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Suppose that, a rank-$2$ tensor transforms as

\begin{align} T'^{ij}=\frac{\partial x'^i}{\partial x^k}\frac{\partial x'^k}{\partial x^l}T^{kl}. \end{align}

How to use this criterion to investigate if the following $2\times2$ matrix (as an example) represents a tensor?

\begin{bmatrix} xy & 1 \\ x^2-y^2 & e^x \\ \end{bmatrix}

Note that:

  • I think this example could be so useful because most textbooks prefer not to illustrate details. In fact, the question is: How (practically) one can find that, when a "thing" is a tensor or not?
  • This could be a good exercise and clarify many ambiguities. Since the focus is on this specific example, there will not be any duplication.
Qmechanic
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Perfect Fluid
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1 Answers1

1

Too long for a comment.

  • In the coordinate system where you label your coordinates by $(x,y)$ any $2\times 2$-matrix whose elements are functions of $(x,y)$ gives rise to the four tensors $$\tag1 a_{ij}\,dx^i\otimes dx^j\,,\quad a_i{}^j\,dx^i\otimes\partial_j\,,\quad a^i{}_j\,\partial_i\otimes dx^j\,,\quad a^{ij}\partial_i\otimes\partial_j $$ (I wrote $(x,y)=(x^1,x^2)\,.$)

  • When you change the coordinates the components $a$ change. In case of $a^{ij}$ they change by a formula like you wrote it in OP.

  • You may want to verify this in the example of the Euclidean metric in Cartesian and polar coordinates: $$ \mathbf{g}=dx\otimes dx+dy\otimes dy=dr\otimes dr+r^2\,d\theta\otimes d\theta. $$

  • For tensors whose components are derivatives it is typically required that their components are derivatives in all coordinate systems. This leads to the concept of covariant derivative which is well-known.

  • As a little exercise you could take the transformation rules \begin{align} \partial_r&=\cos\theta\,\partial_x+\sin\theta\,\partial_y\,,&\partial_\theta&=-r\sin\theta\,\partial_x+r\cos\theta\,\partial_y\,,\\[2mm] dx&=\cos\theta\,dr-r\sin\theta\,d\theta\,, &dy&=\sin\theta\,dr+r\cos\theta\,d\theta \end{align} then transform $dx\otimes \partial_x$ and so on and obtain all four tensors from (1) in polar coordinates.

Kurt G.
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