I'm looking at this wikipedia derivation (link) for the threshold energy of a reaction of the form $$ 1+2 \rightarrow a + b + c $$

but I get to a different result because wikipedia says

$$ E_1 = \frac{E_{CM}^2-(m_1+m_2)^2}{2m_2} $$ but I get

$$ E_1 = \frac{E_{CM}^2-(m_1^2+m_2^2)}{2m_2} $$

who's right and who's wrong?

I get the exact same results for beta and gamma shown in the page but the last step seems to give me a different result. These are my steps $$ \gamma(E_1+m_2-\beta p_1) = E_{CM} $$ $$ \gamma (E_1 - \frac{p_1^2}{E_1+m_2}+m_2) = E_{CM} $$ $$ \gamma \frac{\left[ E_1(E_1+m_2) - p_1^2 + m_2(E_1+m_2)\right]}{E_1+m_2} = E_{CM} $$ $$ \frac{\gamma}{E_1+m_2}\left[ E_1^2-p_1^2 +m_2^2+ 2E_1 m_2\right] = E_{CM} $$ $$ m_1^2+m_2^2 + 2m_2E_1 = E_{CM}^2 $$ $$ E_1 = \frac{E_{CM}^2-m_1^2-m_2^2}{2m_2} $$


1 Answer 1


You are correct; You even had a charitable interpretation of Wikipedia's version, which is actually dimensionally inconsistent. If you look up the history of the article, you will see that there was an old edit where it was correct, and somehow it got corrupted over time.

I think your computation is not optimal. It is far easier to see that many of the steps are really just internal consistency constraints of $$ \begin{pmatrix} \gamma & -\gamma\beta \\ -\gamma\beta & \gamma \end {pmatrix} \left [ \begin{pmatrix} E \\ p \end {pmatrix} + \begin{pmatrix} m_2 \\ 0 \end {pmatrix} \right ] = \begin{pmatrix} m_a + m_b + m_c \\ 0 \end {pmatrix} $$ The cheapest way to get the answer is really to compute the invariant rest energy $$ (E+m_2)^2 - p^2 = (m_a+m_b+m_c)^2\\ E^2 - p^2 + m_2^2 + 2 m_2 E = (m_a+m_b+m_c)^2\\ m_1^2 + m_2^2 + 2 m_2 E = (m_a+m_b+m_c)^2\\ \therefore \qquad E = \frac{(m_a+m_b+m_c)^2-(m_1^2+m_2^2)}{2m_2} $$ Wiki's current version had, instead $$ E = \frac{(m_a+m_b+m_c)^2-(m_1+m_2)^1}{2m_2} $$ which is just plain wrong.

Edit: I have since fixed the error on Wiki, so it should be correct now.

  • $\begingroup$ Thank you for your improved and faster solution. I did feel something was off about taking so many steps but I was biased by following wiki's solution $\endgroup$
    – Crucio
    Commented Apr 28, 2023 at 12:34

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