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Ignoring gravitational focusing, a galaxy sweeps a cylinder of volume $\pi R^2{v}_{\mathrm{rel}}T$. Multiplying this by the number density $N$ of target galaxies and averaging the relative speed gives the mean encounter countFor independent encounters modeled by a Poisson process, the probability of at least one galaxy merger is $1-e^{-\mu }$. The printed expression $P=\mu$ is its rare-encounter approximation $\mu \ll 1$; it is not an exact probability for arbitrary $T$.

For an illustrative present-day field population take an effective merger impact parameter $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N=10^{-2}{\mathrm{Mpc}}^{-3}$, and $⟨v_{\mathrm{rel}}⟩=200\mathrm{km}{\mathrm{s}}^{-1}$. These are assumed order-of-magnitude inputs, not precise observational measurements. Taking a Hubble time of $T\simeq 1.4\times 10^{10}\mathrm{yr}\simeq 4.4\times 10^{17}\mathrm{s}$, the given distance conversion yieldsand henceThus the geometric field estimate is of order $10^{-5}$ to $10^{-4}$ per Hubble time. It scales as $R^{2}N⟨v_{\mathrm{rel}}⟩T$ and is very sensitive to environment and the adopted effective merger radius. Enhanced density in groups, gravitational focusing, and the evolution of the galaxy population are all omitted; this number is not a prediction of the full cosmological merger fraction.

Now use the stipulated rapid, distant encounter. The consistent rectilinear trajectory isWith the stated impact parameter along $x$ and velocity along $z$, this trajectory lies in the $xz$-plane. The original PDF's reference to the $xy$-plane is a typo; the local TeX also corrupts the impact-parameter direction. The original PDF fixes that direction as $x$.

Use the positive potential $G{m}_p/|\mathbf{R}-\mathbf{r}|$, whose gradient is the attractive acceleration under the question's sign convention. Its multipole expansion isThe first term has no force; the linear term accelerates the entire galaxy and disappears in the frame following its centre of mass. The leading internal tidal potential is thereforeThis is a quadrupole approximation valid for $r\ll p$, rather than an exact equality for every $r<p$. Its acceleration tidal tensor acts on $\mathbf{r}$ asIn the impulse approximation, each stellar position is held fixed during the flyby and the velocity kick is the integral of this acceleration. Put $s=vt$; terms odd in $s$ integrate to zero, whileThe integrated diagonal coefficients of the tidal tensor are consequently $(2,-2,0)/(v{p}^2)$, givingThe encounter stretches the galaxy along the impact parameter, compresses it along $y$, and gives no net leading kick along the flyby direction.

Denote the pre-encounter stellar velocity by $\mathbf{u}$, to distinguish it from the relative flyby speed $v$. The instantaneous change of specific kinetic energy isUncorrelated kicks with zero mean cross term imply $⟨\mathbf{u}\cdot \Delta \mathbf{v}⟩=0$, so the phase-averaged heating at a specified position isFor an individual star the cross term need not vanish: the formula is an ensemble or orbital-phase average. Integrating over a spherical galaxy of mass $m_g$, symmetry gives $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. Therefore the total tidal heating isFor two identical galaxies, each receives this heating with $m_p=m_g$. Adding both contributions gives

In the centre of mass frame, equal masses approach with speeds $v/2$ if $v$ is their relative speed at infinity. The initial orbital energy iswhere $m_g/2$ is the reduced mass. The orbital potential energy vanishes at infinite separation. Within the weak-deflection approximation this asymptotic speed is also the nearly constant speed used in the flyby calculation.

Conservation of energy transfers the positive internal heating out of the relative orbit. Tidal capture of galaxies occurs in this model if $E_{\mathrm{orb}}-\Delta E_{\mathrm{int}}<0$, which givesThe equality is marginal capture. A bound pair still needs subsequent evolution to coalesce, so the result is an approximate capture criterion used here as a merger criterion. Its right-hand side has dimensions of length times speed, as required.

The impulse approximation requires the flyby duration $\tau \sim p/v$ to be short compared with the stellar dynamical time ${\Omega }^{-1}$, namely $v/p\gg \Omega$. If $v/p<\Omega$, a star moves substantially during the encounter. Its orbital adiabatic invariance of an orbital action suppresses net heating by a slowly varying tidal field: adiabatic shielding of tidal encounters replaces the frozen-position calculation. The divergent $v^{-2}$ heating predicted by extrapolating the impulsive formula to slow encounters is therefore spurious. Close passages with $p$ comparable to the galaxy size or strong gravitational focusing also lie outside the derivation.

The center of mass isBy Newton's third law, ${\mathbf{F}}_{21}=-{\mathbf{F}}_{12}$, soThus $\dot{\mathbf{R}}$ is constant. For $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$,Thereforewhere $\mu$ is the reduced mass.

A constant-width Breit-Wigner resonance cross section is the simplest nonrelativistic curve; a P-wave scattering width and relativistic kinematics explain important expected distortions. A comparison to specific observed deviations cannot be made without the missing plot.

Write $E$ for the centre-of-mass frame kinetic energy, $\mu =m_p{m}_{\pi }/(m_p+m_{\pi })$ for the reduced mass, and $k$ for the relative wave number. The nonrelativistic relations areFor an isolated particle resonance of spin $J$, averaging over the initial proton spin and using the zero pion spin givesFor the nearly elastic ${\Delta }^{++}$ state, $J=3/2$ and ${\Gamma }_{\pi p}\simeq \Gamma$, soFreezing $k$ at $k_R$ gives a symmetric Lorentzian centered at $E_R$, with full width at half maximum $\Gamma$ and height $8\pi /k_R^2$. This is the local curve to sketch. If the displayed variable is the beam kinetic energy $T$, its width is $\Delta T\simeq (m_p+m_{\pi })\Gamma /m_p$ within this nonrelativistic approximation. Keeping $k(E)$ changes the prefactor and hence the symmetry. Extending a constant-width approximation all the way to $E=0$ produces a spurious divergence, so it is not a valid threshold model.

The pion has intrinsic parity $-1$ and the proton has $+1$. The positive parity and $J=3/2$ of the Delta baryon require odd orbital angular momentum, with the lowest choice $l=1$. The finite-range partial-wave threshold law gives ${\Gamma }_{\pi p}(k)\propto k^{2l+1}=k^3$ near threshold. Inserting this in the elastic numerator makes ${\sigma }_{\mathrm{el}}\propto k^4$ away from the pole as $k\to 0$, rather than diverging. Equivalently the elastic scattering phase shift has ${\delta }_1=O(k^3)$, and ${\sin }^2{\delta }_1/k^2=O(k^4)$. Away from threshold, form factors and other channels modify the simple cubic width.

At the illustrative peak of part (d), the incident pion kinetic energy is comparable to its rest energy, so a nonrelativistic pion treatment is already inaccurate. The exact relative momentum iswith $\sqrt{s}=\sqrt{(m_p+m_{\pi }{)}^2+2m_{p}T}$. Relative to the symmetric local estimate, this changes the horizontal conversion, peak normalization and tails. The illustrative $1232\mathrm{MeV}$ peak has $p_{\ast }\simeq 227\mathrm{MeV}/c$, giving $8\pi (\hslash c{)}^2/(p_{\ast }c{)}^2\simeq 190\mathrm{mb}$ before small inelastic effects.

Finally, nonresonant partial waves can interfere with the resonant scattering amplitude, higher particle resonances and inelastic scattering channels can enter, and instrumental resolution can broaden a measured peak. These are physically distinct possible deviations; the missing data do not justify claiming any particular one was actually observed. Both the missing graphical overlay and its data-specific explanation remain source gaps.

Model alpha decay as a preformed alpha particle making repeated escape attempts from a short-range nuclear potential well, then undergoing quantum tunnelling through the external Coulomb potential. Write the daughter atomic number as $Z_D$, the nuclear radius as $R$, and the relative reduced mass as $\mu$. For zero orbital angular momentum, the WKB approximation givesSet $v=\sqrt{2Q/\mu }$ and $r=r_2{\sin }^2\theta$. ThenWhen $R/r_2\ll 1$, the leading term isHere $P_{\alpha }$ is the alpha preformation probability and ${\nu }_{\mathrm{attempt}}$ the attempt frequency. The stated proportionality retains only the dominant barrier exponent. It assumes a thick classically forbidden barrier, nonrelativistic relative motion, a spherical unscreened external Coulomb potential, and slowly varying prefactors. Finite $R$, changes in alpha preformation probability, and a centrifugal barrier for nonzero orbital angular momentum produce significant corrections. The precise charge in the barrier is $Z_D=Z_P-2$; using the parent charge is an approximation for large $Z_P$. The measured alpha particle speed approximates $v$ only when daughter recoil is negligible.

The gamma width is $0.10\mathrm{eV}=10^{-7}\mathrm{MeV}$, so the remaining proton width is ${\Gamma }_p=0.20-0.15-10^{-7}=0.0499999\mathrm{MeV}$. At the peak of the same resonance the total-width denominator and gamma exit width cancel. alpha particles and carbon-12 have spin zero; the proton and the stated nitrogen level each have spin $1/2$. Hence $g_{\alpha }/g_p=4$, independently of the unspecified resonance spin. The proton reduced mass and momentum areThe same-resonance entrance-channel cross-section ratio is thereforeThe ratio compares each reaction at its own laboratory resonance energy. Factors of ${\hslash }^2$ restoring the cross section's area units cancel. Only the supplied nitrogen spin magnitude is used: its printed positive parity is immaterial to this spin average, so no orbital-parity inference is made from that assignment.

An alpha emitter has an alpha-like cluster coupled to a daughter nucleus. Its positive decay energy is $Q=[M_P-M_D-M_{\alpha }]c^2$; atomic masses may be used consistently because the electron counts balance. The daughter and alpha share this energy according to momentum conservation. The attractive nuclear interior and repulsive exterior Coulomb potential produce a quasibound state with a barrier, so positive $Q$ does not imply rapid classical escape.

For daughter charge $Z_{d}e$, the exterior barrier is approximately $C/r$, $C=2Z_d{e}^2/(4\pi {ϵ}_0)$, with outer turning radius $b=C/Q$. For zero orbital angular momentum, the leading WKB probability isHere $R$ is a nuclear matching radius, $\mu$ the reduced mass, $\nu$ an interior assault frequency and $P_{\mathrm{pre}}$ a preformation factor. The barrier factor alone cannot determine the latter two. Nonzero emitted orbital angular momentum adds a centrifugal barrier and can strongly hinder a transition.

The dominant action varies roughly as $Z_d/\sqrt{Q}$, explaining the Geiger-Nuttall relation and the enormous spread of alpha half-lives. angular momentum and parity constrain the emitted partial wave: a spin-zero positive-parity alpha requires $|J_P-J_D|\le L\le J_P+J_D$ and ${\pi }_P={\pi }_D(-1{)}^L$. Fine-structure alpha groups populate different daughter levels and therefore have different kinetic energies and penetrabilities. Shell structure and alpha clustering also change preformation probabilities, so equal $Q$ values do not guarantee equal lifetimes. Escape conserves the total energy throughout tunnelling; the particle does not borrow a measured energy from the barrier.

The deuteron is the bound proton–neutron system. Its ground state has $J^P=1^{+}$ and isospin $T=0$, with binding energy approximately $2.225\mathrm{MeV}$ and no excited bound state. The small binding compared with characteristic nuclear energy scales produces an extended tail. With reduced mass $\mu \simeq m_N/2$, the asymptotic radial wavefunction decays as $e^{-\kappa r}$, whereThe decay length is not a sharp deuteron radius or the force range. It illustrates weak binding outside the region of strong attraction.

The observed total spin is consistent with aligned nucleon spins, $S=1$, and a predominantly $L=0$ spatial state, ${}^3{S}_1$. A central attraction identical in both spin–isospin channels would not distinguish this channel from a spin-singlet proton–neutron channel. In fact the triplet channel binds while the corresponding singlet channel does not. A realistic nuclear force must therefore distinguish these spin–isospin channels, for example through a term $V_{\sigma }(r){\mathbf{\sigma }}_p\cdot {\mathbf{\sigma }}_n$, whose eigenvalues are $+1$ and $-3$ in triplet and singlet states. The absence of a bound dineutron and diproton reinforces the need to distinguish channels; proton–proton Coulomb repulsion is an additional effect, not a complete explanation of the singlet nuclear interaction.

The clearest evidence against a purely central force is the nonzero electric quadrupole moment. A pure isotropic S-wave charge distribution would give zero. The deuteron quadrupole evidence for a tensor force instead points to a nuclear tensor force,It depends on how the spins point relative to the line joining the particles. It preserves total $J$ but couples orbital angular momenta differing by two, producingBoth terms have $J^P=1^{+}$, and their interference permits a quadrupole moment. The 1939 theoretical analysis explicitly connected the measured quadrupole moment with such directional dependence.^[ref]

The nuclear magnetic moment provides a second check. A pure aligned-spin S wave would give approximately ${\mu }_p+{\mu }_n\simeq 0.880{\mu }_N$, close to but larger than the observed deuteron value approximately $0.857{\mu }_N$. The discrepancy is consistent with non-S-wave structure and interaction-current corrections; it is not by itself a model-independent measurement of a D-state probability. The deuteron D-state admixture depends on the adopted Hamiltonian and electromagnetic current. The magnetic moments and mass differences used for these scales are tabulated physical constants.^[ref]

A useful schematic interaction therefore contains central, spin-dependent and tensor terms, and possibly spin–orbit terms in other channels. Long-range one-pion exchange naturally supplies a tensor component, with its finite range set by the pion mass. The deuteron establishes an attractive triplet channel and a noncentral tensor interaction; a spin-independent central potential is insufficient.

A muonic atom binds a negative muon in the nuclear Coulomb potential. The muon is a spin one-half lepton with the electron's charge but about 207 times its mass. For a point nucleus of mass $M_N$ and charge $Ze$, the reduced mass and nonrelativistic hydrogenic scales, with fine-structure constant $\alpha$, areFor a heavy nucleus ${\mu }_r\simeq m_{\mu }$, giving a radius roughly $1/207$ of the corresponding electronic radius and transition energies roughly 207 times larger. For hydrogen, nuclear recoil matters: $m_{\mu }\simeq 106\mathrm{MeV}/c^2$ gives ${\mu }_r\simeq 95\mathrm{MeV}/c^2$ and $a_{\mu }\simeq 284\mathrm{fm}$ rather than the infinite-nuclear-mass value about $255\mathrm{fm}$.

A slowed negative muon is captured into an excited orbit and undergoes a radiative cascade, ejecting atomic electrons and emitting muonic X-rays. Inner orbits penetrate the nuclear charge distribution, particularly at large $Z$. Replacing a point charge by an extended charge makes the central attraction less singular and shifts states with substantial nuclear overlap. Measured transition energies consequently constrain the nuclear charge radius. Sensitivity is particularly strong for $S$ states, whose contact density scales as $(Z\alpha {\mu }_r{)}^3$.

An instructive experiment is Randolf Pohl's team's muonic-hydrogen spectroscopy. A stopped-muon target prepared a metastable $2S$ population; a tunable pulsed laser drove a $2S\to 2P$ transition, followed by a detectable $2P\to 1S$ X-ray. Scanning laser frequency located a resonance whose interpretation included the proton finite-size contribution to the Lamb shift. Delayed detection selected the long-lived population against prompt cascade backgrounds. Laser-off or detuned measurements provide a background control; energy and time selection alone would not establish a laser-induced resonance.^[ref]

Extracting a radius requires quantum electrodynamics, including vacuum polarization, recoil and nuclear-structure corrections. A free muon decays on a microsecond timescale; a bound negative muon can instead undergo muon capture, for example ${\mu }^{-}p\to n{\nu }_{\mu }$. These processes limit observation time. In heavy atoms, relativistic motion and nuclear overlap make the simple Bohr formulas only qualitative. Muonic spectroscopy converts a smaller orbital scale into enhanced sensitivity to nuclear charge structure, provided atomic and nuclear corrections are controlled.

Use the relative coordinate and reduced massTake $V=-V_0$ inside, $V=0$ outside, and $E=-B$. For $V_0>B>0$, define inverse lengthsThe radial equation becomes $u^{\prime \prime }+k^{2}u=0$ inside and $u^{\prime \prime }-{\kappa }^{2}u=0$ outside. Finiteness of the full wavefunction at the origin imposes $u(0)=0$; normalizability excludes the growing exterior exponential. HenceContinuity of $u$ was built in. Continuity of its derivative at the finite step givesFor the nodeless ground state, $\pi /2<kb<\pi$ selects the first branch. Higher roots would correspond to different wells with more bound states. Normalization is

For the requested weak-binding estimate, $B/V_0\to 0$ makes $\kappa /k\to 0$. On the first branch $kb\to \pi /2$, and $k^2\simeq 2\mu c^2{V}_0/(\hslash c{)}^2$. The leading threshold estimate isThis is the zero-binding threshold of a spherical square well, so nonzero positive binding needs a depth above it. Corrections start at order $\sqrt{B}$, not just at order $B$: setting $kb=\pi /2+ϵ$ gives $ϵ\simeq \kappa b/(\pi /2)$ and

The actual numbers permit an accuracy check: $\kappa =0.231608{\mathrm{fm}}^{-1}$ and $\kappa b=0.486377$ are not extremely small. Retaining the exterior decay, solveThis gives $x=kb=1.830503$, and the full square-well result isNeglecting $B$ only in the interior kinetic relation while retaining the measured exterior decay instead estimates $V_0\simeq 31.50\mathrm{MeV}$. These approximations should be identified: 23.2 MeV is the leading $B\to 0$ threshold, whereas matching the supplied nonzero binding requires a boundary phase shift and a deeper well.

For an isolated approximately spherical mass distribution outside its luminous extent, retain the gravitational force $GMm/r^2$ and use the specified low-acceleration inertial law. circular motion then givesThe radius cancels when enclosed luminous mass is effectively constant. This is the asymptotic modified Newtonian dynamics result associated with the baryonic Tully-Fisher relation; an arbitrary extended disk cannot everywhere be replaced exactly by a point mass.

At the disk edge,Using $G=6.67\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ givesThis is an order-of-magnitude isolated, low-acceleration estimate. The edge acceleration is below $a_0$, but not extremely small compared with it; a full interpolation law, disk geometry and an external gravitational field can affect a more precise mass fit.

For the spherical mass profile for a flat rotation curve, assume spherical symmetry where the halo dominates. A flat rotation curve gives $M(<r)=v^{2}r/G$, and differentiating $M(<r)=4\pi {\int }_0^r\rho (s)s^{2}ds$ yieldsHere $\rho$ is the density of the spherical gravitating model; it is not an exact deprojection of a thin luminous disk or a proof that the central density must diverge. At the outermost stated measured radius,This gives the requested galaxy-plus-halo mass estimate if the model is truncated at the outermost observed radius $2R$. The observation itself constrains enclosed mass within $2R$; any additional outer halo remains unconstrained. Extending $\rho \propto r^{-2}$ to infinity would give divergent mass. Subtracting the MOND-inferred luminous mass to obtain a Newtonian halo mass would mix independent model-dependent estimates without an independently measured luminous mass.

In the Bullet Cluster collision, most observed baryonic mass is hot gas. ram pressure slows and shocks the gas, while the galaxies pass through much more freely. weak gravitational lensing maps the projected gravitational mass without requiring that the gas be in hydrostatic equilibrium. The dominant lensing peaks lie closer to the galaxies than to the displaced X-ray gas. This supports a dominant weakly collisional dark matter component under the usual gravity model. It is stronger evidence than a mass-to-light discrepancy alone because distinct components separate during the same encounter. The inference needs background-source shape/redshift calibration, projection control and uncertainty in the lensing reconstruction; it does not identify a WIMP species or by itself exclude every modified-gravity theory.^[ref]

For dark-matter direct detection, consider XENON100 at Gran Sasso. A halo WIMP can undergo elastic scattering from a xenon atomic nucleus, leaving a recoil of only a few to tens of keV. The expected rate is very low, so a large target, low threshold and long exposure must be combined with background rejection. The nuclear-recoil kinematics of WIMP scattering gives $E_{R,\max }=2{\mu }_{\chi N}^2{v}^2/m_N$ in the nonrelativistic limit, where ${\mu }_{\chi N}$ is the reduced mass and masses and velocities use consistent units. This explains why low-energy response calibration matters.

The dual-phase xenon detector measures prompt particle scintillation S1 and delayed electroluminescence S2 from ionization electrons extracted into gas. Their time separation gives interaction depth; the S2 light pattern supplies transverse position. An inner fiducial volume uses xenon's self-shielding to suppress external backgrounds. The S2/S1 distribution statistically separates nuclear recoils from electron recoils; it is not perfect event-by-event identification. Underground siting reduces cosmic-ray backgrounds, while passive shielding, low-radioactivity components, xenon purification and a surrounding active detector veto reduce other events.^[ref]

Neutrons can also make nuclear recoils, so charge-to-light discrimination alone cannot eliminate them. Multiple-scatter rejection, shielding and background modelling remain necessary. Gamma-source and neutron-source calibrations map electron- and nuclear-recoil distributions, energy scale and selection acceptance. Position-dependent light collection and charge losses need correction; photon-counting fluctuations and uncertain low-energy yields limit threshold sensitivity. Blind event selection and independent sidebands guard against tuning cuts to apparent candidates. The 224.6-live-day, 34 kg search found two benchmark candidates against about one expected background event, consistent with background. Its interaction limit therefore depends on the assumed halo distribution, WIMP mass and nuclear response; it is not a detection or an exclusion of all dark matter.^[ref]

Let $\mu =G{M}_{\ast }$ and use plane polar coordinates in the orbital plane. The Newtonian gravitational potential energy is $U=-\mu M_p/r$, soA central force keeps the orbit in a plane. The star is treated as stationary, appropriate when $M_p\ll M_{\ast }$; otherwise the two-body reduction uses the reduced mass and the relative separation.

Here $G$ is the gravitational constant, $r$ is the separation of the orbiting object from the central body's centre, and $J$ is the magnitude of the object's conserved orbital angular momentum, not angular momentum per unit mass. Approximate the much heavier central body as fixed, neglect other forces, and use the Newtonian gravitational potential energy $U=-GM{m}_0/r$, with zero at infinity.

A central force confines the motion to a fixed orbital plane. In plane polar coordinates its kinetic energy is $T=\frac{1}{2}{m}_0({\dot{r}}^2+r^2{\dot{\varphi }}^2)$. Conservation of angular momentum gives $J=m_0{r}^2\dot{\varphi }$; eliminating $\dot{\varphi }$ yieldsThe centrifugal term belongs to the radial effective potential; it is the eliminated transverse kinetic energy, not an additional physical interaction. If recoil of the central body mattered, the relative-coordinate reduction would use the reduced mass.

For two particles, let $M=m_1+m_2$ and ${\mathbf{V}}_{\mathrm{cm}}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/M$. The centre-of-mass frame velocities are ${\mathbf{v}}_i^{\ast }={\mathbf{v}}_i-{\mathbf{V}}_{\mathrm{cm}}$, so $m_1{\mathbf{v}}_1^{\ast }+m_2{\mathbf{v}}_2^{\ast }=0$. Thus before and after an isolated collision the two momentum vectors are equal and opposite. For an elastic collision their common magnitude is also unchanged: only their direction rotates.

The kinetic energy decomposition about the center of mass separates unchanged bulk translation from relative motion:The reduced mass $\mu$ reduces the calculation to one relative velocity. After solving the geometry in the centre-of-mass frame, add ${\mathbf{V}}_{\mathrm{cm}}$ to each velocity to return to the laboratory. The four panels below illustrate this for the subsequent example.

For equal masses $m$, the reduced mass is $m/2$. The kinetic energy of their centre of mass cannot be removed by an isolated collision, since total momentum is fixed. Therefore the maximum energy available for conversion to internal forms is the initial relative kinetic energy,It is attained by a perfectly inelastic collision in which the final relative velocity is zero. This concerns the convertible initial kinetic energy, not an arbitrary pre-existing internal energy reservoir.

Let $\mathbf{w}={\mathbf{v}}_1-{\mathbf{v}}_2$ be the relative velocity and $\mathbf{V}=(M_1{\mathbf{v}}_1+M_2{\mathbf{v}}_2)/(M_1+M_2)$ the centre-of-mass velocity. Inverting givesSubstitute into the sum of the two kinetic energies. The cross terms cancel, while the coefficient of $|\mathbf{w}{|}^2/2$ is the reduced mass $\mu =M_1{M}_2/(M_1+M_2)$. HenceThe first term is translation of the centre of mass; the second is the kinetic energy measured in the centre-of-momentum frame. For an isolated perfectly inelastic collision, the bodies finish with a common velocity, while their total momentum and thus $\mathbf{V}$ are unchanged. The kinetic energy converted to internal forms is thereforewhere $v$ is their initial relative speed. Total energy remains conserved; it is the translational-plus-relative kinetic energy that decreases.

In the inertial centre of mass frame, the Lagrangian mechanics description starts from kinetic energy minus potential energy:Define the two-particle relative coordinate $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$. The centre of mass condition givesSubstitution therefore givesThe Euler-Lagrange equation is simply $\mu \ddot{\mathbf{r}}=-\nabla V(r)$. Thus center-of-mass reduction turns a two-body problem into a one-particle central force problem with reduced mass $\mu$, reducing six position components to three relative components. In an arbitrary inertial frame one adds the independent centre of mass term $\frac{1}{2}(m_1+m_2)|\dot{\mathbf{R}}{|}^2$; that free motion does not affect the relative solution. Since $V$ depends only on separation, conservation of angular momentum further confines a nonradial relative trajectory to a plane.

Apply the radial momentum operator to a smooth radial test function. More generally $[f(r),p_r]=i\hslash f^{\prime }(r)$, giving the first two labelled identities. Expanding the square gives the third:The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A central potential commutes with $L^2$ and $L_z$, so choose simultaneous angular eigenfunctions $Y_{lm}$, with $L^2=l(l+1){\hslash }^2$ and $l=0,1,\dots$. The radial Hamiltonian consequently has the stated centrifugal term. The radial inner product uses measure $r^{2}dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial momentum is not thereby asserted to be a self-adjoint observable on a half-line.

Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i{p}_r/\hslash$. Since $D^{†}=-D$ formally and $[D,W]=W^{\prime }=q/r^2$,The hydrogen radial Hamiltonian factorization follows by matching the Coulomb coefficient, which fixes the Bohr radiusFor the second pair of labelled relations, the opposite product changes the sign of $W^{\prime }$, henceUsing $[A,A^{†}A]=[A,A^{†}]A$ in the factorization then givesThis Coulomb angular-momentum intertwining means applying $A_l$ to an energy eigenstate either gives zero or an eigenstate of $H_{l+1}$ with the same energy. For a normalized negative-energy state,At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The bound-state ladder termination by norm positivity occurs at some $L$ with $A_L\psi =0$, givingThe terminating radial function is proportional to $r^L{e}^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy scattering states are not subject to this terminating negative-energy argument. Here $m$ is the mass in the supplied Hamiltonian; including nuclear motion replaces it by the reduced mass.

In a classically forbidden region the local wavevector becomes imaginary. For a slowly varying barrier, the WKB transmission probability has leading exponentialwhere $r_1,r_2$ are turning points and $\mu$ is the reduced mass. This is a probability exponent; the wave amplitude has only half that exponent. Near a turning point one uses matching formulas rather than the unmodified local WKB expression.

For alpha decay, an attractive nuclear interior supports a quasibound state behind the repulsive Coulomb barrier of the daughter nucleus. For zero orbital angular momentum, write the exterior potential as $C/r$, $C=2Z_d{e}^2/(4\pi {ϵ}_0)$, with nuclear matching radius $R$ and outer turning point $b=C/Q>R$. The exterior action can be integrated:The decay rate is approximately $\lambda =\nu P_{\mathrm{pre}}{e}^{-2G}$ and the half-life is $\ln 2/\lambda$. Here $\nu$ is the interior assault-frequency scale and $P_{\mathrm{pre}}$ is the alpha preformation probability. They are model-dependent; barrier penetration alone does not determine them. Nonzero orbital angular momentum adds a centrifugal barrier.

For $R/b\ll 1$, the leading dependence is $G\propto Z_d/\sqrt{Q}$. Thus a modest increase in decay energy can cause an enormous increase in decay rate, giving the Geiger-Nuttall law relating logarithmic half-life to inverse square-root energy along a suitable isotope sequence. Tunnelling conserves the total energy: the escaping alpha particle does not acquire energy from the barrier.

Reduced radial equation. A spherical harmonic satisfies ${\hat{L}}^2{Y}_{\ell m}=\ell (\ell +1){\hslash }^2{Y}_{\ell m}$. Substitute $\psi =R{Y}_{\ell m}$ into the time-independent Schrödinger equation, divide out the angular function, and put $R=U/r$. Direct differentiation givesMultiplication by $r$ then gives the Radial Schrodinger equationHere $\mu =m_1{m}_2/(m_1+m_2)$ is the reduced mass, and the extra positive term is the centrifugal potential.

Boundary conditions. At the origin this finite central potential has no point interaction. The physical radial solution is regular there, requiring $U(0)=0$; for $\ell =0$, a nonzero constant $U(0)$ would give $R\sim 1/r$ and a point-source singularity of the three-dimensional Laplacian. Mere square integrability alone would not remove that singular solution. For a bound state, the radial probability integral is ${\int }_0^{\infty }|U{|}^{2}dr$, so the exterior solution must decay and $U(r)\to 0$ as $r\to \infty$.

Solutions for zero orbital angular momentum. For $\ell =0$ and $-V_0<E<0$, defineThe interior equation is $U^{\prime \prime }+k^{2}U=0$; $U(0)=0$ excludes the cosine. The exterior equation is $U^{\prime \prime }-q^{2}U=0$; normalizability excludes the growing exponential. ThusThe constants cannot remain independent once the interface is matched. Since the potential step is finite, both $U$ and $U^{\prime }$ are continuous at $a$:Dividing the equations yields