Although the integrals and for N and U cannot be carried | StudySoup

Textbook Solutions for An Introduction to Thermal Physics

Chapter 7 Problem 32P

Question

Problem 32P

Although the integrals and for N and U cannot be carried out analytically for all T, it’s not difficult to evaluate them numerically using a computer. This calculation has little relevance for electrons in metals (for which the limit kT ≪ ϵF is always sufficient), but it is needed for liquid 3He and for astrophysical systems like the electrons at the center of the sun.

(a) As a warm-up exercise, evaluate the N integral for the case KT = ϵF and µ = 0, and check that your answer is consistent with the graph shown above. (Hint: As always when solving a problem on a computer, it’s best to first put everything in terms of dimensionless variables. So let t = KT/ϵF, c = µ/ϵF, and x = ϵ/ϵF. Rewrite everything in terms of these variables, and then put it on the computer.)

(b). The next step is to vary µ, holding T fixed, until the integral works out to the desired value, N. Do this for values of KT/ϵF ranging from 0.1 up to 2, and plot the results to reproduce Figure. (It’s probably not a good idea to try to use numerical methods when KT/ϵF is much smaller than 0.1, since you can start getting overflow errors from exponentiating large numbers. But this is the region where we’ve already solved the problem analytically.)

(c) Plug your calculated values of µ into the energy integral, and evaluate that integral numerically to obtain the energy as a function of temperature for KT up to 2ϵF. Plot the results, and evaluate the slope to obtain the heat capacity. Check that the heat capacity has the expected behavior at both low and high temperatures.

Equation 1:

Equation 2:

Solution

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The first step in solving 7 problem number 31 trying to solve the problem we have to refer to the textbook question: Problem 32PAlthough the integrals and for N and U cannot be carried out analytically for all T, it’s not difficult to evaluate them numerically using a computer. This calculation has little relevance for electrons in metals (for which the limit kT ≪ ϵF is always sufficient), but it is needed for liquid 3He and for astrophysical systems like the electrons at the center of the sun.(a) As a warm-up exercise, evaluate the N integral for the case KT = ϵF and µ = 0, and check that your answer is consistent with the graph shown above. (Hint: As always when solving a problem on a computer, it’s best to first put everything in terms of dimensionless variables. So let t = KT/ϵF, c = µ/ϵF, and x = ϵ/ϵF. Rewrite everything in terms of these variables, and then put it on the computer.)(b). The next step is to vary µ, holding T fixed, until the integral works out to the desired value, N. Do this for values of KT/ϵF ranging from 0.1 up to 2, and plot the results to reproduce Figure. (It’s probably not a good idea to try to use numerical methods when KT/ϵF is much smaller than 0.1, since you can start getting overflow errors from exponentiating large numbers. But this is the region where we’ve already solved the problem analytically.)(c) Plug your calculated values of µ into the energy integral, and evaluate that integral numerically to obtain the energy as a function of temperature for KT up to 2ϵF. Plot the results, and evaluate the slope to obtain the heat capacity. Check that the heat capacity has the expected behavior at both low and high temperatures.Equation 1:Equation 2:
From the textbook chapter Quantum Statistics you will find a few key concepts needed to solve this.

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Title An Introduction to Thermal Physics  1 
Author Daniel V. Schroeder
ISBN 9780201380279

Although the integrals and for N and U cannot be carried

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