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A kinetic model of a dilute gas. In the Boltzmann equation, $f(t,x,v)\ge 0$ is the particle number density in phase space, so $fdxdv$ counts particles near position $x$ and velocity $v$. For unit particle mass in three dimensions, without external forces, the nonlinear equation isThe left side is the free transport equation. The quadratic right side describes binary elastic collisions occurring locally in space. A prescribed acceleration would add $F\cdot {\nabla }_{v}f$; all conclusions below about energy or entropy must then include the appropriate force terms. Assume enough smoothness, velocity decay and finite moments to justify the calculations, and either periodic space or decay that removes spatial boundary fluxes. These are assumptions on a classical solution, not a general existence theorem.

The dilute-gas approximation neglects simultaneous many-body collisions and treats the interaction region as small relative to macroscopic spatial scales. Molecular chaos for the Boltzmann equation factorizes the distribution of two incoming particles as $f(v)f(v_{\ast })$. It applies to incoming collision pairs; it does not assert that an interacting many-particle system stays exactly independent. This approximation closes the equation at the one-particle level and explains its nonlinearity.

Collision geometry and the gain-loss operator. Let $g=v-v_{\ast }$ and $n\in S^2$. A useful parametrization of an equal-mass elastic collision isThe center velocity $(v+v_{\ast })/2$ is unchanged and the relative velocity is reflected in the plane normal to $n$. ConsequentlyThe fixed-$n$ pre/post-collision transformation is an involutive orthogonal map of ${\mathbb{R}}^6$, so it preserves $dvd{v}_{\ast }$. Take a nonnegative collision kernel $B(g,n)$ invariant under interchange of particles and reversal of the collision; for example $B=B(|g|,|\hat{g}\cdot n|)$. The nonlinear Boltzmann collision operator isAny convention about counting $n$ and $-n$ is absorbed in $B$. For a hard-sphere gas, this reflection parametrization has $B=c|g\cdot n|$, with a positive constant determined by the sphere size and counting convention. Maxwell molecule collision operators instead have a collision frequency independent of $|g|$ after angular integration. A Grad angular cutoff requires a finite angular integral; singular kernels need additional cancellation analysis and cannot be handled by every elementary gain-loss argument below.

Writing $Q=Q^{+}-{\nu }_{f}f$, wheremakes the physical meaning clear: gain into velocity $v$ competes with loss out of $v$. Along a free characteristic curve, put $F(t)=f(t,x+tv,v)$. Then $F^{\prime }+{\nu }_f(t,x+tv,v)F=Q^{+}(t,x+tv,v)$. The integrating factor givesFor finite collision integrals, this representation shows preservation of nonnegativity. It is implicit, because both the gain and damping depend on $f$; it does not alone prove existence.

The weak collision identity and conserved quantities. The weak Boltzmann collision identity follows from particle interchange and the measure-preserving pre/post-collision substitution: it gives, for a velocity test function $\psi$,The factor $1/2$ comes from averaging the two particle labels. Equivalently the right side isA collision invariant has zero square-bracketed change in every collision. The functions $1$, the three velocity coordinates, and $|v{|}^2$ are collision invariants by the geometric identities above. Hence the collision term conserves particle number, momentum and kinetic energy locally in position. Multiplying the full Boltzmann equation by these functions and integrating in velocity gives local conservation laws; integrating in space yields constant total mass, momentum and energy when boundary fluxes vanish.

The smooth collision invariants are exactly their linear span, provided the kernel allows all scattering directions. Here is a direct proof. If $\psi (v)+\psi (v_{\ast })$ is unchanged by every collision, then for fixed center $c$ and relative radius $r$, the sum $\psi (c+r\omega )+\psi (c-r\omega )$ is independent of the unit vector $\omega$. Taylor expansion at $r=0$ forces ${\omega }^T{D}^2\psi (c)\omega$ to be independent of $\omega$, so $D^2\psi (c)=\lambda (c)I$. Mixed derivatives vanish. Differentiating ${\partial }_{ii}\psi =\lambda$ in a direction $j\ne i$ and commuting derivatives gives ${\partial }_j\lambda =0$. Thus $\lambda$ is constant andThis proof can be read for $C^3$ invariants; weaker regularity is handled distributionally. The multidimensional scattering hypothesis matters: in one dimension equal-mass elastic collisions only exchange the two velocities, and every function is then a collision invariant.

Entropy dissipation and the H theorem. Define the Boltzmann H functional $H(f)=\int f\log fdxdv$, with $0\log 0=0$. The physical kinetic entropy has the opposite sign, up to constants and the particle-number normalization. Particle-number conservation removes the $1$ in the derivative of $f\log f$; spatial transport contributes only a boundary flux. With $A=f^{\prime }{f}_{\ast }^{\prime }$ and $B_0=f{f}_{\ast }$, the symmetrized weak collision identity givesThe sign follows from monotonicity of the logarithm: $(a-b)(\log a-\log b)\ge 0$. Zeros are treated by the usual extended nonnegative limit. This is the Boltzmann H theorem for the present kernel normalization, and $D$ is the Boltzmann entropy dissipation.

For a positive smooth $f$ and a kernel positive on all collision directions, equality $D(f)=0$ requires $f^{\prime }{f}_{\ast }^{\prime }=f{f}_{\ast }$ in every collision, so $\log f$ is a collision invariant. Finite velocity mass then forces a negative quadratic coefficient and gives a local Maxwellian:Here $\rho$ is number density, $u$ is mean velocity, and $\theta$ is the temperature expressed in velocity-squared units, $k_{B}T/m$. Completing the square proves the formula; conservation of momentum and kinetic energy makes $M{M}_{\ast }=M^{\prime }{M}_{\ast }^{\prime }$, so $Q(M,M)=0$. Vacuum is another collision equilibrium.

A local Maxwellian annihilates the collision term at each position, but it need not solve the transport part. A spatially uniform time-independent Maxwellian is a full equilibrium on periodic space. On all of ${\mathbb{R}}_x^3$ it has infinite total mass unless it is vacuum. Spatially varying local Maxwellians must satisfy their additional transport equations; collision equilibrium alone is insufficient.

For the spatially homogeneous Boltzmann equation, a Maxwellian with the same mass, mean velocity and energy is selected by these invariants. The relative entropy $\int f\log (f/M)dv$ is nonnegative. Indeed, writing $f=Mr$ and using equal mass givesMoreover $\log M$ is a collision invariant, so the derivative of this relative entropy is $-D(f)$. Thus $M$ minimizes $H$ at fixed conserved moments. Entropy decrease and the characterization of its equality case are structural information; convergence to equilibrium or a decay rate additionally needs compactness or a quantitative coercive estimate.

Moment equations and fluid closure. LetThe tensor $P$ is the kinetic pressure tensor and $q$ the kinetic heat flux. Put $e=(2\rho {)}^{-1}\int |v-u{|}^{2}fdv$. The collision invariants yieldThese identities are exact whenever the moments and derivatives are justified. They are not a closed fluid system: $P$ and $q$ contain information beyond $\rho ,u,e$. For a local Maxwellian, Gaussian integration gives $P=\rho \theta I$, $q=0$ and $e=3\theta /2$. Substitution gives the compressible Euler equations for a monatomic ideal gas, with pressure $p=\rho \theta$ and ratio of specific heats $5/3$.

This connection is naturally organized by the Knudsen number $\epsilon$, the ratio of the mean free path to a macroscopic length. In an appropriate fluid scaling the equation reads ${\partial }_{t}f+v\cdot {\nabla }_{x}f={\epsilon }^{-1}Q(f,f)$. Formally, as $\epsilon \to 0$, the dominant collision term imposes $Q(f,f)=0$, hence local Maxwellian form, and the conserved moments evolve by the Euler equations. The first correction to a local Maxwellian produces viscous stress and thermal conduction, leading to Navier-Stokes equations with a thermal equation. This is a formal hydrodynamic limit of the Boltzmann equation; a rigorous limit requires estimates, preparation of data and treatment of boundaries, and is not justified merely by inserting $\epsilon =0$.

Linearization and its dissipative structure. Around a fixed positive Maxwellian $M$, write $f=M(1+h)$. The linearized collision operator on the relative perturbation isThis follows by expanding the two quadratic products to first order and using $M{M}_{\ast }=M^{\prime }{M}_{\ast }^{\prime }$. In the weighted Hilbert space with inner product $⟨h,k{⟩}_M=\int Mhkdv$, collision symmetrization givesThis dissipation form of the linearized Boltzmann operator and its corresponding polarized identity prove symmetry on a common domain where the integrals converge. Its nullspace is spanned by $1,v_1,v_2,v_3,|v{|}^2$ under the same nondegenerate scattering assumption, exactly the conserved modes. For unbounded collision frequencies, the domain and closed realization must be specified before calling the operator self-adjoint. A spectral gap on the orthogonal complement is an additional theorem depending on the collision kernel; it does not follow just from the displayed nonnegative quadratic form.

Analytic scope and irreversibility. With an angular cutoff and bounded collision kernel, the spatially homogeneous equation has a simple local well-posedness argument in velocity $L^1$. The pre/post-collision substitution gives $\Vert Q(f,g){\Vert }_1\le C\Vert f{\Vert }_1\Vert g{\Vert }_1$, so the quadratic map is locally Lipschitz. The Banach fixed-point theorem applied to the integral equation gives a local unique solution; gain-loss iteration preserves positivity. For nonnegative solutions the conserved mass keeps the $L^1$ norm bounded, allowing continuation for all time. This bounded-kernel homogeneous argument is a precise special case, not a proof of global classical well-posedness for arbitrary inhomogeneous hard-sphere data.

More difficult collision kernels and spatial dependence require different solution frameworks. One should distinguish classical solutions, weak solutions and renormalized solutions rather than infer smoothness from conservation and entropy alone. For instance, a renormalized Boltzmann solution uses nonlinear bounded transforms $\beta (f)$ to make sense of collision and transport terms when their raw integrability is insufficient. No general smooth global existence conclusion for unrestricted large inhomogeneous data follows from the elementary calculations above.

Microscopic elastic collision dynamics are reversible, while the Boltzmann H theorem selects a direction of increasing entropy. The step that introduces this direction is the incoming-pair factorization: time reversal creates correlations in incoming pairs that the same factorization discards. The Boltzmann-Grad limit gives a route from a dilute many-particle gas to a kinetic equation by letting particle size shrink while keeping a finite mean free path. Propagation of an appropriate chaos hypothesis and control of recollisions are substantive parts of that limit; the collision geometry alone is not a derivation.

Let $M_{\mathrm{tot}}={\sum }_i{m}_i$ and introduce positions relative to the centre of mass, ${\mathbf{\rho }}_i={\mathbf{r}}_i-\mathbf{R}$. The definition gives ${\sum }_i{m}_i{\mathbf{\rho }}_i=0$, and differentiation gives ${\sum }_i{m}_i{\dot{\mathbf{\rho }}}_i=0$. Expand the total kinetic energy:The cross term vanishes. This proves the kinetic energy decomposition about the center of mass:The decomposition itself holds without rigidity.

For a rigid body rotating about its centre of mass, ${\dot{\mathbf{\rho }}}_i=\mathbf{\omega }\times {\mathbf{\rho }}_i$. With unit axis vector $\mathbf{n}=\mathbf{\omega }/|\mathbf{\omega }|$, the moment of inertia about that axis iswhere $d_i$ is the perpendicular distance from the axis. Therefore

The total relativistic energy and momentum in the original frame areA Lorentz boost of signed velocity $V$ makes the total spatial momentum $p^{\prime }={\gamma }_V(p-VE/c^2)$. Setting it to zero gives the centre-of-momentum frame velocityIt is a positive-weighted average of subluminal velocities, so $|V|<c$.

In the rest mass of a coalescing relativistic collision, four-momentum conservation gives $P_3=P_1+P_2$ provided no energy or momentum escapes in additional radiation. For the single outgoing particle, $u_3=c^{2}p/E=V$, so its speed in the original frame is $|V|$. Its rest mass is the invariant norm of the total four-momentum:Expanding and using ${\gamma }_i^2(1-u_i^2/c^2)=1$ givesThe positive square root is taken for $m_3$. The factor multiplying $2m_1{m}_2$ is the relative Lorentz factor, at least one. Therefore $m_3\ge m_1+m_2$, with equality when $u_1=u_2$; otherwise the initial relative kinetic energy becomes internal rest energy.

The central force has no transverse component. Its angular equation is thereforeThus the specific angular momentum is the constantThe radial equation isSubstitute $\dot{\theta }=l/r^2$:Consequently the radial motion for $r>0$ is exactly a unit-mass one-dimensional motion in the effective potentialIts conserved energy is $\frac{1}{2}{\dot{r}}^2+V_{\mathrm{eff}}(r)$; the second term in the effective potential accounts for the angular kinetic energy.

Irrotational flow permits $\mathbf{u}=\nabla \varphi$, and incompressible flow imposes $\nabla \cdot \mathbf{u}=0$, hence $\Delta \varphi =0$. In the laboratory frame, the no-penetration condition at the translating sphere is ${\partial }_r\varphi (a,\theta )=U\cos \theta$.

For separation of variables, take the angular factor $g(\theta )=\cos \theta$, which satisfies $\frac{1}{\sin \theta }(\sin \theta g^{\prime }{)}^{\prime }=-2g$. The radial equation iswith solutions $r$ and $r^{-2}$. Rest at infinity excludes the $r$ term. The normal boundary condition fixes the remaining coefficient, giving the potential flow around a translating sphereThere is no azimuthal velocity. With fluid density ${\rho }_f$, the kinetic energy isThe angular integral is $2\pi$, and the radial integral is $1/(3a^3)$, so $K_f={\rho }_f\pi a^3{U}^2/3$. Since $M=(4\pi a^3/3){\rho }_s$ and $k={\rho }_f/{\rho }_s$, $K_f=kM{U}^2/4$. Thus the added mass of a sphere is $kM/2$.

During a fall through $h$, gravity does work $Mgh$ and buoyancy does work $-kMgh$. Starting from rest, conservation of energy givesConsequently $U=\sqrt{4(1-k)gh/(2+k)}$. The mass cancels: the reduced speed comes from both buoyancy and the fluid's added mass.

Let the vertical-velocity amplitude $W$ be real. A convenient inertia-gravity wave polarization, avoiding pressure denominators, isThe physical velocities and buoyancy are their real parts. The period-mean kinetic energy density and available potential energy density areThe latter follows from $A={\rho }_0{b}^2/(2N^2)$, or from $A={\rho }_0{N}^2{\xi }_z^2/2$ for vertical fluid displacement. The factor $1/4$ includes both the energy definition and the mean of a squared harmonic.

Using the dispersion relation,This energy partition of rotating internal waves shows that the printed request for ordinary kinetic/potential equipartition in the rotating case is false in general. An explicit counterexample is $f=1$, $N=2$, $k=m=1$, $W=1$, giving ${\omega }^2=5/2$, $⟨K⟩=3{\rho }_0/5$ and $⟨A⟩=2{\rho }_0/5$.

The valid modified oscillator balance isThe transverse rotational velocity is in quadrature with the in-plane motion, so it supplies an additional positive energy term on the displacement side of this oscillator balance. It remains physically kinetic energy, not gravitational potential energy. Ordinary kinetic/potential equipartition is recovered when $f=0$ or when the transverse amplitude vanishes. For a real harmonic wave the instantaneous total $K+A$ is constant at a fixed point: the coefficient of ${\cos }^2(kx+mz-\omega t)$ from in-plane motion equals the coefficient of ${\sin }^2$ from transverse motion and buoyancy. This consistency does not imply equality of their separate period means.

Integrating the local two-layer quasi-geostrophic energy conservation law over a horizontal domain and using the specified vanishing boundary flux givesPeriodic boundaries, or suitable fixed streamfunction boundary data eliminating the displayed energy flux, provide examples. The gradient terms are the two layer kinetic energies; the last term is available potential energy, equal to $g^{\prime }{\chi }^2/2$ under the interface-displacement relation.

The baroclinic energy ratio and deformation scale for unequal layer depths uses a decomposition uses the depth-weighted barotropic streamfunction $\Psi =(H_1{\psi }_1+H_2{\psi }_2)/(H_1+H_2)$ and $d=2\tilde{\psi }$. Define $H_r=H_1{H}_2/(H_1+H_2)$. ThenFor variations on horizontal scale $l$, this givesFor comparable layer depths $H_r$ is of order either $H_i$, reproducing the requested scale $l^2{f}_0^2/(g^{\prime }{H}_i)$. If one layer is much thinner, its depth controls this ratio; a depth-independent arithmetic barotropic average would leave unwanted cross terms in the energy decomposition.

Baroclinic potential energy dominates at scales much larger than the two-layer internal deformation radius; baroclinic kinetic energy dominates at much smaller scales. They are comparable near $l\sim R_d$. The independent barotropic kinetic energy has no interface-displacement partner, so this scale comparison refers specifically to the baroclinic component.

For any vector $\mathbf{a}$, differentiating its components and the rotating basis gives the rotating-frame derivative formulaApply it twice to the position $\mathbf{x}$, with constant angular velocity $\mathbf{\omega }$. The inertial acceleration iswhere dots denote rotating-frame derivatives. Newton's second law therefore becomesThe last terms are the Coriolis force and the force corresponding to centrifugal acceleration.

For the particle on a uniformly rotating inclined plane, choose orthonormal vectors ${\mathbf{e}}_1$ along the horizontal $\xi$ axis and ${\mathbf{e}}_2$ up the slope, and set ${\mathbf{e}}_3={\mathbf{e}}_1\times {\mathbf{e}}_2$, the upward normal vector. Then ${\mathbf{e}}_z=\sin \theta {\mathbf{e}}_2+\cos \theta {\mathbf{e}}_3$ andHere $N$ is the normal reaction. The marble is modeled as the smooth sliding particle specified by the printed equations, without an additional rolling constraint. The rotating-frame Coriolis acceleration and centrifugal acceleration are respectivelyThe normal rotating-frame acceleration is zero, so the normal reaction is not generally $mg\cos \theta$; it satisfiesProjecting the same equation onto the two tangential axes gives exactly

Multiply the equations by $\dot{\xi }$ and $\dot{\eta }$ and add. The Coriolis force terms cancel, as expected because this force does no work relative to the rotating frame. The normal reaction also does no work along the plane. The conserved quantity is rotating-frame kinetic energy plus gravitational and centrifugal potential energy:This need not be the conserved inertial energy, since the externally driven rotating plane can exchange energy with the particle.

For a massive particle of rest mass $m$, its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$, where the four-velocity is $U^{\mu }=\gamma (c,\mathbf{v})$, $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. With metric signature $(-,+,+,+)$, $P\cdot P=-m^2{c}^2$. A photon has four-momentum $(E/c,(E/c)\hat{\mathbf{n}})$, a future-pointing null vector. Four-momentum conservation states that the total incoming and outgoing four-momenta agree for an isolated interaction; this includes both energy and momentum conservation in every inertial frame.

Take the incoming photon direction as $x$. Its energy is $\hslash \omega =\mu m{c}^2$, so the total initial four-momentum isThe invariant mass available after photon absorption is therefore $M=m\sqrt{1+2\mu }$. In the centre-of-momentum frame, the two equal daughters have opposite momenta. Their total energy is at least their combined rest energy $2\alpha m{c}^2$, with equality precisely when both daughters have zero momentum in that frame. Thus the mass threshold after photon absorption isThe bound is attainable kinematically by taking each daughter four-momentum equal to $P/2$; each then has the required rest mass. It immediately gives ${\alpha }_{\max }\to 1/2$ as $\mu \to 0$.

At this threshold both daughters move with the centre-of-momentum frame. The common laboratory velocity follows from $\mathbf{v}=c\mathbf{P}/P^0$:For $\mu =0$, its continuous limiting value is zero. At smaller allowed daughter rest masses, some of the centre-of-momentum energy appears as their relative kinetic energy, so their four-momenta need not be $P/2$.

For steady spherical polytropic flow, write $u=u_r$ and use the polytropic equation of state $p=K{\rho }^{\gamma }$, with $K>0$. Mass conservation and radial Euler momentum equation areThe mass equation gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Substitute it into momentum balance to obtainAt a sonic point the derivative coefficient vanishes. A smooth finite-slope solution must make the numerator vanish there too:Finally $dp/\rho =d[\gamma K{\rho }^{\gamma -1}/(\gamma -1)]$. Integrating momentum gives the Bernoulli equationThis is kinetic energy plus specific enthalpy plus gravitational potential per unit mass. The polytropic stellar wind selects a particular transonic branch of these equations.

Let $a=({\rho }_b-{\rho }_t)/H={\rho }_0{N}^2/g$. The initial density is $\rho (z)={\rho }_t-az$. Conservation of mass in the material homogenized over $-h<z<0$ fixes its mean density to ${\rho }_u={\rho }_t+ah/2$, while the untouched lower layer has interfacial density ${\rho }_i={\rho }_t+ah$. Hence the reduced gravity difference across the interface isThe reference contribution involving ${\rho }_t$ cancels from the difference.

Only the upper layer changes its potential energy. The potential-energy cost of homogenizing a linear stratification issoThe positive sign reflects work against stable stratification.

The rotating boundary supplies energy on scales set by $\Omega$ and $R$. Turbulent drag and dissipation remove energy, and entrainment adds initially stationary fluid that must be accelerated. If upper-layer energy grows, the increasing turbulent loss provides a restoring tendency; after a short adjustment, input and losses can approximately balance while the interface moves on a slower time scale. This gives a physical rationale for a fixed-energy turbulent mixed layer, not a consequence of mass conservation alone. Its characteristic kinetic energy is $K\sim {\rho }_0\pi R^{2}h{u}_u^2/2$. With the stipulated $N$-independent energy closure,so the characteristic speed decreases as $h^{-1/2}$ rather than remaining constant.

The stress does work at a rate proportional to $\pi R^2\tau u_u=c_D{\rho }_0\pi R^2{u}_u^3$. Combine this with the energy increase and absorb fixed drag/geometrical factors into $C>0$:Here the exponents $\alpha ,\beta$ are unrelated to the plume entrainment coefficient in Question 2. The interfacial Richardson number isSet $q=\Omega /N$, $d=\delta /R$ and $r=h/R$. Then $u_u^2/(N^2{h}^2)=d/(2r{\mathrm{Ri}}_i)$ and $r=q(2A{\mathrm{Ri}}_i/d{)}^{1/2}$. Substitution givesThe assumption that $E$ depends on the local Richardson number alone excludes separate dependences on $q,d$. The entrainment exponent from a local interfacial Richardson closure is consequentlyThis comparison treats $q$ and $d$ as independently variable external controls; fitting only a single apparatus would not by itself determine the exponents.

Finally, use the fixed-energy closure again:Integrating from a positive initial mixed-layer depth $h_0$ at time $t_0$ givesWhen the growth term dominates the initial offset, the rotating-disc mixed-layer depth law isTaking the effective time origin to be zero gives the stated scaling. Its validity is limited to the regime of the closures and $h<H$; the singular velocity predicted at $h=0$ is not a model for the initial boundary-layer formation.

The Lorentz force in the uniform magnetic field giveswhere the last step is the vector triple product identity. HenceOn the other hand, $|\mathbf{r}\times \mathbf{B}{|}^2=B^2{r}^2-(\mathbf{r}\cdot \mathbf{B}{)}^2$, soThe derivatives cancel, proving the magnetic angular-momentum invariantThe magnetic Lorentz force is perpendicular to velocity, so its power is zero:Thus the kinetic energy is constant.

For the alternative kinetic energy expression, work on an interval where $r>0$ so that the radial unit vector $\mathbf{u}=\mathbf{r}/r$ is defined. Differentiating $\mathbf{u}\cdot \mathbf{u}=1$ once and twice givesSince $\mathbf{v}=\dot{r}\mathbf{u}+r\dot{\mathbf{u}}$, it follows that $\mathbf{u}\cdot \mathbf{v}=\dot{r}$ and $v^2={\dot{r}}^2+r^2|\dot{\mathbf{u}}{|}^2$. Therefore the unit-direction kinetic-energy identity isThis identity holds for any twice differentiable trajectory away from the origin, not only for motion in a magnetic field.

In plane polar coordinates, use ${\mathbf{e}}_r=(\cos \theta ,\sin \theta )$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta )$. Their derivatives are ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Differentiating $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and acceleration in polar coordinates:Both the electrostatic inverse-square force and the drag point radially, so the torque about the origin is zero even though the force depends on radial velocity. The vector angular momentum $\mathbf{h}=\mathbf{r}\times \dot{\mathbf{r}}$ is constant. For $\mathbf{h}\ne 0$, every position lies in the fixed plane perpendicular to it; if $\mathbf{h}=0$, the trajectory is radial and still lies in a plane. In the nonzero case the transverse equation is $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$, givingThe signed $h$ depends on the plane orientation. The stipulated ratio $k/|h|$ requires $h\ne 0$.