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Consider the polar coordinate transformation

$$ x = r\cos \theta $$ $$ y = r\sin \theta $$

I am trying to find the most direct way to compute the coordinate basis vectors

$$ \frac{\partial}{\partial x}, \frac{\partial}{\partial y} $$

The exterior derivatives of $x$ and $y$ are given by

$$ dx = \cos \theta dr - r\sin\theta d\theta $$ $$ dy = \sin \theta dr + r \cos \theta d\theta $$ Considering that $dx$ and $dy$ comprise a basis in the cotangent space dual to the tangent space with basis elements $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, in principle, one ought to be able to compute the latter by using the duality relations $$ dx \left(\frac{\partial}{\partial x}\right) = 1, dx \left(\frac{\partial}{\partial y}\right) = 0, dy \left(\frac{\partial}{\partial y}\right) = 1, dy \left(\frac{\partial}{\partial x}\right) = 0 $$

Unfortunately, I am unable to make the algebra for this work out as expected. Is this approach promising, and perhaps I'm simply missing some algebraic manipulations or is there a better approach to this problem?

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3 Answers 3

up vote 4 down vote accepted

Yes, it is promising. Write $$ \frac{\partial}{\partial x} = a_{xr} \frac{\partial}{\partial r} + a_{x \theta} \frac{\partial}{\partial \theta} \qquad \frac{\partial}{\partial y} = a_{yr} \frac{\partial}{\partial r} + a_{y \theta} \frac{\partial}{\partial \theta} $$ Now $$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \langle \mathrm{d} x, \frac{\partial}{\partial x} \rangle & \langle \mathrm{d} y, \frac{\partial}{\partial x} \rangle \\ \langle \mathrm{d} x, \frac{\partial}{\partial y} \rangle & \langle \mathrm{d} y, \frac{\partial}{\partial y} \rangle \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin \theta \\ -r \sin\theta & r \cos \theta \end{pmatrix} \cdot \begin{pmatrix} \langle \mathrm{d} r, \frac{\partial}{\partial r} \rangle & \langle \mathrm{d} \theta, \frac{\partial}{\partial r} \rangle \\ \langle \mathrm{d} r, \frac{\partial}{\partial \theta} \rangle & \langle \mathrm{d} \theta, \frac{\partial}{\partial \theta} \rangle \end{pmatrix} \cdot \begin{pmatrix} a_{x r} & a_{x \theta} \\ a_{y r} & a_{y \theta} \end{pmatrix} $$ Thus $$ \begin{pmatrix} a_{x r} & a_{x \theta} \\ a_{y r} & a_{y \theta} \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin \theta \\ -r \sin\theta & r \cos \theta \end{pmatrix}^{-1} = \begin{pmatrix} \cos\theta & -\frac{1}{r}\sin \theta \\ \sin\theta & \frac{1}{r} \cos \theta \end{pmatrix} $$ Thus $$ \frac{\partial}{\partial x} = \cos(\theta) \frac{\partial}{\partial r} - \sin(\theta) \frac{1}{r} \frac{\partial}{\partial \theta} \qquad \frac{\partial}{\partial y} = \sin(\theta) \frac{\partial}{\partial r} + \cos(\theta) \frac{1}{r} \frac{\partial}{\partial \theta} $$

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Nice! I can see where I went wrong now. –  ItsNotObvious Jan 10 '12 at 20:19

One way is to use the chain rule: $$\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial\theta}$$ and similarly for $\frac{\partial}{\partial y}$.

Your method should also work, however; can you show us what went wrong?

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But this would imply, would it not, that one would have to apply the inverse function theorem to obtain the partials of $r$ and $\theta$? I can see how this would probably work, but I'm hoping for something that doesn't require as much technology; maybe I can't get there without it though.. –  ItsNotObvious Jan 10 '12 at 19:58
    
In general you would use the inverse function theorem, but in this case the inverse functions are easy to find explicitly: $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(y/x)$. –  Skatche Jan 10 '12 at 20:05

using $x=r\cos\theta, y=r\sin\theta, r=\sqrt{x^2+y^2}, \theta=\arctan(y/x)$ we have $$ \frac{\partial r}{\partial x}=\cos\theta, \frac{\partial r}{\partial y}=\sin\theta, \frac{\partial \theta}{\partial x}=-\frac{\sin\theta}{r},\frac{\partial \theta}{\partial y}=\frac{\cos\theta}{r} $$ then using the chain rule $$ \frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial\theta}=\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta} $$ $$ \frac{\partial}{\partial y}=\frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial y}\frac{\partial}{\partial\theta}=\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta} $$

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