# How to obtain the gradient in polar coordinates

I'm not sure on how to find the gradient in polar coordinates. The thing that troubles me the most is how to find the unit vectors $\hat{r}$ and $\hat{\theta}$. My approach for the rest is expressing the partial derivatives in respect of $r$ and $\theta$ using the chain rule.

How can I get around solving this problem?

• The Maple code $$with(Student[VectorCalculus]); Gradient(F(r, theta), 'polar'[r, theta])$$ produces $${\frac {\partial }{\partial r}}F \left( r,\theta \right) \, \mathbf{ \bar{e}_{r}}+{\frac {{\frac {\partial }{\partial \theta}}F \left( r,\theta \right) }{r}}\, \mathbf{ \bar{e}_{\theta}} .$$ – user64494 Nov 30 '13 at 14:29
• Polar coordinates in the form $x=r\cos\theta$ and $y=r\sin\theta$. How can we prove it though? – arcbloom Nov 30 '13 at 14:30
• See that link to this end. It can be found by the "gradient in polar coordinates" googling. – user64494 Nov 30 '13 at 17:58

## 2 Answers

The gradient operator in 2-dimensional Cartesian coordinates is $$\nabla=\hat{\pmb e}_{x}\frac{\partial}{\partial x}+\hat{\pmb e}_{y}\frac{\partial}{\partial y}$$ The most obvious way of converting this into polar coordinates would be to write the basis vectors $\hat{\pmb e}_x$ and $\hat{\pmb e}_{y}$ in terms of $\hat{\pmb e}_{r}$ and $\hat{\pmb e}_{\theta}$ and write the partial derivatives $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ in terms of $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ using the chain rule. So we have: \begin{align} \hat{\pmb e}_{x}&=\cos\theta\, \hat{\pmb e}_{r}-\sin\theta\, \hat{\pmb e}_{\theta} \\ \hat{\pmb e}_{y}&=\sin\theta\, \hat{\pmb e}_{r}+\cos\theta\, \hat{\pmb e}_{\theta} \\ &\\ \frac{\partial}{\partial x}&=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\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} \end{align} Observing that $r=\sqrt{x^2+y^2}$ and $\theta=\arctan\left(\frac{y}{x}\right)$, we have \begin{align} \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} \end{align} and \begin{align} \nabla&=\hat{\pmb e}_{x}\frac{\partial}{\partial x}+\hat{\pmb e}_{y}\frac{\partial}{\partial y}\\ &=(\cos\theta\, \hat{\pmb e}_{r}-\sin\theta\, \hat{\pmb e}_{\theta})\left(\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial \theta}\right)+(\sin\theta\, \hat{\pmb e}_{r}+\cos\theta\, \hat{\pmb e}_{\theta})\left( \frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial y}\frac{\partial}{\partial \theta} \right)\\ &=\ldots\\ &=\hat{\pmb e}_{r}\frac{\partial }{\partial r}+\hat{\pmb e}_{\theta}\frac{1}{r}\frac{\partial }{\partial \theta}. \end{align}

This certainly gives the right answer, but there is a quicker way. Consider a function $f(\pmb r)$ in polar coordinates: this is a function $f(r,\theta)$. The small change in going from the point $\pmb r$ with coordinates $(r,\theta)$ to the point $\pmb r+\operatorname{d}\pmb r$ with coordinates $(r + \operatorname{d}r,\theta + \operatorname{d}\theta)$ is $$\operatorname{d}f=\frac{\partial f}{\partial r}\operatorname{d}r+\frac{\partial f}{\partial \theta}\operatorname{d}\theta \tag 1$$ Observe that $\operatorname{d}f=\operatorname{d}\pmb r\cdot \nabla f$ and $\operatorname{d}\pmb r=\operatorname{d}r\hat{\pmb e}_{r}+r\operatorname{d}\theta \hat{\pmb e}_{\theta}$. Suppose, then, that $$\nabla f=\alpha\hat{\pmb e}_{r}+\beta \hat{\pmb e}_{\theta}$$ where $\alpha$ and $\beta$ are to be found. We get $$\operatorname{d}f=\operatorname{d}\pmb r\cdot \nabla f=\left(\operatorname{d}r\hat{\pmb e}_{r}+r\operatorname{d}\theta \hat{\pmb e}_{\theta}\right)\cdot\left(\alpha\hat{\pmb e}_{r}+\beta \hat{\pmb e}_{\theta}\right)=\alpha\operatorname{d}r+\beta r\operatorname{d}\theta \tag 2$$ because $\hat{\pmb e}_{r}\cdot\hat{\pmb e}_{r}=\hat{\pmb e}_{\theta}\cdot \hat{\pmb e}_{\theta}=1$ and $\hat{\pmb e}_{\theta}\cdot \hat{\pmb e}_{r}=0$. Comparing (1) and (2) we see that $\alpha=\frac{\partial f}{\partial r}$ and $\beta=\frac{1}{r}\frac{\partial f}{\partial \theta}$. Therefore, we get $$\nabla f=\hat{\pmb e}_{r}\frac{\partial f}{\partial r}+\hat{\pmb e}_{\theta}\frac{1}{r}\frac{\partial f}{\partial \theta}$$ and we can identify the gradient operator itself as $$\nabla =\hat{\pmb e}_{r}\frac{\partial }{\partial r}+\hat{\pmb e}_{\theta}\frac{1}{r}\frac{\partial }{\partial \theta}.$$

• That seems pretty alright, however, how did you deduce the angular and radial unit vectors is the part I'm having the most trouble with. Is it a definition or how can you deduce them? – arcbloom Nov 30 '13 at 23:41
• See the figure that I added. – alexjo Dec 1 '13 at 1:10
• You give $r = \sqrt{x^2 + y^2}$, and then say $\frac{\partial r}{\partial x} = \cos\theta$, whereas direct computation yields $\frac{x}{\sqrt{x^2+y^2}$. I see how this is $\cos\theta$, but it effectively requires foreknowledge of the result to express it in this way, no? Any more direct approach? – Rax Adaam Nov 21 '17 at 14:48

This version is for those, who prefer a different notation. $$u \circ \sigma = f, \text { where } \sigma(x,y) = \left(\sqrt{x^2+y^2}, \arctan\frac yx\right)$$ Fix $a=(x,y) = r(\cos\theta,\sin\theta)$. Let $(Df)(a)$ and $(\nabla f)(a)$ denote the linear operator and the vector representing it. By the chain rule, $$(Df)(a) = (Du)(\sigma(a)) \circ (D\sigma)(a).$$ Implying that \begin{align} (\nabla f)(a) &= (\nabla u)(\sigma(a))\cdot \pmatrix{ \cos\theta & \sin\theta \\ -\frac 1 r\sin\theta & \frac 1r\cos\theta} \\ &= u_r(\sigma(a))\cdot(\cos\theta , \sin\theta) + \frac 1r u_{\theta}(\sigma(a))\cdot(-\sin\theta, \cos \theta). \end{align} Let's hold on for a minute and observe our result: The vector $(\nabla f)(a)$ is a linear combination of the vectors $(\cos\theta,\sin\theta)$, and $(-\sin\theta, \cos\theta)$.

In fact, \begin{align}\{\, \boldsymbol{e_r} &= (\cos\theta,\sin\theta)\\ , \boldsymbol{e_{\theta}} &= (-\sin\theta, \cos\theta)\}\end{align} is a basis for $\mathbb R^2$. Using our new notation, the equation above becomes

$$(\nabla f)(a) = u_r(\sigma(a))\cdot \boldsymbol{e_r} + \frac 1r u_{\theta}(\sigma(a))\cdot \boldsymbol{e_{\theta}} = \left(u_r(\sigma(a)), \frac 1r u_{\theta}(\sigma(a))\right) = (\nabla(u))(\sigma(a))\cdot\pmatrix{1 \\ \frac 1r},$$

or equivalently, $$(\nabla f)(r(\sin\theta, \cos\theta)) = \left(u_r(r, \theta), \frac 1r u_{\theta}(r, \theta)\right) = (\nabla u)(r,\theta)\cdot\pmatrix{1 \\ \frac 1r}.$$

If we start omitting a few variables, things become prettier, but also somewhat easier to misunderstand.

$$\nabla f = \nabla u \cdot \pmatrix{1 \\ \frac 1r} = \left(u_r, \frac 1r u_{\theta}\right) = \boldsymbol{e_r} u_r + \boldsymbol{e_{\theta}} \frac 1r u_{\theta} =\boldsymbol{e_r} \frac{\partial u}{\partial r}+ \boldsymbol{e_{\theta}}\frac 1r \frac{\partial u}{\partial \theta}$$ We can also omit function names, leading to

$$\nabla = \boldsymbol{e_r} \frac{\partial}{\partial r}+ \boldsymbol{e_{\theta}} \frac 1r \frac{\partial}{\partial \theta}.$$

## protected by Zev ChonolesAug 26 '16 at 18:20

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