If bunches of protons are being circulated in both directions of the LHC collider with each proton having an energy $E_p=7\ \mathrm{TeV}$, then using the following "Lorentz Invariant Quantity" expression, $s$, for a collider:


I can then take the square root $s$ to get

$$\sqrt{s}=\sqrt{4\times7^2}\ \mathrm{TeV}=14\ \mathrm{TeV}$$

Which is the center of mass energy for proton-proton collisions at the LHC.

I found the expression, $s$ on the top of page 5 of some Oxford university notes. However, I don't really get where $s$ was derived from, so I am not sure why it has a different form for this type of proton-proton collision.

  • $\begingroup$ Different from what? In the LHC context, $s$ is just one of the three Mandelstam variables, I see nothing unusual here. $\endgroup$
    – ACuriousMind
    Dec 15, 2014 at 19:16
  • $\begingroup$ @ACuriousMind different to a fixed target machine for one. Also, that proof does not lead to $4E_p^2$ can you elaborate on what you're talking about? $\endgroup$
    – Magpie
    Dec 15, 2014 at 19:31

1 Answer 1


The definition of $s$ is the following:

$$ s=(p_1+p_2)^2,$$

where $p_1$ and $p_2$ are the 4-momenta of each colliding proton. For head on collision of particles with same energy and momentum (like in LHC) these explicitly read $p_1=(E_p,\vec{p})$ and $p_2=(E_p,-\vec{p})$. Plug this in the definition: \begin{eqnarray} s &=& p_1^2 + p_2^2 +2p_1\cdot p_2\\ &=&(E_p^2-|\vec{p}|^2) + (E_p^2-|\vec{p}|^2)+2 (E_p^2+|\vec{p}|^2)\\ &=& 4E_p^2 \end{eqnarray}

  • 3
    $\begingroup$ It could be shorter, although basically the same as what you've written. $p_1=(E_p, \vec{p})$, $p_2=(E_p,-\vec{p})$, $p_1 + p_2 = (2E_p, \vec{0})$ so $s = 4E_p$. $\endgroup$
    – MMa
    Sep 3, 2017 at 4:59

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