In Sean Carroll's Spacetime and Geometry An Introduction to General Relativity Chapter 2, there is an example of tensor transformation from $x,y$ coordinates to primed ones using $$(x',y') = (\frac{2x}{y},\frac{y}{2}).\tag{1}$$ The given tensor is $$ S_{\mu\nu} = \left(\begin{matrix} 1 & 0 \\ 0 & x^2 \\ \end{matrix}\right)\tag{2} $$ Which is a (0,2) tensor on a 2D manifold.

I want to calculate $S_{\mu'\nu'} $ which from what I understand is

$$S_{\mu'\nu'} = \frac{\partial{x^\mu}}{\partial{x^{\mu'}}} \frac{\partial{x^\nu}}{\partial{x^{\nu'}}} S_{\mu\nu} \tag{3}$$ in the given coordinates. (I mean the indices match right?)

Can someone show (refer) a detailed derivation of this? I essentially want to know what the form of the matrices of the transformation are. This will give a lot of clarification to the framework to me (and hopefully others).

Edit I: I have calculated all partials (correctly), the problem I have is in intuition, as to what the exact form of the matrices are, I have tried a LOT of combinations but I just can't get it right.

Edit II(answer): As a clarification of the notation used and the operations I'll post a detailed answer.

Given a tensor $S_{\mu\nu}$ in a 2 dimensional manifold the transformation to $S_{\mu'\nu'}$ follows from the equation above. The equation is perfectly fine when we consider the indices and the elements of each matrix. In matrix form though the equation is a little bit different.

Assume $x^1,x^2 = x,y$ and $x^{1'},x^{2'} = x',y'$

$$S_{\mu'\nu'} = \left(\begin{matrix} \frac{\partial{x^1}}{\partial{x^{1'}}} & \frac{\partial{x^2}}{\partial{x^{1'}}}\\ \frac{\partial{x^1}}{\partial{x^{2'}}} & \frac{\partial{x^2}}{\partial{x^{2'}}}\\ \end{matrix}\right) \left(\begin{matrix} 1 & 0\\ 0 & x^2\\ \end{matrix}\right) \left(\begin{matrix} \frac{\partial{x^1}}{\partial{x^{1'}}} & \frac{\partial{x^1}}{\partial{x^{2'}}}\\ \frac{\partial{x^2}}{\partial{x^{1'}}} & \frac{\partial{x^2}}{\partial{x^{2'}}}\\ \end{matrix}\right) $$ Where the far right matrix is the transpose of the far left. In this example $x' = \frac{2x}{y}$ and $y' = \frac{y}{2}$ which gives $x = x'y'$ and $y = 2y'$. Substitution of those gives:

$$S_{\mu'\nu'} = \left(\begin{matrix} y' & 0\\ x' & 2\\ \end{matrix}\right) \left(\begin{matrix} 1 & 0\\ 0 & x^2\\ \end{matrix}\right) \left(\begin{matrix} y' & x' \\ 0 & 2 \\ \end{matrix}\right) $$ Which eventually gives $$S_{\mu'\nu'} = \left(\begin{matrix} (y')^2 & y'x'\\ x'y' & (x')^2 + 4(x'y')^2\\ \end{matrix}\right) $$

  • $\begingroup$ Have you computed the partial derivatives? $\endgroup$
    – G. Smith
    Commented Mar 15, 2019 at 16:33

1 Answer 1


You have $y = 2 y'$ and $x = x' y'$ so

$$ \frac{\partial x^0}{\partial x^{0'}} = y' \\ \frac{\partial x^0}{\partial x^{1'}} = 0 \\ \mbox{etc.} $$

With this hint I hope you can find the others.

After that it goes $$ S_{\mu'\nu'} = \sum_{\mu=0}^1 \sum_{\nu=0}^1 K^{\;\mu}_{\mu'} K^{\;\nu}_{\nu'} S_{\mu\nu} $$ where $$ K^{\;\mu}_{\mu'} = \frac{\partial x^\mu}{\partial x^{\mu'}} $$ This can also be seen as the matrix product $$ S' = K S K^T $$ (to check your understanding of this point, note which row/column index is being summed over in the standard rules of matrix multiplication). But you don't have to use matrices if you don't want to.

  • $\begingroup$ Thank you Andrew, although I found the expressions for the partials I don't understand the form of the transformation matrices. That's the main point of my question. $\endgroup$ Commented Mar 15, 2019 at 20:26
  • $\begingroup$ @fielder ok I extended my answer a bit. $\endgroup$ Commented Mar 15, 2019 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.