# Why does a spiral structure appear for this function?

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?

• 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. May 7, 2021 at 22:38

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 $$2\pi$$.

• 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. May 9, 2021 at 9:23
• 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? May 11, 2021 at 2:20
• 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.) May 11, 2021 at 11:24
• When $m=0$ we don't have $V_2$. So the origin-passing circle is not translated. That's what I meant above. May 11, 2021 at 11:48