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Ok so I have a given vector field in Cartesian coordinates, say \begin{align*} \textbf{v}(x,y)=\frac{dx}{dt}\hat{\textbf{e}}_{1}+\frac{dy}{dt}\hat{\textbf{e}}_{2} \end{align*} Where $dx/dt$ and $dy/dt$ are functions of $x$ and $y$. I want to transform this into a polar coordinate field such as \begin{align*} \textbf{u}(r,\theta)=\frac{dr}{dt}\hat{\textbf{r}}+\frac{d\theta}{dt}\hat{\boldsymbol\theta} \end{align*} Where $dr/dt$ and $d\theta/dt$ are functions of $r$ and $\theta$I know the transformation. \begin{align*} \begin{pmatrix} dr/dt \\ rd\theta/dt \end{pmatrix} = \begin{pmatrix} \text{cos}(\theta) & \text{sin}(\theta)\\ -\text{sin}(\theta) & \text{cos}(\theta) \end{pmatrix} \begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix}. \end{align*} Can I just change my Cartesian vector field which is given in $x,y$ to $r,\theta$ and then apply that transformation matrix?

To be clear an example would be \begin{align*} \textbf{v}=(x,0)\\ \\ \text{which becomes}\\\\ \textbf{u}=(r\cos^{2}\theta,-\sin(\theta)\cos(\theta)) \end{align*} Many thanks

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  • $\begingroup$ what is the meaning of each $\hat{\textbf{r}},\hat{\boldsymbol\theta}$? $\endgroup$ – janmarqz Feb 27 '14 at 14:22
  • $\begingroup$ $\hat{\textbf{r}}=cos(\theta)\textbf{i}+sin(\theta)\textbf{j}$ is the unit vector in the radial direction and similarly for $\hat{\boldsymbol\theta}$ $\endgroup$ – Michael Feb 28 '14 at 15:53
  • $\begingroup$ are you requiring that $\hat\theta$ is orthonormal with $\hat{r}$ and oriented? $\endgroup$ – janmarqz Feb 28 '14 at 16:24
  • $\begingroup$ Yes, so $\hat{\boldsymbol\theta}=-sin(\theta)\textbf{i}+cos(\theta)\textbf{j}$ $\endgroup$ – Michael Mar 2 '14 at 9:32
  • $\begingroup$ last question before i could assemble an answer: are they $\{\hat{\boldsymbol e}_1,\hat{\boldsymbol e}_2\}$ and $\{\boldsymbol i,\boldsymbol j\}$ the same basis for you? $\endgroup$ – janmarqz Mar 2 '14 at 16:38
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if $$\hat{\boldsymbol r}=\cos\theta\ \hat{\boldsymbol e}_1+\sin\theta\ \hat{\boldsymbol e}_2,$$ $$\hat{\boldsymbol \theta}=-\sin\theta\ \hat{\boldsymbol e}_1+\cos\theta\ \hat{\boldsymbol e}_2,$$ is the change of basis, then we can solve to find: $$\hat{\boldsymbol e}_1=\cos\theta\ \hat{\boldsymbol r}-\sin\theta\ \hat{\boldsymbol \theta},$$ $$\hat{\boldsymbol e}_2=\sin\theta\ \hat{\boldsymbol r}+\cos\theta\ \hat{\boldsymbol \theta}.$$

So, for $\boldsymbol{\rm v}=\frac{dx}{dt}\hat{\boldsymbol e}_1+\frac{dy}{dt} \hat{\boldsymbol e}_2$ you will get $$\boldsymbol{\rm v}=\left(\frac{dx}{dt}\cos\theta+\frac{dy}{dt}\sin\theta\right)\hat{\boldsymbol r}+ \left(\frac{dy}{dt}\cos\theta-\frac{dx}{dt}\sin\theta\right)\hat{\boldsymbol \theta}.$$

And if you wish to know the value of the functions $\frac{dx}{dt}$ and $\frac{dy}{dt}$ you only need to differentiate the coordinate change $$x=r\cos\theta\quad ,\quad y=r\sin\theta.$$

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