1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Polar Coordinates

  1. Nov 26, 2017 #1
    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?
     
  2. jcsd
  3. Nov 26, 2017 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Nov 26, 2017 #3
    That makes sense. I guess that answers my questions. Is there not a more analytical way though? That seems very geometric.
     
  5. Nov 27, 2017 #4

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    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}}##.
     
    Last edited: Nov 27, 2017
  6. Nov 27, 2017 #5
    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?
     
  7. Nov 27, 2017 #6

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted