advanced36 min studyLesson 603 in Physics

Separate decay directions from laboratory collimation

Compute two-body rest-frame kinematics, identify when isotropy follows from rotational symmetry, and transform energies and angles into a boosted laboratory frame.

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What you’ll understand

  • Compute daughter energy and momentum in a two-body rest frame.
  • Distinguish rotationally unpolarized decays from specified anisotropic angular laws.
  • Transform a massless decay direction and energy under a longitudinal boost.

A fast parent can produce a narrow laboratory cone even when its decay has no preferred direction in its rest frame. Conversely, a nonuniform rest-frame angular distribution can reveal spin alignment and interaction structure. These are different effects. Calculate the two-body kinematics first, state the polarization and dynamical assumptions behind the angular law, and only then apply the boost and the detector's angular acceptance.

Use the rest frame to fix daughter energies

Use natural units and the module's positive-time metric. In a parent's rest frame, the daughters have equal and opposite three-momenta, but their energies need not be equal when their masses differ. Squaring P−p1=p2P-p_1=p_2 determines the first energy directly. The Källén momentum follows by imposing its mass shell. These relations are exact for an on-shell parent decaying to two on-shell daughters and depend only on masses. They do not determine how likely a particular direction is.

For a spin-zero parent, no external direction is available in its rest frame, so the spin-summed decay is isotropic if no external field or correlated selection supplies one. A rotationally unpolarized ensemble likewise has no preferred absolute orientation when unobserved daughter spin states are summed. A polarized parent or a distribution measured relative to an additional spin analyzer can be anisotropic. Conservation laws constrain allowed amplitudes, but assigning a cosine-shaped law requires more information than simply naming the parent spin.

E1∗=M2+m12−m222M,p∗=λ(M2,m12,m22)2M,dΓdΩ∗=p∗32π2M2∣M∣2‾E_1^*=\frac{M^2+m_1^2-m_2^2}{2M},\qquad p^*=\frac{\sqrt{\lambda(M^2,m_1^2,m_2^2)}}{2M},\qquad \frac{d\Gamma}{d\Omega^*}=\frac{p^*}{32\pi^2M^2}\overline{|\mathcal M|^2}
Stars denote the parent rest frame; all masses and momenta use energy units with c = 1. The amplitude is summed over unobserved final spins and averaged over an unpolarized initial ensemble when applicable. Fully labeled identical daughters require their separate counting correction.↔ Scroll wide equations horizontally.

Boost energy and direction together

Take a parent moving along positive z with speed β\beta in the laboratory. To transform a rest-frame daughter into that laboratory frame, the energy and longitudinal momentum receive plus-sign boost terms; the passive transformation into a frame moving along positive z would instead use minus signs. Transverse momentum stays unchanged. Keeping this distinction explicit prevents a common sign error in decay reconstruction. A massless daughter has equal rest-frame energy and momentum magnitude, making its angle transformation especially simple.

The resulting aberration formula maps a rest-frame right angle into a forward laboratory angle. Forward daughters are also more energetic, while backward daughters are redshifted. For a massless isotropic decay, transforming the uniform rest-frame distribution in cos⁡θ∗\cos\theta^* requires a Jacobian, giving a laboratory density proportional to (1−βcos⁡θ)−2(1-\beta\cos\theta)^{-2}. It is therefore incorrect to infer a preferred rest-frame direction from a forward laboratory cone before undoing the boost and acceptance. Massive daughters need their actual energy-to-momentum ratio in the transformation.

E1=γ(E1∗+βp∗cos⁡θ∗),p1z=γ(p∗cos⁡θ∗+βE1∗),cos⁡θ1=cos⁡θ∗+β1+βcos⁡θ∗(m1=0)E_1=\gamma(E_1^*+\beta p^*\cos\theta^*),\quad p_{1z}=\gamma(p^*\cos\theta^*+\beta E_1^*),\quad \cos\theta_1=\frac{\cos\theta^*+\beta}{1+\beta\cos\theta^*}\quad(m_1=0)
This is the active rest-to-laboratory boost for a parent moving in +z, with γ = (1−β²)⁻¹/². The final angular formula assumes a massless daughter; the energy and longitudinal-momentum formulas remain valid for massive daughters.↔ Scroll wide equations horizontally.

Normalize an angular acceptance

An angular distribution is a density in a declared measure. An isotropic distribution is uniform in solid angle, hence uniform in cosine of the polar angle after azimuthal integration. It is not uniform in the polar angle itself, because the solid-angle element contains a sine factor. Normalize the density before integrating a detector region. If the efficiency also depends on energy, boost-induced angle-energy correlations can make a simple angular acceptance insufficient even for a perfectly isotropic rest-frame decay.

The total width is obtained by integrating over all physically distinct directions and spin outcomes. A fitted angular shape can instead use a density normalized to one, which contains no lifetime information. Keeping those quantities separate is essential when an analysis measures polarization from a selected sample and branching fractions from event yields. Detector cuts, a chosen spin axis, and the frame used to define it should accompany every reported angular coefficient. A labeled daughter provides the reference direction; exchanging identical daughters can introduce another counting ambiguity.

Boost a transverse massless decay

A parent of mass 6 GeV6\,\mathrm{GeV} decays to two distinguishable massless daughters. It moves along positive z with β=0.6\beta=0.6. Daughter 1 is emitted at 90∘90^\circ in the parent rest frame. Find its laboratory energy, longitudinal momentum, and polar angle.

  1. Both rest-frame daughter energies are M/2=3 GeVM/2=3\,\mathrm{GeV}. Daughter 1 has pz∗=0p_z^*=0 and transverse momentum pT=3 GeVp_T=3\,\mathrm{GeV}. The parent boost factor is γ=1.25\gamma=1.25.

  2. Apply the rest-to-laboratory boost: E1=1.25(3)=3.75 GeVE_1=1.25(3)=3.75\,\mathrm{GeV} and p1z=1.25(0.6)(3)=2.25 GeVp_{1z}=1.25(0.6)(3)=2.25\,\mathrm{GeV}. The transverse momentum remains three GeV.

  3. The daughter is massless, so ∣p1∣=3.75 GeV|\mathbf p_1|=3.75\,\mathrm{GeV} and cos⁡θ1=2.25/3.75=0.6\cos\theta_1=2.25/3.75=0.6. Thus θ1≈53.13∘\theta_1\approx53.13^\circ, consistent with the aberration formula.

The result

The laboratory daughter has energy 3.75 GeV and longitudinal momentum 2.25 GeV, moving at 53.13 degrees to the boost axis. The other daughter also moves forward here, with opposite transverse momentum.

Explore the model

Interactive lab

Interactive lab · spacetime

Check a Lorentz boost before interpreting angles

Reset and set β = 0.6 and event time ct = 1. The lab's fixed position is x = 0.6; predict x′ = 0 and ct′ = 0.8, with interval 0.64. Inspect the exact readouts. This is a passive transformation of one event with c = 1, not a daughter momentum or an angular decay distribution. The same Lorentz algebra applies to four-momenta, but reversing from passive laboratory-to-rest to active rest-to-laboratory requires the sign convention stated in the lesson.

Check a Lorentz boost before interpreting anglesA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA boost changes components while preserving the invariant, and its direction must be specified.

Check your understanding

What is uniform for an isotropic rest-frame decay after integrating over azimuth?

Try it yourself

A rest-frame selection keeps ∣cos⁡θ∗∣<1/2|\cos\theta^*|<1/2. Compute the accepted fraction for an isotropic decay and for the explicitly specified normalized law f(c)=3(1+c2)/8f(c)=3(1+c^2)/8, where c=cos⁡θ∗c=\cos\theta^*. Assume unit efficiency inside the region.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate each normalized density with respect to c, not with respect to the polar angle.

Compare with the solution

For isotropy, f(c)=1/2f(c)=1/2, so the interval of width one accepts fraction 1/21/2. For the supplied anisotropic law, the fraction is (3/8)∫−1/21/2(1+c2) dc=(3/8)(1+1/12)=13/32≈0.40625(3/8)\int_{-1/2}^{1/2}(1+c^2)\,dc=(3/8)(1+1/12)=13/32\approx0.40625. The full integral of the supplied law over minus one to one equals one. These different acceptances reflect the assumed angular dynamics; neither can be substituted blindly for laboratory acceptance after an unspecified parent boost.

Take it with you

  • Two-body masses fix daughter momentum magnitudes but do not specify an angular law.
  • Isotropic rest-frame decays can be strongly collimated after a boost.
  • Angular densities require a declared frame, spin ensemble, measure, and detector selection.

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