Average distance between two points on circle

[player1_1_1]

Homework Statement

I have very funny thing to do; I must count average distance between two points in circle of R radius:D (hahaha)

Homework Equations

Defined integrals, parametric equations, etc

The Attempt at a Solution

I am describing the circle in two parametres:
\begin{cases}x(t)=R\cos t\\ y(t)=R\sin t\end{cases}
I can choose any two points by choosing any parameters t_1,t_2 in \langle 0;2\pi\rangle limits. Two points A_1,A_2 will be described by coordinates
A_1=\left(R\cos t_1,R\sin t_1\right),A_2=\left(R\cos t_2,R\sin t_2\right)
and distance between them in t_1,t_2 function
s\left(t_1,t_2\right)=\sqrt{\left(\cos t_1-\cos t_2\right)^2+\left(\sin t_1-\sin t_2\right)^2}
now I am making double integral in limits in which t_1,t_2 changes, tzn t_1,t_2\in\langle 0;2\pi\rangle divided by the field where the function is defined
\frac{\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{\left(\cos t_1-\cos t_2\right)^2+\left(\sin t_1-\sin t_2\right)^2}\mbox{d}t_1\mbox{d}t_2}{\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\mbox{d}t_1\mbox{d}t_2}
is it correct answer for average value of this function?
now I am counting integral in this way, bottom integral is 4\pi^2 value, so I have
\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{\left(\cos t_1-\cos t_2\right)^2+\left(\sin t_1-\sin t_2\right)^2}\mbox{d}t_1\mbox{d}t_2=
=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{\cos^2t_1-2\cos t_1\cos t_2+\cos^2t_2+\sin^2t_1-2\sin t_1\sin t_2+\sin^2t_2}\mbox{d}t_1\mbox{d}t_2=
=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{2-2\cos t_1\cos t_2-2\sin t_1\sin t_2}\mbox{d}t_1\mbox{d}t_2=
=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{2\left(1-\cos\left(t_2-t_1\right)\right)}\mbox{d}t_1\mbox{d}t_2=
=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}\sqrt{4\sin^2\left(\frac{t_2-t_1}{2}\right)}\mbox{d}t_1\mbox{d}t_2=\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}2\left|\sin\left(\frac{t_2-t_1}{2}\right)\right|\mbox{d}t_1\mbox{d}t_2
is it good way of thinking? how can I count this double integral (i mean what to do with this abs)? thanks for help!

 

[Dick]

It looks to me like you are doing fine so far. The easiest way to do the last integral is to notice that the integral |sin((t2-t1)/2)|*dt1 is actually independent of the value of t2. |sin(t/2)| is periodic with period pi. It doesn't really matter whether you integrate it over an interval [0,2pi] or [a,a+2pi]. See what I mean? That means you can for example just pick t2=0 and do the integral. If you think about the original problem in the same light, you should realize you never really needed to make both t1 and t2 variable. You could fix one of them and integrate over the other. The circle is symmetric that way.

 

[benorin]

\frac{1}{4\pi^2}\int\limits^{t_1=2\pi}_{t_1=0}\int\limits^{t_2=2\pi}_{t_2=0}2\left|\sin\left(\frac{t_2-t_1}{2}\right)\right|\mbox{d}t_1\mbox{d}t_2

http://img31.imageshack.us/img31/2839/sint2t1over2.jpg

is the plot of z=\sin\left( \frac{t_2-t_1}{2}\right) for t_1,t_2\in [ 0,2\pi].

 

[player1_1_1]

Of course, there should also be R in every integrals:) i see that now, i have finished calculations, thanks you for help! where did you find this thing which is drawing 3d function?

 

[vela]

If I recall correctly, this is one of those problems which has several solutions that will give you equally valid but different answers. Look up Bertrand's paradox.

 

[player1_1_1]

omg what a strange thing:D but when t_1,t_2\in\langle0;2\pi\rangle parameters are describing any possible pair of points on the circle, answer from this integral (or also single integral when one parametr is defined) must be correct for average distance?