Covariant/contravariant transform and metric tensor

[nigelscott]

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 = {(1 0), (cosα sinα)} and get H = {(1 -1/tanα),(0 1/sina)} which is what I expect.

Now I want go from H back to M so I compute M = G^-1H

So by my reckoning G^-1 = 1/sin⁴α{(1 cosα),(cosα 1)}

But when I multiply G^-1 and H I don't get back to M. The is a 1/sin⁴α multiplying the whole thing.
What am I missing?

 

[Muphrid]

When you say G is the metric tensor, the metric tensor converts contravariant to covariant? Or is it the other way around?

Actually, I think it's pretty clear that the issue is with your computed $G^{-1}$. The operator $G$ has determinant 1. You've got an extra factor of $1/\sin^2 \alpha$ for some reason. Did you forget the prefactor from $G$ somewhere?

 

[nigelscott]

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 1/sin²α{(1 0),(0 1)}

So I get the identity matrix multiplied by 1/sin²α which doesn't add up.

Incidentally, this question came up as a result of watching this video http://www.youtube.com/watch?v=qDLmJE2bOy4

 

[Muphrid]

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)}

You're saying $\det G = 1/\sin^2 \alpha$? Because it's not.

 

[nigelscott]

No, I thought I had 1/|detG| where |detG| = 1 - cos²α = sin²α

 

[Muphrid]

It's not $\sin^2 \alpha$ either. Why don't you go through the whole calculation of the determinant?

 

[nigelscott]

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 next post.

 

[Muphrid]

Okay, now that you're getting the identity for $GG^{-1}$, I expect the problems you were having will be resolved.

 

[nigelscott]

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.