Tensors Questions: Seeking Guidance

[peripatein]

Hi,

Homework Statement

I recently started delving into tensor calculus and am quite the stuck with the following:
Given the tensor Aⁱ = (x+y, y-x, z)_i in cartesian coordinates, what would be the second covariant coordinate in cylindrical coordinates?
AND
Given the tensor A_i^j = (-1 0, -1 1)_ij and the metric g_ij = (2 3, 3 4)^ij, what would be A²₁?

Homework Equations

The Attempt at a Solution

First, aren't I actually expected to find y-x in cylindrical coordinates, which is rsinθ - rcosθ? I have found the metric to be (1 0 0, 0 r² 0, 0 0 1), but I am really not sure how to put all the pieces together and how to proceed.
Next, for finding A²₁ won't I actually need to multiply the given matrix by the metric and its inverse, thus yielding a similar matrix as the original?
I could use some guidance, please.

 

[peripatein]

I have managed to answer the first part. Now I am mainly stymied by the second, viz. how to find A²₁. Could anyone please help?

 

[pdxautodidact]

I have managed to answer the first part. Now I am mainly stymied by the second, viz. how to find A²₁. Could anyone please help?

Contract twice?: $$ g^l_i g^j_k A^k_l = A^i_j$$ not entirely sure I understand the problem you've posed.

Which is what I think you said, you'll have $g_{il} g^{jk}$ i.e. the metric and its inverse times the (1,1) tensor. I think I'd write out the summation instead of matrix multiplication, though, because $g_{il}g^{jk} = \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$ seems almost too easy and it's 4 in the morning and I'm ready for sleep, not more math.

 

[peripatein]

What I first got, albeit did not write it here, was g_1jg^2jA_jA_j. Does that agree with what you wrote as the solution?

 

[pdxautodidact]

What I first got, albeit did not write it here, was g_1jg^2jA_jA_j. Does that agree with what you wrote as the solution?

I don't think so, but I'm inclined to say no because you need the metric acting on the (1,1) tensor twice. Your notation alone reverts it to two (0,1) tensors with only one summation. I need sleep but I'm off tomorrow, so I'll give it some thought if you haven't figured it out all the way, I can write write it out..

cheers