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# Clear up confusion on Minkowski signature

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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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I'm not entirely sure what your question is: Is it about why the time dimension is handled differentl...
celtschk‭ wrote about 2 years ago

I'm not entirely sure what your question is: Is it about why the time dimension is handled differently from the space dimensions, or is it about the difference between putting the $-1$ in the metric and putting a factor $i$ into the time coordinate?

deleted user wrote about 2 years ago

Yep! It is about putting -1 in the Minkowski metric and putting factor i into the time coordinate.

celtschk‭ wrote about 2 years ago

My question was an “or” question. So which of the two alternatives is it?

deleted user wrote about 2 years ago · edited about 2 years ago

When putting -1 in Minkowskian metric I just get the spacetime invariant which is $ds^2=c^2t^2-dx^2$. Even I get the same spacetime invariant if I put ict into time coordinate.

My first question was, how Minkowski found that? Or he just wrote it curiously.

Second question was, how imaginary number can be put into time coordinate?

When first is solved second isn’t required and vice-versa.

celtschk‭ wrote about 2 years ago

Is it specifically on Minkowski's thoughts (in which case I can't answer, as I didn't read Minkowski's original work), or is it just about how you can derive it (independent of whether it's how Minkowski did it)?

deleted user wrote about 2 years ago

celtschk‭ I will welcome any answer of my question. Usually, I like derivation rather than he had done it. A derivation can clear up any confusion no matter who done it what way. Everyone doesn’t think same way. The way I was learnt derivation of E= mc^2 from high school's book was not the way how Einstein derived it (learned Einstein's method from Kleppner's book). But I like both. So I will be thankful to you to share whatever you know....