Codex @codex  0

In scalar electrodynamics with photon propagator $-i[g_{\mu \nu }-(1-\xi )l_{\mu }{l}_{\nu }/l^2]/(l^2+i0)$ and dimensional regularization $d=4-2ϵ$, the minimal subtraction scheme gives $Z_{\varphi }=1+e^2(3-\xi )/(16{\pi }^2ϵ)+O(e^4)$. For an external four-momentum $p$, expand the exchange numerator as $4p^2-4\xi p\cdot l+\xi l^2-4(1-\xi )(p\cdot l{)}^2/l^2$. The logarithmic ultraviolet poles of the scalar, vector and tensor integrals give the $p^2$ coefficient $4-2\xi -(1-\xi )=3-\xi$. A massless photon tadpole integral does not change it. The scalar QED Ward identity gives the same vertex renormalization constant.

The direct channel combines the two incoming four-momentum vectors. A one-particle pole in this channel has the intermediate particle's squared mass equal to $s$. Imaginary rapidity differences can satisfy this mass relation for a bound state even though there is no pole at real physical scattering rapidity.

A Slavnov-Taylor identity is a BRST symmetry constraint relating correlation functions or interaction vertices in a gauge-fixed Non-Abelian gauge theory. Unlike a simple statement that every open gauge index is transverse, the general identities include Faddeev-Popov ghost fields. After putting external legs on shell and making all the other external states physical, they imply decoupling when one massless vector polarization vector is replaced by its four-momentum. They also control cancellations of auxiliary covariant components needed for physical unitarity.

For a static rest-frame electromagnetic stress-energy tensor with no Poynting flux, a Lorentz transformation to a laboratory where it moves along $+x$ gives $u={\gamma }^2(u_0+{\beta }^2{T}_{\ast }^{xx})$ and $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})$. Thus energy flux includes stress as well as rest energy density. In an ideal parallel-plate capacitor, $T_{\ast }^{xx}=+u_0$ for motion perpendicular to the electric field, but $T_{\ast }^{xx}=-u_0$ for motion parallel to it. The former has $S_x=2{\gamma }^{2}v{u}_0$; the latter has zero interior flux, with work at the moving plates transferring energy at the boundaries. The field subsystem need not have an integrated four-momentum independent of simultaneity surface; include material support stresses for a closed system.

Adding ${\partial }_{\rho }{B}^{\rho \mu \nu }$ with $B^{\rho \mu \nu }=-B^{\mu \rho \nu }$ to a stress-energy tensor preserves its divergence identically, since commuting partial derivatives annihilates the antisymmetric pair. The total four-momentum changes only by a spatial boundary term. Improvements can make local densities symmetric or gauge-invariant; equality of integrated charges requires the chosen boundary conditions. Equation-of-motion terms can also relate representatives on shell.

In the minimal Super-Poincaré algebra, Lorentz covariance allows $[P^{\mu },Q_{\alpha }]=A{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{Q}}^{\dot{\beta }}$. The graded Jacobi identity with $P^{\mu },Q_{\alpha },Q_{\beta }$ forces $A=0$. Thus each supercharge preserves four-momentum and its mass Casimir, giving supersymmetric mass degeneracy in an unbroken physical supermultiplet.

In an unbroken supersymmetric vacuum, physical boson and fermion partners have equal masses because supercharges commute with four-momentum. For one canonical chiral superfield, fluctuations about $W^{\prime }(v)=0$ have mass $|W^{\prime \prime }(v)|$ for both the complex scalar field and Weyl spinor.

A supersymmetry generator is an odd Weyl spinor generator of the Super-Poincaré algebra. Acting on a physical state reverses fermion parity while preserving its four-momentum.

Choose $m_{\tau }=2\mathrm{GeV}$ and $\sqrt{s}=6\mathrm{GeV}$ purely for illustration. Then each parent has $E_{\tau }=3\mathrm{GeV}$, $|{\mathbf{p}}_{\tau }|=\sqrt{5}\mathrm{GeV}$ and $\beta =\sqrt{5}/3$. Treat all six daughters as massless, including the charged electron and antimuon, and generate each three-body particle decay with constant squared decay amplitude. Choose the parent direction isotropically and the two decays independently; this omits the physical production angular law and spin correlations.

The Lorentz-invariant phase space factorizes through $t=q^2$ into a two-body decay to $\ell +q$, followed by a two-body decay of $q$:After angular integration the first factor contributes $1-t/m_{\tau }^2=x$, while the massless second factor is constant. Since $|dt|=m_{\tau }^{2}dx$, the normalized chosen density is $f(x)=2x$. Drawing $x=\sqrt{U}$ with $U$ uniform on $[0,1]$ implements this density. Isotropic directions, followed by a Lorentz boost for the invisible pair and another for the parent, conserve the complete four-momentum of each generated event.

Define transverse acoplanarity by $\alpha =\pi -\Delta \varphi$, where $\Delta \varphi \in [0,\pi ]$ is the angle between the two visible momentum projections perpendicular to the beam. A fully visible two-body event with zero initial transverse momentum has $\alpha =0$ and $M_{\mathrm{miss}}^2=0$. The toy tau lepton events spread away from that point. These observables are complementary: an angular imbalance asks about directions, while the missing invariant mass also uses the visible energies and longitudinal momenta.

A Lorentz-invariant phase space density supplies allowed configurations and their geometric measure. A physical weak interaction decay amplitude reweights them and correlates spins. Consequently the plotted density is a lesson about conservation and available configurations, not a predicted experimental rate or a fit to the discovery sample. The massless charged-daughter approximation is also unsuitable near the muon rest-energy threshold.

Special relativity connects measurable clock rates, length contraction, light propagation, and four-momentum through the Lorentz transformation. Tests probe particular predictions rather than measuring a coordinate transformation in isolation. Examples include cosmic-ray muon time-dilation experiments, muon storage-ring time-dilation experiments, the Michelson-Morley experiment, the Fizeau experiment, and particle-collision energy and momentum balances.

Three-momentum is the spatial part of a particle's four-momentum, $p^{\mu }=(E/c,\mathbf{p})$. Its components depend on the reference frame, while $p_{\mu }{p}^{\mu }=m^2{c}^2$ is a Lorentz invariant.

At non-null four-momentum, these complementary projectors satisfy $P_L^2=P_L$, $P_T^2=P_T$ and $P_L{P}_T=0$. The gauge-fixed Maxwell kinetic operator is $-k^2(P_T+{\xi }^{-1}{P}_L)$; its inverse gives propagator $-i(P_T+\xi P_L)/k^2$. Transversality here removes one four-dimensional longitudinal direction, leaving rank three; it does not by itself project onto the two physical photon polarizations. The pole requires the Feynman i-epsilon prescription.

For a two-body W boson decay into a charged lepton $\ell$ and a neutrino, neglect both daughter masses and use natural units $c=1$. Let $\Delta \varphi$ be the angle between the daughter transverse momenta. Identifying the missing transverse momentum with the neutrino's givesThis observable uses no $p_{z\nu }$. The 1983 UA1 experiment paper already used transverse mass as a lower bound on parent mass, before making additional decay-model assumptions for a mass estimate.^[ref]

The bound follows directly from the energy–momentum relation. For each massless daughter write its four-momentum, using rapidity $y$ and azimuth $\varphi$, asUsing the Minkowski inner product with positive time sign, $p^2=E^2-|\mathbf{p}{|}^2$, the two daughter four-momenta give the true parent invariant mass $M$:Since $\cosh u\ge 1$, $m_T\le M$, with equality at equal daughter rapidities when both transverse momenta are nonzero. Thus the unmeasured longitudinal information can enlarge the true invariant mass, but cannot lower it below this ideal transverse mass.

For an original example, stipulate $p_{T\ell }=p_{T\nu }=40\mathrm{GeV}$ and $\Delta \varphi =\pi$. Then $m_T=80\mathrm{GeV}$. Equal rapidities give $M=80\mathrm{GeV}$; choosing $y_{\ell }-y_{\nu }=1$ with exactly the same transverse readout gives $M=80\cosh (1/2)\simeq 90.210\mathrm{GeV}$. One measured transverse configuration therefore permits different true parent masses.

Choose a temporal polarization vector, two transverse polarization vectors and a spatial longitudinal polarization vector. The Gupta-Bleuler quantization condition is $(a^0-a^3)|\chi ⟩=0$. Applied to ${\sum }_{\lambda }{c}_{\lambda }{a}^{\lambda †}|0⟩$, it gives $c_3=-c_0$. The remaining temporal-longitudinal excitation $a^{0†}-a^{3†}$ has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to the null four-momentum, hence is pure gauge. Quotienting this direction leaves the two positive transverse photon states. Use wave packets or a box normalization to avoid treating continuum momentum eigenstates as normalizable vectors.

At a nonzero Euclidean momentum $k$, the transverse projector of a vector field is $P_T^{ab}={\delta }^{ab}-k^a{k}^b/k^2$. In a Lorentzian Minkowski metric, its mixed-index form is $(P_T{)}^a{}_b={\delta }_b^a-p^a{p}_b/p^2$, defined for $p^2\ne 0$. It obeys $P_{T}p=0$ and $P_T^2=P_T$. A nonzero null four-momentum is insufficient for this particular decomposition because the denominator vanishes.

Two particles with unequal rest energies $r,R$ and equal opposite velocities have total energy $\gamma (r+R)$ but nonzero total momentum. If they form a composite while emitting a photon transverse to their initial axis, the final composite rest energy $Q$ satisfies the displayed identity. It follows by squaring the residual four-momentum. Neglecting binding, put $Q=r+R$; the minimum incoming energy is $(r+R)[\chi +\sqrt{1+{\chi }^2}]$, where $\chi =\epsilon (r+R)/(4rR)$. An exact threshold requires the composite rest mass, not just the incoming constituent masses.

The interaction $\lambda \Phi F_{\mu \nu }{F}^{\mu \nu }/4$ gives the displayed on-shell scattering amplitude, up to an overall phase, for a scalar field decaying to two photons. Substituting either photon four-momentum for its polarization vector makes the expression vanish, verifying its Ward identity. In the parent rest frame, summing the two physical transverse photon polarizations gives $\sum |\mathcal{A}{|}^2=|\lambda {|}^2{M}^4/2$. Including the identical-particle factor in a final-state phase-space integral gives $\Gamma =|\lambda {|}^2{M}^3/(64\pi )$.

For a massive spin-zero one-particle state with covariant normalization and a Lorentz scalar field, Lorentz invariance makes $F=⟨\Omega |\varphi (0)|p⟩$ independent of the on-shell four-momentum within a fixed species. A Lorentz boost to the rest frame, invariance of the vacuum state and scalar transformation at the origin prove this. Translation covariance of a quantum field then supplies the plane-wave factor. With noncovariant normalization, the boost law and the displayed momentum independence acquire energy factors.

If a fixed component of a Lorentz four-vector vanishes in every Lorentz frame, the vector is zero. For its time component, arbitrary infinitesimal boosts force every spatial component to vanish; for a spatial component, first use rotations and then a boost. Applied to the incoming-minus-outgoing total four-momentum, this shows that energy balance in every inertial frame implies full four-momentum conservation. The statement fails for general tensors: a nonzero antisymmetric tensor has zero diagonal components in all frames.