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The invariant decay amplitude is the transition matrix element for one incoming particle and a specified on-shell final state, with the overall momentum delta function factored out. For covariantly normalized states, write the connected S-matrix element as $i(2\pi {)}^4{\delta }^4(p_i-p_f)\mathcal{M}(i\to f)$. Its overall phase does not affect the decay width. The rate is $\Gamma =(2M_i{)}^{-1}{\sum }_{\mathrm{spins}}\int d{\Phi }_f|\mathcal{M}{|}^2$, with the initial spin average included when the initial state is unpolarized. For a spin-zero parent and an angle-independent two-body amplitude, the two-body Lorentz-invariant phase space gives $\Gamma =|\mathbf{k}|\sum |\mathcal{M}{|}^2/(8\pi M_i^2)$. In leptonic pion decay, contracting the axial hadronic matrix element with the chiral weak charged current makes this amplitude proportional to the charged-lepton mass, proving helicity suppression.

For $x=m_W^2/m_H^2<1/4$, the on-shell longitudinal pair contributes the displayed tree-level partial decay width. In the Higgs boson rest frame, its polarization vectors have contraction $(m_H^2/2-m_W^2)/m_W^2$. Multiplying by the Higgs boson coupling to W bosons gives $|{\mathcal{M}}_{LL}{|}^2=m_H^4(1-2x{)}^2/v^2$; the two-body Lorentz-invariant phase space then gives the width. The sum over transverse polarizations adds $8x^2$ to the longitudinal polynomial. Below threshold, physical decays through virtual W bosons require a different multiparticle calculation.

For the current with $(1-{\gamma }^5)$ rather than $P_L$, the spin-summed symmetric leptonic tensor is $8(q_{1\mu }{q}_{2\nu }+q_{1\nu }{q}_{2\mu }-g_{\mu \nu }{q}_1\cdot q_2)$. Integrating the standard two-body Lorentz-invariant phase space of two massless particles givesIts antisymmetric Levi-Civita symbol term integrates to zero. The normalization changes by a factor of four if the current uses $P_L$ instead. The result separates the leptonic integration from hadronic pseudoscalar-to-pseudoscalar form factors in a three-body decay rate.

In the convention $⟨0|J^{\mu }|P(p)⟩=-i\sqrt{2}{F}_P{p}^{\mu }$, the Fermi interaction gives $\mathcal{M}=G_F{F}_P{p}_{\mu }{\bar{u}}_{\ell }{\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }$. With a massless antineutrino, the external Dirac equations reduce this to $G_F{F}_P{m}_{\ell }{\bar{u}}_{\ell }(1-{\gamma }_5)v_{\nu }$. The spin sum is $8G_F^2{F}_P^2{m}_{\ell }^{2}k\cdot q=4G_F^2{F}_P^2{m}_{\ell }^2(M_P^2-m_{\ell }^2)$. The two-body Lorentz-invariant phase space has integrated value $(1-m_{\ell }^2/M_P^2)/(8\pi )$, and the initial flux factor is $1/(2M_P)$, giving the displayed width. If the decay constant is instead defined without the square root of two, $f_P=\sqrt{2}{F}_P$ and the denominator is $8\pi$. A weak mixing coefficient multiplies the amplitude and its modulus squared multiplies the width.

For the Z boson vertex $g{\gamma }^{\mu }(c_v-c_a{\gamma }^5)/(2c_W)$, sum over the three initial polarizations and final fermion spins. The symmetric Dirac trace is $4(c_v^2+c_a^2)(k_1^{\mu }{k}_2^{\nu }+k_1^{\nu }{k}_2^{\mu }-g^{\mu \nu }{k}_1\cdot k_2)$. The vector-axial interference is antisymmetric and drops out of the massive vector polarization sum. The longitudinal term drops out by current conservation for massless final particles. Contracting the remaining trace gives the displayed result. The massive-vector spin average divides it by three; two-body Lorentz-invariant phase space then gives $\Gamma =g^2{M}_Z(c_v^2+c_a^2)/(48\pi c_W^2)$ per color. These coefficients are twice those used with a vertex prefactor $g/c_W$.

The chiral weak charged current vertex $-ig{\gamma }^{\mu }(1-{\gamma }_5)/(2\sqrt{2})$ gives the spin-averaged squared matrix element $g^2{M}_W^2/3$ for an on-shell W boson decaying to a massless charged lepton and antineutrino. Integrating two-body Lorentz-invariant phase space gives ${\Gamma }_{\ell }=g^2{M}_W/(48\pi )$. The massive-vector spin average divides by three; an extra neutrino-helicity factor would incorrectly double count the chiral projector.

For a vertex $(g/c_W){\gamma }^{\mu }(c_V-c_A{\gamma }^5)$ into one massless color-singlet particle-antiparticle pair, the massive-vector spin average and two-body Lorentz-invariant phase space give $\Gamma =g^2{m}_Z(c_V^2+c_A^2)/(12\pi c_W^2)$. With Standard Model neutral-current vector and axial couplings, one neutrino flavor has ${\Gamma }_{\nu }=g^2{m}_Z/(96\pi c_W^2)$ and one charged-lepton flavor has ${\Gamma }_{\ell }={\Gamma }_{\nu }(1-4s_W^2+8s_W^4)$. Chiral projectors already remove inactive neutrino helicities. These are tree-level massless results.

For a left-handed incoming neutrino, unpolarized electrons, and electron vector/axial couplings $c_V,c_A$, the four-fermion interaction gives $\overline{|\mathcal{M}{|}^2}=4G_F^2[(c_V+c_A{)}^2{s}^2+(c_V-c_A{)}^2{u}^2]$. Integrating massless two-body Lorentz-invariant phase space yields $\sigma =G_F^{2}s(c_V^2+c_V{c}_A+c_A^2)/(3\pi )$. The electron spin average is included, while no sterile neutrino helicity is averaged in.

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. Varying the Proca action with respect to $A_{\nu }$ gives $\partial \mathcal{L}/\partial A_{\nu }=m^2{A}^{\nu }$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{A}_{\nu })=-F^{\mu \nu }$. The Euler-Lagrange equation is thereforeTaking a divergence annihilates the antisymmetric field strength, since commuting derivatives are symmetric in their indices. For $m\ne 0$ this gives the Lorenz constraint in Proca theory ${\partial }_{\nu }{A}^{\nu }=0$, and hence $(\square +m^2)A^{\nu }=0$. The constraint follows from dynamics; it is not an additional gauge fixing condition for this massive theory.

For four-momentum $p^{\mu }=(E,0,0,p)$, longitudinal polarization of a massive vector boson lies in the time–$z$ plane, has $p\cdot \epsilon =0$ and is normalized by $\epsilon \cdot \epsilon =-1$. Takingsatisfies both requirements because $E^2-p^2=m^2$. It can also be obtained by boosting the rest-frame unit spatial polarization along the direction of motion. Together with two transverse polarizations it gives the three physical spin states of a massive spin one particle.

In the Higgs boson rest frame, choose $p^{\mu }=(E,0,0,p^{\ast })$ and $p^{\prime \mu }=(E,0,0,-p^{\ast })$, with $E=m_H/2$ and $p^{\ast 2}=m_H^2/4-m_W^2$. Consistent normalized longitudinal polarization vectors areTheir Lorentz contraction is $(p^{\ast 2}+E^2)/m_W^2=(2p^{\ast 2}+m_W^2)/m_W^2$. The Higgs boson coupling to W bosons consequently givesUsing the two-body Lorentz-invariant phase space, $p^{\ast }=m_H\sqrt{1-4x}/2$ and $v^{-2}=\sqrt{2}{G}_F$ yields the longitudinal partial decay widthThe charged W bosons are distinguishable, so there is no identical-particle factor of $1/2$. This expression includes the entire angular phase space but only the specified longitudinal final polarizations. The Higgs decay to two W bosons summed over all polarizations instead has the factor $1-4x+12x^2$; the additional transverse contribution is $8x^2$. At $m_H\gg m_W$ the longitudinal channel dominates and the leading width grows as $G_F{m}_H^3$.

The Standard Model is a chiral gauge theory: left-chiral quarks and leptons form SU(2) weak doublets, while the corresponding right-chiral charged fields are weak singlets. The covariant kinetic term contains $g_W{\bar{Q}}_L{\gamma }^{\mu }{t}^a{Q}_L{W}_{\mu }^a$. With $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$, its off-diagonal generators give the weak charged currentAn overall interaction sign depends on the convention for the covariant derivative and has no effect on this decay rate. The chiral projector $P_L=(1-{\gamma }^5)/2$ explains the vector-minus-axial combination. The Cabibbo-Kobayashi-Maskawa matrix comes from the different left-handed rotations that diagonalize up- and down-type fermion mass matrices. Set $V_{tb}=1$ and $m_b=0$ as stipulated.

Let $P_{\mu \nu }=-g_{\mu \nu }+p_{3\mu }{p}_{3\nu }/m_W^2$, the massive vector polarization sum. Sum over final spins and average over the two equally populated initial top quark spins. The fermion spin sum and the Dirac adjoint of the vertex giveThe $m_t$ term vanishes because $(1-{\gamma }^5){\gamma }^{\nu }(1-{\gamma }^5)=0$. Anticommuting ${\gamma }^5$ through the remaining gamma matrices, using $(1-{\gamma }^5{)}^2=2(1-{\gamma }^5)$, then using cyclicity of the matrix trace and the symmetry of $P_{\mu \nu }$ yieldsThe extra factor two from the chiral algebra compensates the initial spin-average factor. The term with ${\gamma }^5$ is antisymmetric in $\mu ,\nu$ and vanishes upon contraction with $P_{\mu \nu }$. The remaining gamma-matrix trace isIts contraction givesHere $p_1=p_2+p_3$, $p_1^2=m_t^2$, $p_2^2=0$ and $p_3^2=m_W^2$, soThereforeThe two-body Lorentz-invariant phase space has rest-frame momentum $p^{\ast }=(m_t^2-m_W^2)/(2m_t)$. Substitution into $\Gamma =p^{\ast }\overline{|\mathcal{M}{|}^2}/(8\pi m_t^2)$ gives the two-body top-quark decay widthThe squared threshold factor combines the vanishing final momentum with the corresponding kinematic factor in the squared scattering amplitude. The longitudinal W boson polarization contributes the high-mass enhancement proportional to $m_t^3/m_W^2$. The result is at tree level, for an on-shell W boson and $m_t>m_W$; finite bottom mass and radiative corrections change it. Restoring quark mixing multiplies the width by $|V_{tb}{|}^2$.

Use the current convention in the question, whose vertex prefactor is $g/(2\cos {\theta }_W)$. The neutral-current vector and axial couplings are thenFor a charged lepton, $T_3=-1/2$, $Q=-1$, givingFor an active neutrino, $T_3=1/2$, $Q=0$, so $c_V=c_A=1/2$. The right-handed weak singlet contribution is purely vectorlike in the charge term; combining it with the left-handed contribution gives precisely this Dirac-current form. These coefficients would be halved if the vertex prefactor were $g/\cos {\theta }_W$ instead.

Let $p=k+k^{\prime }$ be the Z boson momentum and $k,k^{\prime }$ the outgoing massless lepton momenta. The decay amplitude isUse the fermion spin sums $\sum u\bar{u}=/k$ and $\sum v\bar{v}=/k^{\prime }$. Anticommutation of ${\gamma }_5$ with the Dirac gamma matrices gives the symmetric part of the trace,The vector-axial cross term is antisymmetric in $\mu ,\nu$, and hence drops out of the massive vector polarization sum. The massless equations also give $p_{\mu }{L}^{\mu \nu }=0$: in the symmetric part, the first two contractions sum to $p^{\nu }k\cdot k^{\prime }$, canceling the last; the antisymmetric term vanishes because $p=k+k^{\prime }$. The longitudinal $p_{\mu }{p}_{\nu }/m_Z^2$ term therefore makes no contribution.

Since $2k\cdot k^{\prime }=m_Z^2$, contraction with $-g_{\mu \nu }$ gives $4(c_V^2+c_A^2)m_Z^2$. Thus the massless fermion Z decay spin sum, including all three initial polarizations, isTo use this fully summed expression for the decay width of one unpolarized Z boson, apply the massive-vector spin average, dividing by three. The width of each fixed polarization is the same after integration over all directions, so that procedure also gives the physical single-particle width. Omitting the division would sum the widths of three initial states.

For two massless distinguishable final particles, the two-body Lorentz-invariant phase space isConsequentlyThe tree-level electroweak symmetry breaking relation is $m_W=m_Z\cos {\theta }_W$. Combining it with the Fermi constant normalization givesFor an active neutrino, the vertex itself projects out inactive chiralities, so no additional helicity factor is needed. This is the tree-level massless leptonic Z decay width; lepton masses and radiative corrections have been excluded.

Take metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and antineutrino momenta are $k$ and $k^{\prime }$. Up to an irrelevant overall phase, the amplitude isA fermion spin sum givesAnticommuting ${\gamma }_5$ through the massless momentum and gamma matrix, and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$, reduces the trace to twice the four-gamma trace minus twice its ${\gamma }_5$ counterpart. In particular its symmetric part isThe epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the neutrino helicity: the chiral vertex has already done so.

The external massless equations imply $p_{\mu }{J}^{\mu }=0$ for $p=k+k^{\prime }$, so $p_{\mu }{T}^{\mu \nu }=0$. With $k\cdot k^{\prime }=M^2/2$, averaging over the three initial spin states givesIndeed $-g_{\mu \nu }{T}_S^{\mu \nu }=16k\cdot k^{\prime }=8M^2$. This is a massive-vector spin average, not an average over the final spins.

In the rest frame, eliminating the momentum delta function and integrating the remaining radial delta function $\delta (M-2|\mathbf{k}|)$ reduces the two-body Lorentz-invariant phase space to $d{\Phi }_2=d\Omega /(32{\pi }^2)$. ThereforeThis derives the massless leptonic W decay width without invoking a precomputed decay formula.

For the specified polarization there is no initial spin average. Put $\mathbf{k}=(M/2)\hat{\mathbf{n}}$, ${\mathbf{k}}^{\prime }=-(M/2)\hat{\mathbf{n}}$ and ${\hat{n}}_z=\cos \theta$. The contraction selectsIt follows thatHere $\int {\sin }^2\theta d\Omega =8\pi /3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/{\Gamma }_z)d{\Gamma }_z/d\cos \theta =3(1-{\cos }^2\theta )/4$ instead of the isotropic unpolarized distribution. This is polarized massless leptonic W decay.

Use the source convention $Q=T_3+Y$, with electric charge measured in units of the positron charge. Expanding the covariant derivative in the fermion kinetic term gives interaction $-\bar{f}{\gamma }_{\mu }(g{T}_3{A}_3^{\mu }+g^{\prime }Y{B}^{\mu })f$. Substituting the given neutral-field rotation, the coefficient of $Z^{\mu }$ isThe photon coefficient is $g{s}_W(T_3+Y)=eQ$, with $e=g{s}_W=g^{\prime }{c}_W$. For a Dirac field, $T_3$ acts only on its left-handed component. ThusThe weak neutral current interaction consequently has the requested form, withThe electroweak representation and hypercharge table gives the following values for one generation:

These neutral-current vector and axial couplings use the source's vertex prefactor $g/(2c_W)$. They are twice the coefficients used with prefactor $g/c_W$; the latter convention must not be mixed into this calculation. For the neutrino, the chiral projector removes the inactive right-handed component automatically.

Set $G=g/(2c_W)$ and suppress the species label. For outgoing momenta $k_1,k_2$, the amplitude is, up to an irrelevant phase,The final-state spin sum is the Dirac trace of $/k_1{\gamma }^{\mu }(c_v-c_a{\gamma }^5)/k_2{\gamma }^{\nu }(c_v-c_a{\gamma }^5)$. Anticommutation of ${\gamma }^5$ with the gamma matrices makes its symmetric partThe vector-axial cross term is proportional to the antisymmetric Levi-Civita symbol, so it contracts to zero with the symmetric massive vector polarization sum $-g_{\mu \nu }+p_{\mu }{p}_{\nu }/M_Z^2$. Since $k_1^2=k_2^2=0$, the symmetric tensor is transverse to $p$ as well. Only $-g_{\mu \nu }$ remains, andMultiplying by $G^2$ proves the massless fermion Z decay spin sum:

For the decay width, average over the three initial polarizations. Rotational invariance makes the integrated width the same for each initial spin state, so the massive-vector spin average is legitimate even for a single prepared polarization. In the rest frame, the delta functions in two-body Lorentz-invariant phase space leaveThis follows by setting ${\mathbf{k}}_2=-{\mathbf{k}}_1$ and integrating $\delta (M_Z-2|{\mathbf{k}}_1|)$; the radial integration supplies a factor $1/2$. HenceFor a quark species multiply by three colors, with strong-interaction radiative corrections when comparing with data. For one effectively massless active neutrino, $c_v=c_a=1/2$, so ${\Gamma }_{\nu }=g^2{M}_Z/(96\pi c_W^2)=G_F{M}_Z^3/(12\sqrt{2}\pi )$.

The total resonance width contains visible charged-lepton and hadron decays plus $N_{\nu }{\Gamma }_{\nu }$. Thus the measured invisible contribution determinesEarly resonance measurements supported three light active neutrino species, with experimental uncertainties; the OPAL analysis of its 1989 data explicitly compares the three- and four-species predictions. Under the question's assumption that every possible family has a massless neutrino with ordinary weak couplings, this constrains the number of such families. The inference does not exclude additional neutrinos heavier than $M_Z/2$ or gauge singlet sterile states with no ordinary $Z$ coupling. This is light-neutrino counting from the Z width.

Put $s=q^2>0$ and neglect electron and final-quark masses compared with $\sqrt{s}$. First fix a final color charge; its multiplicity will be restored for the inclusive rate. The photon-exchange scattering amplitude is, up to an overall phase,Averaging over the two spins of each incident particle and summing final spins givesBy the gamma matrix trace identities, the first trace is $4(p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-g^{\mu \nu }{p}_1\cdot p_2)$, and likewise for the second. Contracting the two tensors cancels the terms proportional to $(p_1\cdot p_2)(k_1\cdot k_2)$, givingIn the centre-of-mass frame all four energies are $\sqrt{s}/2$. The first pair of scalar products equals $s(1-\cos \theta )/4$ each, and the second pair equals $s(1+\cos \theta )/4$ each. Hence $\overline{|\mathcal{M}{|}^2}=e^4{Q}_q^2(1+{\cos }^2\theta )$.

For clarity, the printed flux hint has a factor-of-four inconsistency: when $\mathcal{F}=4\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=2s$, the prefactor multiplying the spin-averaged amplitude is $1/\mathcal{F}$, not $1/(4\mathcal{F})$. Alternatively, $1/(4F_0)$ is correct if $F_0=\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=s/2$. Use this massless two-body scattering flux normalization, keeping the spin average separate.

To obtain the two-body Lorentz-invariant phase space directly, integrate the momentum-conserving spatial delta function, leaving ${\mathbf{k}}_2=-{\mathbf{k}}_1$. The remaining radial integral is $\int dk\delta (\sqrt{s}-2k)=1/2$, so $d{\Phi }_2/d\Omega =1/(32{\pi }^2)$. The relativistic scattering cross-section is consequentlySince $\int d\Omega (1+{\cos }^2\theta )=2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$, the integrated electron-positron annihilation into a quark pair result, for this fixed color, isThe first two printed formulas are thus color-stripped. Distinct final colors are orthogonal states and must be summed, producing $N_c=3$. At leading order each accessible quark flavor contributes independently. Away from thresholds and narrow resonances, quark-hadron duality relates the inclusive parton rate to the observed inclusive electron-positron annihilation into hadrons, givingThis formula concerns the photon channel, assumes $\sqrt{s}$ is large compared with the relevant quark masses and the strong-interaction scale, and does not include Z boson exchange.

Beyond leading order, an exactly exclusive pair of colored quarks is not an observable asymptotic state because of confinement and hadronization. Moreover its virtual corrections have infrared divergences; real soft or collinear gluon emission cannot be excluded if a finite rate is wanted. The inclusive sum includes both the corrected two-parton channel and channels such as $q\bar{q}g$, with their degenerate soft and collinear limits. Real and virtual infrared divergences then cancel for an infrared-safe observable. A specified jet observable can also be meaningful if it defines an infrared-safe observable; a bare two-quark final state is not.

A convenient inclusive calculation uses the optical theorem. For the electromagnetic hadronic current $j_{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }_{\mu }{q}_f$, defineThe hadronic electromagnetic-current spectral density is obtained from its discontinuity. In this convention the hadronic R ratio is $R(s)={\sigma }_{\mathrm{had}}/{\sigma }_{\mu \mu }=12\pi \mathrm{Im}\Pi (s+i0)$, where ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$ at this order. Cutting the forward amplitude sums all allowed hadronic intermediate states, so perturbative Quantum chromodynamics applied to this inclusive quantity implements the required sum of real and virtual contributions.

Finally, write $a={\alpha }_s({\mu }^2)$, $C=1.99-0.11n_f$, $L=\ln (s/{\mu }^2)$, and ${\beta }_0=11-2n_f/3$. The given truncated correction isDifferentiate with respect to $t=\ln {\mu }^2$, using the stated one-loop beta function $da/dt=-{\beta }_0{a}^2/(4\pi )$ and $dL/dt=-1$. The derivative of $a/\pi$ cancels the explicit logarithmic derivative at order $a^2$. Keeping the residual derivative of the truncated expression givesThus the principle of minimal sensitivity, applied to this expression with nonzero coupling and ${\beta }_0\ne 0$, selects the one-loop stationary scale for the hadronic annihilation correctionThis is a prescription based on the residual order-$a^3$ scale dependence of an order-$a^2$ approximation. At strictly order $a^2$, the derivative already vanishes for every renormalization scale; the prescription does not assert exact scale dependence of the observable. It is useful only while the selected running coupling remains perturbative. If $a=0$ or ${\beta }_0=0$, the stated derivative imposes no unique scale.

At momentum transfers small compared with $M_W$, the two weak charged current vertices each carry $g/(2\sqrt{2})$ when written with $(1-{\gamma }_5)$, and the W boson propagator reduces to a local factor proportional to $1/M_W^2$. Matching their product to the Fermi interaction gives, at tree level,The Fermi constant is thus determined by the electroweak coupling and symmetry-breaking scale; the four-fermion description is a low-energy effective interaction.

Only the cross term $J_L^{†}{J}_H$ can annihilate the pion and create the stated charged-lepton/antineutrino pair. Its leptonic matrix element is ${\bar{u}}_e(k){\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }(q)$. Lorentz covariance permits a vacuum-to-spin-zero current matrix element only proportional to $p^{\mu }$. At rest this has only a temporal component. The vacuum is parity-even and the pion is parity-odd, while $V_H^0$ is parity-even, so $⟨0|V_H^0|\pi ⟩$ equals its own negative and vanishes. Lorentz covariance then makes the full vector matrix element vanish. The temporal axial current is parity-odd and is allowed: this is the vacuum-to-pseudoscalar current selection rule.

The minus sign in $J_H=V_H-A_H$ cancels the interaction's minus sign, up to an irrelevant common S-matrix phase. ThereforeThe pion decay constant convention here includes the given $\sqrt{2}$ in its current matrix element. Momentum conservation and the external Dirac equations give $p=k+q$, ${\bar{u}}_e/k=m_e{\bar{u}}_e$, $/q{v}_{\nu }=0$. Since $/q(1-{\gamma }_5)=(1+{\gamma }_5)/q$, the neutrino term vanishes andThis is helicity suppression: the amplitude vanishes in the zero charged-lepton-mass limit despite the larger available phase space.