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I just started a re-reading of the Conformal Field Theory yellow book by Di Francesco et al. In chapter two, after defining imaginary time $\tau$ as $t=-i\tau$, the authors state that the metric tensor, being in real time, e.g., $g_R=\mbox{diag}(1,-1,-1,-1)$, becomes in imaginary time $g_I=\mbox{diag}(1,1,1,1)$.

How can this statement be proven? I know it is a maybe stupid question, but I can't figure it out. For instance, if I consider the covariant four-vector $x^\mu_R=(t,x_0,x_1,x_2)$, the contravariant vector becomes, by application of $g_R$, $x_{R,\mu}=(t,-x_0,-x_1,-x_2)$. Performing Wick rotation, they become $x^\mu_I=(-i\tau,x_0,x_1,x_2)$ and $x_{I,\mu}=(-i\tau,-x_0,-x_1,-x_2)$, that are not transformed into each other by $g_I$! Where does my argument breaks?

Qmechanic
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Luca
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2 Answers2

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  1. By standard convention the Euclidean signature is $(+,+,+,+)$, not $(-,-,-,-)$.

  2. From a pedagogical perspective, it is therefore easier to explain the Wick rotation using the $(-,+,+,+)$ Minkowski signature convention.

  3. The opposite Minkowski signature convention $(+,-,-,-)$ is in principle also possible, but it's likely to confuse the h... out of everyone because somewhere along the line a not-so-natural-looking convention choice must be made.

  4. Related Phys.SE post: Transferring from vectors and tensors in $(-+++)$ signature to $(+---)$ signature?

Qmechanic
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For convenience, the Euclidean metric in this case must be $(----) $. Moreover, $x^{\mu}_I \equiv (\tau ,\vec x) $

Tonduea
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