Polar coordinates, sign ambiguity

[PFuser1232]

The position of a point in cartesian coordinates is given by:
$$\vec{r} = x \hat{\imath} + y \hat{\jmath}$$
In polar coordinates, it is given by:
$$\vec{r} = r \hat{r}$$
Now, $x = r \cos{θ}$ and $y = r \sin{θ}$ assuming $θ$ is measured counterclockwise from the $x$-axis.
Equating the two expressions we obtain:
$$\hat{r} = \hat{\imath} \cos{θ} + \hat{\jmath} \sin{θ}$$
Let's say $\hat{θ} = A \hat{\imath} + B \hat{\jmath}$
We know that:
$$\hat{r} ⋅ \hat{θ} = A\cos{θ} + B\sin{θ} = 0$$
And
$$A^2 + B^2 = 1$$
Since $A = -B \tan{θ}$:
$$B^2 \tan^2{θ} + B^2 = 1$$
$$B^2 = \cos^2{θ}$$
$$B = \pm \cos{θ}$$
And
$$A = \mp \sin{θ}$$
Therefore:
$$\hat{θ} = \mp \sin{θ} \hat{\imath} \pm \cos{θ} \hat{\jmath}$$
On what basis do we choose the correct direction for the tangential unit vector?

 

[Dale]

We choose it so that the cross product is positive.

 

[vanhees71]

To use the arctan formula is not a good idea. You have to adjust the angle by hand, corresponding to the quadrant, where your position vector is located. A more handy formula is
$$\theta=\mathrm{sign}\,y \arccos \left(\frac{x}{\sqrt{x^2+y^2}} \right).$$
Then $\theta \in (-\pi,\pi]$.

 

[PFuser1232]

We choose it so that the cross product is positive.

That is, $\hat{r} × \hat{θ} = \hat{k}$, right?

 

[PFuser1232]

To use the arctan formula is not a good idea. You have to adjust the angle by hand, corresponding to the quadrant, where your position vector is located. A more handy formula is
$$\theta=\mathrm{sign}\,y \arccos \left(\frac{x}{\sqrt{x^2+y^2}} \right).$$
Then $\theta \in (-\pi,\pi]$.

Where have I used the arctangent function?

 

[vanhees71]

Ah, I misunderstood the question. The point is of course, that you choose the vectors of a curvilinear coordinate system $(q_1,q_2)$ as tangents to the coordinate lines. In the case that these are orthogonal coordinates, one usually also normalizes them. In your case you have $q_1=r$ and $q_2=\theta$. The relation to Cartesian coordinates is, as you write
$$\vec{r}=r \cos \theta \hat{i} + r \sin \theta \hat{j}.$$
Then the tangent vectors are
$$\vec{T}_r=\partial_r \vec{r}=\cos \theta \hat{i} + \sin \theta \hat{j},$$
$$\vec{T}_{\theta} = \partial_{\theta} \vec{r} = -r \sin \theta \hat{i} + r \cos \theta \hat{j}.$$
Obviously $\vec{T}_r \cdot \vec{T}_{\theta}=0$, i.e., we have orthogonal coordinates. Thus one uses the normalized basis vectors
$$\hat{r}=\vec{T}_r/|\vec{T}_r|=\vec{T}_r=\cos \theta \hat{i} + \sin \theta \hat{j},$$
$$\hat{\theta}=\vec{T}_{\theta}/|\vec{T}_{\theta}|=\vec{T}_{\theta}/r = -\sin \theta \hat{i} + \cos \theta \hat{j}.$$
A cross product doesn't make too much sense in the planar case. Also one chooses always the vectors in direction of the tangent vectors to the coordinate lines.

For an orientation preserving transformation from Cartesian to curvilinear coordinates you have to make the order of the curvilinear coordinates such that the Jacobian of the transformation is positive. In the case of polar coordinates, the order $(r,\theta)$ is right, because
$$\mathrm{det} \frac{\partial(x,y)}{\partial(r,\theta)}=\mathrm{det} \begin{pmatrix}
\cos \theta & \sin \theta ,\\
-r \sin \theta & \cos \theta
\end{pmatrix}=r>0.$$

 

[Chestermiller]

Just draw a diagram and you will see that when θ=0, θ hat = j, and when θ=π/2, θ hat = -i.

Chet