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by: Dr. Simeon Wiza


Dr. Simeon Wiza
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This 93 page Class Notes was uploaded by Dr. Simeon Wiza on Wednesday September 9, 2015. The Class Notes belongs to PHYS 227 at University of Washington taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/192452/phys-227-university-of-washington in Physics 2 at University of Washington.


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Date Created: 09/09/15
Lecture 6 Review of Vectors Physics in more than one dimension See Chapter 3 in Boas but we try to take a more general approach and in a slightly different order Recall that in the previous two lectures we used the twodimensional vector analog to study complex numbers The concept of a vector is much more general Vectors give us a notation for handling ordered sets of numbers generally real numbers but we can also consider complex valued vectors The rules for handling vectors are rules for manipulating these sets of numbers Making use of our experience with vectors in the real world we can often make use of explicit geometric interpretations of these manipulations to guide our intuition Note that the numbers which represent a specific vector can have different forms depending on our choice of basis vectors this is very similar to the use of the rectilinear representation versus polar representation of complex numbers e g for 3D there are 3 standard choices x yzrectilinear Cgt r6 spherical Cgt p zcylindrical 61 Also note that the quantities preserved during manipulations of vectors are a property of the geometry For example while the individual components of a vector are changed by a rotation the length of the vector is not changed in spherical coordinates r is not changed by rotations but the other coordinates are changed Such invariant quantities specify the symmetry properties of the system ie the geometry Symmetries play an essential role in the understanding of physics An N dimensional vector or Nvector corresponds to an ordered set of N numbers arrayed linearly Thus we can use a single index to label the individual elements or components in this linear array with the index running from 1 to N 2i ltgt AgFM 62 The symbol c is being used here to signify the fact that a vector with the over arrow label 21 is associated with an Ntuple of ordinary numbers and vice versa The two expressions are not strictly equal The lefthand side of this expression is meant to be abstract while the righthand side is concrete Once we have defined how to add and subtract these N dimensional arrays and multiply by a constant we have defined a Ndimensional represented often by ND linear vector space see 310 in Edition 3 and 38 in Edition 2 of Boas Physics 227 Lecture 6 1 Autumn 2008 Let s consider a 3D example 17 lt3 131213 xyz Consider first multiplication by the constant c c ltgt cxcycz 63 Next consider two vectors which we can add or subtract component by component 171i 172 Cgtr11 i rum2 i r2111 i r23x1i x2y1i yzz1 i 22 64 Just as in the case of 1D vectors the familiar scalars objects unchanged by rotations addition and subtraction of vectors of higher dimensionality exhibit the properties of being associative and commutative for addition they can be performed using any grouping and in any order r 7 73 r imir 2 i72 i173associative 71 72 72 71 commutative but Iquot 72 72 i 65 To further define our vector space we need to be able to say something about the lengths of vectors and they relative directions ie we want to say something about the geometry of the space For that purpose we define products of vectors There are two types of multiplication of vectors ie two types of products which find many uses in physics The first is the scalar or inner or dot product which no surprise results in a scalar quantity ie a single number that is unchanged by rotations It is written as 3 r1 39 r2 E xixz yiyz Z122 Z ikr2kgt 66 k and is commutative and distributive but not associative 1 r1 2 r2 r1 commutatlve r 72 r3 5 5 distributive 67 i1 The scalar product provides a simple way to calculate the length or norm of a vector variously written as H l A 1 7 71 not associative V 2 Physics 227 Lecture 6 2 Autumn 2008 392 7 2 W 68 The geometric interpretation of the scalar product is 7 277 7172 7172cos6w 69 where 612 is the polar angle between the directions of the two vectors Thus for vectors with nonzero length the scalar product vanishes 171 172 0 if and only if 912 7r 2 or 3712 ie if the vectors are orthogonal Note that Eq 69 which is valid in any number of Euclidean dimensions is guaranteed to satisfy our expectation that cos 612 S1 by the familiar Schwartz Inequality N N SJZ f Z k 610 kl kl Also note that if we consider complex valued vectors ie the components rlyk are Iii72i72cos612 N 2 7111 2k k1 allowed to take complex values we want to still define the scalar product in such a way that vectors have positive real lengths Hence in the complex case we define I 3 3 v v Ii 3973 2 IlJchJc 0r 2 riJcerc a k1 k1 ASIDE 1 If we want a picture of what is happening with the scalar product we can think of the second vector as being concretely represented by the column vector of its coordinates for some set of basis vectors while the first vector is represented by the complex transpose Hermitian conjugate of its coordinates ie a row vector T r 1 2 1 11 Ha T T r2 Cgt r12 3 l 1 Il 12 1143113 r13 13 Then the scalar product is in the form of usual matrix multiplication of a row times Physics 227 Lecture 6 3 Autumn 2008 a column element by element 21 gt gt 3 i39rz 211131123113 r22 Zrljcrzjc39 613 a 1 k1 39 r23 ASIDE 2 Strictly speaking when we define the scalar product we should consider the general bilinear form see Chapter 10 in Boas defined in N dimensions in terms of an NXN matriX or metric gkl we will introduce a little matriX notation here that will be useful later N N f 72 EZZnkgklru 614 kl l 1 We should think of the metric as defining the geometry of the vector space For the familiar rectilinear coordinates defined by fixed orthogonal unit basis vectors more on this later the metric is just the unit matriX and we obtain the earlier expressions The geometry defined by the unit matriX is called a Euclidean geometry 0 0 0 0 gklz i 615 1 0 0 1 kl As you may know from your studies of special relativity the metric describing 4D spacetime has minus signs either 1 minus and 3 plus signs or 1 plus and 3 minus signs ie there are 2 different conventions in broad usage and the corresponding geometry is called a Minkowski geometry In general relativity the metric has even more structure and gravity arises as a result of the response of the metric to the presence of a nonzero energymomentum density ie the metric is treated as a dynamically varying quantity The second type of useful product is the cross or vector product and results in a new pseudo vector ie the resulting quantity is another Ntuple that changes under Physics 227 Lecture 6 4 Autumn 2008 rotations like the usual vector but does not change under re ection as we will discuss later and is special to 3D also called an aXial vector fix 72 ltgt ylz2 221le2 22x1x1y2 x2y1 616 The vector product is not commutative 71x72 sz71 72X 617 ie it is antisymmetry However it is associative and distributive in the sense that 713972X73Ixf23973723973xf1 gt gt gt gt gt gt gt 618 r1gtltrzr3r1gtltr2r1gtltr3 Note the essential feature that the 123 ordering is maintained in each expression ASIDE A particularly handy notation for the vector product of 3vectors employs the unique completely antisymmetric 3tensor the LeviCevita symbol in 3D which is defined by gklm Slkm 8kml gmlk 5123 1 3 V1 396 X rs Z Sklmrlkr2lr3m9 619 klml 3 2 X 3k Z Sdmr2lr3m E Sklmr2lr3m l m1 The final expression introduces the common notation that repeated indices are understood as being summed over The geometric interpretation of the cross product in Eq 617 is that it defines a vector with magnitude given by compare Eq 69 lixlelfllllesin9m 620 Physics 227 Lecture 6 5 Autumn 2008 where again 612 is the polar angle between the directions of the two vectors The vector direction of 171 x 172 is orthogonal to both I71 and F ie orthogonal to the plane defined by 171 and Recall that any 2 nonparallel vectors de ne a plane in which they both lie The sense of the vector along this direction is given by the infamous righthandrule ie the direction a right handed screw would advance if we tumed it in the direction defined by the cross product rotating 171 into E The cross product of two nonnull vectors vanishes if and only if the two vectors are parallel ie 612 0 or 7 The cross product of a vector with itself vanishes 171 x 171 6 as is clear from the antisymmetric expression in Eq 619 Note that the fact that this cross product defines a unique 3rd vector is special to N 3 In higher dimensions there will be more than one direction orthogonal to the plane defined by the original 2 vectors The socalled scalar triple product in the first line of Eq 618 171 x has a particularly simple geometric interpretation as the volume of the parallelepiped which has the three vectors as contiguous edges This geometric interpretation makes clear why the triple product involving only 2 vectors e g 171 x must vanish because the parallelepiped lies in a plane and has no volume The vector 71 x 172 is orthogonal to both 171 and This reminds us of the essential connection between vectors as Ntuples of numbers and the associated N dimensional linear vector space Pursuing this idea of an N dimensional vector space we note that the strict definition of such a space requires along with Eqs 63 and 64 also the existence of three intuitively reasonable quantities l The null vector 6 such that 21 6 21 for all vectors 11 in the vector space 2 The unit constant 1 such that 121 21 for all vectors 11 in the vector space 3 The opposite or inverse vector l such that 121 zzl 6 for all vectors 11 in the vector space With these definitions we can define the extremely important concept of linearly independent vectors Consider for example three nonnull vectors21l3 in our Physics 227 Lecture 6 6 Autumn 2008 Ndimensional vector space These three vectors are said to be linearly independent if and only if the equation cIZcZEc3C6 621 allows only the trivial solution c1 c2 c3 0 Note that once we pick basis vectors Eq 621 is really a set of N equations one for each of the components clAk csz C3Ck 0 for kl2N ASIDE As a basic first example assume that we can nd 2 orthogonal vectors 1 and B such that 21 3 0 ie the geometric notion of orthogonality being at rightangles means that the scalar product vanishes Then it should be clear that the equation clI cf 6 has only the solution c1 c2 0 To see this explicitly first dot 21 into the equation yielding c1 llzll2 0 3 c1 0 Then dot B in yielding c2 Blz 0 3 c2 0 In fact all we really need for linear independence of the 2 vectors is that they not be parallel 1211 lt Now the really interesting question is for a given vector space what is the maximal number of linearly independent vectors The general approach is to start with some minimal set of vectors that allow only the trivial solution to Eq 621 and then keep adding vectors until a nontrivial solution is unavoidable For any finite dimensional vector space this will happen eventually and in fact serves to define the dimensionality of the space In particular if the set of vectors 17k k l N are a maXimal linearly independent set in a given vector space then for any vector 17 in the space we must be able to find a nontrivial solution to N gt chvk ar 2 0 k1 1 N N gtV chvk Ezrkvk a k1 k1 622 In other words we must be able to write any vector in the space as a linear combination of the vectors in this maXimal linearly independent set of vectors Thus we can associate the Ntuple rk with the vector 17 and this must serve to uniquely Physics 227 Lecture 6 7 Autumn 2008 and concretely define the vector Hence the vectors 17k k 1 N serve as basis vectors for the space ie they can be used to define all vectors in the space Another way to express this fact is to say that the vectors 17k k 1 N span the space It is conventional for a simple geometric picture but not necessary to choose the basic vectors of the N dimensional vector space to be orthogonal and of unit length ie the unit basis vectors you used in introductory physics a Ok l vkv121kzlz5kl 623 This last quantity 51 carries the name Kronecker Delta Function and is just the NxN unit matrix 1 s on the diagonal zeros elsewhere of Eq 615 1000 0100 1 22 22 20 10 01 kl The matrix in Eq 624 which defines the scalar products of the basis vectors and thus the geometry is usually called the metric as noted earlier Orthogonal basis vectors can always be found using for example the GramSchmidt method see Section 310 in Boas It is conventional to define special symbols for the orthonormal orthogonal and normalized basis vectors For example in 3D with a rectilinear basis we define 17 51 iforfjl cgtjl 0j1 IE A A A A A A 625 lgtltjk jgtltl k A xj f xfj A j Physics 227 Lecture 6 8 Autumn 2008 The words labeling the choice of the signs in the cross product are that i jk or AC J92 form a rightedhanded set of unit basis vectors In terms of the corresponding 3tuples that we first discussed we have 3LOJDj c 0L0 2 30JL1 626 The 3tuple ie the coordinates corresponding to any other vector can be found from the appropriate scalar product xzf sz rcgtxyz3y17fI7j 627 zi2f In terms of these explicit unit basis vectors and components we can eXpress the 3D cross product in terms of the familiar determinant of 3X3 matrices We have 12x1xy1yzl rzx2xy2yZ2Z JAC A 2 r1gtltr22 x1 y1 21 x2 y2 22 628 A yl 21 A x1 21 A x1 y1 x y Z yz 22 2 22 x2 y2 ylzz 2122e 21x2 x122 x1y2 y1x22 The 2X2 submatrices are referred to as the cofactors of the unit vectors This form clearly agrees with Eq 616 ASIDE For completeness we recall that the determinant of a 3 X 3 matriX can be expressed as Physics 227 Lecture 6 9 Autumn 2008 a b c a b c f d f d detd fd e fzae b c e hi g 139 g h g 1 g h 1 h aei fh bdi fgcdh eg Note that given a choice of speci c basis vectors and the corresponding components we move from the discussion of relatively abstract vectors to the more concrete discussion of components We can rewrite various properties of vectors in terms of ordinary equations involving their components The statement 171 172 0 is the statement that the 2 vectors are orthogonal without reference to any speci c basis vectors Once we have chosen a basis set ie we have both the vector space and a metric in that space we can write for orthogonal vectors the symbol J stands for perpendicular or orthogonal 171172203171JJ723x1x2y1y221220 629 Likewise 2 parallel vectors are de ned by a vanishing cross product which we can rewrite in terms of components in 3D as fix 3ylzz21y2z1xzx122xly2y1xz0 3 i 1 i Ca constant 630 x2 y2 22 3 12Cr2 We can pictorially represent a vector as the line from the origin 000 to the point de ned by the components xyz where lfl le y2 z2 is the distance from the origin to the point xyz Thus there is a oneto one mapping between points and vectors from the origin to the point once we have chosen an origin Both are speci ed by an Ntuple of numbers More generally we can think of this vector as the difference between two other vectors one from an arbitrary point to the point x yz and the second from the same arbitrary point to the origin ie the point 0 0 0 Xy2 Physics 227 Lecture 6 10 Autumn 2008 Next let s think about how we de ne various geometric gures in a 3dimensional space using the quantities above The simplest as a 0D object the point de ned by x yz which also de nes a vector once we choose an origin Next consider a 1D object the straight line We can uniquely de ne a straight line by specifying 2 ordered points as above or by specifying both the direction of the line modulo the sense and a point through which the line passes For example a straight line passing through the point x0 32020 ltgt I70 and parallel to a vector 17 In general we expect a 1D object to involve a single continuous parameter which we will call I here and which identi es the continuum of points along the line We know from Eq 630 that all vectors of the form t are parallel to the vector 17 We can thus de ne a line passing through the point x0 32020 by de ning a vector as a ll lCthl l of the parameter I in the form Fr70 x7r 631 As the parameter tvaries from 00 to 00 the points de ned by the vector 17t trace out the desired straight line Note that in general neither the vector17tnor the vector I70 are along the line but rather describe the location of points on the line with respect to the origin The obvious physical interpretation is the trajectory of a free particle with initial conditions7 t 0 170 17l O 17l 17 constant the linear dependence on the product t is what guarantees a straight line Note that this de nition in terms of vectors works in any number of dimensions In terms of components in 3D we represent a straight line by the equations compare to Eq 630 the form here ie the number of equations depends on the number of dimensions If For example with 120 ltgt I70 and 17 25c 2 we have 17t l 2IJE 2 t2 If we want to de ne a line in terms of two points E and 17 we can just substitute 17 171 I70 in the above equations so that the analog of Eq 632 is Physics 227 Lecture 6 l 1 Autumn 2008 xx0 yyo ZZo t x1x0 y1y0 21 20 633 The next most interesting but simple object is a at 2D surface or plane In 3D we can define this object in terms of passing through a point x0 y0zo lt3 170 and being orthogonal to a vector JV the normal vector to the plane Note that this latter constraint produces a 2D surface only in a 3D space In N dimensions such a constraint produces an N ldimensional subspace From Eq 629 it follows that the required surface in 3D is given by the equation F FOJ70gtx x0Nx y y0Ny z 20N 0 634 Since this equation is a single constraint on the 3 parameters x yz it clearly defines a 2D 2parameter surface or subspace Since it is a linear constraint the surface is a at plane think about that the normal vector A7 is the same vector everywhere on the surface Similarly we can use three points x1y1zl x2y2zz x3y3z3 to define a plane We just use the 3 points to define 2 vectors in the plane by taking differences and then use their cross product to define the normal vector to the plane Substituting into Eq 634 then yields a triple scalar product form Using the first point as the reference point this is an arbitrary choice we have the defining equation as F aa axa a0 x x1y2 y123 Z1 22 Zly3 y1 y y122 Zlx3 x1 x2 x1Z3 21 Z Z1x2 x1y3 y1 y2 y1x3 x1 The reader is encouraged to verify that this expression is invariant under permutations of the indices 123 These expressions also illustrate the fact that vector notation is typically much more efficient ie simpler to express if not to use than the more concrete component notation recall our motto As a specific example let x1y1zlx2y2zzx3y3z3 be lll230Ol 2 Thus the normal vector to the desired surface is 635 Physics 227 Lecture 6 12 Autumn 2008 2 9 2 Nz agtxa a21 3 1 0 1 01 1 1 2 1 6 19 222 3 4 2 636 The equation de ning the plane is then writing the triple product in determinant form xl y l z l 17 171N0 3 2 1 6x18y 1 2z 1 1 0 3 6x8y 2z l20gt3x 4yz60 637 In physics we are often interested in the minimum or perpendicular distance between a point and a straight line or between a point and a at surface With the point speci ed by the vector 71 with respect to some arbitrary origin and the line speci ed by an equation as in Eq 631 with 170 de ned with respect to the same arbitrary origin we can calculate the perpendicular distance by taking a cross product of the direction 3 with the vector from 71 to any point 7 I along the line recall that the cross product yields the orthogonal component while scalar product yields the parallel component In the gure the desired distance is the length of the unlabeled dashed line which is clearly 171 7sin6 Thus we have the perpendicular distance from a point to a line as l171 I7 lgtlt17 dpoint line H v 638 We take the modulus because the sign is irrelevant here and we divide by M since we only care about the direction of the line and not the magnitude of the vector VA Note that as expected the distance does not depend on the parameter tthat de nes a speci c point along the line For example if the point is de ned by 111 ltgt 71 and Physics 227 Lecture 6 13 Autumn 2008 the line is the one above t l 2t5c 2 t 120 lt3 170 17 2 25 3 we have a perpendicular distance defined by Fl Ft 2t5c 31t2 Sc 3 2 171 tx17 2t 1 1z1 22z 2z922 2 0 1 639 y22f222 m 3 d pointline m J Another useful distance is the perpendicular distance from a point to a at plane In this case we obtain the required distance by projecting onto the normal to the plane If I71 is the vector to the point of interest with respect to an arbitrary origin 17 is the vector to any point on the plane with respect to the same origin and JV is the normal to the plane the required perpendicular distance is yi V d 640 2x m pointplane N Note that we have accounted for the fact that the normal may not be defined as a unit vector Consider for example the point abovelll lt3 171 and the plane de ned in Eq 637 above by the equation 3x 4y z 6 O with normal A7 2 35c 4 2 Since we can choose any point in the plane ie any point that satisfies the equation for the plane we can take 17 lt3 00 6 Thus we have f f272 34 7 12 641 You should convince yourself that you will obtain the same distance starting with any other point on the plane e g 200 Physics 227 Lecture 6 14 Autumn 2008 As a final geometric application of these ideas consider the line defined by the intersection of two planes in 3D Since this line must lie in both planes it must be perpendicular to both normal vectors ie the 2 normal vectors to the planes Thus the direction of the line is provided by the cross product of the 2 normal vectors To fully specify the line we need only 1 point common to both planes and then we can apply Eq 631 For example consider 2 planes defined by the equations x 2y3z4 and 2xy z5 with normalvectors A71 5c 2f232 and A72 2 2 3 2 the reader is encouraged to verify that these definitions are consistent By inspection ie solving 2 simultaneous equations we see that the point 170 lt3 14 5 3 50 lies in both planes and we have R 2 2 1 2 3 5c7j25227 2 1 1 642 Ft t5 7tj 5t where the final expression defines the desired line of intersection Physics 227 Lecture 6 15 Autumn 2008 Lecture 8 Vectors and Matrices 1H Applications of Vectors and Matrices See Sections 32 33 and 38 to 312 in Boas As noted in the previous lecture the notation of vectors and matrices is useful not only for the study of linear transformations but also for the related problem of solving systems of simultaneous linear equations Consider N simultaneous linear equations for Nunknowns xk k 1N A11x1 A12x2 39quot A1N71xN71 A1NxN X1 81 Ale1 AN2x2 39quot ANN71xN71 ANNxN XN Clearly we are encouraged to think in terms of a NxN matriX representation for the coefficients A a N D vector of unknowns xk and a N D vector of known inputs X1 Then using the notation we have developed to simplify our lives we can write A X 82 As long as the matriXA is nonsingular detA 5 0 we can nd its inverse and solve the simultaneous equations in matriX form C A is the cofactor matriX for A as defined in Eq 729 in the previous lecture a N N CAT N XC aZAilX AilX l k x xquot quot detA Edam detA 83 detA xk In the last expression the matrix A is the matriX A with the kth column replaced by the column vector A X A A 11 NH 1 1k1 1N 84 A Nkil X N A Nk1 Physics 227 Lecture 8 1 Autumn 2007 We recognize that Eqs 83 and 84 de ne Cramer s rule see Section 33 in Boas Next let us consider the more general problem where Eq 82 corresponds to M equations for N unknowns The vector X now has M components the vector x has N components and the matrixA is MxN M rows and N columns The question to ask first is when does this system of equations have a solution This corresponds to determining how many linearly independent equations are present in this revised version of Eq 82 To answer this question we perform the process known as row reduction ie add subtract rows times constants tofrom other rows and interchange rows See Boas 32 to find a new version of the matrix A with as many zero elements as possible in the lower left hand corner below the pivot This version of A is called the row echelon form Having performed this process to the maximum extent possible we count the number of rows in the reduced matrix still a MxN matrix with nonzero elements Call this number M 39 and label it the rank of the matrix A where clearly M S M The rank counts the number of linearly independent rows in A and thus the number of linearly independent equations in the original system of equations ASIDE We can apply this same idea of linear independence to a general set of functions not just linear functions Consider the determinant of the matrix defined in terms of N functions fk k 1 N and their derivatives with respect to x fllx f2x fzrvx WXEfISX fZSX fNSX 85 fiN71 ngil fIEINil This determinant is called the Wronskian of the functions At values of x where W x i 0 the functions are linearly independent in the sense that recall our discussion of linearly independent vectors N chfkx030k0klN 86 k1 No linear combination of the functions vanishes except the trivial one all zero coefficients in the range of x where Wx 0 Physics 227 Lecture 8 2 Autumn 2007 Returning to the problem of simultaneous equations we define a new MxN1 matrix the augmented version of A which has the column vector 3 as its N 1 column All AlN X1 AAug 87 AMI AW XM Next we row reduce this matrix to determine its rank Call this last quantity M Aug and note that M Aug reduced to zeros in AAug can certainly be reduced to zeros in A Now we can list the cannot be smaller thanM M S M Lug S M every row that can be various possibilities I If M 39 ltMjmg the rank of A is smaller than the rank of AAug inconsistent with each other and there are no solutions to the set of simultaneous equations the equations are 2 If M M1 mg N SM we can eliminate M M nonindependent equations rows from the system of equations and arrive at the NxN problem we discussed at the beginning of the lecture The resulting NxN matrix of coefficients we label ARed the reduced matrix Since we have eliminated all the rows that have only zero elements we are guaranteed that the determinant of the reduced matrix ARed is nonzero and that an inverse matrix Aged exists Hence we can use this inverse ie Cramer s rule to find a unique solution to the system of equations ie a unique value for the N D column vector 56 AgedX 3 If M M Aug unknowns ie there are not enough independent equations to determine all of the unknowns Rather the solution to this system corresponds to solving for M of the unknowns in terms of the other N M unknowns and the original coefficients and the known vector 3 lt N the reduced problem has fewer equations than there are As an example consider the equations Physics 227 Lecture 8 3 Autumn 2007 x y 22 7 7 1 1 2 2x y 4z 2 2 M 4 a 2 2 1 4 2 X A 5x 4y 102 2 1 N 3 1 5 4 10 88 3x y 62 5 5 3 1 6 Proceeding to row reduce we have 1 2 1 1 2 A 0 3 0 0 3 0 W W R375R1 0 9 R4743R2 0 0 0 8399 R473R1 0 4 0 0 Hence we see that the rank of A isM 2 Proceeding with the augmented matrix we have 1 2 7 1 1 2 7 A 2 1 42 0 30 12 gt g 5 4 10 1 g 0 9 0 36 3 1 6 5 R4731 0 4 0 16 0 3 0 12 W R4743R2 0 0 0 0 0 0 0 0 Thus we have M39 M39 Aug solve for 2 unknowns in terms of the third From the reVised equations we have 2 lt N 3 and the problem is underconstrained We can x y 22 7 x7 y22322 3 3y 2 12 y4 8 The unknown 2 remains a free variable The choice of 2 instead of x to play this role is of course arbitrary Physics 227 Lecture 8 4 Autumn 2007 Now return to Eq 82 and think about it in a slightly different way What if the rightsideside 3 is proportional to the unknown vector 3c ie can we find a vector 7c such that after we perform the linear transform represented by the matrix A we end up back at 55 times a constant As we will see the problem of finding vectors that are invariant except for a overall constant under a given transformation is extremely importantuseful in physics here we move to Section 311 in Boas or Chapter 10 in the previous edition of Boas We want to consider the following new version of Eq 82 Ai2igtA lli6 812 Note that this version is a homogenous equation If the matrix A M is nonsingular and A My1 exists we know that the above equation Eq 812 has only the trivial solution iA M 166 813 Thus to obtain interesting solutions and we want to do that we require that the inverse not exist or that detA ll 0 814 This condition the eigenvalue condition also called the characteristic equation for A is an Nth order equation for the parameter 2 generally having N distinct solutions which may be complex Note that the eigenvalues are a property of the transformation represented by A and not of the vectors being transformed On the other hand for each specific eigenvalue 2k there is a corresponding vector the eigenvector k which satisfies AV Mk 815 Since this is a homogenous equation the length of 7k is not specified if 7k is a solution so is C17k where c is a constant and typically we choose to make it a unit vector to simplify the analysis 17k T 17k 2 17k 17k l Another way to see this point is to note from Eq 814 that only N l of the N equations in 815 are linearly Physics 227 Lecture 8 5 Autumn 2007 independent You should see this explicitly as you solve 815 For complex a T a a a gt a vectors the normalization choice is typically vk vk vk T vk vk vk 1 which leaves an overall phase undetermined ie this constraint is invariant under 17 gt e k k 39 We can derive a few important properties of eigenvalues and vectors by considering a pair of distinct eigenvalues lkjtl and the corresponding eigenvectors rail for a general complex matriX A They satisfy AVk lkvk AV llvl 9T eTn v1 Tl gt v Avk lkvl vk V1sz 37Vsz 816 1 gt l T T T 3Vl Avk va vl 27vkvl Of special interest is the case when the matriX A is Hermitian A 2 AT symmetric if A is real which is typical in physics applications especially Quantum Mechanics For Hermitian matrices the difference between the second eXpression on the second line and the last eXpression in Eq 816 vanishes Thus for HermitianA we have ngv 43 vjA Atv 0 227 t77 kl77k 0gttk l0 817 gtxlt at 39 x17 tk Ogtvkvl 0 The first result tells us that the eigenvalues for a Hermitian matriX are real while the second result says that the eigenvectors for distinct eigenvalues for a Hermitian matrix are orthogonal These results will be very important in your studies in Quantum Mechanics They also suggest a general way to find orthogonal vectors to use as basis vectors It should be clear that our analysis would be simplified if we choose to have our basis vectors aligned with the eigenvectors so that the components of the eigenvectors are of the form 17k 1 001 0 and the matriX A in this basis is diagonal with the eigenvalues along the diagonal We tum now to achieving this goal Physics 227 Lecture 8 6 Autumn 2007 Assuming that we have found the eigenvalues and their eigenvectors in an arbitrary basis we construct the following socalled modal matrix not the cofactor matrix V11 V21 VN1 C 52 V192 3 818 ie each column of the matrix C is composed of the components of one of the eigenvectors in the initial arbitrary basis Next consider the result of multiplying on the left by A 1N vN2 819 If we de ne next the desired diagonal matrix with the eigenvalues along the diagonal ll 0 0 12 DE 39 g 820 0 0 quotLN it follows that Physics 227 Lecture 8 7 Autumn 2007 E E E 821 31V1N 12VZN lNVNN Combining Eqs 819 and 821 while assuming that C is nonsingular ie that the eigenvectors are all independent and they will be for distinct eigenvalues and can be made so by hand even if there are degeneracies in the eigenvalues we can define the inverse of C and arrive at the important relation CD 2 AC gt C IAC D 822 Note that the general transformation of a matrix M defined by C IMC 2 C IMCkl i Cganan 823 mnl is called a similarity transformation where we can think of this transformation as operating separately on both indices of M C 1 from the left and C from the right The corresponding transformation of vectors is 3 gt YC C 1 The specific similarity transformation defined by the matrix C in Eq 818 takes us to the basis system where A is diagonal Note that in the definition of C the choice of the normalization of the eigenvectors plays no special role For any choice of the normalization of the eigenvectors the matrix C will still provide a similarity transformation that diagonalizes the problem However for general normalization this transformation will not be just a rotation 6 3 Example 1 Consider the 2x2 matrix A 2 I which is real but not symmetric and thus not Hermitian AT 2 AT i A Hence we do not expect the eigenvectors of this matrix to be orthogonal to each other We can find the eigenvalues directly from Physics 227 Lecture 8 8 Autumn 2007 6 1 3 2 1 1 For a 2D matrix we can also use the general results that 212 721202 32 4 824 TrA Tr1A TrCC 1A TrC 1AC TrD ilk detA det1A detCC 1A detC 1AC 7 N detD Hlk kl 825 to solve for the eigenvalues ASIDE The interested reader should prove that these results are true in any number of dimensions It follows from simply writing out the definitions in terms of components and noting the cyclic permutation symmetry We have for example TrC 1AC i jg1quotka i kaCgAm flaw1quot klml klml lml N N 826 2A 2 TrA TrD 21 11 11 For the 2X2 matriX A above these relations are sufficient to write TrA72122 213 3 827 detA21221gtlt22 124 The corresponding normalized eigenvectors are found from Physics 227 Lecture 8 9 Autumn 2007 x 1 l 23 6 3 3 0gt 1 1gt 21 xl yl yl V1 3 828 2224 6 4x23y20 3 v L yz 2 2 J5 2 where we have chosen to make them unit vectors As noted earlier since this is a 2 D problem there is really a second equation for each eigenvector which is NOT linearly independent from the first in 828 EXplicitly the second row of the matrix gives us 1 1 ll3gt 2x1l 3y10gti lgtvl1 829 124 2x21 4y20gt 3gtv2i 3 y2 2 JB z ie as expected the same result as the first row There is no independent information here Also note that 171 472 3 2 0 As eXpected the eigenvectors are not orthogonal From the earlier discussion the similarity transformation defined by l 3 2 3 f F 1 F F C 1 2 ac 1 1 830 3 F f 3 will diagonalize the matriX A Note that for a 2X2 matriX nding the inverse matriX from the general formula in terms of the cofactor matriX C d C A 1klz detf A 831 Physics 227 Lecture 8 10 Autumn 2007 is particularly easy Recall that the cofactor is given by 1 1 times the minor the determinant of the matrix with the current row and column removed For a 2X2 matriX the minor is the element diagonally opposite Thus we find A A Ail A22 1 Z 11 Ail 12 Ail detA A22 detA 12 detA 21 21 detA39 832 It is straightforward to con rm that for this definition of C we have 3 0 C IMC O On the other hand as pointed out above the normalization and the phase of the eigenvectors is not an issue here We could as well choose 13 71 2 3 C12C 1 1 833 or with a different phase II 1 3 171 2 3 C 1 2 C 1 1 834 1 l 3 2 3 Cm C1114 J26 1 a i 1 1 83935 All 4 of the implied similarity transformations will result in the same diagonal matrix or even For later reference we note that Physics 227 Lecture 8 11 Autumn 2007 detC detC 1 detC 1 detC 1 detC 1 detCH 836 l detC detC39H 26 26 10 3 Example 2 Let us next consider a matrix that is Herm1t1an A 3 2 j Again it is straightforward to find TrA212ll12gt 211 detllllgtlt L2 AZ 1139 837 Here the eigenvectors are 1 1 1 10 1 3 0 11 xl M w 1 3 838 1211gt10 llx2 3y20gtv2 JE r We see as expected for a Hermitian matrix that the eigenvectors are orthogonal 171 39172 VITVZ 3 3 10 0 Thus a possible similarity transformation to the diagonal form is provided by Physics 227 Lecture 8 12 Autumn 2007 C1 1 3 K 3 1 1 3 1 3 839 C1 CZCTj J 3 1 J 3 1 where in the initial step of finding the inverse Eq 832 above was used Another possible choice differing by a phase is Cl 1 1 3 Cll 1 1 3 CIT 313 31 840 In these two cases we have detC l detC 1 841 detC 1 2 detC 1 For the case of a Hermitian matrix with orthogonal eigenvectors we can interpret the similarity transformation that takes us to the diagonal basis as an orthogonal transformation Orthogonal transformations which as we have discussed include both rotations and re ections have the defining feature that they preserve scalar products In particular the lengths of vectors are unchanged by such transformations and pairs of vectors that are initially orthogonal ie have vanishing scalar products remain orthogonal after the transformation These are of course properties that we expect to be true of rotations and re ections If we define the transformation in terms of components by 7 M kll we have a a a a T a WW 2 WW 2 M17 rTMTW 842 gtMTM1gtMT 2M4 An orthogonal transformation is described by an orthogonal matrix which in turn is defined by M71 MT Looking above we see that the matrices C and C in Example 2 the Hermitian starting matrix have this property where M C 1 C T On the Physics 227 Lecture 8 13 Autumn 2007 other hand none of the various C matrices in Example 1 where the starting matrix is not Hermitian have this property It further follows from the de nition that for orthogonal matrices we have detM 1M 1 det MTJdet M det det MT det 843 3 detM i1 The 1 case corresponds to true rotations For example a rotation in the xy plane by 6 radians in the positive counterclockwise direction corresponds to the matrix cos6 sin6 x xcos6ysin6 M9 ltgt 2 844 sin6 cos6 y xsin6ycos6 and det M 1 independent of the speci c value of 6 In this case we are thinking in terms of rotating the basis vectors while the vector 17 with components x y and x39 y39 in the 2 basis sets remains fixed as suggested in the figure Note that using the results of the Extra Credit problem about the Pauli matrices we can write this 2x2 rotation matrix in terms of the Pauli matrices M 0 61629 On the other hand a re ection through the xaxis changes the sign of all ycomponents and is represented by l 0 Ry 0 1 a 845 with det Ry 1 In general an orthogonal transformation that includes both a rotation and a re ection e g the product of M 6 and R y also has determinant equal to 1 ASIDE In the language of group theory see Section 313 and Lecture 10 orthogonal transformations rotations and reflections in N dimensions correspond Physics 227 Lecture 8 14 Autumn 2007 to the group ON while the special orthogonal group SON includes only transformations with determinant 1 true rotations Going back to Example 2 we see that a Hermitian matrix is diagonalized by an orthogonal matrix Further the arbitrary choice of the phase of the eigenvectors and thus the columns of C determines whether this transformation is a true rotation or includes also a re ection For example the transformation corresponding to C is a pure rotation cos9 1 m while the transformation corresponding to C includes a re ection C C Ry To determine the sign of the angle 9 we should carefully write out the formulae In going from the original frame to the rotated diagonal frame we have C IAC D gt 17 C417 ltgt 7 C7 we C4 846 Thus we see that the corresponding rotation has sin9 3 m 3 9 125 radians ASIDE In general we can think either in terms of rotating the unit basis vectors as here with the actual physics vectors fixed or in terms of rotating the physics vectors with the basis vectors fixed where the rotation is in the opposite direction The former is often called a passive transformation while the later is an active transformation In the next lecture we will see how these methods of matrix eigenvaluesvectors can be employed to analyze coupled harmonic oscillators Coupled harmonic oscillators provide a remarkably accurate description of many complex mechanical systems near equilibrium Physics 227 Lecture 8 15 Autumn 2007 Lecture 3 Appendix B Some sample problems from Boas Chapter 1 Here are some solutions to the sample problems assigned for Chapter 110 113 and 1 1 5 1105 Solution So let s look at some power series and determine the interval of convergence using the tools we have learned Consider xquot xquot S Z gt hme an 11m 0 quot1 n 2 Hoe n 12 Due to the very rapid growth of the factorial function this series passes the preliminary test for any value of x Due to the factorial factors the ratio test will be especially useful ie simple because the ratios of factorials are simple We have n1 2 p 11mw lime i2 0x lt 00 x Wm n1 11025 60 Thus the series converges along the entire real axis oo lt x lt oo all x The 50 corresponding function defined by the series looks like 110 11 Solution Using the same techniques for the new power series we have Physics 227 Lecture 3 Appendix B 1 Autumn 2008 l 2 quot 1 l1 3 n5 limwwrl5 plim LMM1imM1 li l n JeOD xISS Hoon15 5 n2 Thus we conclude that for x lt 5 the series converges absolutely The situation at the endpoints is more interesting For x 5 we have just the Harmonic series with an 1n which we know diverges ie from the ratio test p l buts 0 On the other hand at x 5 we have the Harmonic series with alternating signs which satis es the alternating sign test ie alternating signs and terms that vanish in the limit and converges Hence the interval of convergence for this series is 5 S x lt 5 The function defined this way is lnl XS as you can quickly determine by summing the series are asking Mathematical 11325 Solution We want to the Maclaurin series for the function e 1 We proceed by reorganizing using known series and then expanding Physics 227 Lecture 3 Appendix B 2 Autumn 2008 ezx l izx quot1 n 3 15 n1 2x2 x3 2x4 2x2 x3 2x4 2 1 x x 3 3 15 3 3 15 2x2 x3 2x4 2x x3 2x4 x 3 3 15 3 3 15 1 xx2 1x3 1x4 2 21 15 9 3 x2 x4 1 x 3 45 We nd the same series using the Series command in Mathematical Series2xExp2x 1 x04 x2 x4 5 lrxiri Ox 3 45 11327 Solution Here we play with a series in a series quot0 n 6 120 2 6 120 1 x3 x5 3 1 x3 x5 4 1 x3 x5 5 x x x 6 6 120 24 6 120 120 6 120 x2 3 1 1 4 1 1 5 1 1 1 1x x x x 2 6 6 6 24 120 12 120 2 4 5 x x x 1x 2 8 15 co x2m n 2m11 x3 x5 J 1 x3 x5 J2 es Z 1 x x Physics 227 Lecture 3 Appendix B 3 Autumn 2008 This result is easily con rmed with Mathematical Series Exp2Si141x5 X0 5 x x 6 lxiriri0x 2 8 15 115 17 Solution Here we practice summing a series by recognizing it as a Maclaurin series expansion of a familiar function at a special point Consider the series SiE 1 I n0 This series with all even powers divided by the even factorial but with alternating signs we recognize as the series for the cosine So we have l l 522 cos 0 quot0 2n 2 2 It is informative to ask Mathematical to evaluate this series If you ask Mathematical to evaluate the general power series TE2 gt x it will indeed find cosx But if you go directly to the numerical series above it will find a result in temis of Bessel functions see later in this course which evaluates to a small lOAl7 number but not exactly zero This software is very powerful but sometimes it does not do exactly what you expect and can lead you astray You still need analytic tools to check your answers Physics 227 Lecture 3 Appendix B 4 Autumn 2008 Lecture 5 Complex Variables H Applications in Physics See Chapter 2 in Boas To see why complex variables are so useful consider first the linear mechanics of a single particle described by Newton s equation with Viscous damping as appeared first in Lecture 1 a linear restoring force and a driving force m5c39bfckxFt 51 Along the same line recall Kirchoff s equation describing a series RLC circuit with a voltage source With the current I as the free variable coordinate we have IRJILIVt 52 where the first term is the voltage across the resistor the second term is the voltage across the capacitor VC 2 Q C the third term is the voltage across the inductor and the righthandside is the applied voltage Written in terms of the charge on the capacitor Kirchoff s equation is identical in form to Newton LQRQVt 53 ie in both situations we have a linear second order inhomogeneous differential equation due to the driving term on the RHS Such a situation with multiple similar equations is simplified by application of the rule of Feynman one of many rules with the same name which in this case states that the same equations have the same solutions Only the names of the variables and constants have changed in going from Eq 51 to Eq 53 m gtL b gtR k gt lC and F gt V Here we will study Eq 53 but the discussion applies also to Eq 51 With only a small loss of generality as we will see later in the course we can assume that the driving voltage is a periodic function of time Vt Vt r with period I and we expand the time dependence in a Fourier Series more about this later Vt 20 Vquot cos nwot pn where mo 2 27rr the fundamental frequency This set of functions with both a magnitude and phase to be specified for each term constitutes a complete set of functions with the required periodicity any function with this periodicity can be represented as such a sum We will prove this essential point later in the course Physics 227 Lecture 5 1 Autumn 2007 Note that the n 0 terms allows for the possibility of a constant term while the phases are equivalent to including both sines and cosines We focus first on the fundamental frequency term which we can rewrite as Vt gt V1 cosa0t p1 ReI1ei p1eiw 2 Re VIEW 54 We have defined the complex constant VI Vlei 1 which carries the information on both the magnitude V1 and phase p1 of the fundamental component of the driving voltage By the rule of Feynman we can apply exactly the same decomposition to the driving force in the oscillator problem assuming that it is also periodic Ft 0Fn cosnaot pn The corresponding Ansatz i e educated guess for the charge on the capacitor is QI 20 Qquot cosna0t Q5quot where each term requires us to solve for a magnitude Qquot and a phase n ie for the complex constant Qquot Qnei39lquot Qt ZZORe Qne39m of The corresponding current is then given by 1t 20 Re inonnema O Note the 2 symbols for the 2 phases 1 3 n of the applied voltage and of the charge on the capacitor Magic Point 1 The essential feature here is that both Eq 51 and Eq 53 are linear equations in the sense that the free dynamic variable x and Q respectively appears linearly to the power unity in each term on the lefthandside As a result we can use linear superposition to solve the general equation We break up the righthandside of the equation into bitesized pieces individual frequencies as we are discussing here solve the equation corresponding to each piece of the right handside then sum up all of these individual solutions to find the particular solution to the original equation The general solution to the original equation plus initial conditions Qt 0xt 0 and 039ct 0 can be found by summing the particular solution and the solution to the homogeneous problem with zero righthandside Understand this result and you will have mastered much of this course The next big step is to rewrite the real differential equation as a complex equation We can always take the real part in the end to find the desired physical solution For Physics 227 Lecture 5 2 Autumn 2007 now we focus on a single form of the time dependence ie a single frequency for the driving voltage We find after switching to complex notation taking derivatives and canceling common factors that LQRQVt d2 t d 139 z ReQ1 iw0t 139 t LWReQlew JREReQlew Jf2ReIlew d2 iwt d iwt eiw0t iwt LWQle REQ1e Q1C Vle 0 55 2 39 t 39 I Q 6WD 39 t gt La0 Qle w0 1RcoOQ1e w0 1 Vlem 0 2 Q1 iwot gt La0 Q1 1Ra20Q1 E V1 cancelmg the e factor I Magic Point 2 By using complex notation COS wt 2 Re 6 gt 639 and the special feature of the exponential function deafcit 0560quot we have succeeded in converting the dz39 rential equation into an algebraic equation This is a major step in simplifying our task Recall that we are lazy and smart It is now a simple matter to solve for the complex form of the charge and the current Q 2 V1 1 L002 iRa20 yC 11tQ1tgtI1ia oQ1 56 0171 I7 I 1 1 La02 iRa0 yC R iLa20 wOC This last expression plus our previous experience with DC circuits I V R suggests that we define a complex frequency dependent impedance via Physics 227 Lecture 5 3 Autumn 2007 l 1 Z a ERicoL Ri wL 57 iwC 02C where 1 58 R These expressions suggest that the frequency am 2 1 E where Z a7 R and Z a7 0 must play a special role This is just the natural frequency of the LC circuit when there is no real resistance R 0 in the same way that the natural frequency of the undamped oscillator in Eq 51 b 0 is 6H0 km In terms of this parameter we can write Za R22w2 a7 c2 59 L a 2 62 phaseZ Z tan 1 Then starting with the complex form of the applied voltage of frequency a the corresponding compleX charge and current are I7 Q I7 a 1a Qa 510 zm 2mm In particular for the fundamental frequency we have Physics 227 Lecture 5 4 Autumn 2007 0 emcee V1 eiwzwwz 511 where 512 1La a7L2C mOR q z m0 tan Thus for the single driving voltage Vt gt V1 COSa0t p1 the use of complex variables and the 2 bits of magic noted above allows us to quickly solve for the corresponding current in a general RLC circuit We find 1t Re11moeiwozm V1 cos mot p1 Z 610 513 l2 Owl ie the resulting current differs from the applied voltage 0 in magnitude by 1 over the modulus of the compleX impedance evaluated at the driving frequency 0 in phase from the applied voltage by the negative of the phase of the compleX impedance evaluated at the driving frequency Note that the corresponding charge differs from the current by a factor of la0 in magnitude and 7r 2 in phase the charge lags behind the current by 90 ie it takes time for the current to build up the charge Physics 227 Lecture 5 5 Autumn 2007 Q10 2 Re 1moeiwotiltn WCOS QJ P1 Z m0 j V Wmnwot p1 Z 514 Also note that if the driving voltage is given instead by a sine function ie p1 7r2 the resulting current in Eq 513 is also given by a sine function and the charge in Eq 514 will be given by minus a cosine function with all the functions having the same argument wot Z wo Looking at these results we observe first that as expected the response of the circuit has its maximum amplitude when the driving frequency equals the natural frequency this is how radio tuners work where the impedance has its minimum magnitude and vanishing phase ie the driving voltage and the current are inphase while the charge on the capacitor lags the driving voltage by 90 lzwLC 2R lzwl 2R LCa LC 0 515 ii1a7LC w0l lQ1aLC mo V1 R Next we observe that in the limit where the driving frequency is well below the natural frequency and then goes to zero in the limit with p1 0 V1 t V1 l l Za0 gtw0 C gt 00 w0Za0 gt E 6L0 gtgta20 gt0gt 7 71 952 00 gt tan 1OO 2 E Il gtw0CIlcosw0tplj gt0 516 Q1 gtCV1sina0tplj gtCV1 Physics 227 Lecture 5 6 Autumn 2007 which is the expected result for a DC circuit The charge just builds up on the capacitor to match the applied DC voltage In the opposite limit of a driving frequency well above the natural frequency we have Za0 gt wOL 1 COOL Za0 gttan R j 002600 gt a202L gt 00 BE ltlt 00 gt 00 gt V1 71 V1 11t v lm le EJT wOL wot WI ao 517 V1 Q1t gt ng sin a20tqgt1 gt w02L cosa0tp1 gt0 The amplitudes of both the charge and the current vanish in the limit of very high frequency mo gt 00 The circuit simply cannot respond to a driving voltage that oscillates at a frequency much larger than its natural frequency To see what sets the scale for how rapidly the impedance varies with frequency we consider the impedance in the form 518 Then we ask how far m0 must vary from 5m before lZ m0 varies by a factor J5 Za2 Z6LC sz JER This is given by the equation 2 2 2 2 a a 2 2 MzngmngztAmng zmmzltlta7LC 012 L 6LCAa2 L R 519 Amzzi 2L Physics 227 Lecture 5 7 Autumn 2007 The width of the peak typically called a resonance in the quantity 1 lZ mo is determined by the damping in the system This characteristic resonance shape will appear often in physics and is illustrated in the next gure corresponding to L 1 RL 001 and the xaxis is the scaled frequency mELC 120 60 lZw 40 20 09 095 l 105 11 01ch ASIDE 1 Another way to think about the quantity 1 IZ a is to consider where the complex impedance vanishes R 2 R2 Za030ilii mLC E 520 Thus the quantity 1 Z a has 2 simple poles at the complex frequencies in Eq 520 1 im M Lltw azgtltw azgt39 52 Of course in physical applications with a Z 0 and real we can get close only to the pole at a3 We can easily check that the expression vanishes in the limits 0 gt 0 and a gt 00 Z gt oo see Eqs 516 and 517 and goes to its maximum value for real frequencies lR at a 5m see Eq 515 and the reader is encouraged to perform this check The maximum value for real frequencies is not at Rem due Physics 227 Lecture 5 8 Autumn 2007 to the factor of a in the numerator Note that the distance of the poles from the real aXis R2L is just what sets the width of the peak in Eq 519 It is these poles that characterize the solution to the homogeneous undriven equation as we discuss below ASIDE 2 The time average power consumed in the circuit corresponding to the frequency mo can also be easily calculated in terms of the compleX current and voltage 12 idt1tVt V12 11 d 522 Zw0TJ tcoSwot1 Zwocosm0t l V2 1 1 zmcos z woEReU1 which is very much like the DC result except for the factor of 12 from the time average of cos2 mot and the cosine of the phase difference between the I and V If the current and driving voltage are 7r 2 out of phase as when the R vanishes energy is only stored in the circuit and no power energy is dissipated Now let s apply the rule of Feynman to simply write down the form of the motion of a damped harmonic oscillator driven with the frequency mo F t gt F1 cos wot p1 Changing the names in Eq 56 we have F1 F1 x1 mo 2 b ma0 zbm0 16 2 2 m mHO a0 12 Feiwle WHoWo l bmo 523 2 2 0H0 mo HO m0 tan 1 m Physics 227 Lecture 5 Autumn 2007 Again the maximum amplitude of motion arises for a driving frequency equal to the natural frequency 6H0 km 524 7 2 The corresponding time dependence for general mo is given by see Eq 514 Fl X1 2 005wof P1 Ho 600 525 2 2 2 b 2 m COHO a0 E 00 For a general driving force described by a sum over Fourier components linear superposition yields the complete form of the response to the given driving force called the particular solution xpltrgti F n 710 b 2 cosnaot pn HO nmo 526 nza 2 nzaJO2 In Similar expressions with the names changed arise for the general driving voltage in the RLC circuit problem IpltrgtZL quot0 Z cosnaot pn q z 2 cosnaot Pn z n0 R2 Lj 62 2 22 Qpltrgt V 2 sin 14th pn Z 14600 527 quot0 n2a R2 L2 07210 n2 002 8 Physics 227 Lecture 5 10 Autumn 2007 Although Eqs 526 and 527 appear very similar the astute student will recognize that there is a subtle underlying issue arising from our conventional choice to define the complex impedance to be real at the natural frequency while the denominator in Eq 523 is pure imaginary at the natural frequency of the oscillator this is why the expression for Qt which is the closest analog to xt is a sine function instead of a cosine function We close this discussion of the use of complex variables to solve differential equations by considering the homogeneous version of Eq 51 a similar analysis works for Eq 53 with Vt 0 This will allow us to write down the complete solution to such linear mechanics problems including both the particular and homogenous also called the complementary solutions and to fit any initial conditions In particular consider the equation m56bxkx0 528 and assume an Ansatz of the familiar exponential form xH few with f a constant note that in general we are allowing both E and a to be complex As above this leads to an algebraic equation 2 2 ma2bak02aw iii 529 2m 2m m 2m Note that when we translate to frequencies and RLC circuits these are just the positions of the poles of 1Z in Eq 520 a gtiam gtLb gtRk gt1C 2 3 iaa3i i L i 530 7 7 2L LC 2L For the usual case of small damping where b lt 2Vkm the second term is truly imaginary and the motion corresponds to damped oscillatory motion with frequency H0 2 1 km bZm2 S 6H0 We have Physics 227 Lecture 5 11 Autumn 2007 xH t 271 26 5 10 YZe i z Ha eib quot cosa3HOt 7 531 e 1 f4 cos c HOt is sin OSHOt We can use the 2 constants in either expression Eli or 34 or ENE to fit the initial conditions for xH 0 and 5910 Note that unlike the inhomogeneous driven case we are not explicitly taking any real parts here On the other hand when we fit to the presumably real initial conditions we will find a real xH t ie 34 and ibis will be real while Eli will be appropriately complex 1 f 2 f4 139 f1 f2 is In the opposite limit with b gt 2 called the overdamped case we have instead only damped behavior ie the a are pure real and both negative bzm2 km s b2m xH t 261quot fillquot 532 In the special case of critical damping b 2km with 05 2 a7 a b 2m we find the reader should confirm that the following expression satisfies Eq 528 xH z e 6 El 533 In all three cases with b gt 0 the homogenous solution damps to zero asymptotically xH t gt oo 0 Similar expressions arise for the undriven RLC circuit as you may recall from Physics 122 Finally for the general problem with a driving force and initial conditions we sum linear superposition again the homogeneous solutions and the particular solution xt xH z xpl 534 to obtain a solution that still satisfies the inhomogeneous equation Eq 51 that can be matched to the initial conditions using the 2 constants in xH t and that is just the particular solution due to the driving force at asymptotic times after the homogenous solution has damped out This is an extremely powerful result Physics 227 Lecture 5 12 Autumn 2007 Lecture 13 Appendix B Some sample problems from Boas Here are some solutions to the sample problems assigned for Chapter 68 to 611 68 2 Solution We want to practice doing closed line integrals of the form x 2 y dx Zxdj E C d17 clockwise along a set of curves Let us first check the curl of the vector in the integrand We have 2 f2 2 yx 3 3 32 2 2 42 0 6x 6y 62 x2y 2x 0 Thus we expect nonzero pathdependent closed path integrals a On a circle of radius 1 here we can use cylindrical coordinates see example 2 and find 271 S Sx2ydx 2xdy IdQ cosQ 2sin9 sin9 2cos9cos9 0 2 d9 sin9cos9 2 sin29cos29 2 d9 lsin292 47r 0 2 0 b On the square of side 2 with sides aligned with axes and starting at 11 writing out the contributions from the 4 15 sides only x or y varies on each side we find 1 O 5 715rer5 05 15 r O 5 r l 5 Autumn 2007 Physics 227 Lecture 13 Appendix B gx2ydx 2xdyfdxx2fdy2jdxx 2fdy 2 x722x12y1x72 2x1 2y1 4 4 4 4 16 c On the square of side J2 rotated by 45 with respect to the axis and starting at 01 With constant slope on each side we can express both dx and dy in terms of a single variable say dt d i dt i dt Thus we haVe idtx2y2x xt71y7t 7t y17t 43x2ydx 2xaj zjdt x2y2xx idt x 2y 2xxtyH idt x 2y le iwd jdtUt Zt 3t 2t 3 imam10 22 401 8 0 68 4 Solution Now we want to perform an open line integral Iyzdx 2m dz 5 df along 2 different paths Again we check the curl to see C C if there is path dependence expected We find a nonzero curl and expect path dependence 2 y 2 WEE 3 3 0y022 2y22l y7amp0 ax ay 62 y2 2x 1 Physics 227 Lecture 13 Appendix B 2 Autumn 2007 a First we consider a path composed of straight line segments parallel to the axes Thus we have xzl 1 1 1 Iyzdx2xdydz IdeZ 70 IdZIdy2x C 0 yi 0 0 0 1 2 3 b Now along a path composed of an arc of a circle in the xy plane x2 y2 2y 0gtx02 y12 1and then parallel to the z aXis Using cylindrical coordinates on the former x sin6y l cos6 we have 7r2 l Iyzdx2xdydz Ide cosQ cose2 sin62sin6IdZ C 0 0 Tde cos6lcos2 6 2cos2 6 sin2 6 1 0 Tde cos62 sin2 6 2cos 26 1 0 2 1 8 sin26 l2 l 3 3 0 sin3 6 2sin6 68 8 Solution We want to verify a conservative force and find the potential for F 6 ZJA yf We take the curl and find that it vanishes Physics 227 Lecture 13 Appendix B 3 Autumn 2007 5c f 2 6x16 6 6 6 62 11 0200 1 z y 68 11 Solution Now the same game as in the previous exercise except that now F y Sin 2902 Sill2 X39 So the curl again vanishes 2 2 6 6 6x 62 ysin2x sinzx 0 0f022cosxsinx sin2x 0 6x13 KSQ Klt1 We find the potential from Physics 227 Lecture 13 Appendix B Autumn 2007 6gb sm2x ax y a a 5915 2 2 F V gt gsm x gt ys1n xconstant 0 32 68 14 a yJl xzyz x Il x2y2 we find Solution Finally for F 2 7 2 mi 3 3 2 6x y 62 y Il xzy2 x Il xzy2 0 2 2 2 2 1 xy xy 0 A A A 1 x 0 y 0 2 l x2y2 l x2y2 1x2y23 and this requires recognizing the derivatives of the arcsine function 3x F V gt x 1 x2y2 gt sin 1xyconstant 0 32 Autumn 2007 Physics 227 Lecture 13 Appendix B 69 4 Solution We want to practice using the 2D Green s theorem to perform the indicated contour integral We have I I ex cos ydx ex sin ydy CADB 2 ex cos ydx ex sin ydy I ex cos ydx ex sin ydy C ADBA C BA an PlPeX cosy dx QIQ7eX siny J ex cos yly0 dx C ADBA 71112 2 PlPe cosy dx QIQ7e siny ex Eliz C ADBA 2 de Qdy C ADBA 2 Thus applying Green s theorem we find I 6Q6PMy AreaADBA ax a 2 x x 3 3 e s1ny e s1nydxd 0 2 AreaADBA 2 2 2 69 7 Solution We want to use the result in exercise 696 to calculate the area of an ellipse defined by x acosB y bsinQ with 0 S 6 lt 271 The result in 696 is based on the 2D Green s theorem with the special choices P y 2 Q x 2 With our choices to parameterize the ellipse we have dx a sin 6d6 and ab bcos 6d6 to yield Physics 227 Lecture 13 Appendix B 6 Autumn 2007 A d6 ab cos2 6 ab sin2 6 nab 0 69 10 Solution For the path in the xy plane defined by the 4 points 31 51 53 33 we want to evaluate the line integral ltj2ydx 3xdy This expression suggests that we define P 2y and Q 3x and consider the expression in the 2D Green s theorem 6Q6x 613 y 3 2 5 Since this is a constant Green s theorem tells us that we need only the area of the square defined by the 4 points with sides of length 2 which is just 4 Thus the easy approach use Green is 432ydx 3xdy 5 alxdy 20 square On the other hand proceeding directly and laboriously we find 432ydx 3xdy lama idylt 3xgtm idxmu idylt 3xgtx3 2jdx 15jdy 6jdx9jdy24 30 1218 20 3 l 3 l 69 11 Solution Here is one more contour integral over the indicated triangle in the clockwise direction Using Green s theorem we have 7 05 05 1 1 5 2 Physics 227 Lecture 13 Appendix B 7 Autumn 2007 I q xsinx ydxx y2dy Ctriangle 6x y2 6xsinx y High ax ay dxdy l ldxdy Zarea 22 2 triangle 2 610 4 Solution Now we want to practice using the divergence theorem to relate surface and volume integrals In this case the vector field is I7 x cos2 y x2 2 sin2 y2 and the boundary surface is a sphere of radius 3 Thus we find 17116 I Vd r cosz ysin2 ydr d1 r3 r3 r3 r3 47W3 367139 r3 This approach is clearly simpler than during the surface integral directly 610 7 Solution Next we consider a similar problem where the surface is a cone of height 3 and a base of radius 4 and the vector field is I7 I7 x y 22 Thus we find again this the simpler approach Fdaz jvfdrm3dr3mdr nrzh 3 3 487139 h3r4 Physics 227 Lecture 13 Appendix B 8 Autumn 2007 610 9 Solution Now we want to use the divergence theorem to evaluate the surface integral of F XUACJOA1 on the surface defined by z 4 x2 y2 4 p2 I 616171 247x27y247p2 Note the explicit forms for the surface and the function suggest the use of cylindrical coordinates First check the divergence to see if is usefully simple Since v 173 6 6p 2 recall the previous appendiX or proceed in rectangular coordinates ie a constant the divergence theorem will be useful We can close the surface by adding the disk in the xy plane where the outward normal is just 2 2 and 3 disk 3 2 0 This the desired surface integral is just the volume integral of the divergence H oleF dV01vF2ipdp2fd 4fdz 2472 7y247p2 0ltzlt47p2 2 4 2 4n pdp4 p2 47r2p2 J37 16 0 It is informative if challenging to also try evaluating the original surface integral directly We know that the normal to the surface is given by the normalized gradient of the function defining the surface WHPZ 2pm zp2 J4pzl 2 d zida 4432 1 So the remaining challenge is to usefully eXpress the differential area do To this end it is helpful to read Section 55 in Boas where it is pointed out that for the area of a dc d0 2 Physics 227 Lecture 13 Appendix B 9 Autumn 2007 surface above the x y plane we can express the local differential area as do 2 sec ydxdy 2 sec ypdpd where the angle yis the angle between the local normal to the surface as worked out above and the z direction secy 1 J4p2 1 This factor which is 21 just accounts for how much larger the area of the true surface is compared to its projection onto the x y plane Thus we have in our case fl z4ic2 7y247 p2 A zz 2 2P2 2 U do F avgDoll m 43 1 2 47rjp3dp 7rp4 21671 0 Again we see that the actual integration is much simpler using the divergence theorem 610 12 Solution This exercise concerns an electrostatics problem with concentric charged cylindrical conductors with radii R1 and R2 and with k coulombs per meter and k coulombs per meter respectively We can use Gauss s law applied per unit length to find the electric field as function of the radius and then by integration the corresponding potential Clearly we want to use cylindrical coordinates p z in my notation We begin by using the translational symmetry in the z direction the axis of the cylinders and the cylinders are very long to argue that the electric field has no variation with the z coordinate Likewise the rotational symmetry about the z axis means there is no dependence on the azimuthal angle Since we are assuming that this is a static situation with no moving charges there can be no nonzero components of the electric field in the surface of the conductors otherwise charges would move Hence by general arguments we have only a p component and it varies only with p EFEpp So now we apply Gauss for the three cases p ltR1 R1 lt p ltR2 and p gt R2 In the first and last cases there is no net charge per unit length and thus the electric field must vanish Physics 227 Lecture 13 Appendix B 10 Autumn 2007 J dq dz Ep27rp pltR13l E d5EDPPltR1 dez plt1 ZIlidvollpflRIEd Td Ile0xjd2EpltRl0 pltR1 80 o 0 80 pgtRZIdZ pgt mdvoll f d zqu jdzlz xjdstQgtR20 pltR1 80 o o 80 go dq dz Ep27rp pgtR21Epp PgtRz In the interesting region where the charge inside the surface of integration is nonzero we have R1ltpltR2 Ed6Ep27rp R1 lt pltR2 R1 lt pltR2 I d llRIltpltRzd2 d d15 of H IMMIZS 8sz R1ltpltR2 0 0 0 0 0 B 27180 gtER1ltpltR2 Finally to write the electric as the gradient of a potential E p we can simply integrate For the inner and outer regions where the electric field vanishes the potential is a constant and for simplicity we take the potential to vanish at the center In the interesting region we need only observe that J39 d p p 3 ln 1 p const Thus with the given boundary condition at p R1 pR1 0 we find pltRle0p0 R1ltpltR2E k 4 k 1n 5 27180 27180 p pgtR2E0p k In Physics 227 Lecture 13 Appendix B 11 Autumn 2007 To describe the electric field and the potential right at the surface of the conductor we should be careful about the true thickness of the conductor Here we take the idealized limit that the conducting cylindrical tube is of zero thickness Thus the potential is smooth at the conductors but its gradient the electric field changes discontinuously in this limit We should really be careful about how we define the electric field at the conductors39 we must specify whether we are approaching the conductor from inside or outside 6114 Solution Here we want to practice using curls and Stokes theorem We have the vector field V y 2 3 and a surface composed of the indicated 3 triangles ie the boundary is the remaining triangle in the xy plane To use Stokes theorem we need to specify the vertices of this triangle For 2 0 the plane defines 2x 3y 12 and the vertices are 600 and 040 Thus we have yx 2yyd5 d 0amp2ym7 039 triangle 6 0 0 A A 3 29 Iyy0 dx IZdy Iyx 2yy267x3 3 alx 0 4 6 0 826Tx dx 8 jdx 0 2 8 8 x x 3 3 6 8 l612 12 611 7 Solution Now consider the vector field I x xzz yz3 y2 xzy xz2 with the integral of its curl being over any surface with its boundary in the x y plane 2 0 Thus using Stokes theorem we find Physics 227 Lecture 13 Appendix B 12 Autumn 2007 I HVgtltVd6 g Vdf vXVd6 039 60z0 U z0 Thus using Stokes twice we can write the desired integral as a surface integral in the x y plane where d6 oc 2 This last result suggests that we evaluate the curl of the vector field in the xy plane We have 2 y 2 3 3 3 20 ax ay 62 2 3 2 2 x xz yz y xy xzz 0 20 x2 3y229 x2 2xyz O O So we find thatv gtlt I7 615 surface with a boundary in the x y plane we have IHvxI7d6 WW0160 a U z0 0 0 the curl as no 2 component and thus for any z 611 10 Solution Here we again want to practice using Stokes theorem We have a vector field I7 2102 x2 2xf2 x2222 and a surface defined by z 9 x2 9y2 Note that in this exercise we do not have cylindrical symmetry Instead the intersection of the surface with the xy plane 2 0 is an ellipse x32 y2 1 So first we test for simplifications by evaluating the curl Physics 227 Lecture 13 Appendix B 13 Autumn 2007 QJNgt 23 9 g g a 0 2xzz 2x 2 2x22xzz 22 ny x2 2x xzz2 We also see that the local normal to the specified surface is Wzx29y2 2xv218ij2 n Wltzx2 9y2 J4x2 324 2 1 Thus doing the surface integral directly is going to be a bit messy So instead we use Stokes theorem on the boundary which is the ellipse above On the ellipse we can define the parameterization x 3 cos y sin and write vx d6 j Vds Td I 3sm cos y 297 79y2 9y29 qu 3 sin 23cos sin cosq 9cos2 q 6cos 271 3quot dq 6sin2 cos cos2 3cos 2 0 3Td 6cosq 9COS3 2C052 95 3Td l cosz 6 In obtaining the last line we used the fact that the integral of any odd power of the cosine over a full cycle yields zero plus the double angle formula for the even power ie the average value of 0032 is 12 We will have more to say about this when we study Fourier series To complete this discussion let us try to perform the surface integral directly We have Physics 227 Lecture 13 Appendix B 14 Autumn 2007 va secy ZxZZ 22 m J4x2 324y2 1 sz 324 2 l 36xyz2 2 36xy9 x2 9y22 2 368lxy l8x3y 162xy3 l8x3y3 x5y8lxy5 2 This looks pretty ugly with many places to make an arithmetic error On the other hand the actual integral expressed in the x y plane 3 VlrX3Z dxdyzjdx j dy 3 Mid32 is clearly symmetric in both coordinates Hence all of the terms with odd powers of x andor y everything but the 2 integrate to zero So finally we have recall exercise 697 above 9c24r9y2 X 2 6 X I7 d6 2 dx 1quot 3 dy 2area of ellipse 297Z 79yz 3 7 17x3Z 4 M1 x32 W4i3sin2 6d6 73 n 0 4J393 dg 69 2 6 2 7T So we obtained the same answer as above but we had to work much harder and know more tricks Clearly using Stokes theorem is appropriate for smart but lazy physicists 611 15 Solution As a final example consider the contour integral Physics 227 Lecture 13 Appendix B 15 Autumn 2007 ICJ yafxZdyde C with the contour C defined by the intersection of the surfaces x y 2 and x2 y2 Z2 2x y A little thought yields the result that the second surface is a sphere of radius J2 centered on the point 110 x 12 y 12 Z2 2 Thus the intersection with the first surface a at surface parallel to the z aXis is a circle of radius J2 centered on the point 110 To simplify things we evaluate the curl of the corresponding vector field I7 y 29 x2 to find w lt 93b 8 N Rt 8 N Thus the curl is a simple constant vector and we should be able to Stokes theorem We find 143ydxzdyxdzgS drzmvx da C 60 U 2j2 AU 2j2 27r So the final question is to find the normal to the plane of the contour This is just the normal to the plane x y 2 2 32 where we have used our previous knowledge that the equation of a plane is given by 17 170 15 0 with I70 56 f1 So finally we have note that the sign here is actually ambiguous since the text does not specify the sense of the original contour I E 2 27r 227r J5 Physics 227 Lecture 13 Appendix B 16 Autumn 2007 Lecture 14 Introduction to Review of Differential Equations See Chapter 8 Sections 1 to 7 in Boas We will return for the rest of Chapter 8 shortly As we have noted several times most of the physical systems that we are interested in can be described by systems of differential equations Here we want to begin to develop the tool set we need to systematically consider the issue of solving differential equations To begin let s review some terminology Differential equations are labeled ordinary or partial depending on whether the derivatives that appear in the equation are ordinary e g ddx implying a single independent variable or partial e g Eaax implying more than 1 independent variable the issue is who is being varied and who is held fixed we ll return to this issue The order of a differential equation defines the highest derivative present in the equation For example a first order equation involves only first derivatives y x dydx note the fairly standard notation that a derivative with respect to x is represented by a prime while derivatives with respect to time are represented by dots and a 2nd order equation involves y x dZydxz and possibly y39 A linear differential equation is one where each term in the equation involves the dependent variable or its derivatives to no more than the first power e g y by cy f x is a linear second order ordinary differential equation Note that nonlinear differential equations have the special feature of exhibiting isolated singular solutions that are not contained in the usual general solutions with arbitrary constants Nonlinear equations can also exhibit chaotic behavior If every term in the linear equation has a single power of the dependent variable or its derivatives the equation is said to be homogeneous if you multiply y by a constant you are multiplying every term in the equation by the same constant but note that the label homogeneous is used in several related ways see Eq 8411 and exercise 4131 in Boas Thus y by cy 0 is a linear second order homogeneous ordinary differential equation A first order linear ordinary but inhomogeneous differential equation can often be solved simply by integration y39xfxgtdyy39x0bc 3yxldxyrx 141 3 yx Ialex 0 2 ja fff yx0 Physics 227 Lecture 14 1 Autumn 2007 This equation illustrates the expected result that at least in principle we solve differential equations by doing integrals Note also that in the last expression we have been careful to use a dummy variable of integration that is explicitly different from the independent variable x which now appears as the upper limit of the integral We have also made explicit the point that the constant of integration C is just the desired function evaluated at the lower limit of the integral Since the derivative of an integral with respect to the upper limit is just the integrand evaluated at the upper limit see Leibniz rule in Eq 41213 of Boas Imewym fxa am we are guaranteed to have a solution of the original differential equation satisfying the boundary condition y yx0 at x x0 Thus the middle expression in the last line of Eq 141 can be thought of as the general solution while the last expression is the particular solution satisfying the specific boundary conditions and perhaps other conditions Note that to solve a differential equation by simple integration it is necessary that we can separate all the explicit dependence on the independent variable from the dependence on the dependent variable Thus for a first order linear homogeneous equation of the form y a y where a is a constant we must first divide by y before integrating putting all of the y dependence on the LHS y39x ayx gt d y adx y 4quot x 03 gt I ajdfgth1yx axC 143 gt yx y0e quot The final expression has used the boundary value at x 0 This necessary procedure of separating the variables on the two sides of the equation is called no surprise separation of variables and equations that can be solved this way are called separable The most general separable linear first order equation looks like Physics 227 Lecture 14 2 Autumn 2007 y Px y Qx which we can solve using the above ideas For the homogenous case Q 0 we can proceed as above with a not a constant y39 Pxy 0 gt yx Ceiidxpm y0e 7390 Mm E y 0671 144 As expected there is a single arbitrary constant of integration which we use here to fit the boundary condition at x 0 Next we return to the original inhomogeneous problem This is most easily addressed if we consider the quantity yxel and perform some manipulations In particular we find yxe1x e1xy39yl39 em y39 yP eIXQ x 2 yxe lder Qx 5 145 gt yx e1XIdfeIXQ y0e71x 1x Idin 0 We recognize the second term on the righthandside as the previously found solution to the homogeneous equation including the boundary condition while the first term is a particular solution to the inhomogeneous problem A general noneasilyseparable version of the first order equation can be written in the form note the dependence on both x and y ie the P and Q functions are different from above y39x xi PxydxQxyaj 0 146 This expression is written so as to remind us of the 2D Green s theorem and the curl free theorems of Chapter 6 in Boas In the case that aPay aQax the curlfree case we must be able to find another function F x y such that 6P aQ aFxy p 2 2 8y 6x x y 6x Qx y deQdydFO 6F xy 5y 147 Physics 227 Lecture 14 3 Autumn 2007 In this case the original differential equation is at least implicitly solved by Fxy constant 148 This is a more powerful result than it may first seem because even if it is not initially true that aPay aQax we may be able to find a factor to multiply by in Eq 146 so that the new functions 13 and Q do satisfy this relation For example consider PxydxQxydy03 3 X nyx ydxfxyQ90061 E13JcydxQxydy0 149 Q 3y x The factor f x y is called an integrating factor which is just the role played by em above ASIDE To see this connection explicitly we put the equation from Eq 145 in the language of Eq 146 We have ayyPx Qxdx0Pltx y 2Qx 2Mpx wzo 6y 6x Hence we need an integrating factor which is just the familiar eXponential factor We then have Physics 227 Lecture 14 4 Autumn 2007 Thus the corresponding potential function Fxy is given by Fx y constant 2 y0 2 yx y0e 1x 871XJCRQ81X 0 As expected we obtain again our previous result As in Boas consider the example equation xdy ydx20P yQx l l 1410 If we choose an integrating factor f 1 x2 we have 13912921363 12 Q x x 6y x 6x y Fxycyxcx 1411 xdy ydx xcdx cxdx 0 As discussed in Boas equations which do not fall into one of the above simple forms can often be put in separable integrable form by an appropriate change of variable To illustrate these techniques let us consider familiar problems from mechanics Note that while we typically think of mechanics as described by 2nd order equations ie Newton we can often used conserved quantities ie symmetries to obtain a 1st order differential equation Consider lD motion in a uniform vertical gravitational field motion near the surface of the earth which is a conservative system with constant total energy 2 is the dependent variable and t is the independent one In 2 ETOTTV322mgzzziVETOTng39 1412 Physics 227 Lecture 14 5 Autumn 2007 For a given constant total energy we can solve this problem as above by separation of variables We have dtj dz dz tt 2 2 0 1413 quotET0T mgz zo1 ET0T mgz where we have de ned 2 to 20 Integrating the LHS we find 1414 iETOT mgzo ETOT mgz t t0 We can simplify this expression if we evaluate the total energy at to and define the velocity at that time to be z39t0 2 2390 ETOT mz39g 2 mgz0 Thus we have 2 2 i1lz 2ng0 2ng0 ETOT mgz 20 ETOT mgz gtt0 2 3ETOT mgz 23 2gz ZO ZO gt t02 23 2gz 0t t0g2t t02 1415 gt z 20 20 t tO t t02 This final result should be familiar from introductory physics We can of course obtain it directly by integrating the second order Newton equation using separation of variables twice Physics 227 Lecture 14 Autumn 2007 dz 239 mzzmaz mgszjodzz gjdtsz ZO gt t0 z z to g 2 1416 Idzzz zoIdr20 gr r020r r0 Er r0 ASIDE We can also use similar methods for motion in the 3D conservative gravitation potential of the form Vr GMmr in spherical coordinates where G is Newton s gravitational constant M is the mass of the earth and m is the mass of the test particle For purely radial motion 9 0 conservation of energy now yields ETOT 2722 GMm r Ski szM2Em I m Again this equation can be separated and integrated although with a bit more effort than in 1D But note that we can already see the general structure of the results If ETOT gt 0 then any trajectory with f t 0 gt 0 the plus sign above will continue to have ft gt 0 even as r gt oo ie the mass escapes from the earth For ETOT lt 0 which is now possible and ft 0 gt 0 there will be a turning point where f 0 r GMm ET0T and the mass subsequently falls back towards the earth For the intermediate situation ETOT 0 the mass just escapes from the earth with f gt 0 as tr gt 00 For more general motion nonzero angular motion we can still simplify the problem by using the fact that angular momentum is conserved in a central potential As a final comment on first order equations we note that nonlinear equations like y39 l y2 can eXhibit both general solutions with an arbitrary constant e g y sinx x0 and also singular isolated solutions like y i1 that do not correspond to a value of the constant of integration x0 The singular solutions form the tangent boundary to the general solutions Physics 227 Lecture 14 7 Autumn 2007 Returning to the general issue of differential equations in physics as noted above we are typically interested in 2H order linear differential equations including those that cannot be reduced to a 1St order form We will focus first on systems with only a single independent variable usually time where only ordinary derivatives occur with respect to the independent variable Further we will deal first with the general behavior of solutions to 2nd order linear ordinary differential equations with coefficients that are constants A simple example is 1D motion in gravity as in Eq 1416 This situation becomes somewhat more interesting if we add viscous velocitydependent damping due to air resistance m bz mg 1417 We can often proceed by treating the 2nd order equation as a 1St order equation With the change of variables 239 v we obtain b v v g 1418 m This is an equation amenable to an integrating factor as in Eq 145 above We identify P bm and Q g Thus we find ma a r0 ve1 gel gt vt ge 139 dt e7quot bm voe m 0quotN 4m Voe 1419 0 2 40 emptym 1 3710 V0840 Hence if we release an object at rest v0 0 the large time I gtgt t0e I gt 0 or terminal velocity is just v00 2 mg b qt 00 Due to the damping the falling object Physics 227 Lecture 14 8 Autumn 2007 reaches a maximum velocity The result is in fact true for any initial velocity vt vw v0 vwe71t To find 2t we can integrate once more As mentioned in previous lectures recall especially Lecture 5 more familiar examples of 2nd order linear differential equations with constant coefficients arise in the study of electric circuits and the motion of a mass on a spring in 1D the harmonic oscillator including damping By the rule of Feynman that the same equations have the same solutions we can study the general case and then customize the coefficients to match the specific system Due to the importance of these systems we will once more discuss the solutions systematically and with full generality The simplest and most familiar case is the homogeneous case with no righthandside for the differential equation ie no extemal driving force or voltage Taking x to label the dependent variable or coordinate the location of the mass or the charge on a capacitor the general form of the equation of interest is aj b5ccx0 1420 where a 30 are know constants gt 0 which we take to be real for now and 1 dxdt etc As we have previously discussed we make use of the simple behavior of eXponentials under the derivative operation and try the Ansatz xt x0e Then Eq 1420 becomes 61052 1705 cx0ea 0 251052 bac0 1421 b 172 05 1392 2a 4a2 a With D representing the derivative with respect to the independent variable we can also write this equation in the notation of Boas aj bxcxaD o 1D oz2x0 1422 Thus the general solution to the homogenous equation also called the complementary solution can be written Physics 227 Lecture 14 9 Autumn 2007 XI oct 9 t X16 1 xze 2 a 1423 where xlx2 are constants to be fit to the initial conditions x039c0 Note that the exponents 0512 can be complex and that the 0512 will be pure imaginary if there is no damping b 0 To make a connection to our preVious discussions let us go through this same argument with an explicitly complex Ansatz xt Rezoei 6 The complex analogue of Eq 1421 is a z ib czoei 3 0 D a 2ib c0 1424 c b2 b 2 i i 2139 im n 2a a 4a2 y 0 In this case the complementary solution is written in the familiar form x0 t Rezoe 7 e quot Z0 e 7 cosa0t 0 1425 lzole quot cos wotcos 0 sin wot sin 0 where as usual 0 is the phase of the complex constant 20 Note in particular that as long as mo is real and the exponential is complex the operation of taking the Real part includes the second solution with the a0 exponent and we don t need to explicitly include it to find the general solution On the other hand if mo is imaginary and the exponentials are real as in Eq 1423 we do need to include both exponentials In any case we still have 2 real constants to be determined by the initial conditions We are just writing the same result in a different notation We can make the identifications 0 b2 pal 052 a 1 1426 0 a 4a2 2 a1 2 7 ia0a2 y iw0 Physics 227 Lecture 14 10 Autumn 2007 In either notation the explicit form of the solution will depend on the values of the physics dependent constants a 30 So let us list the various possibilities b2 F gt E With the damping dominant the a are real and negative while a the frequency a 0 is imaginary and smaller in magnitude than 9 This is the overdamped case with the complementary solution exhibiting only exponentially damped behavior as in Eq 1423 b2 c b 2 a y E 050 the critically damped case where the 4a a 2a complementary solution looks like CC I xll x2 eiyt the reader is encouraged to check that this is a solution for any value of the arbitrary constants b2 c b c b2 H lt gym 2 Z i1 H y i two complex conjugates the under damped oscillatory case with CC I do COSUJOI 150 where the constants to fit the initial conditions are now d0 0 Note that mo lt a7 caa02 62 72 where a7 is the natural frequency ofthe system with no damping As a first look at the inhomogeneous problem with a driving term consider the general form we be ox F t We base the general solution on the complementary solution of the homogeneous problem plus any particular solution of the inhomogeneous equation x I x0 I xp I Thus our goal is to find a particular solution for the various forms of F Note these particular solutions are not unique we can always add any amount of the solution of the homogenous problem So we consider some familiar forms for the driving function N o F t 2 Butquot a power series in I Here we use the method of undetermined 110 coefficients to write the particular solution as a corresponding power series Physics 227 Lecture 14 11 Autumn 2007 N xp t antquot and solve for the bquot in terms of the n and abc by equating 110 the coefficients of the powers of t on the two sides of the initial equation For example 39 N0 b0 0c39 l Nl b1B1cb0c 0 b lCZ 39 N 2 b2 Zcb1 c 1 2b 2 c2 b0 cz o 2ac 2 bc 1 2b 2 0339 etc o F t 77639 an exponential where K 7393 051052 With the Ansatz xp t 2 be we find easily that xp t neK a1lt a11lt 052 Note that this works also if K ia ie a sinusoidal driving function F t Re new In this case the maximum response occurs when a m a7 ca ie when you drive the system near its natural frequency To see this we note that the particular solution for a sinusoidal driving function in complex notation zp t 2 206m is given by Z neiwt neiwt P aia a1ia a2 aiaaoyia moy neiwt neiwt ay2a a2 2iym aa72 a2 2iym iwtii ne 6 tanil a 52 a22 432a2 The general form of this solution or more correctly its square is called a Lorentzian shape and it arises so commonly that web pages are dedicated to it From this result we see that the driving force and the response are generally out ofphase ie 6 0 The other question we want to answer is when is the magnitude of the response largest To answer this accurately we need to be careful because there are really Physics 227 Lecture 14 12 Autumn 2007 two questions If we mean for a fixed system ie fixed values of yco0 we ask what value of the driving frequency 60 yields the largest response ie we want to find alzplaw 0 In this case the response is largest ie the magnitude of the denominator is minimum for wz 62 272 a 72 However we could also ask what value of a7 should we tune the system to in order to obtain the largest response from a given driving frequency a this is how radio tuners work Here we want 6 2 Vac 0 and the answer is a7 a In both cases the large response is labeled a resonance F t 77639 but with K 051 or K a we must proceed carefully as we did for the degenerate homogeneous case The trick is to add an extra power of t and try the Ansatz xp t bl 639 Then we find xp t mewa a1 a2 0 F t new but with K a1 a we must proceed even more carefully and try xp t bl2 e Then we find xp t ntzeal 2a The reader is encouraged to check this result N o F t e Z ntquot where we combine our previous results We can write a quot0 N particular solution in terms of undetermined coefficients xp t eK Zb tquot n a and quot0 solve for the coefficients term by term The underlying idea is this last example is the every important concept of linear superposition which is always applicable to a linear equation If the righthandside of the inhomogeneous equation is a sum of terms we find the particular solution as a sum of particular solutions for each of the individual driving terms and sum them up In the next lectures we will look at periodic driving terms in more detail ie use Fourier series techniques Physics 227 Lecture 14 13 Autumn 2007 Lecture 2 Appendix B Some sample problems from Boas Here are some solutions to the sample problems assigned for Chapter 11 and 16 ll 5 Solution We want to use the general expression for the form of a geometric series a S 2 arquot 2 quot2 l r To find the fraction corresponding to S 058333 To you this expression we need to look for a repeating decimal in order the geometric series To this end we rewrite the starting number as S 0583333 2 02503003000300003 Now we can rewrite the repeating decimal as a sum and use the general form to find the desired fraction Sl031ii2i3i4m 4 1010 10 10 1 03 103 11 7 4 1iZ Z E39 10 11 12 Solution Water purification as defined here is another process described by a geometric series If onenth of the current impurity is removed at each stage the total fraction of the impurity removed after N stages of purification is given by 1l lmpuritynNlimizl n n2 nN n1 Physics 227 Lecture 2 Appendix B 1 Autumn 2007 So now we consider the limit N e w for the two cases mentioned n 2 and n 3 too find what fraction of the impurity we can remove in the limit of an in nite number of stages 1 2N 1 2N llmNawlmpunty2N 11mNaw 11mNaw z N 1 1 3N 1 3N1 1 lim Im uri 3N 1im 1m New p New 3N 31 New 3N 2 2 Thus in the first case all of the impurity can be removed but in the second case with only a third removed at each stage the best we can hope for is to remove one half of the impurity 16 3 Solution We want to practice using the comparison test for convergencedivergence Let s proceed by comparing to the convergent geometric series with x 12 quot1112 4 9 16 As suggested in the problem we proceed by performing some manipulations ie resuming on the 2 series We have 1 1 1 1 1 1 1 1 S1 4 9 16 25 36 49 64 225 ie we look at the first term then add together the next 2 terms then add the next 4 terms then add the next 8 terms etc Now compare this to a similar redefinition of the known convergent series Physics 227 Lecture 2 Appendix B 2 Autumn 2007 1 1 1 1 1 1 1 21 8termsof 4 4 16 16 16 16 64 Comparing these two forms for the two series shows that each parenthesis in series S is smaller than the corresponding parenthesis in series C Thus since C converges the comparison test tells us that S must converge also 16 16 Solution This exercise is intended to encourage us to think about the possibly incorrect role of the lower limit in the integral test In considering the sum of terms an 1 n2 we might be tempted to include a lower limit of zero and evaluate the integral This result is misleading since the infinity arises entirely from the lower limit On the other hand we know that the convergencedivergence of the infinite series cannot depend on the first N terms of the series where N is finite By the same token the convergencedivergence of the series cannot depend ie be determined by the behavior of the integral at the lower limit Clearly choosing any lower limit other than the special value zero will lead to a finite integral and the correct conclusion that the series converges The lesson is that you should just ignore the behavior of the integral at the lower limit It tells us nothing useful 16 20 Solution Here we want to practice using the ratio test With the specified series we have Physics 227 Lecture 2 Appendix B 3 Autumn 2007 n S quot n2n aquot 2n hnl 2n n1 1 quot aquot n 2n2 2n22nl 22nl plimHoopnlim 20 H 22nl Thus the ratio test tells us that the series converges absolutely Physics 227 Lecture 2 Appendix B 4 Autumn 2007 Lecture 2 Series See Chapter 1 in Boas A basic and very powerful if pedestrian recall we are lazy AND smart way to solve any differential or integral equation is via a series expansion of the corresponding solution or integrand This is always possible in regions where the expansion exists Since this process can be carried out termbyterm it is ideal for numerical solutions by computer to arbitrary but speci ed accuracy This is how your hand calculator works Recall in particular the familiar Taylor series expansion of a function about the point x a also called a Maclaurin series for a 0 2fquota x an 2 fquota2l n where the primes signify derivatives with respect to the variable X We will return to the subject of power series after we rst consider the properties of plain old numerical series Consider an in nite sequence of numbers labeled by an index n ala2a3 a 39nloo 22 n We de ne the series to be the corresponding sum S 2 2aquot 23 The individual terms an are just numbers or are perhaps expressed as a function of n and any other parameters e g an cx This last expression takes us back to the power series expansion mentioned above In particular this expression an ex de nes a geometric series where the ratio between subsequent terms is a single factor x i e the same factor for each pair of contiguous terms Do such series arise in physics The answer is yes all the time Consider the simple example of dropping a ball from a height h For a real ball the collisions with the oor result in the loss of energy ie the collisions are inelastic some of the kinetic energy of the CM of the ball is converted to heat ie random kinetic energy of the individual molecules in the ball each time the ball collides with the oor If we de ne the fraction of kinetic energy lost in each bounce to be 1 x then the height Physics 227 Lecture 2 1 Autumn 2007 after one bounce is xh recall that the height of the bounce is determined by when all the kinetic energy of the CM is converted into gravitational potential energy 5 mgh1 mghx E0 mgh mgh 24 After n bounces the height is h x h E an Such sequences of numbers that progress by a power correspond to geometric progressions The geometric progression is the simplest progression Now we can ask how far does the ball travel before all kinetic energy is lost Note that in principle this takes in nite time In equations we have Dxh2112h22hn h l2xx2xquot 25 hl2ixquot Ehl25x In these eXpressions the ellipsis the 3 dots represent the implied missing terms For a fairly elastic ball with x 45 we nd 4 4 quot 4 D h2h Eh2hS 5 5 2396 So the question is can we give meaning z39e a numerical value to this in nite geometric series S45 245quot If the sum of the in nite number of terms eXists nl then that sum is the series z39e the value of the series But how can we tell if the sum exists Even if we know that the sum eXists how do we evaluate it Note that it would take an in nite time for a computer to literally add up an in nite number of terms We must be smarter and use more powerful techniques as we will see below First let s de ne some terminology If the sum eXists z39e it is de ned and is less than in nity we say the series converges If the sum does not eXist z39e it is not de ned or is in nite we say the series diverges Physics 227 Lecture 2 2 Autumn 2007 Now consider how to evaluate the sum Since humans like computers do best considering nite things consider only the rst N terms N N N 4 quot SNZanxx2xNZxquot 27 quot1 711 quot1 In an obvious notation this is called a partial sum Since this quantity is guaranteed to be nite assuming an lt 0071 S N we can use normal algebraic manipulations see below for the nonguaranteed in nite case Consider the following true but unmotivated manipulations N 2 3 N1 1 N1 xSNzx x x l E x SNx l x quot1 3SNl xxl xN l xN l x39 28 3 S N x Note that this last eXpression implies the special and very useful relation see below 11xlxx2x3 29 Now consider the limit limNW SN 21x 1 1imNW xN 210 x Evaluating series is closely related to taking limits 7 we must be able to do both For the general case the possibilities are l limNW S N eXists and is nite in which case the series is said to converge and S lim S New N Physics 227 Lecture 2 3 Autumn 2007 2 lim S105 1 51051 l Su 2 l or we have limNW NW S N does not exist because either it exhibits a limit cycle eg S N gt ioo in either case the series is said to diverge It is often helpful to consider the remainder RN E S S N The series converges iff if and only if limNW RN 0 The series diverges for limNW R N 73 0 independent of the precise form of the nonzero righthandside Another way to express convergence is to say that if you pick a small parameter 8 ltlt 1 I can always nd a large integer M such that RNl lt s for all N gtM Note that if you are not comfortable taking these limits N a 00 you should practice doing so It will be very useful below In our simple case above we have the following possibilities note thatx gt 0 here the fraction of energy lost in each bounce is less than 100 l x lt l lim xN 0 and thus the series converges with S 1 x Ngtoo 2 x gt 1 limNW x 00 and thus the series diverges 3 x l limNW x l S 1 which is illde ned but using the original form we nd S 21 00 and the series diverges 111 Clearly we must be careful in this last case For our speci c example we have V Dh2h1 Zh8h9h 211 The distance traveled is nite even for an in nite number of steps Of course when the height of the bounce is less than an Angstrom there is little contribution This series is said to converge quickly The contribution from the bounces after bounce 100 is D 2h w 3 n Zh 31003 3 16 109 212 Ngt100 15 5 5 39 X 39 Physics 227 Lecture 2 4 Autumn 2007 Note that in realistic physics applications where the uncertainty in the measurements is greater than zero we are typically also interested in comparing to a theoretical calculation also with nite accuracy z39e the value of RN for large N Returning to the more general eXpression for the geometric series we de ne S arquot quot0 N71 N N l r 213 5 arquot rquota N quot2 quot21 l r which matches Eq 14 in Boas For lrl lt1 we can take the limit N gt 00 as above to nd a NW SN 214 Slim It is clearly important to be able to tell if a series converges or not Even though divergent series are sometimes useful see below we must be careful when using them In particular we cannot reliably use the same algebraic manipulations on full sums as we did on the partial sum For example we might consider S1248w2quot 2 x 215 22248S 12 1 which is clearly nonsense Consider also the following similar series the Harmonic series 1ll w1 216 2 3 Mn 39 which diverges and Physics 227 Lecture 2 5 Autumn 2007 1 1 w 1n1 1 2 3 2 217 which converges It is clearly essential that we de ne convergence tests that we can generally apply including to cases with alternating signs Consider again the general expression in Eq 23 S 2 with all terms nite n1 an lt 00 n lt 00 To this series sum we can apply the following set of tests 0 The Preliminary Test or Divergence Test if lim a 73 0 S diverges if ngtoo n lim a 0 we must test further z39e this test is necessary but not suf cient ngtoo n For example an 1 n limn a 0 so we test further but for an 2 anon lim a 00 orfor anl lim ngtoo n a 1 we know that the series diverges ASIDE We can apply this test directly if all an 2 0 Otherwise consider 3 E Z n1 absolutely convergent S does not converge S may still converge conditionally if there is enough cancellation We will discuss this case in more detail below If S converges then S converges absolutely If the sum is not an 1 The Comparison Test compare the new series to a series whose convergence properties you already know This test has two parts a Assume we know that the series C ch with all on gt 0 converges If n1 S c for all n n 2 N with N some large integer then S converges absolutely an Note that the rst Nl nite terms cannot affect convergence only the value of S if it convergences b Assume we know that the series D 2d with all d gt 0 diverges If n1 for all n n ZN then S diverges The series S may still converge an 2 d ASIDE If an gt c or an lt d we learn nothing from these comparisons Physics 227 Lecture 2 6 Autumn 2007 1 l Cons1der the example S 2739 an 739 n l 23n ln and compare to the W1 n n 1 c which we know converges We have geometric series with x 2 n S11l1i 2 6 24 120 Cl1 1ii 2 I 16 32 Thus an lt c n gt 3 and S converges very rapidly The downside with the comparison test is that to use it we must rst know the convergence properties of many series So let s develop some other tests 2 The Integral Test consider the eXpression Iandn with the integrand treated as a smooth function of n This will be useful for the case 0 S a 1 S an for n 2 N some large N The series converges or diverges depending on whether the integral Iandn is nite or in nite independent of the nite lower limit You can see this from the following gures an an 2 2 15 15 l l 05 05 n n l 2 6 l 2 4 5 6 00 0 50 w 2 ltl IltZ quot2 1 1 quot1 For the example of the Harmonic series an 1 n which is illustrated in the above gures we learn that Physics 227 Lecture 2 7 Autumn 2007 co n so lnn 00 218 n The Harmonic series with signs diverges as noted earlier in Eq 216 The limitation with this test is that we must be able to perform the integral 3 The Ratio Test rst de ne the ratio of adj acent terms in the series and the asymptotic value of the ratio an1 Pquot E SPElimnw Pn 219 n Since for the geometric series p is just the fixed ratio of terms we learn from our analysis above of the geometric series coupled with the comparison test that lt l S converges absolutely if p 1 may converge use further test 220 gt 1 S diverges As familiar examples consider 1 n 1 an 3 n 3 p 0 3 converges n n1 n1 an13pn n 1p139 23921 n1 17 1 Clearly the case p 1 requires further study In particular we need to specify and understand the implications of how p approaches 1 as n gt oo which we can analyze a step at a time To see what the possible behavior looks like we consider some examples Consider rst the Riemann Zeta function de ned by the series 5P P EZLP 222 Physics 227 Lecture 2 Autumn 2007 and apply the integral test We have that dn 1 oo lt 0Pgt1 p n P l p lipi 00pltl am lnnl oopl where the special case p 1 is just the Harmonic series Thus the Zeta function eXists z39e the series converges for p gtl strictly for Rep gt1 and diverges otherwise Back to the ratio test we have for the Zeta function expanding in powers of ln 1 think about a Taylor series eXpansion of lx7p for small x P L 1 M DH n1 1A WA n 2H2 224 Thus we conclude via the comparison test that for a general series S S converges for p gt1 gb m naw3 am 71 S diverges for p lt1 while for p 1 we must work harder still Next consider the similar series l SMZ mm quot2 nlnns for which the integral test tells us that m oosSl w dn dm lt 00 s gt1 l gtmm I 227 nlnn Physics 227 Lecture 2 9 Autumn 2007 12 the form of the nal integral is just as for the Zeta function On the other hand the ratio test for this series has the form think about how to perform the indicated expansions for large n s 1 Inn 5 1 1 Pquot 1 Hm 1 1 1 n lnln n 1Alnn 228 where we saw from the integral test that this corresponds to a convergent sum iff s gt 1 We can summarize these insights in the following general result for the ratio test Keep the largest terms as n gt 00 and de ne the following sequence of ever more detailed results in this limit gt lS diverges pr lt lS converges elseif 1 d 1 pltlSdiverges 3 an pn aSn oo n p gt1 S converges elseifplplandpnl l s n nlnn aSn oo 229 s lt1 S diverges s gt1 S converges etc As an example consider again our friend the Harmonic series an ln From Eq 221 we have 1 l a 2 3pn l n 230 where the last step used Eq 29 but with x gt x a useful result in it own right Physics 227 Lecture 2 10 Autumn 2007 l l xx2 x3x4 xltl 231 1 x l i In Eq 230 we see that the coef cient of ln is 1 corresponding to p l and that the coef cient of 1 nlnn is zero since the next term is 1172 s 0 Hence by the rules of Eq 229 the Harmonic series diverges as advertised 4 The Combined Test combine test 1 and 3 and 2 again in two steps a Pick a convergent series C 20 c gt 0 and take the ratio to the absolute n value of terms in S 2a an if lime lt 00 then S is absolutely convergent 232 c n This means that the terms in g for large n are proportional to those in C and both series converge if one does In principle we don t care if this proportionality constant is large while in test 1 we were really checking for proportionality constant 1 b Pick a divergent series D 2d d gt 0 and again consider the ratio if an gt 0 and lim new gt 0 then S diverges 233 Again we must be careful in the case of alternating signs see below This last test allows us to simplify the series we are considering by choosing a simpler comparison series with the same asymptotic behavior which we illustrate with some examples Consider the series de ned by a quot2 25 1ngt3 234 quot 4n3 7n22 39 39 For large n we have Physics 227 Lecture 2 11 Autumn 2007 2n2 l n ngtw 413 235 So we don t need to consider the full complexity of an but can consider instead the simpler series given by B 2b b 1 n2 Both the integral test and the ratio test tell us that B converges w dn I I 2 lt 00 n n 2 236 2 Pn 2w1 l1 2gtll n 71 Hence we use B as the convergent comparison series and conclude that S is convergent 0 quot gt 1 lt 00 2 37 bquot ngtoo 39 As another example consider 2 n n 2 2 n n5 5112 quot1113 21nn n e e 3 So try the comparison series d 31175 which we know diverges from the ratio test z 3niln15 3 gt3gt1 gt Pquot 3nn5 quotaw 11n5 239 So the original series diverges since Physics 227 Lecture 2 12 Autumn 2007 Ut n5 3quot n2 l n33quot 1gt0 240 dquot 3n n55n2 15n3 Hoe 39 Now let s return to consider the case of a series with terms alternating in sign which does not converge absolutely It can still converge in which case we say that it exhibits conditional convergence An interesting example is the alternating sign version of the Harmonic series an ln1n The absolute value version of this series the usual Harmonic series we know to diverge Such a situation is addressed by out last test 5 The Alternating Sign Not Absolutely Convergent Test a series with terms that alternate in sign which is not absolutely convergent is conditionally convergent if and only if the terms systematically shrink in magnitude and asymptotically vanish S a n1 an ngtNlimean 0 241 The trick with conditionally convergent series is determining the value to which they converge It is easy to be misled on this issue especially if one just looks at the series The answer can depend on how the series is arranged Note that the test in Eq 241 depends on the order of the terms From our knowledge of the Maclaurin series eXpansion of the logarithm we have 2 4 lnlxx x 242 which we could obtain directly by integrating Eq 231 Thus we know that the standard order alternating sign Harmonic series converges to no n1 El 1 ln20693l47 243 71 quot1 On the other hand we could arrange the terms in the series in the following eXplicitly misleading fashion the righthandside is the running sum Physics 227 Lecture 2 13 Autumn 2007 1ll 15333 3 5 1 10333 2 lliii 15218 244 7 9 11 13 15 l 12718 4 11 1 1i 15144 7 777 17 19 21 23 25 which could be interpreted to mean convergence to 15 if the series is de ned in this way The lesson is that you have to be very careful with in nite series especially when the signs alternate In the later case we are effectively subtracting two divergent series in order to obtain a conditionally convergent series In the case above we have the two series A ill2n 1 and B ill2n S A B 70 711 Clearly the order in which we combine the terms can matter We close this discussion with a summary of useful facts about in nite series which follow from the above considerations 1 Convergence of a series is not affected by overall multiplication z39e multiplying every term by a constant or by changing a nite number of terms 2 Two convergent series A 2a B 2b can be added or subtracted term by terms to obtain another convergent series C Zltan i b A i B z39e convergent series can be treated like ordinary numbers This is not true of divergent series 3 The terms in an absolutely convergent series can be rearranged without fear of changing the convergence or the value of the series Conditionally convergent series depend on the order of the terms both for the question of convergence and for the value of the series 12 the sum Physics 227 Lecture 2 14 Autumn 2007


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