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Since the metric is euclidean in coordinates ##(ict,x)## it can be drawn in a plane, but if the metric is ##diag(1,-1)##, can both axis still be drawn in a plane ?
The ict notation is necessary to confuse people. If students cannot deal with confusing notation, they won't be able to become good theoretical physicists! http://www.staff.science.uu.nl/~hooft101/lectures/genrel_2013.pdfI strongly discourage the use of the ##\mathrm{i} c t## convention in relativity. It's maybe a bit inconvenient first to introduce the Minkowski-pseudometric coefficients ##\eta_{\mu \nu}## and to deal with upper and lower indices for vector and tensor components, but it pays off. At the end at latest when it comes to general relativity the ##\mathrm{i} c t## convention doesn't make any sense anymore!
None of this makes sense. It appears that you do not understand either the rationale for, or the limitations of, the ##ict## convention.i comes too from taking squares when formulating the invariance of speed of light ##′x=ct\Leftrightarrow x'=A(x,t) ct'##, whereas taking plus minus for the A function treats cases separately and has free parameters (unwanted ?), but squaring leads to bilinear forms but also implies the price to pay as a singularity (BTW why is it not called the speed of light catastrophe ?) and imaginary numbers.
I would say it is sufficient to start writing down SR in general coordinate systems to make it no longer work.As @vanhees71 points out, once you move to general relativity, the ict convention no longer works
It might be there, but it is definitely in MTW... not a full chapter, just a few short paragraphs in a named section.I think it was Taylor, in "Space-time physics" who has a chapter heading entitled "Farewell to ict".
I would show another disadvantage to use imaginary time and real space coordinate.Since the metric is euclidean in coordinates (ict,x)(ict,x)(ict,x) it can be drawn in a plane, but if the metric is diag(1,−1)diag(1,−1)diag(1,-1), can both axis still be drawn in a plane ?