= Recoil-free nuclear transition
= Recoilless nuclear transition
{synonym}
A <recoil-free nuclear transition> emits or absorbs a <gamma ray> while leaving the internal vibrational state of the surrounding <solid> unchanged. It belongs to the elastic part of the process: no <phonon> is created or destroyed. The <photon> still carries <momentum> $p=E_\gamma/c$, where $c$ is the <speed of light>. The emitting <solid> takes momentum $-\mathbf p$; the absorbing <solid> takes $+\mathbf p$. <Conservation of momentum> holds for each event.
Why does an isolated <atomic nucleus> pose a problem? Start with an <excited state> at rest and let $E_0$ be its transition <energy>, with nuclear <mass> $M$ and $E_0\ll Mc^2$. <Conservation of momentum> gives a recoil momentum of magnitude $p$; its nonrelativistic <kinetic energy> is therefore
$$
E_R=\frac{p^2}{2M}=\frac{E_\gamma^2}{2Mc^2}\simeq\frac{E_0^2}{2Mc^2}.
$$
By <conservation of energy>, emission spends part of $E_0$ on <nuclear recoil>, whereas absorption must supply both $E_0$ and the final <kinetic energy>. To leading order,
$$
E_{\mathrm{em}}\simeq E_0-E_R,\qquad E_{\mathrm{abs}}\simeq E_0+E_R,\qquad E_{\mathrm{abs}}-E_{\mathrm{em}}\simeq2E_R.
$$
Thus the relevant comparison is $2E_R$ against the overlapping <spectral line> widths, rather than $E_R/E_0$ alone. If $2E_R$ greatly exceeds the widths, the isolated-nucleus lines have negligible overlap. Finite <Lorentzian line shape> tails mean that this is suppression, rather than an absolute prohibition on absorption.
For a bound <atomic nucleus>, the recoil-free channel transfers <momentum> to the <center of mass> of the emitting or absorbing <solid>, with negligible <kinetic energy> $p^2/(2M_{\mathrm{solid}})$. Other channels exchange <phonons> and carry substantial <energy> into or out of the internal vibrations. A large $M_{\mathrm{solid}}$ makes the elastic channel's <energy> cost small; the vibrational <quantum state> determines its probability. Merely replacing $M$ by a large <mass> in the free-particle formula does not calculate that probability.
Here is a deliberately stipulated teaching system, independent of historical isotope data: $E_0=20\,\mathrm{keV}$, $Mc^2=10^{11}\,\mathrm{eV}$, and a single-line <natural linewidth> $\Gamma_E=10^{-7}\,\mathrm{eV}$ measured as an <energy> <full width at half maximum>. These give $E_R=0.002\,\mathrm{eV}$ and $2E_R/\Gamma_E=40{,}000$. A <solid> of $10^{12}$ such nuclear <masses> has elastic recoil <energy> $2\times10^{-15}\,\mathrm{eV}$, far below $\Gamma_E$. In this article $\Gamma_E$ is an <energy> width; the corresponding lifetime is $\tau=\hbar/\Gamma_E$, with $\hbar$ the <Planck constant> divided by $2\pi$.
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