# Ring thrown in space

A ring is thrown randomly in the space and let $A(t)$ be its position at moment $t$. Let us say that the moment $t_{0}$ is "twisted", if the ring $A(t_{0})$ is linked with the rings $A(t)$ for $t$ sufficiently close to $t_{0}$ $(t\neq t_{0})$.

Problem. Find the probability $p$ that a randomly choosen moment $t_{0}$ is twisted.

(easy variant - show that $0<p<1$.)

Do you think that this question makes sense, in particular, is it possible to define correctly such a probability $p$?

P.S. I suppose that the movement of $A(t)$ is smooth.

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If you're assuming continuity then isn't the ring always "twisted"? – EuYu Aug 11 '12 at 11:12
@Eu Yu: No,I define a "twisted" moment of the movement and not every moment is twisted - for example if the ring remains paralel to one and the same plane. – scholar Aug 11 '12 at 11:22
Or, more probably, if it moves perpendicularly to its plane faster than it rotates (so its motion describes a tube) – Ben Millwood Aug 11 '12 at 12:03
@scholar: do gravitational/other forces enter into account? Else, the motion is linear? $p$ is allowed to vary in terms of speed/over $\mathrm{SO}(3)$? Give us more details. – Chindea Filip Aug 11 '12 at 13:04
By ring do you mean a circle? If so the motion can be decomposed into a linear trajectory of the center and rotation in 3d about the center – hardmath Aug 11 '12 at 14:09

The probability is

$$\frac{\arctan\sqrt2}\pi\approx0.3041\;.$$

Seriously. Well, in a sense. This is the probability if the ring is thrown into space by thermal fluctuations. As was indicated in the comments, the proportion of "twisted" moments depends on the ratio between the velocity of the ring perpendicular to its plane and its angular velocity in the plane, so to get a well-defined probability we need a distribution for that ratio. This can hardly come purely from symmetry arguments, so the probability isn't well-defined for a ring thrown into space by a human.

However, we can imagine the ring having become detached from a nanobot, trundling in space, being hit every once in a nanowhile by an atom or two. Then the ring is in thermal equilibrium with its surroundings, so its state of motion follows the Maxwell–Boltzmann distribution, with the thermal energy equipartitioned between the degrees of freedom.

Consider the ring at a time $t$ and place a coordinate system such that its origin is at the centre of the ring and the ring lies in the $x$-$y$ plane. In time $\mathrm dt$, the instantaneous angular velocity $\vec\omega$ displaces the point at position $\vec r$ by $\vec\omega\times\vec r\,\mathrm dt$, whereas the centre-of-mass velocity $\vec v$ displaces it by $\vec v\,\mathrm dt$.

We need to find the two points, if any, at which the displaced ring intersects the $x$-$y$-plane, and determine whether they're both inside or both outside the original ring (untwisted moment) or one inside and one outside (twisted moment). We can choose the $x$ and $y$ axes such that $\vec\omega$ has no $y$ component. Then setting the $z$ coordinate after time $\mathrm dt$ to zero yields $\omega_xy+v_z=0$. With $r=1$ for convenience, this has no solutions if $|v_z|\gt|\omega_x|$, whereas for $|v_z|\lt|\omega_x|$ we get $y=-v_z/\omega_x$ and two corresponding values of $x$ with opposite signs.

To determine whether these points are moving into or out of the original ring, we can multiply their instantaneous velocities by their position vectors; a positive sign means moving outward and a negative sign means moving inward. Rather helpfully, we have $(\vec\omega\times\vec r)\cdot\vec r=0$, so the product is just $\vec v\cdot\vec r$. Since the direction of $\vec v$ is uniformly distributed, averaging the probability of one point moving inward and the other outward over $v_x$ and $v_y$ will yield the proportion of directions that lie between the two solutions for $\vec r$, which is $\frac1\pi\arccos|v_z/\omega_x|$.

Now we need the densities for $v_z$ and $\omega_x$. The density for $v_z$ is proportional to $\exp(-mv_z^2/(2kT))$. The one for $\omega_x$ is only slightly more tricky to determine. First, since we chose the coordinate system to make $\omega_y$ vanish, we actually need the density not for $\omega_x$ but for $\omega=\sqrt{\omega_x^2+\omega_y^2}$. The density for each $\omega_\alpha$ is proportional to $\exp(-I\omega_\alpha^2/(2kT))$, where $I$ is the ring's moment of inertia with respect to an axis in the plane, so the density for $\omega$ is proportional to $\omega\exp(-I\omega^2/(2kT))$. The moment of inertia is readily calculated:

$$I=\int_0^{2\pi}\frac{m\mathrm d\phi}{2\pi}\cos^2x=\frac m2\;.$$

Now we have all the ingredients to set up the integral for the twist probability. Since only the ratio between $v_z$ and $\omega$ matters, we can drop the factors of $m/(2kT)$, leaving us with

$$P(\text{moment is twisted})=\frac1{\pi N}\int_0^\infty\int_0^\omega\omega\exp(-\omega^2/2)\exp(-v_z^2)\arccos\frac{v_z}\omega\mathrm dv_z\mathrm d\omega$$

with the normalization factor

$$N=\int_0^\infty\int_0^\infty\omega\exp(-\omega^2/2)\exp(-v_z^2)\mathrm dv_z\mathrm d\omega=\frac{\sqrt\pi}2\;.$$

According to Wolfram|Alpha, the integral evaluates to

$$\frac{\sqrt\pi}2\arctan\sqrt2\;,$$

and the result follows.

P.S.: I figured out how to solve the integral

$$\int_0^\infty\int_0^\omega\omega\exp(-\omega^2/2)\exp(-v_z^2)\arccos\frac{v_z}\omega\mathrm dv_z\mathrm d\omega\;.$$

Transform to polar coordinates,

$$\int_0^\infty\int_0^{\pi/4}r\cos\phi\exp\left(-r^2\left(\frac12\cos^2\phi+\sin^2\phi\right)\right)\arccos\tan\phi\,\mathrm d\phi\,r\mathrm dr\;,$$

carry out the integration over $r$,

$$\frac{\sqrt\pi}4\int_0^{\pi/4}\cos\phi\left(\frac12\cos^2\phi+\sin^2\phi\right)^{-3/2}\arccos\tan\phi\,\mathrm d\phi\;,$$

substitute $u=\tan\phi$,

$$\frac{\sqrt\pi}4\int_0^1\left(u^2+\frac12\right)^{-3/2}\arccos u\,\mathrm du\;,$$

substitute $t=\arccos u$ (of course both substitutions could be done in one),

$$\frac{\sqrt\pi}4\int_0^{\pi/2}\left(\cos^2t+\frac12\right)^{-3/2}t\sin t\,\mathrm dt\;,$$

integrate by parts, differentiating $t$ and substituting $\cos t$ to integrate the rest,

$$\frac{\sqrt\pi}2\left(-\left[\left(\cos^2t+\frac12\right)^{-1/2}t\cos t\right]_0^{\pi/2}+\int_0^{\pi/2}\left(\cos^2t+\frac12\right)^{-1/2}\cos t\,\mathrm dt\right)\;,$$

evaluate the boundary term to $0$,

$$\frac{\sqrt\pi}2\int_0^{\pi/2}\left(\cos^2t+\frac12\right)^{-1/2}\cos t\,\mathrm dt\;,$$

use $\cos^2t+\sin^2t=1$,

$$\frac{\sqrt\pi}2\int_0^{\pi/2}\left(\frac32-\sin^2 t\right)^{-1/2}\cos t\,\mathrm dt\;,$$

subtitute $\sin t$,

$$\frac{\sqrt\pi}2\left[\arctan\left(\sin t\left(\frac32-\sin^2t\right)^{-1/2}\right)\right]_0^{\pi/2}\;,$$

and evaluate to

$$\frac{\sqrt\pi}2\arctan\sqrt2\;.$$

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