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I was reading the answer from here, which is showing $$\nabla = \partial_r e_r + \frac{1}{r} \partial_\theta e_\theta.$$ The answer actually made the choice $$e_r = (\cos\theta, \sin\theta), \quad e_\theta = (-\sin\theta, \cos\theta)$$ I would like to clarify this step more naturally.

More generally, let $g(x,y) = (u,v)$ be the change of variable map. Given a real valued function $f(x,y)$, we know in the new variable $\tilde f (u,v)= f(g^{-1}(u,v))$.

Now given a vector field $$v(x,y) = v_1\frac{\partial}{\partial x} + v_2\frac{\partial}{\partial y}$$ we would like to calculate $\tilde v(u,v)$ in the new variable, I think this is given by $dg(v)$ following the change of variable formula $$dg\left(\frac{\partial}{\partial x}\right) = \frac{\partial u}{\partial x} \frac{\partial}{\partial u}+\frac{\partial v}{\partial x} \frac{\partial}{\partial v}$$ $$dg\left(\frac{\partial}{\partial y}\right) = \frac{\partial u}{\partial y} \frac{\partial}{\partial u}+\frac{\partial v}{\partial y} \frac{\partial}{\partial v}$$ Let us denote the coordinates of $v$ as $[v] = (v_1, v_2)$, we get $$[\tilde v(u,v)] = (Jg)(g^{-1}(u,v))[v(g^{-1}(u,v))].$$

I tried the following calculation. Given $f(x,y)$, $h(r,\theta) = (x,y)$ the change of variable map, and $\tilde f = f \circ h$.

Given the vector field: gradient of $\tilde f$ $$\tilde v(r, \theta) = \frac{\partial \tilde f}{\partial r}\frac{\partial }{\partial r}+ \frac{1}{r^2}\frac{\partial \tilde f}{\partial \theta}\frac{\partial }{\partial \theta}$$ Then we see that $$dh(\tilde v(r, \theta)) = \frac{\partial \tilde f}{\partial r}dh\left(\frac{\partial }{\partial r}\right)+ \frac{1}{r^2}\frac{\partial \tilde f}{\partial \theta}dh\left(\frac{\partial }{\partial \theta}\right)\\ = \frac{\partial \tilde f}{\partial r}\left(\cos\theta \frac{\partial }{\partial x}+ \sin\theta \frac{\partial }{\partial y}\right)+ \frac{1}{r^2}\frac{\partial \tilde f}{\partial \theta}\left(-{r}\sin\theta \frac{\partial }{\partial x}+ {r}\cos\theta \frac{\partial }{\partial y}\right)\\ = \left(\cos\theta\frac{\partial \tilde f}{\partial r} -\frac{1}{r}\sin\theta \frac{\partial \tilde f}{\partial \theta}\right) \frac{\partial }{\partial x}+ \left(\sin\theta\frac{\partial \tilde f}{\partial r} +\frac{1}{r}\cos\theta \frac{\partial \tilde f}{\partial \theta}\right) \frac{\partial }{\partial y}\\ = \frac{\partial f }{\partial x}\frac{\partial }{\partial x}+ \frac{\partial f }{\partial y}\frac{\partial }{\partial y}. $$

Now $e_r = dh\left(\frac{\partial }{\partial r}\right)$ and $e_\theta = \frac{1}{r} dh\left(\frac{\partial }{\partial \theta}\right)$

Xiao
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    What do you mean you didn't make any choice of basis for the tangent vectors? Once you choose a coordinate system, there's a natural choice for the tangent vectors. For instance in the $(x, y)$- coordinate, the basis vectors are $\partial_x$ and $\partial_y$. Whereas, in polar coordinates, the basis is given by $\partial_r$ and $\partial_\theta$. In the example, the confusion arises from the fact that you were expressing the polar tangent vectors in $\partial_x$ and $\partial_y$. – Jacky Chong Feb 12 '18 at 02:10
  • I meant I did not make an explicit choice, I think $e_r$ should be $\frac{\partial}{\partial r}$ instead of $(\cos\theta, \sin\theta)$ – Xiao Feb 12 '18 at 02:18
  • Okay, I think I kind of understand it now, as you said, in the answer, the $e_r$ is actually $df(\partial_r)$ (where $f(r,\theta) = (x,y)$, when written in $\partial_x, \partial_y$ coordinate, it is $(\cos\theta, \sin\theta)$. – Xiao Feb 12 '18 at 02:36

1 Answers1

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To clarify the confusion, observe that in Cartesian coordinates, we have that \begin{align} [\nabla f]_{\text{Cartesian}} = \frac{\partial f}{\partial x}\mathbf{e}_1+ \frac{\partial f}{\partial y}\mathbf{e}_2 \end{align} where $\mathbf{e}_i$ are the Cartesian basis vectors. However, we also know that \begin{align} \frac{\partial f}{\partial x} =& \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}+\frac{\partial f}{\partial r}\frac{\partial r}{\partial x} = -\frac{\partial f}{\partial \theta}\frac{y}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{x}{\sqrt{x^2+y^2}}\\ \frac{\partial f}{\partial y} =& \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y}+\frac{\partial f}{\partial r}\frac{\partial r}{\partial y} = \frac{\partial f}{\partial \theta}\frac{x}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{y}{\sqrt{x^2+y^2}} \end{align} which means \begin{align} [\nabla f]_{\text{Cartesian}} =&\ \left( -\frac{\partial f}{\partial \theta}\frac{y}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{x}{\sqrt{x^2+y^2}}\right)\mathbf{e}_1+\left( \frac{\partial f}{\partial \theta}\frac{x}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{y}{\sqrt{x^2+y^2}}\right)\mathbf{e}_2\\ =&\ \frac{\partial f}{\partial r}\left( \frac{x}{\sqrt{x^2+y^2}}\mathbf{e}_1+\frac{y}{\sqrt{x^2+y^2}}\mathbf{e}_2\right)+\frac{1}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial \theta}\left( -\frac{y}{\sqrt{x^2+y^2}}\mathbf{e}_1+\frac{x}{\sqrt{x^2+y^2}}\mathbf{e}_2\right) \end{align} where $\partial f/\partial r$ and $\partial f/\partial \theta$ are expressed in terms of $(x, y)$.

However, if you choose to use polar coordinates in space, i.e., replace $x = r\cos\theta$ and $y = r\sin\theta$ then you will see that \begin{align} [\nabla f]_{\text{Cartesian}} = \frac{\partial f}{\partial r}\cdot (\cos\theta \ \mathbf{e}_1+ \sin\theta\ \mathbf{e}_2) + \frac{1}{r}\frac{\partial f}{\partial \theta}\cdot (-\sin\theta\ \mathbf{e}_1+\cos\theta \mathbf{e}_2). \end{align}

Despite the fact that you have changed to polar coordinates, you are still expressing your tangent vectors in terms of the cartesian basis.

Jacky Chong
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