My problem relates to the decay of a neutral pion, more precisely the decay into two photons.

If the pion the rests the momenta of the two photons are contrarily. If the pion still has some momentum at its decay it will be transferred to the photons. (EDIT: The momenta given by the rest enrgy of the pion will always be contrarily. In case the pion is still moving there will be some extra momentum and the photons will most likely not move in opposite directins)

If the photons appear under a certain angle to the pion's direction I can calculate the photons momenta with simple geometry. But what happens if one photon appears in the same direction as the pion and the other one in the opposite direction. If the pion has a mass of 135MeV/c^2 and a momentum of 135MeV/c the photons share them in equal parts (!?). In case of the back photon the momenta would cancel out thus having no momentum.

How do I handle this scenario?

Postscript: (The superscripted numbers are indizes not exponents) Given a pion with mass $m_\pi$ and momentum $\vec{p_\pi}$. The decay will create two photons $\gamma_1$,$\gamma_2$ that share equal parts of $m_\pi$ (follows from conservation of momentum CoM)

$$E_{\pi_{rest}}=m_\pi c^2 \qquad \Rightarrow \qquad E_\gamma=\frac{E_{\pi_{rest}}}{2}$$ $$\left|\vec{p^1_\gamma}\right|=\frac{E_\gamma}{c}=\frac{m_\pi c}{2} \quad (1)$$

CoM leads to

$$\vec{p^1_{\gamma_1}}=-\vec{p^1_{\gamma_2}} \quad (2)$$

$p_\pi$ will be devided onto both photons (Assume the parts are equal)

$$\vec{p^2_{\gamma_1}}=\vec{p^2_{\gamma_2}}=\frac{\vec{p_{\pi}}}{2} \quad (3)$$


$$\vec{p_{\gamma_1}}=\vec{p^1_{\gamma_1}}+\vec{p^2_{\gamma_1}} \\ \vec{p_{\gamma_2}}=\vec{p^1_{\gamma_2}}+\vec{p^2_{\gamma_2}}$$

If now $m_\pi c=\left|\vec{p_\pi}\right| \quad $ and $ \quad \vec{p^1_\gamma} \parallel \vec{p_\pi}$ using (1),(2) and (3)

$$\vec{p_{\gamma_1}}=\frac{\vec{p_{\pi}}}{2}+\frac{\vec{p_{\pi}}}{2}=\vec{p_{\pi}} \\ \vec{p_{\gamma_2}}=\frac{\vec{p_{\pi}}}{2}-\frac{\vec{p_{\pi}}}{2}=0$$

The total momentum is still $p_\pi$ but the second photon has no momentum. How do I interpret this? Is there only one photon or is $p_\pi$ devided unequal?

  • 1
    $\begingroup$ If the photons are going in opposite directions, then the total momentum must zero which means the pion was at rest & thus contradicts your assertion it was moving, no? $\endgroup$ – Kyle Kanos Oct 21 '17 at 13:52
  • $\begingroup$ That was meant in the rest system of the pion. In the labor they won't be opposite. $\endgroup$ – Leeen gold Oct 21 '17 at 14:01
  • $\begingroup$ Right, they won't be opposite in the lab frame. So what's the question then? $\endgroup$ – Kyle Kanos Oct 21 '17 at 14:03
  • $\begingroup$ It is possible that one of the photons will appear in the same direction the pion was moving. The other one has to travel in the opposite direction. They will have different momenta. $\endgroup$ – Leeen gold Oct 21 '17 at 14:07
  • $\begingroup$ Wouldn't that contradict conservation laws? $\endgroup$ – Kyle Kanos Oct 21 '17 at 14:09

Imagine a pion at rest decaying. Then move your frame in the direction of one of the photons. In this frame the energy and momenta of the photons are different. The error is in assuming the energy is equally divided, but it is true only in the rest frame of the pion.

  • $\begingroup$ So $\vec{p^1_{\gamma_1}}=-\vec{p^1_{\gamma_2}}$ is incorrect in the lab frame? $\endgroup$ – Leeen gold Oct 21 '17 at 16:20
  • $\begingroup$ Yes, incorrect. The correct identity is that the vector sum of the photon momenta equals the pion momentum. So if the photons are in the direction of the pion movement, then their momenta is different by the value of the pion momentum. $\endgroup$ – safesphere Oct 21 '17 at 16:30
  • $\begingroup$ Thank you very much. Apparently my reputation is to low to give you credit by upvoting. $\endgroup$ – Leeen gold Oct 21 '17 at 16:44

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