= Solution
\b[Momentum identity.] From the <Lorentz factor>, $v^2/c^2=1-\gamma^{-2}$. The relativistic <three-momentum> therefore satisfies
$$
p^2=\gamma^2m^2v^2=m^2c^2\gamma^2(1-\gamma^{-2}),
\qquad
\boxed{p^2=m^2c^2(\gamma^2-1).}
$$
Together with <relativistic energy> $E=\gamma mc^2$, this gives the <energy–momentum relation> $E^2-p^2c^2=m^2c^4$.
\b[Parent in the laboratory.] The <rest energy> is $m_0c^2=1\,\mathrm{GeV}$, while the total <relativistic energy> is $3\,\mathrm{GeV}$. Consequently
$$
\gamma_0=3,\qquad
\boxed{\frac{v_0}{c}=\sqrt{1-\frac19}=\frac{\sqrt8}{3}=0.942809.}
$$
Its <momentum> magnitude is $p_0c=\sqrt{3^2-1^2}\,\mathrm{GeV}=\sqrt8\,\mathrm{GeV}$.
\b[Daughters in the zero-momentum frame.] The <centre-of-momentum frame> is initially the parent's rest frame. <Conservation of momentum> gives $\mathbf p_1^*=-\mathbf p_2^*$. Equal daughter <rest masses> then imply equal <relativistic energies>. <Conservation of relativistic energy> gives
$$
E_X^*=\frac{m_0c^2}{2}=\frac12\,\mathrm{GeV},\qquad
\gamma_X=\frac{E_X^*}{m_Xc^2}
=\frac{m_0/2}{m_0/3}=\frac32.
$$
Their <speeds> are therefore
$$
\boxed{\gamma_X=\frac32,\qquad \frac{v_X^*}{c}
=\sqrt{1-\frac49}=\frac{\sqrt5}{3}=0.745356.}
$$
The two <velocities> are opposite, and $p_X^*c=\sqrt{(1/2)^2-(1/3)^2}\,\mathrm{GeV}=\sqrt5\,\mathrm{GeV}/6$. Their total <kinetic energy> is $m_0c^2-2m_Xc^2=m_0c^2/3$.
\b[Invariant of the complete system.] Before the decay, in the <centre-of-momentum frame>,
$$
\mathcal I_{\rm before}=(m_0c^2)^2-0=m_0^2c^4.
$$
After the <relativistic two-body decay>,
$$
\mathcal I_{\rm after}=(E_X^*+E_X^*)^2
-c^2|\mathbf p_X^*+(-\mathbf p_X^*)|^2
=(m_0c^2)^2.
$$
In the laboratory, <four-momentum conservation> fixes the summed daughter energy to $3\,\mathrm{GeV}$ and their summed <momentum> to the parent's $\mathbf p_0$. Thus the same calculation is
$$
\boxed{\mathcal I_{\rm before}=\mathcal I_{\rm after}
=(3\,\mathrm{GeV})^2-(\sqrt8\,\mathrm{GeV})^2
=1\,\mathrm{GeV}^2=m_0^2c^4.}
$$
The invariant is formed after summing the daughter <four-momenta>; adding their separate squared rest energies would omit the relative-motion contribution to the invariant <mass> of the pair.
Back to article page
