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In the gravitational potential $-\mu (1/r+a/r^2)$, a regular noncircular bound orbit requires $L^2>2\mu a$ and has $u=\mu (L^2-2\mu a{)}^{-1}[1+e\cos \nu (\psi -{\psi }_0)]$ with $0<e<1$. Each radial period advances the angle by $2\pi /\nu$. The full position and velocity return after finitely many radial periods precisely when $\nu$ is rational. Irrational $\nu$ gives a dense rosette on the two-frequency invariant torus. Rationality permits a fifth orbit label fixing the apsidal phase on that resonant angular momentum family, but does not give global superintegrability for all nearby angular momenta. When $a=0$, every regular bound Kepler orbit is closed and the Laplace-Runge-Lenz vector supplies the globally defined additional integral.

The coefficient of kinetic friction is the dimensionless quantity ${\mu }_k=|f_k|/N$ for sliding contact with positive normal force $N$ in the ideal kinetic friction model. It characterizes the pair of contacting surfaces and their condition. A constant ${\mu }_k$ means the model neglects dependence on contact slip speed; unlike the coefficient of static friction, it specifies a sliding force magnitude rather than a no-slip upper bound.

The ratio compares the tangential kinetic friction magnitude with the perpendicular normal force:Here $\mathbf{w}$ is the tangential relative velocity defining contact slip. Both numerator and denominator have the same physical dimension, so the coefficient of kinetic friction has no units. It belongs to a contact pair under specified conditions, rather than to one material alone. The coefficient of static friction instead describes the threshold $|f_s{|}_{\max }/N$ before contact slip; a pull required to start motion cannot be substituted for a sliding measurement.^[ref]

As an original model example, stipulate ${\mu }_k=0.30$ and gravitational acceleration $g=9.8m/{\mathrm{s}}^2$. An already sliding, nonrotating block of total mass $m=2\mathrm{kg}$ is pulled horizontally across a stationary horizontal surface at constant velocity. With no other vertical force, Newton's second law gives $N=mg=19.6\mathrm{N}$. Zero horizontal acceleration then requires a pulling force $F=|f_k|={\mu }_{k}mg=5.88\mathrm{N}$. Doubling the total mass to $4\mathrm{kg}$ gives $N=39.2\mathrm{N}$ and $F=11.76\mathrm{N}$. The force doubles while the coefficient of kinetic friction stays $0.30$ by assumption. These numbers are model predictions, not observations or a calibration for a particular material pair.

A constant coefficient of kinetic friction is useful because one ratio can summarize many loads, as tested by load dependence of kinetic friction. It is an empirical approximation over a stated range of normal forces, speeds and surface conditions. It implies no universal independence from speed or apparent contact area: a change of face or sliding regime can change the result. Temperature, wear and lubrication are further conditions to report. Newton's second law still applies when the constant-coefficient approximation fails.^[ref]

For a cold equatorial axisymmetric magnetohydrodynamic wind with $B_p=C{R}^{-2}$ in the gravity of a point mass $M$, smooth crossing of the Alfvén surface gives ${\rho }_a={\mu }_0{k}^2$ and $\ell =\omega R_a^2$. Put $x=R/R_a$, $y=\rho /{\rho }_a$, $\alpha ={\omega }^2{R}_a^3/(GM)$ and $\beta =C^2/({\mu }_0{\rho }_{a}GM{R}_a^3)$. The corotating energy, scaled by $GM/R_a$, isThe azimuthal velocity relative to field-line rotation is $\omega R_a(x-x^{-1})/(y-1)$ and the poloidal speed is $C/(\sqrt{{\mu }_0{\rho }_a}{R}_a^2{x}^{2}y)$. The apparent singularity at $x=y=1$ is interpreted through a smooth limiting slope. A nondegenerate magnetosonic crossing on a constant-$f$ curve requires $f_x=f_y=0$ and a real admissible tangent slope.

A one-dimensional collapsible tube model couples area $\alpha$, mean axial velocity $u$ and internal pressure $p$: ${\alpha }_t+(\alpha u{)}_x=0$ and $u_t+u{u}_x=-p_x-R(\alpha )u$. A wall relation specifies internal minus external pressure as a function of area. Decreasing area generally increases the friction coefficient $R$, while positive wall stiffness gives real small-amplitude wave speed.

Linearize the Collisionless Boltzmann equation and the perturbation Poisson equation about a homogeneous velocity distribution $f_0$. A plane wave gives $f_1={\varphi }_1\mathbf{k}\cdot {\nabla }_v{f}_0/(\mathbf{k}\cdot \mathbf{v}-\omega )$. Integrating this density and using $-k^2{\varphi }_1=4\pi G\int f_1{d}^{3}v$ yields $1+(4\pi G/k^2)\int [\mathbf{k}\cdot {\nabla }_v{f}_0/(\mathbf{k}\cdot \mathbf{v}-\omega )]d^{3}v=0$. Growing modes have $\mathrm{Im}\omega >0$ and no real-velocity pole; real and damped frequencies require the causal continuation prescription. The Newtonian homogeneous background requires the Jeans swindle.

Spatial translation symmetry in a flat FRW metric conserves the covariant canonical momentum $q_i=m{a}^{2}d{x}^i/ds$ of a freely falling massive particle. An orthonormal spatial frame gives $p_{\mathrm{phys},i}=q_i/a$, so physical momentum redshifts as $a^{-1}$ even while the particle changes from relativistic to nonrelativistic motion. The local energy is $E=\sqrt{m^2+q^2/a^2}$ and the coordinate velocity is $d\mathbf{x}/dt=\mathbf{q}/[a\sqrt{q^2+m^2{a}^2}]$.

For $i{\psi }_t=-D{\nabla }^2\psi +[V(n)+i\Gamma (n)]\psi$, with real $D,V,\Gamma$ and $\psi =\sqrt{n}{e}^{iS}$, the continuity equation isMultiply the equation by ${\psi }^{\ast }$ and subtract its complex conjugate multiplied by $\psi$ to obtain this identity; real interaction and potential terms cancel. Thus the current velocity is $\mathbf{j}/n=2D\nabla S$. With dimensional kinetic operator $-{\hslash }^2{\nabla }^2/(2m)$ and left side $i\hslash {\psi }_t$, this gives the physical superfluid velocity $(\hslash /m)\nabla S$. In normalized equations with $D=1/2$ or $D=1$, the current velocity is respectively $\nabla S$ or $2\nabla S$. The distinction fixes the radial-flow slope of a driven quantum vortex.

In a unit square with prescribed conductive boundary temperatures and impermeable boundaries, the stream function convention $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$ gives linear equations $\Delta \psi =R{\theta }_x$, ${\theta }_t={\psi }_x+\Delta \theta$, with $\psi =\theta =0$ on all sides. The velocity equation follows by taking the vertical-plane curl of Darcy law, and the temperature equation by linearizing the heat equation about $1-z$. Unlike a laterally unbounded Darcy layer, the conducting side boundaries produce a two-dimensional real marginal eigenspace at $R_c=8{\pi }^2$.

The conduit equation evolves the positive cross-sectional area $A$ of a viscous buoyant conduit. Its scaling uses axial length $a_0/\sqrt{8\lambda }$ and velocity $\Delta \rho g{a}_0^2/(8\lambda {\mu }_o)$, where $\lambda ={\mu }_i/{\mu }_o$. Linearization at $A=1$ gives $\omega =2k/(1+k^2)$: phase velocity is upward but group velocity changes sign at $|k|=1$. Nonlinear elevation waves obey the solitary-wave amplitude-speed relation for the conduit equation.

For the pressure-neglecting confined plume with compensating return flow, define specific plume momentum flux $M=b^2{w}^2$ and area fraction $s=Q^2/(R^{2}M)$. Then $J=M/(1-s)$ andThe branch issuing from a pure plume reaches a fold at $s=1/2$, where plume and ambient have equal areas, opposite velocities and equal specific momentum fluxes. The numerator is nonzero on this branch, so $M^{\prime }$ diverges although $M,Q,w$ remain finite. This signals failure of the steady uniform-profile description, not infinite physical velocity.

In an impermeable cylinder, an axisymmetric turbulent plume with negligible source volume flux requires a downward return flow. With uniform upward plume velocity $w$ and ambient velocity $v$, volume conservation gives $b^{2}w+(R^2-b^2)v=0$. An entrainment coefficient $\alpha$ based on the relative speed gives $Q^{\prime }=2\alpha b(w-v)$. A specified pressure closure is essential for momentum conservation. If the departure from ambient hydrostatic pressure and wall force are neglected in the whole-section integral, the total specific momentum flux $J=b^2{w}^2+(R^2-b^2)v^2$ obeys $J^{\prime }=b^2{g}^{\prime }$. Homogeneous unmodified ambient and no buoyancy loss give $b^{2}w{g}^{\prime }=B_s$. These are leading integral-model closures; restoring the ambient dynamic pressure can change the singularity criterion.

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

For an unmagnetized ideal gas without gravity, the pressure equation gives internal-energy balance ${\partial }_{t}e+{\partial }_x(u_{x}e)=-p{\partial }_x{u}_x$, where $e=p/(\gamma -1)$. The momentum equation gives kinetic-energy balance with source $-u_x{\partial }_{x}p$. Their sum is conservative with the displayed total-energy flux. Tangential velocities contribute to kinetic energy even when all derivatives are in the normal direction. Smooth polytropic flow advects its entropy parameter; a physical shock need not preserve that parameter across the discontinuity.

With fixed inlet thickness $h_0$ and velocity $u_0$, and nose tension $4\mu h_N{u}_x=\rho g^{\prime }{h}_N^2/2$, integrating momentum gives $u_x=\rho g^{\prime }h/(8\mu )$. For a slice injected at $t_0$, write $\tau =t-t_0$ and $T=8\mu /(\rho g^{\prime }{h}_0)$. Its thickness, velocity and position are $h=h_0/(1+\tau /T)$, $u=u_0(1+\tau /T)$, and $x=u_0[\tau +{\tau }^2/(2T)]$. Eliminating its age gives the steady profile behind the nose, while the nose is the slice with the earliest injection time.

With $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$, the half-line heat equation solution $g=C{t}^{-1/2}{e}^{-x^2/(4t)}$ has $g_x(0)=0$ but $g(0)>0$. It gives $\Sigma \propto r^{-3/2}{t}^{-1/2}{e}^{-r/(3At)}$, constant disk mass and angular momentum proportional to $t^{1/2}$. Its radial velocity is $r/t$, and a central torque proportional to $t^{-1/2}$ supplies the increasing angular momentum. It is an outward-spreading, torque-driven disk, not a zero-torque accreting solution.

// Target: astrophysical-fluid-dynamics.bigb

On a connected open interval, if $F^{\prime }=g$, every antiderivative of $g$ has the form $F+C$, where $C$ is a constant function. Indeed, two solutions satisfy $(G-F{)}^{\prime }=0$: the zero derivative on a connected open set principle makes their difference constant. In one dimension, the mean value theorem proves this on any closed subinterval. ^[ref]

An initial condition or boundary condition can determine $C$. If a particular antiderivative is $F_0$ and $F(x_0)=y_0$, thenFor $g(x)=x^2$, $F_0(x)=x^3/3$ and $F(1)=2$, this gives $C=5/3$. Without that reference value, choosing $C=0$ simply selects one member of the family.

The freedom has a physical meaning in a simple kinematic model. A velocity $v(t)=at$, with constant acceleration $a$, determines position only as $x(t)=a{t}^2/2+C$. The initial condition $x(0)=x_0$ selects $C=x_0$; changing the starting position does not change the prescribed velocity.

For the differential equation $T^{\prime }=-B/r^2$ on $r>0$, the antiderivative is $T=B/r+C$. The boundary condition $T(R_s)=T_s$ gives $C=T_s-B/R_s$. Discarding the constant of integration therefore needs a specified condition or an approximation that quantifies its effect. On a domain with several connected components, the freedom is larger: antiderivatives on disconnected domains require a separate constant on each component.