Applied Mathematical Analysis
Applied Mathematical Analysis MATH 321
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F Waleffe 20080901 Vectors These are compact lecture notes for Math 321 at UW Madison Read them carefully ideally before the lecture and complete with your own class notes and pictures Skipping the theory and jumping directly to the exercises is a tried and failed strategy that only leads to the typical question I have no idea how to get started There are many explicit and implicit exercises within the text that complement the theory Many of the proofs are actually good solued exercises The objectiues are to reuiew the key concepts emphasizing geometric understanding and uisualization 1 Vectors Geometric Approach What s a vector in elementary calculus and linear algebra you probably de ned vectors as a list of numbers such as if 4 2 5 with special algebraic manipulations rules but in elementary physics uectors were probably de ned as quantities that have both a magnitude and a direction such as displacements uelocities and forces as opposed to scalars such as mass temperature and pressure which only have magnitude We begin with the latter point of View because the algebraic version hides geometric issues that are important to physics namely that physical laws are invariant under a change of coordinates they do not depend on our particular choice of coordinates and there is no special system of coordinates everything is relative Our other motivation is that to truly understand vectors and math in general you have to be able to uisualize the concepts so rather than developing the geometric interpretation as an after thought we start with it 11 Vector addition and multiplication by a scalar We begin with vectors in 2D and 3D Euclidean spaces E2 and E3 say E3 corresponds to our intuitive notion of the space we live in at human scales E2 is any plane in E3 These are the spaces of classical Euclidean geometry There is no special origin or direction in these spaces All positions are relative to other reference points andor directions Vectors in those spaces are the set of all possible displacements which are oriented line segments whose origin does not matter The vector B E E i is the displacement from point A to point B If we make a the same displacement 139 starting from another point C39gt we Blimp at a another point D but CD i AB Two vectors AB and CD are A K equal if they are opposite legs of a parallelogram and have the same direction We ll denote by E3 the set of all possible vectors in E3 to emphasize that vectors displacements and points are distinct concepts The length of vector i is denoted by lfil It is a positive real number When there is no risk of confusion we simply write a for lfil When writing by hand we use an arrow on top a or a wiggle underneath instead of the boldface i Velocities accelerations and forces are also vectors that can be described by oriented line segments but strictly speaking these are all in different spaces the velocity space acceleration space force space etc All these spaces are physically connected but we do not add displacements to forces for instance so they are mathematically distinct Vectors add according to the parallelogram rule If we move 1 mile North then 1 mile East we end up miles Northeast of the starting point The net displacement would be the same if we move 1mi East rst then 1mi North So vector addition is commutatiue 6 I 3 i It is also associatiue 6E6 fil fi F Wale e Math 321 20080901 2 Note that in general i g and fare not in the same plane so the 2D gure is not general but it is easy enough to visualize associativity in 3D To every vector i we can associate an opposite vector denoted ii that is the displacement exactly opposite to i Vector subtraction 17 i is then de ned as the addition of Hand iii In particular 67fi 0 corresponds 5 to no net displacement This is an important difference between points and displacements there is no special point in our space but there is one special displacement the zero vector 6 such that i iii 6 iii i and fi6fi forany i El The other key operation that characterizes vectors is multiplication by a real number 04 E R Geometrically 3917 oin is a new vector parallel to i but of length l l lallfil The direction of 3917 is the same as i if 04 gt 0 and opposite to i if 04 lt 0 Obviously 71fi iii multiplying i by 71 yields the previously de ned opposite of i Other geometrically obvious properties are 04 mt oin Bi and amt 0436 A more interesting property is distributiuity 04i 04 041 which geometrically corresponds to similarity of tiiangles Generalization of the Vector Space concept Vector addition and multiplication by a realnumber are the two key operations that de ne a Vector Space provided those operations satisfy the following 8 properties Vii b in the vector space and Va 6 in R Required vector addition properties AAAA to VVVV F Wale e Math 321 20080901 3 Required scalar multiplication properties a aaa a 5 and 6436 6 06 I3 046 at 7 10 a 8 When these properties are used to de ne the vector space they are referred to as axioms ie the de ning properties The vector space R Consider the set of ordered n tuplets of real numbers 1 E 12 zn These could correspond to lists of student grades on a particular exam for instance What kind of operations would we want to do on these lists of student grades We ll probably want to add several grades for each student and we ll probably want to Tescale the grades So the natural operations on these n tuplets are addition de ned by adding the respective components 212yEx1017z20277mnyny1 9 and multiplication by a real number 04 E R de ned as 0413 E ax1az2azn 10 The set of n tuplets of real numbers equipped with addition and multiplication by a real number as just de ned is an important vector space called R The vector spaces R2 and R3 will be particularly important to us as they ll soon corresponds to the components of our arrow vectors But we also use R for very large n when studying systems of equations for instance Exercises 1 Show that addition and scalar multiplication of n tuplets satisfy the 8 required properties listed above to De ne addition and scalar multiplication of n tuplets of complex numbers and show that all 8 properties are satis ed That vector space is called C 03 The set of real functions fz is also a vector space De ne addition in the obvious way fz gz E hz another real function and scalar multiplication a m yet another real function Show that all 8 properties are again satis ed q Suppose you de ne addition of n tuplets 1 1 2 zn as usual but de ne scalar multi plication according to 041 0412 zn that is only the rst component is multiplied by 04 Which property is violated What if you de ned 041 am10 0 which property would be violated 01 From the 8 properties show that 06 6 and Eli Ed Vii ie show that multiplica tion by the scalar 0 yields the neutral element for addition and multiplication by El yields the additive inverse F Wale e Math 321 20080901 4 12 Bases and Components of a Vector Addition and scalar multiplication of vectors allow us to de ne the concepts of linear combination basis components and dimension These concepts apply to any vector space A linear combination ofvectors fi and bis an expression ofthe form afi li This linear combination yields another vector 17 The set of all such vectors obtained by taking any a B E R is itself a vector space or more correctly a vector subspace if fi and I are two vectors in E3 for instance We say that fi and I form a basis for that subspace We also say that this is the subspace spanned by fi and For a given vector 17 the unique realnumbers oz and B such that 3917 afi b are called the components of 3917 with respect to the basis fi 7 as E c a 3 51 b fi Oifi Linear Independence The vectors filfig fik for some integer k are linearly independent Ll if the only way to have 041510425204k5k0 is for all the 04 s to be zero a1a2ak0 Dimension The dimension of a vector space is the largest number of linearly independent vectors n say in that space A basis for that space consists of n linearly independent vectors A vector 3917 has n components some of them possibly zero with respect to any basis in that space Examples 0 Two non parallel vectors fi and b in E2 are Ll and these vectors form a basis for E2 Any given vector 3917 in E2 can be written as 3917 afi b for a unique pair 04 B 3917 is the diagonal of the parallelogram afi b Three or more vectors in E2 are linearly dependent 0 Three non coplanar vectors fi b and fi in E3 are Ll and those vectors form a basis for E3 However 4 or more vectors in E3 are linearly dependent Any given vector 3917 can be expanded as 3917 afi Bb yd for a unique triplet of real numbers 04 B y Make sketches to illustrate The 8 properties of addition and scalar multiplication imply that if two vectors ii and 3917 are expanded with respect to the same basis fil fig fig so if U151 U252 143537 17 U161 U262 123637 then 13 17 U1 U1fii U2 U2fiz Us 119537 0417 ow fil 04112fig av3fig so addition and scalar multiplication are performed component by component and the triplets of real components 121122123 are elements of the vector space R3 A basis fil fig fig in E3 provides a one to one correspondence mapping between displacements 3917 in E3 and triplets of real numbers in R 3917 E E3 lt gt 111112113 6 R3 ae e at 5 F W1 If M h 321 20080901 Exercises 1 F93 U a I 00 to HH DH Pick two vectors i I and some arbitrary point A in the plane of your sheet of paper If the A a possible displacements from point A to point B are speci ed by AB ad 0b sketch the region where B can be if 0 040 1 ii 0 la and 71 04 g 1 Given i 0 show that the set of all 3917 Ozfi 03 Va 0 E R is a vector space Show that the set of all vectors 393 Ozfi 0 Va 6 R and xed i I is not a vector space If you de ned addition of ordered pairs 1 1 mg as usual but scalar multiplication by 041 Oz12 would it be possible to represent any vector 1 as a linear combination of two basis vectors a and b Show that the line segment connecting the middle points of two sides of a triangle is parallel to and equal to half of the third side using methods of plane geometry and using vectors Show that the medians of a triangle intersect at the same point which is 23 of the way down from the vertices along each median a median is a line that connects a vertex to the middle of the opposite side Do this using both geometrical methods and vectors Given three point A B C not co linear nd a point 0 such that 34 TB TC 0 Show that the line through A and O cuts BC at its mid point Deduce similar results for the other sides of the triangle ABC and therefore that O is the point of intersection of the medians A A A A Sketch Hint OB OA AB OC A A a Given four points A B C D not co planar nd a point 0 such that mOBOCOD 0 Show that the line through A and O intersects the triangle BCD at its center of area Deduce similar results for the other faces and therefore that the medians of the tetrahedron ABCD de ned as the lines joining each vertex to the center of area of the opposite triangle all intersect at the same point 0 which is 34 of the way down from the vertices along the medians Visualize Find a basis for R consider the natural basis 81 10 0 82 010 0 etc Find a basis for C What is the dimension of that space What is the dimension of the vector space of real continuous functions fz in 0 lt z lt 1 What could be a basis for the vector space of nice functions fz in 01 te 0 lt z lt 1 what s a nice function smooth functions are in nitely differentiable that s nice Partial solutions to problems 5 and 6 Let D and E be the midpoints of segments CB and CA B Geometry consider the triangles ABC and EDC They are similar why so W E and those two segments are parallel Next consider triangles BAC and EDC Those are similar too why so AC 2CD and BC 2CE a a done Vector Algebra Let CA i and CB b t en 1 a a A A ED iii2 b2 AB2 done 2 AD b2 and A A A BE 713 62 Next CG aaAD 13 BEE for some 04 0 Writing this equality in terms of i and I yields 04 B 23 donel Why does this imply that the line through C and C cuts AB at its midpoint7 F Wale e Math 321 20080901 6 No need to introduce cartesian coordinates that would be horrible and full of unnecessary algebra a a The vector solution refers to a vector basis 139 CE and b CB which is perfect for the problem although it is not orthogonal 13 Dot 0160 Scalar or Inner Product The geometric de nition of the dot product of our arrow vectors is at m 3 cos0 11 where 0 g 9 g 7139 is the angle between the vectors 139 and I when their tails coincide The dot product is a real number such that Ei I 0 iff i and I are orthogonal perpendicular The 6 vector is considered orthogonal to any vector The dot product of any vector with itself is the square of its length Ei i lfilz The dot product is directly related to the perpendicular projections of I onto 139 and 139 onto The latter are respectively lIJ lcosdLbIamp lfilcosdfih 12 al b where a ilfil and I are the unit vectors in the i and I directions respectively A unit vector is a vector of length one l l 1 Vi 7 O Curiously unit vectors do not have physical units that is if i is a displacement with physical units of length then a is a pure direction vector For instance if i move northeast 3 miles77 then lfil 3 miles and a northeast ln physics the work W done by a force F on a particle undergoing the displacement Jis equal to distance times the component of F in the direction of d but that is equal to the total force times the component of the displacement Jin the direction of Both of these statements are contained in the symmetric de nition W F d see exercise 1 below Parallel and Perpendicular Components We often want to decompose a vector I into vector components EH and I71 parallel and perpen dicular to a vector i respectively such that b EH by with a c 134 b I bHbaaa J a a I a LI3713HI37Eaal376 7a E m F Wale e Math 321 20080901 7 Properties of the dot product The dot product has the following properties most of which are immediate 1 51755 4 0520 ma0lta6 5 fi 12 Ii i 1 Cauchy Schwartz To verify the distributive property Ei 13 3 Ei 1 Ei geometrically note that the magnitude of 139 drops out so all we need to check is 0 15 0 6 0 13 3 or in other words that the perpendicular projections of 1 and 6 onto a line parallel to i add up to the projection of 1 6 onto that line This is obvious from the gure In general i 1 and dare not in the same plane To visualize the 3D case interpret Fl 1 as the signed distance between two planes perpendicular to i that pass through the tail and head of 1 and likewise for F1 6 and 0 13 The result follows directly since 1 fand 1 form a triangle So the picture is the same but the dotted lines represent planes seen from the sides and 1 6 1 dare not in the plane of the paper in general Exercises 1 A skier slides down an inclined plane with a total vertical drop of h show that the work done by gravity is independent of the slope Use F and d s and sketch the geometry of this result to Visualize the solutions of fi if 04 where i and 04 are known n Sketch 6 i 1 then calculate 6 6 i b 4 If 77 is a unit vector show that ii E 6 1139 is orthogonal to 77 V i Sketch OJ 1i and deduce the law of cosines 5 If 6 i 1 show that 61 ii 171 de ned in previous exercise lnterpret geometrically B is a magnetic eld and 3917 is the velocity of a particle We want to decompose 3917 17L 17H where 17L is perpendicular to the magnetic eld and 17H is parallel to it Derive vector expressions for 17L and 17 a 1 Show that the three normals dropped from the vertices of a triangle perpendicular to their opposite sides intersect at the same point Hint this is similar to problem 6 in section 12 A A A A A but now AD and BE are de ned by AD CB 0 and BE El 0 and the goal is to show A A that CG AB 0 00 A and B are two points on a sphere of radius R speci ed by their longitude and latitude Find the shortest distance between A and B traveling on the sphere If 0 is the center of A A the sphere consider OA OB to determine their angle to Consider 3917 fi t1 where t E R What is the minimum W and for what if Solve two ways geometrically and using calculus F Wale e Math 321 20080901 8 14 Orthonormal basis Given an arbitrary vector 3917 and three non co planar vectors i b and 6 in E3 you can nd the three scalars a B and V such that 17 afi b w by a geometric construction sect 12 The scalars a B and 39y are called the components of 3917 in the basis i b 6 Finding those components is much simpler if the basis is orthogonal ie i b b 6 64139 0 In that case take the dot product of both sides of the equation 3917 oin Bb m with each of the 3 basis vectors and show that a 7 1317 617 04 7 B 7 Y a 1 b b 0 0 An orthonormal basis is even better That s a basis for which the vectors are mutually orthogonal and of unit norm Such a basis is often denoted1 l g 63 Its compact de nition is 14 where ij 1 23 and 61739 is the Kronecker symbol 6 1 ifi j and 0 ifi 7 j The components of a vector 3917 with respect to the orthonormal basis 61 g 63 in E3 are the real numbers 121 v2 123 such that 15 a a If two vectors 139 and b are expanded in terms of 61 62 83 i m a 5a151a252a3537 bbi i 525253537 use the properties of the dot product and the orthonormality of the basis to show that a1b1a2b2agbg D Show that this formula is valid only for orthonormal bases One remarkable property of this formula is that its value is independent of the orthonormal basis The dot product is a geometric property of the vectors 139 and b independent of the basis This is obvious from the geometric de nition 11 but not from its expression in terms of components 16 If 61 g 63 and f g 63 are two distinct orthogonal bases then i a131 1252 1353 1351 4252 12355 but in general the components in the two bases are distinct a1 7 a3 a2 7 a z a3 7 ag and likewise for another vector b yet i I a1b1 azbz agbg Lib1 aZbZ a b The simple algebraic form of the dot product is invariant under a change of orthonormal basis 1Note about notation Forget about the notation i This is old 19th century notation it is unfortunately still Very common in elementary courses but that old notation will get in the way if you stick to it We will NEVER use i i 1 instead we will use 31 32 33 or 70 6y to denote a set of three orthonormal Vectors in 3D euclidean space We will soon use indices i j and k next line alreadyl Those indices are positive integers that can take all the Values from 1 to n the dimension of the space We spend most of our time in 3D space so most of the time the possible Values for these indices i j and k are 1 2 and 3 We will use those indices a lot They should not be confused with those old orthonormal Vectors i i I from elementary calculus F Wale e Math 321 20080901 9 Exercises 1 If 111 21 we calculate 67 111 using 2 notation and 14 to is not true that 21 21 21 wl 102 1037 03 If 3917 21 126 and 1B 21 wi i calculate 17 111 using 2 notation and 14 F If 3917 211216139 and 111 21 wifii where the basis 139 i 123 is not orthonormal calculate 17 11139 Calculate 1 221 517 ii Zia 2215ijajbi7 iii 221 57391 01 De nition of dot product for R The geometric de nition of the dot product 11 is great for oriented line segments as it emphasizes the geometric aspects but the algebraic formula 16 is very useful for calculations It s also the way to de ne the dot product for other vector spaces where the concept of angle between vectors may not be obvious eg what is the angle between the vectors 1234 and 4321 in R4 The dot product aka scalar product or inner product of the vectors 1 and y in R is de ned as suggested by 16 myEm1y1z2y2znyn 17 Verify that this de nition satis es the rst 4 properties of the dot product To show the Cauchy Schwarz property you need a bit of Calculus and a classical trick consider 390 1 y then by prop 4 of the dot product 113 y 1 y 2 O and by props 123 F E 1 y 113 y Zy y 21 y 1 13 For given but arbitrary 1 and y this is a quadratic polynomial in That polynomial F has a single minimum at 721 yy y Find that minimum value of F and deduce the Cauchy Schwarz inequality Once we know that the de nition 17 satis es Cauchy Schwarz 1y2 y y we can de ne the length ofa vector by 1112 this is called the Euclidean length since it corresponds to length in Euclidean geometry by Pythagoras s theorem and the angle 9 between two vectors in R by cos0 1 A vector space for which a dot or inner product is de ned is called a Hilbert space or an inner product space D So what is the angle between 1 2 3 4 and 4 3 2 1 D Can you de ne a dot product for the vector space of real functions fz The bottom line is that for more complex vector spaces the dot or scalar or inner product is a key mathematical construct that allows us to generalize the concept of angle between vectors and most importantly to de ne orthogonal vectors D Find a vector orthogonal to 1 2 34 Find all the vectors orthogonal to 1 2 34 D Decompose 4217 into the sum of two vectors one of which is parallel and the other perpen dicular to 1 23 4 D Show that cos nz with n integer is a set of orthogonal functions on 0 7139 Find formulas for the components of a function fz in terms of that orthogonal basis In particular nd the components of sinz in terms of the cosine basis in that 0 7r interval Norm of a uector The norm of a vector denoted Hall is a positive real number that de nes its size or length but not in the sense of the number of its components For displacement vectors in Euclidean spaces F Wale e Math 321 20080901 10 the norm is the length of the displacement llfill lfil ie the distance between point A and B A if AB a The following properties are geometrically straightforward for length of displacement vectors 1 llall EOand llall 0ltgta6 2 llaall lal Hal 3 lla l bll llall triangle inequality Draw the triangle formed by i I and i l I to see why the latter is called the triangle inequality For more general vector spaces these properties become the de ning properties axioms that a norm must satisfy A vector space for which a norm is de ned is called a Banach space De nition of norm for R For other types of vector space there are many possible de nitions for the norm of a vector as long as those de nitions satisfy the 3 norm properties In R the p norm of vector if is de ned by the positive number 117 7 lep 2lz1l lm2l lznl lt18 where p 2 1 is a real number Commonly used norms are the 2 norm which is the square root of the sum of the squares the 1 norm sum of absolute values and the in nity norm llmlloo de ned as the limit as p a 00 of the above expression Note that the 2 norm 1 1112 and for that reason is also called the Euclidean norm In fact if a dot product is de ned then a norm can always be de ned as the square root of the dot product In other words every Hilbert space is a Banach space but the converse is not true D Show that the in nity norm max D Show that the p norm satis es the three norm properties for p 1 2 00 D De ne a norm for C D De ne the 2 norm for real functions fz in 0 lt z lt 1 15 Cross 01611 Vector or Area Product The cross product is a very useful operation for physical applications mechanics electromag netism but it is particular to 3D space The cross product of two vectors i I is the vector denoted i x I that is 1 orthogonal to both i and E 2 has magnitude equal to the area of the parallelogram with sides 139 and If 3 has direction determined by the right hand rule or the cork screw rule ie ifixII0 l sin0 area of parallelogram 19 a b a x E is right handed where 0 is the angle between i and I as de ned for the dot product ls sin0 2 0 The following gure illustrates the cross product with a perspective view left and a top view right with i x b out of the paper in the top view F Wale e Math 321 20080901 11 m x m l l 9 p x m l l m x p 20 where ii if is the vector component of i perpendicular to band likewise bl I bampamp is the vector component of b perpendicular to i so the meaning of T is relative The cross product has the following properties anti commutativity i x i 0 2 as x 3 fix ab 04Eix E 3 6x 6 6 6x a Ex 13 So we manipulate the cross product as we d expect except for the anti commutativity which is a big difference from our other elementary products The rst 2 properties are geometrically obvious from the de nition To show the third property distributivity let 6 lc l and get rid of by prop 2 in other words we can assume that 0 1 without loss of generality but with gain of simplicity All three cross products give vectors perpendicular to Band furthermore from 20 we have 6x 139 6x ii 6x b x by and 6x 6b 6x fibi where T means vector component perpendicular to 6 ii if Ei 38 etc So the cross products eliminate the components parallel to gand all the action is in the plane peijnendicular to 6 To visualize the distributivity property it suffices to look at that plane from the top with c pointing out of the sheet of paper Then a cross product by 6 assumed of unit norm is equivalent to a rotation of the peijnendicular components by 7r2 counterclockwise Since i I and i b form a triangle their perpendicular projections ii by and fi bi form a triangle and therefore 6x i x b and 6x ii also form a triangle demonstrating distributivity a fibi 0i Orientation of Bases If we pick an arbitrary unit vector 61 then a unit vector g orthogonal to 61 then there are two possible unit vectors g orthogonal to both 61 and 62 One choice gives a right handed basis ie 61 in right thumb direction g in right index direction and 63 in right major direction The other choice gives a left handed basis These two types of bases are mirror images of each other as illustrated in the following gure where 61 61 point straight out of the paper or screen F Wale e Math 321 20080901 12 l 2 82 l l l a l a 83 l 83 l 51 l 51 l Left handed Right handed This gure reveals an interesting subtlety of the cross product For this particular choice of left and right handed bases other arrangements are possible of course 61 61 and 62 g but 63 i g so 61 x 62 63 and 61 x 62 g 763 This indicates that the mirror image of the cross product is notthe cross product of the mirror images On the opposite the mirror image of the cross product 63 is minus the cross product of the images 61 x j We showed this for a special case but this is general the cross product is not invariant under re ection it changes sign Physical laws should not depend on the choice of basis so this implies that they should not be expressed in terms of an odd number of cross products When we write that the velocity of a particle is 3917 43 x 17quot 3917 and Fare good vectors re ecting as they should under mirror symmetry but 43 is not quite a true vector it is a pseudo vector It changes sign under re ection That is because rotation vectors are themselves de ned according to the right hand rule so an expression such as 413 x 17quot actually contains two applications of the right hand rule Likewise in the Lorentz force F q39U x g F and 3917 are good vectors but since the de nition involves a cross product it must be that g is a pseudo vector lndeed g is itself a cross product so the de nition of F actually contains two crossproducts The orientation right handed or left handed did not matter to us before but now that we ve de ned the cross product with the right hand rule we ll typically choose right handed bases We don t have to geometrically speaking but we need to from an algebraic point of view otherwise we d need two sets of algebraic formula one for right handed bases and one for left handed bases In terms of our right handed cross product de nition we can de ne a right handed basis by T1 gtlt 2 gg 21 then deduce geometrically 2 gtlt 11 E1 T3 gtlt T1 gg 22 gg gtlt E1 g gl gtlt gg g E3 gtlt gg 7T1 Note that 22 are cyclic rotations of the basis vectors in 21 ie 616263 a g 3 1 gt 6361 g The orderings of the basis vectors in 23 do not correspond to cyclic rotations For 3 elements a cyclic rotation corresponds to an even number of permutations For instance to go from 616263 to 626361 we can rst permute switch 61 lt gt g to obtain g 6163 then permute 61 and 63 The concept of even and odd number of permutations is more general But for three elements it is useful to think in terms of cyclic and acyclic permutations If we expand i and 0 in terms of the right handed 61 62 63 then apply the 3 properties of the cross product ie in compact summation form aibi 8739 X 57 3 3 3 1 a a a a a E aiei b 578739 gt a gtlt b E i1 j1 i1 j F Waleffe Math 321 20080901 13 we obtain a i gtlt I agbg 7 a3b2 1 a3b1 7 a1b3 2 a1b2 7 a2b1 3 Verify this result explicitly What would the formula be if the basis was left handed That expansion of the cross product with respect to a right handed orthonormal basis 24 is often remembered using the formal determinant ie this is not a true determinant it s just convenient mnemonics 81 82 83 a1 a2 a3 bi 52 53 Double vector product Triple vector product72 gt Exercise Visualize the vector fi x x i Sketch it What are its geometric properties What is its magnitude Double vector products occur frequently in applications eg angular momentum of a rotating body directly or indirectly recall above discussion about mirror re ection and cross products in physics They have simple expressions 17 6 n a7 1 H7 5 we 25 One follows from the other after some manipulations and renaming of vectors but we can remem ber both at once as middle uector times dot product of the other two minus other uector within parentheses times dot product of the other two It is good to visualize the geometry of the double product i x is orthogonal to both i and Ii and 3917 fi x x 6is orthogonal to i x so 3917 is in the i I plane and 3917 afi BI for some scalars oz and 3 Now 3917 is also orthogonal to 6 so 17 e 04i 3 a 0 therefore 04 a and B 7ufi 3 for some scalar u and 3917 Ei 7 ufi This almost gives the formula 25 but we still need to show that u 71 Write 6 6H 61 where 6H is parallel and 61 perpendicular to fiin so 393 fix Iixc i and iii 1561 I36 I361 since parallel to i x I means perpendicular to i and So i Ii 3917 and e are all in the plane perpendicular to i x That s easier to visualize To 4p i determine u in 3917 211 7 lufi 213 take the cross product with hto obtain 3917 x I 216 x Now from the gure and the de nition of we have 17x b 7i x blc il lb Sian and the gure also shows that e loll cos 7 4p therefore i x I Ci u 71 since Sian cos7r2 7 4p with 4p positive clockwise U G l Exercises 1 Show that a x E a E a HP vs 13 2 If three vectors satisfy 139 If 6 0 show algebraically and geometrically that i x I b x 6 8x i Deduce the law of sines7 relating the sines of the angles of a triangle and the lengths of its sides 2The double Vector product is often called triple Vector product there are 3 Vectors but only 2 Vector products 3This is more useful than the confusing BACCAB rule for remembering the 2nd Try applying the BAGCAB mnemonic to I X 639 X it for confusing fun F Wale e Math 321 20080901 14 3 Show by vector algebra and geometry that all the vectors if such that i x if I have the form a b a 3046 5 VQER Hall 4 Show the Jacobi identity 6 x E x a E x 6x a 6x a x E o 5 If 77 is any unit vector show algebraically and geometrically that any vector i can be de composed as iii xd x 2d Hd L 26 The rst component is parallel to 77 the second is perpendicular to 77 6 A particle of charge q moving at velocity 3917 in a magnetic eld E experiences the Lorentz force F q39U x B Show that there is no force in the direction of the magnetic eld 7 A left handed basis 22 g is de ned by 6 6 6H and 61 x 62 ieg Show that 6 x 6 6 has the opposite sign to the corresponding expression for a right handed basis Vij k the de nition of the cross product remaining its right hand rule self Thus deduce that the formula for the components of the crossproduct in the left handed basis would all change sign 8 Prove 25 using the right handed orthonormal basis 61 ilfil g ii x 1lfi x and 2 E3 gtlt gl Then i 04 1 blgl bz a Clgl 02 2 ngg Visualize and explain why this is a general result and therefore a proof of the double cross product identity 9 Prove 25 using the right handed orthonormal basis 61 El61 g ii x 1lfi x and g 63 x 61 In that basis 139 a1 61 a2 2 and I b1 61 1262 but what is 67 Show by direct calculation that i gtlt I 01192 7 a2b1 3 and 17 l61a1b2 7 a2b1 2 gihlgi giwlfi Why is this Ei 3137 Ei and thus a proof of the identity 16 Indicial notation LeviCivita aka alternating or permutation symbol We have used the Kronecker symbol 14 to express all the dot products 67 in a very compact form There is a similar symbol eijk the Leui Ciuita or alternating symbol de ned as 1 if ij k is an even permutation of 123 gigk 71 if ij k is an odd permutation of 123 27 otherwise or explicitly 123 231 312 1 and 6213 132 321 71 all other eijk 0 Recall that for 3 elements an even permutation is the same as a cyclic permutation therefore eijk 67M Ekij Vij k why The eijk symbol provides a compact expression for the components of the cross product of right handed basis vectors gtlt gk Eijk 28 but since this is the k component of x 67 we can also write 3 Q gtlt gj Z Eijkgk 29 k1 F Wale e Math 321 20080901 15 Note that there is only one non zero term in the latter sum but then why can t we drop the sum Verify this result for yourself 161 Sigma notation free and dummy indices a a The expansion of vectors 139 and b in terms of basis 1 g 3 i ale 1 ag 39g ag 39g and b he 1 bg g bg 39g can be written compactly using the sigma E notation 3 3 i1 i391 We have introduced the Kronecker symbol 67 and the LeviCivita symbol gigk in order to write and perform our basic vector operations such as dot and cross products in compact forms when the basis is orthonormal and right handed for instance using 14 and 29 3 3 3 3 i1 i1 j1 i391 j1 3 3 3 Z ail 6 x Z Z Zaiqujk 5k 32 Note that i and j are dummy or summation indices in the sums 30 and 31 they do not have a speci c value they have all the possible values in their range It is their place in the particular expression and their range that matters not their name 3 3 3 3 62646Zaj2ak k 7 zak i 33 i1 j1 k1 k1 lndices come in two kinds the dummies and the free Here s an example gi Z ajbj Ci here j is a dummy summation index but i is free we can pick for it any value 1 2 3 Freedom comes with constraints If we use i on the left hand side of the equation then we have no choice we must use i for oi on the right hand side By convention we try to use i j k l m n to denote indices which are positive integers Greek letters are sometimes used for indices Mathematical operations impose some naming constraints however Although we can use the same index name i in the expansions of i and h when they appear separately as in 30 we cannot use the same index name if we multiply them as in 31 and 32 Bad things will happen if you do for instance 3 3 3 a x E 2 06 x Z 201 e x e 0 WRONG 35 i1 i1 i1 F Wale e Math 321 20080901 16 162 Einstein7s summation convention While he was developing the theory of general relativity Einstein noticed that many of the sums that occur in calculations involve terms where the summation index appears twice For example i appears twice in the single sums in 30 i and j appear twice in the double sum in 31 and i j and h each appear twice in the triple sum in 32 To facilitate such manipulations he dropped the 2 signs and adopted the summation convention that a repeated index implicitly denotes a sum over all values of that index In a letter to a friend he wrote I have made a great discouery in t39 t39 I have p l the quot sign euery time that the summation must be made over an index which occurs twice Thus with Einstein s summation convention we write a men 13 55 5 I m a x 13 ejkamjek 36 and any repeated index implies a sum over all values of that index This is a very useful and widely used notation but you have to use it with care and there are cases where it cannot be used lndices can never be repeated more than twice if they are that s probably a mistake as in 35 if not then you are out of luck and need to use E s or invent your own notation A few common operations in the t t We love to see a 6 involved in a sum since this collapses that sum This is called the substitution rule if 6 appears in a sum we can forget about it and eliminate the summation index for example 51739 agy 51mm 51m 511 37 5ij ijk 6m 0 37 note the second result 6 3 because k is repeated so there is a sum over all values of k and 6M 611 622 633 The last result is because em vanishes whenever two indices are the same That last expression this involves a double sum over i and over j The 61739 collapses one of those sums It doesn t matter which index we choose to eliminate since both are dummy indices Let s compute the 1 component of i x I from 36 We pick 1 because i j and k are already taken The 1 component is 51 ii X 1 Eijkaibjgk 51 Eijkaibj6kl Eijlaibj Elmnambn 38 what happened on that last step rst em 6 because i j k to k i j is a cyclic permutation which corresponds to an even number of permutation in space of odd dimension dimension 3 here and the value of em does not change under even permutations Then i and j are dummies and we renamed them m and n respectively being careful to keep the order The nal result is worth memorizing if 6 i x 1 the 1 component of c is cl qwmambn or switching indices to ij k 6 i gtlt b ltgt Ci Eijkajbk ltgt 6 a Eijkajbk 39 In the spirit of no pain no gain let s write the double cross product identity 6 x x 6 in this indicial notation Let 393 i x hthen the i component of the double cross product 17x is gigwick Now we need the j component of 3917 i x Since i and k are taken we use 1 m as new dummy indices and we have u Ejlmalbm So the i component of the double cross product fi x x 6is EijkEjlmalmek 40 Note that j k l and m are repeated so this expression is a quadruple sum According to our double cross product identity it should be equal to the i component of ii b 7 b E i for any F Wale e Math 321 20080901 17 i 1 6 We want the i component of the latter expression since i is a free index in 40 that i component is 110051 1ch 41 waitl isn t j repeated 4 times no it s not It s repeated twice in separate terms so this is a difference of two sums over j Since 40 and 41 are equal to each other for any i 1 6 this should be telling us something about gigk but to extract that out we need to rewrite 41 in the form albmck How by making use of our ability to rename dummy variables and adding variables using 6 Let s look at the rst term in 41 ajcg39bi here s how to write it in the form altka as in 40 ajCjb39 akckb 6lkalck6imbm 6lk6imalckbm 42 Do similar manipulations to the second term in 41 to obtain bjcja Sildkmalckbm and EijkE mazmek 6lk6im 6il6kmalckbm 43 Since this equality holds for any a1 ck bm we must have gigkegquot 61mm 7 6il6km That s true but it s not written in a nice way so let s clean it up to a form that s easier to reconstruct First note that 677 677 since 677 is invariant under a cyclic permutation of its indices So our identity becomes 677in 61mm 7 il6km We ve done that ipping so the summation indexj is in rst place in both 6 factors Now we prefer the lexicographic order i j k to j ki so let s rename all the indices being careful to rename the correct indices on both sides This yields Eijkeam 5j15km 5jm5ki 44 This takes some digesting Go through it carefully again And again as many times as it takes The identity 44 is actually pretty easy to remember and verify First EijkEl39lm is a sum over i but there is never more than one non zero term why7 Second the only possible values for that expression are 1 0 and 71 why The only way to get 1 is to have j k lm with j l 344 k m why but in that case the right hand side of 44 is also 1 why The only way to get 71 is to have j k ml with j m 344 k 1 why but in that case the right hand side is 71 also why Finally to get 0 we need j k or Z m and the right hand side again vanishes in either case For instance ifj k then we can switch j and k in one of the terms and 5j15km 5jm5ki jl6km 6km6jl 0 Formula 44 has a generalization that does not include summation over one index Eijkqmn 6il6jm6kn 6im6jn6kl 6m6ji5km 45 6im6jl6kn 7 5m5jm5ki 6il6jn6km note that the rst line correspond to ij k and l m 71 matching up to cyclic rotations while the second line corresponds to ij k matching with an odd acyclic rotation of l m Exercises 1 Explain why 677 677 767 for any integer ij k 2 Using 28 and Einstein s notation show that fix 6 eijkaibj and fix eijkaibj k Eijkajbk 03 Show that gig16177 26 by direct deduction and by application of 44 F Deduce 44 from 45 F Wale e Math 321 20080901 18 17 Mixed 0r B0x7 product and Determinant A mixed product4 of three vectors has the form Ii x 6 it involves both a cross and a dot product The result is a scalar We have already encountered mixed products eg eqn 28 but their geometric and algebraic properties are so important that they merit their own subsection 6x13az xaaaxa d0x g6xd 6d x0 46 ivolume of the parallelepiped spanned by i Ii 0 Take 139 and I as the base of the parallelepiped then i x I is perpendicular to the base and has magnitude equal to the base area The height is 2 Ewhere 2 is the unit vector perpendicular to the base 239e parallel to i x So the volume is Ii x 6 Signwise Ii x 6gt 0 if i I and 6 in that order form a right handed basis not orthogonal in general and ii x 6lt 0 if the triplet is left handed Taking I and 6 or Eand i as the base you get the same volume and sign The dot product commutes so x 3 139 fi x a yielding the identity dxg fi5x 47 We can switch the dot and the cross without changing the result We have shown 46 geometrically The properties of the dot and cross products yield many other results such as fix b 7b x fi6 etc We can collect all these results as follows A mixed product is one form of a scalar function of three vectors called the determinant a x B a detfi E a 48 whose value is the signed volume of the parallelepiped with sides i E 6 The determinant has three fundamental properties 1 it changes sign if any two vectors are permuted eg deti 135 idetI a a detI a a 49 2 it is linear in any of its vectors eg V a d detad 4 J 133 a detfi E a detcI 11a 50 4The mixed product is often called triple scalar product F Wale e Math 321 20080901 19 3 if the triplet 61 62 63 is right handed and orthonormal then det 1 2 3 1 51 Deduce these from the properties of the dot and cross products as well as geometrically Property 50 is a combination of the distributivity properties of the dot and cross products with respect to vector addition and multiplication by a scalar For example detad i113 046 d3 Ex a aa3 04 detfi a a I x a a dew 11a From these three properties you deduce easily that the determinant is zero if any two vectors are identical from prop 1 or if any vector is zero from prop 2 with 04 1 and d 6 and that the determinant does not change if we add a multiple of one vector to another for example x 177 detfi 6 a o 52 deti 53 135 detfi E a Geometrically this last one corresponds to a shearing of the parallelepiped with no change in volume or orientation One key application of determinants is detfi 0 6 7 0 ltgt i 1 Eform a basis 53 If detfi 1 a 0 then either one of the vectors is zero or they are co planar and i 0 6 cannot provide a basis for E3 This is how the determinant is introduced in elementary linear algebra But the determinant is so much more It determines the volume of the parallelepiped and its orientation The 3 fundamental properties fully specify the determinant as explored in exercises 5 6 below If the vectors are expanded in terms of a right handed orthonormal basis 239e i we I 13767 6 ck k summation convention then we obtain the following formula for the determinant in terms of the vector components detfi E a a x b a aibjck x 5 7k eijkaibjck 54 Expanding that expression Eijk aibjck a1b263 agbgcl 031ch 7 agblcg 7 agbgcl 7 a1b362 we recognize the familiar algebraic determinants a1 a2 a3 a1 bl C1 deti b a em aibjCk b1 b2 193 a2 b2 02 56 61 02 03 a3 53 03 Note that it does not matter whether we put the vector components along rows or columns This is a non trivial and important property of determinants see section on matrices This familiar determinant has the same three fundamental properties 49 50 51 of course F Wale e Math 321 20080901 20 1 it changes sign if any two columns or rows are permuted eg a1 bi Ci bi a1 Ci a2 b2 32 7 2 a2 32 a3 53 C3 53 a3 C3 2 it is linear in any of its columns or rows eg V 04 d1 d2d3 04011 d1 b1 C1 a1 b1 61 d1 b1 61 0402 d2 b2 32 04 a2 b2 32 d2 b2 32 04013 d3 53 C3 a3 53 C3 d3 53 C3 3 nally the determinant of the natural basis is 1 0 59 0 OHO HOD H H You deduce easily from these three properties that the det vanishes if any column or row is zero or if any two columns or rows is a multiple of another and that the determinant does not change if we add to one column row a linear combination of the other columns rows These properties allow us to calculate determinants by successive shearings and column swapping There is another explicit formula for determinants in addition to the Eijkaibjck formula it is the Laplace or Cofactor expansion in terms of 2 by 2 determinants eg a1 bi Ci a2 192 62 a1 2 CZ a2 1 1 a3 1 Cl 7 60 b 3 C3 3 C3 2 01 a3 3 03 where the 2 by 2 determinants are a1 b1 a2 2 a1b2 7 agbl This formula is nothing but 139 x 3 expressed with respect to a right handed basis To verify that compute the components of x a rst then dot with the components of i This cofactor expansion formula can be applied to any column or row however there are i1 factors that appear We won t go into the straightforward details but all that follows directly from the column swapping property 57 That s essentially the identities 139 x a 6x i Exercises a1b2 7 agbl is the signed area of the par 1 Show that the 2 by 2 determinant 1 b1 52 allelogram with sides 139 al 39l ag 39g 0 bl l bg 39g It is positive if 6137671 is a counterclockwise cycle negative if the cycle is clockwise Sketch of course 2 The determinant detfi 1 a of three oriented line segments i 1 Eis a geometric quantity Show that detfi b a al lbl sin gtcos0 Specify j and 0 Sketch 3 Show that ilfil detfi 1 a lfil When do the equalities apply Sketch H H H 03 H F H 01 F U a I 3500 0 to Wale e Math 321 20080901 Use properties 49 and 50 to show that death Ad 56 ha 5 a a detfi E a apdetfi a a BAdetd E a Apdetd a a Use properties 49 and 51 to show that det 39 k gigk Use property 50 and exercise 5 above to show that if fi we I biei ff ci i summation convention then detfi b a EijkaibjCk Prove the identity fi x and indicial notation fix d fi d 7 fi 3 using both vector identities Express fi x I fi x in terms of dot products of fi and V Show that a 303 E 7 and b fi Iiz is the square of the area of the parallelogram spanned by fi If A is the area the parallelogram with sides fi and Ii show that Tl G l 142 El 9 wt 9i wt 9i lf detfi Ii 3 7 0 then any vector 3917 can be expanded as 3917 mi 00 ya Find explicit expressions for the components 04 B 39y in terms of 3917 and the basis vectors fi b ff in the general case when the latter are not orthogonal Hint project on cross products of the basis vectors then collect the mixed products into determinants and deduce Cramer s rule Given three vectors fil fig fig such that D fil fig x fig 7 0 de ne 62a3gtlt61 figa1 This is the reciprocal basis of the basis fil fig fig i Showthatfifi 6 Vi 4123 7 77 7 77 ii Show that if 77 23 t a and 77 23 h 6 then u 77 a and t 77 6 So the components in one basis are obtained by projecting onto the other basis If fi and I are linearly independent and ffis any arbitrary vector nd 04 B and V such that ff afi 0b 39yfi x b Express 04 B and 39y in terms of dot products only Hint nd 04 and 0 rst then use 6H 6 7 CL Express fi x f2 in terms of dot products of fi I and f2 only Hint solve problem 13 rst a Provide an algorithm to compute the volume of the parallelepiped fi b a by taking only dot products Hint rectify the parallelepiped fi Ii 3 a fiI1f21 a fi I71 61L where I71 and 61 are perpendicular to fi and fin is perpendicular to both fi and Iii Explain geometrically why these transformations do not change the volume Explain why these trans formations do not change the determinant by using the properties of determinants F Wale e Math 321 20080901 22 a 16 If V is the volume of the parallelepiped with sides i b 6 show that V2 0 G l El El El El 0 G l El l l Tl 0l l El ol Oi ol Do this in several ways from problem 13 ii using indicial notation and the formula 45 18 Points Lines Planes etc Points and vectors are different We do not add points but vector addition is de ned However once a reference point has been picked called the origin 0 any point P is uniquely determined by specifying the vector FE This special vector is called the position vector It is often denoted if also Position vectors have a special meaning because they are tied to a speci c origin Representing points by position vectors is extremely useful as all vector operations are then available to us Examples 0 The center of mass F0 of a system of N particles of mass m located at position 17 i 1 N is de ned by M1 mmquot where M mi is the total mass This is a coordinate free expression for the center of mass In particular if all the masses are equal then for N 2 170 171 F22 for N 3170 F1 F2 D Show that the center of gravity of three points of equal mass is at the point of intersection of the medians of the triangle formed by the three points 0 The vector equation of a line parallel to i passing through a point F0 is F7 0xd 0lstgtlf f 0afi VaER 63 o The equation of a plane through F0 parallel to i and I with i x I 7 0 or equivalently perpendicular to 77 E i x b is it 0 0 cgt Fi quot 0afi i Va R 64 o The equation of a sphere of center F0 and radius R is lf if clRltgtFFcR V stl l1 65 D Find vector equations for the line passing through the two points 171 F2 and the plane through the three points 171 F2 173 D What is the distance between the point F1 and the plane through F0 perpendicular to i D What is the distance between the point F1 and the plane through F0 parallel to i and 1 D What is the distance between the line parallel to i that passes through point A and the line parallel to b that passes through point B D A particle was at point P1 at time 251 and is moving at the constant velocity 171 Another particle was at P2 at 252 and is moving at the constant velocity 172 How close did the particles get to each other and at what time What conditions are needed for a collision F Wale e Math 321 20080901 23 19 Vector function of a scalar variable The position vector of a moving particle is a vector function Ft of the scalar time t The derivative of a vector function is de ned as usual at F t h 7 F t fa 11m 66 dt haO h The derivative of the position vector is of course the instantaneous velocity vector 17t dFdt The position vector 17t describes a curve in three dimensional space the particle trajectory and 17t is tangent to that curve The derivative of the velocity vector is the acceleration vector 6t d39Udt We ll often use Newton s notation for time derivatives dFdt E F FFoh2 F etc We need to know how to manipulate derivatives of vector functions It is easy to show that the derivative of a sum of vectors is the sum of the derivatives d a da d5 7 i b 7 7 dt dt dt For the various products we can show by the standard product derivative trick recalled in class that d a dag dfi datidat db aa7 aa 31177Ebaa7 d a a dfi a d5 axb7gxbaxa then dta dat db tdg axb jigxbcaxcaxbdt therefore detfi E a det E a detfi a deti E g All of these are as expected but the formula for the derivative of a determinant is worth noting because it generalizes to any dimension 5 D Show that if t is any vector with constant norm 239e t t Constant V t then da 5a fa o Vt 67 dt The derivative of a constant norm vector is orthogonal to the vector D lfr where F 17t show that drdt E 17 D lf17t dFdt show that dFx 17dt Fx d dt In mechanics 17x m1 is the angular momentum of the particle of mass m and velocity 3917 with respect to the origin We now illustrate all these concepts and results by considering the basic problems of classical mechanics motion of a particle and motion of a rigid body 5For determinants in R3 it reads 511 b1 01 11 b1 i 512 72 02 a1 1 01 a2 72 02 d dt a3 73 ca as 173 as as 173 as and of course we could also take the derivatives along rows instead of columns F Wale e Math 321 20080901 24 110 Motion of a particle In classical mechanics the motion of a particle of mass m is governed by Newton s law 13 mi 68 where F is the resultant of the forces acting on the particle and fit d39Udt FFoh2 is its acceleration with Ft its position vector Newton s law is a vector equation Free motion If F 0 then d39Udt 0 so the velocity of the particle is constant 17t 170 say and its position is given by the vector differential equation dFdt 170 whose solution is 17t F0 t39UO where F0 is a constant of integration which corresponds to the position of the particle at time t O The particle moves in a straight line through F0 parallel to 170 Constant acceleration d2 117 a a w a at 7 a0 69 where io is a time independent vector Integrating we nd t2 1775 iot 170 1775 603 Tot To 70 where 170 and F0 are vector constants of integration They are easily interpreted as the velocity and position at t 0 Uniform rotation If a particle rotates with angular velocity to about an axis parallel to 77 that passes through the point Fa we take 1 and de ne w gt 0 for right handed rotation about 77 w lt 0 for left handed rotation then its velocity is 17t 63 x t 7 Fa where 413 E w is the rotation vector D Show that H t 7 Fa remains constant Calculate the particle acceleration if 413 and Fa are constants and interpret geometrically Find the force required to sustain such a motion Motion due to a central force A force F 7Frig where r that always points toward the origin if Fr gt 0 away if Fr lt 0 and depends only on the distance to the origin is called a central force The gravitational force for planetary motion and the Coulomb force in electromagnetism are of that kind Newton s law for a particle submitted to such a force is 117 ma 7Fr 17quot where Ft M is the position vector of the particle hence both r and i are functions of time and 3917 dFdt Motion due to such a force has two conserved quantities 0 Conservation of angular momentum V Fr F Wale e Math 321 20080901 25 rxi0ltgtiFx 0 gtrx EL0z 71 where L0 gt 0 and 2 are constants So the motion remains in the plane orthogonal to 2 Now Lo dt Fx dt 17 x at 2da2 where la is the triangular area swept by F in time dt This yields Kepler7s law The radius uector sweeps equal areas in equal times 0 Conseruatlon of energy kineth potential 117 m dt where dVrdr E Fr as by the chain rule dVrdt dVdrdrdt dVdr153917 This implies that d 17Fr17 170ltgtampltm lrgt 0 my Vrgt E0 72 where E0 is a constant The rst term mH39UHZ2 is the kinetic energy and the second Vr is the potential energy which is de ned up to an arbitrary constant The constant E0 is the total conserved energy Note that Vr and E0 can be negative but mH39UHZ2 2 0 so the physically admissible r domain is that were Vr is less than E0 111 Motion of a system of particles Consider N particles of mass mi at positions r7 i 1 N The net force acting on particle number i is Fi and Newton s law for each particle reads my Fi Summing over all is yields N N 2mm Zn i1 i1 Great cancellations occur on both sides On the left side let 1quot Fe s where F0 is the center of mass and 39i is the position vector of particle i with respect to the center of mass then Emir ZmiFc i MFCZml39 i 277113 07 as by de nition of the center of mass mmquot M170 where M m is the total mass If the masses mi are constants then migi 0 a misg 0 a misg 0 In that case my m On the right hand side by action reaction all internal forces cancel out and the resultant is therefore the sum of all external forces only F Therefore M45 135 73 where M is the total mass and 138 is the resultant of all external forces acting on all the particles The motion of the center of mass of a system of particles is that of a single particle of mass M with position vector F0 under the action of the sum of all external forces This is a fundamental theorem of mechanics F Wale e Math 321 20080901 26 There are also nice cancellations occurring for the motion about the center of mass This involves considering angular momentum and torques about the center of mass Taking the crossproduct of Newton s law mi F with g for each particle and summing over all particles gives a EgiXmiViE giXFi i 139 On the left hand side 1quot E Fe g and the de nition of center of mass implies mi 0 Therefore EgiXmiViE giXmiVc iE giXm EgiXmigi This last expression is the rate of change of the total angular momentum about the center of mass N E0 E Z x i1 On the right hand side one can argue that the internal forcqexerted by particle j on particle i is in the direction of the relative position ofj with respect to i fij E 17 7 By action reaction the force from i onto j is 7f 704ij17i 7 and the net contribution to the torque from the internal forces will cancel out 17139 x 1 17739 x 0 This is true with respect to any point and in particular with respect to the center of mass 39i x 1 x 0 Hence for the motion about the center of mass we have a disc Tie 74 where Tltegt g x is the net torque about the center of mass due to external forces only This is another fundamental theorem that the rate of change of the total angular momentum about the center of mass is equal to the total torque due to the external forces only D lf 1 Oz 17139 7 17739 and Oz 17739 7 17139 show algebraically and geometrically that g x 1 39j x 0 where is the position vector from the center of mass 112 Motion of a rigid body The two vector differential equations for motion of the center of mass and evolution of the angular momentum about the center of mass are sufficient to fully determine the motion of a rigid body A rigid body is such that all lengths and angles are preserved within the rigid body If A B and C are any three points of the rigid body then B E constant Kinematics of a rigid body Consider a right handed orthonormal basis 1t gt gt tied to the body These vectors are functions of time t because they are frozen into the body so they rotate with it However the basis remains orthonormal as all lengths and angles are preserved Hence t 39jt 67 V ij 1 2 3 and Vt and differentiating with respect to time 7 0 75 F Wale e Math 321 20080901 27 In particular as seen in an earlier exercise the derivative of a unit vector is orthogonal to the vector l e 0 V1 1 23 So we can write 7243M 1 v1123 76 gtlt Bi gtlt 8739 UJJ39 Now let 4313 Zwklgk wu i 012152 w31537 k so 01 is the k component of vector 43 Substituting in 77 gives 2 Eijkwki Z Eijkwkj 78 k k where as before gigk E x 67 k The sums over k have at most one non zero term This yields the three equations i j k 1 2 3 A 0131 0132 2 j k 2 31 a 0112 0113 79 ivjvk 37172 123 W21 7 The second equation for instance says that the rst component ofJJ g is equal to the rst component of 033 Now 01 is arbitrary according to 76 why so we can choose to de ne 0111 the rst component of 4131 for instance equal to the rst components of the other two vectors that are equal to each other 239e 0111 0112 0113 Likewise pick 0122 0123 0121 and 0133 0131 0132 This choice implies that 431032433203 80 The vector 03t is the Poisson vector of the rigid body The Poisson vector 03t gives the rate of change of any vector tied to the body Indeed if A and B are any two points of the body then the vector 6 E E can be expanded with respect to the body basis t 62t gt 675 C1 1lttgt C2 2lttgt C3 3t but the components 0 E 6t t are constants because all lengths and angles and therefore all dot products are timeinvariant Thus d6 3 d 3 EZcd7 zcaxeaxa i1 i1 This is true for any vector tied to the body material vectors implying that the Poisson vector is unique for the body F Wale e Math 321 20080901 28 Dynamics of rigid body The center of mass of a rigid body moves according to the sum of the external forces as for a system of particles A continuous rigid body can be considered as a continuous distribution of in nitesimal masses dm N a a E mis sdm i1 V where the three dimensional integral is over all points g in the domain V of the body dm is the measure of the in nitesimal volume element 1V or in other words dm pdV where p6 is the mass density at point For the motion about the center of mass the position vectors 39i are frozen into the body hence s 413 x 39i for any point of the body The total angular momentum for a rigid system of particles then reads 13 2mg x g 2mg x013 x 3 2m HaH2437 g g m 81 and for a continuous rigid body E 051203781503 m 232 V The Poisson vector is unique for the body so it does not depend on g and we should be able to take it out of the sum or integral That s easy for the Hgllzdi term but how can we get 43 out of the 43 dm term We need to introduce the concepts of tensor product and tensors to do this But we better talk about matrices rst 113 Cartesian Coordinates So far we have avoided using coordinates in order to emphasize the geometric 239e coordinate independent aspects of vectors and vector operations but coordinates are crucial for calculations A cartesian system of coordinates consist of three oriented orthogonal lines the m y and z coordinate axes passing through a point 0 the origin The orientation of the axes is usually chosen to correspond to the right hand rule A point P is then speci ed by its coordinates x y z with respect to the origin along each axis To describe displacements 239e vectors we need a set of basis vectors It makes sense to take a basis that is aligned with the coordinate axes Therefore we pick unit vectors zit g 2 aligned with each of the axes respectively and pointing in the positive direction6 A point with coordinates x y 2 then has position vector 17 133 my 22 or F m m y 39y z z Note that points have coordinates and vectors have components It is often more convenient to use subscripts writing 1 mg mg in lieu of z y In that notation the position vectors reads 7 9511331 9521172 953133 OF 77 95151 90252 90353 Exercises Express the lines planes and spheres of section 18 in terms of Cartesian coordinates 6Although we call i for instance a unit vector in the sense that 1 unit Vectors have no physical units The position Vector F and its components x y z have units of length The unit Vectors in physics are pure numbers indicating directions Direction vector77 is a better term F Wale e Math 321 20080901 29 2 Matrices 21 Orthogonal transformations Consider two orthonormal bases and 6 in 3D euclidean space so i 1 2 3 A vector 3917 can be expanded in terms of each bases as 3917 U161 U262 U363 and 3917 v l 39l U gg vg 3 What are the connections between the two sets of components 121122123 and 11 12 1 We can nd the relations between these coordinates using geometry but it is much easier and sys tematic to use vectors and vector operations Although the components are different 121122123 7 vi 12 1 the geometric vector 3917 is independent of the choice of basis thus 3 a a 1 E vil 3 a Uieiv i1 i1 and likewise where we de ned 85 These QM coefficients are the direction cosmes they equal the cosine of the angle between the direction vectors and A prion there are 9 such coefficients However orthonormality of both bases imply many constraints These constraints follow from eqns 83 84 which must hold for any 121122123 and why24 Substituting 84 into 83 watching out for dummy indices yields 3 3 3 v2 Z akam W a Z Qiko 67 86 131 j1 k1 Likewise substituting 83 into 84 gives 12 3 k1 739 Mo 3 Qkiij39Wv W 2amp2 ij 5ij 87 k1 H These two relationships have simple geometric interpretations Indeed QM 6 j can be inter preted as the j component of 6 in the 1 g 393 basis as well as the i component of 67 in the 1 2 393 basis Therefore we can write 3 3 T 25 5k 5k Z Qik 5k Qilgl 1252 Qi3 37 k 1 w H H 3 3 5739 25739 39 513 513 zij 515 Q1j 1 Q2793 Q3793 k1 k1 F Wale e Math 321 20080901 30 Then Qiijk 5M7 88 1 3 39 j E Z 6139 89 101 These are the orthogonality conditions orthonormality really satis ed by the Q 22 De nitions and basic matrix operations The 9 coe icients QM in 83 are the elements of a 3 by 3 matrix Q Le a 3 by 3 table with the rst index i corresponding to the row index and the second index j to the column index That Q was a very special 239e orthogonal matrix More generally a 3 by 3 real matrix A is a table of 9 real numbers A11 A12 A13 A E Aw E A21 A22 A23 90 A31 A32 A33 Matrices are denoted by a capital letter 69 A and Q and by square brackets By convention vectors in R3 are de ned as 3 by 1 matrices eg although for typographical reasons we ll often write 1 1 mg mg but not 1 m2 33 which would denote a 1 by 3 matrix or row vector The term matrix is similar to vectors in that it implies precise rules for manipulations of these objects for vectors these are the two fundamental addition and scalar multiplication operations with speci c properties see Sect 11 Matrixvector multiply Equation 83 shows how matrix vector multiplication should be de ned The matrix vector product A11 A 3 by 3 1 E R3 is a vector b in R3 whose i th component is the dot product of row i of matrix A with the column 13 Ail b ltgt bi ZAiij 739 The product is performed row by column This product is de ned only if the number of columns of A is equal to the number of rows of 13 A 2 by 1 vector cannot be multiplied by a 3 by 3 matrix Identity Matrix There is a unique matrix such that 13 13 V113 For 1 E R3 show that 100 1010 91 001 F Wale e Math 321 20080901 31 MatrixMatrix multiply Two successive transformation of orthogonal coordinates ie 3 3 x ZAx then xZBx l i ii 77 i i ii 7 j1 7391 can be combined into one transformation from xi to x Mo 3 3 2 ZBMAM 7 E 0 7 j1 k1 j1 where 3 BAMJ39 E CH Z BikAkj 92 k1 This de nes matrix multiplication The product of two matrices BA is a matrix C say whose i j element is the dot product of row i of B with column j of A As for matrix vector multiplication the product of two matrices is done row by column This requires that the length of the rows of the rst matrix equals the length of the columns of the second ie the number of columns of the rst must match the number of rows of the second The product of a 3 by 3 matrix and a 2 by 2 matrix is not de ned In general BA 7 AB matrix multiplication does not commute You can visualize this by considering two successive rotation of axes one by angle 04 about 63 followed by one by 3 about JZ This is not the same as rotating by 3 about 62 then by 04 about 5 3 You can also see it algebraically BA Z BikAk 7i 2 AikBk 2 AB k k Matrix transpose The transformation 84 involves the sum 7 Ajix j that is similar to the matrix vector multiply except that the multiplication is column by columnl To write this as a matrix vector multiply we de ne the transpose matrix AT whose row i correspond to column i of A If the ij element of A is AH then the ij element of AT is A ATM AM Then 3 l ltgt i1 ATilI 93 j1 A symmetric matrix A is such that A AT but an anti symmetric matrix A is such that A iAT Verify as done in class that the transpose of a product is equal to the product of the transposes in reverse order ABT BTAT F Wale e Math 321 20080901 32 Orthogonal Matrices Arbitrary matrices are typically denoted A while orthogonal matrices are typically denoted Q in the literature In matrix notation the orthogonality conditions 88 89 read QTQ QQT I 94 A matrix that satisfy these relationships is called an orthogonal matrix it should have been called orthonormal A proper orthogonal matrix has determinant equal to 1 and corresponds to a pure rotation An improper orthogonal matrix has determinant 1 It corresponds to a combination of rotations and an odd number of re ections The product of orthogonal matrices is an orthogonal matrix This is useful as any 3 by 3 proper orthogonal matrix can be decomposed into the product of three elementary rotations There are several ways to de ne these elementary rotations but a common one that corresponds to spherical coordinates is to 1 rotate by 04 about 63 2 rotate by 0 about g 3 rotate by 39y about 63 The 3 angles 04 B 39y are called Euler angles Hence a general 3 by 3 orthogonal matrix A can always be written as cos 39y sin 39y 0 cos B 0 7 sin 0 cos Oz sin 04 0 Q isin39y cos39y 0 0 1 0 7811104 cosa 0 95 0 0 1 sin 0 0 cos B 0 0 1 To de ne an arbitrary orthogonal matrix we can then simply pick any three arbitrary Euler angles 04 B 39y and construct an orthonormal matrix using 95 Another important procedure to do this is the GramSchmidt procedure pick any three 11 a2 a3 and orthonormalize them te 1 First de ne lt11 alHalH and t1 2 a2 7 a2 q1q17 a as r as mill 2 next de ne lt12 a zHa zH and a a a5 q2q27 3 nally de ne q3 lgHam The vectors q1 q2 q3 form an orthonormal basis This procedure generalizes not only to any dimension but also to other vector spaces 69 to construct orthogonal polynomials Exercises H Give explicit examples of 2 by 2 and 3 by 3 symmetric and antisymmetric matrices to If T plazas calculate 11T11 and 1111T 03 Show that IITZII and mmT are symmetric explicitly and by matrix manipulations F If A is a square matrix of appropriate size what is mTAm U Show that the product of two orthogonal matrices is an orthogonal matrix Interpret geo metrically a What is the general form of a 3 by 3 orthogonal and symmetric matrix 1 What is the orthogonal matrix corresponding to a re ection about the z 7 2 plane What is its determinant 00 What is the most general form of a 2 by 2 orthogonal matrix F Wale e Math 321 20080901 33 9 Suppose that you would like to rotate an object 239e a set of points about a given axis by an angle 39y Can you explain how to do this Hint 1 Translation express coordinates of any point Fwith respect to any point F0 on the rotation axis F7 170 2 Perform two elementary rotations to align the vertical axis with the rotation axis7 239e nd the Euler angles 04 and 3 Express the coordinates of 177 F0 in that new set of coordinates 3 Rotate the vector by 39y this is equivalent to multiplying by the transpose of the matrix corresponding to rotation of axes by 39y Then you need to re express the coordinates in terms of the original axesl that s a few multiplication by transpose of matrices you already have H O What is the rotation matrix corresponding to rotation by 7139 about g H H What is the matrix corresponding to right handed rotation by angle 04 about the direction 51 52 537 12 Find the components of a vector i rotated by angle 39y about the direction 61 262 263 13 Pick three non trivial but arbitrary vectors in R3 eg using Matlab s randn337 then construct an orthonormal basis q1 q2 q3 using the Gram Schmidt procedure Verify that the matrix Q q17q27q3 is orthogonal Note in particular that the rows are orthogonal eventhough you orthogonalized the columns only 14 Pick two arbitrary vectors a1 12 in R3 and orthogonalize them to construct q17 q2 Consider the 3 by 2 matrix Q q17q2 and compute QQT and QTQ Can you explain the results 23 Determinant of a matrix See earlier discussion of determinants section on mixed product The determinant of a matrix has the explicit formula detA eijkAilAJgAkg the only non zero terms are for i7j7 k equal to a permutation of 123 We can deduce several fundamental properties of determinants from that formula We can reorder AilAjzAkg into AllAzmAgn using an even number of permutations if i7j7 k is an even perm of 123 and an odd number for odd permutations So detA eijkA AJZAkg EWAUAZWAM detAT 96 Another useful result is that EijkAilAijkn EijkqmnAnAjzAks 97 Then it is easy to prove that detAB detA detB detAB EijkAilBllAijmgAkang EijkElmnAilAjzAkgBllBszng detA detB 98 One nice thing is that these results and manipulations generalize straightforwardly to any dimen sion 24 Three Views of Am b 241 Column View gt View b as a linear combination of the columns of A Write A as a row of columns7 A 111127 137 where alT 01114121 asil etc7 then b Ail 1011 2012 303 F Wale e Math 321 20080901 34 and b is a linear combination of the columns 11 a2 as If 1 is unknown the linear system of equations A11 b will have a solution for any b if and only if the columns form a basis ie iff deta1 12113 E detA 7 0 If the determinant is zero then the 3 columns are in the same plane and the system will have a solution only if b is also in that plane As seen in earlier exercises we can nd the components m1z2z3 by thinking geometrically and projecting on the reciprocal basis 69 b 12 x as detb a2 a3 E 99 m1 a1a2 x 13 deta1a2a3 L1kew1se deta1 b 13 deta1 12 b 902 7 903 deta1 12113 T deta1 12113 This is a nifty formula Component z equals the determinant where vector i is replaced by b divided by the determinant of the basis vectors You can deduce this directly from the algebraic properties of determinants for example detb 12113 detz1a1 2012 303 12113 1 deta1 a2 13 This is Cramer7s rule and it generalizes to any dimension however computing determinants in higher dimensions can be very costly and the next approach is computationally much more e icient 242 Row View gt View 1 as the intersection of planes perpendicular to the rows of A View A as a column of rows A 1 2 3T where f a11a12 am is the rst row of A etc then n1 11b1 bA1 g 1 ltgt 21b2 g 339bs and 1 is seen as the position vector of the intersection of three planes Recall that 1 C is the equation of a plane perpendicular to and passing through a point 110 such that 110 C for instance the point 110 To nd 1 such that A11 b for given b and A we can combine the equations in order to eliminate unknowns ie 7 7 n11b1 n11 b1 2illb2 ltgt 27a2 113 b27042b1 7 7 7 n31b3 n37043n11 b37043b1 where we pick 042 and 043 such that the new normal vectors Z g 7 042 1 and g g 7 043 1 have a zero 1st component ie 2 Oa 22a 23 g 0 ag2ag3 At the next step one de nes a g g 7 g Z picking 03 so that the 1st and 2nd components of g are zero ie g 00 agg And the resulting system of equations is then easy to solve by backward substitution This is Gaussian Elimination which in general requires swapping of equations to avoid dividing by small numbers We could also pick the 04 s and 0 s to orthogonalize the s just as in the Gram Schmidt procedure That is better in terms of roundoff error and does not require equation swapping but is computationally twice as expensive as Gaussian elimination F Waleffe Math 321 20080901 35 243 Linear Transformation of vectors into vectors gt View b as a linear transformation of 13 Here A is a black box7 that transforms the vector input 1 into the vector output b This is the most general View of A11 b The transformation is linear this means that Aam By aAa BAy V a B E R 11y E R 100 This can be checked directly from the explicit de nition of matrix vector multiply Z Aik04k 5 2 0414mm 2 514mm k k k This linearity property is a key property because if A is really a black box eg the matrix is not actually known it s just a machine that takes a vector and spits out another vector we can gure out the effect of A onto any vector 1 once we know A l A g A n This transformation View of matrices leads to the following extra rules of matrix manipulations Matrix Matrix addition A11 B11 A B1 ltgt ZAlkmk Z Bikxk Biwmk Vzk 101 k k k so matrices are added components by components and A B B A A B C A B C The zero matrix is the matrix whose entries are all zero M atriI scalar multiply Aaa aAa cgt Emma 2am Va 35 102 k k so multiplication by a scalar is also done component by component and 04BA a A 3aA In other words matrices can be seen as elements of a vector space This point of View is also useful in some instances in fact computer languages like C and Fortran typically store matrices as long vectors Fortran stores it column by column and C row by row The set of orthogonal matrices does NOT form a vector space because the sum of two orthogonal matrices is not in general an orthogonal matrix The set of orthogonal matrices is a group the orthogonal group 03 for 3 by 3 matrices The special orthogonal group 503 is the set of all 3 by 3 proper orthogonal matrices ie orthogonal matrices with determinant 1 that correspond to pure rotation not re ections The motion of a rigid body about its center of inertia is a motion in 503 not R3 503 is the con guration space of a rigid body Exercises D Pick a random 3 by 3 matrix A and a vector b ideally in matlab using its Arandn33 brandn31 Solve Am b using Cramer s rule and Gaussian Elimination ldeally again in matlab unless punching numbers into your calculator really turns you on Matlab knows all about matrices and vectors To compute deta1a2a3 detA and detb a2 as in matlab simply use detA det bA 2 A 3 Type help matfun or help elmat and or demos for a peek at all the goodies in matlab F Wale e Math 321 20080901 36 25 Eigenvalues and Eigenvectors Problem Given a matrix A7 nd 1 7 0 and A such that A11 A211 103 These special vectors are eigemectors for A They are simply shrunk or elongated by the transfor mation A The scalar A is the eigenvalue The eigenvalue problem can be rewritten A 7 AI1 0 where I is the identity matrix of the same size as A This will have a non zero solution iff detA 7 AI 0 104 This is the characteiistic equation for A If A is n loy n7 it is a polynomial of degree n in A called the characteristic polynomial 3 Vectors and Matrices in 71 dimensions 31 The vector space R 32 Matrices in R gtlt R 33 Determinants
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