+1)\pi -{\theta }_c$; band-edge reflection follows by a limit. If $k=\omega /u$, multiply the endpoints by $u/l$ for angular frequencies. Identical impedances have no stop band. A finite ladder can transmit by tunnelling, unlike the semi-infinite idealization.

At distance $y$ from the neutral axis, bending strain has magnitude $y/R$ and the corresponding stress is $Ey/R$. Its moment is thereforeThe second moment of area is about the neutral bending axis and has units of length to the fourth power. This assumes a slender linear-elastic beam, small strains and plane sections remaining plane, not necessarily small deflections.

For horizontal compression $F>0$, moment balance across a short arc gives $dB/ds=-Fdy/ds=-F\sin \theta$. Since $B=EI{\theta }^{\prime }$, the exact-angle Euler elastica equation and its first integral areThe first integral follows by multiplying by ${\theta }^{\prime }$ and integrating; no small-angle replacement has been made.

For tension set $F=-T$. The straight far-field ends give $C=T$, so on the increasing-angle tensile elastica loop branchChoosing $s=0$ at $\theta =\pi$ givesThe terminal angles are $0$ and $2\pi$, the same physical tangent direction. At the top, ${\theta }^{\prime }(0)=2\alpha$ and the bending moment is $2EI\alpha$. The horizontal tensile force has lever arm $h$, soIntegrating $(x^{\prime },y^{\prime })=(\cos \theta ,\sin \theta )$ independently gives $x=s-2\tanh (\alpha s)/\alpha$, $y=2\mathrm{sech}(\alpha s)/\alpha$, confirming the height. This ideal loop neglects self-contact and thickness at its crossing.

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.

For the travelling wave, $⟨N_z⟩=\mathrm{Re}(V_0{I}_0^{\ast })/[4\pi r^2\ln (b/a)]$. Integrating over the dielectric annulus yieldsThis is the familiar circuit power, carried here through the dielectric rather than through the ideal metal. In root mean square notation it is $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ for the forward lossless wave; the corresponding static fields give $VI$ without a time-average factor. The average line energy is $(C|V_0{|}^2+L|I_0{|}^2)/4=C|V_0{|}^2/2$, whose product with wave speed agrees with the power.

Finite conductor resistance requires a small longitudinal electric field to drive current. Its cross product with the azimuthal magnetic field produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This Poynting flux into lossy coaxial conductors explains why forward power decreases along the cable as energy becomes heat in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

Let $s(t)$ be the length of the free straight segment from the current contact point to the mass, and let $\theta$ measure the contact point's rotation from first contact. For an inextensible taut rope, winding through angle $\theta$ uses length $R\theta$:The point mass follows an involute of the pole's circular cross-section. Its instantaneous velocity is perpendicular to the free rope. To verify the geometry explicitly, use planar polar unit vectors at the contact point:The rope's tension therefore does no work, and the speed stays at its initial value $v$. The geometry gives $v=s\dot{\theta }$. ConsequentlyThe distance from the pole's axis is $\sqrt{R^2+s^2}$, so a point mass first reaches the pole when $s=0$. The winding time around a circular pole is thusThis has units of time. The model assumes a light inextensible rope, no energy loss at contact and a point mass. The required tension $m{v}^2/s$ diverges at the ideal final contact, so finite strength and size matter very close to impact.

Assume all rings have the same circulation $\Gamma$ and orientation, and that there is no imposed background flow. Vortex lines describe the direction of vorticity, not the direction of velocity. For an incompressible fluid, the velocity induced by a thin vortex filament has the same Biot-Savart integral form as the magnetic field of a current filament:A closely spaced coaxial array of vortex rings is therefore analogous to a solenoid. The velocity field passes along the common axis through the rings and returns outside a finite array. Its orientation reverses if every circulation is reversed. In a long array, far from its ends and vortex cores, the velocity field of a dense coaxial vortex-ring array is approximatelyThe exterior field is exactly zero only in the ideal continuous infinitely long limit, with the background velocity set to zero. In a finite array the streamlines spread out at one end, run back outside, and converge into the other. A large number of closely spaced rings alone does not guarantee that the total array length greatly exceeds $r$; if it does not, the end and return flows remain substantial. Different unspecified ring circulations would require superposing their actual signed contributions.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

An Euler strut is a slender elastic column under axial compression. Let its bending stiffness about the weaker section axis be $E{I}_A$ and its length be $L$. With small lateral deflection $y(x)$ and a dead compressive load $P$, the quadratic total potential energy isThe first term is bending energy; the negative term is work released by the load as bending shortens the end-to-end distance. Stationarity gives $E{I}_A{y}^{\prime \prime \prime \prime }+P{y}^{\prime \prime }=0$. For pinned ends, $y=y^{\prime \prime }=0$ at $0,L$, and $y\propto \sin (n\pi x/L)$ requires $P_n=n^2{\pi }^{2}E{I}_A/L^2$. The first instability is Euler's critical load,Below this load, the straight state has positive quadratic energy against these perturbations. At the threshold, the lowest bending mode has zero quadratic restoring energy; above it the straight state is unstable. For a rectangular section, $I_A=b{h}^3/12$ about the appropriate bending axis, so the smaller second moment of area controls the first buckling direction.

Endpoint constraints matter. Write $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/(KL{)}^2$: $K=1$ for pinned-pinned, $2$ for fixed-free and $1/2$ for fixed-fixed conditions. This elastic instability differs from compressive material failure. Imperfections cause bending before the ideal threshold, and a short column may yield before Euler buckling occurs.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.