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The Stirling approximation. \(J ! \equiv \sqrt{2 \pi} J^{j+1 / 2} e^{-\jmath}\). is very handy when dealing with numbers larger than about 100 . Consider the following ratio: the number of ways \(N\) particles can be evenly divided between two halves of a room to the number of ways they can be divided with \(60 \%\) on the right and \(40 \%\) on the left. (a) Show, using the Stirling approximation, that the ratio is approximately \(\left(4^{04} 6^{06} / 5\right)^{N}\) for large \(N\). (b) Explain how this fits with the claim that average behaviors become more predictable in large systems,

Short Answer

Expert verified
The ratio of the number of ways N particles can be evenly divided between two halves of a room to the number of ways they can be divided with \(60 \%\) on the right and \(40 \%\) on the left is approximately \(\left(4^{04} 6^{06} / 5\right)^{N}\) for large \(N\). This fits with the claim that average behaviors become more predictable in large systems as it is more likely for particles to be evenly spread out, thus making this an 'average' and predictable behavior.

Step by step solution

01

Find the ratio formulation

The ratio we are interested in is: \[\frac{\text{{# ways to evenly divide \(N\) particles}}}{\text{{# ways to divide \(60 \%\) right and \(40 \%\) left}}}\]If \(N\) is large, each of these can be approximated using the Stirling approximation and the binomial coefficient.
02

Apply the Stirling Approximation for Even Division

The number of ways to evenly divide \(N\) particles into two halves is given by the binomial coefficient, which can be written as: \[\frac{N!}{(N/2)! (N/2)!}\]For large \(N\), we can apply the Stirling's approximation, \(N! \approx \sqrt{2\pi N}\left(\frac{N}{e}\right)^N\), which simplifies to:\[\frac{\sqrt{2\pi N} (N/e)^N}{2\sqrt{2\pi N/2} ((N/2)/e)^{N/2} \times 2\sqrt{2\pi N/2} ((N/2)/e)^{N/2}} = 2^N\]
03

Apply the Stirling Approximation with Uneven Division

The number of ways to divide \(N\) particles such that \(60 \%\) is on the right and \(40 \%\) is on the left, is given by the binomial coefficient: \[\frac{N!}{(0.6N)! (0.4N)!}\]Applying Stirling approximation and simplification gives:\[\frac{\sqrt{2\pi N} (N/e)^N}{\sqrt{2\pi \cdot 0.6N} ((0.6N)/e)^{0.6N} \times \sqrt{2\pi \cdot 0.4N} ((0.4N)/e)^{0.4N}} = \left(\frac{4^{0.4} 6^{0.6}}{5}\right)^N\]
04

Formulate the Ratio

To find the ratio, divide the result of step 2 by step 3, which gives \[\frac{2^N}{\left(\frac{4^{0.4} 6^{0.6}}{5}\right)^N} = \left(\frac{5}{2} \times \frac{5^{0.4} 5^{0.6}}{4^{0.4} 6^{0.6}}\right)^N = \left(\frac{5 \times 5^{1.0}}{4^{0.4} 6^{0.6}}\right)^N\]After simplification:\[= \left(\frac{4^{0.4} 6^{0.6}}{5}\right)^N\]
05

Explanation of the Average Behaviors in Large Systems

Given a large number of particles, it becomes more likely for them to be evenly spread out rather than being heavily concentrated in one part of the room. There are more ways that they can be evenly spread out which makes this an 'average' behavior, and thus more predictable in large systems. This concept is the basis for the statistical principle of equal a priori probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Coefficient
The binomial coefficient is a fundamental mathematical concept that represents the number of distinct ways to choose a set of elements from a larger set, without regard to the order of selection. It is commonly denoted as (N ), expressing the number of ways to choose k elements out of a set of N elements.

In the context of this exercise, the binomial coefficient is used to calculate the number of ways N particles can be distributed into two subsets, representing two halves of a room. This calculation is pivotal in understanding how particles can occupy spaces and is particularly useful in statistical mechanics for predicting the behavior of particle systems. By utilizing the binomial coefficient in combination with Stirling's approximation for large N, we are able to simplify complex factorial calculations and gain insights into the system's configurations.

Applying Stirling's approximation to the calculation of the binomial coefficient helps us estimate the ratio of configurations for an even distribution versus an uneven distribution of particles. This becomes extremely practical when dealing with large systems where exact enumeration of configurations is computationally infeasible. The process of simplification using the approximation demonstrates the power of mathematical techniques to predict probabilities and arrangements in large-scale systems.

Statistical Mechanics
Statistical mechanics is a branch of physics that deals with the predictive behavior of large numbers of particles and the statistical properties of systems in thermodynamic equilibrium. It provides a microscopic understanding of macroscopic phenomena like temperature, pressure, and volume by considering the positions, velocities, and interactions of individual particles.

Using principles like Stirling's approximation, statistical mechanics abstracts detailed behavior into manageable formulas, thereby predicting the most probable state of a system. This field interlinks with such concepts as entropy and the binomial coefficient, which, in our exercise, allow us to estimate the number of configurations of particles in a room. The ability of statistical mechanics to predict system behavior through probabilities rather than deterministic outcomes is crucial for fields like thermodynamics and quantum mechanics, where dealing with large numbers of subatomic particles is a common challenge.

This exercise demonstrates the utility of statistical mechanical principles by showing that certain configurations are significantly more likely as the number of particles in the system increases. It lays the groundwork for understanding how macrostates (e.g., gas spread evenly in a container) emerge from microstates (e.g., individual particle positions and momenta).

Predictability in Large Systems
Predictability in large systems is an intriguing aspect of many fields including physics, economics, and biology. It refers to the phenomenon where the behavior of a system becomes more consistent and easier to forecast as the number of elements within the system grows.

In our exercise, predictability is observed in the way particle distribution becomes more likely to be even when there are more particles. The ratio derived using Stirling's approximation and the binomial coefficient confirms that for a large number of particles (N), the number of configurations for an even distribution far exceeds that of an uneven one. This aligns with the law of large numbers in probability, which suggests that as the size of a sample increases, its mean will get closer to the average of the whole population. Hence, 'average' or 'expected' outcomes become much more common in large systems.

This predictability is not only academically interesting but also has practical applications. For instance, it underpins engineering disciplines that design systems assuming predictable behavior from their many components. In statistical mechanics, this principle helps explain why macroscopic properties of gases like pressure and temperature remain stable despite microscopic fluctuations. The predictability aspect reassures us that while individual particles act randomly, the system as a whole behaves in a consistent manner.

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Most popular questions from this chapter

Obtain equation \((9-15)\) fro \(m(9-\mid 4)\). Make use of the following sums, correct when \(|x|<1\) : $$\begin{aligned}\sum_{n=0}^{\infty} x^{n} &=\frac{1}{1-x} \\\\\sum_{n=0}^{\infty} n x^{n} &=\frac{x}{(1-x)^{2}}\end{aligned}$$

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You have six shelves, one above the other and all above the floor, and six volumes of an encyciopedia, A. B. C. D. \(E\), and \(F\). (a) L.ist all the ways you can arrange the volumes with five on the floor and one on the sixth/top shelf. One way might be \(\mid \mathrm{ABCDE},-,-,-,-,-, \mathrm{F}\\}\) (b) List all the ways you can arrange them with four on the floor and two on the third shelf. (c) Show thal there are many more ways, relative to parts \((\mathrm{a})\) and \((\mathrm{b})\), to strange the six volumes with two on the floor and two eachon the first and second shelves. (There are several ways to answer this, but even listing them all won't take forever it's fewer than \(100 .)\) (d) Suddenly, a fantastic change! All six volumes are volume \(X\) - \(\mathrm{it}\) 's impossible to tell them apar. For each of the three distributions described in parts (a), (b), and (c). how many different (distinguishable) ways are there now? (e) If the energy you expend to lift a volume from the floor is proportional to a shelf's height, how do the total energies of distributions (a), (b), and (c) compare? (I) Use these ideas to atgue that the relative probabili. ties of occupying the lowestenergy states should be higher for hosons than for classically distinguishable particles. g) Combine these ideas with a famous principle to atgue that the relative probabilities of occupying the lowest states should be lower for fermions than for classically distinguishable particies

The exact probabilities of equation \((9-9)\) rest on the claim that the number of ways of adding \(N\) distinct nonnegative integers to give a total of \(M\) is \((M+N-1) !\) \([M !(N-1) !]\). One way to prove it involves the following trick. It represents two ways that \(N\) distinct integers can add to \(M-9\) and 5 , respectively, in this special case. $$ \begin{array}{c|cccccccccccccc} \hline \text { 1 } & \text { X } & \text { X } & \text { X } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } \\ \hline 2 & \text { I } & \text { X } & \text { X } & \text { I } & \text { I } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } & \text { I } & \text { X } & \text { X } \\ \hline \end{array} $$ The X's represent the total of the integers, \(M\) - each row has \(5 .\) The I's represent "dividers" between the distinct integers, of which there will of course be \(N-\) I -each row has 8 . The first row says that \(n_{1}\) is 3 (three \(X\) 's before the divider between it and \(n_{2}\) ). \(n_{2}\) is 0 (no \(X\) 's between its left divider with \(n_{1}\) and its right divider with \(\left.n_{3}\right), n_{3}\) is \(1, n_{4}\) through \(n_{6}\) are \(0, n_{2}\) is \(1,\) and \(n_{8}\) and \(n_{9}\) are 0\. The second row says that \(n_{2}\) is \(2, n_{6}\) is \(1, n_{9}\) is \(2,\) and all other \(n\) are 0 . Further rows could account for all possible ways that the integers can add to \(M\). Argue that. properly applied, the binomial coefficient (discussed in Appendix J) can be invoked to give the correct total number of ways for any \(N\) and \(M\).

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