For tangent vectors at a single point, the transformation law is given by

$\tilde{X}^j = \frac{\partial \tilde{x}^j}{\partial x^i} (p) X^i$,

where $\tilde{X}^j$ are the components in the new coordinates, $X^i$ components of the old coordinates, $\frac{\partial \tilde{x}^j}{\partial x^i}$ is the partial derivative of the transition function, and $p$ is the coordinate representation of the point in the old coordinates.

So I figure that to get a vector field in new coordinates, what I do is take a point $\tilde{p}$ in my new coordinates, transform it back using the transition map to get $p$ in old coordinates, and apply the transformation law above.

I try to do a concrete example, and find that there is a mistake somewhere. Suppose I have the vector field $x \partial_x + y \partial_y$ in standard coordinates. I want to write this in polar coordinates as

$f(\theta, r) \partial_r + g(\theta, r) \partial_\theta$.

for some functions $f, g$. Applying the transformation law at some fixed point $(x, y)$, I get that the component of $\partial_r$ is

$2x \cdot \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot x + 2y \cdot \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot y = \sqrt{x^2 + y^2}$

and for $\partial_\theta$, I have

$-yx^{-2} \frac{1}{1+(y/x)^2} \cdot x + x^{-1} \frac{1}{1 + (y/x)^2} \cdot y = 0$

Thus, at any point $(x, y)$, the tangent vector $x \partial_x + y \partial_y$ is given by $\sqrt{x^2 + y^2} \partial_r + 0 \cdot \partial_\theta$. So for any $(r, \theta)$ in polar coordinates, I get $(r \cos \theta, r \sin \theta)$ in standard coordinates, which gives me $r \partial_r$ as my vector field.

But this is clearly wrong. For $x \partial_x + y \partial_y$ is the vector field which attaches the vector $(x, y)$ at the point $(x, y)$, while $r \partial_r$ attaches a horizontal vector of length $r$ at each point. Where have I made my mistake?

EDIT: I guess the computation is correct. I still feel it is a bit lengthy, is there any quick way to compute a change of coordinates from standard to polar? (i.e. substitute $x$ to $r \cos \theta$ in certain places, etc.)

  • 4
    $\begingroup$ Computation is right ! Remember that $\partial_r$ is not horizontal but orientated along $(x,y)$ $\endgroup$
    – marwalix
    Sep 10, 2011 at 22:26

1 Answer 1


Write $\partial_r=\cos{\theta}\partial_x+\sin{\theta}\partial_y$ and $\partial_\theta=-\sin{\theta}\partial_x+\cos{\theta}\partial_y$. Invert it $\partial_x=\cos{\theta}\partial_r-\sin{\theta}\partial_\theta$ and $\partial_y=\sin{\theta}\partial_r+\cos{\theta}\partial_\theta$. Than do the substitutions.


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