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The threshold must conserve both energy and momentum. For head-on photons, the total energy is $E_1+E_2$, while the total momentum has magnitude $|E_2-E_1|/c$. Their combined invariant mass $M$ satisfiesAt the threshold for photon-photon pair production, the electron and Positron have no relative momentum in their centre-of-momentum frame, so $M=2m_e$. ThereforeSimply equating the laboratory energy to $2m_e{c}^2$ would miss the very large laboratory momentum: the newly produced pair must carry it away.

For particles in an inertial frame, total momentum and angular momentum about a fixed origin areThe momentum derivative is the total external force when internal forces cancel pairwise. Angular momentum is conserved when the total external torque vanishes and the internal torques sum to zero, as for pairwise equal-and-opposite central forces. Equal-and-opposite internal forces alone need not cancel internal torque if their lines of action are different. An isolated system with these internal-force properties conserves both quantities; if fields carry momentum, the complete system must include them.

For a straight rocket, let its instantaneous remaining mass be $M$ and let $dM<0$ be ejected during a velocity increase $dv$. The exhaust's inertial velocity is $v-v_e$ to first order. Conservation of momentum givesIntegrating from initial mass $m+m_f$ to dry mass $m$ yields the rocket equationIt is the rocket-plus-exhaust system that is isolated; the rocket alone is a variable-mass open system.

For the firework, the original diagram places one tangentially firing point rocket at the end of each radial rod. The parallel axis theorem changes each rod's central inertia to $m{a}^2/12+m(a/2{)}^2=m{a}^2/3$. Adding all rods and fueled rockets givesWhen fuel is exhausted, $I_f=4m{a}^2$.

Let $M_f$ now denote the total remaining fuel in all three rockets, so $I=4m{a}^2+a^2{M}_f$. Fuel at the rim co-rotates before ejection. Its backward tangential exhaust velocity is $a\omega -v_e$, and its angular momentum per unit mass about the axle is $a(a\omega -v_e)$. For an infinitesimal ejection,Since $dI=a^{2}d{M}_f$, the co-rotation terms cancel and leave $Id\omega =-a{v}_{e}d{M}_f$. The tangential-exhaust rotor with a fixed dry inertia therefore obeysNeglecting the exhaust's initial tangential motion would give an incorrect angular-momentum balance. The rod mass remains in the denominator throughout the burn.

After the burn, the opposing constant axle torque $G>0$ acts on the fixed inertia $I_f$. Its work over the remaining turn equals the initial rotational kinetic energy:

A charge $q$ moving nonrelativistically in prescribed fields obeys the Lorentz force lawThe magnetic part is perpendicular to velocity, so $d(m{v}^2/2)/dt=q\mathbf{E}\cdot \mathbf{v}$: the electric field can change kinetic energy, while the magnetic field alone changes direction without doing work.

In a uniform electric field with no magnetic field, acceleration is constant: $\mathbf{v}={\mathbf{v}}_0+(q\mathbf{E}/m)t$ and $\mathbf{r}={\mathbf{r}}_0+{\mathbf{v}}_{0}t+q\mathbf{E}{t}^2/(2m)$. The trajectory is a parabola when there is initial motion perpendicular to the field, and a line when motion is purely along it. Reversing the charge reverses the electric acceleration.

In a uniform magnetic field with no electric field, resolve velocity into parallel and perpendicular components. The parallel component is constant; the perpendicular component rotates at the cyclotron frequencyThe resulting path is a helix, or a circle if $v_{\parallel }=0$, with pitch $2\pi v_{\parallel }/{\omega }_c$. Charge sign determines its sense of rotation. These relations underlie magnetic bending and mass/charge analysis. At relativistic energies, use momentum $\mathbf{p}=\gamma m\mathbf{v}$; at fixed energy the cyclotron frequency becomes $|q|B/(\gamma m)$.

For uniform crossed fields $\mathbf{E}\perp \mathbf{B}$, the velocity ${\mathbf{v}}_D=\mathbf{E}\times \mathbf{B}/B^2$ satisfies $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$. The orbit is a gyromotion about a guiding center drifting withindependent of mass and charge sign. A velocity selector transmits particles with the appropriate direction and speed $E/B$ without deflection. If the electric field has a component along the magnetic field, that component accelerates the parallel motion instead. Nonuniform fields can add other drifts; the simple helix and uniform crossed-field expressions require the stated uniform-field assumptions.

For a photon, the photon energy formula and the vacuum relation $\nu =c/\lambda$ give $E_{\gamma }=hc/\lambda$. For an electron, the de Broglie wavelength gives the same momentum magnitude $p=h/\lambda$, but the nonrelativistic kinetic energy is $K_e=p^2/(2m_e)$. Using $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$,Thus, to the precision of the wavelength,The electron answer is its kinetic energy, as required for the diffraction experiment; its relativistic energy would additionally include the approximately $511\mathrm{keV}$ rest energy. Since $K_e\ll m_e{c}^2$, the nonrelativistic approximation is well justified. Equal wavelengths imply equal momenta, not equal energies for particles with different energy–momentum relations.

Powered vertical flight. Take upward velocity $v$ as positive, and let $v_e>0$ be the downward exhaust speed relative to the rocket. During $dt$ the rocket's mass changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame velocity $v-v_e$. Compare the momentum of the rocket plus this newly expelled exhaust before and after the interval:Discard the second-order product $dmdv$. The result is $mdv+v_{e}dm=-mgdt$, orThe positive rocket thrust is $-v_e\dot{m}$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the rocket equation with gravitational loss:Using $d(mv)/dt$ for the rocket alone without the exhaust momentum flux would give the wrong equation for this open system.

Orbital speed under the stated gravity assumption. A circular orbit of radius $r_o=R_E+h_o=7.90\times 10^6\mathrm{m}$ requires centripetal acceleration $v_o^2/r_o$. If the magnitude of gravitational acceleration is the stipulated constant $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ even at the orbit,This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.

Burnout height. During the unpowered projectile motion, horizontal velocity is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital speed $v_o$. At burnout, therefore,These follow either by integrating $\ddot{y}=-g$ or by the vertical work-energy theorem. Substituting the preceding constant-$g$ orbital result givesThe vertical component is initially positive, as required to reach the orbit after burnout.

Alternative inverse-square interpretation. A real Earth's exterior Newtonian gravitational field is not constant over a $1500\mathrm{km}$ altitude change. If $9.8\mathrm{m}{\mathrm{s}}^{-2}$ instead means the surface value $g_0$, then Newton's law of universal gravitation gives $GM=g_0{R}_E^2$ andUsing this orbital speed but the paper's constant-$g$ flat projectile motion approximation for the coast would giveThis alternative changes the global constant-gravity assumption; it is not the same model as the $183\mathrm{km}$ result. An exact curved Kepler orbit coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.

Momentum identity. From the Lorentz factor, $v^2/c^2=1-{\gamma }^{-2}$. The relativistic three-momentum therefore satisfiesTogether with relativistic energy $E=\gamma m{c}^2$, this gives the energy–momentum relation $E^2-p^2{c}^2=m^2{c}^4$.

Parent in the laboratory. The rest energy is $m_0{c}^2=1\mathrm{GeV}$, while the total relativistic energy is $3\mathrm{GeV}$. ConsequentlyIts momentum magnitude is $p_{0}c=\sqrt{3^2-1^2}\mathrm{GeV}=\sqrt{8}\mathrm{GeV}$.

Daughters in the zero-momentum frame. The centre-of-momentum frame is initially the parent's rest frame. Conservation of momentum gives ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$. Equal daughter rest masses then imply equal relativistic energies. Conservation of relativistic energy givesTheir speeds are thereforeThe two velocities are opposite, and $p_X^{\ast }c=\sqrt{(1/2{)}^2-(1/3{)}^2}\mathrm{GeV}=\sqrt{5}\mathrm{GeV}/6$. Their total kinetic energy is $m_0{c}^2-2m_X{c}^2=m_0{c}^2/3$.

Invariant of the complete system. Before the decay, in the centre-of-momentum frame,After the relativistic two-body decay,In the laboratory, four-momentum conservation fixes the summed daughter energy to $3\mathrm{GeV}$ and their summed momentum to the parent's ${\mathbf{p}}_0$. Thus the same calculation isThe invariant is formed after summing the daughter four-momenta; adding their separate squared rest energies would omit the relative-motion contribution to the invariant mass of the pair.

Experimental evidence. Tests of special relativity check linked predictions about light propagation, clocks and four-momentum. Several independent kinds of experiment support those predictions.

The observed partial-drag coefficient agrees with this first-order prediction rather than simple Galilean addition of $u$. It tests a different consequence of the Lorentz transformation from a particle-lifetime measurement.^[ref]

Atmospheric muons and the two frame descriptions. Atmospheric cosmic rays create unstable particles whose decay chains produce muons. A stopped-muon experiment measures the time between a stopping signal and the subsequent decay electron, obtaining the proper mean lifetime ${\tau }_0\simeq 2.2\mu \mathrm{s}$. The associated light-travel distance is only $c{\tau }_0\simeq 660\mathrm{m}$, yet substantial numbers of energetic atmospheric muons reach detectors after travelling several kilometres.

For an ideal selected cohort produced height $h$ above a detector and moving downward at constant speed $v$, the flight time in the terrestrial frame is $h/v$. Time dilation predicts a laboratory mean lifetime $\gamma {\tau }_0$, hence survival fractionWithout time dilation, the predicted exponent would be $-h/(v{\tau }_0)$ and the surviving population much smaller. Altitude comparisons constitute a cosmic-ray muon time-dilation experiment: select comparable momenta or energies and account for detector acceptance, atmospheric energy loss and backgrounds before interpreting the count ratio.

For illustration, not as a claimed measurement, take $h=10\mathrm{km}$ and $v=0.998c$. Then $\gamma =15.82$, the laboratory flight time is $33.4\mu \mathrm{s}$, and the predicted surviving fraction is about $0.38$. Omitting time dilation would give only about $2.5\times 10^{-7}$ on the same idealized constant-speed model.

In the muon rest frame, its decay still follows $e^{-\Delta \tau /{\tau }_0}$. The atmosphere and ground move upward. Their separation, measured simultaneously in that frame, is shortened by length contraction:The same survival fraction follows. The terrestrial description uses a longer-lived moving particle; the muon-frame description uses a shorter atmospheric flight path. They describe the same production and detection events. This consistency illustrates both time dilation and length contraction, while the survival measurement is not two independent experimental tests merely because it can be described in two frames.

Energy-momentum invariant. For a particle with rest mass $m$,Since $E=m{c}^2+K$, substitution and cancellation of $m^2{c}^4$ giveThis relativistic energy-momentum relation reduces to $K\simeq p^2/(2m)$ when $K\ll m{c}^2$.

Laboratory beam energies. If both final muons are at rest in the laboratory, their total momentum is zero and their total energy is $2m_{\mu }{c}^2$. Four-momentum conservation therefore requires equal and opposite incoming momenta. Because the electron and Positron have equal rest masses, they then have equal total energies:Subtract each particle's own rest energy:Using the usual handbook rest energies gives approximately $105.1\mathrm{MeV}$ for each beam. These are the ideal kinematic threshold values. Just above threshold, the pair has small, equal and opposite momenta in this laboratory frame.

Positron momentum. Apply the relativistic energy-momentum relation to $E_{+}=m_{\mu }{c}^2$ and use its stipulated negative-$x$ direction:The electron has the opposite momentum, $p_{e,x}=+p_0$ with $p_0=c\sqrt{m_{\mu }^2-m_e^2}$.

Boost to the electron rest frame. The electron's velocity is $v=p_{e,x}{c}^2/E_{-}=p_0/m_{\mu }>0$. Thus $S^{\prime }$ moves in the positive-$x$ direction, and its Lorentz factor satisfiesThe positron has the same laboratory total energy as the electron, so $v{E}_{+}/c^2=v{m}_{\mu }=p_0$. The supplied Lorentz transformation then givesAs a sign check, the same boost gives $p_{e,x}^{\prime }=\gamma (p_0-v{E}_{-}/c^2)=0$, as required for the electron rest frame.

Choose downward velocity $v$ positive, and assume the collected material is initially at rest in the cloud's inertial frame. This assumption is required for the equation in the question. During $dt$, momentum conservation for the droplet plus the material it collects givesup to second-order increments. Thus the variable-mass system satisfiesSince $m\propto r^3$, $\dot{m}/m=3\dot{r}/r=3\mu /(r_0+\mu t)$, givingThe notation $dv/dt$ is appropriate because $v$ here depends only on time. The extra term accounts for the momentum needed to accelerate newly collected mass; using $m\dot{v}=mg$ would omit it. For incoming material with downward velocity $u$, the more general equation is $m\dot{v}=mg+(u-v)\dot{m}$.

Introduce the dimensionless quantities $\tau =\mu t/r_0$ and $w=\mu v/(g{r}_0)$. ThenThe Taylor series at $t=0$ givesThus the initial slope is $g$, as for free fall. At late times,The oblique asymptote is $v=gt/4+g{r}_0/(4\mu )$, approached from below. Direct differentiation givesThe speed increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has $v=gt$: its acceleration remains $g$, whereas the growing drop continually shares its momentum with newly acquired material.

Use energy units $m=m_p{c}^2$, $M=m_A{c}^2$, $e=E_{\gamma }$ and $P=pc=\sqrt{E^2-m^2}$. The soft-photon resonant capture threshold follows from four-momentum conservation:For fixed proton energy the left side is largest at head-on incidence. Its threshold is $E+P=2E_0$. Combining that with $(E-P)(E+P)=m^2$ givesEquality is allowed for a head-on photon. Here $e=k_{B}T=2.3482\times 10^{-4}\mathrm{eV}$, soA very soft photon needs an enormous proton Lorentz factor to reach the specified invariant mass. A true thermal photon distribution and a finite-width resonance would smear this monoenergetic threshold.

For $\theta =0$, $E-P=m^2/(E+P)\le m$, whereas $2E_0\gg m$, so production is impossible. For each nonzero angle, sufficiently large energy is possible, and $P\simeq E$ givesTo find the maximum deflection in soft-photon capture, note that the final particle's transverse momentum comes entirely from the photon:Writing $x=\cos \theta$, maximization gives $(1-x)(2x+1)=0$ and the interior maximum $x=-1/2$. Therefore, to the same extraordinarily accurate ultrarelativistic order,The neglected relative corrections are of order $m^2/E_0^2$ and $e/E_0$, both around $10^{-24}$. The final particle is effectively collinear with the proton despite a large photon incidence angle.

One conduction electron per atom gives number density $n=\rho N_A/M\simeq 8.50\times 10^{28}{\mathrm{m}}^{-3}$. Filling two spin states per momentum state in the three-dimensional Fermi gas gives $k_F=(3{\pi }^{2}n{)}^{1/3}$ andThe Fermi temperature greatly exceeds ordinary temperatures, explaining why these conduction electrons are degenerate rather than a classical Maxwell gas. The numerical result uses the stipulated free-electron band and bare electron mass, rather than a detailed copper band structure.

Zero-point energy is the ground-state energy above the classical minimum, remaining even at zero temperature. For a harmonic oscillator it is $\hslash \omega /2$; simultaneous vanishing position and momentum fluctuations is forbidden by the uncertainty principle.

A concrete test is the Casimir interaction: changing conducting boundaries changes electromagnetic ground-state modes and produces a measurable force. Ideal parallel plates have pressureA torsion-pendulum experiment measured the attraction of closely spaced conducting surfaces, testing this boundary-dependent interaction.^[ref] Such measurements test changes in ground-state energy, not an arbitrary absolute additive vacuum energy; real-material and thermal corrections must be included when comparing precision data.

For proper-time acceleration by reflected laser light, the spacecraft receives momentum $(hf+h{f}_r)/c=2hf/[c(1+\beta )]$ per incident photon. Combining with the impact rate givesBecause $dt/d\tau =\gamma$ and $d(\gamma \beta )/d\beta ={\gamma }^3$,Integrating from rest gives $1/(1-\beta )-1=\tau /{\tau }_0$, henceThe mirror's rest mass is assumed constant and the laser illuminates it continuously; diffraction losses, incomplete interception and microscopic recoil are neglected.

Write $b=\pi k_B^2/(3\hslash c)$, so $U=bL{T}^2$. At fixed length, $dS=C_{L}dT/T$, giving $S=2bLT$ with the entropy constant fixed by the unique photon vacuum. The thermal Helmholtz free energy and generalized pressure follow:The entropy constant can also be checked from the exact discrete photon partition function; the continuum approximation should not be extrapolated blindly below its level-spacing regime. In one dimension $P=U/L$, whereas the isotropic three-dimensional relation is $p=U/(3V)$. Both follow from the number of spatial directions carrying momentum flux.

For the ultrarelativistic magnetic bending radius, the Lorentz force supplies the transverse momentum change, so $r=p/(eB)$. At this energy the proton is ultrarelativistic, giving $pc\simeq E$ andThe electronvolt-to-joule conversion cancels the elementary-charge factor numerically. This uses external CODATA light speed and charge definitions.^[ref] If the stated beam energy means total energy, $pc=\sqrt{E^2-m_p^2{c}^4}$; if it means kinetic energy, $pc=\sqrt{T(T+2m_p{c}^2)}$. Either interpretation gives the quoted radius at the precision of the supplied numbers.

For hyperfine coupling for a spin-one nucleus, let the electron spin be $s=1/2$ and the nuclear spin $I=1$. First couple $\mathbf{j}=\mathbf{L}+\mathbf{s}$, giving $j=L\pm 1/2$, then $\mathbf{F}=\mathbf{j}+\mathbf{I}$, giving $F=|j-I|,|j-I|+1,\dots ,j+I$.

For each $F$, the projection is $m_F=-F,-F+1,\dots ,F$; electron projections similarly satisfy $m_j=-j,\dots ,j$. Summing $2F+1$ gives 18 states for $L=1$ and 30 for $L=2$, agreeing with $(2L+1)(2s+1)(2I+1)$.

Notation qualification: if the printed leading “2” denotes the principal quantum number, a hydrogenic $2d$ bound state does not exist because $L\le n-1$; a physical $d$ state needs $n\ge 3$. If a doublet term symbol ${}^{2}P$ or ${}^{2}D$ was intended, its leading 2 instead means $2s+1$, and the angular-momentum table applies without fixing $n$. The answers above use the explicitly specified $L$ values, while retaining this ambiguity.^[ref]

In the paper's convention $\mathbf{J}$ is measured in units of $\hslash$, and $g$ is dimensionless. For $m=-J,-J+1,\dots ,J$, the magnetic moment and its Zeeman energy areIf $\mathbf{J}$ instead denotes dimensional angular momentum, its coefficient must contain $1/\hslash$. This resolves the units of the printed moment formula.

The Van der Waals equation replaces the freely available volume by $V-Nb$ to represent short-range repulsion and excluded volume. The attractive mean field reduces the pressure by $a{N}^2/V^2$: particles near the boundary are pulled back into the gas, lowering their momentum transfer to the walls. Both are effective corrections, rather than an exact microscopic interaction law.

Nonrelativistic momentum conservation gives $\mathbf{u}={\mathbf{v}}_1+{\mathbf{v}}_2$, while an elastic collision conserves kinetic energy, $u^2=v_1^2+v_2^2$. Squaring the momentum equation yields ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=0$. The two nonzero outgoing velocities are perpendicular. In the degenerate case where one outgoing particle is at rest, its direction is undefined.

For relativistic collisions conserve total four-momentum, with $E=\gamma m{c}^2$, $\mathbf{p}=\gamma m\mathbf{v}$ and $E^2-p^2{c}^2=m^2{c}^4$. Rest energies must be included; for photons $E=pc$.

For the symmetric relativistic equal-mass scattering angle, the incident momentum is $mc\sqrt{\alpha (\alpha +2)}$. Energy conservation gives each final particle $E_f=(1+\alpha /2)m{c}^2$, hence $p_f=(mc/2)\sqrt{\alpha (\alpha +4)}$. Transverse momenta cancel and $2p_f\cos \theta =p_{\mathrm{in}}$, givingThe nonrelativistic limit has $\theta \to 45^{\circ }$, so the outgoing angle is $90^{\circ }$ as above.

For the symmetric two-photon annihilation angle in the specified electron-positron annihilation, each photon has energy $E_{\gamma }=(\alpha +2)m_e{c}^2/2$. Momentum conservation gives $2(E_{\gamma }/c)\cos \varphi =m_{e}c\sqrt{\alpha (\alpha +2)}$, soAs $\alpha \to 0^{+}$ the two photons become back-to-back, transverse to the limiting incident direction; as $\alpha \to \infty$ they are forward collimated. Equal photon energies follow from the specified equal angles, and are not a general property of annihilation in a moving laboratory frame.

The classical minimum is $E_{\mathrm{cl}}=0$, attained at $x=p=0$. The quantum oscillator has $E_0=\hslash \omega /2$, its zero-point energy; simultaneous zero position and momentum uncertainties are forbidden.

Let $\ell =\sqrt{\hslash /(m\omega )}$ and $\xi =x/\ell$. The first three normalized harmonic oscillator eigenfunctions areThey have respectively zero, one and two nodes, with even, odd and even parity. Their phases may be reversed without changing the state.^[ref]