# error propagation and collision in ideal gas

When dealing with gas, a statistical approach is needed because

1. For N particles, you have to solve 6N equations which cant be done analytically. To know our time step for numerical solving, you can estimate the frequency of a collision. For an ideal gas, this frequency is around $10^9$Hz which gives $10^{32}$ equations per second for 1mole of gas at room temperature.

2. Errors propagate quickly and soon it is completely impossible to know anything about the particles you are studying.

To show the second point, consider a collision between 2 molecules as shown on this picture

We have the relation $\frac{sin\theta_1}{l_0} = \frac{sin\theta_0}{2r_0} \approx \frac{\theta_0}{2r_0}$ where $l_0$ is the mean free path and $r_0$ is the radius of the particle

knowing the uncertainty $\delta\theta_0$ , we can estimate the uncertainty on $\delta\theta_1 \approx \delta\theta_0\frac{d\theta_1}{d \theta_0} \approx \delta\theta_0(\frac{l_0}{2r_0}) \approx \delta\theta_0 * 10^3$.

After n collisions, the error is $\delta\theta_0*10^{3n}$ which rapidly explode.

Now see how important this effect was, I tried to apply it to the smallest angle I could find. The observable universe being about $10^{25}m$ in radius and the planck length $10^{-35}m$, the smallest angle one can observe is about $10^{-60} rad = 10^{-3*20} rad$. To get an error of $\pi$, it would therefor take 20 collisions.

This doesn't seem possible, is this result correct? If not, where is the problem? I can't find it but I can't believe that 20 collisions are enough to lose all information about the particle.

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I didn't checked the calculations, but there is an oversimation of the error in assuming "pessimistic" angle deviations correlations, the error should go as the square root of $n$ not as $n$, like the mean distance of a random walker, this raises the collisions to $400$, note also that you are using the linear formula for the error also in the non-linear regime. –  Ikiperu May 27 '13 at 23:34
but isn't it always the case for the linear formula? To calculate the error you make a serie expansion on the parameter you want to evaluate. Even with 400 collisions, isn't that extremely rapid? About 1µs in the simulation and we can't say anything –  Sylvain Blanco May 28 '13 at 20:24
In fact it is rapid, note however that you are using an approximate formula, the correct one is $2r_0\sin \theta_1=(l_0-2r_0 \tan \theta_1)\tan \theta_0$, for which the mean error is lower (you should consider $\theta_0$ uniformly distribuited). The final result will be of order of milliseconds maybe or higher, which is rapid, you are right, I didn't expect that at all. –  Ikiperu May 28 '13 at 21:10