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This is probably a really silly confusion I have about the definition of “coordinate differentials”, which I thought were things like $dx,dy,dz$ etc. The Minkowski line element $$ds^{2}=c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2}$$ defines the Minkowski metric $$\left[\eta_{\mu\nu}\right]=\left(\begin{array}{cccc} c^2 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right).$$ Using index notation, the line element can be written as $ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$. In textbooks I have seen the terms $dx^{\mu},dx^{\nu}$ called “coordinate differentials”, which seems OK except $dx^{0}=cdt$. I realise this is trivial, but is it correct to call $cdt$ a “coordinate differential”? To me it looks like a coordinate differential $dt$ multiplied by $c$.

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I think this is largely just terminology. Strictly speaking the coordinate is $ct$ not $t$, and of course $d(ct) = cdt$. In any case we usually choose units where $c = 1$ and just ignore it.

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  • $\begingroup$ Of course! I should have realised. $\endgroup$ – Peter4075 Jun 5 '14 at 9:42
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Each set of coordinates comes with its own set of coordinate differentials and its own coordinate expression for the metric tensor:

Given coordinates $(t,x,y,z)$, the coordinate differentials are $dt,dx,dy,dz$ and $$ (\eta_{\mu\nu})=\begin{pmatrix} c^2&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix} $$

Given coordinates $(x^0,x^1,x^2,x^3)$, the coordinate differentials are $dx^0,dx^1,dx^2,dx^3$ and $$ (\eta_{\mu\nu})=\begin{pmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix} $$ (Strictly speaking, we should probably use different symbols for the metric tensors, eg $\eta_{\mu\nu}$ and $\tilde\eta_{\mu\nu}$).

As $$ x^0=ct\\x^1=x\\x^2=y\\x^3=z $$ obviously $$ dx^0=cdt\\dx^1=dx\\dx^2=dy\\dx^3=dz $$

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