Mass shell inversion, observer work and variable mass

Let $I\subseteq \mathbb{R}$ be a nonempty open interval, and let $x:I\to A$ be a $C^2$ future-timelike worldline in the flat Minkowski model of Particle spacetime, observers and worldlines. Use one inertial frame and parameterize the curve by that chart's time coordinate $t$, so its coordinates are $(ct,\mathbf{x}(t))$ (Inertial frames and inertial charts of the affine Minkowski model). Identify the observer rest space with ${\mathbb{R}}^3$ by the frame's spatial orthonormal basis and use its Euclidean dot product and norm (The Euclidean spatial structure and cross product of an inertial frame). Write $\mathbf{v}=d\mathbf{x}\mathrm{/}dt$, $\gamma =(1-\mathrm{\mid }\mathbf{v}{\mathrm{\mid }}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$, proper time $\tau$, and $U=dx\mathrm{/}d\tau$. In the constant-mass sector of Adopted massive-particle momentum, observer measurement and isolated-collision conservation, let $m>0$ be constant and $P=mU$. In this chart write $P=(E\mathrm{/}c,\mathbf{p})$, where $E$ and $\mathbf{p}$ are the observer energy and spatial momentum of Mechanical momentum and observer energy. Then $\mathrm{\mid }\mathbf{v}\mathrm{\mid }<c$, $E=\gamma m{c}^2>0$, $\mathbf{p}=\gamma m\mathbf{v}$, and

$E^2=m^2{c}^4+c^2\mathrm{\mid }\mathbf{p}{\mathrm{\mid }}^2,\mathbf{v}=\frac{c^2\mathbf{p}}{E}\mathrm{.}$

Define the three-force by $\mathbf{F}:=d\mathbf{p}\mathrm{/}dt$ and the four-force by $K:=dP\mathrm{/}d\tau$ in this flat inertial chart. In SI units $m$ is in kilograms, $c$ and $\mathbf{v}$ in metres per second, $E$ in joules, $\mathbf{p}$ in kilogram metres per second, and $\mathbf{F},K$ in newtons. Then

$\frac{dE}{dt}=\mathbf{F}\cdot \mathbf{v},K=\gamma (\frac{\mathbf{F}\cdot \mathbf{v}}{c},\mathbf{F})\mathrm{.}$

Separately, let $I_{\tau }\subseteq \mathbb{R}$ be a nonempty open proper-time interval, let $m:I_{\tau }\to (0,\mathrm{\infty })$ be $C^1$, and let $U:I_{\tau }\to V$ be a $C^1$ future timelike vector field with $\eta (U,U)=-c^2$. Measure $m$ in kilograms and $dm\mathrm{/}d\tau$ in kilograms per second, and set $P(\tau ):=m(\tau )U(\tau )$. With $K:=dP\mathrm{/}d\tau$, the kinematic ansatz satisfies

$\eta (P,K)=-m\frac{dm}{d\tau }{c}^2\mathrm{.}$

The last identity is only a differentiated mass-shell relation; it prescribes no evolution for $m$. None of these conclusions assumes a force law.

Let $(A,V,\eta ,C^{+})$ be the oriented time-oriented affine Minkowski model with $c>0$, and fix an inertial frame. For a finite nonempty list $P_1,\dots ,P_N$ of nonzero future-causal vectors in $V$, let $Q:={\sum }_{i=1}^N{P}_i$, let $m_i:=\sqrt{-\eta (P_i,P_i)}\mathrm{/}c$ (so $m_i=0$ exactly for a null leg), and let $M(Q)$ be the invariant mass of the total when $Q$ is timelike, as defined in Total four-momentum, invariant mass and center-of-momentum frame of a finite system.

Then $Q$ is nonzero and future causal. If at least one $P_i$ is timelike, then $Q$ is future timelike. If all $P_i$ are null, then $Q$ is null exactly when all $P_i$ lie on the same future null ray; otherwise $Q$ is future timelike. Whenever $Q$ is timelike, the center-of-momentum frame exists, its time-axis observer $n_Q=Q\mathrm{/}(M(Q)c)$ has zero total spatial momentum and total observer energy $M(Q)c^2$, and

$M(Q)\ge {\sum }_{i=1}^N{m}_i\mathrm{.}$

Equality holds exactly when every $P_i$ is massive and there is one future unit timelike vector $n$ such that $P_i=m_{i}cn$ for every $i$. A nonzero future-null total has no center-of-momentum frame.

Use the flat oriented, time-oriented affine Minkowski model, signature $(-,+,+,+)$, and SI convention $x^0=ct$ of Particle spacetime, observers and worldlines. A relativistic-particle-mechanics collision configuration is a mass-specified causal collision datum of Collision data and four-momentum conservation: an event $p\in A$, a finite nonempty incoming list $P_1,\dots ,P_N\in V$, and a finite nonempty outgoing list $P_1^{\mathrm{\prime }},\dots ,P_K^{\mathrm{\prime }}\in V$, where every leg is a nonzero future-causal vector in the same displacement space $V$. A massive leg has a recorded mass $m>0$ and the massive-sector momentum of Mechanical momentum and observer energy; a null leg has recorded mass $0$ and is a nonzero future-null vector with its scale specified independently. The definition does not identify a null momentum with $mU$ at $m=0$.

The side totals are the finite vector sums $Q_{\mathrm{i}\mathrm{n}}:={\sum }_{i=1}^N{P}_i$ and $Q_{\mathrm{o}\mathrm{u}\mathrm{t}}:={\sum }_{j=1}^K{P}_j^{\mathrm{\prime }}$. All legs are attached to the same event, so the sums compare vectors in one affine displacement space; no transport between different events is part of the datum. The adjective isolated is reserved for a model condition used by the postulate below: no external four-momentum transfer during the idealized event and no omitted incoming or outgoing leg.

At an event, an observer is a future unit timelike vector $n$ with $\eta (n,n)=-1$, and its rest space is $n^{\perp }=\{v\in V:\eta (v,n)=0\}$, as in Particle spacetime, observers and worldlines. Observer energy $E_n(P):=-c\eta (P,n)$ and spatial momentum $q_n(P):=P+\eta (P,n)n\in n^{\perp }$ are those of Mechanical momentum and observer energy. Their units are joules and kilogram metres per second. An inertial laboratory means one named inertial chart and its time-axis observer; it is not a general observer congruence or an arbitrary coordinate chart.

For the relativistic-particle-mechanics collision configurations of Finite event collisions and observer quantities, adopt the isolated-collision balance already stated in clause (iii) of Adopted massive-particle momentum, observer measurement and isolated-collision conservation: when the event is isolated and the incoming and outgoing lists contain every leg, $Q_{\mathrm{i}\mathrm{n}}=Q_{\mathrm{o}\mathrm{u}\mathrm{t}}$. This local postulate records the exact collision interface used on this page; it is a scoped reuse of the existing adopted model assumption, not a second independent conservation assumption.

The balance is an adopted physical premise. It does not show that any specified reaction occurs and supplies no interaction mechanism, selection rule, probability, cross section, or amplitude. The cited treatments motivate and explain the model formulation; they do not turn the postulate into a deductive theorem or a report of an observed collision.

Let a mass-specified isolated collision configuration of Finite event collisions and observer quantities satisfy the balance postulate Isolated event four-momentum balance. Write its common total as $Q:=Q_{\mathrm{i}\mathrm{n}}=Q_{\mathrm{o}\mathrm{u}\mathrm{t}}$. It is nonzero future causal, and it is timelike if either side contains a massive leg. If $Q$ is future timelike, let $M:=\sqrt{-\eta (Q,Q)}\mathrm{/}c$, and let ${\mu }_{\mathrm{i}\mathrm{n}}$ and ${\mu }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ be the sums of the recorded rest masses on the two sides. Then

$M\ge {\mu }_{\mathrm{i}\mathrm{n}},M\ge {\mu }_{\mathrm{o}\mathrm{u}\mathrm{t}}\mathrm{.}$

In particular, a proposed final collection with total rest mass $\mu$ requires the invariant squared energy $\sigma :=-c^2\eta (Q,Q)=M^2{c}^4$ to satisfy $\sigma \ge {\mu }^2{c}^4$. If an actual final collection contains a nonzero null leg, then the inequality is strict. Equality is kinematically attained exactly when every leg on the relevant side is massive and all its four-momenta are proportional to the same future unit timelike vector. This is a necessary kinematic condition; it does not establish that the interaction realizes the channel.

If instead $Q$ is future null, each side consists only of null legs on one common future null ray; no center-of-momentum frame exists and no positive-mass final leg is compatible with that total. In SI units $m_i,M,\mu$ are kilograms, $M{c}^2$ is joules, and $\sigma$ is joules squared.

In a named inertial laboratory frame, consider a mass-specified isolated collision configuration of Finite event collisions and observer quantities whose complete incoming list consists exactly of two massive legs: a projectile of rest mass $m_1>0$ and laboratory momentum $P_1=(E_1\mathrm{/}c,{\mathbf{p}}_1)$, and a target of rest mass $m_2>0$ initially at rest, with momentum $P_2=(m_{2}c,\mathbf{0})$. Assume the adopted balance postulate Isolated event four-momentum balance applies. Let the complete outgoing list be a proposed finite nonempty mass-specified final state, and let $\mu \ge 0$ be the sum of its recorded rest masses. Thus its total momentum equals $Q:=P_1+P_2$. Define

$\sigma :=-c^2\eta (Q,Q)\mathrm{.}$

Then

$\sigma =m_1^2{c}^4+m_2^2{c}^4+2m_2{c}^2{E}_1\mathrm{.}$

The necessary final-state threshold from System invariant mass and final-state thresholds is $\sigma \ge {\mu }^2{c}^4$, hence

$E_1\ge \frac{({\mu }^2-m_1^2-m_2^2)c^2}{2m_2},E_1\ge m_1{c}^2\mathrm{.}$

If the proposed complete final list contains a nonzero null leg, the invariant-mass inequality is strict, $\sigma >{\mu }^2{c}^4$; the displayed invariant threshold can then be an unattained endpoint. These are kinematic necessities, not guarantees that the collision or channel occurs.

In SI units $m_1,m_2,\mu$ are kilograms, $E_1$ is joules, and $\sigma$ is joules squared.

Let a finite isolated collision have future-timelike total $Q$ of invariant mass $M>0$, and set $W:=M{c}^2$. Suppose the complete outgoing list consists of exactly two legs, with recorded rest masses $a,b\ge 0$; by the adopted balance postulate, their four-momenta must sum to $Q$. In a center-of-momentum frame with time-axis observer $n_Q$, a nonzero two-product state with ${\mathbf{p}}_a+{\mathbf{p}}_b=0$ exists exactly when

$W\ge (a+b)c^2\text{if }a,b>0,W>(a+b)c^2\text{if }a=0\text{ or }b=0.$

When it exists, there are $k\ge 0$ and a unit vector $\mathbf{e}\in n_Q^{\perp }$ such that ${\mathbf{p}}_a=k\mathbf{e}$ and ${\mathbf{p}}_b=-k\mathbf{e}$. The energies and momentum magnitude are

$E_a=\frac{W^2+(a^2-b^2)c^4}{2W},E_b=\frac{W^2+(b^2-a^2)c^4}{2W},k^2=\frac{[W^2-(a+b{)}^2{c}^4][W^2-(a-b{)}^2{c}^4]}{4W^2{c}^2}\mathrm{.}$

For $a,b>0$ at threshold, $k=0$ and the direction is immaterial. If a massless product is present, the strict threshold is required because a zero momentum would be the forbidden zero four-vector. These equations describe kinematic balance only. In SI units $a,b,M$ are kilograms, $W,E_a,E_b$ are joules, and $k$ is kilogram metres per second; the norm and dot product on $n_Q^{\perp }$ are induced by $\eta$.

Let $P_1,P_2$ be two nonzero future-null incoming momenta in a finite collision configuration, and let $n$ be the future unit observer of a named inertial laboratory frame. Define $E_i:=E_n(P_i)>0$, ${\mathbf{p}}_i:=q_n(P_i)$, and equip $n^{\perp }$ with the induced Euclidean inner product $⟨u,v{⟩}_n:=\eta (u,v)$ and norm $\mathrm{\mid }u{\mathrm{\mid }}_n:=\sqrt{⟨u,u{⟩}_n}$. Let $\theta \in [0,\pi ]$ be their spatial angle, $\cos \theta =⟨{\mathbf{p}}_1,{\mathbf{p}}_2{⟩}_n\mathrm{/}(\mathrm{\mid }{\mathbf{p}}_1{\mathrm{\mid }}_n\mathrm{\mid }{\mathbf{p}}_2{\mathrm{\mid }}_n)$. For $Q:=P_1+P_2$ and $\sigma :=-c^2\eta (Q,Q)$,

$\sigma =2E_1{E}_2(1-\cos \theta ),M(Q)=\frac{\sqrt{\sigma }}{c^2}\mathrm{.}$

The total $Q$ is null exactly when $\theta =0$ (the two incoming vectors lie on one future null ray); it is timelike for $\theta >0$. In particular, for head-on momenta $\theta =\pi$, $M(Q)=2\sqrt{E_1{E}_2}\mathrm{/}{c}^2$; if $E_1=E_2=E$, this is $2E\mathrm{/}{c}^2$. The angle and the two energies depend on the named observer, while $\sigma$ and $M(Q)$ are invariant.

In SI units $E_1,E_2$ are joules, $\sigma$ is joules squared, and $M(Q)$ is kilograms; $\theta$ is dimensionless.

Consider a two-product elastic channel in a finite isolated collision: the complete incoming and outgoing lists each have exactly two legs, their recorded rest masses agree pairwise, the total four-momentum is timelike, and the balance postulate is adopted. In the center-of-momentum frame, let the incoming momenta be $k{\mathbf{e}}_{\mathrm{i}\mathrm{n}}$ and $-k{\mathbf{e}}_{\mathrm{i}\mathrm{n}}$ in the spatial rest space $n_Q^{\perp }$, where $k\ge 0$ is fixed by Complete two-body rest-frame kinematics. Every admissible outgoing pair has the same momentum magnitude $k$. If $k>0$, any unit vector ${\mathbf{e}}_{\mathrm{o}\mathrm{u}\mathrm{t}}\in n_Q^{\perp }$ gives an on-shell outgoing pair $k{\mathbf{e}}_{\mathrm{o}\mathrm{u}\mathrm{t}},-k{\mathbf{e}}_{\mathrm{o}\mathrm{u}\mathrm{t}}$ with the same total four-momentum; thus kinematics alone leaves the full direction sphere available. If $k=0$, both products are at rest and the direction is undefined. No isotropic or uniform distribution of directions follows.