The Stirling approximation.J!2πJJ+1/2e-J, is very handy when dealing with numbers larger than about100 . Consider the following ratio: the number of ways Nparticles can be evenly divided between two halves of a room to the number of ways they can be divided with60%on the right and40%on the left.

(a) Show, using the Stirling approximation, that the ratio is approximately4046065Nfor largeN.

(b) Explain how this fits with the claim that average behaviours become more predictable in large systems.

Short Answer

Expert verified

a)

The ratio isW1W2=40.460.65N·

b)

The numerical value of the ratio is around unity based on our calculated expression for the ratio, meaning that both circumstances describing and are roughly equally likely to occur. The ratio is considerably raised by making the room very large, meaning that the condition where both parts of the room contain roughly the same number of particles is heavily favoured. This backs with the theory that for very large numbers of people, average behaviours become relatively predictable.

Step by step solution

01

 Step 1: Concept used

Stirling's approximation:

The approximate value for a factorial function (n!) is obtained using the Stirling formula, also known as Stirling's approximation formula. Gamma function can also be performed with this. Applied mathematics also makes use of Stirling's formula. It simplifies calculating the factorial of higher values.

02

Given information from question

In this problem, we are to consider the ratio: the number of ways N particles can be arranged in such a way that both sections of a two-sided room haveN2 particles each W1to the number of ways N particles can be arranged in such a way that the right side has of the total particles and the left has the remaining .40%W2

03

Calculate the number of ways that particles can be arranged

a)

CallingW1 the number of ways N particles can be arranged such that both region contain half of the total number of particles is given by the binomial coefficient:

W1=N!(0.5N)!(N-0.5N)!=N!(0.5N)!2

Contrastingly, calling W2the number of waysN particles can be arranged such that the right region has 60%of the total number of particles and the left region has the remaining 40%is given by:

W2=N!(0.6N)!(N-0.6N)!=N!(0.6N)!(0.4N)!

Taking their ratio we obtain:

W1W2=N!(0.5N)!2N!(0.6N)!(0.4N)!=(0.6N)!(0.4N)!(0.5N)!2

04

 Step 4: Apply Stirling’s approximation to each coefficient

Apply Stirling's approximation to each coefficient with the factorial operation, and simplify the expression:

W2W12π(0.6N)0.6N+1/2e-0.6N2π(0.4N)0.64+1/2e-0.4N2π(0.5N)0.5N+1/2e-0.5N2(0.6N)0.6N+1/2(0.4N)0.4N+1/2e-0.6Ne-0.4N(0.5N)0.5N+1/22e-0.5N2(0.5N)0.5N+1/22(0.6N)0.6N+1/2(0.4N)0.4N+1/2((0.0.6N-0.4N+N(0.6N)0.6N+1/2(0.4N)0.4N+1/2(0.5N)0.5N+1/220.60.6N+1/20.40.4N+1/20.N0.6N+1/2N0.4N+1/20.60.6N+1/20.40.4N+1/20.5NN+1N0.6N+0.4N-N+1/2+1/2-10.60.6N+1/20.40.4N+1/20.5N+1.

Assuming N is very large, we can approximate the above expression further by getting rid of the additive terms in the exponents (i.e. the1/2 's for the 0.6and 0.4, and the 1 for the 0.5. Doing so yields:

W1W20.60.6N0.40.4N0.5N

05

Make minor algebraic manipulations 

Making some minor algebraic manipulations we obtain:

\W1W20.610100.6N0.410100.4N0.51010N40.4N60.6N5N10-0.6N10-0.4N10-N40.4N60.6N5N10N-0.6N-0.4N40.4N60.6N5N40.460.65N·

06

Average behaviours become foreseeable when dealing with very large numbers.

b)

Consider the number N to be a small one. The numerical value of the ratio is around unity based on our calculated expression for the ratio, meaning that both circumstances describing and are roughly equally likely to occur. The ratio is considerably raised by making the room very large, meaning that the condition where both parts of the room contain roughly the same number of particles is heavily favoured. This backs with the theory that for very large numbers of people, average behaviours become relatively predictable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A scientifically untrained but curious friend asks, "When I walk into a room, is there a chance that all the air will be on the other side?" How do you answer this question?

Consider a room divided by imaginary lines into three equal parts. Sketch a two-axis plot of the number of ways of arranging particles versus NleftandNrightfor the caseN=1023, Note that Nmiddleis not independent, being of courseNNnghtNleftYour axes should berole="math" localid="1658331658925" NleftandNright, and the number of ways should be represented by density of shading. (A form for numbers of ways applicable to a three-sided room is given in Appendix I. but the question can be answered without it.)

From elementary electrostatics the total electrostatic potential energy in a sphere of uniform charge Q and radius R is given by

U=3Q25×4π0R

(a) What would be the energy per charge in a lead nucleus if it could be treated as 82 protons distributed uniformly throughout a sphere of radiusrole="math" localid="1659180305412" 7×10-15m

(b) How does this result fit with Exercise 87?

(a) Show that the number of photons per unit volume in a photon gas of temperature Tis approximately(2×107K3m3)T3(Note:0x2(ex1)1dx2.40.)

(b)Combine this with a result derived in Example 9.6 to show that the average photon energy in a cavity at temperatureTis given byE¯2.7kBT.

For a room3.0mtall, by roughly what percent does the probability of an air molecule being found at the ceiling differ from that of an equal speed molecule being found at the floor? Ignore any variation in temperature from floor to ceiling.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free