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A general expression for the classical Hamiltonian is \(H=\alpha p_{i}^{2}+H^{\prime}\) where \(p_{i}\) is the momentum along one dimension for particle \(i, \alpha\) is a constant, and \(H^{\prime}\) are the remaining terms in the Hamiltonian. Substituting this into the equipartition theorem yields \(q=\frac{1}{h^{3 N}} \iint e^{-\beta\left(\alpha p_{i}^{2}+H^{\prime}\right)} d p^{3 N} d x^{3 N}\) a. Starting with this expression, isolate the term involving \(p_{i}\) and determine its contribution to \(q\) b. Given that the average energy \(\langle\varepsilon\rangle\) is related to the partition function as follows, \(\langle\varepsilon\rangle=\frac{-1}{q}\left(\frac{\delta q}{\delta \beta}\right)\) evaluate the expression the contribution from \(p_{i} .\) Is your result consistent with the equipartiton theorem?

Short Answer

Expert verified
In summary, we isolated the term involving \(p_i\) in the partition function and found its contribution to be \(\sqrt{\frac{\pi}{\beta\alpha}}\). Then, we calculated the contribution of \(p_i\) to the average energy as \(\frac{3N}{2\beta\alpha}\), which is consistent with the equipartition theorem.

Step by step solution

01

Rewrite the partition function as a product of individual contributions

The partition function q can be expressed as the product of individual contributions from the terms involved in the Hamiltonian. Rearrange the given expression for the partition function and isolate the term involving pi: \(q = \frac{1}{h^{3N}} \iint e^{-\beta(\alpha p_{i}^2 + H')}\,dp^{3N}\,dx^{3N} = \frac{1}{h^{3N}}\iint e^{-\beta\alpha p_{i}^2}\,dp^{3N} \:e^{-\beta H'}\,dx^{3N}\) The term isolated involving pi is \(e^{-\beta\alpha p_{i}^2}\).
02

Integrate over the momentum term

Now, we need to integrate the isolated term over the momentum: \(\int e^{-\beta\alpha p_{i}^2}\,dp_{i} = \int_{-\infty}^{\infty} e^{-\beta\alpha p_{i}^2}\,dp_{i}\) This integral is a standard Gaussian integral with a result of: \(\sqrt{\frac{\pi}{\beta\alpha}}\) This is the contribution of the momentum term to the partition function q.
03

Calculate the contribution of pi to the average energy

Now, we need to apply the given formula for the average energy: \(\langle\varepsilon\rangle = \frac{-1}{q} \left(\frac{\partial q}{\partial \beta}\right)\) To find the derivative of q, first note that the contribution of the pi term to q is, \(\left(\sum_{i=1}^N \sqrt{\frac{\pi}{\beta\alpha}}\right)^{3} = \left(\sqrt{\frac{\pi}{\beta\alpha}}\right)^{3N}\) Now, differentiate q with respect to beta: \(\frac{\partial q}{\partial \beta} = \frac{-3N}{2}\left(\sqrt{\frac{\pi}{\beta\alpha}}\right)^{3N-1}\frac{1}{\beta^2\alpha}\) Plug this result into the formula for the average energy: \(\langle\varepsilon\rangle = \frac{-1}{\left(\sqrt{\frac{\pi}{\beta\alpha}}\right)^{3N}} \times \frac{-3N}{2}\left(\sqrt{\frac{\pi}{\beta\alpha}}\right)^{3N-1}\frac{1}{\beta^2\alpha}\) Simplify the expression: \(\langle\varepsilon\rangle = \frac{3N}{2\beta\alpha}\) The result shows that the contribution of pi to the average energy is proportional to the number of particles N, which is consistent with the equipartition theorem. The theorem states that each quadratic degree of freedom contributes \(\frac{1}{2}kT\) to the average energy, and the obtained result verifies this.

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Key Concepts

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

Hamiltonian Mechanics
Hamiltonian mechanics is a reformulation of classical mechanics that arose from the work of William Rowan Hamilton and Carl Gustav Jacobi. It represents the physical state of a system in terms of coordinates and conjugate momenta, rather than using forces like in Newtonian mechanics.

In Hamiltonian mechanics, the Hamiltonian function, denoted as \( H \), is a central concept. It represents the total energy of the system, expressed as a sum of kinetic and potential energies. The general form of the Hamiltonian can include terms that describe various interactions within the system, such as electromagnetic forces or the interaction between particles.

In statistical mechanics, the Hamiltonian plays a crucial role in defining the partition function, which is a fundamental quantity that encapsulates all possible states of a system. The partition function is integral to predicting the behavior of systems at the thermodynamic limit and is used to calculate ensemble averages of physical quantities.
Equipartition Theorem
The Equipartition Theorem is a result in statistical mechanics that emerges from classical thermodynamics. According to this theorem, energy is distributed equally among its various 'degrees of freedom.' A degree of freedom is an independent mode of motion, like the translation or vibration of molecules.

For a classical system in thermal equilibrium at a temperature \( T \), the theorem states that each quadratic degree of freedom contributes an average energy of \( \frac{1}{2}kT \), where \( k \) is the Boltzmann constant. When we apply this idea to the Hamiltonian for a system of particles, each term squared in the Hamiltonian, belonging to momentum or position coordinates, provides an average energy of \( \frac{1}{2}kT \) per particle.

Therefore, in the case of a quadratic momentum term in the Hamiltonian, this theorem predicts that its contribution to the total energy of the system should be \( \frac{3N}{2}kT \), considering three spatial dimensions for each of the N particles.
Average Energy in Statistical Mechanics
In the realm of statistical mechanics, the average energy, represented as \( \langle\varepsilon\rangle \), is a statistical average of the energy of a system. It is derived from the partition function \( q \), which is a sum over all possible microstates of the system, each weighted by its Boltzmann factor \( e^{-\beta E_i} \), where \( E_i \) is the energy of the i-th microstate and \( \beta \) is equal to \( \frac{1}{kT} \).

The average energy is especially significant because it links the microscopic properties of individual particles with the macroscopic observables in thermodynamics. By differentiating the partition function with respect to \( \beta \), we obtain an expression for the average energy, and this demonstrates how the energy distribution amongst the particles is influenced by the system's temperature and inherent energies. This theoretical framework is foundational in connecting statistical mechanics with thermodynamic quantities such as heat capacity and entropy.
Gaussian Integral
The Gaussian integral is a key component in the field of statistics and other areas of physics, particularly quantum mechanics and statistical mechanics. It represents the integral of the exponential function of the negative square of a variable and is central to the mathematics that describe systems governed by the normal distribution.

Mathematically, the Gaussian integral is expressed as \( \int_{-\infty}^{\infty} e^{-ax^2} dx \), where \( a \) is a constant. The result of this integral is \( \sqrt{\frac{\pi}{a}} \), which arises often in calculations involving probability and heat distribution.

In the context of statistical mechanics, the Gaussian integral appears when integrating over momentum variables in the partition function. Because the momentum terms in the Hamiltonian often appear squared, their contribution to partition function via the integral yields a predictable and calculable value, which simplifies the calculation of other thermodynamic quantities.

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Most popular questions from this chapter

In general, the high-temperature limit for the rotational partition function is appropriate for almost all molecules at temperatures above the boiling point. Hydrogen is an exception to this generality because the moment of inertia is small due to the small mass of H. Given this, other molecules with \(\mathrm{H}\) may also represent exceptions to this general rule. For example, methane \(\left(\mathrm{CH}_{4}\right)\) has relatively modest moments of inertia \(\left(I_{A}=I_{B}=I_{C}=5.31 \times 10^{-40} \mathrm{g} \mathrm{cm}^{2}\right)\). a. determine \(B_{A}, B_{B},\) and \(B_{C}\) for this molecule. b. Use the answer from part (a) to determine the rotational partition function. Is the high-temperature limit valid? and has a relatively low boiling point of \(T=112 \mathrm{K}\)

Consider para-H \(_{2}\left(B=60.853 \mathrm{cm}^{-1}\right)\) for which only even- \(J\) levels are available. Evaluate the rotational partition function for this species at \(50 \mathrm{K}\). Perform this same calculation for \(\mathrm{HD}\left(B=45.655 \mathrm{cm}^{-1}\right)\).

He has a normal boiling point of 4.2 K. For a mole of gaseous He at \(4.2 \mathrm{K}\) and \(1 \mathrm{atm},\) is the high-temperature limit for translational degrees of freedom applicable?

A. In this chapter, the assumption was made that the harmonic oscillator model is valid such that anharmonicity can be neglected. However, anharmonicity can be included in the expression for vibrational energies. The energy levels for an anharmonic oscillator are given by \(E_{n}=h c \tilde{\nu}\left(n+\frac{1}{2}\right)-h c \widetilde{\chi} \widetilde{\nu}\left(n+\frac{1}{2}\right)^{2}+\ldots\) Neglecting zero point energy, the energy levels become \(E_{n}=h c \widetilde{\nu} n-h c \widetilde{\chi}^{\sim} n^{2}+\ldots .\) Using the preceding expression, demonstrate that the vibrational partition function for the anharmonic oscillator is \(q_{V, \text { anharmonic }}=q_{V, \text { harm }}\left[1+\beta h c \widetilde{\chi} \nu q_{V, \text { harm }}^{2}\left(e^{-2 \beta \widetilde{\nu} h c}+e^{-\beta \widetilde{\nu} h c}\right)\right]\) In deriving the preceding result, the following series relationship will prove useful: \(\sum_{n=0}^{\infty} n^{2} x^{n}=\frac{x^{2}+x}{(1-x)^{3}}\) for \(|x|<1\) b. For \(\mathrm{H}_{2}, \widetilde{\nu}=4401.2 \mathrm{cm}^{-1}\) and \(\widetilde{\chi} \widetilde{\nu}=121.3 \mathrm{cm}^{-1} .\) Use the result from part (a) to determine the percent error in \(q_{V}\) if anharmonicity is ignored.

Evaluate the vibrational partition function for \(\mathrm{NH}_{3}\) at \(1000 .\) K where the vibrational frequencies are \(950 ., 1627.5\) (doubly degenerate), 3335 , and \(3414 \mathrm{cm}^{-1}\) (doubly degenerate). Are there any modes that you can disregard in this calculation? Why or why not?

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