Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In polar coordinate,

$$\nabla U = \frac{\partial U}{\partial r}\hat{\mathbf{r}} + \frac{1}{r}\frac{\partial U}{\partial \theta}\hat{\mathbf{\theta}} .$$

I tried to use chain rule to get this result, but what I got from $\nabla U = \frac{\partial U}{\partial x}\hat{\mathbf{x}}+\frac{\partial U}{\partial y}\hat{\mathbf{y}}$ is some gibberish. Can anyone explain in detail how to get this result?

share|cite|improve this question
up vote 1 down vote accepted

Given that $$ r = \sqrt{x^2 + y^2},\qquad \theta = \arctan\left(\frac{y}{x}\right) $$ and $U = U(x,y)$, using the chain rule in two variables, we have $$ \frac{\partial U}{\partial x} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x} $$ $$ \frac{\partial U}{\partial y} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial y} $$ Now, $$ \frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2+y^2} = \frac{x}{\sqrt{x^2 + y^2}} = \cos \theta $$ $$ \frac{\partial r}{\partial y} = \frac{\partial}{\partial y} \sqrt{x^2+y^2} = \frac{y}{\sqrt{x^2 + y^2}} = \sin \theta $$ and $$ \frac{\partial \theta}{\partial x} = \frac{\partial}{\partial x} \arctan \left(\frac{y}{x}\right) = -\frac{y}{x^2 + y^2} = -\frac{\sin \theta}{r} $$ $$ \frac{\partial \theta}{\partial y} = \frac{\partial}{\partial y} \arctan \left(\frac{y}{x}\right) = \frac{y}{x^2 + y^2} = \frac{\cos \theta}{r} $$ hence $$ \nabla U = U_r \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} + \frac{1}{r} U_\theta \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix} $$ Lastly, the unitary vector in the radial direction $\hat{r}$ is $$ \hat{\textbf{r}} = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} $$ and due orthogonality $$ \hat{\theta} = \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix} $$ which leads to the desired form $$ \nabla U = U_r \hat{\textbf{r}} + \frac{1}{r} U_\theta \hat{\theta} $$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.