Write ${\gamma }^{\prime }=d\gamma /dT<0$. The Maxwell relation givesThus isothermal stretching increases the entropy. For an infinitesimal increase, $dS=-{\gamma }^{\prime }dA$; if the derivative is independent of area, this integrates to $\Delta S=-{\gamma }^{\prime }\Delta A$ at fixed temperature. Mechanical energy used to create extra surface is not, in general, identical to the increase of internal energy at fixed temperature, because the stretching can also absorb heat.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

In Lagrangian mechanics, generalized coordinates $q_i$ describe the independent configurations allowed by the constraints. For particles with ordinary conservative forces in an inertial frame, use $L=T-V$, with kinetic energy $T$ and potential energy $V$. Holonomic constraints can often be built into the coordinates, avoiding explicit constraint forces.

The principle of stationary action makes the action $\mathcal{S}={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$ stationary for variations vanishing at the endpoints. Integrating the velocity variation by parts givesSince each independent variation is arbitrary, the Euler-Lagrange equations areIf nonconservative generalized forces are present, their components appear on the right. The formalism is invariant under suitable coordinate changes and adding a total time derivative to $L$ leaves the fixed-endpoint equations unchanged.

The generalized momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If $q_i$ is cyclic, $\partial L/\partial q_i=0$ and $p_i$ is conserved. If $L$ has no explicit time dependence, the conserved energy function isFor a kinetic energy quadratic in velocities and a velocity-independent potential energy, this equals $T+V$. This connects symmetry to conservation laws.

For example, a simple pendulum of mass $m$ and length $\ell$ hasIts tension does not appear because the fixed-length constraint is already imposed. Expanding for small $\theta$ gives simple harmonic motion with $\omega =\sqrt{g/\ell }$. More generally, expansion of $T$ and $V$ to quadratic order about a stable equilibrium leads directly to the normal-mode eigenvalue problem.

Let a coordinate frame rotate with angular velocity $\mathbf{\Omega }$ relative to an inertial frame. For any vector $\mathbf{u}$,Applying this twice to position, with coincident nonaccelerating origins, givesNewton's second law written using rotating coordinates therefore contains the Coriolis forceThe other inertial corrections are the centrifugal force and, for changing rotation rate, the Euler force. An accelerating origin introduces an additional translational fictitious force.

The Coriolis force depends on velocity and is perpendicular to it, so ${\mathbf{F}}_C\cdot {\mathbf{v}}_{\mathrm{rotating}}=0$: the Coriolis force does no work in the rotating frame. It bends trajectories without directly changing their rotating-frame speed. It is absent for motion parallel to the rotation axis.

For locally horizontal motion on Earth at latitude $\lambda$, take $x$ east, $y$ north and $z$ up. Keeping the vertical component $\Omega \sin \lambda$ givesHorizontal motion is deflected right in the northern hemisphere and left in the southern hemisphere. In the ideal absence of other horizontal forces these equations give inertial circles with angular frequency $|f|$. The same velocity-dependent correction explains the precession of a Foucault pendulum, whose precession-rate magnitude is $\Omega |\sin \lambda |$. Terrestrial vertical motions also involve Earth's horizontal rotation component, so the reduced horizontal equations are not the full force law.

A Hohmann transfer orbit connects two coplanar circular orbits using two tangential, effectively instantaneous velocity changes. Take outward transfer radii $r_2>r_1$, and write ${\mu }_g=GM$ for the central body's gravitational parameter. The intervening ellipse has periapsis $r_1$, apoapsis $r_2$, semimajor axis $a_t=(r_1+r_2)/2$ and orbital eccentricity $e=(r_2-r_1)/(r_1+r_2)$.

The circular speed is $\sqrt{{\mu }_g/r}$. Conservation of orbital energy gives the vis-viva equation $v^2={\mu }_g(2/r-1/a_t)$. Therefore the transfer speeds at the two impulses areThe first prograde impulse raises the far side of the orbit; the second prograde impulse raises the near side to circularize at $r_2$. Their magnitudes areThe transfer takes half an ellipse period. Kepler's third law givesFor inward transfer the same ellipse is traversed from apoapsis to periapsis and both burns are retrograde. A rendezvous also needs the correct initial phase of the target: for outward transfer its initial angular lead is $\pi -\sqrt{{\mu }_g/r_2^3}{t}_t$ modulo $2\pi$. The method assumes a two-body gravitational field, negligible atmospheric drag and burns short compared with an orbit. It is an economical two-impulse transfer between these circles; it is not universally the least-$\Delta v$ transfer if additional impulses and very large intermediate radii are permitted.

The absence of magnetic monopoles is expressed by $\nabla \cdot \mathbf{B}=0$. In a regular simply connected region, a magnetic vector potential may be introduced so thatIts units are $\mathrm{T}\mathrm{m}=\mathrm{Wb}{\mathrm{m}}^{-1}$. It is not unique: adding a gradient leaves the magnetic field unchanged because $\nabla \times \nabla \chi =0$. In time-dependent electromagnetism,is a gauge transformation preserving both physical fields. Stokes theorem gives ${\oint }_C\mathbf{A}\cdot d\mathbf{\ell }={\int }_S\mathbf{B}\cdot d\mathbf{S}={\Phi }_B$ for a contour bounding an appropriate surface.

For localized steady currents, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Substitution into Ampère's law and the curl-of-curl identity givesBoundary conditions are needed in addition to the gauge choice. For a uniform magnetic field one convenient choice is $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. For a magnetic dipole, $\mathbf{A}={\mu }_0\mathbf{m}\times \mathbf{r}/(4\pi r^3)$ outside its source. The magnetic vector potential simplifies field calculations and appears directly in the Lagrangian of a charged particle, $L=\frac{1}{2}m{v}^2+q\mathbf{v}\cdot \mathbf{A}-q\varphi$. In multiply connected regions, a field-free region can retain nontrivial circulation of $\mathbf{A}$, relevant to the Aharonov-Bohm effect; a local gauge argument cannot remove every global circulation.

An electrical transformer transfers alternating electrical power between windings through a shared changing magnetic flux. A core with large magnetic permeability confines most of the flux, so both windings link approximately the same flux $\Phi (t)$. Faraday's law gives induced electromotive forcesFor an ideal electrical transformer, negligible winding resistance and leakage flux imply the terminal-voltage magnitude ratioThe relative phase depends on the winding sense and terminal convention. At a loaded secondary, Lenz's law makes the secondary current oppose the flux change. The primary draws an additional current which approximately cancels the secondary magnetomotive force: $N_p{I}_p\simeq N_s{I}_s$ in magnitude when the magnetizing current is negligible. Consequentlyfor a resistive load, using root mean square values. More generally the ideal device conserves instantaneous power, with the appropriate current signs. A load impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_p/N_s{)}^2{Z}_L$, enabling impedance matching and electrical isolation.

For sinusoidal flux, $\Phi ={\Phi }_{\max }\sin \omega t$ gives $V_{\mathrm{rms}}=\omega N{\Phi }_{\max }/\sqrt{2}$. At a given supply voltage, reducing frequency increases the required peak flux and can drive the core into magnetic saturation. Real losses include winding Joule heating, magnetic hysteresis, eddy currents and leakage flux; a laminated core suppresses eddy-current loops. A steady direct current produces no continuing secondary electromotive force, while its small winding resistance can allow a damaging primary current after switching transients.

Integral conservation. Electric charge cannot be created by ordinary transport. For a fixed volume $V$, outward current reduces the enclosed charge:Here ${\rho }_e$ has units $\mathrm{C}{\mathrm{m}}^{-3}$ and $\mathbf{J}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Applying the divergence theorem to an arbitrary volume gives the local continuity equationIt is local charge conservation: a net outflow must be accompanied by depletion. In a steady state, $\nabla \cdot \mathbf{J}=0$. Integrating over a small circuit junction then gives Kirchhoff's current law. During a transient the junction can accumulate charge and the incoming and outgoing conduction currents need not yet match.

Consistency with Maxwell's equations. In vacuum, take the divergence of the Ampère-Maxwell equation:Gauss's law, $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, reproduces the continuity equation. The displacement current term makes these equations consistent for time-dependent charge distributions, such as a charging capacitor. It is not a stream of free charges across the insulating gap. In matter, $\nabla \cdot \mathbf{D}={\rho }_f$ and $\nabla \times \mathbf{H}={\mathbf{J}}_f+{\partial }_t\mathbf{D}$ give the analogous conservation law for free charge; total charge also includes bound charge.

Magnetic field and force. A steady electric current produces a magnetic field. In vacuum the Biot-Savart law givesThe Lorentz force on the charge carriers transfers force to the material carrying them. The resulting force on a current element of another circuit is $d{\mathbf{F}}_2=I_{2}d{\mathbf{\ell }}_2\times {\mathbf{B}}_1$, or, for a continuous current distribution, the magnetic force per unit volume is $\mathbf{f}=\mathbf{J}\times \mathbf{B}$.

For two long parallel straight wires a distance $r$ apart, Ampère's law gives $B_1={\mu }_0{I}_1/(2\pi r)$. ThereforeThe force is attractive for parallel currents and repulsive for opposite currents, by the cross product rule. Its units are $\mathrm{N}{\mathrm{m}}^{-1}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element forces need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual forces when the entire system is included.

Energy description. For two linear coupled circuits the magnetic energy isHere $q$ is a mechanical coordinate and $M$ the signed mutual inductance. At prescribed currents, the generalized force is $I_1{I}_2\partial M/\partial q$ if the self-inductances do not vary. This positive derivative is obtained from magnetic coenergy, or by including the electrical sources' work; differentiating $W_B$ with a minus sign at fixed currents would omit that source work. At prescribed flux linkages the force instead equals the negative derivative of stored magnetic energy with those linkages fixed. The geometry thus determines both inductance and mechanical force.

The torque difference. Let $\varphi (x,t)$ now be the local angular displacement, and take $G$ and the section constant. The local twist per unit length is ${\partial }_x\varphi$, so the transmitted torque is $C(x,t)=G{J}_p{\partial }_x\varphi$. Across a short slice,Thusto first order in slice length.

The slice's angular acceleration. The slice's axial moment of inertia is $\Delta I=\rho J_p\Delta x$. Its net torque is the difference of the end torques, soCanceling the common geometry gives the torsional waves in a solid circular rod,For the given steel, $c_t\simeq 3.16\times 10^3\mathrm{m}{\mathrm{s}}^{-1}$. The radius cancels because both torsional stiffness and slice inertia contain the same $J_p$. This ideal one-dimensional wave model neglects dissipation and additional elastic modes.

For the uniform solid cylinder, the axial moment of inertia isThe rotational kinetic energy at $\Omega =300\mathrm{rad}{\mathrm{s}}^{-1}$ isThis includes the shaft's bulk steady rotation; the small twist does not appreciably alter its mass moment at this order.

The torque rises linearly with elastic twist, so the stored elastic energy is the work integralUsing the transmitted torque and end twist,This is energy associated with deformation, distinct from the much larger kinetic energy of the whole spinning shaft.

Power is transmitted at rate $P=C\Omega$, soFor $a=0.040\mathrm{m}$, $L=3.0\mathrm{m}$ and $G=80\times 10^9\mathrm{Pa}$, the polar second moment of area is $J_p=4.021\times 10^{-6}{\mathrm{m}}^4$. HenceThe density does not enter this static torsional deformation.

The extra radial term. A circular Kepler orbit has ${\Omega }_{\mathrm{orb}}^2=GM/R^3$. In the frame rotating with the synchronously spinning body, centrifugal acceleration increases by ${\Omega }_{\mathrm{orb}}^{2}a$ across a radial offset $a$. This adds the outward differential accelerationEquivalently, in the inertial frame the farther corotating point needs an extra inward centripetal acceleration ${\Omega }_{\mathrm{orb}}^{2}a$, which gravity alone does not supply. The total radial stretching relevant to the static corotating body is therefore $3GMa/R^3$. This is the corotating point-mass tidal field. The source's “force” expression is per unit mass.

Self-gravitational survival. Denote the asteroid radius by $s$, distinct from the orbital radius $R$. Its mass is $m_a=4\pi \rho s^3/3$ and its surface self-gravitational acceleration is $G{m}_a/s^2=4\pi G\rho s/3$. Compare this restoring acceleration with radial tidal stretching $3GMs/R^3$:The corotating tidal-disruption estimate for a uniform sphere isThis is the requested spherical local force-balance estimate, not an exact fluid Roche figure. It neglects deformation and material cohesion.

Why the asteroid radius cancels. At fixed density, self-gravitational surface acceleration scales as $m_a/s^2\propto s$. The linear tidal acceleration also scales as $s$. Their ratio is therefore independent of $s$, provided $s\ll R$ so that the tidal expansion is valid. Density, rather than asteroid size, controls this estimate.

The terrestrial value. Convert $\rho =2\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ to $2000\mathrm{kg}{\mathrm{m}}^{-3}$ and use $M_E=6.0\times 10^{24}\mathrm{kg}$:This is distance from Earth's centre, not height above its surface.

Take a transverse offset $\mathbf{b}\cdot \mathbf{R}=0$. Its distance from the source changes only at second order, but the direction of the gravitational force changes at first order:in its first-order vector expansion. Subtracting the reference acceleration givesIt compresses the transverse separation. Together with the radial result, this is point-mass tidal stretching and squeezing, with tensor eigenvalues $(2,-1,-1)GM/R^3$.

For a point mass, $\mathbf{g}(\mathbf{R})=-GM\mathbf{R}/R^3$. Along the outward radial direction, $g_R=-GM/R^2$ andThus a small radial displacement $\mathbf{a}=a\hat{\mathbf{R}}$ has differential accelerationThe farther point falls inward less strongly, so its acceleration relative to the reference point points outward. The formula is first order in $|a|/R$.

Both the magnitude $J$ of angular momentum and rotational kinetic energy are conserved after the impulse. At fixed $J$, eliminate $J_3^2=J^2-J_1^2-J_2^2$ to writeThe fixed-angular-momentum energy test for principal-axis rotation makes the stability geometric. The largest moment gives a constrained energy minimum, the smallest a maximum, and the intermediate moment a saddle. Conservation of energy keeps a nearby motion near either isolated extremum, but a saddle permits departure in one direction while compensating in another. Thus both extremes are stable without dissipation; it would be incorrect to demand an energy minimum alone in this conservative problem.

Put $I_1=I_2=I_{\perp }$, with $I_3\ne I_{\perp }$. ThenRotation about the symmetry axis is stable whether that axial moment is larger or smaller than the transverse moment. In fact ${\omega }_3$ and ${\omega }_1^2+{\omega }_2^2$ are exactly constant for a torque-free axisymmetric body: the transverse components rotate into one another at the stated body-frame frequency. Stability here means bounded wobble, not relaxation back to perfect axial rotation.

For three distinct positive principal moments of inertia, the denominator of $p^2$ is positive. If $I_3$ is the largest moment, the two numerator factors have opposite signs, so $p^2<0$. If it is the smallest, they again have opposite signs. If $I_3$ is intermediate, the two factors have the same sign, so $p^2>0$. Thus the intermediate axis theorem givesFor the stable cases the body-frame wobble frequency isThe perturbation does not decay in the torque-free conservative model. For the intermediate moment, a generic small impulse excites the growing branch and eventually leaves the linear regime.

Why the shape is immaterial far away. Partition an arbitrary small planar loop into small rectangular loops with the same oriented current. Currents on internal shared edges cancel, leaving only the original boundary. Their magnetic dipole moments add to $\mathbf{m}=I\mathcal{A}\mathbf{n}$, with $\mathcal{A}$ the enclosed area. Far enough away, the leading fields add as dipoles at the same position. The general magnetic dipole field isHigher spatial multipoles contain shape-dependent corrections. The axial and transverse results above are the special directions of this expression.

The aligned terrestrial model. Take the moment to point toward geographic south. At northern latitude $\lambda$, the magnetic dipole inclination versus latitude gives local northward and downward componentsUsing the specified $\lambda =53^{\circ }$ and $B_{\mathrm{eq}}=35\mu \mathrm{T}$,The direction is geographic north and $69.4^{\circ }$ below horizontal, with no eastward component in this ideal model. These values use the latitude printed in the paper.