Solving a Tensor Problem in 2D Space

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SUMMARY

The discussion focuses on solving a tensor problem in a two-dimensional space defined by the metric tensor field g, represented as g=dq^1 ⊗ dq^1 + 2 dq^1 ⊗ dq^2 + 2dq^2 ⊗ dq^1 + 3dq^2 ⊗ dq^2. The covariant components of the contravariant vector field \textbf{v}=3∂_1 -4∂_2 are calculated using the inverse of the metric tensor G, which is G^{-1}=\left(\begin{array}{ll} {-3} & 2 \\ 2 & {-1} \end{array}\right). The resulting covariant components are a_1=-3 and a_2=-4, leading to the expression \textbf{v}=-3dq^1 -4dq^2. The calculations confirm the relationship a^i=a_k g^{ik} for tensor transformations.

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Homework Statement


A two dimensional space has a metric tensor field
[tex]g=dq^1 \otimes dq^1 + 2 dq^1 \otimes dq^2 + 2dq^2 \otimes dq^1 + 3dq^2 \otimes dq^2[/tex].

With the help of g, find the covariant components of the contravariant vector field
[tex]\textbf{v}=3\partial _1 -4\partial _2[/tex]

The Attempt at a Solution


Not much here as I don't really understand how this works.

g11=1, g12=2, g21=2, g22=3

[tex]\begin{center}G=\left(\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right) \Rightarrow<br /> G^{-1}=\left(\begin{array}{ll} {-3} & \ 2 \\ \ 2 & {-1} \end{array}\right)\end{center}[/tex]

g11=-3, g12=2, g21=2, g22=-1

Now the covariant components ai of v are given from the contravariant components ai like this:
a1=a1g11+a1g12=3*(-3)+3*2=-3
a2=a2g21+a2g22=(-4)*2+(-4)*(-1)=-4

So [tex]\textbf{v}=-3dq^1 -4dq^2[/tex]

Does this make any sense at all?
 
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[itex]a^i=a_k g^{ik}[/itex], yes? So don't you mean [itex]a^1=a_1 g^{11}+a_2 g^{12}[/itex]?
 
Right, of course. I should never ask anything when tired.

Thanks!
 

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