# How velocities transform from Cartesian to Polar coordinates

Consider a transformation from Cartesian to polar coordinates $$(x,y)\rightarrow (r,\theta)$$, $$$$\begin{gathered} x=r\cos\theta,\\ y=r\sin\theta. \end{gathered}$$$$ Here, we denote $$x^{\,\mu}=(x,y)$$ and $$\bar{x}^{\,\mu}=(r,\theta)$$. Now, The question is the following,

In the $$x^{\,\mu}$$ coordinate system, the components of the velocity vector are $$(\dot{x},\dot{y})$$. Find out the components in the polar coordinates using vector/tensor transformation rules.

From the coordinate transformation we have, $$$$\begin{gathered} dx=\cos\theta dr-r\sin\theta d\theta,\\ dy=\sin\theta dr+r\cos\theta d\theta. \end{gathered}$$$$ Thus, $$$$\begin{gathered} \frac{\partial x}{\partial r}=\cos\theta=\frac{x}{r};\quad \frac{\partial x}{\partial \theta}=-r\sin\theta=-y,\\ \frac{\partial y}{\partial r}=\sin\theta=\frac{y}{r};\quad \frac{\partial y}{\partial \theta}=r\cos\theta=x. \end{gathered}$$$$ The transformed components $$\bar{V}^{\,\mu}=\bar{V}^{\,\mu}(x^{\,\alpha})$$ reads, \begin{align} \bar{V}^{\,\mu}=\frac{\partial\, \bar{x}^{\,\mu}}{\partial\, x^{\,\beta}}V^{\,\beta} \end{align} Now, for $$\mu=1$$, \begin{align} \bar{V}^{\,1}&=\frac{\partial\, \bar{x}^{\,1}}{\partial\, x^{\,\beta}}V^{\,\beta}\nonumber\\ &=\frac{\partial\, \bar{x}^{\,1}}{\partial\, x^{\,1}}V^{\,1}+\frac{\partial\, \bar{x}^{\,1}}{\partial\, x^{\,2}}V^{\,2}\nonumber\\ &=\frac{\partial r}{\partial x}V^{\,1}+\frac{\partial\, r}{\partial y}V^{\,2}\nonumber\\ &=\sec\theta V^{\,1}+\csc\theta V^{\,2}\nonumber\\ &=\frac{r}{x} V^{\,1}+\frac{r}{y} V^{\,2} \tag{1}\label{eq:comptransone} \end{align} Now, for $$\mu=2$$, \begin{align} \bar{V}^{\,2}&=\frac{\partial\, \bar{x}^{\,2}}{\partial\, x^{\,\beta}}V^{\,\beta}\nonumber\\ &=\frac{\partial\, \bar{x}^{\,2}}{\partial\, x^{\,1}}V^{\,1}+\frac{\partial\, \bar{x}^{\,2}}{\partial\, x^{\,2}}V^{\,2}\nonumber\\ &=\frac{\partial \theta}{\partial x}V^{\,1}+\frac{\partial\theta}{\partial y}V^{\,2}\nonumber\\ &=-\frac{1}{r}\csc\theta V^{\,1}+\frac{1}{r}\sec\theta V^{\,2}\nonumber\\ &=-\frac{1}{y} V^{\,1}+\frac{1}{x} V^{\,2} \tag{2}\label{eq:comptranstwo} \end{align}

$$$$\begin{gathered} \dot{x}=\cos\theta \dot{r}-r\sin\theta \dot{\theta},\\ \dot{y}=\sin\theta \dot{r}+r\cos\theta \dot{\theta}. \end{gathered}$$$$ Now, we calculate the velocity components in the polar coordinates using equations ($$\ref{eq:comptransone}$$) and ($$\ref{eq:comptranstwo}$$), \begin{align} v^{\,r}&=\sec\theta \dot{x}+\csc\theta\dot{y}\nonumber\\ &=\sec\theta\left(\cos\theta \dot{r}-r\sin\theta \dot{\theta}\right)+\csc\theta\left(\sin\theta \dot{r}+r\cos\theta \dot{\theta}\right)\nonumber\\ &= \dot{r}-r\tan\theta \dot{\theta}+\dot{r}+r\cot\theta \dot{\theta}\nonumber\\ &= 2\dot{r}-r(\tan\theta -\cot\theta) \dot{\theta} \end{align} \begin{align} v^{\,\theta}&=-\frac{1}{r}\csc\theta \dot{x}+\frac{1}{r}\sec\theta \dot{y}\nonumber\\ &=-\frac{1}{r}\csc\theta \left(\cos\theta \dot{r}-r\sin\theta \dot{\theta}\right)+\frac{1}{r}\sec\theta \left(\sin\theta \dot{r}+r\cos\theta \dot{\theta}\right)\nonumber\\ &=-\frac{1}{r}\cot\theta\dot{r}+\dot{\theta}+\frac{1}{r}\tan\theta\dot{r}+\dot{\theta}\nonumber\\ &=2\dot{\theta}+\frac{\dot{r}}{r}(\tan\theta-\cot\theta) \end{align}

Present question: Are the above equations which I derived correct? Shouldn't this be something like $$v^r=\dot{r}$$ and $$v^\theta=r\dot{\theta}$$? Where am I going wrong? Help please.

• You could have avoided some (confusing) math by simply taking the derivative of the first two equations with respect to time. Those results transform from components of polar to rectangular coordinates. If you want to go from rectangular to polar, you need r and θ in terms of x and y. – R.W. Bird Oct 20 '20 at 14:15

The issue with the velocity transformation is resolved if you use the matrix inverse of the Jacobian. In your case, note that the inverse transformation you are using involves terms like $$\frac{\partial r}{\partial x}\Bigg\rvert_\theta = \sec \theta$$ which does not make much sense since $$\theta = \theta(x,y)$$ is also a function of $$x$$ and $$y$$. The issue is resolved by taking derivatives from the inverse functions directly $$r = \sqrt{x^2+y^2} \implies \frac{\partial r}{\partial x}\Bigg\rvert_y = \cos \theta$$ which reproduces the inverse Jacobian matrix elements.
You will get $$\bar{V}^\mu = (\dot r, \dot \theta) = \dot r \partial_r + \dot \theta \partial_\theta$$ as we expect because we should be able to do the time-derivative in any coordinates, and then noting that $$\partial_\theta = r \hat \theta$$ and $$\partial_r = \hat r$$ you recover the usual expression from vector calculus $$\dot r \hat r + r \dot \theta \hat \theta$$