Calculate the divergence of the polar coordinate vector field $\partial_\phi$

Question

I have to solve this problem:

$v=\partial_\phi$ on $M=\mathbb{R}^2\backslash{0}$ where the components of $v$ are in polar coordinates.

Calculate the divergence of $v$.

Even with the help of comments and answers I have problems solving this. Can someone give me a full solution? (I am not trying to dodge working by myself!)

Answer 1

We use the definition of divergence

$$\text{div } u = \frac{1}{\sqrt{g}} \frac{\partial}{\partial x^i} (u^i \sqrt{g})$$

where $g$ here means the determinant of the metric and summation over $i$ is implied.

Here, our vector field is $u = \partial_\phi$. More clearly, write it out in components, in terms of the basis vectors $\partial_r$ and $\partial_\phi$. That is

$$u = u^r \partial_r + u^\phi \partial_\phi$$

That is, $u^r = 0$ and $u^\phi = 1$. These are the numbers that will stand in for $u^i$ in the expression for divergence.

Finally, with $g = r^2$, we get

$$\text{div } u = \frac{1}{r} \left[ \frac{\partial}{\partial r} (0 \cdot r) + \frac{\partial}{\partial \phi} (1 \cdot r) \right] = 0$$

which is zero.

Answer 2

$\partial \phi$ is just the vector often written as $\frac {\partial}{\partial \phi}$. It has a dual $d\phi$, so that $d\phi(\partial \phi)=1$. It is used with polar coordinates $(r, \phi)$, and there you have an orthogonal basis for the tangent space at every point given by $(\partial r, \partial \phi)=(\frac {\partial }{\partial r}, \frac {\partial}{\partial \phi} )$. For a plot, see:

http://mathworld.wolfram.com/CylindricalCoordinates.html

(Ignore the z-coordinate, or take a fixed slice $z=z_0$ and draw $\partial r , \partial \phi$ in the same direction as the $r, \phi$ coordinates in the plot.)

Basically, $\partial r$ at a point $(x,y)$ is drawn by a line of length $\sqrt {x^2+y^2}$ from the origin to the point $(x,y)$ and $\partial \phi$ is drawn perpendicular to $\partial r$.

EDIT: Like Travis pointed out, this is an orthogonal, but not orthonormal basis.

EDIT2: This post was intended to answer the question before it was edited, which was, as I understood, how to plot the vector field $(\partial \phi)$ in polar coordinates.

Answer 3

Edit The original question was changed radically after this answer was posted. The question was (paraphrasing),

In polar coordinates $(r, \phi)$, how do I interpret the vector field $\partial_{\phi}$ and write it in Cartesian coordinates?

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Assuming that $\phi$ is the angular coordinate (typically it is denoted $\theta$), it is the vector field whose action on functions $f(r, \phi)$ is given by the partial derivative with respect to $\phi$, that is, the instantaneous change $\frac{\partial f}{\partial \phi}$ of $f$ under anticlockwise rotation.

If the differentiated function $f$ is given as a function of $x$ and $y$, some elementary trigonometry gives that the value of $f$ at the point given by rotating $(x, y)$ anticlockwise by an angle $\theta$ is $$f(x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta),$$ and the instantaneous change $\frac{\partial f}{\partial \phi}$ of $f$ under anticlockwise rotation is, by definition, the derivative of this quantity with respect to $\theta$ at $\theta = 0$, namely, $$\left.\frac{d}{d \theta} \right\vert_{\theta = 0} f(x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta) = -y \frac{\partial f}{\partial x} + x \frac{\partial f}{\partial y}.$$ But this is just the action of the vector field $$\phantom{(\ast)} \qquad \color{#bf0000}{\boxed{-y \partial_x + x \partial_y}} \qquad (\ast)$$ on the function $f$, so $(\ast)$ is the Cartesian representation of $\partial_{\phi}$.

This computation both hides and resolves a notational abuse and a corresponding technical complication: The expression $\partial_{\phi}$ simply denotes a coordinate vector field in a particular coordinate system, and, strictly speaking, a coordinate system should give unique coordinates for each point in the region the coordinates cover. On the other hand, there is no continuous choice of polar angular coordinate on all of $\Bbb R^2$, or even all of $\Bbb R^2 - \{0\}$, so in order to talk about coordinate vector field $\partial_{\phi}$, we must really restrict our polar coordinates to some (say, open) subset $U \subset \Bbb R^2$ on which there is a continuous choice of angular coordinate. For any such choice, the coordinate vector field $\partial_{\phi}$ is only defined on $U$, rather than all of $\Bbb R^2$.

On the other hand, the above computation shows that, no matter our choice of $U$, $\partial_{\phi}$ is the restriction of the vector field $\color{#bf0000}{-y \partial_x + x \partial_y}$, which is defined on all of $\Bbb R^2$. Because of this, it is suggestive, convenient, and abusive to use the more compact notation $\partial_{\phi}$ for this vector field.

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Edit With the expression $(\ast)$ in hand, it's easy to compute the divergence (though this is certainly slower than using the expression for the divergence in general coordinates given in the comments):

$$\text{div } \partial_{\phi} = \text{div} (-y \partial_x + x \partial_x) = \frac{d}{dx}(-y) + \frac{d}{dy}(x) = 0 .$$