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Let's do some basic calculation. The kinetic energy of the postron 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 have 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.

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.

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.

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.

Let's do some basic calculation. The kinetic energy of the postron 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 two 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 have 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.

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.

Let's do some basic calculation. The kinetic energy of the postron 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 have 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.

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.

Let's do some basic calculation. The kinetic energy of the postron 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 two 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 have 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.

Let's do some basic calculation. The kinetic energy of the postron 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 two 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 have 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.

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.