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Inverting $u+iv=\mathrm{arcosh}z$ gives $z=\cosh (u+iv)$, hence the elliptic coordinatesEliminating $v$ at fixed $u>0$ gives the ellipseIts focal distance is $\sqrt{{\cosh }^{2}u-{\sinh }^{2}u}=1$. At fixed $v$, eliminating $u$ instead gives the hyperbolaIts focal distance is also one, so both families have foci $(\pm 1,0)$. A fixed $v$ traces an appropriate half of the hyperbola; complementary coordinate values complete it. Degenerate values give the focal segment and portions of the coordinate axes. Away from these degeneracies the curves are orthogonal by part (i), or by ${\mathbf{r}}_u\cdot {\mathbf{r}}_v=0$ in the displayed parametrization. A branch of the inverse is used locally; its real part with $u\ge 0$ is single-valued outside the focal segment.

For a repulsive inverse-square force, $V(r)=+\kappa /r$. The effective potential is positive and decreases from infinity to zero, so a trajectory reaching a finite radius has $E>0$. For nonzero angular momentum, it is the accessible branch of a hyperbola, with $e^2=1+2E{J}^2/(m{\kappa }^2)>1$. There are no bound ellipses or finite-distance zero-energy parabolas. At zero angular momentum, the motion is straight inward followed by straight outward after reflection at a radial turning point.

For an attractive inverse-square force, write $V(r)=-\kappa /r$ with $\kappa >0$, and take nonzero conserved angular momentum $J$. The conic section orbit hasThe physically allowed cases are ellipse for $-m{\kappa }^2/(2J^2)\le E<0$, parabola for $E=0$, and hyperbola for $E>0$. The minimum allowed energy gives the circular special case $e=0$. Here $E$ is measured relative to infinity. Zero angular momentum instead gives radial collision or escape trajectories, interpreted as degenerate limits rather than regular closed ellipses.

For two isolated point masses, introduce total mass $M_T=M+m$, reduced mass $\mu =Mm/M_T$ and relative separation $r$. The relative acceleration is $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. Conserved specific angular momentum $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ fixes a plane and a constant areal velocity $h/2$. The specific orbital energy isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.

For a test mass moving around a fixed central mass $M$, gravity is central and conservative. Specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|$ and specific orbital energy $ϵ=v^2/2-GM/r$ are constant. The motion is planar and sweeps equal areas in equal times, $dA/dt=h/2$.

For a nonradial orbit, the inverse-square force gives a conic section,Negative energy gives an ellipse, zero energy a parabola and positive energy a hyperbola. For an ellipse of semimajor axis $a$,A circular orbit is the $e=0$ case, with $v=\sqrt{GM/r}$. Escape from radius $r$ requires $v\ge \sqrt{2GM/r}$ in the absence of further forces. For two comparable masses, apply the relative two-body equation with $GM$ replaced by $G(M+m)$, rather than treating one body as fixed.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

Throughout this solution $c=1$ and $a>0$.

Null metric tensor and boost. Inverting the coordinate definitions gives $t=(p+q)/2$, $x=(q-p)/2$, and henceFor a frame moving at speed $v$ along positive $x$, use $t^{\prime }=\gamma (t-vx)$, $x^{\prime }=\gamma (x-vt)$. Their difference and sum giveThe reciprocal scale factors preserve $dpdq$. They are $e^{\pm {\eta }_v}$ in rapidity notation.

acceleration invariants. Proper-time normalization gives ${\dot{t}}^2-{\dot{x}}^2=1$, so $\dot{p}\dot{q}=1$. The four-acceleration has squared norm ${\ddot{t}}^2-{\ddot{x}}^2=-a^2$; thereforeFor future motion write $\dot{p}=e^{-\eta }$ and $\dot{q}=e^{\eta }$. Then the second equation is $-(\dot{\eta }{)}^2=-a^2$. acceleration toward positive $x$ selects $\eta ={\eta }_0+a\tau$.

hyperbolic motion from an initial event and rapidity. The stated initial event fixes $p(0)=-1/a$, $q(0)=1/a$, but not the initial rapidity ${\eta }_0$. Integrating with positive signed acceleration gives the general familyThe initial event alone does not fix the initial velocity or signed acceleration. The trajectory explicitly requested in the next paragraph selects the additional conditions ${\eta }_0=0$ and positive-$x$ acceleration. With that convention,and consequentlyWith arbitrary ${\eta }_0$ the equivalent expressions are $t=[\sinh ({\eta }_0+a\tau )-\sinh {\eta }_0]/a$ and $x=1/a+[\cosh ({\eta }_0+a\tau )-\cosh {\eta }_0]/a$. Thus position data alone do not select the displayed special hyperbola. The following calculations use that intended hyperbola, which is also explicitly given by the paper.

Diagram and horizon. The selected world line is the right-hand branch $x^2-t^2=a^{-2}$. Its future asymptote $x=t$ is a reception horizon: rightmoving signals have constant $p=t-x$, whereas the observer has $p<0$ and approaches zero only as $\tau \to \infty$. Events with $p\ge 0$ cannot send a future signal to this observer if the acceleration continues indefinitely. This is an observer-dependent causal restriction in flat spacetime, not a curvature singularity.

photon reception by a uniformly accelerated observer. A rightmoving photon from the stationary source has four-momentum $P^{\mu }=h{\nu }_0(1,1)$. Contracting with the observer's four-velocity givesThe same result follows by differentiating its emission-time relation. A photon emitted at $x=L$ and time $t_e$ has $p=t_e-L$, so intersection requiresThus $t_e=L$ is the supremum of receivable emission times, reached only in the limit of infinite receiver proper time. A photon emitted exactly then lies on the horizon and is never received at finite time; later photons cannot catch the ship. Restoring dimensions gives the limiting emission time $L/c$ for this chosen origin and trajectory.

Rindler coordinates. Substitution of the selected trajectory into the coordinate transformation givesDifferentiating the transformation also gives $d{s}^2=e^{2a\zeta }(d{\xi }^2-d{\zeta }^2)$. Curves of fixed $\zeta$ are hyperbolae, while curves of fixed $\xi$ are straight rays $t/x=\tanh (a\xi )$ within the right wedge. The coordinate $\xi$ is the selected ship's proper clock; at another fixed $\zeta$, proper time obeys $d{\tau }_{\zeta }=e^{a\zeta }d\xi$. The spatial coordinate is logarithmic in $\rho =\sqrt{x^2-t^2}=e^{a\zeta }/a$, rather than being proper distance itself. Fixed-$\zeta$ observers have proper acceleration $a{e}^{-a\zeta }$.

Let $e>0$ be the magnitude of the electron's electric charge, $m=m_e$, $F=eE>0$, and $\ell =m{c}^2/F$. The Lorentz force is $d{p}_x/dt=qE$, so the electron and Positron acquire opposite momenta:Using the relativistic energy-momentum relation, $u_x=p{c}^2/\mathcal{E}$, and integrating from rest gives the world linesFor $t<0$, the world lines are vertical at $x=0,d$ on a spacetime diagram. They join smoothly to the accelerated branches with zero initial velocity. At small $t$, $x_e\simeq -F{t}^2/(2m)$ and $x_p-d\simeq F{t}^2/(2m)$; at large $t$, their velocities approach $-c,+c$, respectively. Each accelerated world line is a branch of a hyperbola, since $(x_e-\ell {)}^2-c^2{t}^2={\ell }^2$ and $(x_p-d+\ell {)}^2-c^2{t}^2={\ell }^2$. The electric field separates the opposite electric charges in this orientation. The asymptotes are lightlike, while every finite-time tangent is timelike.

Assume $E,B>0$ for the directions and parametrizations; the boost formulas retain their signed meaning for reversed fields. A Lorentz boost with velocity $w\hat{\mathbf{z}}$ along $\mathbf{E}\times \mathbf{B}$ transforms the transverse electric field and magnetic field asIn the electric-dominated case chooseThe inequality ensures $|w|<c$, so this is an inertial frame. The field-removing boosts for perpendicular electric and magnetic fields are governed by the electromagnetic field invariants $\mathbf{E}\cdot \mathbf{B}=0$ and $E^2-c^2{B}^2>0$ explain why a purely electric inertial frame exists and a purely magnetic one does not.

The electron is not initially at rest in this convenient inertial frame: its initial velocity is $-w\hat{\mathbf{z}}$ and its initial three-momentum is $p_z^{\prime }=-{\gamma }_{w}mw$. Thereafter $p_x^{\prime }=-e{E}^{\prime }{t}^{\prime }$, while $p_z^{\prime }$ stays constant. The electric field accelerates it toward $-x$; as its relativistic energy increases, its negative transverse velocity decreases in magnitude. Defining $a=e{E}^{\prime }/m>0$ and $s=a\tau /c$ gives a useful parametric world line:Here $\tau$ is the proper time. These formulas follow by integrating $U^{\prime 0}={\gamma }_{w}c\cosh s$, $U^{\prime x}=-{\gamma }_{w}c\sinh s$, and constant $U^{\prime z}=-{\gamma }_{w}w$. Its projection onto the $t^{\prime },x^{\prime }$ plane is a hyperbola, but the spatial trajectory includes the initial transverse motion and is not a straight line.

Transforming back with $z={\gamma }_w(z^{\prime }+w{t}^{\prime })$, $t={\gamma }_w(t^{\prime }+w{z}^{\prime }/c^2)$ givesThe laboratory trajectory starts from rest toward $-x$, then bends toward $+z$. It is unbounded in $-x$ and ultimately has $u_z\to w=c^{2}B/E$ and $u_x\to -c/{\gamma }_w$. Thus the late velocity has magnitude $c$ as a limit, consistent with the relativistic energy-momentum relation, rather than undergoing a closed cyclotron orbit.

For the standard right-wedge hyperbola and a stationary source at $x=L$ emitting rightmoving photons, the retarded null coordinate is $p=t-x=-e^{-a\tau }/a$. Equating it to the emission value $t_e-L$ gives the displayed reception relation. Differentiating yields the Doppler frequency ratio $\nu /{\nu }_0=d{t}_e/d\tau =e^{-a\tau }$. The limiting emission time $L$ is a supremum: emission exactly at that time is never received at finite proper time. These expressions use $c=1$ and acceleration continued indefinitely.

A conic section with a focus of a conic at the origin has the polar coordinates equation $r=\ell /(1+e\cos \theta )$, where $\ell >0$ is the semilatus rectum and $e\ge 0$ the eccentricity. The directrix of a conic definition $r=e(d-r\cos \theta )$ gives the formula with $\ell =ed$ when $e>0$; $e=0$ is the circle limit. The cases $0<e<1$, $e=1$, and $e>1$ give an ellipse, parabola, and hyperbola. The physical radius requires a positive denominator. For an attractive inverse-square central force, the Laplace-Runge-Lenz vector gives $\ell =m{h}^2/k$ and $e=A/(mk)$.

Vary the coupling in the opening example. The eigenvectors $e_{+}=(1,1{)}^T/\sqrt{2}$ and $e_{-}=(1,-1{)}^T/\sqrt{2}$ form a fixed orthonormal eigenbasis, with eigenvalues $3+t$ and $3-t$. Writing a vector as $u{e}_{+}+v{e}_{-}$ givesFor $|t|<3$, both directions cost energy and $A_t$ is a positive-definite matrix. At $t=3$, the quadratic form is $6u^2$: the entire line spanned by $e_{-}$ has zero energy, and $A_t$ is a singular positive semidefinite matrix. At $t=-3$, the flat line is instead spanned by $e_{+}$. For $|t|>3$, the quadratic form is an indefinite quadratic form; one direction lowers the energy without bound as its displacement grows. At $t=4$, for instance, $q_t(e_{+})=7$ while $q_t(e_{-})=-1$.

The geometric transition is precise. For $|t|<3$, $q_t=1$ is an ellipse. At $t=3$, it consists of the two parallel lines $u=\pm 1/\sqrt{6}$, extending freely in the zero-energy direction. For $t>3$, it is a hyperbola. The boundary case retains nonnegative energy but loses both the bounded positive level sets and the unique zero-energy vector.

Let $a>0$, $t>0$ and $z=t+ia/t$. The real and imaginary coordinates obey $XY=a$, a hyperbola. Put $s=t^2-a^2/t^2$, which is strictly increasing. Then $z^2=s+2ia$ traces a horizontal line, and $z^4=s^2-4a^2+4ias$ traces a parabola. Eliminating $s$ gives the displayed quadratic relation between real part and imaginary part. The value of $t$ fixes an orientation and determines which segment is traversed; the full real parabola is obtained as $t$ ranges over $(0,\infty )$.

A hyperbola whose two asymptotes are perpendicular. After a suitable choice of coordinates it has equation $xy=c$ for a nonzero constant $c$. In the form $x^2/a^2-y^2/b^2=1$, it is rectangular exactly when $a=b$.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

Let $\theta$ be the scattering angle between incoming and outgoing velocities, so forward scattering is $\theta =0$ and exact reversal is $\theta =\pi$. Define the length $a=\kappa /(2E)$. ThenFor $b>0$, this gives $0<\theta <\pi$; $b=0$ is the head-on limiting case $\theta =\pi$. Large deflections select small impact parameters, rather than requiring every incident alpha particle to reverse.

A short derivation uses conservation of angular momentum and conservation of energy. Choose the polar angle $\varphi =0$ at closest approach, and write $u=1/r$. The Binet equation for the repulsive Coulomb potential givesThe last equality follows by substituting this orbit into $E=m{\dot{r}}^2/2+L^2/(2m{r}^2)+\kappa /r$. The hyperbola reaches infinity at $\varphi =\pm {\varphi }_{\infty }$, with $\cos {\varphi }_{\infty }=1/ϵ$. Its scattering angle is $\theta =\pi -2{\varphi }_{\infty }$, henceThis proves the relation without assuming that the deflection is small. The symbol $ϵ$ here is a dimensionless orbital parameter, distinct from ${\epsilon }_0$.

The semi-major and semi-minor axes are the larger and smaller half-axis lengths of an ellipse. In principal-axis coordinates the ellipse has equation $x^2/a^2+y^2/b^2=1$ with $a\ge b>0$. The same terminology is used for a hyperbola $x^2/a^2-y^2/b^2=1$, where $a$ is half its transverse axis and $b$ half its conjugate axis. Orbital semiaxes apply these geometric quantities to conic sections.

The gradient of this polynomial vanishes exactly when $x=0,\pm \sqrt{a/2}$ and $y=0,\pm \sqrt{b/2}$, giving nine stationary points. Its diagonal Hessian matrix has entries $12x^2-2a$ and $2b-12y^2$. The second-derivative test therefore gives two strict local maxima at $(0,\pm \sqrt{b/2})$, with value $b^2/4$, and two strict local minima at $(\pm \sqrt{a/2},0)$, with value $-a^2/4$. The origin and the four points $(\pm \sqrt{a/2},\pm \sqrt{b/2})$ are saddle points of a scalar function, with values zero and $(b^2-a^2)/4$, respectively. Every stationary point is nondegenerate.

Completing the squares organizes the contour plot:The level set through the four outer saddle points of a scalar function factors asIt is a circle together with a hyperbola when $a\ne b$; for $a=b$, the latter factor's complementary zero set is the pair of lines $y=\pm x$. Neither the local maxima nor the local minima are global, since the restrictions to the two coordinate axes are unbounded in opposite directions. The case $a=1$, $b=4$ gives four outer saddle values $15/4$ and an exact circle-hyperbola contour crossing at all four outer saddle points of a scalar function.

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.