# What is the intuitive meaning of $Q^2$?

In particle physics, $-Q^2$ is defined by the four momentum transfer squared: $$Q^2 = -(p_i - p_f)^2 = (\vec{p}_i-\vec{p}_f)^2-(E_i-E_f)^2$$

For elastic scattering, the meaning of $Q^2$ is clear - it is directly proportional to the energy transfer and the target mass: $$Q^2 = 2 M_T \Delta E$$

But for the general case where the scattering may not be elastic, what is the intuitive meaning of $Q^2$?

What is the intuitive meaning of $Q^2$?
That's at least, with $-Q^2 := (E_{\text{initial}} - E_{\text{final}})^2 - (\| \vec p_{\text{initial}} - \vec p_{\text{final}} \|)^2$,
In this sense, the quantity $-Q^2$ equals the Mandelstam variable $t$, or $u$; corresponding to the (first order) scattering channels t, or u.