Why does vector field <x, -y> have 0 divergence?

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Numerically, the divergence is 0. However, when looking at the vector field, it appears like the flow is away from the origin. Hence, why does it have 0 divergence just by examining the vector field. (Image in link above)

Adding on, why is divergence calculated as a sum of the partial derivatives of every component, instead of solely examining each component and taking the partial derivatives individually.

Breaking up the vector field <x, -y> into its individual components, it has positive divergence in the x-direction, and negative divergence in the y-direction, which seems to give more intuition on the divergence of the vector field than their sum.

TL;DR Why does vector field <x, -y> have 0 divergence and why is divergence calculated as a sum?

Solution 1:

We can look at this vector field $ \ \overrightarrow{F} \ = \ \langle \ x \ , \ -y \ \rangle \ \ $ in a couple of ways. If we look at unit vectors, it becomes $ \ \hat{F} \ = \ \langle \ \frac{x}{r} \ , \ -\frac{y}{r} \ \rangle \ = \ \langle \cos \theta \ , \ -\sin \theta \rangle \ \ . $ This means that on a circle of any radius (centered on the origin), every possible direction for the unit vectors appears. So the sum of these vectors over any circular boundary is zero, indicating that there is no "net flow" across that closed curve.

If we look at the tangent of the angle that the vectors make to the $ \ x-$axis, we find $ \ \tan \theta \ = \ \frac{-y}{x} \ \ , $ which tells us that the direction of "flow" is identical in diagonally-opposed quadrants and exactly reversed in adjacent quadrants. So again, there is the suggestion of "canceling flows". (Somewhat less intuitive -- so requiring a proof -- is that the net flow through any closed curve within the zero-divergence field is equal to zero.) Taking the tangent of the angle above as the slope of a tangent line to a curve, we get a separable differential equation $$ \frac{dy}{dx} \ \ = \ \ \frac{-y}{x} \ \ \Rightarrow \ \ \int \ \frac{dy}{y} \ \ = \ \ \int -\frac{dx}{x} \ \ \Rightarrow \ \ \ln |y| \ = \ -\ln |x| \ + \ C $$ $$ \Rightarrow \ \ y \ \ = \ \ \frac{K}{x} \ \ , \ \ K \neq 0 \ \ . $$ So the flow curves are the family of rectangular hyperbolas centered on the origin, showing the "diagonal symmetry" of the flow and why it is plausible to expect a net flow of zero through at least a closed curve symmetrical about the origin.

The divergence of a vector field is the scalar ("dot") product of the vector differential operator (in two dimensions, $ \ \frac{\partial }{\partial x} + \frac{\partial}{\partial y} \ $) with the components of the vectors, (here $ \ F_x \mathbf{i} \ + \ F_y \mathbf{j} \ ) \ \ , $ which is why it is a (scalar) sum. This is not particularly intuitive in itself. Since the divergence of a field is only meaningful over a closed region, this sum of partial derivatives of corresponding vector components, integrated over a region, gives the sum of the net changes in each component. If this total net change in all of the components is zero $ \ ( \ \mathbf{div} \overrightarrow{F} \ = \ 0 \ ) \ $ in the region, there is no net flow through its boundary.