In the metric convention $(+,-,-,-)$, the spacetime interval is given by $$x^2=x_\mu x^\mu=(x^0)^2-|\textbf{x}|^2=t^2-|\textbf{x}|^2$$ in the units $c=1$. To make the theory Euclidean one considers the change of variable $t\to i\tau$ whic makes the interval $$x^2=-\tau^2-|\textbf{x}|^2=-(\tau^2+|\textbf{x}|^2).\tag{1}$$ But the Euclidean distance is taken as $x^2=\tau^2+|\textbf{x}|^2$. What happens to the minus sign in (1)?
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$\begingroup$ What if I choose (convention you call it) as space time interval $$ x^2=x_\mu x^\mu=-(x^0)^2+|\textbf{x}|^2=-t^2+|\textbf{x}|^2 $$ $\endgroup$– VoulkosCommented Jan 25, 2017 at 15:08
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$\begingroup$ @Frobenius In that metric convention, it works. What about the convention I chose? $\endgroup$– SRSCommented Jan 25, 2017 at 15:12
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$\begingroup$ I think that the main here is your effort to see Minkowski space as Euclidean space. This is impossible because of the indefinite and positive definite of the metric.. That's why I find also the term transition inappropriate in this case. $\endgroup$– VoulkosCommented Jan 25, 2017 at 15:20
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$\begingroup$ @Frobenius This is a widely used trick to perform and make sense of certain integrals. Also used in finding Euclidean solutions of several classical field equations such as instantons. $\endgroup$– SRSCommented Jan 25, 2017 at 15:24
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$\begingroup$ in the metric that you used, the Euclidean distance would be $-\vert x\vert^2$. (sorry can't get the boldface $x$ to work. $\endgroup$– ZeroTheHeroCommented Jan 25, 2017 at 15:34
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1 Answer
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If one uses the metric $(+---)$ then the Euclidean distance would be $-\vert x^2\vert_E$, i.e. $\vert x\vert^2=-\vert x\vert^2_E$. To get the usual Euclidean distance one needs to use the $(-+++)$: as the space components are not affected by the introduction of the imaginary unit, one ought to choose a metric where the space components has the same sign as in the Euclidean distance.
answered Jan 25, 2017 at 15:41