Why do we often need to consider the square of four momentum transfer $Q$ in scattering experiment or particle physics/cosmology instead of just the $Q$ itself?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ If physical quantities were to depend on components of $Q$, the theory wouldn’t have relativistic invariance. $\endgroup$– Prof. LegolasovMay 10, 2019 at 8:24
-
$\begingroup$ Do you mean why not get the length of the four-vector momentum transfer, i.e. the square root of it? the way we use m and not $m^2$ ? $\endgroup$– anna vMay 10, 2019 at 18:31
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
The four-momentum of a physical system is a conserved quantity in a particular reference frame. But we need a relativistic invariant theory which doesn't depend on the choice of coordinates. The squared four-momentum is a scalar invariant, i.e., it doesn't depend on any reference frame. Thus a relativistic invariant theory requires a formulation based on invariant quantities.
-
$\begingroup$ I think the question is why $q^2$ instead of $\sqrt{q^2}$. Both are Lorentz invariant. $\endgroup$– dukwonMay 10, 2019 at 15:01