# Two derivations for rest frame energy (Mandelstam Variable $s$)

In my particle physics course, the Mandelstam variable $s$ is introduced, as the sum of the two 4-vectors in, or out of an interaction: $s=(p_1 + p_2)^2$. One of the problems asked for $s$, for two colliding particles of the same energy, and same mass. (In order to calculate whether the collision was sufficient for creating a $\bar{p}$.) Working in natural units btw.

I said that by going into COM frame, $\vec{p} = -\vec{p}$, and we can create $(p_1+p_2)$ component by component, as:

$$\sqrt{s}=\sqrt{(p_1+p_2)^2} = \sqrt{(\{E_1, \vec{p}\} + \{E_2,-\vec{p}\})^2}=\sqrt{({2E, \vec{0}})^2}=\sqrt{4E^2}=2E.$$

The solution given was:

$$\sqrt{s} = \sqrt{p_1^2+p_2^2 +2p_1\cdot p_2}=\sqrt{m_1^2+m_2^2 +2E_1E_2-2\vec{p}_1\cdot\vec{p}_2},$$ then said since we're dealing with high energies, we can neglect $m$, and then $E\approx p$, so we can write: $$\sqrt{s} \approx \sqrt{4E^2} = 2E.$$

This is the same result that I got, without having to ask that $E \gg m$, or introduce any approximation. Hence my question is, have I made a mistake?

• I see now he needn't have used the approximation, and that settles part of my question. Question remains: is my route to the same answer still valid?
– CDCM
Commented May 28, 2017 at 17:49

Yes, you've made a mistake. The problem only states that the particles' energies are equal; the problem doesn't specify that the particles' masses are the same. If $m_1\ne m_2$ then $E_1\ne E_2$ in the COM frame.
• It seems that whoever wrote the question had a different question in mind when they wrote the solution. If the particles have identical mass and energy, then you're solution is 100% correct and much simpler. But the given solution uses $m_1$ and $m_2$ explicitly, where one would then expect that the solution writer has in their mind that $m_1\ne m_2$ and one needs to go to the high energy limit. Commented May 28, 2017 at 18:38