Why time is not like other dimensions is a real amount? In relativity time axis is $i*c*t$, where $i$ is the imaginary unit and $c$ is light speed in free space. Did science or philosophy reached to any answer of this weird property of time? And does this property have anything to do with the uni direction through time axis?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ Related: Farewell to $ict$ $\endgroup$– G. SmithCommented Oct 26, 2020 at 3:22
-
$\begingroup$ Some would argue(landau, my lie groups,banarch spaces teachers) that (-,+,+,+) makes more sense. Specifiaclly relating it to the quaternions.en.wikipedia.org/wiki/Quaternion $\endgroup$– user220348Commented Oct 26, 2020 at 3:41
-
$\begingroup$ Possible duplicates: physics.stackexchange.com/q/107443/2451 , physics.stackexchange.com/q/121380/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Oct 26, 2020 at 6:46
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
No. The basic invariant is $ds^2-cdt^2$ is a difference so you need an “i” somewhere if you’re gonna take a “usual” scalar product $(ds,icdt)\cdot (ds,icdt)$. It is more convenient to include it with $ct$ because $ds^2=dx^2+dy^2+dz^2$, or $d\vec s=(dx,dy,dz)$ so that’s fewer minuses.
Note that the more “modern” approach define a metric $\eta_{\mu,\nu}=\hbox{diag}(+,+,+,-)$, use $dx^{\mu}=(dx,dy,dz,cdt)$ and form the scalar product using the matrix so that $ds^2-cdt^2= dx^{\mu}\eta_{\mu\nu}d^{\nu}$. The minus sign is then included in the $\eta_{\mu\nu}$ rather than appearing as an “i” that multiplies a coordinate.
answered Oct 26, 2020 at 4:10
-
$\begingroup$ what I mean is that the interval is a difference not a simple sum, hence the need for a minus sign. I see your point though and I fixed the language (hopefully). $\endgroup$ Commented Oct 26, 2020 at 5:44