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 $Q^2$?

To my intuition, for what it's worth, the square of the four momentum transferred (in the course of scattering process) may present the square of the invariant mass of an object which was transferred.

That's at least, with $-Q^2 := (E_{\text{initial}} - E_{\text{final}})^2 - (\| \vec p_{\text{initial}} - \vec p_{\text{final}} \|)^2$,
as far as the differences are taken between initial and final values of the same particle (when considering elastic scattering), or of "the most similar particle" (in case of inelastic scattering; cmp. Peskin/Schroeder, QFT, p. 156).

In this sense, the quantity $-Q^2$ equals the Mandelstam variable $t$, or $u$; corresponding to the (first order) scattering channels t, or u.

(Of course, the invariant mass of such an "object which was transferred" is not necessarily equal to the nominal mass of some particular gauge boson, for instance. The corresponding object would accordingly be called "off-shell", or "virtual".)


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