Center-of-momentum frames of a future timelike four-momentum and their unique time vector

Since $Q$ is future timelike, $-\eta (Q,Q)>0$ and hence $M>0$ and $n_Q=Q\mathrm{/}(Mc)$ is defined; by [F10] and $\eta (Q,Q)=-M^2{c}^2$, $\eta (n_Q,n_Q)=(M^2{c}^2{)}^{-1}\eta (Q,Q)=-1$ and $Q=Mc{n}_Q$ with $Mc>0$, so $Q$ is a positive multiple of the unit vector $n_Q$. Moreover $\eta (Q,n_Q)=Mc\eta (n_Q,n_Q)=-Mc$, whence $h(Q)=Q+\eta (Q,n_Q)n_Q=Mc{n}_Q-Mc{n}_Q=0$ and $E_{n_Q}(Q)=-c\eta (Q,n_Q)=M{c}^2$ by [F2].

By [F9] the future component $C^{+}$ of the timelike set is one of the two convex cones $T_{\pm }$, each closed under multiplication by positive reals; since $Q\in C^{+}$ and $n_Q=(Mc{)}^{-1}Q$ with $(Mc{)}^{-1}>0$, [F10] gives $n_Q\in C^{+}$.

By [F4] the restriction $\delta$ of $\eta$ to $n_Q^{\perp }$ is positive definite, so by [F5] there is a $\delta$-orthonormal basis $(e_1,e_2,e_3)$ of $n_Q^{\perp }$; setting $e_0:=n_Q$, the ordered basis $(e_0,e_1,e_2,e_3)$ is $\eta$-orthonormal, because $\eta (e_0,e_0)=\eta (n_Q,n_Q)=-1$, $\eta (e_0,e_i)=0$ by $e_i\in n_Q^{\perp }$, and $\eta (e_i,e_j)=\delta (e_i,e_j)={\delta }_{ij}$ for $i,j\in \{1,2,3\}$.

For the converse, let $(o;e_0,e_1,e_2,e_3)$ be an inertial frame with $q_{e_0}(Q)=0$. By [F1] and [F2], $Q=-\eta (Q,e_0)e_0$ and $\eta (Q,e_0)=-E_{e_0}(Q)\mathrm{/}c$, and [F3] gives $E_{e_0}(Q)>0$ because $Q$ is nonzero future causal, so $\lambda :=-\eta (Q,e_0)>0$ and $Q=\lambda e_0$. Then $\eta (Q,Q)={\lambda }^2\eta (e_0,e_0)=-{\lambda }^2$ by [F10] and $\eta (e_0,e_0)=-1$, while $\eta (Q,Q)=-M^2{c}^2$ by definition of $M$, so ${\lambda }^2=M^2{c}^2$ and $\lambda =Mc$ because $\lambda >0$ and $Mc>0$. Substituting $Q=\lambda e_0=Mc{e}_0$ into $Q=Mc{n}_Q$ and dividing by $Mc\mathrm{\ne }0$ gives $e_0=n_Q$.

If $Q$ is a nonzero future null vector and an inertial frame satisfied $q_{e_0}(Q)=0$, then [F1] gives $Q=-\eta (Q,e_0)e_0=\lambda e_0$ with $\lambda \mathrm{\ne }0$ because $Q\mathrm{\ne }0$; then $\eta (Q,Q)={\lambda }^2\eta (e_0,e_0)=-{\lambda }^2<0$ by [F10], contradicting $\eta (Q,Q)=0$ for a null vector in [F11]. Hence no inertial frame satisfies $q_{e_0}(Q)=0$, which is (c).

The ordered basis of [step 1.3] has some orientation class; if that class is negative, replace $e_1$ by $-e_1$, which leaves $\eta$-orthonormality and the first vector unchanged and reverses the sign of the determinant of the coordinate matrix, hence changes the orientation class, by [F7] and [F8]. The resulting $(e_0,e_1,e_2,e_3)$ is $\eta$-orthonormal, future (its first vector is $e_0=n_Q\in C^{+}$ by [step 1.2]) and positively oriented, so it is the basis of an inertial frame by [F6].

Two inertial frames whose first vectors both equal $n_Q$ differ by their choice of origin $o$, which is arbitrary, and by their spatial triples $(e_1,e_2,e_3)$ and $(f_1,f_2,f_3)$; both triples are $\delta$-orthonormal and positively oriented bases of the same oriented Euclidean space $n_Q^{\perp }$, so the change-of-basis map between them preserves $\delta$ and the induced orientation, that is it is a spatial rotation. Together with [step 1.4] this gives the uniqueness statement of (b): the time vector is forced to be $n_Q$.

In the frame of [step 2.1] the first vector is $e_0=n_Q$, so $q_{e_0}(Q)=q_{n_Q}(Q)=0$ and $E_{e_0}(Q)=M{c}^2$ by [step 1.1]; by [F1] this frame is a center-of-momentum frame of $Q$, and the components of $Q=Mc{n}_Q=Mc{e}_0$ are $(Mc,0,0,0)$. This proves (a).

Applying the spatial-rotation freedom of [step 2.2] to the frames produced by [step 2.1], two center-of-momentum frames of $Q$ differ only by origin choice and a spatial rotation, and by [step 1.4] no other time vector is possible; this completes (b).

By [step 3.1] clause (a) holds, by [step 3.2] clause (b) holds, and by [step 1.5] clause (c) holds; hence an inertial frame with $e_0=n_Q$ exists, its time vector is unique up to the stated rotations, and a nonzero future null vector has no center-of-momentum frame.