Consider the following 2D vector field on the $xy$-plane $$\vec{V}=\begin{pmatrix} -(m^2-md+x^2)\cos{2d\,t}+xy\sin{2d\,t} \\ -xy\cos{2d\,t}+(m^2-md+y^2)\sin{2d\,t} \end{pmatrix}$$ where $d=\sqrt{x^2+y^2+m^2}$ and a constant $m\geq0$. When plotting the vector's angle $\arctan(V_x,V_y)\in[0,2\pi]$ by color on the $xy$-plane, it always clearly shows a spiral pattern ($t=15,m=0.3$ in the plot below). How can I understand the appearance of the spiral?

enter image description here

  • $\begingroup$ I just wanted to mention that I spent more time than I'd like to admit on this and couldn't reach the desired conclusion. One comment I have is that the spiral shape can look significantly different/less elegant when $t$ and $m$ are of comparable sizes, but it probably always looks like some form of a spiral. $\endgroup$
    – Mark S.
    May 7, 2021 at 22:38

1 Answer 1


Your vector field may be written $$ V(x, y) = \underbrace{(-x\cos 2d\,t + y\sin 2d\,t)(x, y)}_{=:V_{1}(x, y)} - \underbrace{m(m - d)(\cos 2d\,t, -\sin2d\,t)}_{=: -V_{2}(x, y)}. $$ The summand $V_{2}$ is constant (in direction and magnitude) along the level curves of $d$, i.e., circles centered at the origin. For brevity, let $C_{r}$ denote the circle of radius $r$ centered at the origin. Since $d$ grows monotonically with $r$, this component (as a constant-direction field along $C_{r}$) rotates clockwise as $r$ increases.

The summand $V_{1}$ is radial from the origin, and vanishes along the curve \begin{align*} 0 &= -x\cos 2d\,t + y\sin 2d\,t \\ &= -(x, y) \cdot (\cos 2d\,t, -\sin2d\,t), \end{align*} where the position vector $(x, y)$ is orthogonal to $V_{2}(x, y)$. Since $V_{2}$ has constant direction along $C_{r}$, $V_{1}$ vanishes at diametrically opposite points of $C_{r}$, and achieves each direction twice along $C_{r}$. That accounts for the two arms of the spiral, and for the rotation of $V_{1}$, correlated with the rotation of $V_{2}$, as $r$ increases.

In the animation, all three vector fields ($V$ in green, $V_{1}$ in purple, $V_{2}$ in blue) are plotted over circles of varying radius centered at the origin.

Added in edit: The boldfaced phrase above, which explains the two-armed spirals, is true for the original example but not justified in my original answer.

What's happening is, the component $V_{1}(x, y)$ maps the circle $C_{r}$ to a circle $\Gamma$ through the origin and traced twice. The component $V_{2}(x, y)$, which is constant along $C_{r}$, translates $\Gamma$. In the original example, the translated circle winds twice about the origin. In the modified example, the translated circle does not wind about the origin.

To see that $V_{1}$ maps $C_{r}$ to a circle through the origin, we can write $V_{2}(x, y) = (A, B)$ along $C_{r}$, and write $(x, y) = (r\cos t, r\sin t)$. We have $$ V_{1}(r\cos t, r\sin t) = r^{2}(A\cos t + B\sin t)(\cos t, \sin t). $$ In polar coordinates $(\rho, \theta)$, the image is the polar graph $\rho = r^{2}(A\cos\theta + B\sin\theta)$, or $$ x^{2} + y^{2} = \rho^{2} = r^{2}(A\rho\cos\theta + B\rho\sin\theta) = r^{2}(Ax + By), $$ which is a circle passing through the origin, and traced twice as $\theta$ runs from $0$ to $2\pi$.

Decomposition of OP's vector field along circles centered at the origin

  • $\begingroup$ This looks nice. I actually found a similar vector field, which seems to go through the observations above but doesn't exhibit a spiral at all: $$(x\cos 2d\,t + y\sin 2d\,t)(x, y) + m(m + d)(\cos 2d\,t, \sin2d\,t).$$ Am I missing something here? Thanks. $\endgroup$
    – xiaohuamao
    May 9, 2021 at 9:23
  • $\begingroup$ Thanks for the edit! I found that when $m=0$ or very small $m$, both two examples exhibit clear spirals (opposite helicity), but now the trajectories are similar counterclockwise-doubly-traced origin-passing circles. Do you have any insight into this? $\endgroup$
    – xiaohuamao
    May 11, 2021 at 2:20
  • $\begingroup$ Offhand that looks like a consequence of the sign change of the coefficient $m(m + d)$, which controls the direction of $V_{2}$. (Haven't checked carefully, though.) $\endgroup$ May 11, 2021 at 11:24
  • $\begingroup$ When $m=0$ we don't have $V_2$. So the origin-passing circle is not translated. That's what I meant above. $\endgroup$
    – xiaohuamao
    May 11, 2021 at 11:48

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