Deriving Polar Coordinates Without Cartesian System

[Mr Davis 97]

Any point on the plane can be specified with an $r$ and a $\theta$, where $\mathbf{r} = r \hat{\mathbf{r}}(\theta)$. From this, my book derives $\displaystyle \frac{d \mathbf{r}}{dt}$ by making the substitution $\hat{\mathbf{r}}(\theta) = \cos \theta \hat{\mathbf{i}} + \sin \theta \hat{\mathbf{j}}$, and then deriving it from there, concluding that $\mathbf{v} = \dot{r} \hat{\mathbf{r}} + r \dot{\theta} \hat{\mathbf{\theta}}$.

My question is, is is possible to make the derivation without referring to a different coordinate system, the Cartesian system, when the substitution $\hat{\mathbf{r}}(\theta) = \cos \theta \hat{\mathbf{i}} + \sin \theta \hat{\mathbf{j}}$ is made?

 

[andrewkirk]

Sure. Just draw a diagram showing the origin O and the positions P and Q of the particle at times $t$ and $t+\delta t$. Mark point T on the segment OQ that is distance $|OP|$ from O. The curve from P to T is a short part of the circumference of a circle. It has length $|OP|\dot \theta \delta t$ and joins the segment OQ at right angles. The segment $TQ$ has length $\dot r\delta t$.

As $\delta t\to 0$ the shape $TPQ$ approaches a right-angled triangle, and we can derive the formula from that.

 

[Mr Davis 97]

Sure. Just draw a diagram showing the origin O and the positions P and Q of the particle at times $t$ and $t+\delta t$. Mark point T on the segment OQ that is distance $|OP|$ from O. The curve from P to T is a short part of the circumference of a circle. It has length $|OP|\dot \theta \delta t$ and joins the segment OQ at right angles. The segment $TQ$ has length $\dot r\delta t$.

As $\delta t\to 0$ the shape $TPQ$ approaches a right-angled triangle, and we can derive the formula from that.

That makes sense. I guess that answers my questions. Is there not a more analytical way though? That seems very geometric.

 

[DrGreg]

Well,$$
\frac{d}{dt} \left( r \hat{\mathbf{r}} \right) = \frac{dr}{dt} \hat{\mathbf{r}} + r \frac{d\hat{\mathbf{r}}}{dt}
= \frac{dr}{dt} \hat{\mathbf{r}} + r \frac{d\hat{\mathbf{r}}}{d\theta} \frac{d\theta}{dt}
$$So all that remains is to show $d\hat{\mathbf{r}} / d\theta = \hat{\boldsymbol{\theta}}$. How to do that depends on how your book defines $\hat{\boldsymbol{\theta}}$.

 

[Mr Davis 97]

One more question. Going from Cartesian to polar is a change of coordinates. Also, $\hat{i}$ and $\hat{j}$ are basis vectors for $\mathbb{R}^2$. Can we consider $\hat{r}$ and $\hat{\theta}$ to be a basis for $\mathbb{R}^2$ as well? Why or why not?

 

[andrewkirk]

One more question. Going from Cartesian to polar is a change of coordinates. Also, $\hat{i}$ and $\hat{j}$ are basis vectors for $\mathbb{R}^2$. Can we consider $\hat{r}$ and $\hat{\theta}$ to be a basis for $\mathbb{R}^2$ as well? Why or why not?

No, because the direction of $\hat \theta$ and $\hat r$ depend on the location of the point in whose tangent space they belong. Hence the two vectors are not well-defined without also specifying a reference point.

In contrast, $\hat i$ and $\hat j$ always point in the same direction, regardless of the reference location (tangent space).

However, at any given point other than the origin, $\hat\theta$ and $\hat r$ can be used as a basis for the tangent space at that point, which is isomorphic to $\mathbb R^2$. We need to recognise the distinction between $\mathbb R^2$ as a manifold and $\mathbb R^2$ as an isomorph of the tangent space at a point in that manifold. Whether or not those terms mean anything to the reader will depend on whether they have done any differential geometry (calculus on manifolds).