Stuck on the formal derivation of the $ \vec v = \dot r\hat r + r \dot\theta\hat \theta $ formula for velocity in polar coordinates

Question

I'm trying to derive the formula for the velocity of a particle in $ \mathbb R^2 $ in polar coordinates using the framework of differential geometry (charts, tangent vectors, etc.), just to familiarize with this language. (Next step will be to derive the formula for the acceleration by computing some Christoffel symbols).

Consider the manifold $ M = \mathbb R^2\setminus\{x\leqq 0\} $ (i.e., the plane minus the non-positive horizontal semiaxis).

On $ M $ we can put the standard (Cartesian) coordinates $ (x,y) $ or the familiar polar coordinates $ (r,\theta) $. Given $ p\in M $, denote with $ \partial/\partial x\rvert_p $, $ \partial/\partial y\rvert_p $ the basis of $ \mathrm T_pM $ induced by the Cartesian coordinates, and with $ \partial/\partial r\rvert_p $, $ \partial/\partial\theta\rvert_p $ the basis of $ \mathrm T_pM $ induced by the polar ones.

A simple calculation shows that $$ \left.\frac{\partial}{\partial r}\right\rvert_p = \cos\theta(p)\left.\frac{\partial}{\partial x}\right\rvert_p + \sin\theta(p)\left.\frac{\partial}{\partial y}\right\rvert_p\text{,}\qquad \left.\frac{\partial}{\partial \theta}\right\rvert_p = -\sin\theta(p)\left.\frac{\partial}{\partial x}\right\rvert_p + \cos\theta(p)\left.\frac{\partial}{\partial y}\right\rvert_p $$ i.e. that the change of basis matrix $ A_{\mathrm{Cartesian}\to\mathrm{Polar}}(p) $ for $ \mathrm T_pM $ is given by $$ A_{\mathrm{Cartesian}\to\mathrm{Polar}}(p) = R(\theta(p)) $$ where $ R(\theta(p)) $ is the rotation matrix in dimension $ 2 $. We can easily reverse the operation: $$ A_{\mathrm{Polar}\to\mathrm{Cartesian}}(p) = R(-\theta(p)) $$ since $ R(\theta(p))^{-1} = R(-\theta(p)) $.

Now, take a curve $ \gamma\colon \mathbb R\to M $. I want to express the velocity vector $ \dot\gamma(t)\in \mathrm T_{\gamma(t)}M $ of $ \gamma $ at the time $ t $ in polar coordinates, and derive the formula $$ \dot\gamma(t) = r(\gamma (t))^\prime \left.\frac{\partial}{\partial r}\right\rvert_{\gamma(t)} + r(\gamma(t))\theta(\gamma(t))^\prime \left.\frac{\partial}{\partial \theta}\right\rvert_{\gamma(t)}\label{formula}\tag{$ * $} $$ usually seen in freshman Mechanics textbooks. If I'm not mistaken, I can proceed like this. I'll put $ p = \gamma(t) $ in the following.

I can write $$ \dot\gamma(t) = \dot\gamma^1(t)\left.\frac{\partial}{\partial x}\right\rvert_p + \dot\gamma^2(t)\left.\frac{\partial}{\partial y}\right\rvert_p = v^1\left.\frac{\partial}{\partial r}\right\rvert_p + v^2\left.\frac{\partial}{\partial\theta}\right\rvert_p $$ where $ \dot\gamma^i(t) $ is just the derivative of the $ i $-th component of $ \gamma $ seen as a curve in $ \mathbb R^2 $, and at this point I have two ways of calculating the $ v^i $s.

I can go on like in $$ v^i = \dot{\tilde\gamma}(t) $$ where $ \tilde\gamma^i(t) = (r,\theta)\circ \gamma(t) $ and $ (r,\theta)\colon M\to \left]0,+\infty\right[\times \left]-\pi,\pi\right[ $ is the polar chart.

Or I can just do a change of coordinates like in $$ \begin{bmatrix} v^1\\ v^2 \end{bmatrix} = R(-\theta(p)) \begin{bmatrix} \dot\gamma^1(t)\\ \dot\gamma^2(t) \end{bmatrix}\text{.} $$

My problem is that neither way gives me the desired result $ \eqref{formula} $.

Answer 1

The simple calculation is incorrect; the matrix is not a rotation. Check:

$$\partial_\theta = \frac{\partial x}{\partial \theta}\partial_x + \frac{\partial y}{\partial \theta}\partial_y = -r\sin\theta \partial_x + r\cos\theta\partial_y$$

Answer 2

Let radial vector: $\vec r=r\hat r=r(\cos\theta\hat i+\sin\theta\hat j)$,
and unit tangential vector: $\hat\theta=-\sin\theta\hat i+\cos\theta\hat j$ (rotation of the unit radial vector $\hat r$ by $90^\circ$ ccw).
Observe that $\hat\theta=\frac{\mathrm d\hat r}{\mathrm d\theta}$.

Now, by the product rule: $$\Large \color{red}{\vec v=\frac{\mathrm d\vec r}{\mathrm dt}=\frac{\mathrm dr}{\mathrm dt}\hat r+r\frac{\mathrm d\hat r}{\mathrm d\theta}\frac{\mathrm d\theta}{\mathrm dt}=\dot r\hat r+r\dot\theta\hat\theta}$$

If we differentiate $\vec v$ with respect to time $t$, we get the acceleration that includes centrifugal, Coriolis and Euler components.

Answer 3

Too long for a comment. Your (*) looks weird.

We know \begin{align} x&=r\cos\theta\,,&y&=r\sin\theta\,. \end{align} Thus, one and the same curve $\gamma(t)$ can be expressed either way $$ \gamma(t)=\pmatrix{x\\y}(t)=\pmatrix{r\cos\theta\\r\sin\theta}(t)\,. $$ (I pull out the $(t)$ in favour of writing $r(t)\cos(\theta(t))$ etc.)

Likewise, a function $f(x,y)$ can be expressed in polar coordinates as $$ f(r\cos\theta,r\sin\theta)=g(r,\theta)\,. $$ The chain rule gives \begin{align} \partial_rg&=(\partial_xf)(\cos\theta)+(\partial_yf)(\sin\theta)\,,\\[2mm] \partial_\theta g&=(\partial_xf)(-r\sin\theta)+(\partial_yf)(r\cos\theta)\,. \end{align} This leads to the transformation of the Cartesian coordinate basis $\{\partial_x,\partial_y\}$ to the polar coordinate basis $\{\partial_r,\partial_\theta\}$ which seems clear to you now.

Similarly we have by the chain rule $$ \frac d{dt}f(\gamma(t))=(\partial_xf)\,\dot x(t)+(\partial_xf)\,\dot y(t)\,. $$ This shows that the tangent vector to the curve $\gamma(t)$ at $p=\gamma(t)$ is in Cartesian coordinates $$\tag1 \dot x(t)\,\partial_x+\dot y(t)\,\partial_y\,. $$ In polar coordinates we get it from $$ \frac d{dt}g(\gamma(t))=(\partial_rg)\,\dot r(t)+(\partial_\theta g)\,\dot \theta(t) $$ as $$\tag2 \dot r(t)\,\partial_r+\dot\theta(t)\,\partial_\theta\,. $$ Since \begin{align} \dot x&=\dot r\cos\theta-r(\sin\theta)\dot\theta\,,\\[2mm] \dot y&=\dot r\sin\theta+r(\cos\theta)\dot\theta\, \end{align} we can plug into (1) and get \begin{align} \dot x\,\partial_x+\dot y\,\partial_y&=(\dot r\cos\theta-r(\sin\theta)\dot\theta)\,\partial_x+ (\dot r\sin\theta+r(\cos\theta)\dot\theta)\,\partial_y\\ &=\dot r\underbrace{(\cos\theta\,\partial_x+\sin\theta\,\partial_y)}_{\textstyle\partial_r} +\dot\theta\underbrace{(-r\sin\theta\,\partial_x+r\cos\theta\,\partial_y)}_{\textstyle\partial_\theta} \end{align} which shows that (1) and (2) are one and the same tangent vector.

To summarize: It all follows from the chain rule and your (*) should look more like (2).

About the notation in the title: we discussed the metric in polar coordinates by which the vector $\partial_\theta$ is not normalized. It has length $r\,.$

The normalized basis vectors are $\hat r=\partial_r,\hat\theta=\frac1r\partial_\theta\,.$ This turns (2) into $$ \vec v=\dot r\,\hat r+r\,\dot\theta\,\hat\theta $$ as expected.

Answer 4

I confused from the start the vector $ \hat\theta $ from vector calculus with the vector $ \partial_\theta $. The relation between the two in our framework is $$ \hat\theta = \frac1r\partial_\theta\text{.} $$

Now, the relation I was trying to prove, namely $$ \dot\gamma(t) = r(\gamma (t))^\prime \partial_r + r(\gamma(t))\theta(\gamma(t))^\prime \partial_\theta $$ is actually false. What really holds is that $$ \dot\gamma(t) = r(\gamma(t))^\prime \partial_r + \theta(\gamma(t))^\prime \partial_\theta $$ but this is obvious!

What's really interesting in the (correct) change-of-basis formulae $$ \begin{align} \partial_r &=\cos\theta\,\partial_x + \sin\theta\,\partial_x\\ \partial_\theta &=-r\sin\theta\,\partial_x + r\cos\theta\, \partial_y \end{align} $$ is the fact that we can go from the polar coordinate expression $$ \dot\gamma(t) = r(\gamma(t))^\prime \partial_r + \theta(\gamma(t))^\prime \partial_\theta $$ to the Cartesian one $$ \dot\gamma(t) = x(\gamma(t))^\prime \partial_x + y(\gamma(t))^\prime \partial_y $$ and back just multiplying by some change-of-basis matrix: $$ \begin{pmatrix} \dot\gamma^x\\ \dot\gamma^y \end{pmatrix} = \begin{pmatrix}\cos\theta(p) & -r(p)\sin\theta(p)\\ \sin\theta(p) & r(p)\cos\theta(p) \end{pmatrix} \begin{pmatrix} \dot\gamma^r\\ \dot\gamma^\theta \end{pmatrix} \qquad \begin{pmatrix} \dot\gamma^r\\ \dot\gamma^\theta \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta\\ -\frac1r\sin\theta & \frac1r\cos\theta \end{pmatrix} \begin{pmatrix} \dot\gamma^x\\ \dot\gamma^y \end{pmatrix}\text{.} $$

The point is this: I thought that $$ \vec v = \dot r\hat r + r\dot\theta \hat\theta $$ was some obscure formula that had to be derived from some change-of-basis computation. Now I understand that it nothing but the usual $$ \dot\gamma(t) = \dot\gamma^\mu\partial_\mu $$ where one of the $\partial_\mu $s has been substituted with $ \frac1r\partial_\mu $.