= Solution
For <photons> emitted back to back in the measurement frame, their total invariant <mass> obeys $m^2c^4=(E_1+E_2)^2-(E_1-E_2)^2=4E_1E_2$. Thus
$$
\boxed{E_2=\frac{(125\,\mathrm{GeV})^2}{4(30\,\mathrm{GeV})}\simeq130.2\,\mathrm{GeV}.}
$$
The parent's total laboratory <energy> is consequently about $160.2\,\mathrm{GeV}$, exceeding its rest <energy>. The measured-frame opposite-directions condition is essential: opposite directions only in the parent's rest frame would hold automatically and would not determine the second laboratory <energy> by itself.
Back to article page
