Friday 27 July 2018

special relativity - Minkowski spacetime: Is there a signature (+,+,+,+)?


In history there was an attempt to reach (+, +, +, +) by replacing "ct" with "ict", still employed today in form of the "Wick rotation". Wick rotation supposes that time is imaginary. I wonder if there is another way without need to have recourse to imaginary numbers.



Minkowski spacetime is based on the signature (-, +, +, +). In a Minkowski diagram we get the equation: $$ \delta t^2 - \delta x^2 = \tau^2 $$ Tau being the invariant spacetime interval or the proper time.


By replacing time with proper time on the y-axis of the Minkowski diagram, the equation would be $$ \delta x^2 + \tau^2 = \delta t^2$$ In my new diagram this equation would describe a right-angled triangle, and the signature of (proper time, space, space, space) would be (+, +, +, +).


enter image description here


I am aware of the fact that the signature (-, +, +, +) is necessary for the majority of physical calculations and applications (especially Lorentz transforms), and thus the (+, +, +, +) signature would absolutely not be practicable.( Edit: In contrast to some authors on the website about Euclidian spacetime mentioned in alemi’s comment below)


But I wonder if there are some few physical calculations/ applications where this signature is useful in physics (especially when studying the nature of time and of proper time).


Edit (drawing added): Both diagrams (time/space and proper time/space) are observer's views, even if, as it has been pointed out by John Rennie, dt is frame dependent and τ is not.



Answer



The significance of the metric:


$$ d\tau^2 = dt^2 - dx^2 $$


is that $d\tau^2$ is an invarient i.e. every observer in every frame, even accelerated frames, will agree on the value of $d\tau^2$. In contrast $dt$ and $dx$ are coordinate dependant and different observers will disagree about the relative values of $dt$ and $dx$.



So while it is certainly true that:


$$ dt^2 = d\tau^2 + dx^2 $$


this is not (usually) a useful equation because $dt^2$ is frame dependant.


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