Consider the 2-dimensional polar coordinate system describing Euclidean space where points in the space are given by the coordinates $(r,\theta)$. In what sense does the unit vector $\hat{\theta}$ have length 1? If one considers the unit circle centred on the origin of the space, is a step in the $\theta$-direction of unit length just the distance around the circle corresponding to an angle of 1 radian?

Edit for further clarification

My background is in physics, so an example in terms of a physics problem might help highlight my confusion.

From studying physics, I do have some degree of understanding and intuition of vectors defined in a polar coordinate system (or any kind of curvilinear coordinate system, I suppose). I can make sense of the angular velocity vector for circular motion, to pick just one example.

My understanding here is that if the force keeping a particle constrained to a circular trajectory was suddenly removed then the particle would fly off tangent to the circle (i.e. in the instantaneous direction of the angular velocity vector) with a speed given by the magnitude of the angular velocity vector. Essentially, my understanding in this case is that the unit vector in the $\theta$-direction defines one of the two coordinate axes in the tangent space to the circular path, and it is in this direction that the particle would move if suddenly freed of its confining force.

Once freed I'd understand $\hat{\theta}$ to correspond to this tangent space, and some step of unit length in the $\theta$-direction to correspond to a step of unit length in the tangent space to the circle. However, before the particle is freed, it's confined to stay on the circular path and I'm not sure that I can make sense of what it means to make a step of unit length in the $\theta$-direction. My understanding is that if one travels some distance in the $\theta$ direction then they've literally travelled along the circumference of a circle of fixed $r$. This understanding might be wrong. If so, what is the resolution? If not, what does it mean to make a step of unit length in the $\theta$-direction when confined to a circle?

  • $\begingroup$ It is a vector pointing in the angular direction and has length 1. If you do a line integral of it along a circular path I think you would be getting the vector from start point to end point on that circle. If you did a line integral anywhere else than onto the same circle you would be getting the projection onto the same circle. $\endgroup$ – mathreadler Mar 19 '17 at 14:41
  • $\begingroup$ The unit vector encodes the direction to rotate with changing $\theta$ in any given point. The length is of unit such that if we integrate along it at any given point on the unit circle we will create a rotation. Unit circle <-> unit vector, makes sense? If it had another length it would not create a rotation if we did that but a spiral or something third. But then there would be no reason to call the unit circle a "unit" circle. But we do call it a unit circle, so why not call the vector that generates it through integration a "unit vector". $\endgroup$ – mathreadler Mar 21 '17 at 20:11

Maybe one way that will help the understanding is to consider the integral $$\int_0^{\theta_0} {\hat \theta} d{\theta}$$

I am assuming $\hat \theta$ defined in this way: $${\hat \theta} = \begin{bmatrix}-\sin(\theta)\\\cos(\theta)\end{bmatrix}$$

It will be unit vector because of the trigonometric identity: $\sin(\theta)^2+\cos(\theta)^2 = 1$. We can also convince ourselves that it will be orthogonal to $\begin{bmatrix}r\cos(\theta)\\r\sin(\theta)\end{bmatrix}$ which is often defined to be the radial component.

If we approximate the integral using a computer, we see a relation between the angle integrated, the unit circle and the line integral result:

enter image description here The end points of the blue line are angles $0$ and $\theta_0$ on the unit circle. And the red vector is the vector from point $(1,0)$ pointing onto the unit circle at angle $\theta_0$. (I could not find any arrowhead to indicate the direction).

  • $\begingroup$ I'm not sure this answers my question exactly. I still don't understand what a unit step in the $\theta$-direction is. Perhaps I can further elaborate on my confusion (I will edit my original post to – hopefully – further clarify what I'm asking). $\endgroup$ – ScottTheScot Mar 20 '17 at 12:16
  • $\begingroup$ The question was a bit fuzzy. Tried clarifying the answer a bit. $\endgroup$ – mathreadler Mar 20 '17 at 12:24
  • $\begingroup$ I've modified my original post, though I think in gathering my thoughts to try and clarify my confusion I might be close to solving my own problem. I think my problem lies in the fact that the unit vectors are only locally defined in the circular polar coordinate system and that I'm used to being able to freely move vectors around in a Cartesian coordinate system and do all sorts of dodgy things to the vectors that don't work so well in other coordinate systems. $\endgroup$ – ScottTheScot Mar 20 '17 at 13:03

Given polar coordinates $(r, \theta)$ for a vector $$ (x,y) = r \, e_r $$ we have the radial unit vector $$ e_r = (\cos \theta, \sin \theta) \\ $$ and the perpendicular $$ e_\theta = (- \sin \theta, \cos \theta) $$ The length of $e_\theta$ is $$ \lVert e_\theta \rVert = \sqrt{(-\sin \theta)^2 + (\cos \theta)^2} = \sqrt{1} = 1 $$


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