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Assuming a very high energy photon (energy $E$) crosses the atmosphere and produces an electron-positron pair, I would like to know what is the angle between these to leptons produced. I was trying to calculate it by applying the energy-momentum conservation and realized that in this case the angle could be 0 if the momentum $p$ does not need to be conserved.

Question: Does $p$ need to be conserved in the interaction or is it enough that the following relation applies:$$ E^2=2\left(p_\text{e}c\right)^2+2\left(m_\text{e}c^2\right)^2 \,,$$where $p_\text{e}$ is the momentum of the resulting electron/positron and $m_\text{e}$ its mass?

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The center of momentum frame of the resulting electron-positron would have to be a $0$ momentum frame of the starting photon in order for momentum to be conserved, but photons can't have $0$ momentum. This is why pair production must occur near a nucleus or such to receive some recoil. The usual way to derive the angle I think is to consider the recoil, but in the limit where the recoil momentum is small relative to the other momenta (i.e., it's approximately $0$). (So, to answer the question, momentum is of course conserved, but you can take this approximation if you want)

I believe the (special relativity kinematic) angle you get in this case should indeed be $0$, but the angle in reality can be a little bit larger than $0$, depending on how much recoil momentum there was.

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Momentum and energy have to be conserved in pair production. That's why it only occurs near a nucleus: in a perfect vacuum, the simple 'decay' of a sufficiently high-energy photon into an electron-positron pair doesn't conserve momentum.

The consequence of this is that the exact angle between the particles of the pair is not zero. Ideally, we can approximate this angle to $0$ by setting up conditions to ensure that the nucleus has negligible momentum after the collision.

But to predict the actual angle between the leptons, you'll need to know the other forces acting upon the nucleus, the mass of the nucleus, the exact energy of the photon, and something about the motion of the pair, probably velocity..

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  • $\begingroup$ why is momentum not conserved? If both electron and positron are emitted with a 0 deg angle and $p_e$=$p/2$? momentum? $\endgroup$ – Juanjo May 25 '18 at 3:31
  • $\begingroup$ @Juanjo see physics.stackexchange.com/questions/229198/… The nucleus needs to be present to satisfy the law of conservation of momentum. And because of that, we can't accurately say that the angle is 0 degrees and momentum is equally divided, although it's a good approximation for general purposes. $\endgroup$ – user191954 May 25 '18 at 16:31

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