Statistical behaviour of ideal particles in a closed box

[birulami]

Suppose I have N ideal particles in an enclosure, be it a ball or a cube or some other form. The particles shall bounce off the walls of the enclosure and against each other without losing speed. The velocity of each particle $$i$$ shall be such that it fullfills $$|v_i|=\rho$$, where $$\rho$$ is constant, i.e. the speed is always the same, but of course the direction in 3D differs.

Further, on the average, the whole ensemble of points shall not move, i.e.

(*) $$\sum_{i=1}^N v_i = 0$$

or at least the sum is very close to zero.

Apart from this, the velocities' directions shall be completely random. What exactly this would mean may need to be further defined.

My questions are:
1) How can I estimate $$\frac{1}{N^2}\sum_{i<j} (v_i-v_j)^2$$?
2) Is the "boxed" condition used in the derivation or does (*) contain all we need?

Maybe this is not really a physics question, because the setup is too idealised, but I assume that is still closer to physics than to pure math.

Harald.

 

[Naty1]

You might find this interesting:

http://en.wikipedia.org/wiki/Particles_in_a_box

 

[birulami]

You might find this interesting:

http://en.wikipedia.org/wiki/Particles_in_a_box

An interesting link that I will certainly take a closer look at. But on first glance it seems to relate to quantum physics, which is certainly not what I am after here.

Harald.

 

[dgOnPhys]

Hi birulamy,

it looks very much like a micro-canonical ensemble so I would expect you will get the basic results of statistical mechanics.

1) can be simplified by
- extending sum over all i,j and dividing by two
- expanding the square of the binomial
the results I think is rho^2 (or whatever letter you use for the particles' constant speed)
2) not sure what you mean

Let me know if you need details

 

[birulami]

Hi dgOnPhys,

thanks for the hint about the micro-canonical ensemble. In particular the connection to entropy is interesting for me.

It seems like I got a solution myself which is most likely the same as you state. By setting $$d_{ij}=v_i-v_j$$ and assuming that the sum can be approximated by taking $$(N^2-N)/2$$ times the expected value of the $$d_{ij}$$, I got the result $$(1-1/N)\rho^2$$. The discrepancy is most likely that this estimation does not enforce the condition (*), but only converges to it for large N.

Harald.

 

[dgOnPhys]

Hi Harald,

without approximations:
$$\frac{1}{N^2}\sum_{i<j} (v_i-v_j)^2=$$
$$=\frac{1}{2N^2}\sum_{i,j} (v_i-v_j)^2=$$
$$=\frac{1}{2N^2}\sum_{i,j} (v_i^2+v_j^2-2 v_i . v_j)=$$
$$=\frac{1}{2N^2}\sum_{i,j} (2 \rho^2-2 v_i . v_j)=$$
$$=\frac{1}{2N^2}(2 N^2 \rho^2 - 2 \sum_{i,j} (v_i . v_j))=$$
$$=\frac{1}{N^2}( N^2 \rho^2 - \sum_{i} (v_i . \sum_{j}v_j))=\rho^2$$