B Should the 'time' axis of a Minkowski diagram be time's imaginary unit?

jk22

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 ?

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Orodruin

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You can draw anything two-dimensional in a plane. You do not even need a metric at all. What is going to differ is that the interpretation of the physical distance between points on your drawing is going to be irrelevant.

vanhees71

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I 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!

• Dale

atyy

I 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!
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.pdf

• vanhees71

vanhees71

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Hm, ok. The logical conclusion from 't Hooft's many words about his awkward decision to do SRT with the infamous $\mathrm{i} c t$ convention and GRT in the more appropriate real-time formalism is that sign errors are unavoidable in GR. Well, then you have to live with that .

• atyy

jk22

Well the 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.

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PeterDonis

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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.
None of this makes sense. It appears that you do not understand either the rationale for, or the limitations of, the $ict$ convention.

In special relativity, the choice of whether to multiply the $t$ coordinate by $i$ to make the metric look Euclidean (notice I said "look" Euclidean, not "be" Euclidean) is a matter of preference and convention. For some purposes the $ict$ convention can be useful, which is why you sometimes see it in the literature.

However, you can't change the actual physics by changing conventions; the actual, physical spacetime still has timelike, null, and spacelike intervals, and Lorentz transformations still act differently on the three different kinds of intervals (or vectors in a more rigorous formulation) even if you obfuscate that fact by making the metric look Euclidean.

As @vanhees71 points out, once you move to general relativity, the $ict$ convention no longer works; this is its most important limitation, and the reason why GR textbooks will tell you that you have to unlearn it. (MTW, for example, has a clear explanation of the issue.)

Orodruin

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As @vanhees71 points out, once you move to general relativity, the ict convention no longer works
I would say it is sufficient to start writing down SR in general coordinate systems to make it no longer work.

• vanhees71 and pervect

pervect

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I think it was Taylor, in "Space-time physics" who has a chapter heading entitled "Farewell to ict". The takeaway from this should be that one should expect to abandon "ict" at some point, that it's not fundamental.

I haven't verified that I got the source of the "farewell" quote right, this is from memory. But eventually, the ict convention is no longer used, it stops working. Basically, "ict" is a fudge that can delay the need for the introduction of a metric tensor if one uses cartesian coordinates in flat space-time. Cartesian coordinates do not exist in curved space-times, and even in flat space-times one may wish to use other coordinate systems. Using anything other than cartesian coordinates makes the "ict" fudge stop working.

Nugatory

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I think it was Taylor, in "Space-time physics" who has a chapter heading entitled "Farewell to ict".
It might be there, but it is definitely in MTW... not a full chapter, just a few short paragraphs in a named section.

sweet springs

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 ?
I would show another disadvantage to use imaginary time and real space coordinate.
In mathematics we know Gauss plane or complex plane z=x+iy the rules of which include
$$|z|=\sqrt{x^2+y^2}$$
$$z^*=x-iy$$
do not result our $c^2t^2-x^2$ or $x^2-y^2$. To avoid unnecessary confusion it would be better to keep imaginary component for use in Gauss plane that is very useful in all the field of physics.

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haushofer

Just to add: if you write down spacetime intervals for solutions which are not static, (or app,y general coordinate transfo's to static solutions) you can end up wit loose factors of i due to cross terms between time and space

• vanhees71

"Should the 'time' axis of a Minkowski diagram be time's imaginary unit?"

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