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Why is $\pi^0$ created in the high-energy collision $p+p\to p+p+\pi^0$?

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    $\begingroup$ To a significant extant the answer is "Because it can be." There is sufficient energy and the quantum number of the final state agree with those of the initial state. $\endgroup$ – dmckee Jun 2 '13 at 0:38
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    $\begingroup$ Echoing dmckee's comment, the quantum maxim is "everything that can happen does happen". Here "can" means "doesn't violate any physical laws". The rates at which it "does" happen are governed by the laws of quantum theory. $\endgroup$ – twistor59 Jun 2 '13 at 6:48
  • $\begingroup$ I guess my "hiccup" is that I have two protons on the left side and two on the right side plus a pion. Is it energy conservation that forces that? $\endgroup$ – Kyle Kanos Jun 2 '13 at 10:35
  • $\begingroup$ Yes, there is a threshold kinetic energy which the incoming pions must have in order to provide enough for the extra mass of the pion. As the outgoing particles will also have some kinetic energy -for momentum conservation- you have to supply more energy than you'd first think by just doing $E=mc^2$ for the pion. $\endgroup$ – twistor59 Jun 2 '13 at 14:24
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Just summing together all the comments and providing some more explicit calculations, we have conservation of four-momentum (which is the amalgamation of the conservation of energy and conservation of momentum), we have:

$$p^{\mu}_{1}+p_{2}^{\mu}=p_{1}'^{\mu}+p_{2}'^{\mu}+p_{\pi}^{\mu}$$

Taking the inner product of each side with itself, we get:

$$\left\langle p_{1}^{\mu}\middle|p_{1}^{\mu}\right\rangle+2\left\langle p_{1}^{\mu} \middle| p_{2}^{\mu}\right\rangle+\left\langle p_{2}^{\mu}\middle|p_{2}^{\mu}\right\rangle=\left\langle p_{1}'^{\mu} \middle| p_{1}'^{\mu}\right\rangle + 2\left\langle p_{1}'^\mu \middle| p_{2}'^{\mu} \right\rangle + 2\left\langle p_{1}'^{\mu} \middle| p_{\pi}^{\mu}\right\rangle + \left\langle p_{2}'^\mu \middle| p_{2}'^{\mu} \right\rangle + 2\left\langle p_{2}'^{\mu} \middle| p_{\pi}^{\mu} \right\rangle + \left\langle p_{\pi}^{\mu} \middle| p_{\pi}^{\mu}\right\rangle$$

We note that $p^{\mu}p'_{\mu}=\left\langle p^{\mu} \middle| p'^{\mu}\right\rangle$ is invaraiant in all frames of reference and that $p^{\mu}p_{\mu}=m^{2}c^{2}$, we can therefore simplify:

$$2m_{p}^{2}c^{2}+2\left\langle p_{1}^{\mu} \middle| p_{2}^\mu \right\rangle = 4m_{p}^{2}c^{2}+4m_{p}m_{\pi}c^{2}+m_{\pi}^{2}c^{2}$$

If we consider that the second proton is initially at rest we have: $p_{1}^{\mu}=\left(\frac{E}{c},\vec{p}\right)$ and therefore:

$$2m_{p}E=2m_{p}^{2}c^{2}+4m_{p}m_{\pi}c^{2}+m^{2}_{\pi}c^{2}$$

Rearranging we get:

$$E=m_{p}c^{2}+2m_{\pi}c^{2}+\frac{m_{\pi}^{2}c^{2}}{2m_{p}}$$

Using the constants $m_{p}=938\text{ MeV}/c^{2}$ and $m_{\pi}=139.6\text{ MeV}/c^{2}$, we get:

$$E=1.228 \text{ GeV} \implies T = 289 \text{ MeV}$$

So if a proton with 289 MeV of kinetic energy collided with a stationary proton, there is a chance that a pion will be produced.

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