What are the normalized vectors in polar coordinates?

[Bruno Tolentino]

If the vector r is (x,y), so, what is the vector θ? BY THE WAY is (y,-x) ?

 

[HallsofIvy]

I have no idea what you mean by "the vector $\theta$". Could you please explain that? Where did you see a reference to a "vector $\theta$"?

 

[Fredrik]

The question "is (y,-x)?" doesn't make sense either.

 

[HallsofIvy]

If the vector r is (x,y), so, what is the vector θ? BY THE WAY is (y,-x) ?

I didn't notice the "(y, -x)"! If a vector is given by r= (x, y) then its length is |r|= $\sqrt{x^2+ y^2}$ and the angle it makes with the x-axis, it that is what you mean by "$\theta$", is given by $arctan(y/x)$ as long as x is not 0, $\pi/2$ if x= 0 and y is positive, $3\pi/2$ if x= 0 and y is negative.

Given a vector (x, y), the vector (y, -x) is the result of rotating (x, y) through an angle of $pi/2$ radians.

 

[WWGD]

I didn't notice the "(y, -x)"! If a vector is given by r= (x, y) then its length is |r|= $\sqrt{x^2+ y^2}$ and the angle it makes with the x-axis, it that is what you mean by "$\theta$", is given by $arctan(y/x)$ as long as x is not 0, $\pi/2$ if x= 0 and y is positive, $3\pi/2$ if x= 0 and y is negative.

Given a vector (x, y), the vector (y, -x) is the result of rotating (x, y) through an angle of $pi/2$ radians.

The angle $\theta$ depends on your frame of reference : the positive x-axis does not have to represent the angle $0$ , it can represent anything as long as the choices are made consistently, i.e., the angle with the negative x-axis must be $\pi$ larger than the choice on the positive x-axis, as is done, e.g., with branches of the Complex logarithm.

 

[HallsofIvy]

Well, that depends on exactly what Bruno Tolentino means by "the vector $\theta$". I asked that earlier and he still hasn't answered.

 

[WWGD]

Another possible interpretation is that $\theta(x,y)=(-y,x)$ is a _vector field_ , assigning to each point/tangent space at $(x,y)$, the
vector $(-y,x)$.

 

[Bruno Tolentino]

The ideia of vector θ come from following: if the vector dr is the tangent vector to parametric curve and the o vector dn is the normal vector:

And if the UNIT vector r^ is normal to UNIT vector θ^:

So: the vector dn = dθ and therefore θ = (-y,x)!?

EDIT: but confront with the following: if θ = (-y,x), so θ = (- r sin(θ), r cos(θ)) = r (- sin(θ), cos(θ)) = r θ^

Is known that the vector r = r r^

But, is correct to affirm that: θ = r θ^?

The vector θ wouldn't: θ = θ θ^

 

[Fredrik]

In this context (polar coordinates), there's no standard definition of the notation θ. It's definitely non-standard to use that notation for the vector $r\hat{\theta}$.

I think I see what you were thinking now: Since $\hat{\mathbf r}$ is a normalized version of $\mathbf r$, it makes sense to ask if there's a vector that you can normalize to get $\hat\theta$. There is, but it's not denoted by θ.

The vectors $\hat{\mathbf r}$ and $\hat\theta$ are defined as what you get when you normalize $\frac{\partial\mathbf r}{\partial r}$ and $\frac{\partial\mathbf r}{\partial\theta}$.
\begin{align*}
&\mathbf r=(x,y)=(r\cos\theta,r\sin\theta)\\
&\hat{\mathbf r}=\frac{\frac{\partial\mathbf r}{\partial r}}{\left|\frac{\partial\mathbf r}{\partial r}\right|} =\frac{(\cos\theta,\sin\theta)}{1} =\frac{\mathbf r}{r}=\frac{(x,y)}{\sqrt{x^2+y^2}}\\
&\hat{\theta} =\frac{\frac{\partial\mathbf r}{\partial\theta}}{\left|\frac{\partial\mathbf r}{\partial\theta}\right|} =\frac{(-r\sin\theta,r\cos\theta)}{r} =\frac{(-y,x)}{\sqrt{x^2+y^2}}
\end{align*}