Covariant/contravariant transform and metric tensor

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Discussion Overview

The discussion centers around the transformation properties of contravariant and covariant matrices in the context of a metric tensor, specifically examining the calculations involving the metric tensor and its inverse in an oblique coordinate system. Participants explore the relationships between these transformations and the implications of their calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a contravariant transformation matrix H and a covariant transformation matrix M, using a specific metric tensor G and its inverse G-1, and expresses confusion about not retrieving M from H using G-1.
  • Another participant questions the role of the metric tensor in converting between contravariant and covariant forms, suggesting that the computed G-1 may be incorrect due to an extra factor.
  • Several participants engage in calculating the determinant of G, with one asserting that the determinant is not what was previously assumed, leading to further discussion on the correct form of G-1.
  • One participant provides a detailed calculation of the determinant of G, concluding that it leads to a correct form of G-1 that satisfies the identity matrix condition when multiplied by G.
  • Another participant confirms that the issues with the earlier calculations seem to be resolved after correcting the form of G-1.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct computation of the determinant of G and the subsequent form of G-1. While some calculations lead to a consensus on the identity matrix condition, the initial confusion and differing interpretations of the determinant remain unresolved.

Contextual Notes

Limitations include potential errors in the computation of the determinant and the inverse of the metric tensor, as well as dependencies on the definitions of contravariant and covariant transformations. The discussion reflects ongoing refinement of these calculations without reaching a definitive conclusion.

Who May Find This Useful

This discussion may be useful for those studying tensor calculus, particularly in the context of transformations in physics and mathematics, as well as for individuals interested in the application of metric tensors in various coordinate systems.

nigelscott
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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?
 
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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?
 
Last edited:
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
 
nigelscott said:
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.
 
No, I thought I had 1/|detG| where |detG| = 1 - cos2α = sin2α
 
It's not \sin^2 \alpha either. Why don't you go through the whole calculation of the determinant?
 
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.
 
Okay, now that you're getting the identity for GG^{-1}, I expect the problems you were having will be resolved.
 
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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