# Comments on Clear up confusion on Minkowski signature

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# Clear up confusion on Minkowski signature

All given metrics are for orthonormal-basis.

**2 dimensional spacetime :**

I saw that Minkowski Metric looks like this : $$\pmatrix{-1 & 0 \\ 0 & 1}$$ or $$\pmatrix{1 & 0 \\ 0 & -1}$$

I was wondering why it's not written like identity metric. To define a vector in spacetime it's written like this $dS^2 = -(ct)^2+(dx)^2=(ict)^2+(dx)^2$. I know it's same as Minwkowski metric. But in Euclidean metric I was just using identity metric. Is there really any derivation of Minkowski metric? Or he just wrote it curiously. What the negative explains? I know that $ct$ is coordinate. Particle physicist write vector in spacetime (I had read it 3-4 hours ago, I can't find the source again so can't add reference here) like this $ dS^2 = (ct)^2-(dx)^2$.

**4 dimensional spacetime :**

The minkowski signature is $\pmatrix{- & + & + & +}$ or negative for space and positive for time. I read in the question that, the tensor for spacetime can be written like this : $ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}$ while $g_{\mu \nu}$ is Minkowski signature and $dx^{\mu}$ and $dx^{\nu}$ represent space coordinate. But in PSE question he wrote that $dx^0=ict$ where $i$ is imaginary number and $dx^0$ is time coordinate. I wonder what's the main reason of taking $ict$ as coordinate. $ct$ looks good to me to be a coordinate. At first I thought, the tensor got a negative sign for time for Minkowski signature. But when I saw $ict$ as coordinate my mind changed again.

I know the question is too confusing and I am confused about time coordinate, it is $ct$ or $ict$, I don't know. Why Minkowski wrote signature like that?

$g_{xx}=\vec e_x \cdot \vec e_x=1$ That's what we exactly know. But it's different for Minkowski spacetime, when dot product of time basis vector is 1 then dot product of space basis vector is -1 and vice versa, why?

I forgot to mention which actually made me curious to write the question, we know that dot product of any basis vector to itself is $1$ (they always not 1) $\vec e \cdot \vec e = |e| \ |e| cos\theta$ $\theta=0$ hence $\vec e \cdot \vec e = |e| \ |e|$. To write $\vec e \cdot \vec e = -|e|^2$, we must make them perpendicular but which is never true. Angle between dot product of vector of itself is $0$ but in Minkowski spacetime diagram, it is something different.

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