# Why $2$ photons are necessary for conservation of four-momentum?

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Since the p0 is composed of quark-antiquark pairs, it can decay electromagnetically into photons. It requires two photons to conserve momentum

Ok, why 2 and not 1 ?
it is possible, leaving the 2 photons, to make sure that the momentum is not conserved?

Physically, it's impossible.
Why is physically impossible ?

Mathematically, any pair of photons whose sum of four-momentum does not preserve the initial four-momentum satisfies my condition

For example, in the center of mass of the particle that decays the particle velocity is zero, so the 3-momentum is zero and the 4-momentum is p = (Mc, 0, 0, 0) with M the mass of the particle.

If it then decays into two particles we say that these two new momentum are for example (m1c, 0, m1 * 5, 0) and (m2c, m2 * 2, 0, 0), so the first mass particle m1 has speed 5 on the y-axis, the second mass particle m2 has velocity 2 on the x-axis. The sum of these two vectors is ((m1 + m2) c, m2 * 2, m1 * 5, 0) which, as you can see, is very different from our initial p vector.

Go to the rest frame of the pion. In this frame, the total momentum is $0$ and it has energy $E=Mc^2$. If it decays simply to 1 photon, by energy conservation this photon would need to have energy $E$ as well. A photon with energy $E$ has momentum $p=E/c=Mc$ in the direction that it travels. As you can see, there's no way that this photon has $0$ momentum and so there's no way for momentum conservation to happen. Now, if you have 2 photons, then each photon can take half the energy $E/2$ and move in opposite directions, each having momentum with magnitude $E/2c=Mc/2$. Since the two photons are moving in opposite directions their momentum will cancel thereby conserving the total momentum of $0$ that the pion started off with.
• When you say by energy conservation this photon would need to have energy E as well what you exactly mean? a pion to decay into photons need energy ? or for definition photon has kinetic energy because c like speed light is the photon speed ? When you say if you have 2 photons, then each photon can take half the energy $E/2$ from where take this energy? Why moves in opposite direction, this is not clear for me also – jkuzrt6 Sep 13 '18 at 12:07
• @jkuzrt6 You are envisioning a process $\pi^0\rightarrow\gamma$ instead of $\pi^0\rightarrow 2\gamma$ right? If $\pi^0$ has energy $E$ and decays to $\gamma$, then if $\gamma$ doesn't have energy $E$, where would this energy go? That is the very definition of energy conservation. That original energy $E$ has to be given to the daughter particle(s). If there are two photons, the two photons can split it evenly. Moving in opposite directions is the only way to conserve momentum. – enumaris Sep 13 '18 at 16:06
• mm..ok, so $\pi^0$ has energy E, decay in $2\gamma$ but..this seems like a splitting due to a sort of transition that then, I understand, seems to be a decay product just because there is the energy to 'split' donated by decay? I think I understand .. is the energy concept the key, yes, you are right – jkuzrt6 Sep 13 '18 at 19:01