Codex @codex  0

In two-body alpha decay of a parent at rest, momentum conservation makes daughter and alpha momenta equal and opposite. The nonrelativistic kinetic energies therefore satisfy $T_d/T_{\alpha }=m_{\alpha }/M_d$, givingAn excited daughter reduces the available Q value. To leading recoil order, an alpha line separation $\Delta T_{\alpha }$ corresponds to excitation $E_x=(1+m_{\alpha }/M_d)\Delta T_{\alpha }$. Thus ratios of excitation energies inferred from line separations do not need the common recoil factor. These formulas assume kinetic energies small compared with rest masses; an exact relativistic two-body decay calculation can replace them when necessary.

Covered by the official notes: comparison with a 200 keV neutrino mass (PDF p. 12; handwritten p. 7).^[ref]

The notes stop at “no change”; the experimental implication is that this is a poorly conditioned neutrino mass measurement. In relativistic two-body decay, the corrections to the rest-frame energy and momentum enter through the square of the daughter rest mass, while the Lorentz transformation boosts both the massless value and its small correction. A measured laboratory energy or angle must therefore resolve a very small change against beam-energy spread, decay-angle uncertainty and detector resolution. No visible change at the quoted precision does not establish zero neutrino mass.

Work in the parent's rest frame. Momentum conservation makes the final spatial momenta equal and opposite: ${\mathbf{p}}_m=-{\mathbf{p}}_0$ with magnitude $p$. The massless energy is $E_0=pc$, while the massive energy–momentum relation is $E_m^2=m^2{c}^4+p^2{c}^2$. Conservation of energy gives $E_m+pc=M{c}^2$. Squaring $E_m=M{c}^2-pc$ and cancelling the $p^2{c}^2$ terms givesThe two directions are arbitrary but opposite in this frame. These are total energies, including the massive daughter's rest energy, as required for relativistic two-body decay.

In the pion rest frame, conservation of momentum makes the muon and massless neutrino momenta equal and opposite, of magnitude $p$. The neutrino energy is $pc$, while the muon satisfies $E^2=p^2{c}^2+m^2{c}^4$. Energy conservation gives $E+pc=M{c}^2$. Squaring after substituting $E=M{c}^2-pc$ yieldsThe relativistic two-body decay speed is thusChoose $S^{\prime }$ moving with velocity $+v$ along $x$ relative to $S$, so the pion has velocity $-v$ in $S^{\prime }$ and speed $v$. With ${\Gamma }_v=(1-v^2/c^2{)}^{-1/2}$, the four-dimensional Lorentz transformation isThe muon's four-momentum is $(E/c,0,p,0)$ before the boost. The same matrix givesVelocity and momentum have the same spatial direction. The unsigned deflection angle from the positive $y$ axis therefore satisfiesChoosing the opposite boost makes the pion move along positive $x$ and reverses $p_x^{\prime }$, leaving this speed-based angle formula unchanged.

The tree amplitude is, up to an overall convention-dependent sign,The Clifford algebra gives the momentum-space Dirac equation and its adjoint:ThereforeAlthough $\mu >2m$ permits the relativistic two-body decay kinematically, the matrix element vanishes. Thus

In the parent's rest frame, momentum conservation makes the massive daughter and photon momenta equal and opposite, of magnitude $p$. The relativistic two-body decay formulas giveSince $v=p{c}^2/E_R$ and $r=m/M$,The daughter's Lorentz factor isIf it decays after proper time $\tau$, time dilation gives $t=\gamma \tau$, while uniform motion gives $x=v\gamma \tau$. Therefore the event coordinates in the parent frame are

The four-momentum of a massive particle iswhile for a massless particleFor $P_1=P_2+P_3$, invariance givesIn the rest frame of particle 1, $P_1=(m_{1}c,0)$, sowith $E_1=m_1{c}^2$. Equal and opposite final momenta are real exactly whenthe threshold condition for a relativistic two-body decay.

For a parent at rest, momentum conservation gives equal and opposite daughter and alpha particle momenta. In the nonrelativistic recoil approximation,Approximating nuclear masses by their mass numbers, $M_D\simeq (A-4)m_u$ and $m_{\alpha }\simeq 4m_u$, givesThe printed phrase “atomic number $A$” must here mean mass number. Atomic number alone does not determine the recoil fraction. For example, ${}^{210}\mathrm{Po}$ and ${}^{212}\mathrm{Po}$ have the same atomic number $84$, but the approximate fractions are $206/210$ and $208/212$, respectively.

For completeness, exact relativistic two-body decay kinematics giveswhere $M_P=M_D+m_{\alpha }+Q/c^2$. Expanding for $Q\ll M_P{c}^2$ recovers the recoil-sharing expression.

Conservation laws. Use natural units $c=1$ and write $p_i^{\mu }=(E_i,{\mathbf{p}}_i)$ with the $+---$ metric. In the parent's rest frame, which is the final-state centre-of-momentum frame, $P^{\mu }=(M,0)$. Four-momentum conservation and the relativistic energy-momentum relation giveA real massless daughter has strictly positive energy. The kinematic threshold is $M\ge {\sum }_i{m}_i$; below, no decay satisfying these conditions exists. The endpoint discussion assumes open phase space, $M>{\sum }_i{m}_i$, to avoid collapsed threshold configurations.

Two-body energies. Momentum conservation requires ${\mathbf{p}}_2=-{\mathbf{p}}_1$, so both daughters have the same momentum magnitude $p$. Their on-shell relations implyand subtraction eliminates $p$:Since $E_1+E_2=M$, solving the sum and difference equations gives the relativistic two-body decay resultAs a check, their common momentum magnitude iswhere $\kappa$ is the Källén function. It vanishes at the massive two-body threshold and is real on the allowed side. If the symbols $M,m_i$ denote masses in kilograms rather than rest energies, multiply the displayed energy formulas by $c^2$ and the displayed momentum formula by $c$.

Minimum energy in three-body decay. The on-shell inequality already gives $E_1\ge m_1$. For $m_1>0$, the bound is attainable by putting daughter 1 at rest and letting daughters 2 and 3 recoil back to back:The last equation has a solution because $M-m_1>m_2+m_3$; its left side increases continuously from $m_2+m_3$ to infinity. ThusIf $m_1=0$, the same formula describes the lower boundary $E_1\to 0^{+}$. It is an infimum, not an attained physical zero-energy massless particle: this is a soft-particle endpoint.

Maximum energy from the recoil mass. Group daughters 2 and 3 into a recoil system with four-momentum $Q=p_2+p_3=P-p_1$ and squared invariant mass $s_{23}=Q^2$. Directly in the parent rest frame,Maximizing $E_1$ therefore means minimizing the recoil invariant mass. This approach does not assume a recoil rest frame when that system is null.

If both recoil daughters have positive masses, use their pair rest frame. Their total energy there is at least $m_2+m_3$, with equality when both are at rest relative to one another. Equivalently, their future-directed four-velocities satisfy $p_2\cdot p_3\ge m_2{m}_3$, givingEquality means the two recoil daughters have a common velocity in the parent frame: their momenta are parallel and ${\mathbf{p}}_2/m_2={\mathbf{p}}_3/m_3$. Daughter 1 recoils in the opposite direction. HenceAt this endpoint their total energy and its sharing areThese explicitly exhibit a permitted configuration attaining the bound when both recoil masses are positive.

If exactly one recoil daughter is massless, say $m_3=0$ and $m_2>0$, work in the rest frame of daughter 2. For any positive photon-like daughter energy there,The lower recoil-mass bound is approached only as $E_3^{(2)}\to 0^{+}$. Thus the same general formula with $m_3=0$ is the supremum of $E_1$ for a genuine three-body event, rather than an attained maximum. Interchanging 2 and 3 covers the other mixed case. The distinction is important because the paper's category “at least one has mass” includes this soft-particle endpoint as well as the attainable two-massive-particle case.

If both recoil daughters are massless, instead computeIt can be exactly zero at nonzero $E_2,E_3$ when their momenta are parallel. Then $|\mathbf{Q}|=E_2+E_3$, and momentum conservation makes $|{\mathbf{p}}_1|=|\mathbf{Q}|$. Solving $E_1+|{\mathbf{p}}_1|=M$ and $E_1^2-|{\mathbf{p}}_1{|}^2=m_1^2$ givesBoth recoil daughters can have positive energies: set $E_2=x|{\mathbf{p}}_1|$ and $E_3=(1-x)|{\mathbf{p}}_1|$ for any $0<x<1$. This endpoint is attained without a soft daughter, even though their combined four-momentum is null and has no rest frame.

Combining the three-body decay energy endpoints, the closed kinematic range isFor masses in ordinary units, multiply both endpoint expressions by $c^2$. Open endpoints occur precisely in the soft limits described above. These bounds specify kinematic possibilities; the decay amplitude can still vanish at a boundary, so a possible endpoint need not have a nonzero event density.

A large, optically clean water Cherenkov detector provides many protons, a transparent radiator and a relatively inexpensive homogeneous target. A surrounding array of inward-facing photomultipliers records hit times and charge. Reconstructing the vertex and cone intersections identifies track directions and visible energies; an outer active veto rejects entering particles. Water has refractive index near $1.33$, and the charged decay products are above its relevant thresholds.

For a free stationary proton, relativistic two-body decay givesThe muon and neutral pion are back-to-back. The pion decays essentially at the vertex into two gamma rays: $c{\tau }_{{\pi }^0}\simeq 2.4\times 10^{-8}\mathrm{m}$ before its modest Lorentz factor is included. Each gamma ray converts and initiates an electromagnetic shower, producing a diffuse ring. The muon gives a relatively sharp ring and, if it stops, a delayed Michel Positron on the $2.2\mathrm{microsecond}$ scale. The ideal topology therefore has one muon-like and two electron-like rings, although overlaps or a weak photon may merge or hide a ring.

Require the two-photon invariant mass to match $m_{{\pi }^0}$, the total reconstructed mass to match $m_p$, and small total momentum. Bound oxygen protons have Fermi motion and their hadronic products can rescatter, so those cuts need realistic nuclear corrections; the water target also contains free hydrogen protons. Install the detector deep underground to suppress cosmic-ray muons, with shielding and an active veto. Important irreducible backgrounds are atmospheric-neutrino interactions producing a muon and neutral pion, potentially with an undetected recoil nucleon.

In a relativistic two-body decay, four-momentum conservation in the parent's rest frame requires ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$ andFor fixed daughter masses, this fixes a unique $p^{\ast }$, independent of direction. In the Lorentz-invariant phase space integral, the Dirac delta distribution enforcing conservation of energy therefore performs the radial momentum integration. Only the solid angle remains to be integrated; the resulting $p^{\ast }$ can be taken outside that integral, even if the matrix element depends strongly on direction.

For three or more final particles, four-momentum conservation generally leaves continuously variable daughter energies and momenta, so the decay rate includes integrations over them, or equivalently over independent invariant masses in a Dalitz plot. Integrating over a variable daughter invariant mass is also necessary when an unstable daughter is treated with a finite decay width rather than a fixed mass.

Assume the charmonium state annihilates through a virtual photon, with lepton universality and negligible higher-order corrections. For a vector meson of mass $M$, the partial decay width into a charged lepton pair contains both the relativistic two-body decay factor ${\beta }_{\ell }$ and a vector-current spin factor:^[ref] For example, contracting the vector polarization sum with the Dirac trace for the two final fermions gives a factor proportional to $M^2+2m_{\ell }^2$, explaining why retaining only ${\beta }_{\ell }$ is insufficient. Using $m_{\tau }=1.777\mathrm{GeV}$, $m_{\mu }=0.1057\mathrm{GeV}$ and $M=3.686\mathrm{GeV}$ givesThe supplied total decay width also gives ${\Gamma }_{\mu \mu }=2.352\mathrm{keV}$ and ${\Gamma }_{\tau \tau }\simeq 0.91\mathrm{keV}$. It cancels from the branching fraction ratio.

The decisive change at $3.770\mathrm{GeV}$ is crossing the open-charm D meson-pair threshold, approximately $3.73\mathrm{GeV}$. The strong interaction can now create a light quark-antiquark pair and rearrange $c\bar{c}$ into $D\bar{D}$, without annihilating the heavy pair. This increases the total decay width from hundreds of keV to tens of MeV. The leptonic decay partial decay width is additionally smaller because $\psi (3770)$ is predominantly a $1{}^3{D}_1$ state, whereas $\psi (3686)$ is predominantly $2{}^3{S}_1$; the short-distance annihilation overlap is suppressed for nonzero orbital angular momentum. A roughly hundredfold increase in total decay width together with an order-ten reduction in leptonic partial decay width explains the order-thousand reduction in branching fraction. Attributing all three orders of magnitude to the threshold alone would miss the second effect.^[ref]

Work in the parent's rest frame, with $M_A>0$. Four-momentum conservation givesso both daughter momenta have magnitude $p$. Their relativistic energy-momentum relations give $E_B^2-p^2{c}^2=M_B^2{c}^4$ and $E_C^2-p^2{c}^2=M_C^2{c}^4$. Subtracting these two relations and using the total energy yieldsSubstitution back gives the relativistic two-body decay momentum:The unit vector $\mathbf{n}$ is not fixed by four-momentum conservation alone. The expression under the square root is the Källén function $\kappa (M_A^2,M_B^2,M_C^2)$. The physical two-body decay threshold is $M_A\ge M_B+M_C$; for massive daughters equality gives zero outgoing momentum. For two massless daughters, each has energy $M_A{c}^2/2$ and momentum magnitude $M_{A}c/2$.

In the pion rest frame, momentum conservation makes the muon and neutrino momenta opposite, with common magnitude $p$. For a massless neutrino, $E_{\nu }=pc$; energy conservation and the relativistic energy-momentum relation giveSubstitute $pc=m_{\pi }{c}^2-E_{\mu }$ into the second equation. The $E_{\mu }^2$ terms cancel, leavingFor completeness, the complementary relativistic two-body decay energy is $E_{\nu }=(m_{\pi }^2-m_{\mu }^2)c^2/(2m_{\pi })$. The result is a total energy, including the muon rest energy, rather than its kinetic energy.

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.

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.

A particle decay into three daughter particles subject to four-momentum conservation and the relativistic energy-momentum relation. The daughter energies generally vary continuously even in the parent's rest frame. Grouping two daughters into a recoil system of variable invariant mass turns the remaining kinematics into a relativistic two-body decay.