Interesting differential equation

[birulami]

Comparing the average speed of a bunch of points with the individual points' speeds, I came across the following equation:

$\left(\frac{dx(t)}{dt}\right)^2 <br /> = \frac{1}{N^2} c^2 \sum_{i\neq j} \left(1-\frac{v_i(t) v_j(t)}{c^2}\right)$

where the $v_i$ are the velocity vectors (3 dimensions) of the N points. They fulfil the equation $|v_i(t)|^2 = c^2$. If I didn't loose some constant factor, the equation above should be the same as

$\left(\frac{dx(t)}{dt}\right)^2 <br /> = \frac{1}{N^2} \sum_{i< j} (v_i - v_j)^2$

Any chance to solve one or the other for $x(t)$? I hesitate to take the square root and try to integrate the square root of the sum. Are there better ways to solve this?

Harald.

 

[fresh_42]

There is a mistake in the second formula which should be $\left( \frac{dx(t)}{dt}\right)^2 = \frac{1-c^2}{N^2} \sum_{i< j} (v_i - v_j)^2$. It can be rewritten as
$$
\dfrac{N^2}{1-c^2}\, \ddot{x}(t)^2 = \left|\left|\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\times \begin{pmatrix}1\\1\\1\end{pmatrix}\right|\right|^2 =\left|\left|\begin{pmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{pmatrix}.\mathbf{v}\right|\right|^2
$$
So the entire equation looks a bit like $\gamma \,||\dot{\mathbf{v}}||^2 = ||\mathbf{A.v}||^2$ or $\mathbf{\dot v} = \dfrac{\sqrt{1-c^2}}{N}\mathbf{A.v}$ with an exponential function as solution.

This leads to another idea, namely to consider $x(t)$ as path in a Lie group, presumably $\operatorname{SO}(3)$.