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An unstable particle of rest lifetime ${\tau }_0$ traversing distance $H$ at constant speed $v$ has survival probability $p=\exp [-H/(\gamma v{\tau }_0)]$. Requiring a flight time no greater than one dilated mean lifetime selects $p\ge e^{-1}$. In the ultrarelativistic approximation $v\simeq c$, this gives $\gamma \gtrsim H/(c{\tau }_0)$ and $E\gtrsim m{c}^{2}H/(c{\tau }_0)$. It is a specified survival-probability criterion, not a sharp cutoff for individual random decays. Continuous energy loss requires integrating the decay probability along the path.

For a uniform fixed-end mechanical string of length $L$, linear mass density $\mu$ and wave speed $c$, let its $n$th normal mode be $y_n=[A_n\cos {\omega }_{n}t+B_n\sin {\omega }_{n}t]\sin (n\pi x/L)$, where ${\omega }_n=cn\pi /L$. The conserved energy is the sum of kinetic energy and elastic potential energy,The sine and cosine squared integrals are $L/2$, so substitution proves the title formula. Orthogonality makes energies of distinct normal modes add. For zero initial displacement and initial modal velocity $V_n$, $A_n=0$ and $B_n=V_n/{\omega }_n$, giving $E_n=\mu L{V}_n^2/4$.

A directed stream of molecules used to prepare samples for spectroscopy.

A molecular beam also includes a stream of chemical atoms, then called an atomic beam. An oven supplies particles; aligned slits restrict their directions, and a vacuum limits subsequent collisions. Narrowing the directions does not by itself make every particle's speed equal. This gives a useful separation of tasks: prepare a path, select or measure speed, apply an interaction, and read the resulting particle counts or deflection.

The advantage is unusually tangible. A magnetic field can separate spin components in the Stern-Gerlach experiment; even gravity can sort a beam by flight time. The effusive beam speed distribution explains which particles leave the source, while gravity deflection of an atomic beam explains how that distribution becomes a pattern on a detector.

For a nonrelativistic point mass of fixed mass $m>0$, measured in an inertial frame, its momentum isMomentum is a vector pointing along the velocity, with International System of Units unit $\mathrm{kg}m/s$. Doubling either mass or velocity doubles momentum. In one dimension, a chosen positive direction makes $p=mv$ a signed quantity: a $2\mathrm{kg}$ body moving right at $3m/s$ has $p=+6\mathrm{kg}m/s$; moving left at the same speed gives $p=-6\mathrm{kg}m/s$. ^[ref]

For a collection, total momentum is the vector sum $\mathbf{P}={\sum }_i{m}_i{\mathbf{v}}_i$. Two equal masses moving with opposite velocities can have zero total momentum while both have positive kinetic energy. Direction is therefore essential when using momentum conservation; adding speeds instead of signed velocities gives the wrong balance.

Newton's second law connects the net force on one fixed-mass point mass to its momentum through ${\mathbf{F}}_{\mathrm{net}}=d\mathbf{p}/dt$. Integrating that change gives impulse. Here $\mathbf{p}$ denotes mechanical momentum; when canonical momentum differs from mass times velocity, the distinction in mechanical momentum below applies. The nonrelativistic formula also has a limited regime: special relativity is needed at speeds comparable with the speed of light.

The most probable speed is the mode, or density maximum, of the Maxwell speed distribution:For small equal-width speed bins, the bin near $v_{\mathrm{mp}}$ contains the largest expected fraction of particles. This does not mean that any exact speed has positive probability, or that every particle has this speed. The three common summaries obeyso $v_{\mathrm{mp}}<⟨v⟩<v_{\mathrm{rms}}$. The first describes a density peak, the second the mean speed, and the third the root mean square associated with kinetic energy. ^[ref]

For a concrete scale $s=300\mathrm{m}{\mathrm{s}}^{-1}$, these are approximately $424$, $479$, and $520\mathrm{m}{\mathrm{s}}^{-1}$. The most probable velocity vector and each most probable component remain zero in the gas rest frame. A zero vector labels one small region of velocity space, whereas a positive speed labels an entire spherical shell.

A short check uses the derivative of the logarithm of the Maxwell speed distribution for $v>0$:It is positive below $\sqrt{2}s$ and negative above it, locating the unique peak. The factor $v^2$ from the spherical shell supplies the term $2/v$; omitting it would incorrectly move the speed peak to zero.

For an ideal infinite filament with zero line charge density and current $I$ in a laboratory frame, the line-integrated four-current is $(0,I)$ in temporal/axial components. An axial boost of speed $u$ gives the title density and current. The laboratory magnetic field ${\mu }_{0}I\hat{\mathbf{\phi }}/(2\pi r)$ therefore becomes both an electric field from ${\lambda }^{\prime }$ and a magnetic field from $I^{\prime }$. Charge neutrality is frame-dependent for a current-carrying system. A model built from oppositely charged moving and resting populations makes this result explicit; the charge of each fixed material particle set remains invariant.

Let $n(E)dE$ be the number of neutrons per unit volume in an energy interval. The scalar neutron flux per unit energy is $\varphi (E)=v(E)n(E)$, with $v(E)$ the speed. Its integral has units of inverse area per unit time. It counts path length traversed per volume per time, including all directions; it is not the signed current through an oriented surface.

For a small uniform specimen containing $N_T$ target atomic nuclei, a reaction-specific collision cross-section ${\sigma }_a(E)$ gives the expected production rateAssume the incident field is little changed within the specimen, target depletion is negligible and each counted reaction creates the product of interest. If that product has a production branching fraction below one, include it in ${\sigma }_a$ or as an additional factor. A single neutron activation measurement supplies one weighted integral of an energy distribution, not a unique reconstruction of that distribution.

For an original two-group example, stipulate response collision cross-sections ${\sigma }_s=10{\sigma }_f$. With the same total scalar neutron flux $\Phi$, changing from all fast neutrons to half slow and half fast changes $R/(N_T{\sigma }_f\Phi )$ from $1$ to $5.5$. No additional neutrons need be emitted by the source. The chosen factor ten is not a historical silver measurement.

For some capture reactions over a limited low-energy interval, ${\sigma }_a\propto 1/v$ is a useful approximation. At fixed neutron flux, it gives increasing reaction probability per path length as speed decreases. At fixed neutron number density, ${\sigma }_{a}v$ is constant in this approximation, so the rate per target is unchanged. A real neutron moderator changes the local density, energy distribution, escape and absorption together. Neither all collision cross-sections nor all neutron activation rates vary monotonically as $1/v$.

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

Nonlinear acoustics retains finite-amplitude effects in sound waves, including amplitude-dependent propagation speed, distortion and possible shock waves. For a one-dimensional polytropic flow with $p\propto {\rho }^{\gamma }$, a rightgoing simple wave has characteristic speed $u+c=c_0+(\gamma +1)u/2$ relative to a quiescent reference. Thus small-amplitude linear acoustics need not be uniform when a background velocity nearly cancels the speed of sound.

A local scalar nonlinearity preserves Lorentz covariance of the wave operator when the wave speed is one. If $f=F^{\prime }$, the equation follows from $\mathcal{L}=({\partial }_t\varphi {)}^2/2-({\partial }_x\varphi {)}^2/2+F(\varphi )$. The physical potential is $-F$, so its sign must be distinguished from the sign written in the Lagrangian.

Robert Andrews Millikan's 1911 protocol alternated electrically driven ascent with timed descent after short-circuiting the plates; a changed ascent speed signalled ion capture.^[ref]

His 1913 refinements checked voltage before and after runs, calibrated optical distance and the clock, compared independent air-dynamic viscosity measurements, and varied pressure to test slip.^[ref] His 1924 lecture describes suppressing convection, switching the electric field, neutralizing drops, and reproducing smallest observed electrical speed increments.^[ref]

Harvey Fletcher's posthumous account recalls building the first oil apparatus and measuring with Robert Andrews Millikan; priority for suggesting oil remains uncertain.^[ref]

To interpret the readout, keep one drop and one electric field polarity. Let $u_i$ be its signed upward terminal velocity in charge state $i$, and let $v_f>0$ be its field-off falling speed. Use the linear drag coefficient $k$ already defined in elementary charge, with any required gas slip correction to Stokes drag included. Provided the drop's radius, surrounding air and electric field remain fixed, its electrical speed increment isThese relations apply while the electric force points upward in every state. A drop may still descend, with $u_i<0$, when that force is too weak to support it. A held drop has $u_i=0$ and therefore $S_i=v_f$, rather than zero electrical response. A neutral drop has $S_i=0$: its signed terminal velocity agrees with field-off descent. Thus neither raw upward speed nor the ability to hold a drop is itself a count of electrons.

The ratios cancel a fixed $k/E$, even if its value is poorly known. This is why relative charge quantization and absolute measurement calibration are different questions. A changed field-off $v_f$ warns that evaporation, air motion or another change has spoiled comparison of charge states. A stable $v_f$ is useful evidence but cannot exclude every compensating change. Repeat field-off checks around a proposed jump, allow the terminal velocity under linear drag to settle after switching, and propagate measurement uncertainty in both the field-on and field-off readings. Reusing a single estimate of $v_f$ gives the different $S_i$ a shared covariance.

Orbital speed is the magnitude of the velocity of an orbiting body in a specified reference frame. For a Newtonian circular orbit around a point source of mass $M$, equating $v^2/r$ to $GM/r^2$ gives $v=\sqrt{GM/r}$. In a curved spacetime, a speed inferred from coordinate radius and coordinate time generally differs from a speed measured by a local observer. For a circular orbit in the Schwarzschild metric, a static local observer measures $v_{\mathrm{local}}=r|d\varphi /dt|/\sqrt{1-2GM/(c^{2}r)}$; $r|d\varphi /dt|$ uses the clock normalization at infinity instead.

Two broad perfect conductors of width $w$, separated by gap $d\ll w$, carry opposite steady electric currents $I$ and connect through a terminating resistor $R$. Away from edges, Ampère's circuital law and the superposition principle give gap magnetic field magnitude ${\mu }_{0}I/w$. Each conductor is an equipotential, and Ohm's law sets their electric potential difference to $IR$. Planar symmetry and charge-free Gauss's law then give uniform gap electric field magnitude $IR/d$. Injection parallel to the upper current makes the electric and magnetic Lorentz forces opposite. Their cancellation selects speed $Rw/({\mu }_{0}d)$ independent of nonzero electric charge and of operating current; nonrelativistic kinetic energy is $m[Rw/({\mu }_{0}d){]}^2/2$.

Take the trace of the statistically isotropic magnetic second-moment tensor. There are three spatial directions, soChoosing the nonnegative speed,Thus $v_A$ is the root-mean-square value of any one Cartesian magnetic field component in Alfvén velocity units. It is not the magnitude of the mean Alfvén velocity, which is zero here. Statistical isotropy fixes a second moment even when the first moment vanishes.

Write the centre position as $\mathbf{x}$ and the bead positions as ${\mathbf{r}}_{1,2}=\mathbf{x}\mp \mathbf{R}/2$. In the affine velocity field, $\mathbf{u}({\mathbf{r}}_{1,2})=\mathbf{u}(\mathbf{x})\mp K\mathbf{R}/2$. The midpoint follows the flow, so its two slip velocities areEach bead loses mechanical power $\zeta$ times its squared slip speed. Adding the two powers proves the drag dissipation of a bead-spring pair:The factor one half comes from two equal-and-opposite bead slips, not from assigning only one bead a drag force.

The imaginary part of the complex potential of a two-dimensional potential flow is the stream function. Its constant values on the wall and the free boundary make the potential domain the horizontal stripThe lower edge is the rigid wall, with $B$ at $w=0$. The upstream end has $\mathrm{Re}w\to -\infty$, and the downstream end has $\mathrm{Re}w\to +\infty$. The upper edge is the free streamline. Thus $AB$ and $BC$ are the negative and positive portions of the real $w$ axis, respectively.

Write $q=w^{\prime }=u-iv$ for the complex velocity. Downstream $q\to 1$. Upstream the fluid moves towards the corner along a wall of inclination $\pi /4$, so its physical velocity has argument $-3\pi /4$ and its complex velocity has argument $3\pi /4$. Along each straight wall the velocity direction is fixed. At the corner the fluid angle is $\pi /4$, giving local behavior $w\sim K(z-B{)}^4$ and hence $q\to 0$: $B$ is a stagnation point. On the free streamline, the speed is one, so $|q|=1$.

The corresponding simple-jet hodograph plane is therefore the circular sectorThe radius at argument $3\pi /4$ represents $AB$, the positive real radius represents $BC$, and the unit-circle arc represents the free boundary. The upstream and downstream ends are $e^{3\pi i/4}$ and one, respectively; the corner is the sector's origin.

Work in the source rest frame and neglect the source's small cosmological redshift. Over an emission interval $\Delta t$, a ballistic relativistic jet feature travels $\beta c\Delta t$, advancing $\beta c\Delta t\cos \theta$ toward the observer and $\beta c\Delta t\sin \theta$ across the sky. The second photon has a shorter journey, so its arrival is delayed by onlyDividing the projected displacement by this arrival interval givesApparent superluminal motion is a combination of projection and light-travel delay. The feature's local speed remains $\beta c<c$; its inferred transverse displacement divided by the observer's arrival interval is not a local signal velocity.

For fixed $\beta$, differentiate the apparent speed with respect to angle:The maximum lies at $\cos \theta =\beta$ and has value ${\beta }_{\mathrm{app},\max }=\beta /\sqrt{1-{\beta }^2}=\gamma \beta$. Hence strict apparent superluminal motion requiresThe threshold is an infimum: equality yields exactly ${\beta }_{\mathrm{app}}=1$ at $\theta =\pi /4=45^{\circ }$, rather than a strictly superluminal value. For a chosen larger $\beta$, the optimal inclination is $\theta =\arccos \beta$, and the full superluminal interval satisfies $\sin \theta +\cos \theta >1/\beta$.

For intrinsically symmetric oppositely directed relativistic jets, their proper motions areThe ratio removes $D$ and yieldsTaking the product instead givesCausality requires $\beta \le 1$, so the limiting distance and inclination areHere and in e, “maximum” denotes the limiting $\beta \to 1$ bound; massive ejecta obey strict inequalities. The conversion is $1\mathrm{mas}\mathrm{da}{\mathrm{y}}^{-1}=5.6113\times 10^{-14}\mathrm{rad}{\mathrm{s}}^{-1}$, henceBoth bounds occur together in the light-speed limit. At any specified smaller distance, ${\beta }^2=x^2+(1-x^2)(D/D_{\max }{)}^2$ determines the intrinsic speed.