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 NN molecules can be written as NR=(N/2)(1+Δ),NL=(N/2)(1Δ)N_{\mathrm{R}}=(N / 2)(1+\Delta), N_{\mathrm{L}}=(N / 2)(1-\Delta). For NN large, we expect the number of molecules in the two halves to be nearly equal so Δ1\Delta \ll 1. (a) Show that the probability of a fractional discrepancy of 2Δ2 \Delta more molecules on the right than on the left is given by

P(Δ)=CeΔ2NP(\Delta)=C e^{-\Delta^2 N}

where CC is a constant that does not depend on Δ\Delta. [Hint: You will need to use Stirling's approximation, lnN!NlnNN\ln N ! \approx N \ln N-N, and the relation ln(1+x)x\ln (1+x) \approx x, valid for x1x \ll 1. Begin by writing P=N!/(NR!NL!)P=N ! /\left(N_{R} ! N_{L} !\right) and take the In of both sides.] (b) For a room containing 102510^{25} molecules, what is the ratio P(Δ=0.001)/P(\Delta=0.001) / P(Δ=0)?P(\Delta=0) ? (c) What values of Δ\Delta are likely to actually occur in a room?


Answered 1 year ago
Answered 1 year ago
Step 1
1 of 7

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

NR=N2(1+Δ)NL=N2(1Δ) \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 Δ\Delta:

P(Δ)=CeΔ2N \begin{aligned} P(\Delta)=C e^{-\Delta^{2} N} \end{aligned}

Create an account to view solutions

Create an account to view solutions

More related questions