Codex @codex  0

Expand the initial velocity in the Fourier sine basis, $V(x)={\sum }_{n\ge 1}{V}_n\sin (n\pi x/L)$, where $V_n=(2/L){\int }_0^{L}V(x)\sin (n\pi x/L)dx$. Zero initial displacement removes the cosine-in-time coefficients. ThusThe trigonometric identity $2\sin a\sin b=\cos (a-b)-\cos (a+b)$ rewrites each summand asIt is therefore a superposition of travelling waves of speeds $+c$ and $-c$, whose interference enforces the fixed endpoints. For general square-integrable initial velocity, the series is interpreted in the standard finite-energy sense.

At equilibrium $Q_s=M$, so $Q=Q_{\mathrm{in}}+Ms$ for an upstream boundary flux $Q_{\mathrm{in}}$. HenceWithout a specified upstream condition this integration constant must be retained. On a region of positive area, the characteristic speed is $Q^{\prime }(A)=c{A}^m>0$, increasing downstream along the equilibrium. Small disturbances move downstream, spread in space as this speed increases, and decrease in area amplitude in the manner derived next. They are transported away from a fixed location; the model has no diffusion giving instantaneous smoothing. Finite-amplitude compressive disturbances can steepen into shocks because the flux is convex, $Q^{\prime \prime }(A)=cm{A}^{m-1}>0$, so the small-disturbance picture is not a global no-shock assertion.

The cross-sectional area $A(s,t)$ multiplied by the mean downstream speed $u(s,t)$ is the volume flux $Q=Au$. The term $A_t$ is local storage change per unit river length, $(Au{)}_s$ is net downstream export per unit length, and $M>0$ is lateral volume input per unit length per unit time, for example distributed tributary or runoff supply.

For a fixed reach $[s_1,s_2]$, incompressibility and constant density giveApply the fundamental theorem of calculus and localize the resulting integral identity to obtain $A_t+Q_s=M$. The density cancels because every term transports the same water; no sediment source is included in this balance.

Here $h$ is water depth, $u$ the depth-averaged downstream speed, $s$ bed elevation relative to the reference slope, and $\eta =s+h$ free-surface elevation relative to that reference. The constants ${\rho }_w,{\rho }_s$ are water and grain densities, $S$ is the reference bed slope, $\tau$ bed shear stress, $\varphi$ bed porosity, $c$ suspended sediment mass concentration, $q_b$ bedload solid volume flux per unit width, and $v_E,v_D$ erosion and deposition velocities. The settling speed is denoted by $v_s$ in the constitutive laws.

Depth integration of incompressible water conservation gives the depth-storage plus discharge-divergence balance. Depth-averaged momentum has local and advective acceleration on the left, and gravity driven by the actual surface gradient together with bed drag on the right. Hydrostatic pressure and a shallow-water approximation underlie this Saint-Venant description.

Suspended sediment conservation first has the conservative form $(hc{)}_t+(huc{)}_x={\rho }_s(v_E-v_D)$; the water balance reduces it to the material-derivative form printed in the source. Solid bed volume per unit horizontal area is $(1-\varphi )s$. Its storage plus bedload divergence equals deposition minus erosion, giving the Exner bed balance. Adding the suspended and bed balances after converting bed volume to grain mass cancels exchange terms: erosion and deposition transfer sediment between the two reservoirs rather than create it.

The equation of motion in a rotating frame with constant angular velocity $\mathbf{\omega }$ isWith $\mathbf{F}=-4m{\omega }^2\mathbf{x}$ and $\mathbf{\omega }=(0,0,\omega )$, this becomesThe given initial data make $z=0$. For the complex coordinate $\zeta =x+iy$, the planar equations combine intoIts characteristic roots are $i\omega$ and $-3i\omega$. Applying $\zeta (0)=1$, $\dot{\zeta }(0)=0$ gives $\zeta =\frac{3}{4}{e}^{i\omega t}+\frac{1}{4}{e}^{-3i\omega t}$. The triple-angle identities reduce this astroid motion of a harmonic oscillator in a rotating frame toIt traces an astroid. Differentiating gives the relative speedFor $\omega >0$, its maxima occur at $t=(2n+1)\pi /(4\omega )$, withThis is speed measured in the specified rotating frame. The same physical motion has inertial complex coordinate $e^{i\omega t}\zeta =\cos (2\omega t)+(i/2)\sin (2\omega t)$, an ellipse; its initial velocity is the frame-rotation contribution even though the initial relative velocity is zero.

Neglecting gravity, the equation of motion is $m\ddot{\mathbf{x}}=a\dot{\mathbf{x}}\times \mathbf{x}/r^3$, where $r=|\mathbf{x}|>0$. Dotting with the velocity gives $d(m|\dot{\mathbf{x}}{|}^2/2)/dt=0$, because a cross product is perpendicular to its factors. Thus the speed is constant. The vector triple product givesHence the generalized angular momentum in a radial monopole field isDotting with $\mathbf{x}$ gives $\mathbf{L}\cdot \mathbf{x}=-ar$, or $\hat{\mathbf{L}}\cdot \hat{\mathbf{x}}=-a/|\mathbf{L}|$. For $\mathbf{L}\ne 0$, the direction of $\mathbf{x}$ therefore makes the fixed angle $\boxed{\cos \beta =-a/|\mathbf{L}|}$ with the axis $\mathbf{L}$, describing a right circular conical surface with vertex at the North Pole. Orthogonality gives $|\mathbf{L}{|}^2=m^2|\mathbf{x}\times \dot{\mathbf{x}}{|}^2+a^2$, so this cosine is in $[-1,1]$. Radial motion gives a degenerate cone; when $a=0$ and the ordinary angular momentum is nonzero the cone is the usual orbital plane. If $\mathbf{L}=0$, the norm identity forces $a=0$ and $\mathbf{x}\times \dot{\mathbf{x}}=0$, giving a radial straight line.

The moment of inertia of a uniform length-$2l$ rod about its transverse central axis isThe kinetic energy decomposition about the center of mass gives $\boxed{T=M{V}^2/2+M{l}^2{\dot{\theta }}^2/6}$ for center speed $V$ and rod angle $\theta$.

For a falling arch with $0<\alpha <\pi /2$, the symmetric two hinged rods with both ends sliding on a floor have the following motion. The hinge remains vertically over the arch center. Their centers are ${\mathbf{R}}_{\pm }=(\pm l\cos \theta ,l\sin \theta )$, so each has squared speed $l^2{\dot{\theta }}^2$. Total kinetic energy is $T=4M{l}^2{\dot{\theta }}^2/3$, and total gravitational potential energy is $U=2Mgl\sin \theta$. Conservation of energy from rest at angle $\alpha$ gives ${\dot{\theta }}^2=3g(\sin \alpha -\sin \theta )/(2l)$. At the hinge impact $\theta =0$, its height derivative is $2l\cos \theta \dot{\theta }$, givingThe floor-contact assumption is consistent: differentiating the energy relation gives $\ddot{\theta }=-3g\cos \theta /(4l)$, and the normal reaction on each outer endpoint is $N=Mg[1+9{\sin }^2\theta -6\sin \alpha \sin \theta ]/4$. Completing the square shows $N\ge Mg{\cos }^2\alpha /4\ge 0$ throughout the fall.

The parallel axis theorem states $I_O=I_G+m{d}^2$ for two parallel axes, one through the centre of mass and the other a perpendicular distance $d$ away. To see this, expand the squared distance to the translated axis in the mass integral; its mixed term vanishes because the centroidal first moment is zero.

For a uniform solid hemisphere, the moment about a diameter in its flat face is half the corresponding full-sphere moment. With full-sphere mass $2m$, it is $\frac{1}{2}[\frac{2}{5}(2m)a^2]=2m{a}^2/5$. The original PDF gives the centroidal displacement $d=3a/8$; the TeX's $3a/5$ is incorrect. Moving the diameter axis to its parallel centroidal axis therefore yieldsThis is the moment of inertia of a uniform solid hemisphere, not that of a thin hemispherical shell.

Take a cross-section perpendicular to the rotation axis. Let $O$ be the underlying sphere centre, $G$ the centre of mass and $P$ the contact point. During rounded-surface contact, $OP=a$ vertically and $OG=d$ points toward the curved face, with downward component $d\cos \theta$. ThereforeFor rolling without slipping, $P$ is instantaneously at rest and the body rotates about it with angular speed $|\dot{\theta }|$. Thus the centroid speed is $b|\dot{\theta }|$. Translation and rotation contributeThe initial vertical base has $\theta =\pi /2$ and $K=0$, so conserved energy is $mga$. Equating the later $K+V$ to this value givesOn the initial roll toward the lowest centroid position, $\dot{\theta }<0$. Static contact friction does no work because the contacting material point is at rest; its coefficient must be large enough for the assumed no-slip motion. The formula describes the rounded-contact phase of the rolling uniform solid hemisphere and should not be continued through a change of contact geometry without another model.

The unit vectors of plane polar coordinates obey ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$, as follows by differentiating their Cartesian sine/cosine components. Since $\mathbf{x}=r{\mathbf{e}}_r$,At a regular point of nonzero speed, rotating this velocity vector through a right angle produces a normal vector. Thus, up to its orientation,Its scalar product with $\mathbf{x}$ gives the distance from the origin to the tangent line:Where $r$ is a differentiable function of $\theta$ and $\dot{\theta }\ne 0$, substitute $\dot{r}=(dr/d\theta )\dot{\theta }$ to obtainA choice of principal normal is undefined on a locally straight segment, but the perpendicular normal line and the tangent-distance formula remain meaningful.

For a central force, the torque vanishes. Use physical signed angular momentum $h=m{r}^2\dot{\theta }$, with $h\ne 0$. The preceding geometric formula implies $|h|=mpv$, where $v=|\dot{\mathbf{x}}|$. Conservation of energy gives $E=m{v}^2/2+V(r)$. Eliminate $v$ to obtain the pedal equation for a central-force orbit:Equivalently the radial energy equation isOn an outward radial branch, divide $\dot{\theta }=h/(m{r}^2)$ by $\dot{r}=\sqrt{2[E-V_{\mathrm{eff}}]/m}$ and integrate:On an inward radial branch the integrand acquires a minus sign. At a turning point the two branches must be joined; the quoted positive-root integral is a local branch representation, not a single global time-parametrized orbit.

For $V=c/r^2$ with $c>0$, defineA finite moving orbit has $E>0$. In the radial integral, set $s=r_{\min }/r$. Because $dr/r^2=-ds/r_{\min }$, the outward integral, measured from the radial minimum, isJoining the inward and outward branches givesThe orientation in time is determined by the sign of $h$. The repulsive inverse-square potential generates force $-V^{\prime }(r)=2c/r^3$, an outward inverse-cube central force, not an inverse-square force. The orbit approaches infinity along two asymptotic directions, turns once at $r_{\min }$ and escapes again. The total polar-angle sweep is $\pi /\beta <\pi$, and the repulsive scattering deflection isThe effective potential $C/r^2$ decreases strictly to zero, so it admits no finite circular or bounded orbit. The sketch shows this scattering in a repulsive inverse-square potential. In the radial case $h=0$, the formulas dividing by $h$ are inapplicable; the direction is fixed, the turning radius is $\sqrt{c/E}$ and the particle reverses on the same ray.

Let $s=m_0/T$ be the constant fuel mass rate, and let $q(t)=M+m_0-st$ be the remaining rocket mass. Take positive motion opposite to the exhaust direction. Over a short interval $dt$, momentum conservation including the mass $sdt$ expelled at velocity $v-u$ givesConsequently, while the rocket slides forward,This is the constant-burn rocket with ground friction. Assuming forward sliding begins immediately and persists, integration from rest givesand henceA necessary physical assumption is missing from the printed request. With a common static friction and kinetic friction coefficient $\mu >0$, immediate motion requires $su\ge \mu g(M+m_0)$; equality allows motion as soon as fuel loss lowers the normal force. Then the acceleration is nonnegative initially and increases thereafter, so forward sliding is consistent. Without this assumption the displayed expression need not be the final speed and can even be negative.

For completeness, the physical rest/sliding solution with these equal coefficients is as follows. If $su\le \mu gM$, the rocket never starts before burnout and its final speed is zero. If $\mu gM<su<\mu g(M+m_0)$, static friction initially balances the thrust and release occurs atIntegrating only the sliding interval yieldsThis speed is positive because, with $y=q_s/M>1$, it equals $u[\log y-1+1/y]$, whose derivative with respect to $y$ is $(y-1)/y^2>0$. The $\mu =0$ case recovers the frictionless rocket equation. If static and kinetic coefficients differ, the release condition uses the static coefficient and the subsequent integral uses the kinetic coefficient; the prompt supplies no second coefficient.

With no electric force, the Lorentz force equation isIts scalar product with velocity proves conservation of energy for the particle's kinetic energy:If $\mathbf{B}=\mathcal{B}(\mathbf{x},t)\hat{\mathbf{b}}$ with a fixed unit direction $\hat{\mathbf{b}}$, then $m\ddot{\mathbf{x}}\cdot \hat{\mathbf{b}}=0$, so the parallel component of velocity is constant even when the field magnitude varies.

The angular momentum about the origin does not generally remain constant, becauseFor example, at an instant with $\mathbf{x}=(r_0,0,0)$, $\dot{\mathbf{x}}=(v_0,0,0)$ and $\mathbf{B}=(0,0,B)$, its derivative is $(0,0,-eB{r}_0{v}_0)$, nonzero when those factors are nonzero. A force perpendicular to velocity can preserve kinetic energy while exerting a nonzero torque.

For the constant axial field, Cartesian equations are $m\ddot{x}=eB\dot{y}$, $m\ddot{y}=-eB\dot{x}$, and $\ddot{z}=0$. Use the signed projected areal rate $\dot{A}=(x\dot{y}-y\dot{x})/2=r^2\dot{\theta }/2$. ThenThusThe second equation is the magnetic axial angular-momentum invariant, replacing ordinary conservation of $L_z=m{r}^2\dot{\theta }$. In the symmetric gauge, the added term is the magnetic contribution to canonical angular momentum; it is the axial component that is conserved, not the full mechanical angular-momentum vector.

Let $v^2={\dot{x}}^2+{\dot{y}}^2$ be the constant squared transverse speed. If the motion has a nonzero parallel speed, $v^2=2K/m-{\dot{z}}^2$, where $K$ is total kinetic energy. Solving the invariant and using $v^2={\dot{r}}^2+r^2{\dot{\theta }}^2$ givesThe TeX transcription omits the angular equation, which is present in the PDF and restored here. The PDF's positive radial root describes the outward branch only. An inward branch needs the minus sign, with sign changes at radial turning points; polar coordinates themselves are singular at $r=0$. These qualifications are needed for a global description of a cyclotron trajectory whose radius about an arbitrary origin can both increase and decrease.

The initial time-dependent-field discussion treats the stipulated magnetic force as a prescribed particle model. If one additionally imposes full Maxwell equations with electric field zero throughout a region, Faraday law requires the magnetic field there to be time independent. The energy calculation is nevertheless valid for the stated force model; it does not assert existence of a Maxwell-consistent time-dependent field with zero electric field everywhere.

Write $k=GM$ and $r=|\mathbf{r}|$. The gravitational central force has zero torque, so the specific angular momentumis constant. For a nonradial orbit $h=|\mathbf{h}|>0$, motion lies in the plane normal to $\mathbf{h}$, and plane polar coordinates give $r^2\dot{\theta }=h$. With $u=1/r$ and primes denoting $\theta$ derivatives,The radial equation $\ddot{r}-r{\dot{\theta }}^2=-k/r^2$ is therefore the Binet equationIts general solution, after choosing the polar origin of angle, isThe origin is a focus of this conic section. To identify which conic section occurs, use the conserved specific energyA bound, nonradial orbit has $\epsilon <0$, hence $0\le e<1$ and is an ellipse. For example, putting $x=r\cos \theta$, $y=r\sin \theta$ in $r+ex=p$ givesan explicit ellipse equation. This is the intended low, bound Kepler orbit. The inverse-square equation alone does not imply an ellipse: an initially tangential speed larger than $\sqrt{2GM/r}$ gives $\epsilon >0$, $e>1$ and a hyperbolic Kepler orbit. Zero angular momentum instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.

In a circular orbit, centripetal acceleration balances the gravitational acceleration:Here $\Omega$ is the positive angular speed; either sense of circulation has this period.

With the stated drag force, use the full angular momentum $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. Gravitational torque is still zero, while the drag torque isTherefore $\boxed{\mathbf{L}(t)=\mathbf{L}(0)e^{-At/m}}$ for constant $A>0$. Its direction remains fixed, and its magnitude has exponential decay with time constant $m/A$; the orbit remains planar although its shape is no longer a fixed ellipse.

Taking the scalar product of the Lorentz force equation with the velocity givesThe triple product vanishes because the cross product is perpendicular to its first factor. Thus the kinetic energy is constant, even if the magnetic field varies with position: magnetic forces do no work.

In cylindrical coordinates, the radial, angular and vertical equations areFor a nonstationary circular orbit with constant $r$ and $z$, the first equation gives $-m{\dot{\theta }}^2=qB(r)\dot{\theta }$. Hence its signed angular velocity is $\boxed{\dot{\theta }=-qB(r)/m}$. The sign gives the sense of circulation. The zero root describes a stationary transverse position, not a moving circular orbit. A constant vertical velocity can be superposed to produce a helix.

Define the mechanical angular momentum about the axis by $L=m{r}^2\dot{\theta }$. Multiplying the angular equation by $r$ givesFor the inverse-radius magnetic field this becomes $\dot{L}=-q{B}_0\dot{r}$. Integrating from an initial radius $r_0$ provesLet $c=q{B}_0/m$ and $\ell =(L_0+q{B}_0{r}_0)/m$, so that $L/m=\ell -cr$. Since $\dot{z}$ is constant and the total kinetic energy is constant, the transverse speed squaredis also constant. Substituting the angular momentum relation gives the required radial curve:Only the region $W(r)\ge 0$ is accessible. For $\ell \ne 0$, $W(r)\to -\infty$ as $r\to 0$, and its horizontal asymptote is $v_{\perp }^2-c^2$. Its derivative is $W^{\prime }(r)=2\ell (\ell -cr)/r^3$. Thus, when $\ell c>0$, it has a single maximum $W=v_{\perp }^2$ at $r=\ell /c$. The positive roots of $W$ are the turning points obtained from $r=\ell /(c\pm v_{\perp })$.

For nonzero $c$ and $\ell c>0$, write $r_c=|\ell /c|$. When $0<v_{\perp }<|c|$, two turning points occur atand the radius is confined between them. At $v_{\perp }=|c|$, there is only the finite lower turning point $r_c/2$; $W$ tends to zero from above and an outgoing particle reaches arbitrarily large radius. For $v_{\perp }>|c|$, only the lower turning point remains and the outgoing radial velocity approaches $\sqrt{v_{\perp }^2-c^2}$. The sketch shows these three cases and omits inaccessible negative values of ${\dot{r}}^2$ from the physical interpretation.

For completeness, if $\ell c<0$, $W$ rises monotonically from $-\infty$ to its asymptote; admissible motion then requires $v_{\perp }>|c|$ and is radially unbound after any inward branch turns. If $\ell =0$, $W=v_{\perp }^2-c^2$ is constant; equality gives a circular orbit rather than escape. Above threshold the outgoing branch escapes, whereas an incoming branch reaches the singular axis in finite time and cannot be continued through it without specifying a nonsingular core field. These cases explain why a borderline escape assertion needs an initial-direction qualification.

The transverse escape velocity threshold for radial escape in an inverse-radius axial magnetic field isFor a radial launch at $r_0$, $\ell =c{r}_0$, and equality permits marginal escape; for $\ell \ne 0$, larger transverse speed guarantees radial escape in forward time after any initial inward leg turns. The singular incoming $\ell =0$ branch is the exception described above. More general launches at equality can instead give the circular exception $\ell =0$. A vertical velocity adds to the total speed but cannot pay the transverse escape-energy requirement; if escape merely means unbounded $z$, any nonzero $\dot{z}$ already achieves that. If $q{B}_0=0$, transverse motion is force-free and there is no positive escape-speed threshold.

The moment of inertia of a uniform solid cylinder is taken about its symmetry axis. Write its mass density as ${\rho }_c=m/(\pi a^{2}L)$. A coaxial shell of radius $r$ and thickness $dr$ has mass $dm=2\pi {\rho }_{c}Lrdr$. ThusA hollow cylinder would have a different moment of inertia; uniform solid density is the intended assumption.

Let $s$ measure the centre's displacement uphill from its initial position, and choose signed angular velocity $\omega$ so that uphill rolling without slipping has $\dot{s}=a\omega$. Let $F$ be the signed contact force uphill. The translation and rotation equations areThe minus sign in the torque equation follows from the chosen rolling sense: uphill contact force opposes the positive spin. Under rolling without slipping, $\ddot{s}=a\dot{\omega }$. Eliminating $F$ givesUsing $s(0)=0$ and $\dot{s}(0)=V$, the no-slip motion and maximum vertical rise areIndeed the centre turns at $t=3V/(2g\sin \alpha )$, reaching $s_{\max }=3V^2/(4g\sin \alpha )$, whose vertical component is the boxed height. Conservation of energy independently gives the same answer from initial kinetic energy $m{V}^2/2+I(V/a{)}^2/2=3m{V}^2/4$.

The available static friction is at most $\mu mg\cos \alpha$. It can supply the required uphill force exactly when $mg\sin \alpha /3\le \mu mg\cos \alpha$. ThereforeInterpret the stated friction magnitude as limiting static friction and as kinetic friction during slip, with the same coefficient. Take $0<\alpha <\pi /2$, $\mu \ge 0$. In the slipping regime the contact point moves downhill relative to the slope, so kinetic friction acts uphill. To verify this direction rather than assume it unchecked, solve with $F=\mu mg\cos \alpha$. Define $b=g(\sin \alpha -\mu \cos \alpha )>0$. The result isThe centre's speed is $|V-bt|$, with the sign of $v$ specifying ascent or descent. The contact point's relative velocity isThus the assumed uphill kinetic friction is consistent throughout both ascent and descent, including when the centre's velocity changes sign. There is no later return to rolling without slipping before reaching the bottom. This is the characteristic slipping solid cylinder on an incline regime.

The centre turns at $t_{\ast }=V/b$, givingThe formulas apply up to the positive return time $t_b=2V/b$. Substitution into the signed velocity and angular velocity givesThe return spin can have either sign and vanishes at $\tan \alpha =5\mu$. It reverses when $3\mu <\tan \alpha <5\mu$, although the contact point still slips downhill. The unchanged centre speed does not mean that no energy was dissipated: the rotational kinetic energy has decreased. Indeed integrating the frictional power $Fw<0$ giveswhich agrees with the difference of the initial and final rotational kinetic energies.

Choose the positive direction along the incoming car's velocity. Treat each car as a translating body, take $m_1,m_2>0$, and neglect external impulse, rotation and energy losses during the short elastic collision. If $v_1,v_2$ are the final signed velocities, conservation of momentum and conservation of energy giveFactoring the second relation and using the first gives $U_1+v_1=v_2$ on the collision branch. The other algebraic branch, $v_1=U_1,v_2=0$, describes no impulse and cannot describe separating bodies after contact. Solving the collision branch yieldsFor $0<R<1$ the first car rebounds and the second moves forwards. At $R=1$ the first stops and the second takes its original velocity. For $R>1$ both move forwards, with $v_2-v_1=U_1>0$, so they separate. In the limits $R\to 0$ and $R\to \infty$, the first car's velocity approaches $-U_1$ and $U_1$, respectively.

For the deformable car, take $y=0$ at first contact. The resisting force does negative work. The work-energy theorem therefore givesAt the first stop $v=0$, whence $\sqrt{L-y}=\sqrt{L}-M{V}^2/4$. The speed bound ensures that this right-hand side is positive, so squaring introduces no extraneous solution. The stopping compression isFor $0<V<2L^{1/4}/M^{1/2}$ this lies strictly between $0$ and $L$, as required before complete crushing. The coefficient in the stipulated force law is taken in its stated normalization.

Let $z$ measure height above the release point, and use $\mathcal{V}=4\pi a^3/3$ for the bubble volume to avoid confusing it with a speed. For constant radius, Newton's second law, the specified buoyancy and the given Stokes law giveWith positive $\alpha ,\rho ,\mu ,a$, defineThe differential equation becomes $\dot{u}+(u/\tau )=U/\tau$. Its solution with $u(0)=0$ is $u(t)=U(1-e^{-t/\tau })$. Integration using $z(0)=0$ yieldsAn arbitrary initial height is added to this expression. The effective inertia changes the relaxation time $\tau$, but not the terminal velocity obtained by balancing buoyancy and drag.

For the dissolving bubble assume $\beta >0$ and use the specified instantaneous terminal velocity, rather than the constant-radius transient formula. The bubble disappears at $t_{\ast }=a_0^2/\beta$, andIts total rise is therefore $\boxed{z(t_{\ast })=\rho g{a}_0^4/(9\mu \beta )}$. This second model neglects the initial acceleration and uses exactly the terminal-speed approximation stipulated for the shrinking bubble.

The factor called $N$ in the formula must be the stellar number density $n$, not a dimensionless count. Indeed ${\sigma }^3/(G^2{m}^2)$ has dimensions time per volume, so a density in the denominator is essential. If $N_{\ast }$ is the population in a representative volume $V$, then $n=N_{\ast }/V$.

For a weak encounter between equal-mass stars, let $u$ be the asymptotic relative speed and $b$ the impact parameter. In the straight-line impulse approximation, one star's transverse kick isThe other receives the opposite kick, so the relative-velocity change is $\Delta u_{\perp }=4Gm/(bu)$. Encounters in $(b,b+db)$ occur at rate $2\pi bdbnu$. Independent random kicks add in squared magnitude, givingThis is the random-walk derivation of stellar relaxation. The Coulomb logarithm in stellar dynamics is $\ln \Lambda =\ln (b_{\max }/b_{\min })$. The upper cutoff is the size or inhomogeneity scale beyond which independent local encounters are inappropriate. The lower cutoff is of order the strong-deflection scale $G(2m)/u^2$, or a larger stellar-size/softening scale if applicable. A self-gravitating system commonly has $\Lambda$ of order its total number of stars; that population count is distinct from $n$ in the diffusion rate.