[Physics] energy threshold of proton-proton collision

collisioninvariantslorentz-symmetryspecial-relativity

Could you please help me understand what's wrong with the following calculation?

When a proton moves toward a fixed proton with very high speed and collides with it, then the following interaction can occur.
$p + p \rightarrow p + p + p + \overline{p}$

In CM frame, the two proton moves toward each other and the initial energy of the system is ${E}=2\gamma m_{p} c^2$

Next, final energy is $E=\gamma_{1}m_{p}c^2 + \gamma_{2}m_{p}c^2 + \gamma_{3}m_{p}c^2 + \gamma_{4}m_{\overline{p}}c^2$. When the initial energy is just enough to make the interaction happen, then every $\gamma$ here would be 1. Thus from $E_{i}=E_{f}$ you have $2\gamma m_{p} c^2 = 4 m_{\overline{p}} c^2 $. Therefore $\gamma = 2$ and $E_{i}=4mc^2$

However, Griffith's introduction to elementary particle says that it should be $E=7mc^2$. (Example 3.3) But I don't understand what I did wrong in the above solution…

Best Answer

If you do the calculation completely in the CM system, you must transform back to the lab to answer the question. Kinetic energy is depends on the frame of reference.

An easier direction to proceed is to use a Lorentz-invariant quantity:

$$\left[E_{lab}^2-p_{lab}^2c^2\right]_{before} = \left[E_{CM}^2-p_{CM}^2c^2\right]_{after}.$$

Of course, $p_{CM}=0$, and $E_{CM}$ is simply the total mass-energy. The real work comes with determining the total lab energy and total lab momentum before the collision. That's not tremendously hard, but it does require some thought. Two massive particles, one at rest and the other moving.

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