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I am was reading the wikipedia page on metric tensors, when I saw something that was hard to grasp in the coordinate transformation section. This topic is a little bit uncomfortable to me, so maybe I have missed something, but there appears to be a factor of 4 that appears when working everything out by hand?

With r being a vector valued function $\vec{r}(u,\,v) = \bigl( x(u,\,v),\, y(u,\,v),\, z(u,\,v) \bigr)$, and with the following identity $$ \begin{bmatrix} \frac{\partial r}{\partial u}\frac{\partial r}{\partial u} & \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} \\ \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} & \frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \end{bmatrix} = \begin{bmatrix} E & F \\ F & G \end{bmatrix} $$

the coordinate transformation is given by,

$$ \begin{aligned} \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix}^\top \begin{bmatrix} E & F \\ F & G \end{bmatrix} \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix} \end{aligned} $$

With the following substitution in the coordinate transformation matrix,

$$ \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} $$

The transformation then becomes,

$$ \begin{aligned} \begin{bmatrix} A & B \\ C & D \end{bmatrix}^\top \begin{bmatrix} E & F \\ F & G \end{bmatrix} \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} \underbrace{A^2 E + 2 ACF + C^2G}_{E^\prime} & \underbrace{ABE + BCF + AFD + CDG}_{F^\prime} \\ \underbrace{ABE + BCF + AFD + CDG}_{F^\prime} & \underbrace{B^2E + 2BFD + D^2G}_{G^\prime} \end{bmatrix} \end{aligned} $$

Plugging the values into the variables give the following expressions,

$$ \begin{aligned} E^\prime &= \frac{\partial u}{\partial u^\prime}\frac{\partial u}{\partial u^\prime} \frac{\partial r}{\partial u}\frac{\partial r}{\partial u} + 2 \frac{\partial u}{\partial u^\prime}\frac{\partial v}{\partial u^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v} + \frac{\partial v}{\partial u^\prime}\frac{\partial v}{\partial u^\prime}\frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \\ F^\prime &= \frac{\partial u}{\partial u^\prime}\frac{\partial u}{\partial v^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial u} + \frac{\partial u}{\partial v^\prime}\frac{\partial v}{\partial u^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v} + \frac{\partial u}{\partial u^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v}\frac{\partial v}{\partial v^\prime} + \frac{\partial v}{\partial u^\prime}\frac{\partial v}{\partial v^\prime}\frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \\ G^\prime &= \frac{\partial u}{\partial v^\prime}\frac{\partial u}{\partial v^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial u} + 2\frac{\partial u}{\partial v^\prime}\frac{\partial r}{\partial u}\frac{\partial r}{\partial v}\frac{\partial v}{\partial v^\prime} + \frac{\partial v}{\partial v^\prime}\frac{\partial v}{\partial v^\prime}\frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \end{aligned} $$

after simplifying by cancelling out similar factors in the numerator and denominator and summing the result, everything comes out to,

$$ \require{cancel} \begin{aligned} E^\prime &= \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\cancel{\partial u}}{\partial u^\prime} \frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial u}} + 2 \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\cancel{\partial v}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}} + \frac{\cancel{\partial v}}{\partial u^\prime}\frac{\cancel{\partial v}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial v}}\frac{\partial r}{\cancel{\partial v}} \\ F^\prime &= \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\cancel{\partial u}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial u}} + \frac{\cancel{\partial u}}{\partial v^\prime}\frac{\cancel{\partial v}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}} + \frac{\cancel{\partial u}}{\partial u^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}}\frac{\cancel{\partial v}}{\partial v^\prime} + \frac{\cancel{\partial v}}{\partial u^\prime}\frac{\cancel{\partial v}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial v}}\frac{\partial r}{\cancel{\partial v}} \\ G^\prime &= \frac{\cancel{\partial u}}{\partial v^\prime}\frac{\cancel{\partial u}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial u}} + 2\frac{\cancel{\partial u}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial u}}\frac{\partial r}{\cancel{\partial v}}\frac{\cancel{\partial v}}{\partial v^\prime} + \frac{\cancel{\partial v}}{\partial v^\prime}\frac{\cancel{\partial v}}{\partial v^\prime}\frac{\partial r}{\cancel{\partial v}}\frac{\partial r}{\cancel{\partial v}} \end{aligned} $$

which then reduces to the following by adding the remaining terms,

$$ \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} = 4\begin{bmatrix} \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial u^\prime} & \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} \\ \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} & \frac{\partial r}{\partial v^\prime}\frac{\partial r}{\partial v^\prime} \end{bmatrix} $$

The Wikipedia article states (above equation 2') that the values of $E^\prime, F^\prime, G^\prime$ are in fact $ E^\prime = \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial u^\prime}, \;\; F^\prime = \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime}, \;\; G^\prime = \frac{\partial r}{\partial v^\prime}\frac{\partial r}{\partial v^\prime} $

which then means that I get the following expression after simplifying my manual transformation above, and comparing it with the definition of $E^\prime, F^\prime, G^\prime$ from Wikipedia.

$$ \begin{aligned} \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} \neq 4\begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} \end{aligned} $$

Questions

  1. How can I reconcile that this factor of 4 comes out? Is it just because the factor of 4 becomes irrelevant for an infinitesimal difference? Or have I made a terrible error somewhere?

  2. Generally, how can I understand coordinate transformations when dealing with a Jacobian matrix, are they the same thing as a change of basis in linear algebra when we see the form $P^{-1}AP$? What is the significance here that this form is $P^\top AP$ with a transpose instead of an inverse? The Jacobian is highly unlikely to be orthonormal (right?), so the transpose is definitely not the inverse.

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  • $\begingroup$ What following equality do you obtain? P.S. You are assuming $A,B,C,D$ all symmetric, which is beyond false. $\endgroup$ Commented Oct 14, 2022 at 1:17
  • $\begingroup$ I corrected the grammar to say that I am comparing the definition of $E^\prime, F^\prime, G^\prime$ from Wikipedia to my own derived expression, which are not equal. As far as I know, I am not assuming anything about $A, B, C, D$ being symmetric, the notation comes from the Wikipedia article and I believe they are all scalars $\endgroup$
    – Joff
    Commented Oct 14, 2022 at 1:29
  • $\begingroup$ OK, you’re right that this is surfaces only and these are scalars. You’re missing dot products, but that doesn’t account for the 4. I’ll need paper and pencil to check your work. The transpose appears instead of inverse because this is the transformation of a bilinear form, not linear transformation. $\endgroup$ Commented Oct 14, 2022 at 2:54
  • $\begingroup$ Everything looks fine once you put in dot products , but there are no 4s. I don’t understand where you got the 4s. $\endgroup$ Commented Oct 14, 2022 at 3:06
  • $\begingroup$ Which dot products are you referring to that I am missing? The fours come from the fact that the partial derivative cancel out and the terms I made that sum over all the partials ends up being equal to 4 $\endgroup$
    – Joff
    Commented Oct 14, 2022 at 3:10

1 Answer 1

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With r being a vector valued function $\vec{r}(u,\,v) = \bigl( x(u,\,v),\, y(u,\,v),\, z(u,\,v) \bigr)$, the partial derivatives $\frac{\partial r}{\partial u} \,\text{and}\, \frac{\partial r}{\partial v} \in \mathbb{R}^{1 \times 3}$, and with the following identity $$ \begin{bmatrix} \frac{\partial r}{\partial u}\frac{\partial r}{\partial u} & \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} \\ \frac{\partial r}{\partial u}\frac{\partial r}{\partial v} & \frac{\partial r}{\partial v}\frac{\partial r}{\partial v} \end{bmatrix} = \begin{bmatrix} E & F \\ F & G \end{bmatrix} = \begin{bmatrix} \frac{\partial r}{\partial u} \\ \frac{\partial r}{\partial v} \end{bmatrix} \begin{bmatrix} \frac{\partial r}{\partial u} & \frac{\partial r}{\partial v} \end{bmatrix} $$

It is convenient to view the matrix as an outer product as we will see, the coordinate transformation is given by,

$$ \begin{aligned} \begin{bmatrix} E^\prime & F^\prime \\ F^\prime & G^\prime \end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix}^\top \begin{bmatrix} \frac{\partial r}{\partial u} \\ \frac{\partial r}{\partial v} \end{bmatrix} \begin{bmatrix} \frac{\partial r}{\partial u} & \frac{\partial r}{\partial v} \end{bmatrix} \begin{bmatrix} \frac{\partial u}{\partial u^\prime} & \frac{\partial u}{\partial v^\prime} \\ \frac{\partial v}{\partial u^\prime} & \frac{\partial v}{\partial v^\prime} \\ \end{bmatrix} \end{aligned} = P^\top P $$

$P$ then becomes the total derivative w.r.t. the new variables $u^\prime$ and $v^\prime$

$$ \begin{aligned} P = \begin{bmatrix} \frac{\partial r}{\partial u}\frac{\partial u}{\partial u^\prime} + \frac{\partial r}{\partial v}\frac{\partial v}{\partial u^\prime} & \frac{\partial r}{\partial u}\frac{\partial u}{\partial v^\prime} + \frac{\partial r}{\partial v} \frac{\partial v}{\partial v^\prime} \end{bmatrix} = \begin{bmatrix} \frac{\partial r}{\partial u^\prime} & \frac{\partial r}{\partial v^\prime} \end{bmatrix} \end{aligned} \implies P^\top P = \begin{bmatrix} \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial u^\prime} & \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} \\ \frac{\partial r}{\partial u^\prime}\frac{\partial r}{\partial v^\prime} & \frac{\partial r}{\partial v^\prime}\frac{\partial r}{\partial v^\prime} \end{bmatrix} $$

which gives the answer as specified in the Wikpedia page. The problem with the initial version is that I was accumulating terms of the total derivative which were not in fact like terms, but they were contributions to the total derivative from changes in the new variables $u^\prime$ and $v^\prime$.

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