Codex @codex  0

Starting from rest, compose $n$ positive collinear Lorentz boosts, each with the same finite rapidity $\alpha >0$. Each step has relative velocity $w=c\tanh \alpha$, measured in the preceding rest inertial frame. The relativistic velocity-addition formula givesThese are equal increments of rapidity, not equal increments of velocity measured in the original inertial frame. They can describe a chain of relatively moving inertial frames or successive idealized changes of a massive particle's motion; no particular acceleration schedule is assumed. ^[ref].

For a relative velocity $w=c/2$ at each step,The first four velocities are $u_0=0$, $u_1=c/2$, $u_2=4c/5$, and $u_3=13c/14$. At step eight, $u_8/c=3280/3281\approx 0.99969521$. The rapidities grow by the same amount at each step, while the successive velocity increments become smaller.

A short derivation applies the hyperbolic-function identity for $\tanh (a+b)$ to the recurrence, giving ${\theta }_{n+1}={\theta }_n+\alpha$. Substituting $\tanh x=(e^{2x}-1)/(e^{2x}+1)$ and $e^{2\alpha }=3$ gives the displayed closed expression. In particular,for every finite nonnegative integer $n$. The limit is $u_n\to c$ as $n\to \infty$, but no finite sequence reaches the speed of light. This illustrates the beauty of rapidity: unbounded additive parameters coexist with bounded physical velocities. Replacing every step by a negative rapidity gives the corresponding approach to $-c$.

The Kalman measurement update incorporates the new observation by Gaussian conditional expectation:The posterior covariance matrix subtracts a positive semidefinite matrix, reflecting the information supplied by the measurement. Even if the innovation is zero and the mean does not move, the uncertainty can decrease. In this linear Gaussian model the posterior covariance matrix depends on the observation geometry and its uncertainty, rather than on the numerical observed value.

The derivation is one application of Gaussian conditional expectation per measurement. Before observing $y_t$, the state and observation have joint covariance matrixConditioning the upper block on the lower block gives exactly the two update equations. The upper-right block explains the Kalman gain: a measured component corrects a state component only to the extent that their prediction errors have covariance. The measurement can therefore improve an unobserved velocity when a state-space model correlates its error with an observed position.

Two deep layers of incompressible flow with upper mass density ${\rho }_1$, lower mass density ${\rho }_2$, and tangential velocities $U_1,U_2$ have this dispersion relation when surface tension is absent. Decaying velocity potentials and the linearized kinematic boundary condition determine their amplitudes; the dynamic boundary condition for an inviscid interface gives the displayed relation. For $k>0$ the squared departure of the phase velocity $c$ from the density-weighted mean velocity is $g({\rho }_2-{\rho }_1)/[k({\rho }_1+{\rho }_2)]-{\rho }_1{\rho }_2(U_1-U_2{)}^2/({\rho }_1+{\rho }_2{)}^2$. A negative value gives Kelvin-Helmholtz instability despite stable density stratification.

For unit mass in $\Phi =-GM/r$, conservation of angular momentum splits the velocity norm into ${\dot{r}}^2+L^2/r^2$. Conservation of energy then gives $\frac{1}{2}{\dot{r}}^2+L^2/(2r^2)=E+GM/r$. It separates radial dynamics from the reconstruction of the orbital angle.

The relative displacement of a Kepler orbit sweeps equal areas about the attracting focus of a conic in equal times. A short interval near periapsis covers a wide angle at small separation; the same interval near apoapsis covers a narrow angle at large separation. The swept areas are equal even though the travelled distances are different. In particular, the sectors have curved ellipse boundaries: triangles formed by joining their endpoints only approximate the swept areas for short intervals.

Let $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ be the relative specific angular momentum. The areal velocity isA central force exerts zero torque about its centre, so conservation of angular momentum makes this rate constant. This explanation also applies to other central forces; the ellipse shape specifically comes from Newtonian gravity.

In the $a=1$, $e=0.6$, $\mu =1$ example, specific angular momentum is $h=\sqrt{\mu a(1-e^2)}=\sqrt{0.64}=0.8$, giving areal velocity $0.4$. At periapsis and apoapsis, the velocity is perpendicular to the relative displacement, so $h=rv$. The endpoint speeds are consequentlyThe fourfold speed difference is consistent with the fourfold separation difference. Away from these endpoints, $h/r$ gives only the transverse component of velocity; the full speed follows from the vis-viva equation.

To actually place equal-time markers, use Kepler's equation. Write mean anomaly as $\mathcal{M}=nt$, with $t=0$ at periapsis and mean motion $n=\sqrt{\mu /a^3}=1$. The different symbol $\mathcal{M}$ avoids confusing this angle with the mass $M$. Solve for eccentric anomaly $E$ in radians, then compute the relative position:This timing procedure is explained in NASA educational archive, How Orbital Motion is Calculated.

For the first interval $0\le t\le \pi /6$, the final eccentric anomaly is $E\approx 1.041495$. For the equal interval $\pi \le t\le 7\pi /6$ starting at apoapsis, it ends at $E\approx 3.471065$. Both swept areas are $\Delta A=(h/2)\Delta t=0.4\pi /6=\pi /15\approx 0.209440$, one twelfth of the entire ellipse's area $0.8\pi$. The diagram computes its markers from Kepler's equation, making visible how uniform time steps become uneven spatial steps.

In axes directed along the periapsis and transverse to it, a Kepler orbit with gravitational parameter ${\mu }_g$ and specific angular momentum $h$ has the displayed velocity. The velocity tip traces a circle of radius ${\mu }_g/h$ centred at $(0,e{\mu }_g/h)$. To derive it, differentiate $r=h^2/[{\mu }_g(1+e\cos f)]$, use $r^2\dot{f}=h$, and resolve $\dot{r}\hat{\mathbf{r}}+r\dot{f}\hat{\mathbf{\theta }}$. A rotation of the spatial axes rotates the entire hodograph by the same angle.

Dividing a pressure disturbance by the constant reference mass density ${\rho }_0$ gives a quantity whose gradient enters the Boussinesq equations directly as an acceleration. Its units are squared velocity. Distinguish this convention from physical pressure when calculating energy flux or comparing wave polarization formulae.

For a fixed geometry, the steady Stokes equation maps $(\mathbf{u},p^{\prime },\mathbf{f})$ to $(-\mathbf{u},-p^{\prime },-\mathbf{f})$ when all imposed boundary velocities and disturbance forces are reversed; $p^{\prime }$ is pressure relative to any unchanged hydrostatic pressure. Linearity of Stokes flow and Uniqueness of Stokes flow therefore reverse the entire velocity field. If the driving schedule and evolving geometry are retraced, fluid elements retrace their paths too. Reversing just one drive while leaving another applied force unchanged does not meet these conditions.

This reversal concerns the paths, while viscous dissipation remains positive: for a sphere, the required power $6\pi \mu a{U}^2$ is unchanged by $U\mapsto -U$. A reversed experiment must supply fresh energy. Inertia, Brownian motion and irreversible contact can prevent perfect retracing. Edward M. Purcell illustrated the mechanism with a viscous fluid flow demonstration in Life at low Reynolds number, American Journal of Physics 45, 3–11 (1977).

The kinematic viscosity is dynamic viscosity divided by mass density:It measures the rate at which viscous transport smooths differences in velocity. For constant $\mu$ and $\rho$ in incompressible flow, dividing the Navier-Stokes equation by mass density makes the viscous acceleration $\nu {\nabla }^2\mathbf{u}$, where ${\nabla }^2$ is the Laplacian. ^[ref]

For example, stipulate two Newtonian fluids with the same dynamic viscosity $0.10\mathrm{Pa}\mathrm{s}$ but different mass densities. In the same steady Couette flow as above, both have shear stress $2.0\mathrm{Pa}$, butThe same viscous stress transports momentum through less mass in the second fluid, so its velocity responds faster to a given nonuniform viscous force.

To see the diffusion equation directly, consider an unsteady parallel velocity field $\mathbf{u}=(u(y,t),0,0)$ with constant material coefficients and no streamwise pressure gradient or body force. Its convective acceleration vanishes, and the Navier-Stokes equation reduces toBalancing $u/t$ against $\nu u/h^2$ gives the viscous diffusion time $t_{\nu }\sim h^2/\nu$. For the $1.0\mathrm{mm}$ gap, the two fluids have scales $0.010\mathrm{s}$ and $0.0050\mathrm{s}$. This is a characteristic scale, not an exact settling time. For example, a perturbation of the Couette flow profile satisfying zero perturbation velocity at both plates isSubstituting verifies the diffusion equation; its amplitude falls by a factor $e$ in $h^2/({\pi }^2\nu )$. Kinematic viscosity is therefore useful for predicting how quickly a moving wall influences the interior, while dynamic viscosity gives the shear stress at a prescribed shear rate.

Kinetic energy is the energy associated with motion. For a nonrelativistic point mass of mass $m>0$, measured in a specified inertial frame,Here $\mathbf{v}$ is velocity and $v$ is speed; $T$ is another common symbol for kinetic energy. The dot product makes $K$ nonnegative regardless of the direction of velocity. At fixed mass, doubling speed quadruples kinetic energy: a $2\mathrm{kg}$ point mass at $2m/s$ has $4\mathrm{J}$, whereas at $4m/s$ it has $16\mathrm{J}$. The SI units are $\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$. Different reference frames can assign different velocities and therefore different kinetic energies. ^[ref]

A short outline explains why kinetic energy has this formula. In a nonrelativistic inertial frame, constant mass and Newton's second law give, using the chain rule,Integrating over time proves the work-energy theorem, $W_{\mathrm{net}}=K_f-K_i$. The net force is the sum of all forces, and $W_{\mathrm{net}}$ is their total work, not merely the work of a chosen driving force. This route to final speed is useful when the forces as functions of position are known but the full motion as a function of time would be harder to find. ^[ref]

In the running example, a $2\mathrm{kg}$ point mass starts at rest and receives a constant net $3\mathrm{N}$ force through $4\mathrm{m}$. Its net work is $12\mathrm{J}$, soAt any intermediate distance $x\ge 0$, $K(x)=(3\mathrm{N})x$ and $v(x)=\sqrt{2Fx/m}$: kinetic energy is a linear function of distance, while speed is proportional to the square root of distance. If a net opposing $3\mathrm{N}$ braking force subsequently acts for $4\mathrm{m}$, it does $-12\mathrm{J}$ of work and brings this point mass to rest. Negative net work decreases kinetic energy; it does not make kinetic energy negative. Beyond the stopping point the motion must be reconsidered, since the point mass cannot keep moving forward with a negative $K$.

Kinetic energy also makes the exchange with potential energy intuitive: a falling body's loss of gravitational potential energy can appear as increased kinetic energy. The formula above applies when speed is small compared with the speed of light; at higher speeds, special relativity changes the expression. ^[ref]