Vectors And Tensors Solutions D. Fleisch 6cgz

A Student's Guide to Vectors and Tensors

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Dan Fleisch is a Professor in the Department of Physics at Wittenberg University, where he specializes in electromagnetics and space physics. He is the author of the internationally best-selling book A Student’s Guide to Maxwell’s Equations, published by Cambridge University Press in January 2008. This book, currently in its 11th printing, has been translated into Japanese, Korean, and Chinese. Dr. Fleisch’s latest book, A Student’s Guide to Vectors and

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Tensors, was published by Cambridge Press in September of 2011. Fleisch is also the co-author

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with the late Prof. John Kraus of The Ohio State University of the McGraw-Hill textbook Electromagnetics with Applications.

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Prof. Fleisch has published technical articles in the IEEE Transactions, The Journal of Atmospheric and Terrestrial Physics, and Microwave Journal, and has presented more than a dozen

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professional papers on topics related to high-speed microwave instrumentation and radar crosssection measurement. He has been a regular contributor of science commentary to PBS station WYSO of Yellow Springs, and in 2006 he appeared in the documentary "The Dayton Codebreakers" shown on Public Television. Prof. Fleisch was named Outstanding Faculty Member at the Wittenberg Greek scholarship awards in 2000, and in 2002 he won the Omicron Delta Kappa award for Excellence in Teaching. In 2003 and 2005 he was recognized for Faculty Excellence and Innovation by the Southwestern Ohio Council for Higher Education (SOCHE), and in 2004 he received Wittenberg’s Distinguished Teaching Award, the university’s highest faculty award. In November, 2010 Prof. Fleisch was named the Ohio Professor of the Year by the Carnegie Foundation and the Council for the Advancement and of Education. Fleisch received his B.S. in Physics from Georgetown University in 1974 and his M.S. and Ph.D. in Space Physics and Astronomy from Rice University in 1976 and 1980, respectively. To Dan Fleisch, send e-mail to A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press

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Chapter 1

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Home > Chapter 1 Problems


Problem 1.3 Imagine that the y-axis points north and the x-axis points east. (a) If you travel a distance r = 22 km in a straight line from the origin in a direction 35 degrees south of west, what is your position in Cartesian (x,y) coordinates? (b) If you travel 6 miles due south from the origin and then turn west

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and travel 2 miles, how far from the origin and in what direction is your

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final position?

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Problem 1.3 Imagine that the y-axis points north and the x-axis points east. (a) If you travel a distance r = 22 km in a straight line from the origin in a direction 35 degrees south of west, what is your position in Cartesian (x,y) coordinates? (b) If you travel 6 miles due south from the origin and then turn west

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and travel 2 miles, how far from the origin and in what direction is your

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final position?


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Full solution for part (b): From a sketch such as Figure P1.3b, you can see that point P has coordinates x = -2 miles and y = -6 miles. Thus

and

where 180° has been added to the result of the arctan operation since the denominator was negative. Continue to Problem 1.4

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Problem 1.5 Cylindrical coordinates

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Full solution for part (b):

Continue to Problem 1.6


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Chapter 1

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Full solution: To find the unit basis vectors, use

Thus


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Full solution: The definition of the scalar product in Cartesian coordinates is

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which in this case gives

or

and

so

and


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or

Thus in this case


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so

or

Likewise, for the vector product

so

or


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Problem 2.4


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which is

So


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Chapter 2 Chapter 3

Problem 2.5


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Full Solution: Using the BAC-CAB rule,

Inserting values for the vectors that appear in the dot products gives

So


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Chapter 2 Chapter 3

Problem 2.6

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Alternative forms of

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Full Solution: Using the rules of partial differentiation, the partial derivative with respect to x is

and the partial derivative with respect to y is


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Chapter 2 Chapter 3

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Problem 2.9 What is the curl of the vector field given in the previous problem?

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Chapter 1




Problem 2.10 Find the Laplacian of the function given in Problem 2.6.




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Taking the first partial derivatives makes this

Taking the second partial derivatives makes this


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At the optimum angle of 90°, for a force of 25 Newtons and a distance


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Chapter 1



Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Matrix Algebra

Problem 3.1 Solve the box-on-a-ramp problem (that is, find the acceleration of the box) for the frictionless case using a Cartesian coordinate system for which the y-axis points vertically upward and the x-axis points horizontally to the right.

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Full Solution: Begin by making a free-body diagram such as Fig. P3.3 showing all the forces acting on the box along with the coordinate axes. The equations for the sum of the forces in the x- and y-directions are

and the y-components are

This means that Newton's 2nd law for this case is

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To find the velocity after the deliverywoman has pushed the box for 2 meters, use the constant-acceleration kinematic equation

So, for example, if the mass of the box is 1 kg, the speed of the box after the deliverywoman has pushed with a force of 10 Newtons over a distance of 2 meters will be


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Chapter 1




Problem 3.4 If the hammer thrower shown on the cover of this book wishes to launch a hammer of mass 7.26 kg on a cable of length 1.22 meters with a speed of 22 m/s, what is the magnitude of the centripetal force he must supply?


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and the magnitude of the centripetal acceleration is related to the velocity and radius of curvature by

thus

Plugging in values gives


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Chapter 1




Problem 3.5 Imagine a Formula 1 car going around a curve with radius of 10 meters while slowing down from a speed of 180 mph to a speed of 120 mph in 2 seconds. What are the magnitude and direction of the car's acceleration at the instant the car's speed is 150 mph?



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Chapter 2 Chapter 3

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Thus

Taking the partial derivatives gives

Thus


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Chapter 1



Chapter 2 Chapter 3

Problem 3.8


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and in cylindrical coordinates, the gradient is given by

Taking the partial derivatives makes this

and


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which in this case

or

Thus

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Chapter 1




Problem 4.1 Write the inverse transformation matrix for a 70° rotation of the 2D Cartesian coordinate axes and the direct transformation matrix for the rotation of a vector through an angle of 70° degrees. Show that the product of these two transformation matrices is the identity matrix.



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which is the identity matrix. Continue to Problem 4.2 A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press

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or

which means


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Chapter 2 Chapter 3

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and

and

that these are the components of the rotated unit vectors in the original (non-rotated) coordinate system. Continue to Problem 4.4 A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press

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and

This is consistent with the results of Problems 4.2 and 4.4. Continue to Problem 4.6 A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press

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and

which in this case is

and

Solving the first of these two simultaneous equations for A1 gives

and substituting this into the second equation gives

Thus

and

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Another approach to finding the contravariant components is to use the matrix method and Cramer's Rule. To do this, write

in matrix form as

or

Now use Cramer's Rule to find A1 and A2:

and

in agreement with the results of the substitution method. Continue to Problem 4.7 A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press

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which may be written as

or

You can now use Cramer's Rule:

and

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Chapter 1




Problem 4.9 Use the substitution method and the elimination method to solve the two simultaneous equations that result from vector Equation 4.27.

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Chapter 1



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Problem 4.10 Show that the elements of the Cartesian-to-polar transformation matrix are the components of the basis vectors tangent to the original (Cartesian) coordinate axes.

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Problem 5.1 Show that the process of subtracting one tensor from another results in a quantity that is also a tensor.

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Chapter 1




Problem 5.2 Find the elements of the metric tensor for spherical coordinates by forming the dot products of the relevant basis vectors.


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You can determine the elements of the metric tensor by finding the scalar products of these basis vectors:

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Chapter 1




Problem 5.3 Show how the derivatives given after Equation 5.16 lead to the elements of the metric tensor for spherical polar coordinates (Equation 5.17).



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Full Solution: The steps in getting from Equation 5.16 to Equation 5.17 can be found in the solution for Problem 5.2. But in case you skipped that problem, here it is again. The relevant derivatives are

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Problem 5.4 Use the scale factors for spherical polar coordinates to the expressions given in Chapter 2 for the gradient, divergence, curl, and Laplacian in spherical coordinates.



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in agreement with Equation 2.34 in the text. For the divergence, inserting the spherical polar scale factors gives

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in agreement with Equation 2.39 in the text. The curl is given by

Inserting the scale factors gives

which simplified to

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http://www.danfleisch.com/sgvt/chapter5/problem4/ or

in agreement with Equation 2.42 in the text. The Laplacian can be found as

which is

in agreement with Equation 2.54 in the text.



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Full Solution: For diagonal matrices, the inverse matrix is also diagonal, and each (diagonal) element of the inverse matrix is the inverse of the corresponding element of the original matrix. Since

the inverse matrix is



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Problem 5.9 What are the Christoffel symbols for the 2-D spherical surface of Problem 5.8?



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Full Solution: To find Christoffel symbols, use

and from the solution for Problem 5.8, you know that



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Chapter 1




Problem 5.10 Show that the covariant derivative of the metric tensor equals zero.




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Full Solution: The covariant derivative of a second-rank tensor is

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Applying this to the metric tensor gives

Based on Equation 5.23 in the text, the two Christoffel symbols in this expression are related to the metric tensor as

in which the dummy index is written as m since the letter k is being used as the index with respect to which the covariant derivative is being taken. Thus


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Chapter 1




Problem 6.1 Find the inertia tensor for a cubical arrangement of eight identical masses with the origin of coordinates at one of the corners and the co-ordinate axes along the edges of the cube.



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Chapter 1




Problem 6.2 How would the moment of inertia tensor of Problem 6.1 change if one of the eight masses is removed?



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Problem 6.3 Find the moment of inertia tensor for the arrangement of masses of Problem 6.2 if the coordinate system is rotated by 20 degrees about one of the coordinate axes (do this by finding the locations of the masses in the rotated coordinate system).


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Full Solution: A sketch such as Figure P6.3(a) will help you determine the impact of rotating the coordinate system. In this case, the Cartesian axes are rotated 20° about the z-axis, so the x and y coordinates of the masses will change (excepting the mass at the origin). One approach to determining the values of x' and y' for each mass is to make a sketch such as Figure P6.3(b), which is the view you would have if you looked directly down the z-axis (toward the origin). To make it a bit easier to determine the x' and y' values for each mass, "picture has been tilted" in Figure 6.3(c) so that the x' axis is horizontal and the y' axis is vertical. Applying geometry to a figure such as this allows you can determine the coordinates of each of the masses in the primed coordinate system. For example, the x' coordinate of masses 2 and 6 may be seen from the figure to equal s sin 20° = 0.342s, and the y' coordinate of masses 2 and 6 is to equal s cos 20° = 0.940s.

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An alternative approach is to use a rotation matrix to convert the

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Problem 6.4 Use the similarity-transform approach to the moment of inertia tensor you found in Problem 6.3.



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Problem 6.5 Show how the vector wave equation results from taking the curl of both sides of Faraday's Law and inserting the curl of the magnetic field from the Ampere-Maxwell Law.



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Problem 6.6 If an observer in one coordinate system measures an electric field of 5 volts per meter in the z-direction and zero magnetic field, what electric and magnetic fields would be measured by a second observer moving at 1/4 the speed of light along the x-axis?



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Chapter 1




Problem 6.7 If an observer in one coordinate system measures a magnetic field of 1.5 Tesla in the z-direction and zero electric field, what electric and magnetic fields would be measured by a second observer moving at 1/4 the speed of light along the x-axis?



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Problem 6.10 Find the Ricci tensor and scalar for the 2-sphere of Section 6.3.



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1 EM Tensor Comparison

This document compares the electromagnetic field tensor presented in A Student’s Guide to Vectors and Tensors (SGVT) to the equivalent tensor in several other texts. It is not meant to provide additional information about this tensor, but rather to help you understand why the tensor may appear different when you encounter it elsewhere. This initial document will compare the EM field tensor as presented in SGVT to the versions presented in Griffiths’ Introduction to Electrodynamics, Jackson’s Classical Electrodynamics, and Shadowitz’s The Electromagnetic Field ; I’ll add the versions of other texts based on readers’ suggestions. The electromagnetic field tensor presented in Section 6.2 of SGVT has the following form: ⎛ ⎞ 0 −Ex /c −Ey /c −Ez /c ⎜ Ex /c 0 −Bz By ⎟ ⎟. F αβ = ⎜ ⎝ Ey /c Bz 0 −Bx ⎠ Ez /c −By Bx 0 Notice that this is the contravariant form of the EM field tensor (in both indices), so the first thing to check when making comparisons between texts is that you’re using the same form with respect to contravariant and covariant indices. In Griffiths’ Introduction to Electrodynamics, the doubly contravariant EM field tensor is presented as ⎛ ⎞ 0 Ex /c Ey /c Ez /c ⎜ −Ex /c 0 Bz −By ⎟ ⎟ F µν = ⎜ ⎝ −Ey /c −Bz 0 Bx ⎠ −Ez /c By −Bx 0

2

EM Tensor Comparison

which differs by a minus sign from the version presented in SGVT. This difference is attributable to the fact that Griffiths uses the sign convention of placing a minus sign on the time component and a plus sign on the spatial components of four-vectors, and the opposite convention is used in SGVT. In Jackson’s Classical Electrodynamics, the field tensor is given as ⎛ 0 −Ex −Ey ⎜ Ex 0 −Bz αβ F =⎜ ⎝ Ey Bz 0 Ez −By Bx

doubly contravariant EM ⎞ −Ez By ⎟ ⎟ −Bx ⎠ 0

In this case, the sign convention is the same as that used in SGVT, but the electric-field do not contain the factor of 1/c included in the SGVT version. The reason for this difference is that Jackson uses Gaussian rather than SI units for all chapters after Chapter 10 of the latest (third) edition of his book (that’s what the little “G” means after the chapter headings on the top of the even pages). You can understand the impact of those units by looking at Table 3 in the Appendix of the Jackson text. There he indicates that you should make the following substitutions to convert his equations between systems: in SI E

→

in SI B

→

J in SI

→

E √ in Gaussian system 4π0 mu0 B in Gaussian system 4π √ 4π0 J in Gaussian system

If you make the appropriate conversions in the EM field tensor (as well as in the expression relating the derivatives of the field-tensor elements to the four-current), you can get to Maxwell’s Equations in the same manner described in Section 6.2 of SGVT.

EM Tensor Comparison

3

In Shadowitz’s The Electromagnetic Field, the doubly contravariant EM field tensor is written as ⎛ ⎞ 0 −cBz −cBy −Ex ⎜ −cBz 0 cBx −Ey ⎟ ⎟ F JK = ⎜ ⎝ cBy −cBx 0 −Ez ⎠ Ex

Ey

Ez

0

This appears quite different from the other versions of this tensor discussed above, but notice that a large part of this difference comes about because the electric-field appear in the fourth row and column rather than in the first row and column as in the previous versions. The reason for this difference is that Shadowitz writes four-vectors with the time component in the last position (after the spatial components) rather than in the first position. Additionally, the EM field tensor presented by Shadowitz is essentially c times the EM field tensor in SGVT, so Ex /c in SGVT becomes Ex in Shadowitz, and Bx in SGVT becomes cBx in Shadowitz. To compensate for this difference when deriving Maxwell’s Equations from the EM field tensor, Shadowitz includes a factor of µ0 /0 in front of his four-current (J) term: µ0 ∂Fij = Ji ∂xj 0 so if you pull the multiplicative factor of c out of Shadowitz’s EM tensor and divide it into the four-current term, this becomes µ0 1 µ0 √ Ji = µ0 0 Ji c 0 0 µ20 0 = Ji = µ0 Ji 0 √ where the fact that c = 1/ µ0 0 has been used, making the final expression consistent with the four-current term in SGVT. References Griffiths, D., Introduction to Electrodynamics, Benjamin-Cummings 1999. Jackson, J.D., Classical Electrodynamics, John Wiley 1999. Shadowitz, A., The Electromagnetic Field, McGraw-Hill 1975.

1 Matrix Algebra Review

Introduction This is the matrix review for A Student’s Guide to Vectors and Tensors (SGVT). It is not meant to be a thorough introduction to the theory and practice of matrix algebra, but rather a review of the matrix concepts you should understand when learning about vectors and tensors. A matrix is defined as an array of numbers. Some authors restrict the term “matrix” to refer only to square arrays of real numbers, but in common usage a matrix is a rectangular (m x n) array of complex numbers or “elements,” where “m” represents the number of rows, “n” represents the number of columns, and “complex” means that the numbers may be purely real, purely imaginary, or mixed. Thus ⎡ ⎤ 8 ⎢ 0 ⎥ ⎢ ⎥

3 −2 A= C = 9 15 3 −8 12 −2 B=⎢ −2 ⎥ ⎢ ⎥ 1 0 ⎣ 1 ⎦ 5 and

⎡

⎤ 5 − 2i 3i ⎢ 0 15 ⎥ ⎥ D=⎢ ⎣ −4i 2 + 8i ⎦ 12 i−1

⎡

1 E=⎣ 0 0

0 1 0

⎤ 0 0 ⎦ 1

F =

0 0 0 0

are all matrices. In these examples, matrix A has “dimensions” or “order” (rows x columns) of 2x2, B of 5x1, C of 1x6, D of 4x2, E of 3x3, and F of 2x2.

2

Matrix Algebra Review

If both m and n for a matrix are equal to one, that matrix is called a scalar (because it’s just a single value), and if either (but not both) of m and n are one, that matrix may be called a vector. So in the examples shown above, B and C are vectors; in some texts you’ll see B called a column vector and C called a row vector. A square matrix (that is, a matrix with m=n) with ones on the diagonal and zeroes in all off-diagonal elements (such as matrix E above) is called the “unit” or “identity” matrix, and a matrix with all elements equal to zero is called the “null” or “zero” matrix. Two matrices are said to be equal only if they have the same number of rows as well as the same number of columns and if every element in one matrix is equal in value to the corresponding element in the other matrix. Matrices are usually denoted upercase letters or bold font, often surrounded by square brackets (such as [A]), and the elements are often written using lowercase letters with subscripts. So you may see the elements of matrix [A] written as aij , although some authors use Aij or [A]ij to refer to the elements of A (but be careful to note that Aij may also refer to the “Matrix of Cofactors” of matrix A, which you can read about in Section E of this document). So does terminology such as “row vector” mean that matrices, vectors, and tensors are the same thing? Not really. It’s certainly true that the vectors and tensors in SGVT are often represented using matrices, but it’s important to what those matrices stand for. Those arrays of values represent the components of vectors and tensors, and those components have meaning only when associated with the basis vectors of a particular coordinate system. Since the basis vectors are not always shown, it’s convenient to think of the matrix as representing the vector or tensor itself, and that’s fine as long as you that the actual vector or tensor has existence independent of any coordinate system. A) Matrix addition, multiplication by a scalar, and subtraction Matrices may be added only if both the row dimension (m) and their column dimension (n) are equal (such matrices are said to be “of the same order”). The addition is accomplished simply by adding each element of one matrix to the corresponding element of the other matrix. For example:

5 −3

2 0

+

−3 1 0 4

=

2 3 −3 4

Matrix Algebra Review or in general b11 a11 a12 + a21 a22 b21

b12 b22

=

3

a11 + b11 a21 + b21

a12 + b12 a22 + b22

.

Note that the result is a matrix of the same order as the matrices being added. Note also that matrix addition is commutative, so [A]+[B]=[B]+[A]. Multiplication of a matrix by a scalar is straightforward; you simply multiply each element of the matrix by the scalar. Thus 3 −2 9 −6 3A = 3 = 1 0 3 0 and generally

kA = k

a11 a21

a12 a22

ka11 ka21

=

ka12 ka22

You can use the rules for addition of matrices and scalar multiplication to see that subtraction of matrices is accomplished simply by subtracting the corresponding elements. Thus [A]-[B]=[A]+(-1)[B]. So if a11 a12 A= a21 a22 and

B=

then A−B =

a11 + (−1)b11 a21 + (−1)b21

b11 b21

b12 b22

a12 + (−1)b12 a22 + (−1)b22

=

a11 − b11 a21 − b21

a12 − b12 a22 − b22

.

Just as for addition, subtraction of matrices only works for matrices of the same order. B) Matrix multiplication There are several different ways to multiply matrices; the most common and most relevant to the vector and tensor concepts in SGVT is to multiply two matrices (call them A and B) by multiplying the elements of each row in A by the elements of each column of B and then summing the results. So if matrix A is given by a11 a12 a13 a21 a22 a23

4


and matrix B is given by ⎡

b11 ⎣ b21 b31

⎤ b13 b23 ⎦ b33

b12 b22 b32

then the product A times B is given by ⎡ b11 a11 a12 a13 × ⎣ b21 a21 a22 a23 b31 =

a11 b11 + a12 b21 + a13 b31 a21 b11 + a22 b21 + a23 b31

b12 b22 b32

⎤ b13 b23 ⎦ b33

a11 b12 + a12 b22 + a13 b32 a21 b12 + a22 b22 + a23 b32

a11 b13 + a12 b23 + a13 b33 a21 b13 + a22 b23 + a23 b33

This result is achieved by multiplying the elements of the first row of A by the elements of the first column of B, summing the results, and placing the sum in the first row and first column of the result matrix, as shown in Figure 1.1.

Sum products Multiply first row by first column

{

Place result in first row and first column

a11 a12 a13 ( ) ( )

( )

x

b11 ( ) b21 ( ) b31 ( )

( ) ( ) ( )

=

a11b11+ a12b21+ a13b31 ( )

( ) ( )

( ) ( )

Figure 1.1 Multiplying first row by first column

The next step is to multiply the elements of the first row of A by the elements of the second column of B, summing the products, and placing the sum in the first row and second column of the result matrix, as shown in Figure 1.2. After multiplying the first row of A by the third column of B and placing the sum in the first row and third column of the result matrix, the same procedure is done with the second row of A - those elements are multiplied by the first column of B, summed, and placed in the second

.


5

Sum products Multiply first row by second column

{

Place result in first row and second column

a11 a12 a13

x

( ) ( ) ( )

( ) b12

( )

( ) b22

( ) ( )

( ) b32

=

( )

a11b12+ a12b22+ a13b32

( )

( )

( )

( )

Figure 1.2 Multiplying first row by second column

Multiply second row by first column

( ) ( ) ( ) x a21 a22 a23

b11 ( ) b21 ( ) b31 ( )

( ) ( ) ( )

=

( )

( )

( )

a21b11+ a22b21+ a23b31

( )

( )

{

Place result in second row and first column Sum products

Figure 1.3 Multiplying second row by first column

row and first column of the result matrix, as shown in Figure 1.3. Notice that the matrix that results from multipling a 2x3 matrix (A) by a 3x3 matrix (B) is a 2x3 matrix - the result has the number of rows of the first matrix and the number of columns of the second matrix. You should also note that this type of matrix multiplication only works when the number of columns of the first matrix (3 in the case of A) equals the number of rows of the second matrix (also 3 in the case of B). To understand why matrices are multiplied in this way, consider the sorting of five types of toy marbles (Bowlers, Cat’s Eyes, Steelies, Aggies, and Commons) into four sizes of package (Small, Medium, Large, and Extra Large). The number of each type of marble in each size package

6


is shown in the following table: Bowlers

Cat’s Eyes

Steelies

Aggies

Commons

3 5 10 15

2 3 5 12

5 7 12 16

4 6 10 14

Small package 1 Medium package 2 Large package 5 Extra Large package 10

This array of numbers can be put into a matrix - call it “P” the package matrix: ⎡ ⎤ 1 3 2 5 4 ⎢ 2 5 3 7 6 ⎥ ⎥ P =⎢ ⎣ 5 10 5 12 10 ⎦ 10

15

12

16

14

Now imagine three containers (Cans, Boxes, and Crates) with different numbers of packages in each type:

Can Box Crate

Small Pkgs

Medium Pkgs

Large Pkgs

10 15 40

8 12 25

5 7 15

Extra-Large Pkgs 2 3 10

Put this array into a matrix called “C” the container matrix: ⎡ ⎤ 10 8 5 2 C = ⎣ 15 12 7 3 ⎦ 40 25 15 10 If you wish to find the number of each type of marble in each type of container, you could do it like this: Bowlers per Can = 10 small packages × 1 Bowler per small package + 8 medium packages × 2 Bowlers per medium package + 5 large packages × 5 Bowlers per large package + 2 extra-large packages × 10 Bowlers per extra-large package = (10)(1)+(8)(2)+(5)(5)+(2)(10)=71 Bowlers per Can


7

Likewise, you can find the number of Cat’s Eyes per can using Cat’s Eyes per Can=10 small packages×3 Cat’s Eyes per small package + 8 medium packages × 5 Cat’s Eyes per medium package + 5 large packages × 10 Cat’s Eyes per large package + 2 extra-large packages × 15 Cat’s Eyes per extra-large package = (10)(3)+(8)(5)+(5)(10)+(2)(15)= 150 Cat’s Eyes per Can And if you wished to find the number of Bowlers per Box, you could use Bowlers per Box = 15 small packages × 1 Bowler per small package + 12 medium packages × 2 Bowlers per medium package + 7 large packages × 5 Bowlers per large package + 3 extra-large packages × 10 Bowlers per extra-large package = (15)(1)+(12)(2)+(7)(5)+(3)(10)=104 Bowlers per Box If you compare the numbers in these calculations to the values in the P and C matrices, you’ll see that in each case you’re multiplying the row elements of C by the column elements of P, which is exactly how matrix multiplication works: ⎡ ⎤ ⎡ ⎡ ⎤ ⎤ 1 3 2 5 4 10 8 5 2 71 150 93 198 166 ⎢ 2 5 3 7 6 ⎥ ⎥ ⎣ ⎦ = ⎣ 15 12 7 3 ⎦⎢ ⎣ 5 10 5 12 10 ⎦ = 104 220 137 291 244 40 25 15 10 265 545 350 715 600 10 15 12 16 14 So the product matrix shows you how many of each type of marble are in each type of container:

Can Box Crate

Bowlers

Cat’s Eyes

Steelies

Aggies

71 104 265

150 220 545

93 137 350

198 291 715

Commons 166 244 600

In addition to showing why the rows of the first matrix are multiplied by the columns of the second matrix, this example also illustrates the point made earlier: when you multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. So in this case, multiplication works only if you multiply C by P, not P by C. The larger point is that matrix multiplication is not

8


commutative, so even if you’re able to multiply two matrices in either order (which you can do, for example, with square matrices), the answer is not the same. So in matrix world, AB is in general not equal to BA. Another difference between matrix multiplication and multiplication of numbers is that it’s possible to get zero as the result of multiplying two matrices, even if neither matrix is zero. For example, 2 1 1 0 0 0 × = . 0 0 −2 0 0 0 If you study vectors and tensors, you’re likely to come across the multiplication of a row vector times a column vector or vice versa, and you should understand that such products use the same process described above for matrix multiplication, although that may not be immediately obvious when you look at the result. For example, multiplying a row vector A by a column vector B gives ⎡ ⎤ b11

a11 a12 a13 × ⎣ b21 ⎦ = [a11 b11 + a12 b21 + a13 b31 ] . b31 This scalar result (essentially the dot product between A and B) comes from multiplying the first (and only) row of A by the first (and only) column of B and adding the sums. Likewise, if A is a column vector and B a row vector, the product AB is ⎡ ⎤ ⎤ ⎡ a11 b11 a11 b12 a11 b13 a11

⎣ a21 ⎦ × b11 b12 b13 = ⎣ a21 b11 a21 b12 a21 b13 ⎦ a31 a31 b11 a31 b12 a31 b13 Once again, the same rule of matrix multiplication has been applied. The first row of A (just a11 in this case) is multiplied by the first column of B (just b11 in this case), and since there are no other elements in the first row of A or the first column of B, there is nothing to add, and the result (a11 b11 ) is written in the first row and column of the result matrix. Then the first row of A (again, only a11 ) is multiplied by the second row of B (only b12 ), and the result is written in the first row, second column of the result matrix. After doing the same for the first row of A and the third column of B, you then multiply the second row of A (which is just a21 in this case) by the first column of B (again just b11 ) and write the result in the second row, first column of the result matrix. So although you get a scalar when you multiply a row vector by a column vector (sometimes called the “inner product” of the matrics) and


9

a matrix when you multiply a column vector by a row vector (sometimes called the “outer product” of the matrices), the process is exactly the same in both cases. Although matrix multiplication is not commutative (so AB is not necessarily equal to BA), matrix multiplication is the associative and distributive over addition, so for matrices A, B, and C (AB)C = A(BC) and A(B + C) = AB + AC as long as you not to reverse the order of any of the products. You may be wondering if there’s ever a need to multiply each element of one matrix by the corresponding element of an equal-size matrix. There certainly is (for example, when applying a two-dimensional filter funtion to an image). This process is called “element-by-element” or “entrywise” matrix multiplication and the result is sometimes called the “Hadamard product.” In MATLAB, such multiplication is denoted A.*B, where the decimal point before the multiplication symbol signifies element-by-element multiplication. C) Transpose and trace of a matrix The transpose of a matrix is accomplished by interchanging the rows with the columns of that matrix. This is usually denoted by placing the superscript “T” after the matrix name, so the transpose of matrix A is written as AT (MATLAB uses A ). So if the matrix A has elements a11 a12 a13 a14 A= a21 a22 a23 a24 then the transpose of A is given by ⎡ a11 ⎢ a12 T A =⎢ ⎣ a13 a14

⎤ a21 a22 ⎥ ⎥. a23 ⎦ a24

So (Aij )T = Aji ; notice that the indices have been switched. The transpose of the product of two matrices is equal to the transpose of each matrix multiplied in reverse order, so (AB)T = B T AT .

10


so long as the dimensions of the matrices allow such products to be formed. For square matrices, the trace of the matrix is given by the sum of the diagonal elements. The trace is usually denoted by “Tr” so the trace of matrix A is written as Tr(A). So if the matrix A is given by ⎤ ⎡ a11 a12 a13 a14 ⎢ a21 a22 a23 a24 ⎥ ⎥ A=⎢ ⎣ a31 a32 a33 a34 ⎦ a41

a42

a43

a44

the trace of A is T r(A) = a11 + a22 + a33 + a44 =

aii .

D) Determinant, minors, and cofactors of a matrix The determinant of a square matrix is a scalar calculated by multiplying, adding, and subtracting various elements of the matrix. If A is a 2x2 matrix a11 a12 A= a21 a22 the determinant of A (denoted using vertical bars on each side of A) is found by cross-multiplying the upper-left element times the lower-right element and then subtracting the product of the lower-left element times the upper-right element: a a12 = a11 a22 − a21 a12 . |A| = 11 a21 a22 Hence if A is

A=

4 −2 3 1

the determinant of A is 4 −2 = (4)(1) − (3)(−2) = 4 − (−6) = 10. |A| = 3 1 To find the determinant of higher-order matrices, you must use the “minors” and “cofactors” of the determinant of the matrix. The minor for each element of the determinant of a square matrix is found by


11

eliminating the entire row and the entire column in which the element appears and then writing the remaining elements as a new determinant, with row and column dimensions reduced by one (so the minor of each element of a 3x3 determinant is a 2x2 determinant). So for the 3x3 determinant of matrix A a11 a12 a13 |A| = a21 a22 a23 a a a 31

the minor of element

a11 is − − − a22 − a32

− a23 a33

32

33

a22 = a32

a23 a33

obtained by elminating the first row and column of the determinant of matrix A (since the element for which this minor is taken (a11 ) is in the first row and the first column). Likewise, the minor for element a12 is found by eliminating the row and column in which a12 appears (first row and second column): − − − a21 − a23 = a21 a23 . a31 a33 a − a33 31 For element a22 , the

minor is a11 − a31

− a13 − − − a33

a11 = a31

a13 . a33

Before you can use minors to find the determinant of a higher-order matrix, you have turn each minor into a “cofactor.” The cofactor of an element is just the minor of that element multiplied by either +1 or -1. To know which of these to use, just determine whether the sum of the element’s row index and its column index is even or odd. If even (as it is, for example, for the element in the first row and first column, since 1+1=2 which is even), the cofactor equals the minor of that element mutiplied by +1. If odd (for example, for the element in the first row and second column, since 1+2=3 which is odd) the cofactor equals the minor of that element multiplied by -1. So for element aij with row index of i and column index of j, Cofactor of (aij ) = (−1)(i+j) Minor of (aij ).

12


So for the minors shown above, the cofactors are a a23 = (1)(a22 a33 − a32 a23 ) Cofactor of (a11 ) = (−1)(1+1) 22 a32 a33 a Cofactor of (a12 ) = (−1)(1+2) 21 a31 and (2+2)

Cofactor of (a22 ) = (−1)

a11 a31

a23 = (−1)(a21 a33 − a31 a23 ) a33 a13 = (1)(a11 a33 − a31 a13 ). a33

With cofactors in hand, the determinant of a 3x3 matrix is straightforward to find. Simply pick one row or column, and multiply each of the elements of that row or column by its cofactor and sum the results. So, for example, if you choose to use the first row of the 3x3 matrix A, the determinant is given by |A| = a11 (Cofactor of a11 ) + a12 (Cofactor of a12 ) + a13 (Cofactor of a13 ). or |A| = a11 (1)(a22 a33 −a32 a23 )+a12 (−1)(a21 a33 −a31 a23 )+a13 (1)(a21 a32 −a31 a22 ). The value of the determinant would have been the same if you had used a different row (or any of the columns) of matrix A, as long as you multiply each element in your selected row or column by its cofactor and sum the results. The same approach can be used to find the determinant of 4x4 and higher-order matrices; you just have to expand each of the determinants using elements and cofactors until you get down to the 2x2 level. Of course, the calculation becomes a bit tedious, so it’s worth your time to learn to find determinants using a program like MATLAB. Some useful characteristics of determinants are that the determinant of the transpose of a matrix equals the determinant of the matrix (so |AT | = |A|) and the determinant of the product of two matrices is the same as the determinant of the reverse product (so |AB| = |BA|) provided that the dimensions of A and B allow both of these products to be made. You may also find it useful to know that if a matrix has two identical rows or columns, or if one row or column is an integer multiple of another row or column, the determinant of that matrix must be zero.


13

E) Inverse of a matrix As indicated in Section C of this document, matrix multiplication is quite different from ordinary multiplication of numbers. One example of this is that the product of two matrices may be zero even if both of the matrices are non-zero (so you can’t simply divide both sides of a matrix equation such as AB = 0 by A to get B = 0). Additionally, the matrix equation AB = AC does not mean that matrix B equals matrix C (so dividing both sides of this equation by A does not work for matrices). Differences such as this suggest that matrix division has little in common with ordinary division of numbers. About the closest you can get to a process similar to division for matrices comes about by considering the matrix equation AX = B where A, X, and B are all matrices. If you wish to find matrix X from this equation, here’s one thing you definitely cannot do: B AX = A A B X= A because this type of division does not work for matrices. What you can do to find X is this: you can try to find a matrix A−1 (called the “inverse” or “reciprocal” of A) with the following property: A−1 A = I where I represents the identity matrix (ones on the diagonal and zeroes everywhere else). Now you can find X rather easily: AX = B A

−1

(AX) = A−1 B

(A−1 A)X = A−1 B IX = A−1 B X = A−1 B since IX = X (multiplying the identity matrix times any matrix does not change that matrix). So although you haven’t really divided matrix B by matrix A, you’ve used the matrix equivalent of multiplying by the reciprocal of a number to achieve the same result as dividing.

14


The question is, does every matrix have an inverse (that is, another matrix with which it multiplies to give the identiry matrix), and if so, how do you find it? The first part of this question is easy to answer: any matrix that has a non-zero determinant (|A| = 0) has an inverse. Matrices that have no inverse ((|A| = 0) are called “singular” matrices, and matrices that have an inverse are called “non-singular” matrices. But if a certain matrix has an inverse, how do you find that inverse? There are several ways to go about this, but one approach is to use the concepts of cofactors, transpose, and determinant described above. Those concepts appear in the following equation for the inverse of matrix A: A−1 =

(Matrix of cofactors of A)T |A|

in which the “Matrix of cofactors of A” is a matrix in which each element of A is replaced by that element’s cofactor. Here’s how that works for a 3x3 matrix. If matrix A is the usual ⎡ ⎤ a11 a12 a13 A = ⎣ a21 a22 a23 ⎦ a31 a32 a33 then the matrix of cofactors (MoC) for A is ⎡

⎤ a22 a33 − a32 a23 −(a21 a33 − a31 a23 ) a21 a32 − a31 a22 MoC (A) = ⎣ −(a12 a33 − a32 a13 ) a11 a33 − a31 a13 −(a11 a32 − a31 a12 ) ⎦ . −(a11 a23 − a21 a13 ) a11 a22 − a21 a12 a12 a23 − a22 a13 The equation for the inverse of A shown above requires that you take the transpose of this matrix, which is ⎤ −(a12 a33 − a32 a13 ) a12 a23 − a22 a13 a22 a33 − a32 a23 [MoC (A)]T = ⎣ −(a21 a33 − a31 a23 ) a11 a33 − a31 a13 −(a11 a23 − a21 a13 ) ⎦ . −(a11 a32 − a31 a12 ) a11 a22 − a21 a12 a21 a32 − a31 a22 ⎡

Dividing this matrix by the determinant of A provides the inverse of A. Fortunately, the inverse of non-singular matrices is easily found using most scientific calculators or computers, usually by defining a matrix A and then raising A to the power of minus one. For a diagonal matrix, the inverse is simply another diagonal matrix in which each diagonal element is the reciprocal of the corresponding

Matrix Algebra Review element in the original matrix. So ⎤−1 ⎡ 1 ⎡ 0 a11 0 a11 ⎣ 0 a22 0 ⎦ = ⎢ ⎣ 0 0 0 a33 0

15

0 1 a22

0

⎤ 0 ⎥ 0 ⎦.

1 a33

F) Simultaneous linear equations and Cramer’s Rule The relationship between matrices and simultaneous linear equations can be understood by considering equations such as 2x + 5y − z = 12 −3x − 3y + z = −1 x + y = 1. Using the rules of matrix multiplication, this system of equations can be written as a single matrix equaton: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 2 5 −1 x 12 ⎣ −3 −3 1 ⎦ × ⎣ y ⎦ = ⎣ −1 ⎦ . 1 1 0 z 1 In general, three linear equations in three unknowns (x1 , x2 , x3 ) can be written as a11 x1 + a12 x2 + a13 x3 = b1 a21 x1 + a22 x2 + a23 x3 = b2 a31 x1 + a32 x2 + a33 x3 = b3 or

⎡

a11 ⎣ a21 a31

a12 a22 a32

If you define the matrices

⎤ ⎡ ⎤ ⎡ ⎤ x1 b1 a13 a23 ⎦ × ⎣ x2 ⎦ = ⎣ b2 ⎦ . a33 x3 b3 ⎡

a11 A = ⎣ a21 a31 and

⎡

a12 a22 a32

⎤ x1 x = ⎣ x2 ⎦ x3

⎤ a13 a23 ⎦ a33

16


and ⎤ b1 b = ⎣ b2 ⎦ b3 ⎡

then the system of equations can be written as a single matrix equation: Ax = b which can sometimes be solved using the inverse of matrix A: x = A−1 b. The reason that the word “sometimes” appears in the previous sentence is that there are several conditions which will prevent this approach from succeeding. For example, the system of equations may be inconsistent that is, there may be no set of values for x1 , x2 , x3 that satisfy all three equations. In that case, you will not be able to find the inverse of A because it will be a singular matrix. And if matrix b equals zero, you have a system of homogeneous linear equations, which means there will be only the trivial solution (x1 = x2 = x3 = 0) if A is non-singular or an infinite number of solutions if A is singular. In cases for which A is non-singular and b does not equal zero, you can find the values of x1 , x2 , and x3 by finding the inverse of matrix A and multiplying that inverse by matrix b, or you can use Cramer’s Rule. In that approach, the first unknown (x1 in this case) is found by replacing the values in the first column of the coefficient matrix (A) with the elements of matrix b and dividing the determinant of that matrix by the determinant of A. Here’s how that looks: b1 a12 a13 b2 a22 a23 b a a33 3 32 x1 = a11 a12 a13 a21 a22 a23 a a32 a33 31

.

Likewise, to find the second unknonwn (x2 in this case), replace the

Matrix Algebra Review values in the second column of A with the elements of b: a11 b1 a13 a21 b2 a23 a b3 a33 31 x2 = a11 a12 a13 a21 a22 a23 a a32 a33 31 and to find x3 use

a11 a12 b1 a21 a22 b2 a a32 b3 31 x3 = a11 a12 a13 a21 a22 a23 a a32 a33 31

.

Thus for the equations given at the start of this section, ⎡

2 A = ⎣ −3 1

5 −3 1

⎤ −1 1 ⎦ 0

⎡

⎤ ⎡ ⎤ x1 x x = ⎣ x2 ⎦ = ⎣ y ⎦ x3 z and

Hence

⎤ ⎡ ⎤ 12 b1 b = ⎣ b2 ⎦ = ⎣ −1 ⎦ . b3 1 ⎡

x = x1 =

5 −1 −3 1 1 0 2 5 −1 −3 −3 1 1 1 0

12 −1 1

17

18


so 12[(−3)(0) − (1)(1)] + 5(−1)[(−1)(0) − (1)(1)] − 1[(−1)(1) − (1)(−3)] 2[(−3)(0) − (1)(1)] + 5(−1)[(−3)(0) − (1)(1)] − 1[(−3)(1) − (1)(−3)] −12 + 5 − 2 = −3. = −2 + 5 − 0

x=

Proceeding in the same way for y and 2 −3 1 y = x2 = 2 −3 1 so y= and

z=

12 −1 −1 1 1 0 5 −1 −3 1 1 0

−2 + 12 + 2 =4 −2 + 5 − 0

z = x3 =

so

z gives

2 5 12 −3 −3 −1 1 1 1 2 5 −1 −3 −3 1 1 1 0

−4 + 10 + 0 = 2. −2 + 5 − 0

G) Matrix diagonalization using eigenvectors and eigenvalues In the matrix equation discussed in the previous section Ax = b the matrix A operates on matrix x to produce matrix b. If x and b are vectors, this equation represents the transformation by matrix A of vector x into vector b. In some cases, the operation of A on the components of x produces the components of a vector b that is a scaled (but not rotated) version of x. In such cases, the equation becomes Ax = λx


19

where λ represents a scalar multiplier (and scalar multipliers can change the length but not the direction of a vector). Any vector x which satisfy this equation for a matrix A is called an “eigenvector” of matrix A, and the scalar λ is called the “eigenvalue” associated with that eigenvector. The eigenvalues of a matrix can be very useful in finding a diagonal version of that matrix, so you may wish to understand how to find the eigenvalues of a given matrix. To do that, write the previous equation as Ax − λx = 0 which, since Ix = x, can be written as Ax − λ(Ix) = 0 or (A − λI)x = 0. which means that either x = 0 (which is a the trivial case) or |A − λI| = 0. This equation is called the “characteristic for a 3x3 matrix it looks like this: 1 a11 a12 a13 a21 a22 a23 − λ 0 0 a a32 a33 31 or

a11 − λ a21 a 31

a12 a22 − λ a32

equation” for matrix A, and 0 1 0

a13 a23 a33 − λ

0 0 1

=0

= 0.

This expands to (a11 − λ)[(a22 − λ)(a33 − λ) − a32 a23 ] + a12 (−1)[a21 (a33 − λ) − a31 a23 ] + a13 [a21 a32 − a31 (a22 − λ)] = 0. Finding the roots of this polynomial provides the eigenvalues (λ) for matrix A, and substituting those values back into the matrix equation Ax = λx allows you to find eigenvectors corresponding to each eigenvalue. The process of finding the roots is less daunting than it may

20


appear, as you can see by considering the following example. For the 3x3 matrix A given by ⎡ ⎤ 4 −2 −2 A = ⎣ −7 5 8 ⎦ 5 −1 −4 the characteristic equation is 4−λ −2 −7 5−λ 5 −1

−2 8 −4 − λ

=0

or (4 − λ)[(5 − λ)(−4 − λ) − (−1)(8)] − 2(−1)[(−7)(−4 − λ) − (5)(8)] − 2[(−7)(−1) − (5)(5 − λ)] = 0. Multiplying through and subtracting within the square brackets makes this (4 − λ)(λ2 − λ − 12) + 2(7λ − 12) − 2(5λ − 18) = 0 or −λ3 + 5λ2 + 12λ − 36 = 0. Finding the roots of a polynomial like this is probably best left to a computer, but if you’re lucky enough to have a polynomial with integer roots, you know that each root must be a factor of the term not involving λ (36 in this case). So (+/-) 2,3,4,6,9,12,and 18 are possibilities, and it turns out that +2 works just fine: −(2)3 + 5(22 ) + 12(2) − 36 = −8 + 20 + 24 − 36 = 0. So you know that one root of the characteristic equation (and hence one eigenvalue) must be +2. That means you can divide a factor of (λ − 2) out of the equation and try to see other roots in the remainder. That division yields this: −λ3 + 5λ2 + 12λ − 36 = −λ2 + 3λ + 18. (λ − 2) The roots remaining polynomial on the right-hand side of this equation are +6 and -3, so you now have −λ3 + 5λ2 + 12λ − 36 = (λ − 2)(6 − λ)(λ + 3) = 0.


21

So matrix A has three distinct eigenvalues with values +6, -3, and +2; these are the factors by which matrix A scales its eigenvectors. You could find the eigenvectors of A by plugging each of the eigenvalues back into the characteristic equation for A, but as long as you can find N distinct eigenvalues for an NxN matrix, you can be sure that A can be diagonalized simply by constructing a new diagonal matrix with the eigenvalues as the diagonal elements. So in this case, the diagonal matrix (call it A ) associated with matrix A is ⎡ ⎤ 6 0 0 A = ⎣ 0 −3 0 ⎦ . 0 0 2 To see why this is true, consider the operation of matrix A on each of its three eigenvectors (call them e, f , and g): Ae = λ1 e Af = λ2 f Ag = λ3 g Now imagine a matrix E whose columns are made up of the eigenvectors of matrix A: ⎤ ⎡ e1 f1 g1 E = ⎣ e2 f2 g2 ⎦ e3 f3 g3 where the components of eigenvector e are (e1 , e2 , e3 ), the components of eigenvector f are (f1 , f2 , f3 ), and the components of eigenvector g are (g1 , g2 , g3 ). Mutiplying matrix A (the original matrix) by E (the matrix made up of the eigenvectors of A), you get ⎤ ⎡ ⎤ ⎡ e1 f1 g1 a11 a12 a13 AE = ⎣ a21 a22 a23 ⎦ × ⎣ e2 f2 g2 ⎦ a31 a32 a33 e3 f3 g3 which is ⎡

a11 e1 + a12 e2 + a13 e3 AE = ⎣ a21 e1 + a22 e2 + a23 e3 a31 e1 + a32 e2 + a33 e3

a11 f1 + a12 f2 + a13 f3 a21 f1 + a22 f2 + a23 f3 a31 f1 + a32 f2 + a33 f3

⎤ a11 g1 + a12 g2 + a13 g3 a21 g1 + a22 g2 + a23 g3 ⎦ a31 g1 + a32 g2 + a33 g3

The columns of this AE matrix are the result of multiplying A by each of the eigenvectors. But you know from the definition of eigenvectors

22


and eigenvalues that

⎡

⎤ ⎡ ⎤ e1 λ1 e1 Ae = λ1 e = λ1 ⎣ e2 ⎦ = ⎣ λ1 e2 ⎦ e3 λ1 e3

and

and

⎡

⎤ ⎡ ⎤ f1 λ2 f1 Af = λ2 f = λ2 ⎣ f2 ⎦ = ⎣ λ2 f2 ⎦ f3 λ2 f3 ⎡

⎤ ⎡ ⎤ g1 λ3 g1 Ag = λ3 g = λ3 ⎣ g2 ⎦ = ⎣ λ3 g2 ⎦ . g3 λ3 g3

This means that the product AE can be written ⎤ ⎡ λ1 e1 λ2 f1 λ3 g1 AE = ⎣ λ1 e2 λ2 f2 λ3 g2 ⎦ . λ1 e3 λ2 f3 λ3 g3 But the matrix on the right-hand side can also be written like this: ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ e1 f1 g1 λ1 0 0 λ1 e1 λ2 f1 λ3 g1 ⎣ λ1 e2 λ2 f2 λ3 g2 ⎦ = ⎣ e2 f2 g2 ⎦ × ⎣ 0 λ2 0 ⎦ e3 f3 g3 λ1 e3 λ2 f3 λ3 g3 0 0 λ3 This means that you can write

⎡

λ1 AE = E ⎣ 0 0

0 λ2 0

⎤ 0 0 ⎦ λ3

and multiplying both sides by the inverse of matrix ⎤ ⎡ ⎡ λ1 0 λ1 0 −1 −1 ⎣ E AE = E E 0 λ2 0 ⎦ = ⎣ 0 0 0 λ3 0

E (E −1 ) gives ⎤ 0 0 λ2 0 ⎦ . 0 λ3

If you tracked the discussion of the similarity transform in Section 6.1, you’ll recognize the expression E −1 AE as the similarity transform of matrix A to a coordinate system with basis vectors that are the columns of matrix E. Those columns are the eigenvectors of matrix A, and the matrix that results from the similarity transform (call it A ) is diagonal and has the eigenvalues of A as its diagonal elements.


23

References Pettofrezzo, A., Matrices and Transformations, Prentice-Hall 1966. Riley, K., Hobson, M.P., and Bence, S.J. Mathematical Methods for Physics and Enginnering, Cambridge 2006. Schwartz, J.T., Introduction to Matrices and Vectors, McGraw-Hill 1961.

Errata and Clarifications (September 2015) 1) On page 18, the radial line from the origin to point P should be labeled “r” rather than “ρ”. ˆ 2) On page 19, the last of the three equations of Eq. 1.14 should say zˆ = k. 3) On page 24 in the second line of Problem 1.9, the word “spherical” should be replaced by the word “cylindrical”. Hence the phrase should be “...a steady current I is given in cylindrical coordinates by the expression....” 4) On the top of page 35 in the second line of the right-hand side of the equation ~ × (B ~ × C), ~ the term Ax Cy should be Ax Cx . for A ~ ◦ A, ~ there 5) On the top of page 48 in the right-hand side of the equation for ∇ should be no vector symbols over the components Ax , Ay , and Az . 6) On page 57, the paragraph beginning with ”And how does the difference...” should read ”And how does the difference between a function’s value at a point and the average value at neighboring points relate to the divergence of the gradient of that function? To understand that, think about a point which is a local maximum of the function - at such at point, the function’s value is greater than the surrounding average. Likewise, at a point which is a local minimum, the function’s value is less than the surrounding average. This is the reason you may find the Laplacian described as a ”concavity detector” - it finds points at which the value of the function sticks above or falls below the average value of the surrounding points.” 7) On page 60 in Problem 2.3, in two places vertical-line symbols are printed as slanting lines (this occurs after the letter “B”, once before “cos” and once before “sin”). Here is the correct text: ~ ◦B ~ = Ax Bx + Ay By + Az Bz = |A|| ~ B| ~ cos(θ) and that |A ~ ◦ B| ~ = 2.3 Show that A ~ B| ~ sin(θ). |A|| 8) On page 68, the sentence beginning “According to this equation...” should say that the y-component of the box’s acceleration is proportional to (rather than equal to) the difference between the magnitude of the normal force and the y-component of the gravitational force. 9) On page 71, 77, and 80 the word “accelerate” is misspelled. 10) On page 87, the units of the permittivity of free space should be C2 /(Nm2 ). 11) On page 88, the equation on the second line should contain the term kre2q

1

rather than

ke q r

.

12) On page 90, the equation for the divergence of the magnetic field near the bottom of the page (after Eq. 3.36) should read µ0 I ~ ◦B ~ = 1 ∂Bφ = 1 ∂ ∇ r ∂φ r ∂φ 2πr

13) On page 114 in the right portion of Figure 4.13, the basis vector along the y axis labeled ~e 2 should be ~e2 . Here’s a corrected version of the figure:

14) Clarification: On page 126, the paragraph beginning with “And here’s the important insight...” should make it clear that the columns of the transformation matrix are the components of the basis vectors. 15) On the bottom of page 129 in the right side of Equation 4.62, Ai should be Aj . 16) In the statement for Problem 4.1 on page 130, the word “indirect” should be “direct.” 17) In the statements for Problems 4.6 and 4.8 on page 131, the vector components A1 , A2 , A1 and A2 should not have vector arrows. 18) On pages 133 to 136, in Equations 5.1 through 5.9 and the text in-between, 0 the prime marks should appear before the superscripts. So on these pages, ∂xi

2

should be ∂x0i . 19) On page 134, the statement beginning with “And just as in Equation 5.1...” should read “And just as in Equation 5.1, the elements of the inverse transformation matrix also represent the basis vectors tangent to the original coordinate axes.” 20) On page 135, the statement beginning with “In this case, the weighting fac∂xj tors...” should read “In this case, the weighting factors ( ∂x i0 ) are the elements of the direct transformation matrix from the unprimed to the primed coordinate systems...” 21) On page 140, in the sentence “...or using covariant components and dual basis vectors (~ei )...” the subscript ‘i‘ should be a superscript. So the corrected sentence reads “...or using covariant components and dual basis vectors (~e i )...”. 22) On page 141, the final equation in the second block of equations (after the sentence “A third option is to use contravariant components on one side of the dot and covariant components on the other:”) should say = dxi dxi . ~ 23) On p the top of page 142 √ in the second line of the equation for |A, the expression Ai Aj should be Ai Ai . 24) On page 148, in the final equation on the page, the subscript 1 in the basis vector ~e1 should be an i. So the corrected equation is ~ ∂A ∂(A1 e~1 + A2 e~2 + A3 e~3 ) = ∂x1 ∂x1 ∂(Ai e~i ) = ∂x1 ∂ e~i ∂Ai e~i + Ai 1 = ∂x1 ∂x

25) On page 149, in the fourth line from the top of the page, the word “differentiation” is misspelled as “defferentiation”. 26) On page 151, the word “of” is missing from the sentence beginning “Now it’s just a matter of pulling out...” 27) On page 153, in the expression for Γ122 , the multiplying factor for the third term should be 12 g 31 . 28) On page 155 in the expression for Aφ;φ on the right side of the equation the second term should contain Ar instead of Aφ and the third term should contain 3

Aφ instead of Ar . The correct equation is: ∂Aφ + Ar Γφrφ + Aφ Γφφφ + Az Γφzφ ∂φ ∂Aφ 1 r +A +0+0 = ∂φ r

Aφ;φ =

29) On page 155, the final equation should be ~ ∂A ∂Ar ∂Aφ 1 ∂Az =( − rAφ )~er + ( + Ar )~eφ + ~ez ∂φ ∂φ ∂φ r ∂φ 30) On page 176, in the expression for the square of the distance covered by a wavefront of the light wave in the unprimed coordinate system, the right side should be (ct)2 , and in the primed coordinate system, the right side should be (c0 t0 )2 . 31) On page 182 in the matrix at the top of the page, at the bottom row of the second column, the expression (By )(−γ) should be (−By )(γ), and in the ~ expression for F~ 0 , the element in the third column of the first row should be −γ(Ey /c − βBz ). 32) On page 183 near the end of the second paragraph, the second-last sentence should ask“So does the electric field exist or not?” 33) On page 190, in the expression 1 ∂gφφ Γθφφ = ( ) − g θθ 2 ∂θ the minus sign should be in front of the (1/2) factor. So the correct expression is 1 ∂gφφ Γθφφ = −( )g θθ 2 ∂θ

34) On page 191, in the second equation from the top, which is Rθφθθ =

∂Γθφθ ∂Γθφθ − + Γθφθ Γθθθ + Γφφθ Γθφθ − Γθφθ Γθθθ − Γφφθ Γθφθ ∂θ ∂φ

the second term after the equals sign should have θ instead of φ in the denominator. The correct expression is Rθφθθ =

∂Γθφθ ∂Γθφθ − + Γθφθ Γθθθ + Γφφθ Γθφθ − Γθφθ Γθθθ − Γφφθ Γθφθ ∂θ ∂θ

35) On page 193, the statement for Problem 6.3 should say“Find the moment of inertia tensor for the arrangement of masses in Problem 6.1....” 4

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