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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...

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  • $\begingroup$ What is the initial energy of the system in the lab frame? $\endgroup$ – Cosmas Zachos Sep 30 '16 at 15:23
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    $\begingroup$ Griffith (and indeed almost every text, and this professor) is asking you to compare the beam energy needed for a collider and that needed for a fixed-target experiment. You've worked the easy half of the problem. The higher energy is the answer to the fixed-target version. $\endgroup$ – dmckee Sep 30 '16 at 17:15
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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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Momentum also has to be conserved in the collision, so you cannot have all the particles at rest at the end of the collision. The minimum kinetic energy will be with all four particles moving in the same direction with the same speed. I haven't done the maths but I'm pretty sure that is the problem.

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  • $\begingroup$ He's performed the work in the CoM frame. He starts with zero momentum and therefore must end with zero momentum. $\endgroup$ – dmckee Sep 30 '16 at 17:13
  • $\begingroup$ @dmckee - yes I thought that was completely obvious. $\endgroup$ – Suzu Hirose Oct 1 '16 at 0:37

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