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\begin{document}
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\heading{E010: Average distance between two particles in a box}
\auname{Elkana Porat}
{\bf The problem: }
\Dn
In a one dimensional box with length ${L}$, two particles have
random positions $x_1$, $x_2$ . The particles don't know about each other. The
probability function for finding a particle in a specific place in
the box is uniform.
Let ${r = x_1 - x_2}$ be the relative location of the particles.
Find ${\langle\hat{r}\rangle}$ and the dispertion relation ${\sigma_{r}}$:
\Dn
(1) By using the adding rules for the expectation values and variances of the independent positions.
(2) By calculating the probability function: ${f\left(r\right)dr={P}\left(r<\hat{r} & = & \int^{L}_{0}{\frac{xdx}{L}} = L/2\nonumber\\
\left<\hat{x}_{1,2}^2\right> & = & \int^{L}_{0}{\frac{x^2dx}{L}} = L^2/3\nonumber\\
\sigma^2_{1,2} & = & \left - \left^2 = L^2/12\nonumber
\end{eqnarray}
Now, the relative location is defined as $\hat{r} = \hat{x}_1 - \hat{x}_2$, so by using the rules for adding expectation values and variances we get:
$$ \left<\hat{r}\right> = \left<\hat{x}_1\right> - \left<\hat{x}_2\right> = 0 $$
$$ \sigma_{r}^2 = \sigma^2_1 + \sigma^2_2 = L^2/6 $$
\Dn\Dn
(2) We need to calculate $P\left(r<\hat{r} 0$}:
\ \ $ x_2 + r = x_1 < L\ \Rightarrow\ x_2 < L - r$
\begin{eqnarray}
&&\ \ \ \ P\left(r<\hat{r} 0\ \Rightarrow\ x_2 > -r$
\begin{eqnarray}
&&\ \ \ \ P\left(r<\hat{r} = \int_{-L}^{L}{\left(L-\left|r\right|\right)r\frac{dr}{L^2}} = 0 $$
$$\sigma^2_r =\left<\hat{r}^2\right> = 2\int_{0}^{L}{\left(L - r\right)r^2\frac{dr}{L^2}} = L^2/6 $$
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