- #31
- 24,488
- 15,057
Well, the correct way to explain it is as follows:
A photon hits a very massive particle like an atomic nucleus with mass ##m_A##. Let's assume that the photon energy is very small compared to the mass of the nucleus (I'm using natural units with ##\hbar=c=1## so that masses, energies and momenta all have the same units; the use of kg is very inconvenient in particle physics, so let's use electron volts or in this case MeV).
Now if the photon hits the nucleus under these circumstances the nucleus will almost get momentum transferred (consider a ball hitting a wall). Thus the energy-momentum balance for making an electron-positron pair at rest reads
$$E_{\gamma}=2 m_e,$$
and that's the minimum photon energy needed for this reaction.
A single photon cannot decay to an electron positron-pair in the vacuum, because this is forbidden by energy-momentum conservation. That's easy to see by considering the relativistic invariants of the corresponding kinematics. If this would happen, you'd have
$$p_{\gamma}=p_{e^-}+p_{e^+},$$
where the ##p_j## are the four-momenta of the particles. Squaring the equation in the sense of the Minkowski product you'd get
$$p_\gamma^2=0=(p_{e^-}+p_{e^+})^2=2 m_e^2 + p_{e^+} \cdot p_{e^-}.$$
Now
$$p_{e^+} \cdot p_{e^-}=E_{e^+} E_{e^-}-\vec{p}_{e^+} \cdot \vec{p}_{e^-}.$$
Since now ##E_{e^{+}}=\sqrt{\vec{p}_{e^+}^2+m_e^2} > E_{E^{+}}## you can never make the right-hand side vanish in contradiction to the result on the left-hand side.
For a process to occur you must conserve energy and momentum and obey the "on-shell conditions", ##p^2=m^2##, for all particles in the initial and the final state! That's not possible for a single photon in empty space, as just shown.
A photon hits a very massive particle like an atomic nucleus with mass ##m_A##. Let's assume that the photon energy is very small compared to the mass of the nucleus (I'm using natural units with ##\hbar=c=1## so that masses, energies and momenta all have the same units; the use of kg is very inconvenient in particle physics, so let's use electron volts or in this case MeV).
Now if the photon hits the nucleus under these circumstances the nucleus will almost get momentum transferred (consider a ball hitting a wall). Thus the energy-momentum balance for making an electron-positron pair at rest reads
$$E_{\gamma}=2 m_e,$$
and that's the minimum photon energy needed for this reaction.
A single photon cannot decay to an electron positron-pair in the vacuum, because this is forbidden by energy-momentum conservation. That's easy to see by considering the relativistic invariants of the corresponding kinematics. If this would happen, you'd have
$$p_{\gamma}=p_{e^-}+p_{e^+},$$
where the ##p_j## are the four-momenta of the particles. Squaring the equation in the sense of the Minkowski product you'd get
$$p_\gamma^2=0=(p_{e^-}+p_{e^+})^2=2 m_e^2 + p_{e^+} \cdot p_{e^-}.$$
Now
$$p_{e^+} \cdot p_{e^-}=E_{e^+} E_{e^-}-\vec{p}_{e^+} \cdot \vec{p}_{e^-}.$$
Since now ##E_{e^{+}}=\sqrt{\vec{p}_{e^+}^2+m_e^2} > E_{E^{+}}## you can never make the right-hand side vanish in contradiction to the result on the left-hand side.
For a process to occur you must conserve energy and momentum and obey the "on-shell conditions", ##p^2=m^2##, for all particles in the initial and the final state! That's not possible for a single photon in empty space, as just shown.