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Momentum conservation states that total momentum stays constant when the net force exerted from outside the chosen collection is zero. Use a closed collection, meaning the same fixed-mass point masses are followed throughout, and an inertial frame. ThenTo see why, sum Newton's second law over the members. If ${\mathbf{F}}_{ij}$ is the force on member $i$ from member $j$, Newton's third law assumes ${\mathbf{F}}_{ij}=-{\mathbf{F}}_{ji}$. Each pair cancels in the total:The internal forces act on different bodies and can change their individual momenta; their contributions cancel only in the collection's sum. An isolated system has no external interaction, but balanced external forces also suffice for momentum conservation. A closed system merely excluding matter exchange need not have conserved momentum. ^[ref]

The usefulness is that the detailed internal forces during a collision need not be known. Both an elastic collision and a perfectly inelastic collision obey the same isolated momentum balance; kinetic energy conservation is an additional condition for the elastic collision. If the external impulse is small but nonzero, the total momentum change is correspondingly small. For a variable-mass system exchanging matter, the entering or leaving momentum must also be included.

In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.

Newtonian mechanics describes motion using forces, masses and acceleration in inertial frames with absolute time. It is the low-speed limit of special relativity; for constant proper acceleration with $|at|\ll c$, relativistic hyperbolic motion reduces to $x\simeq a{t}^2/2$.

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]

Newton's laws of motion consist of inertia in an inertial frame, the relation between net force and change of momentum, and equal-and-opposite pair forces. The following three children state the individual laws.

In an inertial frame, a point mass of constant mass $m>0$ obeysThe net force is the sum of all forces acting on that body; momentum, velocity and acceleration refer to the same reference frame. A force changes velocity, rather than sustaining it: zero net force implies constant velocity, including a nonzero one. Doubling the net force at fixed mass doubles acceleration; doubling mass at fixed net force halves it. ^[ref]

For example, an otherwise balanced block of mass $0.5\mathrm{kg}$ receives a constant horizontal net force $2\mathrm{N}$ for $0.5\mathrm{s}$. Its acceleration is $4m/{\mathrm{s}}^2$, so its velocity changes by $2m/s$. Equivalently, the impulse is $F\Delta t=1\mathrm{N}\mathrm{s}$, and $\Delta v=\Delta p/m$. Starting from rest, its displacement during the pulse is $a(\Delta t{)}^2/2=0.5\mathrm{m}$. After the pulse, zero net force leaves it moving at constant velocity.

Applying the law therefore requires both a specification of the net force and initial conditions. In the harmonic oscillator example, the force is $-kx$, so Newton's second law becomes a differential equation for position. In an accelerating reference frame, the measured relative acceleration includes a contribution from the frame's own motion; the equation of motion in a rotating frame illustrates the extra terms required.

Constant mass is essential to the displayed $m\mathbf{a}$ form. A variable-mass system exchanging matter also exchanges momentum; its balance includes the entering or leaving matter's velocity. Simply differentiating $m\mathbf{v}$ and equating it to external force on that open body can omit this flux. The existing variable-mass system equation specifies the needed correction.

In the 1687 Principia, Newton's second law was stated in terms of change of motion proportional to impressed force, directed along that force. Its original statement is verbal; the derivative equation above is modern notation. ^[ref]

In the pair-interaction model of Newtonian mechanics, two bodies exert equal and opposite forces on each other:Both vectors are expressed in the same inertial frame at the same time. The subscripts name the agent and the recipient: $A$ acts on $B$, and $B$ acts on $A$. These are forces on different bodies, so they cannot cancel in the net force on either body. Newton's second law still uses all the forces on the particular body whose acceleration is sought. Identifying this third-law interaction pair does not require the bodies to be isolated: additional external forces can act at the same time.

The distinction makes Newton's third law useful both for drawing a free-body diagram and for deriving momentum conservation. A body-by-body calculation retains one member of each third-law interaction pair; a calculation for a collection containing both bodies cancels that pair in the total momentum balance. Equal interaction forces consequently coexist with nonzero, generally unequal accelerations.

Isaac Newton stated the equality and opposition of mutual actions in Law III of his 1687 Philosophiae Naturalis Principia Mathematica. The explanation distinguishes equal opposite changes of momentum from changes of velocity, which are inversely proportional to the bodies' masses when other influences are absent. Corollary III then explains why mutual interactions preserve the signed total momentum. The equations here are modern notation for that distinction. ^[ref]

The instantaneous body-pair equation has a scope: use it for the mechanical interactions assumed by this Newtonian mechanics model. In electromagnetism, bodies can exchange momentum with an electromagnetic field; their instantaneous mutual forces need not sum to zero. Momentum conservation then accounts for electromagnetic field momentum as well as matter, and for momentum crossing the chosen boundary. This complication occurs in classical electromagnetism, without invoking quantum mechanics.^[ref]

A worldline whose tangent is a null vector. In flat Minkowski spacetime, a freely propagating light signal follows a straight null geodesic. Its coordinate speed is $c$ in an inertial frame, even when another observer moves relative to that frame.

A one-way speed of light measurement assigns a travel time between spatially separated emission and reception events and therefore requires a distant-clock synchronization convention. In an inertial frame, Einstein synchronisation assigns the vacuum value $c$ in either direction. A completed-loop comparison such as the Sagnac effect is a different observable.

Conservation of momentum in the parent's rest frame gives $p_A+p_B=0$. Equal daughter masses and the relativistic energy-momentum relation then give equal daughter energies. Energy conservation fixesUsing $E^2=p^2{c}^2+m^2{c}^4$ and the specified direction of A,The velocities follow from $v=p{c}^2/E$:These are spatial momenta in the original inertial frame. For positive daughter rest mass, $m<M/2$ gives $0<V<c$.

The Einstein's principle of relativity says that the laws of physics take the same form in every inertial frame: no inertial frame is physically privileged, and uniform absolute motion cannot be detected by an internal experiment.

Assume $|u|<c$. Direct expansion of the transformed coordinates givesSince $y^{\prime }=y$ and $z^{\prime }=z$, the full Minkowski interval is invariant. Points of the outgoing light shell therefore satisfyAt each fixed $t^{\prime }>0$ this is a sphere of radius $c{t}^{\prime }$. The future branch remains future-directed: for a shell point with $t>0$, $|x|\le ct$ gives $t^{\prime }=\gamma (t-ux/c^2)>0$.

A Lorentz transformation is a linear coordinate transformation preserving the Minkowski metric, equivalently ${\Lambda }^T\eta \Lambda =\eta$. The calculation establishes this property for all events, not only points on the light cone. The displayed map has determinant one and preserves time orientation, so it is the proper orthochronous boost along the $x$ direction.

Let ${\mathbf{e}}_1^{\prime }(t),{\mathbf{e}}_2^{\prime }(t),{\mathbf{e}}_3^{\prime }(t)$ be an orthonormal basis fixed in the rotating reference frame $S^{\prime }$. A small rotation through $\mathbf{\omega }dt$ changes each basis vector by $\mathbf{\omega }\times {\mathbf{e}}_j^{\prime }dt$, so its derivative in the inertial frame isWrite the arbitrary vector as $\mathbf{a}={\sum }_j{a}_j(t){\mathbf{e}}_j^{\prime }(t)$. Differentiating the components and the moving basis separately givesThis proves the rotating-frame derivative formula, including when the angular velocity varies with time.

Use signed velocities in Skaro's inertial frame; the returning velocity $w$ must be negative. For the first two rockets, the Lorentz transformation of infinitesimal separations is $dx={\gamma }_u(d{x}^{\prime }+ud{t}^{\prime })$, $dt={\gamma }_u(d{t}^{\prime }+ud{x}^{\prime }/c^2)$. Dividing with $d{x}^{\prime }/d{t}^{\prime }=v^{\prime }$ gives the relativistic velocity-addition formulaFor the third rocket, its signed velocity in the second frame is $-w^{\prime \prime }$, so another application givesSince $0<v<c$ and $0<w^{\prime \prime }<c$, the denominator $1-v{w}^{\prime \prime }/c^2$ is positive. The third rocket returns from positive $x$ to Skaro if and only if $w<0$, henceEquality leaves it stationary in Skaro's frame, so it never returns from $x=2L$; a smaller $w^{\prime \prime }$ leaves it travelling outward.

Let $O$ denote departure, $A$ boarding the second rocket, $B$ boarding the third, and $C$ reunion with Skaro. In coordinates $(t,x)$, their positions on the worldline areIndeed, the first two coordinate durations are $L/u$ and $L/v$, and the returning leg has duration $-2L/w>0$. Equivalently, their $(ct,x)$ coordinates are obtained by multiplying each displayed time by $c$; these are the coordinates used for the spacetime diagram.

The diagram uses representative velocities satisfying the return condition; its event labels apply to the general formulas above. Skaro's elapsed time isOn a constant-velocity leg the traveller's proper time is $\Delta t/{\gamma }_V=\Delta t\sqrt{1-V^2/c^2}$, by time dilation. Neglecting the specified brief accelerations, Davros therefore ages byEvery contribution is positive because $w<0$. Each square-root factor is smaller than one, so $\Delta \tau <T$: Davros returns younger than a companion who remained on Skaro. The elapsed times compare the same departure and reunion events; they cannot be obtained by applying one time-dilation factor to the whole journey, whose worldline has three different velocities.

Work in an inertial frame, take all masses to be constant, and assume the internal forces obey the pairwise relation from Newton's third law $F_{ij}=-F_{ji}$. For the angular momentum result, also assume these pair forces are central, so $F_{ij}$ is parallel to $r_i-r_j$ and its pair torque vanishes. The equations from Newton's second law areSumming them and pairing the internal forces gives

About a fixed point $a$, define the total angular momentum and external torque byDifferentiation produces no velocity term because ${\dot{r}}_i\times m_i{\dot{r}}_i=0$. Grouping internal terms into pairs givesEach final cross product is zero by centrality, proving $\boxed{d{L}_a/dt=G_a}$. Equal and opposite internal forces give momentum conservation when the total external force is zero, but do not in general make their internal torques cancel; the central-force assumption supplies that extra step.

Let $M={\sum }_i{m}_i$ and let the centre of mass be $R=M^{-1}{\sum }_i{m}_i{r}_i$, so $P=M\dot{R}$. For an arbitrary moving reference point $a(t)$, the same differentiation givesTaking $a=R$, the extra term is $-\dot{R}\times M\dot{R}=0$. Therefore the corresponding angular momentum about the centre of mass satisfiesIts definition using velocities relative to $\dot{R}$ agrees with the one above, since ${\sum }_i{m}_i(r_i-R)=0$ makes ${\sum }_i(r_i-R)\times m_i\dot{R}=0$. Thus this result remains true even when the centre of mass accelerates.

Finally take the common mass to be $m$ and $F_i=-k{\dot{r}}_i$, with constant $k$. About either a fixed point or the centre of mass,The angular-momentum equation becomes $\dot{L}=-(k/m)L$. Multiplying by the integrating factor $e^{kt/m}$ gives $\frac{d}{dt}(e^{kt/m}L)=0$, hence the angular momentum decay under uniform linear drag isThe common mass and common drag coefficient ensure that the same decay factor applies to every term.

Let $p$ be the magnitude of the incident particle's momentum. The relativistic energy-momentum relation givesThe stationary particle contributes rest energy $m{c}^2$ and zero momentum. Conservation of four-momentum therefore gives equal photon energies $\epsilon =(E+m{c}^2)/2$. A photon has momentum magnitude $\epsilon /c$, so if the two momentum vectors meet at angle $\theta$,Substitution gives the angular relation for two-photon electron-positron annihilation, whose derivation depends only on the stipulated equal rest masses and equal photon energies:All energies and the angle here are in the laboratory inertial frame.

For perpendicular photon trajectories, $\cos \theta =0$ and hence $E=3m{c}^2$. Since $E=\gamma m{c}^2$ with Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For the small-speed limit, set $\beta =v/c\ge 0$ and $\delta =\pi -\theta$. The angle formula impliesUsing $\arcsin z=z+O(z^3)$ therefore gives $\delta =\beta +O({\beta }^3)$, soAt zero incident speed the momenta are equal and opposite, so $\theta =\pi$, consistent with the expansion.

Choose the two inertial frames to have coincident origins at $t=t^{\prime }=0$, and let the origin of $S^{\prime }$ have $x=vt$. The Lorentz transformation isHere ${\gamma }_v$ is the Lorentz factor. Along the particle's trajectory, $dx/dt=u$, soDividing gives the relativistic velocity-addition formulaThe denominator is positive because $|uv|<c^2$. Direct algebra yieldssince both factors in the numerator are positive. Consequently $\boxed{|u^{\prime }|<c}$: a subluminal velocity remains subluminal in every such inertial frame.

Apply the rotating-frame derivative formula first to position and then to its inertial velocity:The angular velocity is constant, so there is no term involving $\dot{\mathbf{\omega }}$. By Newton's second law in the inertial frame, the inertial acceleration is $\mathbf{F}/m$. Rearrangement gives the equation of motion in a rotating frame:The two additional accelerations are the Coriolis acceleration and centrifugal acceleration.

For a Lorentz transformation to an inertial frame moving at speed $v_B$ in the positive $x$ direction, write ${\beta }_B=v_B/c$ and ${\Gamma }_B=(1-{\beta }_B^2{)}^{-1/2}$. Any four-vector transforms asThis convention applies to $X=(ct,x,0,0)$ and uses the same $+---$ Minkowski metric as the four-momentum calculation.

The definition of proper time gives $d\tau =dt\sqrt{1-u^2/c^2}=dt/\gamma$ for $|u|<c$. Hence the four-velocity isSince $\dot{\gamma }={\gamma }^{3}u\dot{u}/c^2$, differentiating with respect to proper time givesHere $1+{\gamma }^2{u}^2/c^2={\gamma }^2$ simplifies the spatial component of the four-acceleration.

At each event, boost with $v_B=u$ to the instantaneous rest inertial frame. Then $A^{\prime 0}=0$ and $A^{\prime 1}=\gamma {\gamma }^4\dot{u}(1-u^2/c^2)={\gamma }^3\dot{u}$. Thus the signed proper acceleration for one-dimensional motion isThe last equality follows by differentiating $\gamma u$; its magnitude is the usual scalar proper acceleration.

If this signed proper acceleration is constant and the particle starts from rest, integration gives $\gamma u=at$, soIntegrating once more and imposing $x(0)=0$ yields the hyperbolic motion worldlineFor $a=0$ the continuous limiting solution is $x=0$. Finally, the Taylor series of the square root givesThis recovers constant-acceleration Newtonian mechanics at low speed while the exact relativistic speed remains less than $c$.

In the center of mass frame, the velocities are ${\mathbf{v}}_1=(M_2/M)\mathbf{v}$ and ${\mathbf{v}}_2=-(M_1/M)\mathbf{v}$. Hence the kinetic energy is $T=\frac{1}{2}\mu v^2$, while the Newtonian gravitational potential energy is $U=-G{M}_1{M}_2/r=-GM\mu /r$. The two-body orbital energy is thereforeThe relative equation of Newtonian gravity is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. It is a central force, so conservation of angular momentum confines the motion to a plane and preserves the specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|=r^2\dot{\theta }$.

For completeness, derive the polar equation of a Kepler orbit. Let $w=1/r$ and use primes for $\theta$ derivatives. Then $\dot{r}=-h{w}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{w}^2(w^{\prime \prime }+w)$. The radial equation gives the Binet equationChoose the angular origin at closest approach. Its solution is $w=(GM/h^2)(1+e\cos \theta )$, soFor the ellipse, the orbital eccentricity satisfies $0\le e<1$. Its extreme separations are $r_p=l/(1+e)$ and $r_a=l/(1-e)$. The semi-major axis is half their sum, giving $a=l/(1-e^2)$ and thus $l=a(1-e^2)$. The true anomaly $\theta$ runs through a full $2\pi$; the endpoints are the same position, and for a circular Kepler orbit the angular origin is arbitrary.

The radial and transverse relative speeds are $\dot{r}=(GM/h)e\sin \theta$ and $r\dot{\theta }=h/r$. Substituting these into the specific orbital energy givesTherefore the conserved orbital energy isThe negative specific orbital energy and fixed angular momentum characterize the bound Kepler orbit; neither should be assumed unchanged through a mass-ejecting explosion.

Initially the circular Kepler orbit has $r=a$ and $v^2=GM/a$. Treat the supernova kick in a binary star as impulsive: the relative position does not change, the companion's velocity is unchanged during the impulse, and the new neutron star receives $\mathbf{u}$ with $|\mathbf{u}|=\alpha v$. Thus ${\mathbf{v}}^{\prime }=\mathbf{v}+\mathbf{u}$, and with $\psi$ the angle between these vectors before the kick,Write $M^{\prime }=M_1^{\prime }+M_2$, $\beta =M^{\prime }/M$ and ${\mu }^{\prime }=M_1^{\prime }{M}_2/M^{\prime }$. The post-explosion specific orbital energy isUsing ${\epsilon }^{\prime }=-G{M}^{\prime }/(2a^{\prime })$ givesHere $a^{\prime }>0$ for a bound Kepler orbit. For a hyperbolic Kepler orbit, this formula uses the signed energy parameter $a^{\prime }<0$, rather than the positive geometric magnitude of the hyperbola's semi-major axis. At the parabolic Kepler orbit boundary, $1/a^{\prime }=0$.

The kick-direction binary survival criterion is bound if $S<2\beta$ and unbound with positive asymptotic speed if $S>2\beta$. To prove the printed sufficient disruption condition, choose a perpendicular kick, $\cos \psi =0$: then $S=1+{\alpha }^2$, so $M^{\prime }<\frac{1}{2}(1+{\alpha }^2)M$ produces positive specific orbital energy. This is sufficient, not the sharp existence threshold. Sincethe complete kick-direction binary survival criterion isThe second inequality follows by taking a kick directly opposite to $\mathbf{v}$. All directions remain bound if $2\beta >(1+\alpha {)}^2$, while every direction has positive escape energy if $2\beta <(1-\alpha {)}^2$. For $\alpha >0$, the marginal direction obeys $\cos \psi =(2\beta -1-{\alpha }^2)/(2\alpha )$ when the right-hand side lies in $[-1,1]$. Equalities give marginal parabolic Kepler orbits for the relevant extreme direction. With zero kick, losing more than half the original total mass unbinds the circular Kepler orbit.

For an escaping pair, the Newtonian gravitational potential energy approaches zero at infinite separation. Conservation of the post-explosion two-body orbital energy then gives $\frac{1}{2}{\mu }^{\prime }{V}^2={\mu }^{\prime }{\epsilon }^{\prime }$, or the asymptotic relative speed of a disrupted binaryThis is the relative recession speed, not the velocity of the new center of mass or either star's individual velocity in the original inertial frame. A marginal parabolic Kepler orbit separates with $V=0$ at infinity.

In the Earth inertial frame, the return journey takes $2d/v$. Bob is at rest in this frame, while Alice's proper time is reduced by the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Neglecting the short turnaround interval, their increases in age are

The twin paradox is the apparent contradiction from applying reciprocal time dilation to claim that each twin must be younger than the other. Alice changes inertial frames at the turnaround, so a single comoving inertial frame cannot describe her whole journey; Bob uses one inertial frame throughout. The twins follow different world lines, and integrating proper time along those world lines gives the displayed, consistent ages. The change in Alice's notion of simultaneity at turnaround removes the apparent contradiction. Acceleration distinguishes the paths; its magnitude is not an additional local rule for aging.

The meetings occur where Bob's constant position $x=\alpha c{t}_0$ intersects Alice's positive-$x$ branch of the hyperbolic motion:Bob is at rest in this inertial frame, so his elapsed proper time is

Parametrize Alice's world line bySince $t$ increases with $\eta$, the proper time element isAt the meetings, $\cosh \eta =\alpha$, giving endpoints $\eta =\pm \mathrm{arcosh}\alpha$. Thus Alice's elapsed proper time isHere $\mathrm{arcosh}$ is the nonnegative real inverse hyperbolic cosine. As a check, if $\alpha =\cosh {\eta }_{\ast }$ with ${\eta }_{\ast }>0$, then Bob ages by $2t_0\sinh {\eta }_{\ast }$ and Alice by $2t_0{\eta }_{\ast }$; $\sinh {\eta }_{\ast }>{\eta }_{\ast }$ confirms that Bob again ages more.