# Wick rotation - time and what else changes?

For aid of example consider two quantities the four-momentum $\tilde P$ and a time-independent four potential $\tilde A$. Now if a wick's rotation was carried out by simply replacing $it$ with $\tau$ then under a Wick's rotation we would get: $$\tilde P'=i \tilde P$$ $$\tilde A'=\tilde A$$ whilst if it was carried out as a rotation by $\pi/2$ in the complex plane of the $0$th component we would get: $$P_0'=-iP_0,\quad A_0'=-iA_0$$ with all other components remaining the same. Which of these (if either) is the correct interpretation of a Wick's rotation - if either? and why?

1. OP's second option is correct: The zero-components $$V^0$$ of all contravariant $$4$$-vectors $$V^{\mu}$$ do Wick-rotate $$V^0_E~=~iV^0_M ;\tag{A}$$ not just time $$x^0$$ in the spacetime position 4-vector $$x^{\mu}$$: $$x^0_E~=~ix^0_M. \tag{B}$$ That is how inner products remains invariant $$V\cdot V~=~V^{\mu}~\eta_{\mu\nu}~ V^{\nu} \tag{C}$$ when going from Minkowski$$^1$$ signature $$(-,+,+,+)$$ to Euclidean signature $$(+,+,+,+)$$.
2. Similarly, the zero-components $$V_0$$ of all covariant $$4$$-vectors $$V_{\mu}$$ do Wick-rotate in the opposite direction: $$V_0^M~=~iV_0^E.\tag{D}$$ In particular, the zero-component $$p_0$$ of the energy-momentum 4-covector $$p_{\mu}$$ Wick-rotate as$$^2$$ $$p_0^M~=~ip_0^E.\tag{E}$$ (The latter is related to the fact that the Fourier-integral representation $$\delta^4(x)~=~\int_{\mathbb{R}^4} \frac{d^4p}{(2\pi\hbar)^4}~\exp\left(\frac{ip\cdot x}{\hbar} \right)\tag{F}$$ of the Dirac delta distribution cannot be analytically continued to the ambient complexified spacetime: The real integration region can at most be deformed, i.e. the $$x^0$$ and $$p_0$$ Wick-rotations must be balanced, cf. this Phys.SE post.)
$$^1$$ The speed of light $$c=1$$ is set to one in this Phys.SE answer for simplicity. Concerning the reason for the choice of Minkowski signature, see my Phys.SE answer here.
$$^2$$ Warning: Traditionally we assign the energy variable $$E=p^0$$ to behave contravariantly, so that $$E_E=iE_M$$. However some authors effectively define the energy variable $$E=-p_0$$ to behave covariantly, so that $$E_M=iE_E$$ and $$E_E$$ is effectively the negative Euclidean energy! See also my Phys.SE answer here.