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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The question for today
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 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.
Boost energy and direction together
Take a parent moving along positive z with speed 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 requires a Jacobian, giving a laboratory density proportional to . 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.
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.
Work it through
Boost a transverse massless decay
A parent of mass decays to two distinguishable massless daughters. It moves along positive z with . Daughter 1 is emitted at in the parent rest frame. Find its laboratory energy, longitudinal momentum, and polar angle.
Both rest-frame daughter energies are . Daughter 1 has and transverse momentum . The parent boost factor is .
Apply the rest-to-laboratory boost: and . The transverse momentum remains three GeV.
The daughter is massless, so and . Thus , consistent with the aberration formula.
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.
The living notebook
Explore the model
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.
- Lorentz factor γ
- 1.12
- Event x′
- 0.168
- Event ct′
- 0.817
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.
Pause & reason
Check your understanding
Your turn
Try it yourself
A rest-frame selection keeps . Compute the accepted fraction for an isotropic decay and for the explicitly specified normalized law , where . 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, , so the interval of width one accepts fraction . For the supplied anisotropic law, the fraction is . 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.
Keep the idea
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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