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Posted

A metric is always a distance. Or do you mean metric tensor? I gather that's what you could mean by your reference to "components". If that is the case then all components of the metric tensor are dimensionless, as you can see from the equation of the invariant interval:

 

[math]ds^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/math]

 

[math]ds^2[/math] has dimension [math][L]^2[/math]. [math]dx^{\mu}[/math] and [math]dx^{\nu}[/math] have dimension [math][L][/math]. So [math]\eta_{\mu\nu}[/math] (and hence all of its components) are dimensionless.

Posted

Yes, I meant tensor

 

Suppose we had spherical coordinates [math](r,\theta,\phi)[/math] : then is [math]d\theta[/math] dimensionless, the euclidean metric diag(1,1,1) transforms into [math] (1,r^2,r^2\sin(\theta))[/math], so the units should be 1,L^2,L^2 (?)

Posted

Yes, if you change to spherical coordinates then not all of the components of the [tex]dx^{\mu}[/tex] have the same dimension. Consequently, not all of the components of the metric tensor do.

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