Mathematical Physics PHYS 2920
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Phys 2920 MathPhys Notes David Murdock7 TTU May 57 2009 Contents 1 D 3 Preliminaries and Vector Algebra 1 11 Plane Polar Coordinates 1 12 Complex Numbers 1 13 Scalars and Vectors 2 14 Vectors Basics 2 15 Basic Operations 3 16 Components7 Basis Vectors 3 17 Products of Vectors 4 18 Vector Algebra Relations 4 19 Some Physics 4 110 The Symbols 6 and eijk 5 Vector Spaces and Matrices 7 21 Vectors Spaces 7 22 Basis Vectors 7 23 lnner Product 7 24 Linear Operators 8 25 Matrices 9 26 Basic Operations on Matrices 10 27 Inverse of a Matrix 11 28 Trace of a Matrix 11 29 Determinant of a Matrix 12 210 More On the Inverse of a Matrix 12 211 Special Types of Matrices 13 212 Eigenvectors and Eigenvalues 13 213 Determination of Eigenvectors and Eigenvalues 14 214 Similarity Transformations 15 2141 Unitary Transforms 15 215 Diagonalization of Matrices 16 216 Simultaneous Linear Equations 16 Vector Analysis 19 31 Vector Functions 19 32 Differentiation of Vectors 19 321 Product Rules 20 33 Space Curves 21 34 Scalar and Vector elds 21 4 CONTENTS 35 Vector Operators 21 351 Gradient 21 352 Divergence 22 353 Curl 22 36 Vector Operator Formulae 22 37 Double Operations 23 38 Cylindrical Polar Coordinates 23 39 Vector Operators in Cylindrical Coordinates 25 310 Spherical Coordinates 25 311 Vector Operators in Spherical Coordinates 26 312 Line7 Surface and Volume lntegrals 27 3121 Line lntegrals 27 3122 Example of Line lntegrals 28 3123 More On Line lntegrals 29 3124 Line lntegrals in Physics 29 3125 Green s Theorem in a Plane 30 3126 Conservative Fields 30 313 Surface lntegrals 30 3131 A General Method 31 3132 lntegrals on Coordinate Surfaces 33 3133 Applications in Physics 34 3134 Surface lntegrals on Spherical Surfaces 34 314 Volume lntegrals 34 315 lntegration Theorems 36 3151 The Divergence Theorem 36 3152 The Divergence Theorem and Physics 36 3153 Stokes7 Theorem 37 3154 Stokes7 Theorem and Physics 37 316 The Dirac Delta Function 38 3161 A Paradox 38 3162 A Sequence of Functions 39 3163 ThreekDimensional Delta Function 40 3164 Resolving the Paradox 40 Complex Variables 41 41 Complex Numbers Basic Operations 41 411 Polar Representation Relation to Trig Functions 42 412 Applications to Basic lntegrals 43 413 Hyperbolic Functions 7 A Review7 Maybe 44 414 Complex Powers and Logarithms 44 42 Functions of Complex Variables 45 421 Notation 45 422 The Elementary Functions 45 C ON TEN TS 5 44 45 46 47 433 Singular Points 48 Complex Integration Cauchy s Theorem 48 Cauchy s Integral Formula 49 In nite Series Taylor and Laurent Series 49 The Residue Theorem 50 C ON TEN TS Chapter 1 Preliminaries and Vector Algebra 11 Plane Polar Coordinates Points in the z y plane are described by the coordinates p and 1 If we connect the point P to the origin with a straight segment7 the segment has length p and is inclined at an angle 1 from the m axis7 as shown in Fig 11 zpcos gt ypsin gt and we can go backwards Via pxm2y2 tan gta 1 where we have to choose 1 appropriately to be in the correct quadrant This system will be extended to three dimensions in two different ways later on in the course 12 Complex Numbers Complex numbers use the imaginary number i 71 They have the form zabi where a and b are real numbers Figure 11 Plane polar coordinates 2 CHAPTER 1 PRELIMINARIES AND VECTOR ALGEBRA They occur in many branches of mathematics and theoretical physics While they make calcu lations easier in some instances they are essential in the study of quantum mechanics because the central element in QM the wave function is a complex function of space and time The magnitude or modulus of z is given by z a2b2 ll v One represent the complex numbers as points in a plane illustrated in an Argand diagram lf 2 a bl then a and I give the horizontal and vertical coordinates of the point P corresponding to The polar coordinates of P are given by b plzlva2b2 tan97 a One can take the exponential function of a complex number Here the fundamental relation is e cos0 lsin0 It follows that any complex number can be written in the form 2 pew where p and j are as given above 13 Scalars and Vectors Quantities which can be given by a single number are called scalars Quantities which must be speci ed by giving a magnitude and a direction are known as vectors Examples of scalars in physics are Temperature Energy the Electric Potential and Density Examples of vectors in physics are velocity force and the electric eld Vectors can be represented on paper by an arrow The length of the arrow gives an indication of the magnitude and the direction of the arrow gives the direction of the vector 14 Vectors Basics Vectors the 2 and 3 dimensional ones can be represented on paper by an arrow pointing in the appropriate direction with a length indicative of its magnitude It does not matter where on the page the vector is drawn only the magnitude and direction matter When we give a vector a name it is good to distinguish it from a scalar as our problems will involve both kinds of quantities In print like here I will put the name in boldface thus a is the name of a vector On paper and on a blackboard this is hard to do so there one should put an arrow on top of the symbol like 71 Practices vary for this nowadays textbooks are also putting little arrows on top of the letters to emphasize to students that the things they re talking about have directionl So you might also see a in printed material 15 BASIC OPERATIONS 3 15 Basic Operations Vectors are useful mathematical entities because we can add them together On paper7 the addition of vectors is done by joining them headitoitail keeping their directions and magnitudes the same Vector addition has the properties 0 a b b a Commutative o a b c a b c Associative o 7b is a vector equal in magnitude but opposite in direction to b Subtracting vectors means a 7 b a 7b Vectors can be multiplied by scalars The vector a points in the same direction as a opposite if is negative but has A times the magnitude Scalar multiplication follows the rules MW M021 MM 0 a b a b Distributive 16 Components Basis Vectors In three dimensions7 using the unit vectors along the m y and z axes called i7 j and k here we can express any other vector as a linear combination a ami ayj azk The notation for these unit vectors varies between books also used are 2 y 2 and em ey ez so be prepared for different notation In N dimensions which we will be considering before long7 we will generally use a set of N fundamental of basis vectors to construct all the others aa1 1a2 2aN N We will want to use a set of basis vectors which spans the entire space that is7 can give any vector of interest by taking linear combinations but which is not redundant7 meaning speci cally that the i s are linearly independent A set of vectors i is linearly independent if 01 162 2CN N0 implies that all the 01 s are zero CHAPTER 1 PRELIMINARIES AND VECTOR ALGEBRA 17 Products of Vectors ab abcos0 11 abambmaybyazbz 12 a X b 0Lbe 7 azbyi azbm 7 ambzj amby 7 aybmk 13 If you are familiar with evaluating the determinant of a 3 x 3 matrix7 this can be written 1 j k axb am ay aZ 14 5w by bz 18 Vector Algebra Relations The scalar triple product77 formed from the vectors a7 b and c a b x c is of course7 a scalar One can show abxcaxbc Note7 the order is important in these expressions This product can be written as sub7 c and equals the determinant given by 11 ay aZ abx c a7b7c bm by bz cm Cy CZ Another useful vector relation is axbcxdacbd7adbc Finally we consider a triple product of vectors7 a x b x c Since the cross product is not associative7 we note axbxc7 axbxc so lots of care is needed with the order and grouping when cross products are involved Two useful formulae are ax bx c 7 acb7abc ax b xc acb7bca 19 Some Physics The dot product shows up in elementary physics in the de nition of work If a particle moves by some small displacement ds while a force F acts on it7 the work done by the force is dWFds 110 THE SYMBOLS 6U AND WK 5 The cross product shows up in the de nition of torque also called moment of the force by some people Torque is measured with respect to some origin if a force F acts at point r then the torque exerted by that force is 7 r x F It is also appears in the de nition of angular momentum also de ned with respect to a given origin If a particle of mass m has velocity V at point r its angular momentum is Lrxprxmv One sees a cross product in the expression for the magnetic force on a charged particle moving in a magnetic eld If a charge q has velocity V as it experiences a magnetic eld B the magnetic force on the charge is Fmag qV X B 110 The Symbols 6 and eijk Before moving on to the next chapter I want to introduce some notational tricks which help in doing derivations with vectors Recall the meaning of Kronecker delta 5 1 0 2 For example if your basis vectors are orthonormal then 8139 j 6H and the usual dot product is 3 a b Z aib l j ij1 Another useful symbol with the name LeviCevita tensor is denoted by gigk has the value 1 71 or 0 depending on the values of where the indices have the values 1 2 or 3 We have 6123 1 6132 1 6312 1 etc where we get a change of sign if we switch two indices If any two indices are the same it is zero hence 133 0 6121 0 6222 0 etc With gigk we can write the cross product more compactly If e stands for the m y and 2 unit vectors then 3 a gtlt b Z aiqujk k ij1 3 I a gtlt b gt Ck Z aiqujk ij1 6 CHAPTER 1 PRELIMINARIES AND VECTOR ALGEBRA To save on writing in our derivations of vector relations there is another convention which is common If an index appears twice on one side of an equation in a product we can agree that we will sum on that index Then with this convention we would have written I a X b gt Ck aiqujk where in the second expression we know that we should sum on i and j With this notation the dot product is simply abab Several useful theorems help us to use the 6 s in proving theorems In both we are using the new summation convention Eijkqu 6il6jm 6im6jl 15 Eijkqji 261d 16 Eijkeijk 6 17 As example of the use this new notation we will prove the triple product rule aX chacb7abc lmagine doing this proof longhand you d spend a lot of time writing out all the components which occur in many combinations Smart people use the 6 s and 6 s Start with the left side and use the as to write out the cross products a X b X C b X Cjqjk k aiblcmqmjqjk k Switching indices on one of the 6 s gives a X b X C aiblcmqqujk k aiblCmElijikj k Using Eq 15 gives aiblcm6il6km 5ki5m k liaibl6il6km6m aiblcm6km6iml k Use aibl6il a b etc and get 7a bCk a Cbk k a cbk k 7 a bck k a cb 7 a bc Chapter 2 Vector Spaces and Matrices 21 Vectors Spaces A set of objects a b c forms a vector space if c The set is closed under addition addition is commutative and associative o The set is closed under multiplication by a scalar which is also distributive and associative 1 a b a b 2 ua a ua 3 Mira M021 0 There is a null vector 0 such that a 0 a for all a o All vectors have a negative 7a such that a 7a 0 And 7a 71a The scalars which multiply the vectors can be restricted to real numbers giving a real vector space but in general they are complex giving a complex vector space 22 Basis Vectors 2 3 Inner Product It is useful to add the concept of an inner product to a vector space The inner product of a and b denoted alb gives a scalar that is a complex number The inner product has the properties 0 alb blagt it isn t quite commutative o It is distributive that is ltalAb m Altalbgt Mltalcgt From the rst property it follows that ltAa Mblcgt mam mltblcgt 8 CHAPTER 2 VECTOR SPACES AND MATRICES This is a generalization of the familiar dot product of threedimensional vectors Two vectors a and b are orthogonal if ltalbgt O The norm of a vector is given by Hall ltalagt12 so that we hereby restrict attention to vector spaces for which ltalagt 2 0 If ltalagt 0 the a 0 Now suppose the Nidimensional vector space has a basis e1 2 N which has the desirable property that il 6 Then we say the basis is orthonormal If the vectors a and b expressed in this basis are N N i1 i1 and if the basis is orthonormal then we get N gala a and ltalbgt Zam i1 Two inequalities follow from the de nition of the inner product The Schwarz Inequality is lltalbgtl S Hall Hbll 21 The triangle inequality is Habll S Hall MW 2 That s all for math of the vectors within a vector space Now we introduce the idea of linear operators leading to the objects known as matrices 24 Linear Operators A linear operator changes one vector into another but in a speci c way For the operator A for every vector X it associates a vector y in the same vector space written AX y A linear operator has the property Aa ub AAa MAb Now we introduce a speci c basis i If A acts on ej we have to get some linear combination of all the i s We write N A Z A i1 Then if y AX then N N y A Zj j Zj Aij i 23 71 71 11 2 5 MATRICES 9 Eq 23 de nes the matrix elements of the matrix A for the linear transformation A for the basis l Matching the coef cients of gives N yl ZAljm 2394 j1 Eq 24 shows how the components of vectors transform for the linear transformation A One can also talk about linear transformations between vector spaces with different dimension ality but we won t Properties of linear operators ABXAx8x AAx MAX ABX ABx Note that while the addition of linear operators is commutative the multiplication that is successive application of operators is not We also have the null and identity operators which for all x give Ox 0 Ix x 25 Matrices In a particular basis i both vectors and linear operators are described in terms oftheir components As we ve seen the components transform as N y AX gt yl39 ZAijj j1 and we write the set of numbers A as A11 A12 39 39 39 A1N A A A A 21 22 2N ANl AN2 ANN We will assume that a matrix is square same number of rows as columns though if the linear transformation is between spaces of different dimensionality it won t be Likewise we write out the components of the vector x as a column 901v One can show 10 CHAPTER 2 VECTOR SPACES AND MATRICES A 3 AM BM 0 AA j AA ABM 2k AikBkj The last of these gives the rule for matrix multiplication There is always a matrix such that A O A It s simply a matrix full of zeroes There is also always a matrix 1 for which 1A A1 A It is a matrix with 1 s along the diagonal but zeroes everywhere else Thus we have the unit matrix7 10 0 01 0 1 0 0 1 26 Basic Operations on Matrices In working with matrices7 we will have reason to carry out the following operations on their elements Transpose lnterchange rows and columns Denote the transpose of A by AT Then AT ij Ajl39 The transpose of a product of matrices is related to a product of the transposes of the individual matrices you have to reverse the order ABT BTAT Complex Conjugate Then the complex conjugate of all elements denote by A Thus ATM Alf 7 7 The complex conjugate of a product doesn t change the order AB A 8 Hermitian Conjugate Take the complex conjugate and the transpose Denote by Al Thus Al ART ATV ATM All The Hermitian conjugate of a product also gives a reversed order AE3T BTAT Note that when we take a vector inner product using components of the vectors we are really doing a sort of matrix multiplication for real vectors it is b1 52 ltalbgt a1 a2 IN bN 27 RV VERSE OF A MATRIX 11 while for a complex vector space it is ltalbgtai at aTb Of course the second case includes the rst one for the real vector space the Hermitian conjugate is the transpose 27 Inverse of a Matrix A matrix A can have an inverse that is there may be some matrix A 1 for which A 1 A A A 1 1 More on which matrices have an inverse shortlyl If we have the matrix equation AB C and the matrix B is unknown but matrix A has an inverse we can nd B by multiplying both sides of the equation on the left by A l AB C gt B A lC But we note that since in general matrices do not commute it would be wrong to write B CA 1 here One must be careful about the order of matrix multiplication 28 Trace of a Matrix The trace of a square matrix is the sum of the diagonal elements N TFAA11 A22ANN 2AM i1 Thus the Trace operation takes a matrix and returns a number Some properties are TrA i B TrA i m3 TrAB TrBA More generally the trace does not change if we do a cyclic rotation of the product of matrices For three matrices this gives TrABC TrCAB TrBCA And remember the product of the matrices itself may change if we switch the order in this way but its trace does not 12 CHAPTER 2 VECTOR SPACES AND MATRICES 29 Determinant of a Matrix The determinant is another operation which takes a matrix and returns a number It is denoted detA and is indicated with vertical bars on either side of the matrix entries A11 A12 AlN mam 521 522 ii 52 ANl AN2 ANN Recall for the matrix itself we enclose the entries with parentheses It is a rather simple operation for a 2 x 2 or a 3 x 3 matrix but the simple pattern for these cases does not hold for bigger sizes Properties of the determinant 1 The determinant of a transpose is the same that of the original matrix lATl W 2 The determinant of a complex conjugate and Hermitian conjugate satisfy Wl lATl Wl W 3 If any two rows or columns of the matrix are interchanged the determinant changes sign 4 lAl ANlAl where N is the dimension of matrix A 5 If any two rows or columns of A are the same lAl 0 6 If you add a constant multiple of one row or column to another row or column the value of the determinant is unchanged 7 For square matrices lABl W W lBAl so the value of the determinant is unchanged by any permutation of the order of the matrices in a product 210 More On the Inverse of a Matrix It follows from det1 1 that the determinants of A and A 1 are inverses of each other in the normal numerical sense This must mean that if the determinant of A is zero then A cannot have an inverse In that case we say that A is singular But for a non singular matrix A we can always nd an inverse Computing the inverse of a matrix can be done by any computer and many calculators but it could be done by hand It can be shown that CT C A 1 4 l l W W where CM is the cofactor of the ik element 211 SPECIAL TYPES OF MATRICES 13 And there other ways to calculate it Explanation to be given in class on an algorithm which will produce the inverse from a given matrixlt turns out that it requires a bit of care to program a computer to compute it accurately Properties of the inverse 1 AB 1 B lA l ABG 1 Gil B1A1 and as already mentioned7 1 W11 i W 211 Special Types of Matrices A diagonal matrix only has non zero entries along the diagonal that is7 Aij 0 ifi 7 j A symmetric matrix is equal to its transpose A AT So Aij Aji for all i j An antisymmetric matrix equals the negative of its transpose A iAT So Aij 7A7 for all i j An antisymmetric matrix must have zeroes along the diagonal lf AT A 1 then the matrix A is orthogonal And the inverse of an orthogonal matrix is also orthogonall The determinant of on orthogonal matrix must be i1 A special property of the transformation effected b an orthogonal matrix is that the norm of the vector is unchanged If A A then the matrix A is Hermitian If A 7A then A is antiHermitian If A A l7 then the matrix A is unitary If A is real7 then it is also orthogonal If A is unitary7 then 1A is in general a complex number with magnitude 1 A unitary matrix has the property that it leaves the magnitude of a complex vector unchanged lf AAf ATA then A is normal 212 Eigenvectors and Eigenvalues Suppose a linear operator A transforms vectors x in an Nidimensinal space into other vectors in the same space our normal situation It could happen that x is transformed into a vector parallel to itself that is it could hbe the case that Axx for some vectors x and the accompanying numbers Then x is called an eigenvector of A and is an eigenvalue of A To the uninitiated7 these might seems like odd things to look for7 but 14 CHAPTER 2 VECTOR SPACES AND MATRICES they of fundamental importance in quantum mechanics They also arise in a few areas of classical mechanics To compute these vectors we choose a particular basis for which the operator A has the repre sentation A But we can give a few facts about them before doing any computations o If A is nonsingular and A has eigenvectors xi and corresponding eigenvalues A then A 1 has the same eigenvectors with eigenvalues 1A o If A is Hermitian its eigenvalues are real and eigenvectors corresponding to di erent eigenvalues are orthogonal 213 Determination of Eigenvectors and Eigenvalues AxAx gt AxiAx0 gt A7A1x0 Now if the last of these is written out longhand we get A11 901 A12902 A1NN 0 A12901 A22 MM A1NN 0 25 AN19011 AN2902 ANN AN 0 What do we have here This is a set of N homogeneous linear equations that is the constant term is zero in each of them If for some A there is a solution then we will have an eigenvalue and an eigenvector Now we get a bit ahead of ourselves since we haven t yet discussed systems of linear equations but turns out that such a system will have a solution if the determinant of the coef cients is zero A11 A12 AlN A21 A22 A2N lA 7 All 0 ANl AN2 m ANN For a 2 x 2 matrix this will give a quadratic equation for A which of course is easy to solve and we get at most two possible values for A but a 3 x 3 matrix gives a cubic equation for A and large matrices are even messier Fortunately the computer packages are quite good at nding eigenvalues Having the eigenvalues A we can go on to solve the equations in 25 to get the eigenvector for each one One can show that the sum of the eigenvalues gives the trace N Z A TrA i1 where A are the eigenvalues of A This can give a check on on a computation of the eigenvalues One can also show that if matrices A and B commute that is AB BA then their eigenvalues are the same It may happen that a matrix has two or more eigenvalues which are the same that is the algebraic equation for A has multiple roots Then the corresponding eigenvectors are said to be degenerate and we can nd linearly independent eigenvectors with the same eigenvalue 214 SHVHLARITY TRANSFORMATIONS 15 214 Similarity Transformations Recall that a vector space is a set of abstract objects as are the linear operators which act upon them We can express them as a set of numbers only when we make a particular choice of a basis If we have some new basis vectors 3 related to some old basis vectors ei by N e Z Sige i1 then one can show that the new components of the vector x for the two choices of basis are related by N z Z Sigm or x Sx or x S lx j1 S is the transformation matrix associated with the change in basis Since the e are linearly independent S is non singular and so it does have an inverse S71 We also need to transform the the matrix expression of linear operators One can show A S lAS The change of basis is an important operation in quantum mechanics where can use coordinates or momenta to label the components of the quantum mechanical vectors ie the quantum states S is called a similarity transform Note that after a transformation with S the vectors and operators still have the same relations to one another Some facts about a transformation with S o lfA1thenA 1 o A l Al the value of the determinant is unchanged 0 The eigenvalues of A are the same as those of A o Tr A Tr A 2141 Unitary Transforms An important class of similarity transforms is one for which S is unitary SJr S l Then A S lAS szs These are important because if the original basis is orthonormal then the transformed basis is also orthonormal lt il jgt 51quot lt l 2gt 6H For unitary transforms we have o If A is Hermitian then A is Hermitian o If A is unitary then A is unitary 16 CHAPTER 2 VECTOR SPACES AND MATRICES 215 Diagonalization 0f Matrices A diagonal matrix is clearly a very simple kind of matrix If we begin with a more complicated matrix A it is useful procedure to nd a similarity transformation which can transform it into a diagonal matrix A And the eigenvectors of the matrix A are used for this Suppose we begin with a matrix A and choose the eigenvectors of A as the new basis Thus for ij Aij we want to nd an S such that N Xj Z Sijei i1 But this equation tells us that the columns of S are the r t of the 39 o u With S constructed this way one can show Alj STIASM M21 Then A is diagonal and its elements are the eigenvalues of the original matrix and of course the matrix A itself Thus with this S 1 0 0 A 0 A2 0 7 0 0 0 AN Can we always do this If A has N distinct eigenvalues then it can be diagonalized by this procedure But in general if a square matrix has degenerate eigenvalues it may not have N independent eigenvectors if it doesn t then it can t be diagonalized For matrices which are Hermitian or unitary the N eigenvectors are linearly independent and so one can choose an orthonormal set If the columns of S are made of orthonormal vectors one can show 515 6 sis 1 and then A S lAS szs 216 Simultaneous Linear Equations This is probably the topic where you rst heard about matrices and determinants We ve left it for last ln physical applications we often have a set of M linear equations for N unknowns ml mN The set would look like 1411961 A122A1N96N bi 1421961 A222A2N96N 52 AM11 AM22AMNN bM 216 SHVIULTANEOUS LINEAR EQUATIONS 17 There can be a unique solution no solution or an in nite number of solutions In any case the set of equations is conveniently expressed in matrixvector form as A11 A12 Am 951 bi A A AN m b 12 22 2 2 or Axb 26 AM AMN mN 5M Generally speaking if M lt N there is an in nity of solutions If M gt N then the system is over determined and there is no solution We will only consider the simple but common case M N With M N the matrix A in 26 is square and we can nd its determinant lf lAl 7 0 then A had an inverse and we can get x from x Ailb but if lAl 0 then there are in nitely many solutions for x This is probably the preferred method of solution if you have a computer handy If not then due to the computation demands of nding A 1 we would need an easier method The method that may be familiar to you from high school or someplace is called Cramer7s Rule it involves taking the ratios of determinants of matrices made from the coef cients in the problem and thse ratios give the solution for the unknowns For clarity we ll show why it works with a set of 3 equations for 3 unknowns the pattern will then be obvious Start with 1411951 1412962 1413963 bi 1421951 1422962 1423963 52 1431951 1432962 1433963 53 If we form the matrix A from the coef cients we nd A11 A12 A13 1 bi A12 A13 lAl A21 A22 A23 a 52 A22 A23 A31 A32 A33 1 53 A32 A33 How did we get the last equality First we multiplied the rst colmun by 1 whatever it turns out to be and then divided by the same with no net change Then we took 2 times the second column and add it to the rst column Then we took 3 times the last column and added it to the rst column Both of these have no net effect on the determinant from the properties of the determinant But now by the original set of equations the rst column is equal to 1 b2 3 And thus we get the last expression If we make the de nitions b1 A12 A13 A11 b1 A13 A11 A12 b1 A1 52 A22 A23 A2 A21 52 A23 A3 A21 A22 52 7 53 A32 A33 A31 53 A33 A31 A32 53 then we have 1 1 1 W A1 W A2 W A3 1 2 3 which as if by magic can be rearranged to give A1 A2 A3 7 x7 m7 27 W 2 W 3 W l m1 18 CHAPTER 2 VECTOR SPACES AND MATRICES which gives a straightforward way to get the solution as long as we can take all the determinants The extension to more equations and more variables as long as there are the same number of each is clear Eqs 27 can not be evaluated if lAl 0 and indeed in that case there is no unique solution Chapter 3 Vector Analysis We now consider only the mundane vectors we used in basic physics the threedimensional ones made out of i j and k and which are represented by arrows But we will now use them in function relations and analyze those functions with the ideas of calculus 31 Vector Functions Vectors may themselves be functions of numbers or indeed functions of other vectors such as the coordinate r in a given coordinate system In the rst case we evaluate say the vector a at each value of t written at The most familiar example of this is the evaluation of the location r at each time t the function rt is the trajectory of the moving particle Or it could be the case that at each point in space speci ed by r or z y we can give the value a scalar 1 or a vector a These are respectively scalar or vector elds written as gtr and ar With this concept we have a dependence of j on three independent variables 1 gtm y z for the vector eld we have three components depending on three coordinates ar39 a x y zi Lgx y azz y zk Whoa Lots of opportunities for fun here In this framework we are still interested in the calculus ideas of rates of change77 and of integrals in nite summations but they need to be made more elaborate To start with we will consider the case that a vector is a function of a single scalar parameter which we might think of as the time 25 but which we ll call For this case we have altugt mm ayltugtj azltugtk 32 Differentiation of Vectors The derivative of the vector au with respect to the parameter u is j 7 hm au A107 au 7 dam day daz du 7 AuaO Au du du du k 31 19 20 CHAPTER 3 VECTOR ANALYSIS Of course this operation is familiar from kinematics where we all learned dr39 dm dy dz 7 a 7 1EJEk UmlUyJUzk dV dvm dvy allZ dt 1251 dt dt k amiayj azk Old stuff But we should appreciate here that when we took the derivatives of V and a we used the fact that the unit vectors 1 j and k did not change with time Eq 31 above assumes that the unit vectors 1 j k do not depend on u or if the time t is the parameter on the time This seems obvious but if we had another scheme for a set of standard unit vectors this would not be the case In plane polar coordinates we can consider the unit vectors p and egg which point outward from the origin and perpendicular to this in the counterclockwise direction These The particle s location in the plane is given by r p p where the particle s polar coordinates are p gt The new unit vectors are given by pcos gtisin gtj 7sin gticos j and now if we take the time derivative of these vectors we get d p d dt e dt we where This gives dr dpA d ii A V a d8ppdj PepP e p p p gt This is the expression for velocity using polar coordinates for both the coordinates and the unit vectors 321 Product Rules When a scalar j and vector a are functions of the parameter u we have 1 da d gt z a 7 t a 1 db da i b 7 ib dulta a du du 1 db da duaxb ax xb In the last of these note how the order is preserved this is important since the cross product is not commutative 33 SPACE CURVES 21 33 Space Curves A general curve in space can be given parametrically by 1W 90101 9003 200k This form of a curve is useful for analyzing a trajectory in mechanics it is also useful when we want to evaluate in integral along a path more on that later While the parameter u could be anything7 it is often useful to use the arclength of the curve as a parameter The arclength 3 starting from the point speci ed by ul is 2 dr dr 3 i i du ul du du There are lots of juicy mathematical issues in the analysis of curves in space This material falls under the heading of classical differential geometry which does have some use in physics but not enough that we need to go there right now Similarly7 a surface can be speci ed by ru7 U mu7 Ul yu7 2u7 Uk 34 Scalar and Vector elds Now we return to the concept that a scalar say7 gt or a vector say7 a can be de ned at each point in spaceso that we deal with the functions gtr and ar which are scalar and vector elds Actually7 a eld might depend on the space and time coordinate ltjgtr7 25 That does arise in physics but here we will ignore any possible if dependence We are immediately interested in space derivatives of some kind for these functions7 and as already mentioned there are lots of choices 35 Vector Operators Now we consider the idea of a scalar or vector eld7 where the scalar 1 or the vector a depends on the vector r7 that is7 on the 3 coordinates z7 y The common vector derivative operators all use the V operator called del or nabla de ned by i 3kg 7 8m Jay 82 V 351 Gradient The gradient of the scalar eld Mm7 y z is de ned by 7 71 J9 19 grad 7V 71J87ykE 32 22 CHAPTER 3 VECTOR ANALYSIS What does one do with the gradient This operation produces the change in j in a certain direction In calculus the rate of change is unambiguously written as loosely speaking step by dz and get a change 01 but here we need to specify the direction in which we step by some distance ds One can show that if we step by a small distance ds in the direction of the unit vector a then d gt 7 A E 7 V gt a To see this write 8 8 8 7 gt gt gt 7 d gt7 adm87ydy Edzivdrdr so with dr ds 5 we get the advertised result From this it follows that at a given point the direction in 1 gives the greatest change per unit displacement is the direction of the gradient and the magnitude of that rate of change is d if W 352 Divergence The divergence of a vector eld az y z is given by 8am 8011 8m 8y 82 In words the divergence tells how much a vector function spreads out or well diverges from a given point divaVa 33 353 Curl The curl of a vector eld am y z is given by 8a 8a 8a 8a 8a 8a 1 7179 39 7 1 73771 k 34 cura vxa lt81 82gt1lt82 8xgtJlt8m 8y a formula which is more easily remembered with the schematic expression 1 j k curlaan QQm 881 882 35 L1 ay aZ 36 Vector Operator Formulae If we perform a vector operation on a product of two scalar or vector functions we can show that the result can be written in a form similar to the product rule77 of elementary calculus Here though there are more possibilities for the product and the derivative operator One can show vow 7 gtWW Vab axbebxVxaaVbbVa V gta gtVaaV gt Vaxb baniabe Vx gta V gtxa gtan Vxaxb aVb7bVabVa7aVb 3 7 DOUBLE OPERATIONS 23 37 Double Operations There ve ways to combine the grad div and curl operations operating on a scalar or vector eld and the results are interesting Two of them in fact give zero For any sensible elds 1 and a we have VXV gt 0 Van 0 ln words the curl of a gradient is zero and the divergence of a curl is zero The other three possible relations need to done individually First after taking the gradient of a scalar eld one can then take the divergence We get 8 82 82 v v v2 7 7 7 8x2 8342 822 The operator V2 which here operates on the scalar j and often pronounced as del squared77 is of great importance in physics It appears in classical wave equations and in the quantum Schrodinger wave equation You sometimes see V2 operating on a vector When the vector is expressed in Cartesian coordinates the meaning of this is Vza Vzam Vzay Wank where we note that for example V20m really contains three terms so there are actually nine second derivatives contained here The gradient of a divergence VV a is not necessarily zero but it doesn t come up very much in physics For the record in Cartesian coordinates it is given by 82am 8202 3201 82am 8202 8201 VltViagt 7 lt8z2 amay 8m82gt lt13sz 8342 8y82gtJ 82am 8202 8201 7 k 828 8283 822 gt Finally one can take the curl of the curl of a vector eld One gets Vx an VVa7V2a 36 This isn t so bad but some caution is needed we will soon look at these vector operators as they are used in other coordinate systems where the unit vectors depend on the coordinatse and then the expression for Vza given above is not useful in that case Eq 36 is still true but we need another way of nding Vza 38 Cylindrical Polar Coordinates An alternative to the usual system of Cartesian coordinates x y z is shown in Fig 31 Instead we uniquely specify a point by joining it to the z axis with a line segment the length of that segment is p the angle it makes with the m axis is j and its 2 coordinate is used as before the point is speci ed with the set 0 j z the cylindrical polar coordinates of the point 24 CHAPTER 3 VECTOR ANALYSIS Figure 31 Cylindrical polar coordinates p d and z The Cartesian and cylindrical polar coordinates are related by zpcos gt ypsin gt 22 37 with0 p 0070 gtlt27r7andfooltzltoo We could then write a coordinate vector as r pcos gti psin gtj 2k but it is useful to new unit vectors to go along with the new coordinates The unit vector associ ated with a coordinate must point in the direction we go when when change only that particular coordinate We get these by taking 8r 8r d 8r i 7 an i 377 13 82 and making a unit vector of the result One nds pcos gtisin j 7sin gticos j ik 38 These three unit vectors are mutually orthogonal7 which was assure by the choice ofthe coordinate system itself But the vectors depend on the coordinates themselves a situation we did not have with our Cartesian unit vectors i7 j7 k We must keep this in imnd when we work with the new unit vectors Eqs 38 can be inverted to give the Cartesian unit vectors in terms of the cylindrical ones icos gt psin gt jsin gt pcos gt k z 39 We will want to evaluate integrals on lines7 surfaces and volumes using these coordinates7 and we will need expressions for the line7 surface and volume elements in terms of them In Cartesian coordinates we had the relatively simple ideas that a small displacement path element is drdmidyjdzk 39 VECTOR OPERATORS IN CYLHVDRICAL COORDHVATES 25 An area element that is perpendicular to the z direction for example is given by da dm dy i and a small element of volume is given by dV dz dy d2 drdpeppd gte dz z 310 dV pdpd gtdz 311 A surface of constant p a cylinder everywhere has its normal parallel to p A small area element on this surface is given by dap pd gt 12 Similarly if we consider surfaces of constant 1 a half plane and z a plane we nd that the vector area elements have components dap pdzj dz da dp dz daz pdp d gt 312 39 Vector Operators in Cylindrical Coordinates A vector eld expressed completely in terms of cylindrical coordinates and the cylindrical units vectors has the form a am gt7 2 p 037 gt7 2 qs LA97 gt7 2 z where there s nary an x y or z to be seen We would like to have expressions for the vector derivative operators which can be used with a vector eld given in this form 8ltIgtA 18ltIgtA 8ltIgtA 87W 5 1 aaz p 13 82 1 8 81 1821 821 2 P87 P872 W VltIgt 18 Va if a p8pPJ 310 Spherical Coordinates Another common alternative to the Cartesian coordinates x y z is shown in Fig 32 Here we uniquely specify a point by joining it to the origin with a line segment the length of that segment is r the angle it makes with the 2 axis is 9 and if we join the point to the z axis with a shortest line segment the angle which that segment makes with the 1 axis is 1 then the point is speci ed with the set r 0 gt the spherical polar coordinates of the point The Cartesian and spherical coordinates are related by mrsin0cos gt yrsin0sin gt zrcos0 313 As with cylindrical coordinates we can form unit vectors for spherical coordinates unit vectors pointing in the direction of displacement if only one of the coordinates changes The relations are 26 CHAPTER 3 VECTOR ANALYSIS Figure 32 Spherical polar coordinates r 6 and er sin0cos gtisin0sin jcos0k e c0s0cos gticos0sin jisin0k isin gtic0s gtj with inverse relations 1 sin0c0s gt Tcos0cos gt 97sin gt j sin0sin gt Tcos0sin gt 9cos gt k c0s0 T isin0 9 The line elements and the volume element in spherical coordinates are given by drdr rrd0 9rsin0d gte 314 rsm r 5 W 2 0d d0d 31 A surface of constant T a sphere everywhere has its normal parallel to r A small area element on this surface is given by dar r2 sin0d0 01 Similarly7 if we consider surfaces of constant 9 a cone and j a halfiplane we nd that the vector area elements have components dar 72 sin0d0 d gt dGQ r sin0dr d gt da 7 dr 10 316 311 Vector Operators in Spherical Coordinates And now we consider scalar and vector eld expressed completely in terms of spherical coordinates and the spherical unit vectors7 in the form 1 1gtT707 gt a aw 07 fquot T7 07 e MW 07 as We would like to have expressions for the vector derivative operators which can be used with a vector eld given in this form One can show that they are 312 LHVE SURFACE AND VOLUME HVTEGRALS 27 V I ag r 183 3 r 80 rs1n0 19 Va ltrza7 sin0a9 V X a lltsm0a gt 7 2 r 23 7 gm g g 2 vzq T227 8111027 rzsiln20g7q 312 Line Surface and Volume Integrals Opening up our space to three dimensions and our objects to vectors gives more options for per forming integrations in nite summations of vanishingly small entities 3121 Line Integrals Possibilities are gtdr adr axdr C C C For example the second of these means to consider the limiting process as we split up the curve C into N displacements 11quot located at around the coordinate f 3f 2 and then take N Z am 3f 2 11quot i in the limit of large N and small intervals But how we evaluate such a thing In basic calculus we have the Fundamental Theorem 17 b dF 35 dz Fa Fb 7 Fa where 7 35 a 1 dm Here the step dr can be written dr dzi dyj dzk but more care is needed because the dm dy and dz are related as we step along the curve Likewise the argument is a function of three variables given by the particular path So it s not so simple Generally we have to the reduce the integral down to one variable which will be some parameter giving the point along the curve C Before giving an example we state some obvious facts about line integrals B A adr7 adr A B B P B adr adr adr A A P where the integral are all done along the same curve C and the point P is on the curve between A and B 28 CHAPTER 3 VECTOR ANALYSIS 239 M 11 iii Figure 33 Line integral evaluated on three di erent paths 3122 Example of Line Integrals RHB give a useful example of evaluating a line integral by different methods Example Here we want to evaluate Iadr where amyiyizj C along each of the paths in Fig 33 which are 2 the parabola yz z from 11 to 42 22 the curve z 2222 22 1 y 1 222 from 11 to 42 222 the line y 1 from 11 to 41 followed by the line z 4 from 41 to 4 2 In all the cases we have dr 1271 dyj so that in general Iadrzydmyizdy 317 C C but the different paths necessitate different steps to go further 2 Along the parabola yz m the differentials are related by 2y dy dm Substitute for z and dz in 317 for the both integrals isince everything is now in terms of yi the limits are now y 1 to y 2 This gives 2 2 11242ydy197222dy374 318 where I assumed that in the nal step you can work out such a simple standard integral 22 With the parametric representation of the path m 2222 22 1 and y 1 222 we have dm 4u1du dy2udu and on the curve the parameter 22 goes from 0 to 1 Substituting all of this 317 becomes 1 I 2u2u11u24u11u272u27u712u du is 319 0 where again I omit the steps by which we get the nal answer from a simple standard integral 312 LHVE SURFACE AND VOLUME HVTEGRALS 29 iii For the third path each section is simple since either z or y is held constant but we have to do them separately One the rst segment y is constant at y 1 and dy 0 so that the second term in 317 is zero The integration limits are x 1 to z 4 so this part gives 4 hnm 1 On the second section z is constant at z 4 and dz 0 so that the rst term in 317 is zero The integration limits are y 1 to y 2 so for this part we have 2 12 9 4dy 1 and the sum of the two parts gives Ihh7 8 3123 More On Line Integrals If the curve is very kinky it may not be representable as a simple function or In that case it can be subdivided or else one should use a parametric representation We note that in general as we saw in the example above the value of the line integral depends on the particular path taken between the endpoints Physicists often deal with the case where it does not depend on it more on that later but one can t assume that Some notation One can consider a line integral which ends up at the starting point Then there really is not beginning and end point of the curve it is just one path but the sign of the result does depend on the direction of the path the convention is the counter clockwise sense whatever that may mean for the particular problem I ii a dr For a closed path we write eg There are situations where an integral on a closed path necessarily gives zero 3124 Line Integrals in Physics So far you have seen the de nition of the work done by a force F of a particle which moves on a path on the curve C W Fdr C For some forces ie the conservative ones this is equal to the negative of the change in potential energy Ampere s law in electromagnetism involves a line integral of the magnetic eld over a closed path jig B dr HO O 30 CHAPTER 3 VECTOR ANALYSIS where I is the total electric current passing through the loop And ao is a constant If there is no net current though the loop7 then f B dr O The force on a nite length of a currenticarrying wire is FI 1pr C and the force on a closed current loop is FIji 1pr C 3125 Green7s Theorem in a Plane A possibly useful theorem is the case the a closed path C in a plane say the my plane encloses a region R Then we have 86 8P jl lPdm Qdy 7 7 dzdy 3126 Conservative Fields For a certain class of vector elds7 the line integral ffa dr has an important property It is independent of the path C taken to get from A to B7 and as a consequence7 the integral 3 a dr around closed loop C is zero A very important theorem tells us that the following conditions on a vector eld a are equivalent 0 The integral ff a dr is independent of the path taken from A to B Equivalently7 the integral 3 a dr around any closed loop C is zero 0 There exists a function 1 of position such that a V gt oVXa0 313 Surface Integrals ds adS ade S S S While the meaning of these is pretty clear7 we realize that there s a remaining ambiguity about the direction of the surface element 18 Making one choice or the other amounts to a change in overall sign for the surface integral which is not really profound but it s useful to have some sort of convention For a closed surface7 there is an outward and an inward direction for the surface vector and the choice will be clear For an open surface7 we are often also thinking of evaluating a line integral on the boundary of the surface7 which is a closed path In that case7 when we choose the direction in which we want to go around the closed path and the direction of the surface vector 18 is given by the right hand rule7 as shown in Fig 34 Possibilities are 313 SURFACE HVTEGRALS 31 Figure 34 Direction of surface Vector goes with the sense of the line integral on the boundary by the righthand rule Figure 35 Setting up surface integral for an arbitrary surface 3131 A General Method While a surface integral is fairly easy to do if the surface in question is one where a coordinate is constant for a curvilinear system for example7 the conical surface where 9 is constant in spherical coordinates we would like to have a method good for an arbitrary surface Consider the situation shown in Fig 35 The strategy is to project the surface S down to the my plane where a region R is swept out eventually we want to do the integral over R A path of surface S has surface element 18 which makes an angle 04 with the z direction The projection of this area element in the my plane is 1A These in nitesimal areas are related by dA cosadS If n is the unit normal to the surface for this patch that is7 dS dS then COSOz k or dA AdA n m k and d8 7 k From our study ofthe gradient7 we know that if the surface is given by the equation m y 2 0 then the unit normal at any point in given by 15 A Vf n 7 lVf l With this7 the rst of the expression for 15 can be written dA lV dA W dA d5 A k W T n a 32 CHAPTER 3 VECTOR ANALYSIS The last expression can using the de ning relation for the surface be written as a function of z and y and in this way the original integral involving a vector eld and the surface S can be made into an integral over the region R in the my plane As a simple maybe too simple example of the general method we nd the area of a hemisphere of radius a The hemisphere is centered at the origin and we consider the part with z gt 0 On this surface S we want to evaluate f5 dS Of course we have to get half the area of a sphere 27ra2l The simplest way to do the integral is to use spherical coordinates as S is a surface on constant T The surface element for a constant r surface is dwa r2 sin 0 d0 d gt which on the surface where r a has magnitude a2 sin0d0 01 For the upper hemisphere 9 goes from 0 to 7r2 and 1 goes from 0 to 27139 The area is then 27r 7r2 7r2 Sda a2s1n0d0d 27ra2 sin0d0 S 0 0 0 The sin integral is easy and we get S 27139a22 47ra2 Now we work the example by projecting onto the my plane The function de ning the surface is fmyzm2y2227a2 0 which gives Vf2mi2yj22k2r gt lV 2lrl2a Also g222 a27m27y2 gt dSMA 82 ltgt 2 a22y2 And using this the integral over S becomes an integral over the region R a circle of radius a in the my plane AreadS lV dA S R 81082 Rxazizziyz At this point we realize that this integral will be much easier to work using plane polar coordinates p gt for which the area element is pdp 01 This gives pdpdtj 27ra71 a2 7 p2 Area a 27ra7107 a 271112 0 27r 1 0W 313 SURFACE HVTEGRALS 33 X Figure 36 lee cream cone surface for the example 3132 Integrals 0n Coordinate Surfaces If the surface of integration is one where a particular coordinate of a curvilinear system is constant the integral is usually much easier and we make use the area elements dai for the surface in question An example that will give us lots of practice is part of a problem from our EM text Example For the vector eld V r2 sin0 473 cos0 9 rztan0 evaluate the integral 3 V 18 for the closed ice cream cone77 shown in Fig 36 The top surface is spherical with radius R and centered at the origin The surface here has two parts the cone part is a surface of constant 0 namely 9 7r6 on which r runs from 0 to R and 1 runs from 0 to 27139 The area element for this part is dae rsin0drd gt e 97r6 r sin7r6 dr d e The unit vector e does point outward for this surface which is what we want The surface integral picks out only the e part of the eld V we get Amnev39ds fwdersinltw6gtlt4r2cosltw6gtgt R 4 4 R 3R 427rsin7r6 cos7r6 rgdr 87139 g T i 0 Now do the ice cream77 part of the surface which is a surface of constant T ie r R on which 9 runs from 0 to 7r6 and 1 runs from 0 to 27139 Here the area element is da r2 sin0 d0 d gt H R2s1n0d0d 34 CHAPTER 3 VECTOR ANALYSIS Then the surface integral for the cap is 27r 7r6 v dS d d0r2 sin 0R2 sin 0 S cap 0 0 7r6 R427r sin20d0 27rR4 7 cos0 sin0 0 0 7 7r6 0 41 7quot 27rR i g The integral over the whole surface is the sum of the integrals from the two parts7 thus x3 x3W 4 8 12 LEAwa 3V3 dS2 R4 iv 7 12 3133 Applications in Physics Possibly the most famous appearance of a surface integral is in Gauss7 law7 a relation between the electric eld integrated over a closed surface and the total charge contained within that surface Eds s 60 ln general7 a vector eld dotted with the surface area element and integrated over a surface is known as a ux for a general surface f5 E 18 is called electric ux 3134 Surface Integrals 0n Spherical Surfaces ln general7 for problems in spherical coordinates we are evaluating integrals of the form 27r 7r fr0 gtr2dr sin0d0d gt 0 0 314 Volume Integrals Possibilities are adV V V The last case can be split up as adViadejadekade V V V V Again7 note that we can factor out the unit vectors for Cartesian coordinates because they are constant Obvious examples of volume integrals in physics are the total mass and total charge of an object M messltrgtdv c2 Vphltrgtdv 314 VOLUME INTEGRALS 35 Another is the moment of inertia of a mass about a particular axis if 31quot is the distance of the mass point at r from the axis then the moment of inertia is 7 I39SI39Z Iimessm ltgtdv 320 Example Find the moment of inertia of a uniform sphere of radius R about an axis through its center For a mass element at r 0 gt the distance from the axis is 31quot rsin 0 The volume element in spherical coordinates is 1V r2 sin0 d0 d gt and for the integral over the volume of the sphere r goes from 0 to R 9 goes from 0 to 7139 and 1 goes from 0 to 27139 Then if the uniform mass density of the material is p then using Eq 320 we have 27r 7r R 7r R I rsin02r2sin0d0d 27rp r4sin30drd0 0 0 0 0 0 For the second step we have brought the constant p outside the integral and done the j integral which gives a factor of 27139 For the t9 integral we have 7r 1 sin20sin0d0 1735mm 0 71 where we have changed variables to z cos0 and then used sin2 9 1 7 2 This is then 9 2 4 gt zig 127gg Ther integral is just R 5R 5 z4dr R 0 5 0 5 Putting all of these factors together we have 7 87rpR5 I 15 but to express it in terms of the mass M of the sphere use 7 M p 7 g nRg which gives 5 I 87rR M gMRZ 15 gwRS 36 CHAPTER 3 VECTOR ANALYSIS 315 Integration Theorems 3151 The Divergence Theorem Given some vector eld ar and a volume V bounded by a closed surface S we can evaluate the volume integral fV V adV and the surface integral 3553p 18 These give the same result VadVjia 18 321 V S lf 1 and 1 are two scalar elds one can show i gtW 7 we ds My 7 WM W 322 S V One can also show for a vector eld b V gtdVjli gtdS bedVjlideb V S V S 3152 The Divergence Theorem and Physics Suppose we know that for some elds 1 and a and for an arbitrary volume V gtdVjziadSVVadV 323 where the second equality is necessarily true from the divergence theorem Now if the rst and last expressions in 323 are equal for any volume V then we are free to choose a very small volume V centered at the point r Over such a small volume the values of the integrands are essentially constant giving us gtdV V a or gtrVa In this way an integral relation between scalar and vector elds has become a di erential relation In this way the Maxwell equations of electrodynamics have both integral and differential ex pressions One of those equations is Eds lprdv s 60 60 V where qtot is the total charge enclosed by the surface But the divergence theorem allows us the write the surface as a volume integral and so 1 VEdVMi prdV V 60 60 V Since this is true for any volume we can equate the arguments of the volume integrals at any value of r r V E 60 Also from electromagnetism is the principle of the conservation of charge If J is the current density then the integral 3 J 18 gives the outward rate of passage of electric charge through a 315 INTEGRATION THE OREMS 37 closed surface S Because of charge conservation this rate of the charge ow is equal to the rate at which charge is lost from the interior of S Again if qtot is the total amount of charge inside S then this is stated as flds idqmt 1 prtdV 7 dV 5 dt dt V Vat Using the divergence theorem the surface integral is made into a volume integral giving VJdV7dv V V13t and since is true for any volume V we can give the differential relation 80 VJa 3153 Stokes7 Theorem For this theorem we consider an open surface S bounded by a closed curve C and a vector eld a As discussed before when doing integrals on both the surface and its bounding curve we want to be consistent with our choice of direction for the area element 18 and the sense in which we go around the curve C We agree that we ll follow the right hand rule Having agreed to this we have an equality between a surface integral and and a line integral andSjl adr 324 One can also show the related theorems with scalar eld gt SdeV gtjziC gtdr SdeVxajziCdrxa A special case of Stokes7 Theorem is for a plane surface S and its planar bounding curve C Here the surface elements 18 only have a 2 component so that the surface integral in Stokes7 Theorem is fSV x azdS Also the curve C is limited to line elements dr dz 1 dyj so that Stokes7 Theorem gives 802 8am 7 7 7 d5 d 1 Max 8y if may 3quot But this is exactly the form of Green s theorem in a plane given earlier with P replaced by am and Q replaced by ay 3154 Stokes7 Theorem and Physics The most famous example ofthe use of Stokes7 Theorem in physics is probably the relation between magnetic elds and currents known as Ampere7s Law To use this law consider a set of currents such as might be owing though a set of wires and consider a closed path C which might loop around some of those wires and a surface bounded by that path The currents give rise to a magnetic eld B and on the curve C we can calculate the line integral f B dr lf em is the total current enclosed by the loop then we have B dr More 325 C 38 CHAPTER 3 VECTOR ANALYSIS But if we draw a simple surface S which is bounded by the curve C then the current passing through C is the same as the integral of the current density J over this surface 15mJds S Also by Stokes7 Theorem the line integral in 325 can be made into a surface integral on S icBdrSVXBdS VxBdS MOJus S S The fact that this holds for any surface S lets us use an argument similar to when we used for Gauss7 law to conclude that the integrands here are themselves equal at all points in space thus Combining these gives VXBLLOJ which we might call Ampere s law in differential form A word of caution is needed here the preceding discussion is only valid for static ie constant currents and charge densities ie what we call electrostatics If either of these changes and then original expression of Ampere s law needs changing and then so does the differential relation The more general version is one of the true Maxwell equations 316 The Dirac Delta Function 1 had this section here because the delta function is now becoming a standard part of the under graduate curriculum and because one of its uses is to resolve an inconsistency in the use of the divergence theorem 1 will follow the approach of Grif ths in him EM book So you ll see it again but it will be easier then 3161 A Paradox Consider the vector eld in spherical coordinates if we integrate this eld over the surface of sphere of radius R centered at the origin we get 27r 7r1 iFdS jrzdrsin0d0d gt S 0 0 7 But if we take the divergence and throw away all caution we get VF718ltr21gti00 m 7 72 37 72 72 47139 326 R T and the integral of the divergence over the interior gives VFdV0 m 327 V 316 THE DIRAC DELTA FUNCTION 39 But by the divergence theorem the two integrals in 326 and 327 should be equal shouldn t they So is 47139 really equal to zero The problem is in the volume integral in 327 The divergence of F is indeed zero everywhere except the origin where the original function and its derivatives are really unde ned That s only one point but it s enough to render the divergence theorem unusable the functions have to be nicely behaved everywhere in the interior so this is not really a paradox merely a bad experience from not listening to the mathematicians We will nd that we can get out divergence theorem back with a bit of unconventional mathematics 3162 A Sequence of Functions We consider the set of functions de ned by 0 l fam gti lllt This function is a step of width g and height so that the area under the curve is 1 for any value of a but as we consider increasing values of a the function becomes taller and thinner What happens in the limit as a 7 oo Loosely speaking the function turns into a spike which has in nite height zero width but maintains its unit normalization actually the function has no limit But when the function is used inside an integral it does Consider the integral of our faz s times some well7behaued function 900 00 lim dm 1700 700 As a gets very large and faz becomes very thin the only contributions to the integral come from values of x close to 0 that is from 90 In the limit the factor 9z can be taken outside the integral as the constant 90 leaving 00 lim 00 dm 90 7 dm 90 1700 7 If we give the limit of the set of functions a name M 7 11330 12100 then we can write We can put the long thin spike anywhere else 6m 7 c fam 7 c and we then nd 00 gltzgt6ltz 7 c dz m 00 so that the function 6z 7 0 picks out the value of 9z at z c More generally b 9 fl g67cdm 90 1 lt clt a I 0 otherwise 40 CHAPTER 3 VECTOR ANALYSIS 3163 Three7Dimensional Delta Function Generally we will be integrating in more than one dimension with functions of r 3 coordinates In Cartesian coordinates the function which picks out the value a well7behaved function at the position c is 63r39 7 C 6m 7 cm6y 7 cy6z 7 CZ which gives 0 otherwise Agonwgoa 7 CdV 9c ifV includes c 3164 Resolving the Paradox Now we return to the suspect volume integral of Eq 327 We want to patch it up so that it gives 47139 in accordance with the divergence theorem We really need to patch up the result for V It is zero everywhere but at the origin where it is not de ned But we can using the delta function we make a de nition for all points of the interior If we de ne 47r6gr 328 then we get for the sphere of radius R or in fact any volume V which includes the origin VV72dV47r63rdV47r Chapter 4 Complex Variables We have covered the most basic facts about complex numbers in the Preliminaries section In this part of the class we explore functions of complex variables and then doing calculus derivatives and integrals of various sorts with these functions One might think that we just have a simple extension of the basic calculus of real functions7 but that turns out not to be the case Again we need to re do rstiyear calculus 41 Complex Numbers Basic Operations A complex number is formed using the imaginary number i which has the property 2 2 71 It is also true that 702 71 For real numbers z and y a complex number is z m iy and we write x Rez and y lmz The complex conjugate of a complex number 2 denoted 2 and in some books by E is the same as 2 but with the sign of the imaginary part changed zziy gt 2z7iy The modulus of z denoted by 2 is zmiy gt jzjvx2y2 The sum and product of the complex numbers 21 1 HM and 22 2 iyg are 2 1 2 2 951 952 2242 242 2122 961952 241242 i1241 902242 which gives 22 90 224W 7 2 24 902 242 M2 The complex conjugate operation distributes over sums and products at 21 i 22V 2T i 2 212W 2T2 1 2 2 2 2 All complex numbers except for 0 have a multiplicative inverse so that as usual division just means multiplying by the inverse When dividing7 it is often useful to get the result in the form x iy This can be done by multiplying top and bottom by the complex conjugate of the denominator 4 4 4 7 901 2241 902 i 2242 7 901902 241242 Z902241 901242 2 2 902 2242 902 i 2242 952 92 42 CHAPTER 4 COMPLEX VARIABLES 411 Polar Representation Relation to Trig Functions The exponential function has a important role in the theory of complex variables It is de ned by n 51 expz Z 7 41 n0 39 where O 1 One can show that the sum in 41 converges for any 2 Two more properties of real exponentials can also be shown to hold here 511512 511 and 51 em for positive integers 71 If 2 is purely imaginary say 2 2 0 with 9 real one can show that the sum in 41 gives ei9lt 7 mgtilt 7 wgt which we de ne as 9 el cos0isin0 so that cos 9 and sin0 now have rm mathematical de nitions as 3 5 to to 2 4 l cos0217 s1n0E P an From this multiplying by the real number r gives rem rcos0 isin0 rcos0 irsin0 z iy so that any complex number can be represented as i9 27 with rz and tan0a 13 and we choose the proper quadrant for 0 for now choose 77139 lt 9 g 7139 We note that without some kind of restriction there is no single value for the exponent since 2 rem Tei92mr 7 this ambiguity will give us major headaches shortly 13992 7 In polar form multiplication and division are easy for 21 rlewl and 22 r25 i9192 and ei9192 22 72 2122 71725 With the above properties of the complex exponential we very easily get for real 9 and integer ewe cos 9 isin0 cos 710 ism 710 a result which sometimes is called De Moivre s theorem For n 2 it gives us cos2 9 7 sin2 0 i2 sin0 cos 0 cos 20 i sin 20 which gives the doubleiangle identities cos 20 cos2 9 7 sin2 9 sin 20 2 sin0cos0 41 COMPLEX NUMBERS BASIC OPERATIONS 43 In fact7 many of the identities of trigonometry can be shown much more easily using complex exponentials It is true that for real x7 and complex 0 one can show 1 dx 50x 660x a fact which we ll use below in working out an integral but as for taking a derivative with respect number 2 as in digs we will have to be much more careful about what this operation means more on that later One can show that the equation 2 for positive integer n has n roots7 given by 252mn where k012n71 412 Applications to Basic Integrals Example Evaluate the integral I f5 cos bx dx While the standard techniques can be of use here7 we will use complex variable to get it into a form so that we just need to know one of the elementary forms We note that we can write the cosine in the integrand as the Real part ofa complex exponential this gives 5 cos bx Ree cos bx i sin bx Ree eib Ree ibm With this substitution7 the integral is ax aib 61 I e cosbxdxRe e xdxRe 4 0 a 2b and we re done But we want to do some algebra to get a more convenient form aib 7 4 I Re 5 4 a 1 a 2b a 7 2b a2 b2 Re a 7 ibe wcos bx i sinbx from which we can read off the real part of the stuff inside the parentheses to get 6041 I W acosbx bs1nbx C Had we taken the lmaginary part of the part we would have arrived at 041 5 sin bx dx 26 2 7b cos bx asin bx a 44 CHAPTER 4 COMPLEX VARIABLES 413 Hyperbolic Functions 7 A Review Maybe Since we will see a close relation between the trig and hyperbolic functions when we consider functions of complex variables we will do a review of the hyperbolic functions of real variables lf scant attention was paid to them in your calculus class this review will be useful The hyperbolic functions sinhm cosh z etc are de ned as m 7w m 7 7w h m 7 7w coshm i sinhz 1 tanhm M 1 42 2 2 cosh z em 5 with the others given by 1 1 1 cosh x h 7 h 7 th 7 sec m coshm CSC m sinhz CO m tanhm sinhz The hyperbolic functions obey some identities which look very much like corresponding ones for the trig functions watch out though because there is usually a difference of sign someplace Some of these include cosh2 z 7 sinhz z 1 1 7 tanh2 z secth cochz 7 1 cscth sinhz1 i 2 sinh 1 cosh mg i cosh 1 sinh 2 coshz1 i 2 cosh 1 cosh mg i sinh 1 sinh 2 t h i tanh ml i tanh 2 an x x 1 2 1 i tanh 1 tanh 2 The inverses of the hyperbolic functions are denoted by sinh 1 m cosh 1 z tanh 1 z etc Here y sinh 1 z means z sinhy Note for the inverse cosh function one must make a choice of sign we choose cosh 1 z 2 0 It is possible to get a closed form77 for the inverse hyperbolic functions One can show cosh 1 z lnm sz 71 sinh 1 z lnm sz 1 tanh 1 z lnlt1 gt 17x The derivatives of the hyperbolic functions are similar to those of the trig functions but as always watch the signs coshz sinhz sinh z coshm anhz sechzm lt608h 1 gt ltsinh 1 gt w lttanh 1 gta232 414 Complex Powers and Logarithms We note that for real variables the equation y 51 can be inverted so that for a given y it has one solution x lny But we found out that different complex exponents of 5 can give the same number namely those which differ by multiples of 2m 2 rem rei927r rei94 Because of this the complex exponential does not invert so easily 42 FUNCTIONS OF COMPLEX VARIABLES 45 42 Functions of Complex Variables 421 Notation Just as with functions of real variables we let the independent variable be called 2 the dependent variable called 3 with y m the most common choice of letters for functions of complex variables is 2 for the independent variable and the dependent variable called w with w As usual7 2 ziy and w will also have real and imaginary parts7 for which we write w uiv 422 The Elementary Functions sini2 isinh2 cosi2 cosh2 tani2 itanh2 sinhi2 isin2 coshi2 cos 2 tanhi2 itan2 2i 1722 sin 12 lni2 V17 22 cos 12 1n2 V 22 71 tan 12 l ln lt1 Examples Evaluating functions of complex variables Some examples I came up with to illustrate how your calculator knows what to do with complex variables i Find cos27r Use the de nition of the cosine function7 i2 7i2 i27ri 7i27ri cos2 E gt cos27r In the numerator of the last expression we can pull out the factors 27ri 1 e and 5727 1 and combining the factors ofi in the exponents gives e 1el 1 1 521 7 Ti lteggti 2e 1543 ii Find cosh27r Here it might be easiest to use the angle addition77 rule for cosh7 cosh21 22 cosh 21 cosh 22 sinh 21 sinh 22 to get cosh27r cosh27r coshi sinh27r sinhi cos1 cosh27r i sin1 sinh27r where in the last step we ve used cosi2 7 cos2 and sinhi2 isin2 46 CHAPTER 4 COMPLEX VARIABLES In the last form we can use a 10 calculator to get cosh27r z145 i 225 iii Find lni 1 Write 1 H in polar form 1 7 mwew so the multi valued ln function is ln lt 5i2mgt Ln V a 27m k 0 1 2 As our calculator will choose the branch with k 0 this gives ln1i Ln xE 2 3 x 03466 2 07854 7 iv Find sin 117 Here use the closediform form formula for the sin 1 function Get sin 117 7 llni17 7 M17 mm 7 llni17 7 1697 Z Z You ll note here that chose the positive sign to go with i for the square root ofthe negative number The other choice would give another branch this choice is the one our calculator takes Reducing it further 1 1 1 gt 71n3397 71n3397e 2 7 Ln 3397 Z Z Z where for the log we choose the principal value again as our calculator will do Finally gt gLn 3397 x 157 76352 V Find 27 Comment on the difference with 23 8 From the de nition of powers we have 27f 57ran But ln 2 is multivalued 1n2 1n2emk Ln2 23927rk k 0 1 2 This gives 1 I 2 2 EWlLDZHWl em zem k z 882cos27r2k isin27r2k which has many different messy values So why doesn t 23 give the same problems Solving that problem in the same way gives 23 63Ln25i237rk 63Ln25i67rk eBLnZ 1 8 The integer power gives only one answer 43 DERIVATIVES OF COMPLEX FUNCTIONS 47 423 Basic Types of Functions 424 Branch Points and Branch Cuts The simplest example of the complications which arise in multi valued functions and what to do about it is the function f2 212 which among friends can be written f2 We already know that here there is a rather arbitrary choice to make for the sign of the function For real variables we specify that we take the positive root of the equation yz 2 when we write y but for complex variables we always want to consider the entire complex plane so the choice is not so clear We want a function of 2 to change in a continuous way when 2 is changed continuously We consider a trip around the origin in the 2 plane and the corresponding path in the w plane 43 Derivatives of Complex Functions What does it mean to take the derivative of the function f2 that is what do we mean by f 27 As with real variables it does mean to take the limit of the ratio f2 A2 e W T 43 as A2 gets small but more precision in the de nition is required for complex variables because A2 can get small in many ways whereas the real Am has only one way to get small It might have only a real part and shrink that or just an imaginary part or shrink both parts at once If the limit in 43 exists irregardless I say irregardless of the way that A2 goes to zero then we say that f2 is differentiable at 2 If f2 is differentiable at all points in some region R of the complex plane we say that f2 is analytic in R 431 CauchyiRiemann Equations One can show that if w f2 142 y ivz y is analytic in some region R then in R u and U will satisfy the CauchyiRiemann equations 814 81 814 81 13 8y 89 78 44 432 Rules for Differentiation Fortunately if we have a set of functions which are differentiable we can use the familiar rules for derivatives that we know from elementary calculus These now look like maww gmggo f zg z gem 21242 c z d gimp2 miclt2 Mfgre impz govlt2 If wltltgt and ltgltzgt then die dCf ltgltzgtgtcltzgt 127g 48 CHAPTER 4 COMPLEX VARIABLES Figure 41 A complex line integral It is also true that the derivatives of the elementary functions of complex variables have the same form as the ones we know from elementary realivariable calculus For example7 d 7 d d izn nzn 1 fez eZ idz dzlnd dz dz dz d i d i d 2 7s1nzcosz icosz7s1nz itanzsec z dz dz dz 433 Singular Points If a function fz is not analytic at a point7 the point is called a singular point or singularity of the function We can distinguish various types of singularities that is7 give the reasons why the function fails to be analytic For isolated singularities the types are c Poles These occur basically because there is a polynomial in a denominator which is zero at some point We can distinguish the order of the pole if we nd an integer n such that lim z 7 z0nfz A 74 0 2420 then the point zo is a pole of order 71 If n 17 the point is called a simple pole 44 Complex Integration Cauchy s Theorem We consider a curve C in the complex plane with endpoints a and b With zO E a and zn E b7 This process is illustrated in Fig 41 From this curve we select a sequence of n points 20a7217227Zn7172nb and evaluate the sum f k2k 21H Z f kAzk 1 k1 M w H 45 CAUCHY S INTEGRAL FORMULA 49 We then consider the limit where n a 00 and the size of the largest increment Azk goes to zero If this limit exists7 it de nes the line integral of fz along the path C from a to b fzdz Tler Zf kAzk 0 mac k1 Cauchy7s theorem says that if fz is analytic in a regionR and on its boundary C7 then fzdz0 Yet one more surprising property of analyticityl It follows from Cauchy s theorem that if a and b are any two points in R then ab 102 dz is independent of the path in R joining z and b It also follows that if a and z are any two points in R and Z Glt2gt no dc 1 then Cz is analytic in R and G z A more useful statement ofthis property is If a and b are any two points in R7 and F z z7 then b 102 dz Fan a M 1 so that when the integrand is nice and analytic in the region of interest7 integration is just as easy as it is for real variables 45 Cauchy s Integral Formula fa i 2 dz 45 27110270 x fWa 012 n123 46 46 In nite Series Taylor and Laurent Series 23 25 22 24 sinz27 7 cosz 75Ei 2 3 3 5 ln1zzii tan 12277 1 1 7 1 1MP1p222mwyrm 50 CHAPTER 4 COMPLEX VARIABLES Figure 42 Contour for the integral f0 dz Contour encloses a pole at z i 47 The Residue Theorem The residue associated with the singularity a which in our examples is nearly always a pole of some order is the coef cient of Zia in the Laurent series about a In general we can nd the residue using the general formula kil 1 l zsak z 47 For a pole of order 1 we can use the simpler formula L1 lim 2 7 afz 48 24111 The residue theorem says that if fz is single valued and analytic inside a simple closed curve C except at the singularities a b c inside C which have residues 11 b4 01 then fl 102 dz 2ma1 51 124 49 C Example Do the integral 00 1 7 dm 0 m2 1 This one of course is elementary and can done without complex variables but it will provide a check on the method First we note that since the integrand is an even function of m we have 1 7 1 1 0 m2172 700m21 and the latter integral can be expanded into a contour integral 3 1211 dz which runs along the real axis from 7R to R and then returns along a huge semi circle of radius R as shown in Fig 42 The big semi circle is not part of the original integral and will have to be dealt with later The integrand 2211 has poles at ii and clearly they are simple poles Only the pole at z i lies inside our contour so we just need to nd the residue there The simple pole residue formula gives using the residue theorem i 4 1 1 Fl h lm l 2 47 THE RESIDUE THEOREM 51 then the residue theorem gives 1 4 41 if dz 2ma1 7 27112 7 7139 But we now have to deal with the curved path we will do it in detail here and not go into as much detail in the other examples On the curvy part we have 2 R519 with 9 0 7 7139 On this part of the path7 1 Rzezw 1 1 1 f2R252i91 R262i91 In the denominator of the last expression7 use an inequality for complex numbers 121 2212 12111221 to make the denominator smaller and hence make to whole expression bigger 1 1 1 lt lt 7 7 3252191 7 Rzezle 1 321 At this point we realize that in the limit R 7 00 big numbers 1322 is certainly less than R2 7 17 so that our inequality is now w gt1 lt 1 2 z 7 7 R22 R2 and7 calling the curvy path P the length of P is 7rR and the integral on P is bounded by 2 27139 F W dz lfzlmaer FR E which vanishes as R 7 00 Thus the integral on the curved part contributes nothing and we are done with the problem We have shown md 7 00d 7 diil 1 0 m21 2 Ooz21 2 22 2w 2 Is that right You bet it is 00 1 d t 1 0 7 an 7 1 0 27171m 0 2 2 But of course the new method will give results where we don t know how to do the integral by elementary means Example Do the integral Here the the integrand is not an elementary form and though one might be able to do a complex factorization of the denominator and use partial fractions 7 this may not be worth the effort7 and in fact this one is probably easier to do by contour integration 52 CHAPTER 4 COMPLEX VARIABLES iure 2 onour or einera 77 z onourencoses 0esaze39 anze 39 Pg 430 t f th tglel d C t 1 pl t quot24 d 24 Again since the integrand is an even function of x we will use 1 1 1 7 dm 7 7 dm 0 m4 l 1 2 700 m4 1 with the intention of extending the second integral to a contour in the complex plane The integrand has poles at the values of z where the denominator is zero This occurs where 4 z 71 or Z emAl7 637139147 657147 e77rl4 of which the rst two are in the upper halfiplane If we enlarge the contour to include a large semi circle in the upper halfiplane to close the path we note that it encloses two poles This is shown in Fig 43 Again if we can deal with or better ignore the integral along the curvy part then the residue theorem will give us the answer Calculate the residues at the two poles Pole at 57 It is clearly a simple pole so use 39 1 1 1 7 7 7rl4 1 7 ail T ALISAZ 5 gt24 1 13314 423 453mm Pole at calm4 Again a simple pole so use 39 1 1 1 1 7 7 37rl4 1 7 ail 7 JigAZ 5 gt24 1 21324 423 4591174 45114 Note in both cases l Hospital s rule was used for the limit Using our results for the residues at the two poles the residue theorem gives the value of the contour integral on C jig24 dz 2mza1 2m 5 3quoti4 aim4 Then a little trig gives 1 7 41 1 4 1 1 4 1 7 M 4 7 7quot A dz gml z 1 z gt KW E We have to worry about what happens on the curvy part but in fact because the denominator contains an even higher power of 2 than the last example the integral will vanish in the limit 47 THE RESIDUE THEOREM 53 Figure 44 Contour for integration example and the poles of the integrand The poles are at z O z 7 and z 72 R a 00 for the same reason So the integral on the whole contour equals the integral just along the real axis Finally the value of the original integral is 00 1 1 0 1 1 17139 7r 0 mm Emmdggil f dzi m e m Example Do the integral 27r cos 30 d0 0 5 4 c0s0 The trick here is to make the real integral into a contour integral by changing the integration over 9 0 a 27139 into a contour around the unit circle via the substitution 25 gt dziei9d0izd0 We will also need 1 3m 73w 3 73 c0s0 and c0s30 and the path of integration C in the complex plane is the unit circle as shown in Fig 44 Making these substitutions we get 2 cos 30 4 23 EU2 dz 1 23 2 3 dz 0 C 0 7 7 54c0s0 52z2 1i2 2i C522222 Then factoring the denominator and multiplying top and bottom by 23 gives 2 cos30 d0 1 261d2 0 54c0s0 22 023221z2 which can be evaluated by the residue theorem
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