The Missing Mass of a reaction

[etotheipi]

Wikipedia says this about the missing mass of a reaction:

The term invariant mass is also used in inelastic scattering experiments. Given an inelastic reaction with total incoming energy larger than the total detected energy (i.e. not all outgoing particles are detected in the experiment), the invariant mass (also known as the "missing mass") W of the reaction is defined as follows (in natural units): $$W^2 = \left(\sum E_{in} - \sum E_{out} \right)^2 - ||\sum \mathbf{p}_{in} - \sum \mathbf{p}_{out}||^2$$

I wondered where such an expression is coming from? The invariant mass of a system $M$, in natural units, satisfies $$M^2 = \left(\sum E \right)^2 - ||\sum \mathbf{p}||^2$$ If anything, then the "missing" mass (which they also, confusingly, term as invariant mass) should go as $$\Delta [M^2] = \Delta \left[ \left(\sum E \right)^2 \right] - \Delta \left[ ||\sum \mathbf{p}||^2 \right]$$ I'm not sure why the equation they give is valid, since surely it is incorrect to bring the $\Delta$ inside the squared terms?

Is this just then an arbitrary definition, which is not directly related to $M^2 = E^2 - p^2$? Thank you .

 

[robphy]

That section from Wikipedia maybe be referring to scattering as in this passage... and likely many others like it.

"Deep Inelastic Scattering: Experiments on the Proton and the Observation of Scaling"
Nobel Lecture, December 8, 1990 by Henry Kendall
https://www.nobelprize.org/uploads/2018/06/kendall-lecture-1.pdf#page=9 (see page 684 and onward)

describes an electron scattering off a stationary proton, where the scattered electron has lost some energy to a [hadronic] particle that was not being detected. Conservation of 4-momentum would read
$$\tilde e_1 + \tilde p = \tilde e_2 + \tilde W$$
so
$\begin{align*}<br /> \tilde W<br /> &= \tilde p +\tilde e_1 - \tilde e_2\\<br /> &= \tilde p +\tilde Q\\<br /> \end{align*}$
where $\tilde Q = \tilde e_1 - \tilde e_2$ is the spacelike 4-momentum transferred to the particle.
So, the invariant mass of this particle that wasn't being detected must have been
$W^2= M_p^2 + Q^2 + 2 M_p (e_1\cosh\theta_1 - e_2\cosh\theta_2) = M_p^2 - q^2 + 2 M_p (E_1 - E_2)$
If one expresses the undetected-particle's 4-momentum $\tilde W$ in terms of the energy and momentum components of the given and detected particles, one would get the expression given by Wikipedia.