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by: Malvina Orn


Malvina Orn
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This 12 page Class Notes was uploaded by Malvina Orn on Monday September 7, 2015. The Class Notes belongs to PHY 315 at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/181833/phy-315-university-of-texas-at-austin in Physics 2 at University of Texas at Austin.




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Date Created: 09/07/15
82500 A Review of Vector Calculus with Exercises TOPICS I Introduction II Integrals Line Surface and Volume III Gradient IV Divergence V Laplace and Poisson Equations VI Curl and Stokes Law These notes provide a quick review and summary of the concepts of vector calculus as used in electromagnetism They include a number of exercises with answers to illustrate the applications and provide familiarity with the manipulations Since a vector is naturally a spatial and geometrical object it is extremely useful to make sketches of the various functions and vector fields in the exercisesA The notation is conventional Vectors are denoted by boldface rA unit vectors as x and components either by subscript AX or as a triplet AXAyAZ I Introduction Since electromagnetism is inherently threedimensional the mathematical description is inevitably in threespace Relativity comes later Although one could consider any system of coordinates and some unusual ones are advantageous in certain problems we shall consider only the conventional Cartesian Xyz cylindrical r9z and spherical r9p systems cf Figure attached Since the fundamental concepts are phrased in terms of invariants they can all be generalized to coordinate systems of arbitrary dimension and form but it is more efficient to defer that treatment to tensor calculus which provides a more natural and thorough formalism A vector is a geometrical object with magnitude and direction independent of any particular coordinate system A representation in terms of components or unit vectors may be important for calculation and application but is not intrinsic to the concept of vector An equation A B states an equality independent of coordinates and thus requires that the representation of A in any coordinate system be identical to that of B in that system If an object were specified in some manner such this were not true the object would not be a vector Vectors satisfy addition D A B C A C BABCABC These properties are sufficient to justify the form A Ala where A is the invariant length and a shares the invariant direction but has unit length One can introduce another invariant the scalar product ArB AHB a b Here a b is interpreted as the invariant length of a in the direction b or vise versa and is geometrically the cosine of the angle between a and b The operation is commutative and distributive Any vector can therefore be represented in Cartesian components A AXAyAZ AXx Ay y AZ z where AX Arx etc In fact any set of orthogonal unit vectors e1 that is eirej 0 for i j can provide a unique representation of a vector with the usual property that 3 A39B 2 AiBi il One can also introduce a quotvector produc quot A X B but there are some surprising subtleties It is normally defined geometrically as C A X B with C AHB sin GAB and direction perpendicular to the plane of A B with the right hand rule With varying degrees of rigor and effort one then obtains the mnemonic using a determinant for a representation in orthogonal components Review of Vector Calculus 2 e1 e2 93 A1 A2 A3 B1 B2 B3 C Ironically this form is closer to the fundamental de nition Recalling or introducing two functions from linear algebra 51 1 if ij 0 otherwise Kronecker delta Silk l ifijk an even permutation of 123 1 if an odd permutation eg2l3 0 otherwise ie if any index is repeated as in 122 Permutation symbol the determinant expression is identical to the following definition 3 3 Ci 2 2 81JkAjBk j1 k1 This is actually the best definition of vector product it is easily generalized to tensors It is relatively straightforward to show that it also has the expected properties C is perpendicular to A and B because CrA 0 and CB 0 simple consequnces of the properties of permutation symbol The definition follows the righthand rule assuming ei are right handed and the equation for the magnitude can be established quickly from some identities below The fact that the sign of AxB changes between right and left handed coordinate systems raises an interesting point for vectors were defined as being invariant and independent of coordinate system The resolution of this paradox is that the vector product is not a true vector it is sometimes called a pseudovector It behaves like a vector otherwise but it is mathematically a tensor Tij AiBj AJ39Bi technically an antisymmetric tensor of rank two This detail has no effect on the discussion here but may relieve some doubts that physics is making some subtle mathematical error which might have unforeseen consequences Permutation symbols are quite useful once you are accustomed to them An important result of which you must convince yourself is that 3 2 eijkemnk 81 8 3981 5 k1 m n n m but a little consideration of the permutation symbol definition should suffice It becomes very clear in retrospect This result makes it easy to prove the standard identity A x BxC ArC B AB C The relation for the magnitude of the vector product can be proven as BxCrBxC 81JkBjCk elmn Ban 83H 5n Si 5m BjCkBan where the quottensorquot convention of summing over repeated indices is understood All indices in this case B2C2 BC2 B2C21 c0s29 B2C2sin29 Review of Vector Calculus 3 the required result 11 Integrals Line Surface and Volume A common integral which arises in several physical contexts is the line integral which is equivalent to a onedimensional integral b b E b IErdl 1E1 cos 9 d1 a a E taken along some specified path E between a and b If one introduces a special coordinate s which measures distance along the path and assumes that all quantities can be parameterized by that coordinate the integral becomes a simple integral L J Es cos 9s ds If the path follows a coordinate line or if one can choose a coordinate system for which the lines coincide with the path this form is easily constructed and the problem has probably been chosen so that the resulting integral is not too difficult For more complex paths however the explicit evaluation of s which one needs to perform the actual calculation may be awkward More often the path is specified by some parameterization rt xt yt zt where t might even be one of the coordinates and Exyz is also given Then and one can compute either from components or magnitudes and the cosine depending on convenience and again arrive at a simple integral t2 j ft dt t1 The procedure can be applied to other coordinate systems but the equation for dl must be modified appropriately dl dr r d9 dz dl dr r d9 r sin 9 dp Exercise 1 Evaluate the following line integrals JErdl for a E Ax2y Byz szz along the axes 000 gt100 gt110 gt 111 b E Ax2y Byz szz along the curve 2t1 t2 4t21 from t0 to t 1 c E Ar sin2 9 Bcos p Csin 9 cos p along spherical coordinate lines 17E20 gt 1n0a2n0 gt2n3nZ Review of Vector Calculus 4 d E Ar2 sin 0 B2 sin 0 cos 0 Cr along cylindrical coordinate lines from 201 gt 40 14JE1 Twodimensional integrals are surface integrals but the generalization of surface integrals in three dimensions is somewhat more complicated The integrals of concern for physics have the form 11 ErdS over some speci ed surface and the vector dS is defined as 1A1 dA 1A1 dX dy for example where n is the vector normal to the surface the only unique direction that can be associated with a surface and some convention is specified for which direction is positive These surface integrals are defined for all sorts of complicated surfaces To evaluate the integrals some mapping or parameterization of the surface in terms of two convenient variables for integration must be found Although such general surfaces are essential for proofs and general arguments the integrals which must generally be evaluated are much simpler A coordinate system can usually be chosen such that the surface of integration has one of the coordinates constant e g a sphere of r a and the other two provide natural variables on the surface This kind of integral is easily formulated as a conventional integral in two variables Exercise 2 Evaluate the following surface integrals a E AXZ By C24 over a rectangle in the yz plane 122 124132134 b E Az2y BXZ39 Cyzz over the simple unit cube between 000 and 111 c E Acos2 pr Bsin2 0 0 over the surface ofthe sphere r 2 d E Ay BXZZ39 Cz Cartesian over the surface of the sphere r 4 e E Ar3 cos 0 Bsin2 0 C22 over the cylindrical wedge OSrSl OSGSTE4 03232 f E Ar2 Br sin 0 Ccos p over the outside conical surface ISrS2 07E3 this is an open surface excluding the end faces Volume integrals are actually the least complicated variety of integral in three dimensions They have the basic form fr d3r with d3r dx dy dz r dr d0 dz r2 sin 9 dr d9 dp Review of Vector Calculus 5 for the volume element The only possible complication is the limits of integration the expression of the boundary of the volume Again one can usually choose the coordinate system so that the boundaries are easily represented often as sections of the coordinate surfaces Exercise 3 Evaluate the following volume integrals a p Axyz for a rectangular volume with sides along coordinate planes between 112 and 356 b p Ae39OLr for the cylinder 23235 r34 C p Ax2 for the sphere r32 d p Ae39OLr sin2 9 for the sphere r33 III Gradient The gradient arises from the generalization of a derivative of a function r that depends on a position that requires several variables to specify The derivative in any direction 1 is given by d A a f V r1 where the directional derivative in the direction 1 is given by the usual limit and this equation serves as the definition of gradient grad I E V In particular by choosing l to be the unit vectors in any speci ed system of coordinates explicit expressions for VQ in that system may be obtained If the coordinate is E dq aq dE WE alfg 3 Using this expression verify that the expressions for gradient in the three coordinate systems are B B B Vq Big g Cartes1an B 1 B B if J Cy11ndr1cal Bl 1 Bi 1 Bi Spherical Br r 39 a r sin 9 3p I n t w 0 dimensions one can visualize Qxy as a contour map hxy for which h is the altitude of the land at the point xy The vertical dashed lines indicate the elevation h 39 drawing Lines constant h are the contours of constant elevation shown as ovals above and the maxima and minima are peaks and holes One maximum shown The gradient vector is two dimensional in the xy plane for this case Vh is shown as an arrow for several representative points above The direction of Vh at each point indicates the direction of steepest rise and its magnitude is the steepness Note that Vh 0 at the summit and in general at any maximum or minimum The principal application of gradient in electromagnetism begins with the electric potential and Review of Vector Calculus the electrostatic relation E VQ spherical Exercise 4 a I NF b q A1 x2y222 c I Axy d e Ax2 ByZ f Q Arl 8 cos 9 g Qsin9cosp Not all vectors can be obtained as the gradient of a scalar Q The condition is that V x E 0 as discussed in Section IV If E VQ Q can be obtained in several ways one of which is simply to integrate the partial derivatives in sequence For example if E EXrEyrEZr lExrdx fyz fyZ lEyr t BIEXrdx By For the following Q nd E VQ cylindrical 8lt l spherical Ey1 dy gz Q Ax By Sketch for ABl and for Al B4 BIEXrdx 3y Review of Vector Calculus 7 This expression for fxy may next be substituted into I IEXrdx fyz and the result used in EZ1 EXP32 to obtain an equation for dgzdz which may then be integrated to nd xyz to within a constant The actual process is almost easier than this description but it is important to remember that integrating partial derivatives allows not constants of integration but the functions fyz gz Exercise 5 For the following quotEquot determine ifE V and if so nd I c E 6yz 6XZ 6Xy d E 4x2y 2x3z 3y2z e E cos2 0r2 2 sin 0 cos 0r2 0 cylindrical f E cos 0 cos pr3 sin 0 cos p2r3 cot 0 sin p2r3 spherical Potential differences can also be determined by directly inverting the de nition of directional derivative as a a bbIE39d1 Since the result is independent of the path chosen between a and b one can use any path It is often most convenient to take segments along the coordinate axes but one could follow any curve Exercise 6 For E 6xy 3x2 0 nd the potential of 110 with respect to the origin 000 by a Integrating the partial derivatives as in Exercise 2 to nd r and evaluating the difference b Evaluating the line integral along the axes on the path 000 gtl00 gtll0 and also along the path 000 gt0l0 gtll0 c Evaluating the line integral along the curve y x d Evaluating the line integral along an arbitrary yfx 20 where one need only require that f0 0 and fl 1 Exercise 7 For E 2r sin2 0 sin p 2r sin 0 cos 0sin p r sin 0 cos p in spherical coordinates nd the potential of 5 with respect to 000 and 200 by a Integrating the partial derivatives to nd I as in Exercise 2 and evaluating the differences b Evaluating the line integral along a radial line r 72 E from 000 to 53 for the rst answer note that 00000p a degeneracy c Evaluating the line integral along the radial line r00 from 200 to 500 along 0 to 530 and then along p to 5 for the second IV Divergence The term divergence as most of the terminology in vector calculus derives from the application to uid mechanics Those origins are useful in obtaining a physical feeling for the meaning Review of Vector Calculus 8 of the mathematics If Vr represents the vector uid velocity the divergence of V div V or VrV indicates the extent to which the ow is diverging from the point 1 it suggests that there is a source of uid there The divergence operator is de ned to satisfy the divergence theorem H ErdS er d3r where it is understood that the surface integral covers the closed surface that encloses the volume over which the second integral is taken and the normal dS points outward By considering small volumes with sides along the coordinate axes in each coordinate system expressions for the divergence may be obtained Bx By 1m 1919 92 r Br r 39 32 m m 9Eltp r2 Br r sin 9 39 r sin 9 3p For spherical coordinates this is illustrated in the figure The sides of the peculiar solid volume element are Ar in the radial direction and rAG in the 9 direction and r sin 9 Ap in the p direction at the base Considering in turn the pairs of r 9 and p Ecp faces in the surface integral one obtains the complicated express1ons ErrArrArA9rArsin 9 Ap ErrrA9 r sin 9 A p Ee9A9Ar r sin9A9 Ap Ee9Ar r sin 9 Ap EppAp EppAr rAG because the lengths of the edges are along 9 and p differ for different surfaces If one expands fxAx in a Taylor series this expression for the surface integral becomes 2 r2 sin GArAG Ap 1398 sin GArAG Np 1 a E E r2sin9ArA9Ap r sin 9 3p Since the common factor r2 sin 9 Ar A9 Ap is the spherical volume element r2 sin 9 dr d9 dp its coef cient should be the divergence in order to make the divergence theorem correct One can quickly verify that the terms are indeed simply the expansion of the expression for divergence in spherical coordinates listed above Review of Vector Calculus 9 Exercise 8 Compute the divergence of the following vector elds Be sure to sketch the vectors and divergence and consider the physical interpretation if the vector is uid velocity or an electric eld charge density a E3x2 o 0 b E 6Xy 3X2 0 c E y X6y22 Zyz d E x y zx2y222g e E r392 0 0 spherical f E r 0 0 for rltb and E b2r 0 0 elsewhere cylindrical g E e39OLrZrz 0 0 spherical Exercise 9 Confirm the divergence theorem by evaluating both the surface and volume integrals over a unit cube with sides along the axes between 000 and 111 for the following vector fields a E 2Xyz X22 Xzy b E yz X z y Exercise 10 Confirm the divergence theorem by evaluating both the surface and volume integrals over a cylinder centered on the z axis of radius b for 03231 for E r cos2 0 r sin2 0 AZ Exercise 1 Confirm the divergence theorem by evaluating both the surface and volume integrals over the unit sphere centered on the origin for E Ar Dr sin 0 0 Certain aspects of this are subtle Since the normal to the surface is only in the radial direction only the contribution to the divergence from Er is in some sense quotcountedquot in the integral Other contributions to the divergence may occur as in this calculation but they vanish by symmetry in the volume integral You might think you could defeat this by choice of E9 e g E9 Dr cos 0 but then the divergence diverges at 00TE You have actually introduced line charges along the z axis exactly analogous to the case E r392 0 0 which appears superficially to have div E0 although HErdS 47E because the divergence is singular at r0 and is actually a delta function point charge The divergence theorem must be used very carefully if div E diverges at any point in the volume Exercise 12 Confirm the divergence theorem by evaluating both the surface and volume integrals over the unit northern hemisphere centered on the origin with its bottom plane defined by 07E 2 for E Ar Dr sin 0 0 V Laplace and Poisson Equations The combination E V and VE pl 80 define the Poisson equation V2 preo where the Laplacian operator V2 is defined as the divergence of the gradient using the operators already cited and may be summarized as ra r 2 7 V P 3x2 By2 322 Review of Vector Calculus 10 12 31 LE 327 irdradr r2392 322 1 a a 1 a a 1 32 r Z gaza fyr r2 sin 9 Slne r2 sin2 9 B Q Exercise 13 Determine the charge density required to support the following potentials Note that a particular solution for Qr depends not only on pr but also on the boundary conditions or symmetry a I Axyz b Axyz c Q Ar2 both cylindrical and spherical d I AX2 e I Ayzz f Q r2 sin 39 cylindrical g Q r3 cos 9 sin2 p Much of electrostatics as well as a number of other physics problems reduce to solving the Laplace or Poisson equations given pr and boundary conditions It is the boundary conditions which actually cause the greatest complexity VI Curl and Stokes Law The vector operator curl or circulation is constructed to satisfy Stokes Law as 3111 vards where the surface S is any surface bounded by the closed curve around which the line integral is taken and the positive normal is determine by the right hand rule following the direction of the line integral By considering differential areas defined by increments along the coordinate axes formulas can be obtained in various systems 3amp 6131an 9B2 6131 613 VXB ay 39 az ax 39 By ragBZ BZBr r Br r39 7 1313Z BB9BBr 3131130139 1 339 9B BB9 BB 3B 3Be BB 1 E11m1 13 3 1 Review of Vector Calculus 11 An example of this calculation curl Br for cylindrical coordinates is provided in the figure The 9 component of curl B9 curl is obtained b considering the Ar AZ loop The Brdl BZrAz BrzAzAr BZrArAz BrzAr BBZBr BBrBzArAz which is indeed the component of curl times the element of area ArAz as in Stokes law T e radial component is similar with Ed BezrA9 BZ9A9AZ BezAzrA9 7 BB9 1 BB2 BZ9AZ 7 W g I AGAZ again the component of curl times the element of area The z corstmnent is the most complicated with B d Br9Ar BerArrArA9 7 1 BB1 Br9A9Ar BerrA9 7 F g curl BZ 33 rArAG which is again the expansion of the expression for curl times the element of area Note that one must use a righthanded coordinate system and obey the sign conventions to obtain the correct result Exercise 14 The significance and terminology of curl can be seen by considering a uid rotating about the z axis as a rigid body The velocity is simply v 0 or 0 in cylindrical coordinates Con rm that V x v is the angular velocity 0 Verify Stokes law using a circle ra An important consequence of Stokes law which was used in Section I is its implication for path independence of integrals a SlEgtdl0 ltgt IErdl a b b If the line integral is pathindependent the integral around a closed path must be zero and vise versa This may be easily seen because the difference between any two path integrals between a and b is a closed path integral Therefore VxE 0 is a necessary and sufficient condition for the existence of a potential E V Actual calculations with Stokes law are primarily associated with the two curl equations of electromagnetism Exercise 15 For each of the following magnetic fields compute the curl the associated current distribution and con rm Stokes law by computing each integral as suggested a b 0 Exercise 16 Review of Vector Calculus 12 B Axy Ay22 CX for the rectangle lll32l23 20 B 0 Ar2 0 cylindrical for the circle r 6 and computing the surface integral both for the plane surface and for a hemisphere bounded by the circle B 0 0 Ar sin 9 spherical for the equator r 2 and the equatorial plane As an example of a vector eld with both divergence and curl consider a velocity eld in cylinrical coordinates VOr VOOL 0 This describes a uid ow with streamlines r 1990 which are spirals the reverse of ow down a drain Sketch and check this statement Con rm that the divergence is zero except for the deltafunction source at r0 and that the curl is VGoar z 5 a b C d e f ANSWERS C3 2111A 4666B 3877C TEE2 6A C 2 47cA 2567cC3 47cC 8 77cB2 840A 67cA 1287cA15 87EA32OL3 90L 60L2 20L3e3930 Q Mr Q N X2y222 I Axy Q AX By Q AX2 By2 Q Ar1 8 cos 9 Qsin9cosp spherical cylindrical 8lt l spherical Q 3X22xy None I 6Xyz Q 2X2 2xy 3yz 22 Q cos2 9r Q cos 9 cos p2r2 6 Q 3X2y AQ 3


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