First, sorry for an almost similar post (that I could delete).

I have a tensor of rank 2 with contravariant components expressed in a basis and noticed $S^{\mu \nu}$.

I just want to check how to justify the formula for the transformation of coordinates between the original basis and a new basis :

For example, we should have the relation :

$$S^{\mu^{\prime} v^{\prime}}=\frac{\partial x^{\mu^{\prime}}}{\partial x^\mu} \frac{\partial x^{\nu^{\prime}}}{\partial x^\nu} S^{\mu \nu}$$

But I wonder why we couldn't write rather :

$$S^{\mu^{\prime} v^{\prime}}=\frac{\partial x^{\mu}}{\partial x^\mu{^{\prime}}} \frac{\partial x^{\nu}}{\partial x^\nu{^{\prime}}} S^{\mu \nu}$$

Indeed, in General Relativity, we can write :

$$ds^2 = g_{\mu\nu} \text{dx}^{\mu} \text{dx}^{\nu} = \eta_{\mu '\nu '} \text{dx}^{\mu '} \text{dx}^{\nu'}$$

In my case, could we assimilate $\eta_{\mu' \nu'}$ to $g_{\mu'\nu'}$ to gt the following formula :

$$ds^2 = g_{\mu\nu} \text{dx}^{\mu} \text{dx}^{\nu} = g_{\mu '\nu '} \text{dx}^{\mu '} \text{dx}^{\nu '}$$

  • $\begingroup$ In the equation$$g_{\mu\nu} \text{dx}^{\mu} \text{dx}^{\nu} = g_{\mu '\nu '} \text{dx}^{\mu '} \text{dx}^{\nu '}$$you simply renamed the indices. What you probably actually meant is$$g(e_{\mu},e_{\nu})e^{\mu} e^{\nu} = g(\tilde e_{\mu},\tilde e_{\nu})\tilde e^{\mu} \tilde e^{\nu}$$ $\endgroup$
    – Filippo
    Jun 27 at 20:35
  • $\begingroup$ @Filippo : yes exactly, sorry . So what is the link with my issue about transformation of contravariant coordinates ? $\endgroup$
    – guizmo133
    Jun 27 at 20:38
  • $\begingroup$ I am trying to figure out whether I can help you without using terminology from differential geometry :D $\endgroup$
    – Filippo
    Jun 27 at 20:40
  • $\begingroup$ @Filippo I have a few notions about tensor calculus but it begins by doing a long time ... $\endgroup$
    – guizmo133
    Jun 27 at 20:42
  • $\begingroup$ @guizmo133 You can't in general relate $\eta$ to an arbitrary metric tensor $g$ like this. $\eta$ is the flat Minkowski metric which is always diagonal (and has the Lorentzian signature $(-+++)$ or the opposite...). Maybe you can explain why you are considering $\eta$ in the context of GR in the first place. $\endgroup$
    – Amit
    Jun 27 at 20:48

2 Answers 2


I would like to understand why we have this equivalence between the both equalities

If you accept that $\mathrm{d}(x')^\mu=\frac{\partial (x')^\mu}{\partial x^\nu}\mathrm{d}x^\nu$ and $S^{\mu\nu}=S(\mathrm{d}x^\mu,\mathrm{d}x^\nu)$, then we obtain $$(S')^{\mu\nu}=S(\mathrm{d}(x')^\mu,\mathrm{d}(x')^\nu)=\frac{\partial (x')^{\mu}}{\partial x^k} \frac{\partial (x')^{\nu}}{\partial x^l} S(\mathrm{d}x^k,\mathrm{d}x^l)=\frac{\partial (x')^{\mu}}{\partial x^k} \frac{\partial (x')^{\nu}}{\partial x^l} S^{kl}$$since $S$ is assumed to be bilinear.


$$\newcommand{r}[1]{\color{red}{#1}} \newcommand{g}[1]{\color{green}{#1}} \newcommand{b}[1]{\color{blue}{#1}}$$ The expressions $$S^{\mu^{\prime} v^{\prime}}=\frac{\partial x^{\r{\mu^{\prime}}}}{\partial x^\mu} \frac{\partial x^{\r{\nu^{\prime}}}}{\partial x^\nu} S^{\mu \nu}$$ and $$S^{\mu^{\prime} v^{\prime}}=\frac{\partial x^{\mu}}{\partial x^\r{\mu{^{\prime}}}} \frac{\partial x^{\nu}}{\partial x^\r{\nu{^{\prime}}}} S^{\mu \nu}$$ are not the same, and the latter is incorrect. The Einstein summation convention dictates that the repeated indices $(\mu,\nu)$ should appear exactly twice - once upstairs and once downstairs; futhermore, the free indices $(\mu',\nu')$ should appear exactly once, and are in the same position (upstairs or downstairs) on either side of an equality. Since the latter expression violates these rules, it is not covariant.

I think you're getting confused by the primes and overloading of the symbol $x$. If you have two coordinate systems $(x^1,x^2)$ and $(y^1,y^2)$, then you can express the components of $S$ in the $x$-basis or the $y$-basis. The relationship between them is

$$S_{(y)}^{\r{\mu \nu}} = \sum_{\g a=1}^2 \sum_{\b b=1}^2 \frac{\partial y^\r\mu}{\partial x^\g a} \frac{\partial y^\r\nu}{\partial x^\b b} S_{(x)}^{\g a\b b}$$ So for example, $$S^{\r{11}}_{(y)} = \frac{\partial y^\r1}{\partial x^\g1} \frac{\partial y^\r1}{\partial x^\b1} S^{\g1\b1}_{(x)}+\frac{\partial y^\r1}{\partial x^\g1} \frac{\partial y^\r1}{\partial x^\b2} S^{\g1\b2}_{(x)}+\frac{\partial y^\r1}{\partial x^\g2} \frac{\partial y^\r1}{\partial x^\b1} S^{\g2\b1}_{(x)}+\frac{\partial y^\r1}{\partial x^\g2} \frac{\partial y^\r1}{\partial x^\b2} S^{\g2\b2}_{(x)}$$

  • $\begingroup$ Thanks a lot for your quick answer. I would like to understand why we have this equivalence between the both equalities : $$g(e_{\mu},e_{\nu})e^{\mu} e^{\nu} = g(\tilde e_{\mu},\tilde e_{\nu})\tilde e^{\mu} \tilde e^{\nu} \quad \text{and}\quad S^{\mu^{\prime} v^{\prime}}\text{dx}^{\mu}\text{dx}^{\nu}= S^{\mu\nu} \text{dx}^{\mu'}\text{dx}^{\nu'}$$ $\endgroup$
    – guizmo133
    Jun 27 at 21:08
  • $\begingroup$ The first says that the tensor $g$ can be expressed either in the $\{e_\mu\}$ basis or the $\{\tilde e_\mu\}$ basis. You can prove that they're equal by noting that $\tilde e_\mu = T^\rho_{\ \ \mu} e_\rho$ and $\tilde e^\mu = (T^{-1})^\mu_{ \ \ \rho} e^\rho$ (by the definition of the dual basis $e^\mu$); plugging that in to your expression yields the equality. Your second expression is false. $\endgroup$
    – J. Murray
    Jun 27 at 22:06
  • $\begingroup$ Could you take a look please at my update ? I try to better understand where is my error. $\endgroup$
    – guizmo133
    Jun 27 at 22:23
  • $\begingroup$ @guizmo133 I may be misreading it, but I don't see a major error in your (now-deleted) update. The transformation rule for the metric (and any $(0,2)$-tensor) is $g'_{\mu\nu} = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} g_{\alpha\beta}$. $\endgroup$
    – J. Murray
    Jun 28 at 5:07
  • $\begingroup$ oh you are killing me, I have had to post another question on : physics.stackexchange.com/questions/769950/… . So finally here in this post, the expression $S^{\mu^{\prime} v^{\prime}}=\frac{\partial x^{\mu^{\prime}}}{\partial x^\mu} \frac{\partial x^{\nu^{\prime}}}{\partial x^\nu} S^{\mu \nu}$ is right since we are working with contravariant coordinates ( tensor(2,0) tensor) ? $\endgroup$
    – guizmo133
    Jun 28 at 5:16

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