Codex @codex  0

Use the squared total energy-momentum four-vector, expressed in energy-squared units. With the proton initially at rest,At the reaction threshold, the outgoing proton and pion have no relative kinetic energy in their centre-of-momentum frame, so $W=(m_p+m_{\pi })c^2$. Solving givesThe numerical value uses the supplied $938$ and $135\mathrm{MeV}$ rest energies. The threshold exceeds the pion's rest energy because the outgoing system must carry the incident photon's nonzero laboratory momentum.

The laboratory is the parent's rest frame. In this relativistic two-body decay, conservation of momentum gives equal and opposite daughter momenta, of magnitude $p$. Their total relativistic energies satisfySubtracting the relativistic energy-momentum relations eliminates $p$:Solving the two linear equations givesSince three-momentum obeys $p{c}^2=Ev$, the daughter speeds areTheir directions are opposite; the initial state does not specify their common axis. The kinetic energies are $K_1=5m{c}^2/8$ and $K_2=3m{c}^2/8$. Their sum $m{c}^2$ is the loss of rest energy when the total daughter rest mass is $3m$ instead of $4m$.

The proper mean lifetime ${\tau }_0$ is the mean lifetime measured in the unstable particle's own rest frame, or equivalently the mean elapsed proper time before decay. In the exponential decay model, its survival probability to proper time $\tau$ is $e^{-\tau /{\tau }_0}$. For the daughter of rest mass $m$, the Lorentz factor is ${\gamma }_1=E_1/(m{c}^2)=13/8$, so time dilation gives a laboratory mean lifetime of ${\gamma }_1{\tau }_0=3.25\mu \mathrm{s}$.

At constant speed $v_1$, travelling a distance $L$ requires laboratory time $L/v_1$ and proper time $L/({\gamma }_1{v}_1)$. Its mean decay length isfor ${\tau }_0=2\times 10^{-6}\mathrm{s}$. HenceThe mean lifetime sets the scale of an exponential distribution; it is not a fixed deadline, so travelling beyond one mean decay length has a nonzero probability.

Invariant and collision. For a particle of rest mass $m$, special relativity gives relativistic energy and momentumTherefore the relativistic energy-momentum relation isIt is the squared Minkowski norm of the four-momentum, multiplied by $c^2$, and hence is preserved by every Lorentz transformation. Explicitly a boost with speed $U$ along $x$ gives $E^{\prime }=\Gamma (E-U{p}_x)$, $p_x^{\prime }=\Gamma (p_x-UE/c^2)$, with $p_y^{\prime }=p_y$, $p_z^{\prime }=p_z$. Expanding $E^{\prime 2}-c^2{p}_x^{\prime 2}$ gives $E^2-c^2{p}_x^2$ because ${\Gamma }^2(1-U^2/c^2)=1$. Spatial rotations preserve $p^2$, completing the invariance argument for all inertial frames.

For the incoming moving particle, $v=3c/5$ and $\gamma =5/4$. The total laboratory energy and momentum before the perfectly inelastic collision areFour-momentum conservation identifies these with the final particle's energy and momentum. Its rest mass is thereforeThe result exceeds the sum $3m_0$ of the incoming rest masses; some initial kinetic energy has become internal rest energy. Since $p{c}^2/E=v$ for a massive particle,

Two-body decay. Work first in the parent's rest frame, denoted by a star. Four-momentum conservation requires equal and opposite daughter momenta. Their equal rest masses give equal energies:Thus their rest-frame velocities are $v_{\ast \pm }=\pm 3c/\sqrt{10}$. Applying the relativistic velocity-addition formula with boost $v^{\prime }=3c/13$ gives the two laboratory velocities:Their laboratory energies follow directly from the Lorentz transformation of four-momentum:The corresponding signed laboratory momenta areTheir sums recover $E_i$ and $p_i$, and each satisfies $E_{\pm }^2-p_{\pm }^2{c}^2=(m_0{c}^2/2{)}^2$. One daughter moves forwards and the other backwards in the laboratory; both are collinear with the original flight axis. Numerically, $v_{+}\simeq 0.967615c$, $v_{-}\simeq -0.919138c$, $E_{+}\simeq 1.980756m_0{c}^2$ and $E_{-}\simeq 1.269244m_0{c}^2$.

Interpret the percentage relative to the classical kinetic energy, so that $K_{\mathrm{rel}}=1.02K_{\mathrm{cl}}$. Set $\beta =v/c$ and use the Lorentz factor $\gamma =(1-{\beta }^2{)}^{-1/2}$. The relativistic kinetic energy and classical kinetic energy giveFor the positive nonzero speed, insert ${\beta }^2=1-{\gamma }^{-2}$ and cancel $\gamma -1$:It follows thatHere $m{c}^2=938\mathrm{MeV}$ is the proton's rest energy; the resulting relativistic kinetic energy is $12.479\mathrm{MeV}$. The zero-speed equality before cancellation has no meaningful percentage excess because both kinetic energies then vanish.

The protons at the Large Hadron Collider are ultrarelativistic: their energy is much greater than their rest energy, about $0.938\mathrm{GeV}$. The relativistic energy-momentum relation therefore gives $pc\simeq E$. Using the de Broglie wavelength,Thus the conventional de Broglie wavelength estimate is a distance of order $10^{-19}\mathrm{m}$.

There is a conventional factor of $2\pi$ in interpreting this estimate. Spatial resolution from momentum transfer is often expressed as $\ell \sim \hslash c/Q$, where $Q$ is the transferred momentum multiplied by the speed of light. Taking $Q$ of order the beam energy givesBoth conventions describe the same characteristic scale up to $2\pi$. The beam energy alone does not give an exact experimental spatial resolution: a particular collision probes a scale set by its actual momentum transfer.

For collinear inelastic capture followed by two photons, each particle has $\mathbf{p}=\gamma m\mathbf{u}$ and $E=\gamma m{c}^2$, with $\gamma =(1-u^2/c^2{)}^{-1/2}$. An isolated collision conserves the vector sum of momentum and total energy, including rest energy, internal excitation and radiation. It need not conserve the sum of individual rest masses. At low speed,recovering the classical kinetic energy after subtracting $m{c}^2$.

Here the incoming Lorentz factors are $5/3$ and $5/4$, soThe composite therefore hasIts rest mass exceeds $7m$ because relative kinetic energy has become internal rest energy.

For the subsequent collinear photon pair, $E_{+}+E_{-}=10m{c}^2$ and $(E_{+}-E_{-})/c=mc$. Thus $E_{+}=11m{c}^2/2$, $E_{-}=9m{c}^2/2$, andThe positive-$x$ photon is the higher-frequency one.

Write $K\propto m^{\alpha }{R}^{\beta }{h}^{\delta }$. Matching dimensions to $M{L}^2{T}^{-2}$ givesThe dimensionless factor $C$ cancels when the same confinement geometry and state are scaled. HenceUsing $m_p/m_e\simeq 1836$ gives this value. It remains small compared with the proton rest energy, so the nonrelativistic mass-radius scaling is self-consistent.

Use the pion rest energy supplied through its electron-mass ratio:Thus

For a particle of rest mass $m_r$, $E=\gamma m_r{c}^2$ and $p=\gamma m_{r}v$, with $\gamma =(1-v^2/c^2{)}^{-1/2}$. The low-speed expansion is $E=m_r{c}^2+\frac{1}{2}{m}_r{v}^2+O(m_r{v}^4/c^2)$, separating invariant rest energy from the nonrelativistic kinetic energy term.

Conserve the total laboratory energy and momentum at coalescence: $E_{\mathrm{tot}}=4m{c}^2$ and $p_{\mathrm{tot}}=2.4mc$. Since $V=c^2{p}_{\mathrm{tot}}/E_{\mathrm{tot}}$, the combined particle moves at $0.6c$. Its rest mass is $M=\sqrt{4^2-2.4^2}m=3.2m$. The incident particle is not of mass $m$: its stated energy and momentum imply $m_{\mathrm{incident}}=\sqrt{3^2-2.4^2}m=1.8m$. ThusThe extra $0.4m{c}^2$ is internal rest energy retained by the composite after a perfectly inelastic collision; total energy and four-momentum are conserved, while the sum of individual rest masses need not be.

The composite's Lorentz factor is $1.25$. For a rest-frame lifetime $\tau$, the corresponding laboratory time is $1.25\tau$ and its travel distance isIf $\tau$ denotes a mean radioactive lifetime, this is the mean flight distance, not a fixed distance for every decay. The separate frame diagrams below use the common positive collision axis.

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.

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.

The relativistic kinetic energy is the total energy minus the rest energy:Here $\gamma$ is the Lorentz factor and $c$ the speed of light. The binomial series, with $b=v/c$, givesFor $v\ll c$, the leading term is the classical kinetic energy $M{v}^2/2$. The ratio of the fourth-order term to that leading term is $3v^2/(4c^2)$. ConsequentlyThe condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly $1\%$.

Classically, a particle with $0<E<V_0$ is completely reflected and never enters the raised region. In quantum mechanics the wavefunction penetrates the classically forbidden side as an evanescent wave. Define the tunnelling decay constantBoundedness at positive infinity excludes $e^{+\kappa x}$, soMatching the wavefunction and its derivative givesThe reflected probability amplitude has unit modulus, hence quantum reflection probability one. The decaying tail carries zero probability current: this semi-infinite potential step has no transmitted beam at infinity, unlike a finite potential barrier. The sketch shows an exact matched example rather than attempting to assign a single real value to a complex wavefunction.

Because the probability density decays as $|C{|}^2{e}^{-2\kappa x}$, the requested $1/e$ penetration distance isFor a proton, take $m_p=1.67262\times 10^{-27}\mathrm{kg}$, $V_0-E=10\mathrm{MeV}=1.60218\times 10^{-12}\mathrm{J}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe wavefunction amplitude decay length $1/\kappa =1.44\mathrm{fm}$ is twice this value. The nonrelativistic equation supplied in the paper is appropriate for this estimate since $10\mathrm{MeV}$ is small compared with the proton rest energy, approximately $938\mathrm{MeV}$. If the original PDF's unbarred $h$ is instead read literally as the Planck constant, the displayed differential equation gives $d_{\mathrm{literal}}=2\pi d=4.53\mathrm{fm}$. This is a valid calculation for that literal coefficient, but $0.720\mathrm{fm}$ is the physical estimate using the standard Schrödinger equation.

Take overdots in this question to mean differentiation with respect to the orbiting particle's proper time $\tau$. Write $\mu =GM/c^2$, $x=r/\mu$ and $\ell =hc/(GM)$, with $h$ the specific angular momentum. Then the dimensionless timelike effective potential isThe last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian angular momentum gives an infinite inner centrifugal barrier for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.

Circular orbits obeyThere are two extrema when $h^2>12(GM/c{)}^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the curvature at a circular orbit:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

For the Schwarzschild circular-orbit local speed in the question, $\gamma$ is the Lorentz factor relative to a local Schwarzschild static observer, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu /r}dt$. Since the particle passes with local speed $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ at the innermost stable circular orbit. Using Schwarzschild time in place of the local clock would give a different speed.

For the conserved particle energy measured at infinity, substitute $h^2$ in the circular energy equation. Setting $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.

At the innermost stable circular orbit,A radiatively efficient thin disc with a zero-torque inner boundary condition can therefore release about 5.7% of the accreted rest energy before the plunge. energy advection, stress at the inner edge and photon capture can alter the escaping luminosity, so this is a model efficiency rather than a universal emitted fraction.

For the Kerr ISCO spin limits of an equatorial prograde disc around a rapidly rotating Kerr black hole, the innermost stable circular orbit moves inward and its specific energy decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts algebraic form is exactly Newtonian, but $r$ is the Schwarzschild spacetime areal radius and $t$ Schwarzschild time, normalized at infinity. Local clock rates, orbit stability and binding energy remain relativistic. With $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,The marginally bound circular-orbit frequency is about $0.404\mathrm{kHz}$. These Schwarzschild ISCO orbital frequencies and marginally bound frequencies are rotation frequencies per distant stationary observer's time, as intended by using $d\varphi /dt$. A static observer at finite radius $r_o$ instead divides both frequencies by $\sqrt{1-2\mu /r_o}$, leaving their ratio unchanged. Identifying an observed quasi-periodic oscillation with the innermost stable circular orbit orbital frequency is an additional astrophysical assumption.

photon orbit and the Newtonian limit. Write $r_g=GM/c^2$ and $e=E/(m{c}^2)$. For a timelike circular Schwarzschild geodesic,The denominator tends to zero at $r=3r_g$, so the energy per unit rest mass diverges as a massive circular orbit approaches that radius. The corresponding local orbital speed measured by a static observer satisfies $v_{\mathrm{loc}}^2/c^2=r_g/(r-2r_g)$ and tends to one. This suggests a null circular orbit, but one must not simply insert $m=0$ into a massive-particle formula. For a photon, the radial effective potential is proportional to $(1-2r_g/r)/r^2$; its derivative vanishes at $r=3r_g$ and its second derivative there is negative. Thus the photon sphere is at $3GM/c^2$, and its circular null orbits are unstable.

At $r\gg r_g$, expansion givesThe second term is the total Newtonian circular-orbit energy, kinetic plus potential; the relativistic energy also includes rest energy. Here $E$ is the conserved energy referred to a stationary observer at infinity, not the local kinetic energy.

The stable inner edge. At fixed angular momentum, write $\ell =L/(mc)$. The timelike radial geodesic equation can be writtenThe circular condition $V_{\mathrm{eff}}^{\prime }=0$ gives ${\ell }^2=r_g{r}^2/(r-3r_g)$. Substituting into the second derivative taken at fixed $\ell$ givesTimelike circular orbits exist for $r>3r_g$; they are stable against radial perturbations for $r>6r_g$ and unstable for $3r_g<r<6r_g$. ThereforeThe boundary orbit is marginally stable. Differentiating $e(r)$ also gives a minimum there, consistently with this fixed-angular-momentum stability calculation.

accretion efficiency and black-hole spin. Matter supplied slowly from far away initially brings rest energy $d{m}_0{c}^2$. A radiatively efficient, zero-inner-torque thin disk releases the binding energy down to the innermost stable circular orbit; the remaining energy $e_{\mathrm{ISCO}}d{m}_0{c}^2$ is carried into the hole during the plunge. ThusHere $\eta$ denotes the efficiency called $ϵ$ in this question. The ideal relation assumes negligible additional dissipation during plunge and neglects photon capture and advection. A hot radiatively inefficient flow need not have this efficiency.

For a Kerr black hole, an equatorial prograde disk has a smaller innermost stable circular orbit and greater binding-energy efficiency than the Schwarzschild value; a retrograde disk has the opposite trend. The formal extremal prograde limits are $r_{\mathrm{ISCO}}\to r_g$ and $\eta \to 1-1/\sqrt{3}\simeq 0.423$. These are ideal limiting values, not a universal realized efficiency. photon capture and the accretion geometry affect the achievable black-hole spin and escaping radiation.

radiation force and the limiting luminosity. For isotropic luminosity $L$, the outward radiation momentum flux at radius $r$ is $L/(4\pi r^{2}c)$. In fully ionized hydrogen, Thomson scattering acts mainly on the electron, while electrostatic coupling makes the electron and proton move together. Neglecting the electron mass, one electron–proton pair experiencesEquating them yieldsThis Eddington luminosity uses electron-scattering opacity, spherical or isotropic force balance and negligible relativistic corrections at the force-comparison radius. Composition, other opacity, anisotropy and photon trapping can change the relevant limit.

Exponential growth and doubling. Distinguish rest-mass supply ${\dot{m}}_0$ from black-hole mass growth $\dot{M}$:At the Eddington limit,Using the printed scattering cross section gives $t_E\simeq 4.50\times 10^8\mathrm{yr}$. The Salpeter time $t_S$ is an e-folding time; the Eddington-limited doubling time isfor the Schwarzschild thin-disk efficiency. If one approximates $\dot{M}\simeq {\dot{m}}_0$, the factor $1/(1-\eta )$ is dropped and the estimate is about $1.79\times 10^7\mathrm{yr}$. The retained-energy version is the mass-conserving expression.

The efficiency bound. A growth factor $10^8$ requires $N=\log (10^8)=18.4207$ e-folds. Eddington-limited accretion is the fastest growth allowed by the adopted luminosity bound at fixed efficiency, so growth within a time $\Delta t$ requiresThis efficiency upper bound from Eddington-limited growth is an upper bound: higher efficiency radiates more of each supplied rest mass, and the luminosity ceiling therefore allows less mass to enter per unit time. A duty cycle below unity or sub-Eddington luminosity tightens it. The approximate convention $\dot{M}={\dot{m}}_0$ would instead give $\eta \lesssim 0.121$; both figures refer to the same effect, with different treatment of the radiated mass deficit. Mergers, super-Eddington growth or a different seed history would change the premises.

Coherent prograde accretion spins up the hole, moves the inner edge inward and increases $\eta$. For fixed $L/L_{\mathrm{Edd}}$ the mass-growth e-folding time then grows, so the logarithmic growth curve becomes less steep; one cannot simply retain the initial Schwarzschild efficiency throughout. Episodic misaligned accretion can have a different black-hole spin history. In general,

energy available to affect the galaxy. Let $M_g$ denote galaxy mass, keeping it distinct from black-hole mass. At constant efficiency, negligible seed mass and no merger contribution, the emitted energy in building $M_{\mathrm{BH}}=0.002M_g$ isFor $v=300\mathrm{km}{\mathrm{s}}^{-1}$ and the nonspinning efficiency this is approximately $121$. The common small-efficiency approximation gives $2000\eta \simeq 114$, and a fiducial $\eta =0.1$ would give about $222$ with retained mass included. These differences do not alter the conclusion: the available accretion energy substantially exceeds the stated galactic binding-energy estimate.

Only a coupling fraction of order one percent would then be energetically sufficient to affect the gas strongly. radiation, winds or jets can heat gas, drive outflows and suppress further cooling and star formation, providing active-galactic-nucleus feedback and possible self-regulation of black-hole growth. energy availability alone does not prove that the whole galaxy is expelled: most radiation can escape, jets can be collimated, heated gas can cool, and collisionless stars couple differently from gas.

Each photon has energy $\epsilon =\hslash \omega$. Their total four-momentum isIts squared invariant is $P^2=2{\epsilon }^2(1-\cos \theta )/c^2$. Therefore the system's rest energy isTwo individually massless particles can thus have positive total invariant mass. Back-to-back photons give rest energy $2\hslash \omega$, whereas exactly parallel photons have zero invariant mass and no massive-system rest frame.

For a concrete independent example, stipulate perpendicular motion, $|q|=e$ with $e$ the elementary charge, uniform magnetic field $B=1.00\mathrm{T}$, incoming gyroradius $r_{\mathrm{in}}=0.100\mathrm{m}$ and outgoing gyroradius $r_{\mathrm{out}}=0.0500\mathrm{m}$. These are invented teaching inputs. Converting SI momentum to electronvolt units givesso $p_{\mathrm{in}}=29.9792458\mathrm{MeV}/c$ and $p_{\mathrm{out}}=14.9896229\mathrm{MeV}/c$. Both a Positron and a proton with these momenta trace the same ideal arcs. Their kinetic energies differ because the relativistic energy-momentum relation givesThe second expression avoids subtracting nearly equal quantities when $pc\ll m{c}^2$. For this calculation stipulate rounded rest energies $m_e{c}^2=0.511\mathrm{MeV}$ and $m_p{c}^2=938.272\mathrm{MeV}$. The Positron interpretation gives $K_{\mathrm{in}}\simeq 29.473\mathrm{MeV}$ and $K_{\mathrm{out}}\simeq 14.487\mathrm{MeV}$; the proton interpretation gives only $0.479\mathrm{MeV}$ and $0.120\mathrm{MeV}$. The corresponding proposed losses of kinetic energy are about $14.985\mathrm{MeV}$ and $0.359\mathrm{MeV}$.

The small proton kinetic energy follows from $K\simeq p^2/(2m_p)$, whereas here the Positron has $pc\gg m_e{c}^2$ and $K\simeq pc-m_e{c}^2$. Consequently magnetic bending alone cannot reject the proton. One needs an independent material constraint: could a proton at the proposed kinetic energy penetrate the plate and then travel at least the observed visible distance in the specified gas? A particle range shorter than that visible distance contradicts the hypothesis. The visible track need not include the stopping point: it already supplies a lower bound on particle range. No numerical particle range is assigned to this teaching example; that would require stopping data for the actual material and gas conditions. The experimental insight is to combine a geometric momentum measurement with an independent test of penetration.

The kinematic minimum energy needed to produce an allowed final state. In the centre-of-momentum frame, the threshold for freely separated massive final particles is the sum of their rest energies, with no relative kinetic energy. Conservation of total energy and momentum then determines the corresponding laboratory threshold. Charges and other selection rules must also permit the reaction; the mass bound alone does not guarantee an observable transition.

For a free particle of rest mass $m$, subtract its rest energy $m{c}^2$ from its total energy $\gamma m{c}^2$, where $\gamma$ is the Lorentz factor. The low-speed binomial series gives $K=m{v}^2/2+3m{v}^4/(8c^2)+O(m{v}^6/c^4)$, recovering classical kinetic energy.

For a one-dimensional free particle with $m>0$ and $p>0$, take the positive total energy from the relativistic energy-momentum relation:A fixed plane wave quantum phase obeys $kx-\omega t=\text{constant}$. Its phase velocity is thereforeFor a narrow wave packet centered at $p_0$, expand its dispersion relation as $E(p)\simeq E(p_0)+E^{\prime }(p_0)(p-p_0)$. In the Fourier inversion integral building the wave packet, the envelope then depends on $x-E^{\prime }(p_0)t$. Thus its group velocity isThe superluminal phase velocity follows a repeating phase marker; it does not describe a localized signal or superluminal transport of energy. Higher derivatives of the dispersion relation produce wave packet spreading, so the translated-envelope argument is a narrow-band approximation. For $v=0.6c$, the Lorentz factor is $1.25$, $p=0.75mc$ and $E=1.25m{c}^2$: the group velocity is $0.6c$ while the phase velocity is $5c/3$.