kleinwolf Posted February 9, 2009 Posted February 9, 2009 What are the units of the metric : It seems components can have different ones ?
Tom Mattson Posted February 10, 2009 Posted February 10, 2009 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.
kleinwolf Posted February 11, 2009 Author Posted February 11, 2009 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 (?)
Tom Mattson Posted February 14, 2009 Posted February 14, 2009 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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