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Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.

By Newton's second law, force is mass times acceleration. Hence its dimensions are

Measure displacement downward from the release point and take downward speed $v\ge 0$. Newton's second law with quadratic drag givesWritingseparation of variables gives the terminal velocity solutionThe distance fallen by time $t$ is thereforeSetting $y(t)=h$ and solving for $t$ yields

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $r=|\mathbf{r}|$. Newton's second law for the two planetesimals givesSubtracting yields the forced relative two-body problemThe internal gravitational forces cancel from the centre of mass equation but add in the relative equation; external forces affect the internal orbit through their difference in acceleration.

The work done along the trajectory isBy Newton's second law, $F=m\ddot{x}$, soThis is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with $V(0)=0$ isConservation of energy for $x>0$ givesThus

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the escape velocity $v_0=v_e$, it decreases as $v_e{e}^{-x/(2\lambda )}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the turning pointwhere $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the period iswhere $u=e^{-x/\lambda }$. Therefore

Equating the satellite's mean density inside $r_t$ with the host's mean density inside orbital radius $R$ givesThus a measured tidal radius and independently estimated satellite mass infer $M(<R)\simeq m(R/r_t{)}^3$. Measurements from satellites over a range of $R$ trace the host's enclosed-mass profile and hence its gravitational potential. Globular clusters are more often tidally limited: their stellar extent can approach the Jacobi boundary, whereas dwarf galaxies commonly occupy extended dark-matter haloes and their observed stars may substantially underfill it. Orbital eccentricity, mass loss, and nonequilibrium structure complicate this inference.

Let the separation vector point from the host to the satellite. Subtracting the two Newton's second law equations givesThe relative orbit is therefore a one-body Kepler orbit with gravitational parameter $G(M+m)$. A circular orbit of separation $R_0$ hasThe centre-of-mass condition gives $x_m=M{R}_0/(M+m)$ for the satellite's distance from the barycentre.

Consider the inner collinear equilibrium a distance $r_J$ toward the host from the satellite. Taking the positive axis from the host toward the satellite, differentiation of the gravitational plus centrifugal effective potential givesSince $r_J\ll R_0$,The constant host-gravity term cancels ${\Omega }^2{x}_m=GM/R_0^2$. The remaining equation isFor a low-mass satellite, the term proportional to $m{r}_J$ is negligible compared with $3M{r}_J$, and the Jacobi tidal radius isThe outer collinear point gives the same leading result.

Stars that cross the two nearby Lagrange points are no longer bound to the satellite. Small energy and angular-momentum offsets place them on slightly different host orbits, producing one leading and one trailing tidal tail. Differential orbital frequency stretches these streams around the host, while epicyclic motion can create density clumps. The tail's position and velocity structure retain information about the satellite orbit and host potential.

Take positive $y$ vertically downward. For a short string element, the transverse tension resultant is $T{y}_{xx}dx$, while gravity and linear drag contribute $\mu gdx$ and $-2k\mu y_{t}dx$. Newton's second law givesSince $T=\mu c^2$, the damped wave equation isChoosing upward displacement reverses the sign of the gravity term.

Assume Newton's second law, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summingcancels internal pairs and gives $dP/dt=F:={\sum }_i{F}_i$. Taking moments about fixed $a$ cancels the internal torques pairwise and gives

The angular momentum about the centre of mass result follows similarly: differentiating ${\sum }_i(r_i-R)\times m_i({\dot{r}}_i-\dot{R})$ introduces no extra term because both total relative position weighted by mass and total relative momentum vanish. Thus its derivative is the external torque about $R$.

Finally, if every mass is $m$ and $F_i=-k{\dot{r}}_i$, then about a fixed pointTherefore

For a particle in the scalar potential energy $U(x)$,The canonical commutation relation $[x,p]=i\hslash$ givesThe Ehrenfest theorem with $O=x$ therefore givesSince $[p^2,p]=0$ and direct action on a wavefunction gives $[U(x),p]=i\hslash U^{\prime }(x)$,Finally $H$ has no explicit time dependence and $[H,H]=0$, so

The first two equations are the expectation-value counterparts of momentum equals mass times velocity and Newton's second law; the last is conservation of energy. They become the classical equations directly when the force varies negligibly across the wave packet, so that $⟨U^{\prime }(x)⟩\simeq U^{\prime }(⟨x⟩)$. These conclusions are summarized by the classical equations from Ehrenfest theorem.