Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

We claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles. It should be a state of maximum entropy! The proof involves mathematical techniques beyond the scope of the text, but finding support is good exercise and not difficult. Consider a system of 11 oscillators sharing a total energy of just \(5 h \omega_{0}\). In the symbols of Section \(9.3, N=11\) and \(M=5\). (a) Using equation \((9-9),\) calculate the probabilities of \(n\) being \(0.1,2 .\) and \(3 .\) (b) How many particles. \(N_{n}\), would be expected in each level? Round each other nearest integer. (Happily. the number is still 11 . and the energy still \(\operatorname{she}_{0}\) ) What you have is a distribution of the energy that is as close to expectations is possible, given that numbers at each level in a real case are integers. (c) Entropy is related to the number of microscopic ways the macrostate can be obtained. and the number of ways of permuting particle labels with \(N_{0}\). \(N_{1}, N_{2},\) an \(d N_{1}\) fixed and totaling 11 is \(11 ! /\left(N_{0} ! N_{1} !\right.\) \(\left.N_{2} ! N_{3}^{\prime}\right) .\) (See Appendix J for the proof.) Calculate the number of ways for your distribution. (d) Calculate the number of ways if there were 6 particles in \(n=0.5\) in \(n=1\), and none higher. Note that this also has the same total energy. (e) Find at least one other distribution in which the 11 oscillators share the same energy, and calculate the number of ways. (f) What do your findings suggest?

Short Answer

Expert verified
The conclusions would depend on the calculations that have been made in the steps. The goal is to see whether the most probable distributions yield the most disordered state, a state with maximum entropy. The short answer could therefore be 'the most probable distribution of particles across energy levels is the one which maximises the entropy of the system.'

Step by step solution

01

Calculate the probabilities for each energy level

The probability is calculated using equation 9-9 described in the textbook: \( P(n)=\frac{\omega(n)}{\Omega(M,N)} \) where \(\omega(n)\) represents the number of microstates corresponding to an energy level (n) and \(\Omega(M,N)\) is the total number of microstates of the system. The values for \(M=5\) and \(N=11\) can be substituted in. This equation can be used to find the probabilities for \(n=0\), \(n=1\), \(n=2\), and \(n=3\).
02

Calculate the expected number of particles at each level

The expected number of particles at each energy level, or \(N_n\), can be calculated as \(N_n = N*P(n)\), where N is the total number of particles (11) and P(n) is the probability for that energy level. Again, this must be calculated for \(n=0\), \(n=1\), \(n=2\), and \(n=3\). The results can be rounded to the nearest integer.
03

Calculate the number of ways the distribution can be achieved

The number of ways of achieving this distribution can be calculated using permutations. The number of permutable ways equals \(N! / (N_0! N_1! N_2! N_3!)\), where \(N = 11\), and are calculated from Step 2. This can be calculated by substituting in the numerical values.
04

Recalculate with different distribution

Now we recalculate the number of ways for a different distribution where we have 6 particles in n=0, 5 in n=1, with none higher. This can be calculated using a similar process to Step 3 with their respective scenarios.
05

Find and evaluate different distribution

We need to find at least one more distribution in which the 11 oscillators share the same energy, and calculate the number of ways for this distribution using similar process.
06

Conclude findings

Finally, we examine and summarize the results of the calculations, and consider what implications these might have for the distribution of particles across energy levels in the oscillators system.

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!

Key Concepts

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

Entropy
Entropy in statistical mechanics is a measure of the disorder or randomness of a system. It's connected to the number of ways we can arrange a system without altering its overall energy state. In simpler terms, entropy gives us an idea of the degree of chaos in a system. Higher entropy corresponds to more disorder, where systems have many possible arrangements or microstates that result in the same macroscopic condition.

When we say that a state has maximum entropy, we're referring to the fact that it's the state with the greatest number of microstates, making it the most probable due to its many possible arrangements. For our 11-oscillator system, each arrangement where energy is distributed across different energy levels results in a different microstate. However, some distributions have more microstates than others, resulting in higher entropy. By calculating these for different energy distributions, we can determine the state of maximum entropy.
Microstates
Microstates represent the specific arrangements of particles or energy in a system, while producing the same macroscopic state. Each possible arrangement that fits within the constraints of the system's total energy and particles count is a microstate.

In the exercise, we look at a system of 11 oscillators sharing a fixed total energy, where different microstates reflect varying distributions of energy among the oscillators. For each potential energy level—say one oscillator having more energy while another has less—the pattern creates unique microstates.

The total number of microstates informs us about how numerous and diverse energy distributions can be, all maintaining the same total energy. A system's macrostate, such as having a specific total energy level or particle number, arises from summing these microstates, demonstrating their relevance in calculating probabilities and understanding entropy.
Energy Distribution
Energy distribution encompasses how energy is allotted among particles in a system, in this case, oscillators. We need to determine how energy levels are populated in terms of number of particles at different energy levels.

Calculating energy distribution involves using probabilities determined from microstate counts. By examining various energy levels, we consider how likely each configuration occurs in terms of particle distribution. This information helps us understand what typical or probable configurations look like under specified conditions.

In the exercise, energy distribution calculations reveal how 11 oscillators share a specific amount of energy, applying probabilities to anticipate the number of particles at each energy level. This analysis supports insights into why certain distributions are more probable, owing to their higher entropy and more extensive microstate count.
Oscillators
Oscillators in statistical mechanics refer to systems—often simplified as particles or units—that have specific energy levels they may occupy. These can be thought of as vibrating particles, where quantized energy states allow these oscillators to reside in different energy levels.

Our problem involves 11 such oscillators distributing a quantized energy amount, typically depicted in physics by the term \(h\omega_0\). The exercise asks us to consider how these oscillators, given a set quantum of energy, might distribute themselves across allowable energy levels.

These oscillators behave under the constraints of statistical mechanics, where their arrangement across energy levels becomes significantly important. The task involves calculating how these particles distribute, understanding their probabilities, and what configuration results in maximized entropy. By studying this distribution through computation of probabilities, microstates, and entropic states, we unlock insights into the state behavior of collective oscillators.

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

Somehow you have a two-dimensional solid. a sheet of atoms in a square lattice, each atom linked to its four closest neighbors by four springs oriented along the two perpendicular axes. (a) What would you expect the molar heat capacity to be at very low temperature and at very thigh temperature? (b) What quantity would determine, roughly, the line between low and high?

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

When a star has nearly bumed up its intemal fuel, it may become a white dwarf. It is crushed under its own enonnous gravitational forces to the point at which the exclusion principle for the electrons becomes a factor. A smaller size would decrease the gravitational potential energy, but assuming the electrons to be packed into the lowest energy states consistent with the exclusion principle. "squeezing" the potential well necessarily increases the ener gies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is a mini. mum total energy and corresponding equilibrium radius. (a) Treat the electrons in a white dwarf as a quantum gas. The minimum energy allowed by the exclusion principle (see Exercise 67 ) is $$ U_{\text {elecimns }}=\frac{3}{10}\left(\frac{3 \pi^{2} \hbar^{3}}{m_{e}^{3 / 2} V}\right)^{2 / 3} N^{5 / 3} $$ Note that as the volume \(V\) is decreased, the energy does increase. For a neutral star. the number of electrons, \(N\), equals the number of protons. Assuming that protons account for half of the white dwarf's mass \(M\) (neutrons accounting for the other half). show that the minimum electron energy may be written $$ U_{\text {electrons }}=\frac{9 \hbar^{2}}{80 m_{e}}\left(\frac{3 \pi^{2} M^{5}}{m_{\mathrm{p}}^{5}}\right)^{1 / 3} \frac{1}{R^{2}} $$ where \(R\) is the star's radius. (b) The gravitational potential energy of a sphere of mass \(M\) and radius \(R\) is given by $$ U_{\operatorname{mav}}=-\frac{3}{5} \frac{G M^{2}}{R} $$ Taking both factors into account, show that the minimum total energy occurs when $$ R=\frac{3 h^{2}}{8 G}\left(\frac{3 \pi^{2}}{m^{3} m_{p}^{5} M}\right)^{1 / 3} $$ (c) Evaluate this radius for a star whose mass is equal to that of our Sun, \(\sim 2 \times 10^{30} \mathrm{~kg}\). (d) White dwarfs are comparable to the size of Eath. Does the value in part (c) agree?

Given an arbitrary thermodynanic systemn, which is larger. the number of possible macrostates, or the number of possible microstates, or is it impossible to say? Explain your answer. (For most systems, both are infinite, but it is still possible to answer the question)

A two-sided room contains six particles, \(a, b, c, d\). \(e .\) and \(f\), with two on the left and four on the right. (a) Describe the macrostate. (b) Identify the possible microstates. (Note: With only six particles, this isn't a thermodynamic system, but the general idea stitl applies. and the number of combinations is tractable.)

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