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A massive vector in $D$ spacetime dimensions has $D-1$ polarizations perpendicular to its timelike momentum. A massless vector has transverse polarizations modulo its null longitudinal polarization, leaving $D-2$ physical degrees of freedom.

Here $\Delta \psi =\psi (v^{\prime })+\psi (v_{\ast }^{\prime })-\psi (v)-\psi (v_{\ast })$ is the pairwise collision change, not a Laplace operator. Average over the two particle labels and use collision reversal to obtain the formula. Every collision invariant has $\Delta \psi =0$, giving mass, momentum and kinetic energy conservation. A second symmetrization replaces $f{f}_{\ast }\Delta \psi /2$ by $(f^{\prime }{f}_{\ast }^{\prime }-f{f}_{\ast })(-\Delta \psi )/4$.

An einbein specifies a one-dimensional metric tensor along a worldline. In the relativistic particle phase-space action it is the Lagrange multiplier for the mass-shell condition and transforms as $\delta e=\dot{ϵ}$ under the corresponding canonical gauge transformation. After eliminating the momentum, its geometric interpretation depends on the normalization of the particle action.

When matter leaves a moving body with zero instantaneous relative velocity, it carries the body's current momentum per unit mass but creates no additional recoil term. A short-interval momentum balance including the newly released material gives $q\dot{v}=F_{\mathrm{ext}}$ for the remaining body's mass $q(t)$. The expression $\frac{d}{dt}(qv)=F_{\mathrm{ext}}$ omits the outgoing momentum flux. Nonzero relative ejection velocity instead produces the thrust term of the rocket equation.