I am following the procedure outlined in the book "Spacetime and Geometry" by Carroll. The objective is to transform a tensor. In the book, he has this example:

We have a (0,2) tensor of the following form: \begin{equation} S_{\mu \nu}= \left( \begin{array}{cc} 1 & 0\\ 0 & x^2 \end{array} \right) \end{equation}

We impose the following change of coordinates: \begin{equation} x' = \frac{2 x}{y} \end{equation} \begin{equation} y' = \frac{y}{2} \end{equation} which we can invert to give: \begin{equation} x = x'y' \end{equation} \begin{equation} y = 2y' \end{equation} We also have the following transformation relation for a tensor: \begin{equation} T^{\mu_1'\dots\mu_k'}_{\;\;\;\;\;\;\;\;\nu_1'\dots\nu_b' } = \frac{\partial x^{\mu_1'}}{\partial x^{\mu_1}}\dots \frac{\partial x^{\mu_k'}}{\partial x^{\mu_k}}\frac{\partial x^{\nu_1}}{\partial x^{\nu_1'}}\dots\frac{\partial x^{\nu_b}}{\partial x^{\nu_b'}}T^{\mu_1\dots\mu_k}_{\;\;\;\;\;\;\;\;\nu_1\dots\nu_b } \end{equation} In the book, he obtains the answer using another method, but says the reader should check to see that using this transformation equation will yield the same answer. So this is what I did.

So first off, in this case, $k=1$ and $b=1$, so we have : \begin{equation} T^{\mu'}_{\;\;\nu'}=\frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial x^{\nu}}{\partial x^{\nu'}} T^{\mu}_{\;\;\nu} \end{equation} Also in our case, we have a (0,2) tensor, while this is a (1,1) tensor. So we need to lower $\mu$. \begin{equation} T_{\mu\nu}=\eta_{\mu\sigma}T^{\sigma}_{\;\;\nu} \end{equation} I don't think this has any bearing on the transformation equation. We have: \begin{equation} \frac{\partial x^{\nu}}{\partial x^{\nu'}} = \left( \begin{array}{cc} y' & x'\\ 0 & 2 \end{array} \right) \end{equation} \begin{equation} \frac{\partial x^{\mu'}}{\partial x^{\mu}} = \left( \begin{array}{cc} \frac{2}{y} & \frac{-2x}{y^2}\\ 0 & \frac{1}{2} \end{array} \right) \end{equation}

This is where I get confused. These matrices are not in the same terms. One is in primed coordinates and one is in unprimed coordinates. I am guessing I just write everything in terms of primed coordinates, meaning I would get:

\begin{equation} S_{\mu'\nu'}= \left( \begin{array}{cc} \frac{1}{y'} & \frac{-x'}{2y'}\\ 0 & \frac{1}{2} \end{array} \right) \left( \begin{array}{cc} 1 & 0\\ 0 & (x'y')^2 \end{array} \right) \left( \begin{array}{cc} y' & x'\\ 0 & 2 \end{array} \right) \end{equation}

where the center matrix is just $S_{\mu\nu}$. This gives me: \begin{equation} S_{\mu'\nu'}= \left( \begin{array}{cc} 1 & 0\\ 0 & (x'y')^2 \end{array} \right) \end{equation}

Which is just $S_{\mu\nu}$ and not the right answer. The right answer is: \begin{equation} S_{\mu'\nu'}= \left( \begin{array}{cc} (y')^2 & x'y'\\ x'y' & (x')^2+4(x'y')^2 \end{array} \right) \end{equation}

Where did I go wrong?????? What am I not getting??

  • $\begingroup$ $k=0, b=2$ (not $1,1$) is the first mistake I can see. $\endgroup$
    – user107153
    Commented Sep 12, 2016 at 21:46
  • $\begingroup$ Ah yes! Good point! $\endgroup$
    – user41178
    Commented Sep 12, 2016 at 21:52

3 Answers 3


I find the notation that you use in your steps a bit confusing, so I suspect that your mistake comes from a wrong interpretation of the tensor transformation formulae. Anyway, this is how I calculate the first component of $S_{\mu'\nu'}$, which I will call $S_{x'x'}$:

$$S_{x'x'}=\frac{\partial x}{\partial x'}\frac{\partial x}{\partial x'}S_{xx}+\frac{\partial x}{\partial x'}\frac{\partial y}{\partial x'}S_{xy}+\frac{\partial y}{\partial x'}\frac{\partial x}{\partial x'}S_{yx}+\frac{\partial y}{\partial x'}\frac{\partial y}{\partial x'}S_{yy}$$

In your exercise:

$$\frac{\partial x}{\partial x'}=y'$$ $$\frac{\partial y}{\partial x'}=0$$ $$S_{xx}=1$$

and so

$$S_{x'x'}=y'^2 + 0+0+0= y'^2$$

With this example, you should be able to calculate the other three components correctly.

  • $\begingroup$ So what is up with the notation in the book for the tensor transformation formula? I copied it down exactly how it is written. Is it because the formula is for a mixed tensor, where the tensor I have has both indices lowered?? $\endgroup$
    – user41178
    Commented Sep 13, 2016 at 0:14
  • $\begingroup$ @user41178: That formula is for the most general tensor, with any number of up and down indices. You should have used the case of no up indices and two down indices. $\endgroup$
    – Javier
    Commented Sep 13, 2016 at 0:55
  • $\begingroup$ WThat is, you should not have used $\eta{_{\mu\nu}}$, but either do what @Javier says in his comment or calculate the $g{_{\mu\nu}}$ $\endgroup$
    – Bob Bee
    Commented Sep 13, 2016 at 1:22

Your tensor has two indices down, so just leave it like that. The transformation law would be

$$S_{\mu' \nu'} = \Lambda^\mu_{\ \ \mu'} \Lambda^\nu_{\ \ \nu'} S_{\mu\nu}$$


$$\Lambda^\mu_{\ \ \mu'} = \frac{\partial x^\mu}{\partial x^{\mu'}}$$

or, in matrix, notation,

$$S' = \Lambda^T S \Lambda$$

This way is usually the clearest, at least for two-index tensors. With more indices things get hairy.


You are using the minkowski metric to manipulate the indices. This is only valid in flat spacetime.


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