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In a Cartesian inertial frame, let $\mathbf{v}$ and $\mathbf{a}=d\mathbf{v}/dt$ be the coordinate velocity and acceleration of a massive particle. With $\gamma =(1-v^2/c^2{)}^{-1/2}$, differentiating its four-velocity givesIn the $+---$ metric signature, its invariant Minkowski norm is $A^{\mu }{A}_{\mu }=-{\alpha }^2$. Expanding that contraction gives the displayed proper acceleration. Resolving the coordinate acceleration into components parallel and perpendicular to $\mathbf{v}$ yields the equivalent formula ${\alpha }^2={\gamma }^6{a}_{\parallel }^2+{\gamma }^4{a}_{\perp }^2$. In particular, constant coordinate acceleration and constant proper acceleration are different prescriptions.

Rapidity is a dimensionless parameter for a signed collinear velocity $u$ with $|u|<c$, defined using the inverse hyperbolic functions byHere $c$ is the speed of light, and $\tanh$ is the hyperbolic tangent. Zero velocity has zero rapidity; negative velocity has negative rapidity. Every finite real rapidity gives $|u|<c$, while $u\to \pm c$ requires $\theta \to \pm \infty$. Thus a light ray has no finite rapidity in this parametrization. The header's formula uses natural units $c=1$: its $v$ is the numerical velocity $u/c$.

For the inertial frames in the velocity-addition formula, $S^{\prime }$ moves at signed velocity $V$ along the $x$-axis of $S$. Set $\varphi =\mathrm{artanh}(V/c)$. Re-expressing the same particle in the moving inertial frame givesThe subtraction is a passive change of reference frame; the inverse adds the particle's rapidity in $S^{\prime }$ to the frame's rapidity relative to $S$. For example, $V=0.6c$ and $u^{\prime }=0.6c$ give $\varphi ={\theta }^{\prime }=\ln 2$, so $\theta =\ln 4$ and $u=c\tanh (\ln 4)=15c/17$, exactly as in the velocity-addition formula.

For a massive particle of rest mass $m>0$ moving in one spatial dimension, its energy and signed momentum areThe hyperbolic cosine is the particle's Lorentz factor, and the hyperbolic sine is $\gamma u/c$. The hyperbolic Pythagorean identity immediately gives the relativistic energy-momentum relation $E^2-c^2{p}^2=m^2{c}^4$. In the older $c=1$ notation, writing the rest mass as $M$ recovers $v=\tanh \theta$, $E=M\cosh \theta$, and $p=M\sinh \theta$. These expressions use natural units in which the factors of $c$ are numerically one; the preceding formulas restore those factors explicitly. ^[ref].

The useful simplification follows from the hyperbolic-function identity $\tanh (a+b)=(\tanh a+\tanh b)/(1+\tanh a\tanh b)$: the nonlinear velocity-addition formula becomes ordinary addition of rapidities. See iterated collinear boosts with equal rapidity for a plotted example. A common collinear Lorentz boost shifts every rapidity by the same constant, so a rapidity difference is invariant. This makes rapidity useful for relativistic factorized scattering in one spatial dimension, where two-body S-matrices depend on the difference of the incoming rapidities, with particle species fixed.

For rectilinear motion write the four-velocity in an inertial frame as $U=(c\cosh \eta ,c\sinh \eta )$, where $\eta$ is the rapidity and $\tau$ is proper time. Differentiation gives $A=c\dot{\eta }(\sinh \eta ,\cosh \eta )$, with invariant $A\cdot A=-c^2{\dot{\eta }}^2$ in the $+-$ metric signature. The signed proper acceleration is therefore $a=c\dot{\eta }$. Constant signed acceleration gives $\eta ={\eta }_0+a\tau /c$; specifying initial position alone leaves the initial rapidity ${\eta }_0$ undetermined.

For future-directed unit timelike four-velocities in metric signature $(+---)$, $U_1\cdot U_2/c^2$ is the Lorentz factor of either observer measured in the rest frame of the other. In a common inertial frame it equals ${\gamma }_1{\gamma }_2(1-{\mathbf{\beta }}_1\cdot {\mathbf{\beta }}_2)$. ThereforeExpanding the numerator converts this to the dot-and-cross-product form of relativistic velocity addition. The invariant construction explains why the relative speed is symmetric even when the spatial velocity components belong to different rest frames.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.

Consider an ideal circular vacuum path of laboratory radius $R$, a limit of closely spaced mirrors constraining light around the circle at laboratory speed $c$. The source and detector rotate uniformly together with signed angular velocity $\Omega$, where $|\Omega |R<c$, in flat spacetime. Let $L=2\pi R$ be the laboratory circumference and $A=\pi R^2$ the positive area. The $+$ pulse travels counterclockwise and the $-$ pulse clockwise, independently of the sign of $\Omega$. Both depart from the detector at $t=0$.

Unwrap the circle into a laboratory arc coordinate. The two first-return conditions areConsequently the exact laboratory arrival times and their difference areThe denominators $c\mp \Omega R$ are closing rates in these laboratory equations; they are not local speed of light measurements by a rotating observer. The two arrival spacetime events lie on the same detector world line. Its constant speed is $|\Omega |R$, so proper time along that world line givesThus laboratory coordinate-time and rotating-detector proper time delays agree only to leading order at small $|\Omega |R/c$. The radius and circumference here are laboratory quantities, not lengths obtained by laying comoving rods around the rim. Finite mirror polygons, transparent media, and optical dispersion require their own path and propagation assumptions before using an exact formula.

An exact dimensionless check takes $\Omega R/c=1/5$. Relative to $L/c$, the laboratory arrival times are $5/4$ and $5/6$, giving $\Delta t/(L/c)=5/12$. The detector instead records $\Delta \tau /(L/c)=\sqrt{6}/6\simeq 0.408248$. The leading Sagnac area law would give $\Delta t/(L/c)\simeq 2/5$. These are model values chosen to make the relativistic distinction visible, not attainable parameters of the historical apparatus.

The rotating loop does not constitute a universal measurement of the one-way speed of light. It compares completed circuits at a single readout; assigning a velocity to each remote leg additionally requires a clock synchronization convention. Local Einstein synchronisation around a rotating rim does not close into a single globally synchronized clock network: the accumulated mismatch is another expression of the rotating-path asymmetry. Local comoving vacuum measurements still give $c$ in both directions. Special relativity predicts a nonzero closed-loop delay because the receiver moves during propagation, while no single comoving inertial frame covers the whole circuit.

Spatially separated simultaneous events in one inertial frame are generally not simultaneous in a relatively moving frame. The Lorentz transformation explicitly mixes position into time. This is essential when comparing distances between moving light pulses: one must compare their positions at the same time in the measuring frame.

At $v=0.6c$, the rod's simultaneous laboratory endpoint measurements in length contraction have $(ct,x)=(0,0)$ and $(0,8\mathrm{m})$. They become $(c{t}^{\prime },x^{\prime })=(0,0)$ and $(-6\mathrm{m},10\mathrm{m})$. In contrast, simultaneous rod-frame measurements at $t^{\prime }=0$ use $(ct,x)=(0,0)$ and $(7.5\mathrm{m},12.5\mathrm{m})$. The latter front-end event occurs later in the laboratory, after the front has moved. The sloping $t^{\prime }=0$ line in the figure is a set of simultaneous spacetime events, not a signal's trajectory. ^[ref].

The rest frame of a massive particle is an inertial frame in which its spatial momentum is zero. Its four-momentum is then $(mc,0)$ and its energy is $m{c}^2$. Such a frame exists for a future-directed timelike four-momentum: the Lorentz transformation with boost velocity $\mathbf{v}=c^2\mathbf{p}/E$ makes the spatial momentum zero, and the mass shell relation implies $|\mathbf{v}|<c$. A massless particle has no rest frame, because $E=c|\mathbf{p}|$ would require a boost with speed $c$.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]

The Sagnac effect is a difference in return times, or an associated interference fringe displacement, for light sent in opposite directions around a closed optical path in a rotating apparatus. The light leaves one beam splitter, travels around the same loop in opposite senses, and returns to the rotating beam splitter and readout. For a slowly rotating apparatus in flat spacetime, the leading laboratory arrival-time difference isHere $c$ is the vacuum speed of light, $\mathbf{\Omega }$ is the apparatus's angular velocity relative to an inertial frame, and $\mathbf{A}$ is the vector area oriented by the chosen positive direction around the closed curve $C$. With positive rotation about the upward normal vector, the counterclockwise beam is labelled $+$ and arrives later. Reversing rotation, or reversing the orientation convention alone, changes the sign of $\Delta t$.

The surprise is that a small loop can reveal its rotation by comparing two beams locally, even though each local vacuum speed of light measurement gives $c$. A rotating reference frame is not a single inertial frame: a beam must catch the moving receiver after a full circuit. The Sagnac area law makes this difference a practical way to measure angular velocity, while the Relativistic Sagnac delay on a circular ring distinguishes laboratory coordinate time from the receiver's proper time.

A secular tidal torque is the torque remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned tidal deformation lets the companion's gravity change the spin angular momentum; the orbit receives the opposite torque. Tidal dissipation converts mechanical energy into heat, while total angular momentum is conserved in an isolated system. Thus the spin-orbit coupling exchanges angular momentum rather than destroying it. ^[ref].

For intuition near synchronous rotation on a circular orbit, hold mean motion $n$ and moment of inertia $I$ fixed and adopt the illustrative linear dissipative lawHere $\omega$ is spin angular velocity, $\theta$ is a marked surface direction in an inertial frame, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss areThe orbital contribution to this power is $-n⟨\tau ⟩$, and the spin contribution is $\omega ⟨\tau ⟩$; their sum becomes heat. Holding $n$ fixed is a local approximation to the coupled evolution. Actual secular tidal torque depends on the deformation response and orbital eccentricity, so this simple law does not establish which spin-orbit resonance an eccentric body captures into.

Special relativity describes physics in Minkowski spacetime, where inertial frames assign different space and time coordinates to the same spacetime events. Its two postulates are Einstein's principle of relativity and the same vacuum speed of light $c$ in every inertial frame, independent of the source's motion. Together they replace the Galilean transformation with the Lorentz transformation, coupling space and time rather than providing a universal clock. Einstein's 1905 paper developed this account through measuring rods, clocks, and light signals. ^[ref].

A spacetime event is an occurrence at a definite place and time: a pulse reaching a detector, for example. An inertial frame supplies a Cartesian coordinate system and clocks at rest throughout it. Einstein synchronisation sets a distant clock's reflection reading to the midpoint between a light signal's departure and return readings at the sending clock. The resulting coordinate time records the local synchronized clock at each spacetime event; it differs from the time when a distant observer receives its image. This distinction makes time dilation, length contraction, and the relativity of simultaneity precise measurements. ^[ref].

The beauty of special relativity is that these different measurements fit one geometry: all inertial frames agree on the Minkowski interval. It lets us translate an experiment between a laboratory and a moving apparatus while preserving the light-signal relations between spacetime events.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

Synchronous rotation means one prograde turn about the body's own axis per orbital period, measured against fixed directions in an inertial frame. This $1:1$ spin-orbit resonance approximately describes the Moon's synchronous rotation toward Earth. ^[ref].

The striking geometric idea is that one turn in space can keep one face toward the companion: a fixed orientation in the frame turning with the orbit is a changing orientation in an inertial frame. This distinction helps us predict illumination: a synchronously rotating exoplanet orbiting its illuminating star can have persistent day and night hemispheres, whereas the Moon, synchronized to Earth, receives changing illumination from a different body, the Sun.

If one clock measures a proper time interval $\Delta \tau$ while moving at constant speed $v$ in an inertial frame, that frame measures $\Delta t=\gamma \Delta \tau$, where $\gamma$ is the Lorentz factor. The two spacetime events must lie on that clock's world line.

Suppose a clock moving at $0.6c$ advances by $4\mu \mathrm{s}$ between two ticks. Set $t=x=t^{\prime }=x^{\prime }=0$ at the first tick. The laboratory's synchronized clocks assign $\Delta t=1.25\times 4=5\mu \mathrm{s}$. During that interval the clock travels $\Delta x=0.6c\times 5\mu \mathrm{s}=3c\mu \mathrm{s}$, approximately $899.38\mathrm{m}$. The Minkowski interval checks the result:Both frames therefore recover the same $4\mu \mathrm{s}$ of proper time. This calculation is useful whenever a process is timed on a moving apparatus but described using laboratory clocks.

A light clock makes the mechanism visible. Two mirrors separated by $h$ perpendicular to the motion exchange a pulse; define one tick as its complete return to the lower mirror. In the clock's rest frame the tick takes $\Delta \tau =2h/c$. In the laboratory each half-tick lasts $T$ and the light traverses a diagonal, so the Pythagorean theorem givesFor $h=1\mathrm{m}$ and $v=0.6c$, each leg covers $1.25\mathrm{m}$ while the mirrors move $0.75\mathrm{m}$. The return tick lasts $2.5\mathrm{m}/c$ instead of $2\mathrm{m}/c$: the light has the same speed of light in both accounts. ^[ref].

Reciprocal time dilation is consistent: each inertial frame times the other frame's single moving clock using its own synchronized clock network. These comparisons use different simultaneous spacetime events. Comparing reunited clocks instead requires their complete world lines, as in the twin paradox.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

For an axially symmetric rigid body with $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$, no external torque means that angular momentum $\mathbf{L}$ is fixed in an inertial frame. The axial angular velocity ${\omega }_3$ is constant by Euler equations for a torque-free rigid body, and the symmetry axis obeysIts angle to $\mathbf{L}$ is constant and it precesses in space at $L/I_{\perp }$. The transverse angular-velocity components instead rotate in body coordinates at signed rate $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. These are different frame descriptions and must not be substituted for one another.

The twin paradox compares proper time along the different world lines of a travelling observer and a stay-at-home observer who reunite. Reciprocal inertial time dilation produces an apparent contradiction only if the traveller's change of inertial frame is ignored.

The decisive experiment is local: put two clocks together at departure, separate them, then compare their readings together at reunion. Everyone agrees on those two spacetime events and on the final difference. The twin paradox asks why two clocks can accumulate different proper times between the same spacetime events, despite reciprocal time dilation during uniform motion. Its answer is the geometry of their complete world lines, together with the relativity of simultaneity when comparing distant clocks.