Mass shell inversion, observer work and variable mass — Examples

For any massive leg $P=aU$ of [F1], write $U=(U^0,\mathbf{U})$ in the laboratory, with $U^0>0$ by future orientation. The auxiliary curve $y:\mathbb{R}\to A$, $y(\tau )=p+\tau U$, is smooth and future timelike, and $\tau$ is proper time since $\eta (y^{\mathrm{\prime }},y^{\mathrm{\prime }})=-c^2$. Its laboratory time is $t=t_p+\tau U^0\mathrm{/}c$, where $t_p$ is the event's time coordinate, so it admits a smooth time parametrization on the open interval $\mathbb{R}$ with velocity ${\mathbf{v}}_P=c\mathbf{U}\mathrm{/}{U}^0$. Assign constant mass $a$ and momentum $ady\mathrm{/}d\tau =P$; [F2] therefore applies. This velocity depends only on the leg's normalized tangent and agrees with the coordinate velocity of any massive trajectory with that tangent; the auxiliary curve assumes no smooth continuation of the collision itself. Applying these identities to the specified incoming velocities gives $P_1=(\gamma mc,\gamma m\mathbf{v})$ and $P_2=(\gamma mc,-\gamma m\mathbf{v})$. Thus $Q_{\mathrm{i}\mathrm{n}}=P_1+P_2=(2\gamma mc,\mathbf{0})$, which is future timelike and has invariant mass $\sqrt{-\eta (Q_{\mathrm{i}\mathrm{n}},Q_{\mathrm{i}\mathrm{n}})}\mathrm{/}c=2m\gamma$.

The output is one massive leg at rest, so the pointwise identities established in step 1.1 give $P_{\mathrm{o}\mathrm{u}\mathrm{t}}=(Mc,\mathbf{0})$. Balance [A1] gives $P_{\mathrm{o}\mathrm{u}\mathrm{t}}=Q_{\mathrm{i}\mathrm{n}}$ and therefore $M=2m\gamma$. Subtracting the two incoming rest masses gives $M-2m=2m(\gamma -1)$; multiplying by $c^2$ gives the rest-energy increase. The incoming total energy is $2\gamma m{c}^2$ and its rest-energy baseline is $2m{c}^2$, so the same difference is the incoming kinetic energy, defined as energy minus rest energy. Since $\gamma =1$ exactly at $\mathbf{v}=0$, the zero-speed boundary gives zero excess.

For either massive leg $P=aU$ of [F1], write $U=(U^0,\mathbf{U})$ in the parent rest chart, with $U^0>0$ by future orientation. The auxiliary affine curve $y:\mathbb{R}\to A$, $y(\tau )=p+\tau U$, is smooth and future timelike with proper-time parameter $\tau$ since $\eta (y^{\mathrm{\prime }},y^{\mathrm{\prime }})=-c^2$. Its chart time is $t=t_p+\tau U^0\mathrm{/}c$, where $t_p$ is the event's time coordinate, so it has a smooth time parametrization on the open interval $\mathbb{R}$. Assign constant mass $a$ and momentum $ady\mathrm{/}d\tau =P$. This meets every hypothesis of [F3], giving the shell and ${\mathbf{v}}_P=c\mathbf{U}\mathrm{/}{U}^0=c^2\mathbf{p}\mathrm{/}E$. This velocity depends only on the leg's normalized tangent and agrees with any trajectory's coordinate velocity at that tangent; no smooth continuation through the decay is assumed. For the rest parent, $\mathbf{U}=0$ and normalization gives $U^0=c$, hence its total is $(Mc,\mathbf{0})$ with $W=M{c}^2$. Balance [A1] therefore makes this rest frame the final total's center-of-momentum frame. Apply [F2] with outgoing masses $a=m$ and $b=0$. Since $M>m$, the strict condition $W>m{c}^2$ holds, so the null output is nonzero. The formula for $k$ gives $k=\sqrt{[W^2-m^2{c}^4{]}^2}\mathrm{/}(2Wc)=(M^2-m^2)c\mathrm{/}(2M)$; the energy formulas give the stated $E_{\mathrm{d}}$ and $E_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}$ and opposite spatial momenta.

The daughter velocity identity established in step 1.1 gives $\mathrm{\mid }{\mathbf{v}}_{\mathrm{d}}\mathrm{\mid }=c^{2}k\mathrm{/}{E}_{\mathrm{d}}=(M^2-m^2)c\mathrm{/}(M^2+m^2)<c$, since $0<M^2-m^2<M^2+m^2$. Also $E_{\mathrm{d}}-m{c}^2=(M-m{)}^2{c}^2\mathrm{/}(2M)$, so $E_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}+(E_{\mathrm{d}}-m{c}^2)=(M-m)c^2$; for $0<m<M$, this daughter kinetic energy, defined as energy minus rest energy, is positive and $E_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}<(M-m)c^2$.