This question refers to example 3.4 of Griffiths Introduction to Elementary Particles Second addition.

In the example, a moving proton is collided into a proton at rest, producing 3 protons and 1 anti-proton. "Lab" refers to the lab frame and "CM" refers to the center of mass frame.

enter image description here

In this text, Griffiths discusses that some things are invariant (i.e. they are the same under Lorentz transformations) and that some things are conserved (i.e. they are the same before and after some sort of event).

It is clear in the picture above that "conserved" corresponds to moving left and right across the 2 rows and "invariant" corresponds to moving up and down across the two columns.

When Griffiths solves this problem, however, he uses the fact that

$$p_\mu p^\mu = {p_\mu}' {p^\mu}'$$

where $p$ is the four-momentum of the entire system and the non-primed product refers to the top left picture and the primed product refers to the bottom right picture. This leads to my question:

Is it correct to say that $p_\mu p^\mu$ is both conserved and invariant under particle collisions?

  • $\begingroup$ The answer is yes, assuming that $p^\mu$ refers to the total energy-momentum of the system. It's not clear to me why you're in doubt, or why you're not satisfied with Avantgarde's answer. Could you edit the question to explain why you're in doubt? $\endgroup$
    – user4552
    Commented Dec 30, 2018 at 16:14

1 Answer 1


You should define what $p^\mu$ is, both before and after scattering.

Suppose you take $p^\mu$ to be the total 4-momentum of the system. Before collision, it is $p^\mu_{in}$. After collision, it is $p^\mu_{out}$. But, from conservation of 4-momentum in a scattering process, we know that $p^\mu_{in} = p^\mu_{out}$. So it is then trivial that $p^\mu p_\mu$ is the same for both in and out stages of the scattering process.

However, if you take $p^2=p^\mu p_\mu$ to mean $m^2$, where $m$ is the mass of one of the particles (with momentum $p^\mu$) in the scattering event, then $p^2$ is not necessarily conserved, for instance, electron positron annihilating to produce 2 photons. Here, the mass of electron (or positron) is not conserved.

  • $\begingroup$ Thanks, I edited the problem above. In this example, then, my question still stands. $\endgroup$ Commented Dec 29, 2018 at 20:29
  • $\begingroup$ @LukePolson The answer to your question is in the second paragraph of my answer. $\endgroup$
    – Avantgarde
    Commented Dec 30, 2018 at 8:16
  • $\begingroup$ If I understand correctly then, you're saying if we take $p$ to be the four-momentum of the entire system then $p^{\mu}$ is conserved. Finally, since the scalar product is invariant, it must be that $p^2$ must be both conserved and invariant under collisions. $\endgroup$ Commented Dec 31, 2018 at 4:53
  • $\begingroup$ @LukePolson Yes; except it's better to say that $p^2$ is invariant under Lorentz transformations, and then,through conservation of momentum, we can evaluate $p^2$ before or after collisions, and they will be equal. In fig 3.6, Lorenz invariance relates figures vertically, while momentum conservation relates things horizontally. Griffiths uses both to equate the figure on top left to the figure on bottom right. $\endgroup$
    – Avantgarde
    Commented Dec 31, 2018 at 11:25

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