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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?

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  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.)

  3. To see how the Wick rotation works in gauge theory, see e.g. this related Phys.SE post.

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$^1$ 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.

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