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I am not professional physicist; but I am curious about Stephen Hawking's "imaginary time". It would be better to elaborate exactly what it is. I am not confused because of the word "imaginary" but I find it confusing to imagine a two dimensional "plane time". If we express time in a plane instead of a one dimensional axis, then what does the movement of an observer along the imaginary axis signify physically?

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marked as duplicate by Chris White, rob, Ali, Brandon Enright, Kyle Kanos Jul 5 '14 at 1:15

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possible duplicate of The meaning of imaginary time – ACuriousMind Jul 4 '14 at 22:41
Obligatory – PPR Aug 8 '14 at 17:14
up vote 4 down vote accepted

Imaginary time has no physical meaning. It is an assumption physicists make, namely, that the math will endure the analytic continuation of the time variable onto the complex plane, which makes some calculations easier, but it is not absolutely necessary (e.g. you can do instantons without imaginary time).

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In special relativity, the metric on spacetime is

$$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 - \mathrm{d}t^2$$

(or with inverted signs). If you now formally transform $t \mapsto \mathrm{i}t$, this becomes the familiar Euclidean metric on $\mathbb{R}^4$

$$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 + \mathrm{d}t^2$$

which has several advantages over the Minkowski one in explicit calculations (especially when considering convergence of integrals in QFT). One then does the caclulation in the Euclidean case and resubstitutes $t\mapsto -\mathrm{i}t$ to get the Minkowski result. This process is known as Wick rotation.

It should be noted that Wick rotation can fail to provide the correct result, this is the case when the results considered don't possess analytic continuations that allow us to consider them as unambiguous functions of complex variables.

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