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For a complete entry and exit of a small square loop, integrating the drag over two distances $a$ gives impulse magnitude $2B^2{a}^3/R$. This is independent of speed in the ideal sharp-edge, zero-inductance model, though dissipated energy is not.

For an infinitely long thin strip of width $w$ carrying uniform longitudinal current $I$, the magnetic field a perpendicular distance $z>0$ above its centre has magnitude $B={\mu }_{0}I\arctan (w/(2z))/(\pi w)$. Its near-sheet limit is ${\mu }_{0}I/(2w)$ and its distant line-current limit is ${\mu }_{0}I/(2\pi z)$.

Take a long strip along $y$, occupying $-w/2<x<w/2$ in $z=0$, with uniform surface current $K=I/w$. Above the strip's centre, the contributions odd in $x$ cancel, while the Biot-Savart law gives the magnetic field of a uniformly conducting stripThe field reverses below the strip. Its field lines wrap around the longitudinal current: nearly parallel to the faces near the centre, then curving round the edges and approaching circles at large distances. The sketch is an original field integration; the conductor thickness is idealized as negligible.

At perpendicular distance $z\ll w$ from the middle of a face, but outside the conductor, the strip approximates an infinite current sheet. The tangential field jump is ${\mu }_{0}K$ and symmetry gives equal opposite fields on the two sides. Hence $B\simeq {\mu }_{0}I/(2w)$ on either side. The centreline integral gives the same limit as ${\tan }^{-1}(w/(2z))\to \pi /2$. Uniform current across the width and distance from the strip edges are implicit in this sheet approximation; a nonuniform surface current would require its local value instead.

At distance $r\gg w$, the strip is indistinguishable to leading order from a straight line current. Ampère's law on a large circular contour gives $B\simeq {\mu }_{0}I/(2\pi r)$, directed tangentially around the strip's longitudinal axis by the right-hand rule. On the perpendicular centreline this also follows by expanding ${\tan }^{-1}(w/(2z))\simeq w/(2z)$.

Faraday's law for the moving loop is $\mathcal{E}=-d{\Phi }_B/dt$, equivalently $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }$ for its moving circuit. Lenz's law fixes its orientation: the induced current opposes the change in linked magnetic flux. The minus sign represents this opposition, not opposition to the existing field in every phase of motion.

Unroll the small loop's motion locally into translation to the right at speed $v\simeq b\omega$, where $b$ is the radius marked in the source figure. Take the imposed field into the page. On entry the linked into-page flux increases, so the current is counterclockwise. The leading vertical side carries current upward and feels a leftward magnetic force $iBa$. The in-field parts of the horizontal sides feel equal opposite vertical forces; the trailing side is outside the field. On exit the current is clockwise. The trailing vertical side is now the one in the field and again carries upward current, producing leftward drag; the horizontal forces reverse and still cancel. When the entire loop is within the uniform region, its flux is constant, so $i=0$ and every side force is zero under the stated negligible-self-inductance approximation.

For a loop crossing a magnetic field boundary, overlap changes at rate $av$, henceFor approximately constant speed this gives two negative rectangular force pulses of duration $a/v$, separated by a short zero-force interval while the loop is fully inside. Their time integral isThe quoted impulse is its magnitude; this impulse of magnetic braking opposes the motion. In fact its independence of speed remains exact within this locally translational sharp-boundary model even if speed changes: each crossing contributes $-(B^2{a}^2/R)\int vdt=-(B^2{a}^2/R)\int dx$ over distance $a$.

Write the system's moment of inertia as $\mathcal{I}$. Since $d{K}_{\mathrm{rot}}/dt=\tau \omega =F_{x}b\omega =F_{x}v$, the local energy balance during overlap isThus rotational energy lost is exactly resistive heat in this model. At nearly constant $v$, each pass givesFor a finite speed reduction the exact relation is the integral of $i(t{)}^{2}R$, rather than initial speed times impulse; the equality of work and heating still holds. Fringing fields, loop curvature and finite self-inductance smooth the ideal pulses. A practical user control is to vary the electromagnet current, hence $B$: braking scales as $B^2$.

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.

In the stipulated constant-speed model, the Earth-frame speed is $v=D/t=3.844\times 10^8/(76\times 3600)\simeq 1.405\times 10^3$ m/s. The Earth and Moon are stationary endpoints in this model, so their Earth-frame separation $D$ is the proper length. The spacecraft-frame separation, measured simultaneously in that frame, is $D^{\prime }=D/\gamma =D\sqrt{1-v^2/c^2}$. ThusThe spacecraft-frame distance is shorter by about 4.2 mm. This compares simultaneous endpoint measurements in two frames, not the coordinate difference between a launch event and an arrival event. The small-speed length contraction estimate also avoids numerical cancellation when evaluating the tiny contraction.

For a particle of rest mass $m_r$, $E=\gamma m_r{c}^2$ and $p=\gamma m_{r}v$, with $\gamma =(1-v^2/c^2{)}^{-1/2}$. The low-speed expansion is $E=m_r{c}^2+\frac{1}{2}{m}_r{v}^2+O(m_r{v}^4/c^2)$, separating invariant rest energy from the nonrelativistic kinetic energy term.

Conserve the total laboratory energy and momentum at coalescence: $E_{\mathrm{tot}}=4m{c}^2$ and $p_{\mathrm{tot}}=2.4mc$. Since $V=c^2{p}_{\mathrm{tot}}/E_{\mathrm{tot}}$, the combined particle moves at $0.6c$. Its rest mass is $M=\sqrt{4^2-2.4^2}m=3.2m$. The incident particle is not of mass $m$: its stated energy and momentum imply $m_{\mathrm{incident}}=\sqrt{3^2-2.4^2}m=1.8m$. ThusThe extra $0.4m{c}^2$ is internal rest energy retained by the composite after a perfectly inelastic collision; total energy and four-momentum are conserved, while the sum of individual rest masses need not be.

The composite's Lorentz factor is $1.25$. For a rest-frame lifetime $\tau$, the corresponding laboratory time is $1.25\tau$ and its travel distance isIf $\tau$ denotes a mean radioactive lifetime, this is the mean flight distance, not a fixed distance for every decay. The separate frame diagrams below use the common positive collision axis.

Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

A small longitudinal electric field drives conductor current in a lossy coaxial line. Its cross product with the azimuthal magnetic field sends energy radially into both conductors, where it becomes heat; forward power decreases with distance.