# Does the arc length constant of the sine function interval occur anywhere in physics?

π occurs in many formulae, but does the arc length of a single sine function interval have any meaning in physics? Does it occur in any formula?

According to What is the length of a sine wave from 0 to 2π? its value is $$\approx 7.640395578055424$$, approximately 21.6% larger than 2π.

• You can move along any curve, so why not on the graph of a sin- function. Aug 28, 2022 at 14:35
• That constant comes from an elliptic integral (of the 2nd kind), and elliptic integrals arise in various contexts in physics & astronomy. Eg, geodesic calculations on an ellipsoid require an elliptic integral of the 2nd kind. See en.wikipedia.org/wiki/Geodesics_on_an_ellipsoid Aug 28, 2022 at 17:04
• Here's a fast arbitrary precision calculator for that constant, using the AGM: sagecell.sagemath.org/… Aug 28, 2022 at 17:06

Does the arc length constant of the sine function interval occur anywhere in physics?

Yes!

Using the definition

$$E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2\theta}\,d\theta$$

for the complete elliptic integral of the second kind, your sine-arc-length constant is

$$C=4\sqrt2\,E(1/\sqrt2)\approx 7.64\dots.$$

(See the derivation in the Math SE question that you linked to.)

The electrostatic potential on a uniformly charged disk of radius $$R$$ with surface charge density $$\sigma$$, at distance $$\rho$$ from the center, is given by

$$\varphi(\rho)=\frac{\sigma R}{\pi\varepsilon_0}E(\rho/R)$$

where $$\varepsilon_0$$ is the electric constant. (See equation 7 and figure 2 in the linked paper.)

Thus at radial distance $$\rho=R/\sqrt{2}$$, the value of the potential involves your constant $$C$$:

$$\varphi(R/\sqrt2)=\frac{\sigma R}{\pi\varepsilon_0}E(1/\sqrt2)=\frac{C}{4\sqrt2\pi}\frac{\sigma R}{\varepsilon_0}.$$

Note that the derivation of the potential does not involve computing the arc length of one period of a sine curve. Instead, it is an example of how “special functions” such as elliptic integrals can turn up in a wide variety of circumstances. This is of course why they become standardized and are given names.

The parametric equations of a circle of radius R are obviously of the form: $$x=R \cos\left(\frac{s}{R}\right) , y= R \sin \left(\frac{s}{R}\right)$$

where $$s$$ is the natural parameter (length of arc).

we have $$\ddot{x}=-\frac{1}{R}\cos\left(\frac{s}{R}\right)\;\;\;,\;\ddot{y}=-\frac{1}{R}\sin\left(\frac{s}{R}\right)$$ We have:$$k(s)=\frac{1}{R}\;\;\;$$for all $$s$$

therefore, the curvature of the circle is constant and equal to the inverse of its radius.

Along any curve, the curvature is a determined function of the length of the arc:$$\frac{1}{\rho}=f(s) \;\;\tag{1}$$ one shows reciprocally, that to any equation equation of the form (1) corresponds a determined curve. Indeed, we choose any direction for the X axis and let $$\varphi$$ be the angle formed by the tengent to the curve with this axis. we know that: $$\;\frac{1}{\rho}=\pm\frac{d\varphi}{ds}$$ and equation (1) becomes:$$\pm\frac{d\varphi}{ds}=f(s)$$ from where:$$\varphi=\pm\int_{0}^{s}f(s)ds+C$$ We can consider that the direction of the X axis corresponds to the direction of the tengent for $$s=0$$, so that we can consider $$C=0$$ in the last formula, i.e. we have for the angle $$\varphi$$:$$\varphi=\pm F(s) \;\;{where} \;\;F(s)=\int_{0}^{s}f(s)ds$$ we know that $$\frac{dx}{ds}=\cos(\varphi)\;\;,\;\;\frac{dy}{ds}=\sin(\varphi)$$ whence, given the previous equality $$x=\int_{0}^{s}\cos[F(s)]ds+C_{1}\;\;,\;\;y=\int_{0}^{s}\sin[F(s)]ds+C_{2}$$

If we take the origin of the coordinates at the point of the curve where $$s=0$$, we must take $$C_{1}=C_{2}=0$$ and we obtain a well-defined curve:$$x=\int_{0}^{s}\cos[F(s)]ds\;\;,\;\;y=\int_{0}^{s}\sin[F(s)]ds$$ Equation (1) is the natural equation of the curve in the sense that this curve is in no way linked to the arbitrary choice of the coordinate axes and that it corresponds to a well-determined curve 'up to a symmetry'.