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Wick Rotation, interpretation of $\bar{p}^2$ vs the usual $p^2=m^2$

Suppose we use the metric $(+,-,-,-)$ thus the momentum squared is $p^2 = p_0^2-\vec{p}^2 = m^2>0$ Defining $p_E:=\mathrm{i}\cdot p_0$ and $\bar{p}:=(\,p_E,\vec{p})$ with Euclidean norm ...