A new curve whose arc length is given by the elliptic integral of the first kind has been developed. I'm seeking help in determining properties beyond the arc length.

The roles of the elliptic integrals in the analysis of ellipses are well established and will not be repeated here. The basic results are that the average radius and the arc length of the elliptic curve for $\theta\in[0,\pi/2]$ are given by the elliptic integrals of the first and second kinds, respectively. Specifically,

$$ {\overline r}=\frac{2a}{\pi}~\text{K}(-A)\\ s=b~\text{E}(-A)\\ A=\frac{a^2-b^2}{b^2} $$

where $a,b$ are the semi-major and -minor axes.

We wondered if the there is a curve whose arc length is given by the elliptic integral of the first kind. To that end, we consider the arc length in the complex plane and reverse-engineered it to get the curve we want.

$$ s=\int |\dot z(u)| du\\ \text{K}(k^2)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}\\ |\dot z(u)|=\frac{1}{\sqrt{1+a^2\sin^2\theta}}\\ \dot z(u)=\frac{1}{1-ai\sin\theta}\\ z=\frac{2}{\sqrt{1+a^2}}\tan^{-1}\left[\frac{ai+\tan\frac{\theta}{2}}{\sqrt{1+a^2}} \right] $$

I'll confess to using Mathematica to get indefinite integral, but I did verify its derivative manually. This is a single-parameter family of sigmoidal curves for $\theta\in[0,\pi]$ that can form closed curves when continued into the remaining quadrants. The arc length of the sigmoidal curve is $s=2\text{K}(-a^2)$. This result has been verified numerically, as well.

Some example closed curves are shown below. The top figure shows random solutions for $a<\sqrt{2}$ and $a>\sqrt{2}$ on the left and right respectively. Those on the left overlap, those on the right do not. The lower figure demonstrates that these curves tessellate.

So finally, the outstanding questions are:

(1) The solution for $z$ seems to be intractable; it is important to simplify it if we are to have any chance of finding the area under the curve or its centroid. These will allow us to get at the bodies of revolution, as well. The problem is here is that these depend on the product $z^*\dot z$. The centroid of the curve itself is probably at its center point due to its symmetry.

(2) A proper formula for extending the equation all the way around to a closed form is needed, in much the same way as the following equation describes a complete ellipse.

$$z=a\cos\theta~\text{sign}(\cos\theta)+b\sin\theta~\text{sign}(\sin\theta),\quad \theta\in[0,2\pi]$$

(3) The significance of $a=\sqrt{2}$, which, by the way, is empirical, and may only be approximate.

What I've tried so far: I can express $z(\theta)$ in terms of a logarithm instead of the arctangnet (by way of the arc hyperbolic tangent), but that does not improve matters. Here it is

$$z=-\frac{i}{\sqrt{1+a^2}}\log\left[\frac{\sqrt{1+a^2}-a+i\tan\frac{\theta}{2}}{\sqrt{1+a^2}+a-i\tan\frac{\theta}{2}} \right]$$

Clearly, we express the tangent as sine and cosine, or exponentials, but it just gets worse and worse. Comparison of EllipticK curves Tessellation of EllipticK curves

  • $\begingroup$ wouldn't be awesome if mathjax supported tikz ? $\endgroup$ – Anonymous Jun 29 '17 at 16:41
  • $\begingroup$ @Anonymous Alright, I'll bite. What's that? $\endgroup$ – Cye Waldman Jun 29 '17 at 16:45
  • $\begingroup$ mathjax is what we use here to write latex, and TiKz is package for creating graphics using latex syntax. ( bit.ly/2unSm6h ) $\endgroup$ – Anonymous Jun 29 '17 at 16:48
  • $\begingroup$ Slightly related: you might be interested in Serret's curves. $\endgroup$ – J. M. is a poor mathematician Jul 26 '17 at 5:25

This is a partial answer, specifically attempting to address question (1).

I have converted the expression for $z(\theta)$ into cartesian form. Starting from

$$ z(\theta) = \dfrac{i}{\sqrt{a^2+1}}\log\left[\dfrac{\sqrt{a^2+1}-a - i\tan\frac{\theta}{2}}{\sqrt{a^2+1}+a+i\tan\frac{\theta}{2}}\right]$$ which has the derivative


I managed to get this cartesian form

$$z(\theta)= -\dfrac{1}{\sqrt{a^2+1}}\left[\tan^{-1}\left(\dfrac{-\sqrt{a^2+1}\sin\theta}{\cos\theta}\right)+2\pi n\right]+\dfrac{i}{\sqrt{a^2+1}}\dfrac{1}{2}\ln\left(1-\dfrac{4a\sqrt{a^2+1}}{\left(\sqrt{a^2+1}+a\right)^2+\tan^2\frac{\theta}{2}}\right)$$

From this, it is straightforward to get a horrendous expression for polar form, so I'm not going to add it here.

Finding the conjuate of $z(\theta)$ is now straightforward. However the conjugate is still not likely amenable to integration.

The real part of $z$ is somewhat satisfying: a scaled arctanget of an inversely scaled tanget (if we take the $n=0$ branch). I left the $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$, so that it was obvious what needs to be fed into an atan2() function to avoid ambiguities.

The only thing I can say about the imaginary part, is that it is easy to show that it is always $ \le 0$

  • $\begingroup$ Andy, it's nice to see you again. Thanks for your reply. $\endgroup$ – Cye Waldman Dec 17 '17 at 0:58
  • $\begingroup$ Hi Cye. I should mention for the (circular?) case of a=0, this form makes it easy to see that $z(\theta)$ degenerates to a real straight line $z(\theta) = \theta$. If that provides any insight. $\endgroup$ – Andy Walls Dec 17 '17 at 3:12

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