- #1
lonelyphysicist
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I understand it is always possible to diagonalize a metric to the form
diag[itex][1,-1,\dots,-1][/itex]
at any given point in spacetime because the metric is symmetric and we can always re-scale our eigenvectors.
But is this achievable via a coordinate transformation? That is, would the basis vectors in such a diagonalized metric always be coordinate vectors [itex]\{ \partial / \partial x^i \}[/itex]? More explicitly, if we start with the coordinates [itex]\{ y^i \} [/itex], can we always find [itex]\{ x^i \}[/itex] such that
[itex]diag[1,-1,\dots,-1]_{ij} = \frac{\partial y^{a'}}{\partial x^i} \frac{\partial y^{b'}}{\partial x^j} g_{a'b'}[/itex]
diag[itex][1,-1,\dots,-1][/itex]
at any given point in spacetime because the metric is symmetric and we can always re-scale our eigenvectors.
But is this achievable via a coordinate transformation? That is, would the basis vectors in such a diagonalized metric always be coordinate vectors [itex]\{ \partial / \partial x^i \}[/itex]? More explicitly, if we start with the coordinates [itex]\{ y^i \} [/itex], can we always find [itex]\{ x^i \}[/itex] such that
[itex]diag[1,-1,\dots,-1]_{ij} = \frac{\partial y^{a'}}{\partial x^i} \frac{\partial y^{b'}}{\partial x^j} g_{a'b'}[/itex]
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