Say I have a vector field expressed in Cartesian coordinates: $$\mathbf{A} = \sum_i A_i \mathbf{\hat{e}}_i$$ where the $\hat{\mathbf{e}}_i$ are the generalisation of the unit vectors $\mathbf{\hat i}, \mathbf{\hat j}, \mathbf{\hat k}$ from 3 dimensions.

I want to know how to determine the components $A_i'$ of the same vector field $\mathbf{A}$ expressed in terms of another orthonormal coordinate system: $$\mathbf{A} = \sum_i A_i' \mathbf{\hat{e}}_i'.$$ Of course, since $\mathbf{\hat{e}}_i'\cdot\mathbf{\hat{e}}_j' = \delta_{ij}$, we can determine these components as follows: $$A_j' = \mathbf{A}\cdot\mathbf{\hat{e}}_j' = \sum_i A_i \mathbf{\hat{e}}_i\cdot\mathbf{\hat{e}}_j'.$$ Now I think calculating $\mathbf{\hat{e}}_i\cdot\mathbf{\hat{e}}_j'$ in general is tedious. However, I've seen it written in various places (like in the book by Arfken, Weber and Harris) that for linear coordinate transformations, the new components can be calculated by $$A_j' = \sum_i A_i \frac{\partial x_j'}{\partial x_i}$$ where $x_i$ are the Cartesian coordinates and $x_j'$ are the new coordinates. For linear coordinate transformations, this makes sense, but I've also seen it used for general coordinate transformations, such as from Cartesian to curvilinear coordinates.

Is this in fact correct for transformations to curvilinear coordinates, i.e. does $\mathbf{\hat{e}}_i\cdot\mathbf{\hat{e}}_j' = \partial x_j' / \partial x_i$? And, if so, why?


1 Answer 1


You inadvertently stumbled upon the whole idea of curvilinear coordinates and its deep connection with differential geometry. This is the basic idea, in the basis $\{{\bf e}_i \}_i$ you can write the position vector of a point as

$$ {\bf x} = \sum_i x^i {\bf e}_i $$

Now imagine a smooth transformation of coordinates of the form $x^i = f^i({\bf q})$, and its inverse $q^i = g^i({\bf x})$. Examples include spherical coordinate $(q^1, q^2, q^3) = (r, \theta, \phi)$, cylindrical coordinates $(q^1, q^2, q^3) = (R, \phi, z)$, ... and of course linear transformations.

Now, the condition $q^i = f^i({\bf x}) = {\rm const}$ defines a surface. For example in the spherical case $q^1 = {\rm const}$ defines a sphere, $q^2 = {\rm const}$ defines a cone and $q^3 = {\rm const}$ a plane. You can think of a point as the intersection of these surfaces. And the unitary vectors associated with these new coordinates as the tangents to that surface along each coordinate.

So the notion of base vector is now location dependent, they point in different directions depending where you are standing, but what is important is to remember that they are tangent to the surfaces $f^i({\bf q}) = {\rm const}$. With this piece of information at hand you can now build them

$$ {\bf e}'_i \sim \frac{\partial {\bf x}}{\partial q^i} $$

Where I have omitted the sign "$=$" to emphasize that you'd need to normalize the result to ensure $|{\bf e}'_i| = 1$. Now let's go back to you question, assuming that the base $\{e_i \}_i$ is coordinate independent, e.g. the cartesian coordinate system

$$ {\bf e}_i \cdot {\bf e}'_j = {\bf e}_i \cdot \frac{\partial}{\partial q^j}_k\sum x^k({\bf q}) {\bf e}_k = \frac{\partial x^i({\bf q})}{\partial q^j} $$

  • $\begingroup$ This is great, thank you! I just have one question. If you include normalisation, $\mathbf{\hat{e}}_i' = \frac{\partial\mathbf{x}/\partial q^i}{|\partial\mathbf{x}/\partial q^i|}$, and so the final result isn't an equality but a proportionality. Is there any way to get around this? $\endgroup$
    – J-J
    Dec 8, 2018 at 21:10
  • 2
    $\begingroup$ @Jean-Jacq Not really, but actually you don't want. The normalization plays a huge role in the transformation. It defines the metric, which turns out to be a very fundamental concept $\endgroup$
    – caverac
    Dec 8, 2018 at 21:36
  • $\begingroup$ @caverac: Hi, your answer helped me out a lot. I did want to understand your answer in terms of tangent/cotangent spaces and asked a follow-up question (to your answer) here: math.stackexchange.com/questions/3490345/… I'd be super grateful if you could have a look and help me out with that! $\endgroup$
    – Shirish
    Dec 30, 2019 at 17:11
  • $\begingroup$ @ShirishKulhari Great! Happy it helped you out. Will look at it later tonight $\endgroup$
    – caverac
    Dec 30, 2019 at 17:12

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