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For a wavefunction $\Psi (x,t)$, the probability density and probability current areMultiply the Schrodinger equation by ${\Psi }^{\ast }$, multiply its complex conjugate by $\Psi$, and subtract. The real potential energy terms cancel, leavingwhich is the probability continuity equation ${\rho }_t+j_x=0$.

For a stationary state, ${\rho }_t=0$, so $j_x=0$ and $j$ is constant in space. A normalizable wavefunction has vanishing current at spatial infinity, forcing this constant to be zero. By contrast the nonnormalizable plane wavehas stationary density $|A{|}^2$ and nonzero currentwhen $k\ne 0$.

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $\mathbf{v}=\dot{\mathbf{r}}$. Relative to the centre of mass, the two positions are ${\mathbf{r}}_1=-(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=(M_1/M)\mathbf{r}$. Their velocities have the same mass factors. Adding their kinetic energies and angular momenta gives, with reduced mass $\mu =M_1{M}_2/M$,These are the internal orbital quantities, excluding any uniform centre-of-mass motion. Newton's law of universal gravitation givesFor constant masses, differentiating $E$ makes the gravitational work cancel the derivative of the potential energy, so $\dot{E}=0$. Also $\dot{\mathbf{h}}=\mathbf{r}\times \ddot{\mathbf{r}}=0$, since the force is central; hence conservation of angular momentum gives $\dot{\mathbf{J}}=0$.

The eccentricity vector, the normalized attractive Laplace-Runge-Lenz vector, isUsing $\dot{\hat{\mathbf{r}}}=\mathbf{v}/r-\mathbf{r}(\mathbf{r}\cdot \mathbf{v})/r^3$ andshows that its derivative also vanishes. Adding a relative perturbing acceleration $\mathbf{f}$ changes these cancellations only by the perturbing terms. The perturbed Kepler-orbit conservation laws are therefore

For the distant Newtonian gravitational potential, expand each source term about the binary's centre of mass:The dipole vanishes because ${\sum }_i{M}_i{\mathbf{r}}_i=0$. With $\mathbf{a}=\mathbf{r}$, the quadratic mass moment is ${\sum }_i{M}_i{\mathbf{r}}_i{\mathbf{r}}_i=\mu \mathbf{aa}$, so the quadrupole potential of a circular binary isOn the inner binary's fast timescale, the distant companion's position is approximately fixed. Averaging over the inner circular orbit in the common plane gives $⟨{\cos }^2\theta ⟩=1/2$ and henceThis averaged potential function is independent of both time and azimuth. Consequently the companion's energy and axial angular momentum are conserved in the averaged problem. The exact motion has small fast variations; the statement is a secular approximation for a hierarchical system away from an orbital resonance.

For a circular outer orbit, force balance givesso the circumbinary orbital-period correction isConservative binary mass transfer leaves $M$ unchanged, so it does not change the leading monopole force. It changes $\mu$ and $a$, and therefore the quadrupole correction. If the inner orbital angular momentum is conserved, $J_{\mathrm{in}}=\mu \sqrt{GMa}$ gives $a\propto {\mu }^{-2}$ and $Q=\mu a^2\propto {\mu }^{-3}$. The correction is thus largest as the binary becomes strongly unequal in mass and widens, within the hierarchy $a\ll x$.

One must also allow the outer radius to respond. For slow axisymmetric evolution its specific angular momentum $h_3$ remains constant, while its energy need not: the averaged potential now depends on time. A nearly circular outer orbit has $h_3^2=GMx+3GQ/(4x)$ and $P_3=2\pi x^2/h_3$. At fixed $M,h_3$, differentiation givesThus mass exchange inside the binary can produce a small measurable outer-period change even without total mass loss. Nonconservative mass loss additionally changes the leading monopole term.

The shell equation is conservative because its radial acceleration is the negative derivative of the specific potential energyMultiplying the equation of motion by $\dot{r}$ givesHenceThis is energy per unit shell mass. The total energy of a uniform sphere requires integrating the energies of its shells.

For a sphere of physical radius $R$, the shell theorem gives $M(r)=M(r/R{)}^3$ and $dM=3M{r}^{2}dr/R^3$. Assemble the sphere shell by shell to obtain its Newtonian gravitational potential energy:The cosmological constant contributes a prescribed one-body potential $-\Lambda r^2/6$, rather than a pair interaction, so there is no additional factor $1/2$:Evaluating at the turnaround radius givesAs in the question, $\Lambda$ here includes the factor $c^2$ that would multiply the geometrical cosmological constant in SI units.

For the final equilibrium apply the scalar virial theorem. The gravitational virial is $W_G$, whereas the cosmological constant force contributesThus, neglecting a surface-pressure term,At fixed $R$, a positive cosmological constant reduces the kinetic energy needed for virial equilibrium because its outward acceleration partially offsets self-gravity.

For the uniform-sphere collapse with a cosmological constant, assume a uniform final sphere of the same mass, with no energy loss or escaping matter. PutThe sphere has no bulk kinetic energy at turnaround, so $E_{\mathrm{ta}}=-A-B$. The final virial theorem givesSince ${\rho }_{\mathrm{ta}}=3M/(4\pi r_{\mathrm{ta}}^3)$,Conservation of energy therefore givesSelect the collapsing branch continuous from $x=1/2$ at $\eta =0$, rather than an unrelated root of the cubic equation.

For $|\eta |\ll 1$, the root lies close to $1/2$. Linearize $x^3$ there:Substitution into the cubic equation gives $(\eta /2-2)x+1-\eta /2\simeq 0$, whenceThe rational expression is an approximation, not an exact solution of the cubic equation. In fact both it and the exact collapsing root agree through order ${\eta }^2$; their first difference is of order ${\eta }^3$.

With zero cosmological constant, spherical collapse gives $r_f=r_{\mathrm{ta}}/2$. For small positive $\Lambda$, the same sphere with the same turnaround radius reaches a smaller final radius. This does not mean that repulsion promotes collapse: as the sphere contracts, $W_{\Lambda }$ becomes less negative, so some released gravitational binding energy pays the energetic cost of moving inward against the repulsive force. Combined with the modified virial theorem, energy conservation requires a more compact endpoint. Positive $\Lambda$ can prevent turnaround altogether; an inward acceleration at $r_{\mathrm{ta}}$ requiresThe perturbative radius formula applies well inside this limit, for spheres that actually turn around. A small negative cosmological constant instead gives $r_f/r_{\mathrm{ta}}>1/2$ at fixed turnaround data.

Ignoring gravitational focusing, a galaxy sweeps a cylinder of volume $\pi R^2{v}_{\mathrm{rel}}T$. Multiplying this by the number density $N$ of target galaxies and averaging the relative speed gives the mean encounter countFor independent encounters modeled by a Poisson process, the probability of at least one galaxy merger is $1-e^{-\mu }$. The printed expression $P=\mu$ is its rare-encounter approximation $\mu \ll 1$; it is not an exact probability for arbitrary $T$.

For an illustrative present-day field population take an effective merger impact parameter $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N=10^{-2}{\mathrm{Mpc}}^{-3}$, and $⟨v_{\mathrm{rel}}⟩=200\mathrm{km}{\mathrm{s}}^{-1}$. These are assumed order-of-magnitude inputs, not precise observational measurements. Taking a Hubble time of $T\simeq 1.4\times 10^{10}\mathrm{yr}\simeq 4.4\times 10^{17}\mathrm{s}$, the given distance conversion yieldsand henceThus the geometric field estimate is of order $10^{-5}$ to $10^{-4}$ per Hubble time. It scales as $R^{2}N⟨v_{\mathrm{rel}}⟩T$ and is very sensitive to environment and the adopted effective merger radius. Enhanced density in groups, gravitational focusing, and the evolution of the galaxy population are all omitted; this number is not a prediction of the full cosmological merger fraction.

Now use the stipulated rapid, distant encounter. The consistent rectilinear trajectory isWith the stated impact parameter along $x$ and velocity along $z$, this trajectory lies in the $xz$-plane. The original PDF's reference to the $xy$-plane is a typo; the local TeX also corrupts the impact-parameter direction. The original PDF fixes that direction as $x$.

Use the positive potential $G{m}_p/|\mathbf{R}-\mathbf{r}|$, whose gradient is the attractive acceleration under the question's sign convention. Its multipole expansion isThe first term has no force; the linear term accelerates the entire galaxy and disappears in the frame following its centre of mass. The leading internal tidal potential is thereforeThis is a quadrupole approximation valid for $r\ll p$, rather than an exact equality for every $r<p$. Its acceleration tidal tensor acts on $\mathbf{r}$ asIn the impulse approximation, each stellar position is held fixed during the flyby and the velocity kick is the integral of this acceleration. Put $s=vt$; terms odd in $s$ integrate to zero, whileThe integrated diagonal coefficients of the tidal tensor are consequently $(2,-2,0)/(v{p}^2)$, givingThe encounter stretches the galaxy along the impact parameter, compresses it along $y$, and gives no net leading kick along the flyby direction.

Denote the pre-encounter stellar velocity by $\mathbf{u}$, to distinguish it from the relative flyby speed $v$. The instantaneous change of specific kinetic energy isUncorrelated kicks with zero mean cross term imply $⟨\mathbf{u}\cdot \Delta \mathbf{v}⟩=0$, so the phase-averaged heating at a specified position isFor an individual star the cross term need not vanish: the formula is an ensemble or orbital-phase average. Integrating over a spherical galaxy of mass $m_g$, symmetry gives $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. Therefore the total tidal heating isFor two identical galaxies, each receives this heating with $m_p=m_g$. Adding both contributions gives

In the centre of mass frame, equal masses approach with speeds $v/2$ if $v$ is their relative speed at infinity. The initial orbital energy iswhere $m_g/2$ is the reduced mass. The orbital potential energy vanishes at infinite separation. Within the weak-deflection approximation this asymptotic speed is also the nearly constant speed used in the flyby calculation.

Conservation of energy transfers the positive internal heating out of the relative orbit. Tidal capture of galaxies occurs in this model if $E_{\mathrm{orb}}-\Delta E_{\mathrm{int}}<0$, which givesThe equality is marginal capture. A bound pair still needs subsequent evolution to coalesce, so the result is an approximate capture criterion used here as a merger criterion. Its right-hand side has dimensions of length times speed, as required.

The impulse approximation requires the flyby duration $\tau \sim p/v$ to be short compared with the stellar dynamical time ${\Omega }^{-1}$, namely $v/p\gg \Omega$. If $v/p<\Omega$, a star moves substantially during the encounter. Its orbital adiabatic invariance of an orbital action suppresses net heating by a slowly varying tidal field: adiabatic shielding of tidal encounters replaces the frozen-position calculation. The divergent $v^{-2}$ heating predicted by extrapolating the impulsive formula to slow encounters is therefore spurious. Close passages with $p$ comparable to the galaxy size or strong gravitational focusing also lie outside the derivation.

If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are

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

The kinetic energy of the two equal masses isThe extensions of the left, coupling, and right springs are respectively $x_1$, $x_2-x_1$, and $x_2$. Their potential energy is thereforeThus the mechanical Lagrangian $L=T-U$ has the requested quadratic form with the symmetric small-oscillation mass and stiffness matrices

With a nonconstant potential energy, momentum no longer diagonalizes the Hamiltonian operator. The operator derivation must use the energy eigenfunctions of the full Hamiltonian, a Dyson series, or a time-sliced Trotter product formula. In the classical derivation, the straight paths are replaced by every solution of the nonlinear Euler-Lagrange equation with the specified endpoints. The semiclassical propagator becomes a sumwhere the prefactor is the Van Vleck determinant and ${\mu }_{\mathrm{cl}}$ is a Maslov index. Unlike a quadratic theory, the classical-path sum is generally only an asymptotic approximation: the exact path integral includes fluctuations of every order. On the circle, the sum must still include all winding number sectors.

The shell feels only radial gravity and the radial force due to the cosmological constant, so its torque vanishes and its specific angular momentum $L=r^2\dot{\theta }$ is conserved. Multiplyingby $\dot{r}$ and integrating gives the conserved specific orbital energy

For a uniform sphere, assembling concentric shells gives its gravitational potential energyThe cosmological-constant potential per unit mass is $-\Lambda r^2/6$. Since $⟨r^2⟩=3r_{\mathrm{ta}}^2/5$ in a uniform sphere,

The scalar virial theorem weights a potential homogeneous of degree $n$ by $-nW$. Gravity has degree $-1$ and the $\Lambda$ potential degree $2$, so the final state obeysAt turnaround $T=0$, while the virial relation gives $E_f=W_{G,f}/2+2W_{\Lambda ,f}$. Conservation of energy, together with $M=4\pi {\rho }_{\mathrm{ta}}{r}_{\mathrm{ta}}^3/3$, then gives, for $x=r_f/r_{\mathrm{ta}}$ and $\eta =\Lambda /(4\pi G{\rho }_{\mathrm{ta}})$,The root connected continuously to the $\Lambda =0$ solution has $x=1/2-\eta /8+O({\eta }^2)$, equivalentlyto first order in $\eta$. With $\Lambda =0$, virialization occurs at half the turnaround radius. At fixed turnaround state, positive $\Lambda$ makes this equilibrium root slightly smaller because its repulsive quadratic potential enters both energy conservation and the virial relation; sufficiently strong repulsion instead prevents a bound virialized state. Negative $\Lambda$ shifts the root in the opposite direction.

The gravitational three-body Lagrangian is

Put $M_{12}=m_1+m_2$ and $M=M_{12}+m_3$. Solving the definitions of the Jacobi coordinates for three particles givesSubstitution into the kinetic energy makes all cross terms cancel:where

The pair separations areHence the gravitational potential energy iswhich is independent of $\mathbf{c}$. Thus $\mathbf{c}$ is an ignorable coordinate andThe center of mass moves uniformly because the isolated system has no external force.

Finally, substituting the inverse coordinate transformation intoagain cancels every cross term and yieldsThis is the angular momentum decomposition in three-body Jacobi coordinates.

The spatial wavefunction $\chi (x)$ is the position amplitude of a stationary quantum state, so $|\chi (x){|}^2$ is its probability density after normalization. The real function $U(x)$ is the potential energy, $E$ is the energy eigenvalue, $m$ is the particle mass, and $\hslash$ is the reduced Planck constant.

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.

Choose ordered particle angles $q_1,q_2,q_3$ withand put $\Phi =(q_1+q_2+q_3)/3$. Solving for the individual angles givesThe kinetic energy is thereforeThe collective angle is an ignorable coordinate and decouples from the gaps, so its term may be omitted from the relative Lagrangian. Substituting $\gamma =2\pi -\alpha -\beta$ into the potential energy gives

Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos \theta /r^2$. The no-penetration condition in the laboratory frame isIt fixes $A=-U{a}^3/2$, so the potential flow around a translating sphere isThe laboratory-frame velocity components are

The fluid kinetic energy iswhere $V=4\pi a^3/3$. Thus the added mass of a sphere is $\rho V/2$.

When the sphere falls, replacing fluid of density $\rho$ by material of density ${\rho }_s$ lowers the gravitational potential energy at rateThe total kinetic energy of sphere and fluid isConservation of total energy givesFor $U\ne 0$, the acceleration of a freely falling sphere with added mass isThe same formula holds at release by continuity.

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.