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The time integral of torque about a specified origin equals the change in angular momentum about that origin under the usual fixed-origin inertial frame balance. Its units are $\mathrm{N}\mathrm{m}\mathrm{s}$, equivalently $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. For rotation about a fixed principal axis, the magnitude is $I\Delta \omega$.

Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

Move the fixed origin from $O$ to $O^{\prime }=O+\mathbf{a}$, keeping the axes parallel and the inertial frame's velocity measurements unchanged. The new positions are ${\mathbf{r}}_i^{\prime }={\mathbf{r}}_i-\mathbf{a}$, so distributivity of the cross product givesHere $\mathbf{P}$ is total linear momentum. For the angular momentum example with $\mathbf{a}=(1,0,0)\mathrm{m}$, the new position is $(1,0,0)\mathrm{m}$ and $L_{O^{\prime },z}=6-3=3\mathrm{kg}{\mathrm{m}}^2/s$. If total momentum is zero, angular momentum is independent of this origin shift. A moving origin additionally contributes $-\dot{\mathbf{a}}\times \mathbf{P}$ when this expression is differentiated while retaining the original momenta; see angular momentum about the centre of mass for an important case where that term vanishes.

Areal velocity measures how fast a position vector from a chosen origin sweeps signed swept area. It connects a geometric picture of motion to angular momentum: a long position vector with slow transverse motion can sweep the same area per unit time as a short one with fast transverse motion. Constancy requires a dynamical condition, given by the torque criterion for constant areal velocity.

Choose a fixed origin $O$ in an inertial frame and a fixed plane through $O$, with constant unit vector $\mathbf{n}$ normal to it. For a differentiable trajectory in that plane, let $\mathbf{r}(t)$ be position relative to $O$ and $\mathbf{v}=\dot{\mathbf{r}}$ its velocity. Positive swept area follows the right-hand rule about $\mathbf{n}$. In positively oriented Cartesian coordinates in the plane, and in plane polar coordinates with angle $\theta$ in radians,Here the dot product selects the normal component of the cross product. Motion away from or towards $O$ alone contributes no area. The signed specific angular momentum component $h_n=\mathbf{n}\cdot (\mathbf{r}\times \mathbf{v})$ obeysfor a constant mass $m$. If $\mathbf{n}$ is chosen along a nonzero conserved specific angular momentum vector, then $h_n=h=|\mathbf{r}\times \mathbf{v}|$ and the familiar positive formula is $\dot{A}=h/2$. Reversing the orientation changes the sign; the instantaneous unsigned swept-area rate is $|\dot{A}|$.

Consider a stipulated ideal cart modeled as a point mass of mass $2\mathrm{kg}$ on a horizontal track in a laboratory inertial frame. Its initial position is $x(0)=0$ and its initial velocity is $v_x(0)=1.5m/s$ to the right. Independent devices exert horizontal forces $+6\mathrm{N}$ and $-6\mathrm{N}$ on this same cart. Its upward support force balances its downward gravitational weight; no other forces contribute. Rotation and deformation are excluded from this model.After $4\mathrm{s}$, the cart is at $6\mathrm{m}$, still moving right at $1.5m/s$. Starting the same model at rest instead would leave it at rest. Force balance determines the zero acceleration, while the initial velocity determines whether the cart moves.

For the plotted model, both horizontal devices are switched off simultaneously at $t=2\mathrm{s}$, with no impulse. The support and gravitational weight remain balanced. The individual horizontal forces change, but their sum stays zero on both sides of the switch, so the velocity stays continuous and unchanged. Switching off just one device would be a different model with a nonzero net external force.

For a fixed spherical gravitating source, place a test particle at $(a,0)$ with initial velocity $(0,\sqrt{\mu /a})$, where $a$ is outside the source. Its exact circular orbit isDifferentiating twice gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}=-\mu \mathbf{r}/a^3$, precisely the Newtonian gravitational field at each point of that circle. There is no outward physical force to cancel gravity in this inertial frame. Gravity supplies the inward centripetal acceleration that continually changes the velocity's direction.

Over a short interval, the displacement from the straight tangent path begins inward with magnitude $\frac{1}{2}(\mu /a^2)t^2$. The curvature of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius $a$ rather than striking the source.

Its speed is $\sqrt{\mu /a}$, its oscillation period is $2\pi \sqrt{a^3/\mu }$, and its specific orbital energy is $-\mu /(2a)$. Removing gravity while keeping the initial velocity would give the straight path $(a,a\omega t)$, rather than a stationary particle. This contrast explains why orbital weightlessness is compatible with a substantial gravitational force: a freely coasting spacecraft and its contents fall together. ^[ref]

Compton scattering redistributes photon energy and momentum in a collision with a charged subatomic particle, here an electron. For a free electron initially at rest in the chosen inertial frame, an incident photon of energy $E=hc/\lambda$ leaves with energy $E^{\prime }=hc/{\lambda }^{\prime }$ after deflection through $\theta$. The angle $\theta \in [0,\pi ]$ is between the incident and outgoing photon directions; it is not the electron recoil angle. Four-momentum conservation giveswhere $m_e$ is the electron rest mass and $T_e$ its final kinetic energy. The photon energy loss pays for the electron recoil: neither energy nor momentum disappears.

For an original worked example, stipulate $E=70\mathrm{keV}$, the rounded electron rest energy $m_e{c}^2=511\mathrm{keV}$, and the rounded Compton wavelength ${\lambda }_C=2.426\mathrm{pm}$; $\mathrm{keV}$ means $10^3$ electronvolts. Treat these rounded inputs as the example's constants. They give the following calculated values, with energies rounded to three decimal places:

Since $0\le 1-\cos \theta \le 2$, every outgoing photon has $0<E^{\prime }\le E$ and $T_e\ge 0$. At zero deflection there is no recoil; backscattering transfers the most energy. At fixed $\theta$, the absolute wavelength increase is independent of incident energy, while the fractional energy loss grows with $E/(m_e{c}^2)$. This makes Compton scattering useful for inferring energy transfers from directions in photon detectors.

The free-electron-at-rest assumption matters. An electron in a bound state has binding energy and an initial momentum distribution. Neglecting these is useful when the transferred energy is large compared with the relevant binding energy and the initial momentum is small compared with the resolved momentum transfer. Otherwise the outgoing photon energies can broaden or depart from the stationary-target formula; an initially moving electron can even give energy to the photon.

A material may also scatter X-rays through a coherent elastic scattering channel with an effectively unshifted component. When an intact chemical atom of mass $M$ takes the recoil, the corresponding Compton wavelength $h/(Mc)$ is much smaller than the electron value. The single free-electron curve does not predict that component's weight. More generally, energy versus angle is a kinematic constraint; angular wave intensity requires a differential scattering cross-section, together with target and detector information.

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

The Coriolis acceleration is the Coriolis force divided by the particle's mass:It is therefore independent of mass for a specified relative velocity and angular velocity. This is one contribution to the acceleration measured in a rotating reference frame. For a fixed common origin with an inertial frame and constant $\mathbf{\Omega }$, the full equation of motion in a rotating frame isAll vectors here are expressed in the rotating axes; ${\mathbf{F}}_{\mathrm{real}}$ is the sum of actual forces. An accelerating origin adds a translational apparent force, and time-dependent angular velocity also adds the Euler acceleration.

The factor two follows directly from the rotating-frame derivative formula. First ${\mathbf{v}}_{\mathrm{lab}}={\dot{\mathbf{r}}}_{\mathrm{rot}}+\mathbf{\Omega }\times \mathbf{r}$. Differentiate again in the inertial frame: differentiating ${\dot{\mathbf{r}}}_{\mathrm{rot}}$ contributes one $\mathbf{\Omega }\times {\dot{\mathbf{r}}}_{\mathrm{rot}}$, and differentiating $\mathbf{\Omega }\times \mathbf{r}$ contributes another. Move both terms to the rotating side of Newton's second law to obtain the minus sign and the factor two. The centrifugal acceleration comes from the remaining double cross product.

The Coriolis force is the apparent force proportional to a particle's velocity relative to a rotating reference frame. For constant angular velocity $\mathbf{\Omega }$ and particle mass $m$,Here $\mathbf{v}$ in the title formula means ${\mathbf{v}}_{\mathrm{rot}}$, not the velocity measured in an inertial frame. The Coriolis force belongs to the chosen reference frame: a particle stationary in that rotating reference frame has no Coriolis force, even though it can have a nonzero velocity in the laboratory. Other reference frames assign different angular velocities and relative velocities, and hence different Coriolis forces to the same motion.

The cross product makes the Coriolis force perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}_{\mathrm{rot}}$, with magnitude $2m\Omega |{\mathbf{v}}_{\mathrm{rot}}|\sin \alpha$, where $\alpha$ is the angle between them. Apply the right-hand rule to $\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}$ and then reverse its direction. Motion parallel to the rotation axis gives zero Coriolis force; reversing the relative velocity reverses the Coriolis force.

The appeal of this description is practical: axes attached to the Earth or a turning apparatus can be kept fixed on the page, provided Newton's second law includes the extra terms. A free particle in rotating coordinates shows exactly how a straight laboratory trajectory can become curved, while Coriolis acceleration in local Earth coordinates gives a useful terrestrial scale. The rotating-table demonstration of Coriolis deflection makes the two viewpoints visible in one experiment.

For a block sliding right on a stationary horizontal plane, take rightward acceleration $a$ as positive. A gauge reading $F$ for a pull at angle $\alpha$ above horizontal has horizontal component $F\cos \alpha$ and upward component $F\sin \alpha$. With no vertical acceleration or other forces, Newton's second law givesand thereforeThis inference requires continued contact with $N>0$ and rightward contact slip, so its numerator must be nonnegative. An upward pull reduces normal force; a downward pull, represented by $\alpha <0$, increases it. The horizontal constant-velocity formula is the special case $\alpha =0$, $a=0$.

For the original $2\mathrm{kg}$, ${\mu }_k=0.30$ model, an upward pull at $\alpha =30^{\circ }$ maintaining constant velocity needsDividing this gauge reading by $mg$ would give about $0.2953$, rather than $0.30$: the gauge reads the full inclined force, and the pull has changed the normal force.

Even with a horizontal pull, an accelerating measurement needs the $ma$ correction: ${\mu }_k=(F-ma)/(mg)$. In the same model, a horizontal $7.88\mathrm{N}$ pull produces $a=1m/{\mathrm{s}}^2$ while the block slides right; subtracting $ma=2\mathrm{N}$ recovers the $5.88\mathrm{N}$ kinetic friction. Dividing the uncorrected reading by $19.6\mathrm{N}$ instead gives about $0.4020$. Measure acceleration in an inertial frame, along the sliding direction, over the same interval as the gauge reading. These corrections follow from the force balance, independently of whether the resulting coefficient of kinetic friction is constant across runs.

// Target: classical-mechanics.bigb

For a source-free plane electromagnetic wave in vacuum, let $n^{\mu }=(1,\mathbf{s})$ be its null propagation Lorentz four-vector. In signature $(+---)$, take $F^{0i}=-E_i/c$, $F^{ij}=-{ϵ}_{ijk}{B}_k$, and define the dual electromagnetic field tensor by ${\tilde{F}}^{0i}=-B_i$, ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. Lower both indices with the Minkowski metric. Direct contraction givesBoth contractions vanish exactly when the transverse plane electromagnetic wave Maxwell equations hold. Being zero covectors, they remain zero in every inertial frame. Thus the transformed spatial wavevector is perpendicular to both transformed fields, even though the separate three-dimensional dot products are not themselves Lorentz scalars for arbitrary fields.

One contraction alone is insufficient without the other Maxwell equations. In particular, adding a magnetic field parallel to $\mathbf{s}$ leaves $n^{\mu }{F}_{\mu \nu }$ unchanged but violates $\mathbf{s}\cdot \mathbf{B}=0$. Conversely the two scalar perpendicularity conditions alone do not impose the required relation $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$ between field strengths and directions.

Earth's axial rotation relative to an inertial frame. The Foucault pendulum detects its component along the local vertical through the Latitude law of Foucault precession.

The laws of physics have the same form in all inertial frames; uniform absolute motion cannot be identified by experiments confined to an inertial laboratory. The invariant vacuum speed of light is a separate postulate. Together they motivate the Lorentz transformations of special relativity.

Einstein synchronisation sets clocks at rest in one inertial frame using exchanged light signals. If the sending clock records departure at $t_A$ and return at $t_A^{\prime }$, the distant clock's reflection reading is set to $t_B=(t_A+t_A^{\prime })/2$. This convention assigns equal outgoing and incoming travel times and defines the frame's distant simultaneity. It does not set the arrival time of an image equal to the time of the remote spacetime event.

For a circular orbit, gravity supplies centripetal acceleration, giving $v_c^2/r=\mu /r^2$. ThusA body in a circular orbit is bound despite its continuous free fall. At the same $r$, escape needs twice its kinetic energy per unit mass. An instantaneous forward tangential velocity increment from the circular orbit requires $\Delta v=(\sqrt{2}-1)v_c$, rather than the full escape speed from rest. In the synthetic example this is exactly $2(\sqrt{2}-1)m/s\approx 0.828427m/s$.^[ref]

Escape speed is a coasting threshold, not a required continuously maintained speed or a measure of rocket propellant. A powered vehicle can start at rest and acquire energy gradually. The rocket equation relates an ideal velocity increment to rocket exhaust speed and rocket mass ratio; rocket thrust, drag, and the timing and direction of propulsion affect a real ascent. Rotation also changes the relation between surface-relative velocity and the inertial frame used in the calculation.

Reaching another body is a finite-destination spacecraft trajectory problem. In the synthetic radial example, a launch at $2m/s$ reaches $r=4\mathrm{m}$ while remaining bound. A spacecraft can similarly encounter the Moon on a bound geocentric transfer; actual planning must include the Moon's gravity, target motion, arrival velocity, and any propulsion needed for capture. Escaping Earth alone does not imply escaping the Sun, because the relevant reference frame and attracting mass have changed. The isolated formula describes one gravitational well; a full mission requires more than its scalar threshold.

The escape velocity is the least initial speed that permits a freely coasting body to move arbitrarily far from an isolated attracting body. Despite its conventional name, it is a speed, not a unique velocity vector. For a fixed point mass $M$, or outside a body with spherical symmetry, set Newtonian gravitational potential energy zero at infinity. An exterior test particle of constant mass $m\ll M$ then hasHere $\mu =GM$ is the standard gravitational parameter, and $r$ is distance from the centre, not height above the surface. Set $r=R$ for surface escape velocity. Evaluate velocity in the body's rest inertial frame. Neglect drag, later rocket thrust, other bodies' gravity, and relativistic effects.^[ref]

Conservation of energy gives the threshold by requiring $E\ge 0$: at infinity the potential energy tends to zero and the kinetic energy cannot be negative. If $E=0$, the outward motion loses speed continuously toward zero. If $E>0$, the asymptotic speed is $v_{\infty }=\sqrt{2E/m}=\sqrt{v_0^2-v_{\mathrm{esc}}(r_0{)}^2}$. If $E<0$, infinity is inaccessible. For a radial outward launch, its outer turning point is $r_{\max }=-GMm/E$. For nonradial motion use angular momentum and the escape energy threshold to distinguish zero radial velocity from zero total speed.

Direction still determines the orbit and whether the body hits an obstacle. An initially inward velocity can cause collision even when $E\ge 0$. For an isolated spherical body, an exterior outward launch, or a tangential launch at or above escape velocity, has an outward escape branch without intersecting the body. A parabolic trajectory is the nonradial threshold path; a hyperbolic Kepler orbit has positive energy. The formula specifies their required local speed, not their orientation.^[ref]

A spacetime event is an idealized occurrence at one point of spacetime, such as a light pulse reaching a detector. An inertial frame assigns spatial coordinates and a time coordinate to it. Different frames can assign different coordinates to the same spacetime event; the event itself is not changed by a coordinate transformation.

For coordinates $x^{\mu }=(ct,x,y,z)$, the four-gradient of a scalar field is its covector of coordinate derivatives. Under a constant Lorentz transformation, ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. In metric signature $(+,-,-,-)$, raising its index gives ${\partial }^{\mu }=(c^{-1}{\partial }_t,-\nabla )$. Contracting it with a four-vector field produces the Lorentz scalar four-divergence. In particular ${\partial }_{\mu }{J}^{\mu }=0$ is charge continuity in every inertial frame.

A four-vector is a vector whose four vector components transform under a Lorentz transformation as $U^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{U}^{\nu }$. Its individual vector components generally change between inertial frames; the Minkowski metric combines them into invariant scalar products. Listing four quantities is not enough to make a four-vector: their transformation law matters.

The most direct example is the displacement between two spacetime events, written $(c\Delta t,\Delta x,\Delta y,\Delta z)$. Multiplying the time difference by the speed of light gives all four vector components the same length units. A position four-vector describes an event relative to a chosen origin, while a displacement does not depend on a common translation of the origin. Four-momentum and four-velocity use the same transformation rule for different physical quantities.