Can annihilation in flight produce higher energy γ-rays?

[ksambourship]

I was wondering wether the γ-ray energy produced from the annihilation of a positron with a electron is always given only by the rest mass E=mc^2, that is 1022keV in total. I've read that the positron will slow down almost to rest before annihilation takes place. I've also read in http://prola.aps.org/abstract/PR/v77/i2/p205_1" that the thermalization time for positrons is small compared to the annihilation process. Nevertheless this does not prove that annihilation cannot take place in higher kinetic energies of the electron-positron pair. If that happens is it possible to get γ-rays with energies given by E=sqrt(p^2c^2 + m^2c^4). Experimentally this does not seem to be the case bacause only the photopeak coming from the pair annihilation appears exactly at 511keV whereas if the the γ-ray could also get part of the kinetic energie of the positron and the electron one would see a whole continuum of energies above the cutoff of 511keV.

 

[diazona]

No, in general the total energy of the photons produced by annihilation is the sum of the kinetic energies of the electron and positron, i.e.
$$E = \sqrt{p^2 c^2 + m^2 c^4}$$
If you've seen a distribution of energies which exhibits a peak at 511 keV, it's probably from a low-energy collision experiment.

I don't know what you've read about the positron slowing down almost to rest before annihilation takes place, but I can't think of any physical reason that would spontaneously happen. It would have to be colliding with something. Regardless, nothing prevents high-energy positrons and electrons from annihilating with each other; it happens all the time in particle accelerators these days. For example, the http://keyhole.web.cern.ch/keyhole/accelerators/LEP.html.

 

[enotstrebor]

No, in general the total energy of the photons produced by annihilation is the sum of the kinetic energies of the electron and positron, ...

This is correct. The question then becomes, if except for the invariant mass energy, the relativistic energy is reference frame energy, does this additional energy come from the reference frames smashing together?

What is a reference frame and how does it hold energy? How does the reference frame energy get accellerated allong with the particle and follow the particle so it to can collide?

This is another reason to believe that the reference frame view of the mathematics is wrong.

The reference frame is part of the matheology of physics. One no longer has to explain what it is or how it can be, if its mathematically correct then any view of the mathematics is ``reality'' even if this view is bad physics.

 

[jtbell]

No, in general the total energy of the photons produced by annihilation is the sum of the kinetic energies of the electron and positron, i.e.
$$E = \sqrt{p^2 c^2 + m^2 c^4}$$

That's what I call simply the energy of an electron (or positron), or sometimes the total energy if I want to be explicit. The kinetic energy is $K = E - mc^2$. (Here I take m to mean the invariant mass a.k.a. "rest mass"; I assume you do also.)

 

[ksambourship]

No, in general the total energy of the photons produced by annihilation is the sum of the kinetic energies of the electron and positron, i.e.
$$E = \sqrt{p^2 c^2 + m^2 c^4}$$

Although it seems reasonable I can't find any papaer stating that the energy of the 2 photons will exceed the total rest mass energy. Can you give me a citation on that??

 

[Bob S]

Positrons do annihilate in flight with a reasonably large cross section. Although the angular distribution of the annihilation gammas is isotropic in the center of mass, it is highly peaked in the forward direction in the Lab. The annihilation gammas appear as a peak at the top end of the bremsstrahlung spectrum. Experimenters sometimes use annihilation in flight to measure photo-production cross sections by doing a positron yield minus electron yield (photon difference method) to get photoproduction yield from the annihilation gamma. All the necessary formulas are available in papers by kendall and Deutsch about 1956. Also see Heitler "Quantum Theory of Radiation" (book) about positron annihilation in flight cross section.
Bob S

 

[Bob S]

The probability of positron annihilation in flight (annihilating before stopping) rises from about 5% at 1 MeV to about 18% at 50 MeV. See plot on page 385 of Heitler "Quantum Theory of Radiation", 3rd Edition (Oxford, 1954).
Bob S

 

[evilcman]

I think the question is supposed correspond to situations when a positron is passing through matter, and annihilates with one of the electrons electrons of the media, like how PET works. Right, ksambourship?

I think a good way of looking at this would be to look at the low energy cross section(if i remember correctly it is quite flat so maybe you can call it constant, well maybe not, you's better look it up...), which together with the elctron density gives you a mean free path without annihilation, and then to estimate how long it takes the positron to slow down by colliding with particles from the matter(maybe you can simply estimate with the Drude-model). The second estimate is supposed to be smaller then the first, so the electron first slows down, and then eliminates, meaning you will have a peak at 511keV at almost 180 degrees if you do coincidence mesurement. I never actually calculated this...

 

[Bob S]

As stated earlier, the probability of positron annihilation in flight in matter is plotted in Heitler's book on page 385. At 1 MeV, the probability of NOT annihilating in flight is about 95%, and at 50 KeV the probability of NOT annihilating in flight is over 99%. Because the Bethe-Bloch dE/dx ionization energy loss is also based on the electron density, this probability should be independent of matter (type of tissue).
Bob S

 

[ksambourship]

Thanks for all the answers especially Bob S for the citations. As I understand the in flight anihhilation is not prohibited but it's just improbable. I haven't found the Heitler book yet but I will look for it.

 

[Bob S]

Thanks for all the answers especially Bob S for the citations. As I understand the in flight anihhilation is not prohibited but it's just improbable. I haven't found the Heitler book yet but I will look for it.

An annihilation in flight probability of 18% at 50 MeV is high enough to make the photon difference method (see earlier post) with positrons practical. I think SLAC at one time had a 1-milliamp pulsed positron beam.

[Added] See Fig. 2 in
http://www.jetpletters.ac.ru/ps/1810/article_27668.pdf
for bremsstrahlung spectra for 800 MeV positrons.
Bob S