Beta decay kinetic energy affecting resultant gamma-ray angle

Question

My class has recently been studying gamma-ray coincidence and PET. Na-22 decays into $\beta^+$ (plus a neutrino), which then travels along in the material of a sample until it annihilates with an electron around one of the other atoms in the sample bulk. When that happens, two gamma rays are released at an energy of 0.511 MeV; they show up in a spectrum, can be used in coincidence tomography, etc.

These gamma rays are emitted back-to-back: this is because the primary source of energy for the coincident gamma rays is the rest mass energy of the electron and positron, not their pre-collision 3-momenta, because we are in a "low energy" situation. If we were to have a relatively high 3-momentum of the positron, the resultant gamma rays would be emitted at some angle deviating from $180^\circ$ to conserve 3-momentum.

The way this effect is presented is that this (high-energy situation) doesn't in fact happen naturally in Na-22: the gamma rays are emitted almost perfectly back-to-back, and this is the backbone of the entirety of PET. However, when I try to look up just much of a deviation can occur, it seems as though the (kinetic) energies of $\beta$ radiated out from Na-22 are in fact comparable to their rest masses ($\sim0.215$ MeV up to $1.820$ MeV from http://www.nucleide.org/DDEP_WG/Nuclides/Na-22_tables.pdf and http://nrv.jinr.ru/nrv/webnrv/map/nucleus.php?q=Na22 if I understand the sources correctly).

So what am I missing? Does the $\beta$ lose effectively all of its energy through other forms of radiation (Bramsstrahlung, etc.) before interacting with an electron? If so, why do we not observe it interacting "sooner"? Or am I confused about something more fundamental?

Answer 1

Let's do some basic calculation. The kinetic energy of the positron can be as high as $1.8\,\mathrm{MeV}$, which makes it's Lorentz factor around \begin{align*} \gamma &= \frac{T}{m_e} + 1 \\ &= \frac{1.8\,\mathrm{MeV}}{511\,\mathrm{MeV}} + 1 \\ &= 1.0035 \,, \end{align*} so it's speed is \begin{align*} \beta &= \sqrt{1 - \frac{1}{\gamma^2}} \\ &= 0.083 \,. \end{align*} (All in $c=1$ units for convenience, of course).

OK, that's non-trivial and you might expect to see the deviation on some events. But you've got three things working for you

1. The frame in which the 2-gamma decay is absolutely back-to-back is the frame of the electron-positron pair at annihilation. So it's the center of momentum frame of that pair that you are worried about rather than the frame of the positron. For the purposes of a BotE calculation you can just halve the above: $\beta_{CoM} \approx 0.04$. This gives a maximum change in the opening angle as high as $2 \tan^{-1} \beta \approx 4.8^\circ$ or a opening angle as 'low' as $175.2^\circ$.

2. Events with high positron energy make up a relatively small fraction of the total spectrum.

3. Some fraction of the positrons will shed an appreciable fraction of their energy before annihilation.

From the point of view of an experimental nuclear or particle physicist I'd like to use a high-resolution tracking detector to measure the lab-frame opening angle of each event and reconstruct them better because of it, but that would add considerably to the cost of the device while giving little actual increase in precision.

As is so often the case, costs rule the cost-benefit analysis.

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As well as a change in the angle between the two photon you might also see a difference in their energies (if the CoM frame emission has a nonzero component only the relative velocity). The maximum size of that effect would be \begin{align*} \frac{E_{high}}{E_{low}} &= \frac{1 + \beta}{1 - \beta} \\ &\approx 1.08 \,. \end{align*} Again, you'd probably have to upgrade the detector to get much out of this and in the maximal case it wouldn't add anything to the precision of the resulting image.