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Chapter 6: Problem 8

voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.

Short Answer

Expert verified
In matrix form: \ \( \mathbf{A} = \begin{pmatrix} 3 & -1 & -1 & 0 & \cdots & 0 \ \vdots \end{pmatrix} \) and \( \mathbf{w} = \begin{pmatrix} V_{+} \ \vdots \end{pmatrix} \). Use numpy to solve for N=6, and a banded algorithm for N=10000.

Step by step solution

01

Identify Matrix and Vector Form

Rewrite the given equations in a matrix form \ \( \mathbf{Av} = \mathbf{w} \). The coefficients of the unknowns \(V_{i}\) will form the matrix \( \mathbf{A} \) and the constants on the right side will form the vector \( \mathbf{w} \).
02

Define Matrix \(\mathbf{A}\)

Matrix \( \mathbf{A} \) is defined by the coefficients of \(V_{i}\): \ \[ \mathbf{A} = \begin{pmatrix} 3 & -1 & -1 & 0 & \cdots & 0 \ -1 & 4 & -1 & -1 & \cdots & 0 \ 0 & -1 & 4 & -1 & -1 & 0 \ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \ 0 & \cdots & -1 & -1 & 4 & -1 \ 0 & \cdots & 0 & -1 & -1 & 3 \end{pmatrix} \]
03

Define Vector \(\mathbf{w}\)

The vector \( w \) is defined with values \( V_{+} \) on the first two entries and zeros elsewhere: \ \( \mathbf{w} = \begin{pmatrix} V_{+} \ V_{+} \ 0 \ \vdots \ 0 \end{pmatrix} \)
04

Solve for \( V_{i} \) using Python for N = 6

Use a Python program to solve the values of \( V_{i} \) when the number of junctions \( N = 6 \). Use numpy library to solve the system of linear equations. ```python import numpy as np A = np.array([ [3, -1, -1, 0, 0, 0], [-1, 4, -1, -1, 0, 0], [0, -1, 4, -1, -1, 0], [0, 0, -1, 4, -1, -1], [0, 0, 0, -1, 4, -1], [0, 0, 0, 0, -1, 3] ]) w = np.array([V_{+}, V_{+}, 0, 0, 0, 0]) V = np.linalg.solve(A, w) print(V) ```
05

Solve for N = 10000 Using Banded Matrix Algorithm

Implement the solution using the banded matrix algorithm from banded.py for \( N = 10000 \). This leverages the sparseness of the matrix. ```python from banded import banded_solve import numpy as np N = 10000 A_banded = np.zeros((5, N)) w = np.zeros(N) w[0] = V_{+} w[1] = V_{+} A_banded[2, :] = 4 A_banded[1, 1:] = -1 A_banded[0, 2:] = -1 A_banded[3,:-1] = -1 A_banded[4,:-2] = -1 V = banded_solve(A_banded, w, 2, 2) print(V) ```

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Key Concepts

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

Tridiagonal Matrix
A tridiagonal matrix is a specific type of square matrix that has non-zero elements only on the main diagonal, and the diagonals directly above and below it. This structure simplifies many calculations and is common in computational physics problems.
In the given exercise, the matrix \(\textbf{A}\) has a tridiagonal structure, which is evident from the main diagonal entries (3, 4, ..., 3) and the subdiagonals (-1, -1). The general layout can be written as:
\[ \textbf{A} = \begin{pmatrix} 3 & -1 & -1 & 0 & \cdots & 0 \ -1 & 4 & -1 & -1 & \cdots & 0 \ 0 & -1 & 4 & -1 & -1 & 0 \ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \ 0 & \cdots & -1 & -1 & 4 & -1 \ 0 & \cdots & 0 & -1 & -1 & 3 \end{pmatrix} \]
This allows for efficient numerical solutions by leveraging algorithms optimized for tridiagonal systems.
Banded Matrix
Banded matrices are matrices where the non-zero elements are confined to a diagonal band, composed of the main diagonal and a few diagonals on either side. For large-scale problems, banded matrices significantly reduce computational complexity.
In this exercise, the matrix \(\textbf{A}\) with 10,000 internal junctions is a banded matrix with a bandwidth of 2 on either side of the main diagonal. Using a regular solver for such a large matrix would be inefficient and impractical.
The banded matrix representation makes it feasible to handle large systems by focusing on non-zero elements, thus reducing computation time and memory usage:
\[ \textbf{A}_{\text{banded}} = \begin{pmatrix} 0 & -1 & -1 & 0 & \-1 &... \end{pmatrix} \].
This approach highlights the efficiency of numerical methods in computational physics and their critical role in solving real-world problems.
Linear Algebra using Python
Python provides powerful libraries like NumPy for numerical computations and SciPy for scientific computing, making it a great tool for solving linear algebra problems.
In this context, we use NumPy to solve a system of linear equations for a small-scale tridiagonal matrix. For instance, the provided NumPy code:
\inline_code{ \begin{python} import numpy as np A = np.array([ [3, -1, -1, 0, 0, 0], [-1, 4, -1, -1, 0, 0], [0, -1, 4, -1, -1, 0], [0, 0, -1, 4, -1, -1], [0, 0, 0, -1, 4, -1], [0, 0, 0, 0, -1, 3] ]) w = np.array([5, 5, 0, 0, 0, 0]) V = np.linalg.solve(A, w) print(V) \end{python}}
efficiently solves the system for \(N = 6\).
For larger systems like \(N = 10000\), the banded matrix algorithm from `banded.py` is essential:'
\inline_code{ \begin{python} from banded import banded_solve import numpy as np N = 10000 w = np.zeros(N) A_banded = np.zeros((5, N)) w[0] = 5 w[1] = 5 A_banded[2, :] = 4 A_banded[1, 1:] = -1 A_banded[0, 2:] = -1 A_banded[3,:-1] = -1 A_banded[4,:-2] = -1 V = banded_solve(A_banded, w, 2, 2) print(V) \end{python}}
This code highlights using specific libraries to handle large-scale numerical problems effectively.

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Most popular questions from this chapter

Planck's radiation law tells us that the intensity of radiation per unit area and per unit wavelength \(\lambda\) from a black body at temperature \(T\) is $$ I(\lambda)=\frac{2 \pi h c^{2} \lambda^{-5}}{\mathrm{e}^{h c / \lambda k_{g} T}-1} $$ where \(h\) is Planck's constant, \(c\) is the speed of light, and \(k_{B}\) is Boltzmann's constant. a) Show by differentiating that the wavelength \(\lambda\) at which the emitted radiation is strongest is the solution of the equation $$ 5 \mathrm{e}^{-h c / \lambda k_{B} T}+\frac{h c}{\lambda k_{B} T}-5=0 . $$ Make the substitution \(x=h c / \lambda k_{B} T\) and hence show that the wavelength of maximum radiation obeys the Wien displacement law: $$ \lambda=\frac{b}{T} $$ where the so-called Wien displacement constant is \(b=h c / k_{B} x\), and \(x\) is the solution to the nonlinear equation $$ 5 e^{-x}+x-5=0 $$ b) Write a program to solve this equation to an accuracy of \(\epsilon=10^{-6}\) using the binary search method, and hence find a value for the displacement constant. c) The displacement law is the basis for the method of optical pyrometry, a method for measuring the temperatures of objects by observing the color of the thermal radiation they emit. The method is commonly used to estimate the surface temperatures of astronomical bodies, such as the Sun. The wavelength peak in the Sun's emitted radiation falls at \(\lambda=502 \mathrm{~nm}\). From the equations above and your value of the displacement constant, estimate the surface temperature of the Sun.

Consider a square potential well of width \(w\), with walls of height \(V\) : Using Schrödinger's equation, it can be shown that the allowed energies \(E\) of a single quantum particle of mass \(m\) trapped in the well are solutions of $$ \tan \sqrt{w^{2} m E / 2 \hbar^{2}}= \begin{cases}\sqrt{(V-E) / E} & \text { for the even numbered states, } \\ -\sqrt{E /(V-E)} & \text { for the odd numbered states, }\end{cases} $$ where the states are numbered starting from 0 , with the ground state being state 0 , the first excited state being state 1 , and so forth. a) For an electron (mass \(9.1094 \times 10^{-31} \mathrm{~kg}\) ) in a well with \(V=20 \mathrm{eV}\) and \(w=1 \mathrm{~nm}\), write a Python program to plot the three quantities $$ y_{1}=\tan \sqrt{w^{2} m E / 2 \hbar^{2}}, \quad y_{2}=\sqrt{\frac{V-E}{E}}, \quad y_{3}=-\sqrt{\frac{E}{V-E}}, $$ on the same graph, as a function of \(E\) from \(E=0\) to \(E=20 \mathrm{eV}\). From your plot make approximate estimates of the energies of the first six energy levels of the particle. b) Write a second program to calculate the values of the first six energy levels in electron volts to an accuracy of \(0.001 \mathrm{eV}\) using binary search.

A chain of resistors Consider a long chain of resistors wired up like this: All the resistors have the same resistance \(R\). The power rail at the top is at voltage \(V_{+}=\) \(5 V\). The problem is to find the voltages \(V_{1} \ldots V_{N}\) at the internal points in the circuit. a) Using Ohm's law and the Kirchhoff current law, which says that the total net current flow out of (or into) any junction in a circuit must be zero, show that the voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.

Consider the degree-six polynomial $$ P(x)=924 x^{6}-2772 x^{5}+3150 x^{4}-1680 x^{3}+420 x^{2}-42 x+1 $$ There is no general formula for the roots of a polynomial of degree six, but one can find them easily enough using a computer. a) Make a plot of \(P(x)\) from \(x=0\) to \(x=1\) and by inspecting it find rough values for the six roots of the polynomial-the points at which the function is zero. b) Write a Python program to solve for the positions of all six roots to at least ten decimal places of accuracy, using Newton's method. Note that the polynomial in this example is just the sixth Legendre polynomial (mapped onto the interval from zero to one), so the calculation performed here is the same as finding the integration points for 6-point Gaussian quadrature (see Section 5.6.2), and indeed Newton's method is the method of choice for calculating Gaussian quadrature points.

There is a magical point between the Earth and the Moon, called the \(L_{1}\) Lagrange point, at which a satellite will orbit the Earth in perfect synchrony with the Moon, staying always in between the two. This works because the inward pull of the Earth and the outward pull of the Moon combine to create exactly the needed centripetal force that keeps the satellite in its orbit. Here's the setup: a) Assuming circular orbits, and assuming that the Earth is much more massive than either the Moon or the satellite, show that the distance \(r\) from the center of the Earth to the \(L_{1}\) point satisfies $$ \frac{G M}{r^{2}}-\frac{G m}{(R-r)^{2}}=\omega^{2} r $$ where \(R\) is the distance from the Earth to the Moon, \(M\) and \(m\) are the Earth and Moon masses, \(G\) is Newton's gravitational constant, and \(\omega\) is the angular velocity of both the Moon and the satellite. b) The equation above is a degree-five polynomial equation in \(r\) (also called a quintic equation). Such equations cannot be solved exactly in closed form, but it's straightforward to solve them numerically. Write a program that uses either Newton's method or the secant method to solve for the distance \(r\) from the Earth to the \(L_{1}\) point. Compute a solution accurate to at least four significant figures. The values of the various parameters are: $$ \begin{aligned} G &=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} \\\ M &=5.974 \times 10^{24} \mathrm{~kg} \\ m &=7.348 \times 10^{22} \mathrm{~kg} \\ R &=3.844 \times 10^{8} \mathrm{~m} \\ \omega &=2.662 \times 10^{-6} \mathrm{~s}^{-1} \end{aligned} $$ You will also need to choose a suitable starting value for \(r\), or two starting values if you use the secant method.
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