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In the line element

$ds=\frac{\partial s}{\partial x^1} dx^1+\frac{\partial s}{\partial x^2} dx^2$

(superscripts are indices, not powers)

the basis vectors are defined as $e_1=\frac{\partial s}{\partial x^1}$ and $e_2=\frac{\partial s}{\partial x^2}$

The metric is then said to be obtained by multiplying these basis vectors together in each of their possible combinations. In this case there are two basis vectors so there will be four elements in the metric:





$\implies g=\begin{bmatrix} g_{11} &g_{12} \\ g_{21}& g_{22} \end{bmatrix} = \begin{bmatrix} (e_1)^2 & e_1e_2 \\ e_1e_2& (e_2)^2 \end{bmatrix}$

This much I follow. Now, as I understand it, the metric is a diagonal matrix. Take the Euclidian metric for eg.

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

So in what way does $e_1e_2$ equal $0$?

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I don't understand "in what way does $e_1e_2$ equal $0$?". In what sorts of different ways can things equal $0$? Perhaps you mean "why"? – joriki Nov 24 '11 at 13:44
I mean I don't see how $\frac{\partial s}{\partial x^1}\frac{\partial s}{\partial x^2}=0$, which is what has to happen for the metric to be a diagonal matrix. – ben Nov 24 '11 at 14:00
Why do you insist on writing "how"? Do you really mean "how"? If so, please explain what that means. If you mean "why", that would be clearer. – joriki Nov 24 '11 at 14:04

It's not true in general that the matrix of the metric tensor is a diagonal matrix. This is only the case in orthogonal coordinates, which can in fact be defined as coordinates in which the matrix of the metric tensor is diagonal.

It's also not true that the matrix of the Euclidean metric tensor is always diagonal; again, this is only the case in orthogonal coordinates, and Cartesian coordinates happen to be orthogonal with respect to the Euclidean metric.

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Ah ok so you would have to multiply by the reciprocal basis to get the diagonal metric? Why does the Minkowski metric always remain a diagonal metric? Are its basis vectors multiplied by their reciprocals? – ben Nov 24 '11 at 14:31

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