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    Covariant/contravariant transform and metric tensor

    Yes, I got it to work when I used the form G-1 = sin2α(1 +cosα),(+cosα 1)}. If I use G-1 = {(sin2α +cosα.sin2α),(cosα.sin2α,1)} it doesn't.
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    Covariant/contravariant transform and metric tensor

    G = 1/sin2a{(1 -cosa),(-cosa 1)} = {(1/sin2a -cosa/sin2a),(-cosa/sin2a 1/sin2a)} detG = 1/sin4a - cos2a/sin4a = 1/sin2a Thus, 1/|detG| = sin2a So G-1 = sin2a{(1 +cosa),(+cosa 1)} Now GG-1 does equal the identity matrix. Am I good so far? If so I will return to my original question in the...
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    Covariant/contravariant transform and metric tensor

    No, I thought I had 1/|detG| where |detG| = 1 - cos2α = sin2α
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    Covariant/contravariant transform and metric tensor

    contravariant = G * covariant covariant = G-1 * contravariant If G = 1/sin2α{(1 -cosα),(-cosα 1)} then wouldn't G-1 = (1/sin2α)(1/|detG|){(1 +cosα),(+cosα 1)} which gives 1/sin4α{(1 +cosα),(+cosα 1)} Now if I compute GG-1 I get 1/sin4{(sin2α 0),(0 sin2α)} which is equivalent to...
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    Covariant/contravariant transform and metric tensor

    H is a contravariant transformation matrix, M is a covariant transformation matrix, G is the metric tensor and G-1 is its inverse. Consider an oblique coordinates system with angle between the axes = α I have G = 1/sin2α{(1 -cosα),(-cosα 1)} <- 2 x 2 matrix I compute H = G*M where M =...
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