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I have $\Gamma$ a manifold and $f:\Gamma \to \mathbb R$. We have in $(x,y)$ that $f(x,y)=x^2+y^2$. Now, in my lecture, they denote $(r,\theta )$ the polar coordinates. And after, they wrote : since $$\partial _r =\cos\theta \partial _x +\sin(\theta )\partial _y$$ we have $$\partial _rf(r,\theta )=\cos(\theta )\partial _x f(r,\theta )+\sin(\theta )\partial _y f(r,\theta ).$$

  1. First, the notation $f(r,\theta )$ look strange. Does $f(r,\theta )$ means $f(r\cos \theta ,r\sin \theta )$ ? If yes, is it a common notation ?

  2. Moreover, the notation $\partial _xf(r,\theta )$ looks strange. Shouldn't it be $\partial _x f(x,y)$ ? (same with $\partial _y f(r,\theta )$).

  3. Finally, I have to give the gradient in $(r,\theta )$. My teacher wrote $$\nabla _{(x,y)}f(x,y)=\frac{1}{r}\begin{pmatrix}r\cos \theta &-\sin \theta\\ -r\sin \theta &\cos \theta \end{pmatrix}\nabla _{r,\theta }f(r,\theta ).$$

For me it doesn't make sense. Why don't we simply don't have : set $h(r,\theta )=f(r\cos \theta ,r\sin \theta )$. We have that $f(r\cos \theta ,r\sin \theta )=r^2$. Therefore
$$\nabla _{r,\theta }h(r,\theta )=(2r,0).$$ Is this wrong ? If yes, why ?

user841366
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  • (1) Using the same name is a rather common (bad) habit of physicists. (2) Is the chain rule using the bad notation (1). $\partial_x f(r,\theta )$ is actually $(\partial_x f)(r\cos \theta,r\sin \theta)$. – Martín-Blas Pérez Pinilla Feb 20 '21 at 21:31
  • @Martín-BlasPérezPinilla: could you please be more precise for (1) ? Because I'm not a physicist, and I would have written $f(r,\theta )$ for $f(r\cos\theta ,r\sin\theta )$ (and I think that the latter notation is never used, unless for beginner students in differential geometry) – Surb Feb 26 '21 at 00:56
  • @Surb, they are different functions. See my answer. – Martín-Blas Pérez Pinilla Feb 26 '21 at 05:21
  • @Martín-BlasPérezPinilla : as state in the OP question, $f$ is not defined on $\mathbb R^2$ but on a manifold $\Gamma $. Therefore, $f(r,\theta ):=f(\varphi ^{-1}(u))$ , $u\in \Gamma $ for some chart $\varphi $ as $f(x,y):=f(\psi^{-1}(v))$, $v\in \Gamma $ for some chart $\psi$. When you write $\varphi (r,\theta )=f(r\cos\theta ,r\sin \theta )$ you just express $f$ in some coordinate system $(r,\theta )$ that is normally written as $f(r,\theta )$... – Surb Feb 26 '21 at 12:34

1 Answers1

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Let be $\varphi(r,\theta) = f(r\cos\theta,r\sin\theta)$. Writing the vectors as column vectors and using the chain rule: $$ \begin{pmatrix} \partial_r\varphi(r,\theta)\\ \partial_\theta\varphi(r,\theta) \end{pmatrix} = \nabla_{r,\theta}\varphi(r,\theta) = (\nabla f)(r\cos\theta,r\sin\theta) \nabla_{r,\theta} \begin{pmatrix} r\cos\theta\\ r\sin\theta \end{pmatrix} = (\nabla f)(r\cos\theta,r\sin\theta) \begin{pmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta \end{pmatrix}. $$ Can you continue?

Martín-Blas Pérez Pinilla
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  • My problem is not to understand how to do the calculation. For me, the gradient in polar would be $\partial _{r,\theta }g(r,\theta )$ where $g(r,\theta )=f(r\cos\theta ,r\sin\theta )$. Why we have to express $\partial _xf$ and $\partial _fy$ as function of $\partial _r$ and $\partial _\theta $ ? – user841366 Feb 21 '21 at 09:42
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    @user841366, again, the problem is the bad and confusing notation. I think that your teacher wanted $\nabla f$ "in the new coordinates" in general (for any $f$). BTW, my $\varphi$ is your $h$ as column vector. – Martín-Blas Pérez Pinilla Feb 21 '21 at 09:51