## Related questions with answers

Continuing with the situation in the previous problem, the number of molecules in the right and left halves of a room containing $N$ molecules can be written as $N_{\mathrm{R}}=(N / 2)(1+\Delta), N_{\mathrm{L}}=(N / 2)(1-\Delta)$. For $N$ large, we expect the number of molecules in the two halves to be nearly equal so $\Delta \ll 1$. (a) Show that the probability of a fractional discrepancy of $2 \Delta$ more molecules on the right than on the left is given by

$P(\Delta)=C e^{-\Delta^2 N}$

where $C$ is a constant that does not depend on $\Delta$. [Hint: You will need to use Stirling's approximation, $\ln N ! \approx N \ln N-N$, and the relation $\ln (1+x) \approx x$, valid for $x \ll 1$. Begin by writing $P=N ! /\left(N_{R} ! N_{L} !\right)$ and take the In of both sides.] (b) For a room containing $10^{25}$ molecules, what is the ratio $P(\Delta=0.001) /$ $P(\Delta=0) ?$ (c) What values of $\Delta$ are likely to actually occur in a room?

Solution

VerifiedThe number of particles inside the left and right halves can be defined as:

$\begin{aligned} N_{R}=\frac{N}{2} \cdot(1+\Delta) \quad N_{L}=\frac{N}{2}(1-\Delta) \end{aligned}$

It's necessary to write the probability in the following form where the constant C doesn't depend on $\Delta$:

$\begin{aligned} P(\Delta)=C e^{-\Delta^{2} N} \end{aligned}$

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