2.4 Orthogonal Coordinate Systems (pp.16-33)

Transcription

1 8/26/2004 sec 2_4 blank.doc 1/6 2.4 Orthogonal Coordinate Sstems (pp.16-33) 1) 2) Q: A: Definition: ).

2 8/26/2004 sec 2_4 blank.doc 2/6 A. Coordinates * * * Point P(0,0,0) is alwas the origin. HO: Cartesian Coordinates HO: Clindrical Coordinates HO: Spherical Coordinates B. Coordinate Transformations

3 8/26/2004 sec 2_4 blank.doc 3/6 HO: Coordinate Transformations Eample: Coordinate Transformations C. Base Vectors * * HO: Base Vectors HO: Cartesian Base Vectors

4 8/26/2004 sec 2_4 blank.doc 4/6 D. Vector Epansion using Base Vectors Q: A: e.g., or B = B ˆ a + B ˆ a + B ˆ a C = C ˆ a + C ˆ a + C ˆ a HO: Vector Epansion using Base Vectors E. Spherical and Clindrical Base Vectors HO: Spherical Base Vectors HO: Clindrical Base Vectors

5 8/26/2004 sec 2_4 blank.doc 5/6 F. Vector Algebra and Vector Epansions HO: Vector Algebra using Orthonormal Base Vectors G. The Vector Field * * * * This means that the 3 scalar components of vector field are each a scalar field! HO: Vector Fields

6 8/26/2004 sec 2_4 blank.doc 6/6 HO: Epressing Vector Fields with Coordinate Sstems H. The Position Vector We call this directed distance the position vector. HO: The Position Vector HO: Applications of the Position Vector HO: Vector Field Notation HO: A Galler of Vector Fields

7 8/16/2004 Cartesian Coordinates.doc 1/2 Cartesian Coordinates You re probabl familiar with Cartesian coordinates. In twodimensions, we can specif a point on a plane using two scalar values, generall called and. - ais P(,) origin - ais We can etend this to three-dimensions, b adding a third scalar value. - ais P(,,) origin - ais - ais Jim Stiles The Universit of Kansas Dept. of EECS

8 8/16/2004 Cartesian Coordinates.doc 2/2 Note the coordinate values in the Cartesian sstem effectivel represent the distance from a plane intersecting the origin. For eample, =3 means that the point is 3 units from the - plane (i.e., the = 0 plane). Likewise, the coordinate provides the distance from the - (=0) plane, and the coordinate provides the distance from the - ( =0) plane. Once all three distances are specified, the position of a point is uniquel identified. P(0,0,0) 3 2 P(2,3,2.5) 2.5 Jim Stiles The Universit of Kansas Dept. of EECS

9 8/16/2004 Clindrical Coordinates.doc 1/2 Clindrical Coordinates You re probabl also familiar with polar coordinates. In twodimensions, we can also specif a point with two scalar values, generall called ρ and φ. ρ P(ρ,φ) φ We can etend this to three-dimensions, b adding a third scalar value. This method for identifing the position of a point is referred to as clindrical coordinates. P(ρ,φ,)

10 8/16/2004 Clindrical Coordinates.doc 2/2 Note the phsical significance of each parameter of clindrical coordinates: 1. The value ρ indicates the distance of the point from the - ais ( 0 ρ < ). 2. The value φ indicates the rotation angle around the -ais ( 0 φ < 2 π ), precisel the same as the angle φ used in spherical coordinates. 3. The value indicates the distance of the point from the - ( = 0) plane ( < < ), precisel the same as the coordinate used in Cartesian coordinates Once all three values are specified, the position of a point is uniquel identified. P(0,φ,0) 60 3 P(3, 60,2.5) 2.5

11 8/16/2004 Spherical Coordinates.doc 1/2 Spherical Coordinates * Geographers specif a location on the Earth s surface using three scalar values: longitude, latitude, and altitude. * Both longitude and latitude are angular measures, while altitude is a measure of distance. * Latitude, longitude, and altitude are similar to spherical coordinates. * Spherical coordinates consist of one scalar value (r), with units of distance, while the other two scalar values ( θ, φ ) have angular units (degrees or radians). P( r, θ, φ )

12 8/16/2004 Spherical Coordinates.doc 2/2 1. For spherical coordinates, r (0 r < ) epresses the distance of the point from the origin (i.e., similar to altitude). 2. Angle θ (0 θ π ) represents the angle formed with the -ais (i.e., similar to latitude). 3. Angle φ (0 φ < 2π ) represents the rotation angle around the -ais, precisel the same as the clindrical coordinate φ (i.e., similar to longitude). θ = 45 P(3.0,45,60 ) P(0,θ,φ) r =3.0 φ = 60 Thus, using spherical coordinates, a point in space can be unambiguousl defined b one distance and two angles.

13 8/16/2004 Coordinate Transformations.doc 1/4 Coordinate Transformations Sa we know the location of a point, or the description of some scalar field in terms of Cartesian coordinates (e.g., T (,,)). What if we decide to epress this point or this scalar field in terms of clindrical or spherical coordinates instead? Q: How do we accomplish this coordinate transformation? A: Eas! We simpl appl our knowledge of trigonometr. We see that the coordinate values, ρ, r, and θ are all variables of a right triangle! We can use our knowledge of trigonometr to relate them to each other. In fact, we can completel derive the relationship between all si independent coordinate values b considering just two ver important right triangles! Hint: Memorie these 2 triangles!!!

14 8/16/2004 Coordinate Transformations.doc 2/4 Ver Important Right Triangle #1 P It is evident from the triangle that, for eample: = ρ = r = θ =

15 8/16/2004 Coordinate Transformations.doc 3/4 Likewise, the coordinate values,, ρ, and φ are also related b a right triangle! Ver Important Right Triangle #2 P From the resulting triangle, it is evident that: = ρ cosφ = ρ sinφ ρ = φ = tan cos sin = = ρ ρ Combining the results of the two triangles allows us to each coordinate set in terms of each other:

16 8/16/2004 Coordinate Transformations.doc 4/4 Cartesian and Clindrical = ρ cosφ = ρ sin φ = ρ = + φ = tan = 1 (be careful!) Cartesian and Spherical = r sin θ cosφ = r sin θ sinφ = r cosθ 2 r = + + θ = cos φ = tan Clindrical and Spherical ρ = r sinθ φ = φ = r cosθ r = ρ + θ = φ = φ -1 tan ρ

17 8/16/2004 Eample Coordinate Transformations.doc 1/4 Eample: Coordinate Transformations Sa we have denoted a point in space (using Cartesian Coordinates) as P ( =-3, =-3, =2). Let s instead define this same point using clindrical coordinates ρ, φ, : ρ ( ) ( ) = + = + = φ = 3 = = 3 [] = = tan tan tan 1 45 Therefore, the location of this point can perhaps be defined also as P ( ρ = 3 2, φ = 45, = 2). Q: Wait! Something has gone horribl wrong. Coordinate φ = 45 indicates that point P is located in quadrant I, whereas the coordinates =-3, =-3 tell us it is in fact in quadrant III! Jim Stiles The Universit of Kansas Dept. of EECS

18 8/16/2004 Eample Coordinate Transformations.doc 2/4 A: The problem is our interpretation of the inverse tangent! Remember that 0 φ < 360, so that we must do a four quadrant inverse tangent. Your calculator likel onl does a two quadrant inverse tangent (i.e., 90 φ 90 ), so be careful! Therefore, if we correctl find the coordinate φ : φ = tan = tan = II φ = 225 I φ = 45 III IV P The location of point P can be epressed as either P (=-3, =-3, =2) or P ( ρ = 3 2, φ = 225, = 2). Jim Stiles The Universit of Kansas Dept. of EECS

19 8/16/2004 Eample Coordinate Transformations.doc 3/4 We can also perform a coordinate transformation on a scalar field. For eample, consider the scalar field (i.e., scalar function): g ( ρ, φ, ) = ρ 3 sinφ Lets tr to rewrite this function in terms of Cartesian coordinates. We first note that since ρ = +, ( ) / ρ = Now, what about sinφ? We know that φ tan [ ] might be tempted to write: 32 =, thus we sinφ -1 = sin tan Although technicall correct, this is one ugl epression. We can instead turn to one of the ver important right triangles that we discussed earlier: P ρ From this triangle, it is apparent that: φ sinφ = = ρ + Jim Stiles The Universit of Kansas Dept. of EECS

20 8/16/2004 Eample Coordinate Transformations.doc 4/4 As a result, the scalar field can be written in Cartesian coordinates as: 3 2 g (,, ) = ( + ) + ( ) = + Remember, although the scalar fields: and: (,, ) = ( + ) g g ( ρ, φ, ) = ρ 3 sinφ look ver different, the are in fact eactl the same functions onl epressed using different coordinate variables. For eample, if ou evaluate each of the scalar fields at the point described earlier in the handout, ou will get eactl the same result! (,, ) g = 3 = 3 = 2 = 108 g( ρ = 3 2, φ = 225, = 2) = 108 Jim Stiles The Universit of Kansas Dept. of EECS

21 8/18/2004 Base Vectors.doc 1/4 Base Vectors Q: You said earlier that vector quantities (either discrete or field) have both and magnitude and direction. But how do we specif direction in 3-D space? Do we use coordinate values (e.g.,,, )?? A: It is ver important that ou understand that coordinates onl allow us to specif position in 3-D space. The cannot be used to specif direction! The most convenient wa for us to specif the direction of a vector quantit is b using a well-defined orthornormal set of vectors known as base vectors. Recall that an orthonormal set of vectors, sa ˆ a 1, ˆ a 2, ˆ a 3, have the following properties: 1. Each vector is a unit vector: ˆ a ˆ a = ˆ a ˆ a = ˆ a ˆ a = Each vector is mutuall orthogonal: ˆ a ˆ a = ˆ a ˆ a = ˆ a ˆ a =

22 8/18/2004 Base Vectors.doc 2/4 Additionall, a set of base vectors ˆ a 1, ˆ a 2, ˆ a 3 must be arranged such that: ˆ a ˆ a = ˆ a, ˆ a ˆ a = ˆ a, ˆ a ˆ a = ˆ a â â 2 â 1 An orthonormal set with this propert is known as a righthanded sstem. All base vectors ˆ a 1, ˆ a 2, ˆ a 3 must form a right-handed, orthonormal set. Recall that we use unit vectors to define direction. Thus, a set of base vectors defines three distinct directions in our 3- D space! Q: But, what three directions do we use?? I remember that ou said that there are an infinite number of possible orientations of an orthonormal set!!

23 8/18/2004 Base Vectors.doc 3/4 A: We will define several sstematic, mathematicall precise methods for defining the orientation of base vectors. Generall speaking, we will find that the orientation of these base vectors will not be fied, but will in fact var with position in space (i.e., as a function of coordinate values)! Essentiall, we will define at each and ever point in space a different set of basis vectors, which can be used to uniquel define the direction of an vector quantit at that point! Q: Good goll! Defining a different set of base vectors for ever point in space just seems dad-gum confusing. Wh can t we just fi a set of base vectors such that their orientation is the same at all points in space? A: We will in fact stud one method for defining base vectors that does in fact result in an othonormal set whose orientation is fied the same at all points in space (Cartesian base vectors). However, we will stud two other methods where the orientation of base vectors is different at all points in space (spherical and clindrical base vectors). We use these two methods to define base vectors because for man phsical problems, it is actuall easier and wiser to do so!

24 8/18/2004 Base Vectors.doc 4/4 For eample, consider how we define direction on Earth: North/South, East/West, Up/Down. Each of these directions can be represented b a unit vector, and the three unit vectors together form a set of base vectors. Think about, however, how these base vectors are oriented! Since we live on the surface of a sphere (i.e., the Earth), it makes sense for us to orient the base vectors with respect to the spherical surface. What this means, of course, is that each location on the Earth will orient its base vectors differentl. This orientation is thus different for ever point on Earth a method that makes perfect sense!

25 8/26/2004 Cartesian Base Vectors.doc 1/1 Cartesian Base Vectors As the name implies, the Cartesian base vectors are related to the Cartesian coordinates. Specificall, the unit vector a ˆ points in the direction of increasing. In other words, it points awa from the - (=0) plane. Similarl, aˆ and a ˆ point in the direction of increasing and, respectfull. aˆ aˆ aˆ We said that the directions of base vectors generall var with location in space Cartesian base vectors are the eception! Their directions are the same regardless of where ou are in space.

26 8/26/2004 Vector Epansion using Base Vectors.doc 1/6 Vector Epansion using Base Vectors Having defined an orthonormal set of base vectors, we can epress an vector in terms of these unit vectors: A = A ˆ a + A ˆ a + A ˆ a Note therefore that an vector can be written as a sum of three vectors! * Each of these three vectors point in each of the three orthogonal directions a ˆ, a ˆ, a ˆ. * The magnitude of each of these three vectors are determined b the scalar values A, A, and A. * The values A, A, and A are called the scalar components of vector A. * The vectors A ˆ a, components of A. A ˆ a, A ˆ a are called the vector

27 8/26/2004 Vector Epansion using Base Vectors.doc 2/6 Q: What the heck are scalar the components A, A, and A, and how do we determine them?? A: Use the dot product to evaluate the epression above! Begin b taking the dot product of the above epression with unit vector a ˆ : ( ) A ˆ a = A ˆ a + A ˆ a + A ˆ a ˆ a = Aaˆ ˆ a + Aaˆ ˆ a + Aaˆ ˆ a But, since the unit vectors are orthogonal, we know that: ˆ a ˆ a = 1 ˆ a ˆ a = 0 ˆ a ˆ a = 0 Thus, the epression above becomes: A = A ˆ a In other words, the scalar component A is just the value of the dot product of vector A and base vector a ˆ. Similarl, we find that: A = A ˆ a and A = A ˆ a

28 8/26/2004 Vector Epansion using Base Vectors.doc 3/6 Thus, an vector can be epressed specificall as: ( ) ( ) ( ) A= A ˆ a ˆ a + A ˆ a ˆ a + A ˆ a ˆ a = Aaˆ + Aaˆ + Aaˆ We can demonstrate this vector epression geometricall. Aa ˆ A = A ˆ a + Aaˆ a ˆ aˆ A ˆ a Note the length (i.e., magnitude) of vector A can be related to the length of vector Aa ˆ using trigonometr: A = A cosθ A aˆ Aaˆ θ A A = A ˆ a + Aaˆ aˆ

29 8/26/2004 Vector Epansion using Base Vectors.doc 4/6 Likewise, we find that the scalar component A is related to A as: A = A ˆ a + Aaˆ aˆ θ A A ˆ a aˆ A = Α cosθ A From this geometric interpretation, we can see wh we often refer to the scalar component A as the scalar projection of vector A onto vector (direction) a ˆ. Likewise, we often refer to the vector component A ˆ a as the vector projection of vector A onto vector (direction) a ˆ. As ou ma have alread noticed, the scalar component A, which we determined geometricall, can likewise be epressed in terms of a dot product! A = = Α Α = Α ˆ a cosθ A ˆ a cosθ A

30 8/26/2004 Vector Epansion using Base Vectors.doc 5/6 Accordingl, we find that the scalar component of vector A are determined b doting vector A with each of the three base vectors ˆ a, ˆ a, ˆ a : A = A ˆ a A = A ˆ a A = A ˆ a Said another wa, we project vector A onto the directions ˆ a, ˆ a, ˆ a. Either wa, the result is the same as determined earlier: ever vector A can be epressed as a sum of three orthogonal components: ( ) ( ) ( ) A= A ˆ a ˆ a + A ˆ a ˆ a + A ˆ a ˆ a = Aaˆ + Aaˆ + Aaˆ For eample, consider a vector A, along with two different sets of orthonormal base vectors: A aˆ ˆ2 a aˆ a ˆ1

31 8/26/2004 Vector Epansion using Base Vectors.doc 6/6 The scalar components of vector A, in the direction of each base vector are: A = A ˆ a = 20. A = A ˆ a = 15. A = A ˆ a = 00. A = A ˆ a = A = A ˆ a = 25. A = A ˆ a = Using the first set of base vectors, we can write the vector A as: A A = A ˆ a + A ˆ a + A ˆ a = 20. ˆ a ˆ a 15. ˆ a 20. ˆ a Or, using the second set, we find that: A = Aaˆ + Aaˆ + Aaˆ = 25. ˆ a 2 A 25a. ˆ2 It is ver important to realie that: A = 20. ˆ a ˆ a = 25. ˆ a2 In other words, both epressions represent eactl the same vector! The difference in the representations is a result of using different base vectors, not because vector A is somehow different for each representation.

32 8/26/2004 Spherical Base Vectors.doc 1/3 Spherical Base Vectors Spherical base vectors are the natural base vectors of a sphere. â r points in the direction of increasing r. In other words â r points awa from the origin. This is analogous to the direction we call up. â points in the direction of increasing θ. This is θ analogous to the direction we call south. â points in the direction of increasing φ. This is φ analogous to the direction we call east. aˆr aˆφ aˆθ

33 8/26/2004 Spherical Base Vectors.doc 2/3 IMPORTANT NOTE: The directions of spherical base vectors are dependent on position. First ou must determine where ou are in space (using coordinate values), then ou can define the directions of ˆar, ˆa ˆ θ, aφ. Note Cartesian base vectors are special, in that their directions are independent of location the have the same directions throughout all space. Thus, it is helpful to define spherical base vectors in terms of Cartesian base vectors. It can be shown that: ˆa ˆa = sinθ cosφ r ˆa ˆa = sinθ sinφ r ˆa ˆa = cosθ r ˆa ˆa = cosθ cosφ θ ˆa ˆa = cosθ sinφ θ ˆa ˆa = sinθ θ ˆa ˆa = sinφ φ ˆa ˆa = cosφ φ ˆa ˆa = 0 φ Recall that an vector A can be written as: ( ) ( ) ( ) A = A ˆa ˆa + A ˆa ˆa + A ˆa ˆa. Therefore, we can write â r as, for eample: ( ) ( ) ( ) ˆa = ˆa ˆa ˆa + ˆa ˆa ˆa + ˆa ˆa ˆa r r r r = sin θcos φ ˆa + sin θ sin φ ˆa + cos θ ˆa This result eplicitl shows that â r is a function of θ and φ.

34 8/26/2004 Spherical Base Vectors.doc 3/3 For eample, at the point in space r = 7.239, θ = 90 and φ = 0, we find that ˆa = ˆa. In other words, at this point in space, the r direction â r points in the -direction. Or, at the point in space r = 2.735, θ = 90 and φ = 90, we find that ˆar = ˆa. In other words, at this point in space, â r points in the -direction. Additionall, we can write ˆa θ and ˆa as: φ ( ) ( ) ( ) ˆa = ˆa ˆa ˆa + ˆa ˆa ˆa + ˆa ˆa ˆa θ θ θ θ ( ) ( ) ( ) ˆa = ˆa ˆa ˆa + ˆa ˆa ˆa + ˆa ˆa ˆa φ φ φ φ Alternativel, we can write Cartesian base vectors in terms of spherical base vectors, i.e., ( ) ( ) ( ) ˆa = ˆa ˆa ˆa + ˆa ˆa ˆa + ˆa ˆa ˆa r r θ θ φ φ ( ) ( ) ( ) ˆa = ˆa ˆa ˆa + ˆa ˆa ˆa + ˆa ˆa ˆa r r θ θ φ φ ( ) ( ) ( ) ˆa = ˆa ˆa ˆa + ˆa ˆa ˆa + ˆa ˆa ˆa r r θ θ φ φ Using the table on the previous page, we can insert the result of each dot product to epress each base vector in terms of spherical coordinates.

35 08/26/04 Clindrical Base Vectors.doc 1/3 Clindrical Base Vectors Clindrical base vectors are the natural base vectors of a clinder. ˆ a ρ points in the direction of increasing ρ. In other words, ˆ a ρ points awa from the -ais. ˆ a φ points in the direction of increasing φ. This is precisel the same base vector we described for spherical base vectors. a ˆ points in the direction of increasing. This is precisel the same base vector we described for Cartesian base vectors. aˆ ˆ a φ ˆ a ρ

36 08/26/04 Clindrical Base Vectors.doc 2/3 It is evident, that like spherical base vectors, the clindrical base vectors are dependent on position. A vector that points awa from the -ais (e.g., â ρ ), will point in a direction that is dependent on where we are in space! We can epress clindrical base vectors in terms of Cartesian base vectors. First, we find that: ˆ a ˆ a = cosφ ρ ˆ a ˆ a = sinφ ρ ˆ a ˆ a = 0 ρ ˆ a ˆ a = sinφ φ ˆ a ˆ a = cosφ φ ˆ a ˆ a = 0 φ ˆ a ˆ a = 0 ˆ a ˆ a = 0 ˆ a ˆ a = 1 We can use these results to write clindrical base vectors in terms of Cartesian base vectors, or vice versa! For eample, or, ( ) ( ) ( ) ˆ a = ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a p p p p = cosφ ˆ a + sinφ ˆ a ( ρ) ρ ( φ) φ ( ) ˆ a = ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a = cosφ ˆ a sinφ ˆ a ρ φ

37 08/26/04 Clindrical Base Vectors.doc 3/3 Finall, we can write clindrical base vectors in terms of spherical base vectors, or vice versa, using the following relationships: ˆ a ˆ ρ a r = sinθ ˆ a ˆ a = cosθ ρ θ ˆ a ˆ a = 0 ρ φ ˆ a ˆ φ a r = 0 ˆ a ˆ a = 0 φ θ ˆ a ˆ a = 1 φ φ ˆ a ˆ a = cosθ r ˆ a ˆ a = sinθ ˆ a ˆ a = 0 θ φ e.g., ( ) ( ) ( ) ˆ a = ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a p p r r p θ θ p φ φ = sinθ ˆ a + cosθ ˆ a r θ ( ) ( ) ( ) ˆ a = ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a + ˆ a ˆ a ˆ a θ θ ρ ρ θ φ φ θ = cosθ ˆ a sinθ ˆ a ρ

38 8/18/2004 Vector Algebra using Orthonormal Base Vectors.doc 1/7 Vector Algebra using Orthonormal Base Vectors Q: Wh epress a vector using orthonormal base vectors? Doesn t this just make things more complicated?? A: Actuall, it makes things much simpler. The evaluation of vector operations such as addition, subtraction, multiplication, dot product, and cross product all become straightforward if all vectors are epressed using the same set of base vectors. Consider two vectors A and B, each epressed using the same set of base vectors ˆ a, ˆ a, ˆ a : A = A ˆ a + A ˆ a + A ˆ a B = B ˆ a + B ˆ a + B ˆ a

39 8/18/2004 Vector Algebra using Orthonormal Base Vectors.doc 2/7 1. Addition and Subtraction If we add these two vectors together, we find: ( A ˆ ˆ ˆ ) ( ˆ ˆ ˆ a A a A a B a B a B a ) A+ B= = A ˆ a + B ˆ a + A ˆ a + B ˆ a + A ˆ a + B ˆ a ( ) ( ) ( ) = A + B ˆ a + A + B ˆ a + A + B ˆ a In other words, each component of the sum of two vectors is equal to the sum of each component. Similarl, we find for subtraction: ( A ˆ ˆ ˆ ) ( ˆ ˆ ˆ a A a A a B a B a B a ) A B= = A ˆ a B ˆ a + A ˆ a B ˆ a + A ˆ a B ˆ a ( ) ( ) ( ) = A B ˆ a + A B ˆ a + A B ˆ a 2. Vector/Scalar Multiplication Sa we multipl a scalar a and a vector B, i.e., a B:

40 8/18/2004 Vector Algebra using Orthonormal Base Vectors.doc 3/7 ( ˆ ˆ ˆ ) ab = a B a + B a + B a = ab ˆ a + ab ˆ a + ab ˆ a ( ) ( ) ( ) = ab ˆ a + ab ˆ a + ab ˆ a In other words, each component of the product of a scalar and a vector are equal to the product of the scalar and each component. 3. Dot Product Sa we take the dot product of A and B: ( A ˆ ˆ ˆ ) ( ˆ ˆ ˆ a A a A a B a B a B a ) A ˆ ( ˆ ˆ ˆ a B a B a B a ) A ˆ ( ˆ ˆ ˆ a B a B a B a ) A ˆ ( ˆ ˆ ˆ a B a B a B a ) A ( ˆ ˆ ) ( ˆ ˆ ) ( ˆ ˆ B a a A B a a A B a a ) A ( ˆ ˆ ) ( ˆ ˆ ) ( ˆ ˆ B a a A B a a A B a a ) A ( ˆ ˆ B a a ) A B ( ˆ a ˆ ) ( ˆ ˆ a + A B a a ) A B = = = A: Be patient! Recall that these are orthonormal base vectors, therefore: ˆ a ˆ a = ˆ a ˆ a = ˆ a ˆ a = 1 and ˆ a ˆ a = ˆ a ˆ a = ˆ a ˆ a = 0 Q: I thought this was suppose to make things easier!?!

41 8/18/2004 Vector Algebra using Orthonormal Base Vectors.doc 4/7 As a result, our dot product epression reduces to this simple epression: A B = AB + AB + AB We can appl this to the epression for determining the magnitude of a vector: Therefore: = = A + A + A A A A A = A A = A + A + A 2 For eample, consider a previous handout, where we epressed a vector using two different sets of basis vectors: or, A = 20. ˆa ˆa A = 25ˆ. b Therefore, the magnitude of A is determined to be:

42 8/18/2004 Vector Algebra using Orthonormal Base Vectors.doc 5/7 or, A = + = = A = = = Q: He! We get the same answer from both epressions; is this a coincidence? A: No! Remember, both epressions represent the same vector, onl using different sets of base vectors. The magnitude of vector A is 2.5, regardless of how we choose to epress A. 4. Cross Product Now lets take the cross product AB: ( A ˆ ˆ ˆ ) ( ˆ ˆ ˆ a A a A a B a B a B a ) A ˆ ( ˆ ˆ ˆ a B a B a B a ) A ˆ ( ˆ ˆ ˆ a B a B a B a ) A ˆ ( ˆ ˆ ˆ a B a B a B a ) AB ( ˆ ˆ ) ( ˆ ˆ ) ( ˆ ˆ a a AB a a AB a a ) AB ( ˆ ˆ ) ( ˆ ˆ ) ( ˆ ˆ a a AB a a AB a a ) AB ( ˆ ˆ a a ) AB ( ˆ a ˆ ) ( ˆ ˆ a + AB a a ) AB= = =

43 8/18/2004 Vector Algebra using Orthonormal Base Vectors.doc 6/7 Remember, we know that: ˆ a ˆ a = ˆ a ˆ a = ˆ a ˆ a = 0 also, since base vectors form a right-handed sstem: ˆ a ˆ a = ˆ a ˆ a ˆ a = ˆ a ˆ a ˆ a = ˆ a Remember also that AB = -(BA), therefore: ˆ a ˆ a = ˆ a ˆ a ˆ a = ˆ a ˆ a ˆ a = ˆ a Combining all the equations above, we get: ( ) ( ) ( ) A B = AB AB ˆ a + AB AB ˆ a + AB AB ˆ a Y Z Z Y X Z X X Z Y X Y Y X Z 5. Triple Product Combining the results of the dot product and the cross product, we find that the triple product can be epressed as: ( A B C AB C AB C ) ( A B C A B C AB C ) A B C =

44 8/18/2004 Vector Algebra using Orthonormal Base Vectors.doc 7/7 IMPORTANT NOTES: In addition to all that we have discussed here, it is critical that ou understand the following points about vector algebra using orthonormal base vectors! * The results provided in this handout were given for Cartesian base vectors ( ˆ a, ˆ a, ˆ a ). However, the are equall valid for an right-handed set of base vectors ˆ a ˆ ˆ 1, a2, a 3 (e.g., ˆ a, ˆ a, ˆ a ˆ a, ˆ a, ˆ a r θ φ ). ρ φ or * These results are algorithms for evaluating various vector algebraic operations. The are not definitions of the operations. The definitions of these operations were covered in Section 2-3. * The scalar components A, A, and A represent either discrete scalar (e.g., A = 4 2. ) or scalar field quantities (e.g., 2 Aθ r sin cos = θ φ.

45 8/18/2004 Vector Fields.doc 1/3 Vector Fields Base vectors give us a convenient wa to epress vector fields! You will recall that a vector field is a vector quantit that is a function of other scalar values. In this class, we will stud Α,, ). vector fields that are a function of position (e.g., ( ) We earlier considered an eample of a vector field of this tpe: v, across the upper Midwest. the wind velocit ( ) When we epress a vector field using orthonormal base vectors, the scalar component of each direction is a scalar field a scalar function of position! Jim Stiles The Universit of Kansas Dept. of EECS

46 8/18/2004 Vector Fields.doc 2/3 In other words, a vector field can have the form: A (,, ) = A (,, ) ˆ a + A (,, ) ˆ a + A (,, ) ˆ a We therefore can epress a vector field A (,, ) in terms of 3 scalar fields: A(,, ), A(,, ), and A(,, ), which epress each of the 3 scalar components as a function of position (,,). For eample, we might encounter this vector field: A (,, ) = ( + ) ˆ a + ˆ a + ( 3 ) ˆ a In this case it is evident that: A (,, ) = ( + ) A (,, ) = ( ) A (,, ) = 3 The vector algebraic rules that we discussed in previous handouts are just as valid for vector fields and scalar field components as the are for discrete vectors and discrete scalar components. Jim Stiles The Universit of Kansas Dept. of EECS

47 8/18/2004 Vector Fields.doc 3/3 For eample, consider these two vector fields, epressed in terms of orthonormal base vectors ˆ a, ˆ a, ˆ a : 2 A(,, ) = ˆ a + ( ) ˆ a + ˆ a B(,, ) = ( + 2) ˆ a + ˆ a + ˆ a The dot product of these two vector fields is a scalar field: A(,, ) B(,, ) = A B + A B + A B 2 = ( + 2) + ( ) + Likewise, the sum of these two vector fields is a vector field: A(,, ) + B(,, ) = ( A + B ) ˆ a + ( A + B ) ˆ a + ( A + B ) ˆ a 2 ( + 1) = ( + + 2) ˆ a + ˆ a + ˆ a 2 Jim Stiles The Universit of Kansas Dept. of EECS

48 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 1/8 Eample: Epressing Vector Fields with Coordinate Sstems Consider the vector field: ˆ ( ) ˆ A = a + + a + ˆ a Let s tr to accomplish three things: 1. Epress A using spherical coordinates and Cartesian base vectors. 2. Epress A using Cartesian coordinates and spherical base vectors. 3. Epress A using clindrical coordinates and clindrical base vectors. 1. The vector field is alread epressed with Cartesian base vectors, therefore we onl need to change the Cartesian coordinates in each scalar component into spherical coordinates.

49 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 2/8 The scalar component of A in the -direction is: A = = = ( r sin θ cos φ)( r cosθ) 2 r sin θ cos θ cosφ The scalar component of A in the -direction is: A = + ( r sin θ cos φ) ( r sin θ sinφ) = + ( ) = r sin θ cos φ + sin φ = r sin θ The scalar component of A in the -direction is: A = r sin θ cosφ = r cosθ = tanθ cos φ Therefore, the vector field can be epressed using spherical coordinates as: 2 A = r sin θ cos θ cosφ ˆ a + r sin θ ˆ a + tanθcosφ ˆ a

50 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 3/8 2. Now, let s epress A using spherical base vectors. We cannot simpl change the coordinates of each component. Rather, we must determine new scalar components, since we are using a new set of base vectors. We begin b stating: ( r ) r ( ) ( ) A= A ˆ a ˆ a + A ˆ a ˆ a + A ˆ a ˆ a θ θ φ φ The scalar component A r is therefore: ˆ ˆ ˆ ( ) ˆ ˆ A a ˆ ˆ r = a ar + + a ar + a ar = = ( sinθcosφ) ( )( sinθsinφ) ( cosθ ) ( ) ( + ) 2 = = Likewise, the scalar component A θ is:

51 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 4/8 ˆ ˆ ˆ ( ) ˆ ˆ A a ˆ ˆ θ = a aθ + + a aθ + a aθ = + + = ( cosθcosφ ) ( )( cosθsinφ ) ( sinθ ) ( ) ( + ) ( + ) = = And finall, the scalar component A φ is: ˆ ˆ ˆ ( ) ˆ ˆ a = a ˆ ˆ a φ + + a a + a a φ = (-sinφ ) + ( + )( cosφ ) + 0 = + ( + ) + + A φ φ = +

52 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 5/8 Whew! We re finished! The vector A is epressed using spherical base vectors as: A = ar ˆ ˆ a 2 θ ˆ a φ + 3. Now, let s write A in terms of clindrical coordinates and clindrical base vectors (i.e., in terms of the clindrical coordinate sstem). First, A ρ is: ( ρ) ρ ( φ) φ ( ) A= A ˆ a ˆ a + A ˆ a ˆ a + A ˆ a ˆ a ˆ ˆ ˆ ( ) ˆ ˆ A a ˆ ˆ ρ = a aρ + + a aρ + a a = ( cosφ ) + ( + )( sinφ ) + ( 0) = ρcosφ cosφ + ρ s φ = cos + 2 ( ) ( in ) ρ φ ρ φ sin ρ

53 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 6/8 And A φ is: ˆ ˆ ˆ ( ) ˆ ˆ A a ˆ ˆ φ = a aφ + + a aφ + a aφ = (-sinφ ) ( )( cosφ ) ( 0) 2 ( ) ( ) ( sin ) = ρcosφ sinφ + ρ cosφ = ρcosφ ρ φ And finall, A is: ˆ ˆ ˆ ( ) ˆ ˆ A a ˆ ˆ = a a + + a a + a a = ( 0) + ( + )( 0) + ( 1) = ρcosφ = We can therefore epress the vector field A using both clindrical coordinates and clindrical base vectors: ( ) ˆ ( ) ˆ ρ cosφ A = ρcos φ + ρ sinφ a cos sin a ˆ ρ + ρ φ ρ φ φ + a

54 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 7/8 Thus, we have determined three possible was (there are man other was!) to epress the vector field A: 1. 2 A = r sin θ cos θ cosφ ˆ a + r sin θ ˆ a + tanθcosφ ˆ a A = ar ˆ ˆ a 2 θ ˆ a φ + 3. ( ) ˆ ( ) ˆ ρ cosφ A = ρcos φ + ρ sinφ a cos sin a ˆ ρ + ρ φ ρ φ φ + a Please note: * The three epressions for vector field A provided in this handout each look ver different. However, the are just three different methods for describing the same vector field. An one of the three is correct, and will result in the same result for an phsical problem. * We can epress a vector field using an set of coordinate variables and an set of base vectors.

55 8/18/2004 Eample Epressing Vector Fields with Coordinate Sstems.doc 8/8 * Generall speaking, however, we use one coordinate sstem to describe a vector field. For eample, we use both spherical coordinates and spherical base vectors. Q: So, which coordinate sstem (Cartesian, clindrical, spherical) should we use? How can we decide between the three? A: Ideall, we select that sstem that most simplifies the mathematics. This depends on the phsical problem we are solving. For eample, if we are determining the fields resulting from a sphericall smmetric charge densit, we will find that using the spherical coordinate sstem will make our analsis the easiest and most straightforward.

56 8/26/2004 The Position Vector.doc 1/7 The Position Vector Consider a point whose location in space is specified with Cartesian coordinates (e.g., P(,,)). Now consider the directed distance (a vector quantit!) etending from the origin to this point. r P(,,) This particular directed distance a vector beginning at the origin and etending outward to a point is a ver important and fundamental directed distance known as the position vector r. Using the Cartesian coordinate sstem, the position vector can be eplicitl written as: r ˆ a + ˆ a + ˆ a =

57 8/26/2004 The Position Vector.doc 2/7 * Note that given the coordinates of some point (e.g., =1, =2, =-3), we can easil determine the corresponding position vector (e.g., r = ˆa + 2ˆa 3ˆa ). * Moreover, given some specific position vector (e.g., r = 4ˆa 2ˆa ), we can easil determine the corresponding coordinates of that point (e.g., =0, =4, =-2). In other words, a position vector r is an alternative wa to denote the location of a point in space! We can use three coordinate values to specif a point s location, or we can use a single position vector r. I see! The position vector is essentiall a pointer. Look at the end of the vector, and ou will find the point specified! P(r ) r

58 8/26/2004 The Position Vector.doc 3/7 The magnitude of r Note the magnitude of an and all position vectors is: 2 r = r r = + + = r The magnitude of the position vector is equal to the coordinate value r of the point the position vector is pointing to! Q: He, this makes perfect sense! Doesn t the coordinate value r have a phsical interpretation as the distance between the point and the origin? A: That s right! The magnitude of a directed distance vector is equal to the distance between the two points in this case the distance between the specified point and the origin! Alternative forms of the position vector Be careful! Although the position vector is correctl epressed as: r = aˆ ˆ ˆ + a + a

59 8/26/2004 The Position Vector.doc 4/7 It is NOT CORRECT to epress the position vector as: nor ρ ˆ a + φ ˆ a + ˆ a r p φ r raˆ + θ ˆ a + φ ˆ a r θ φ NEVER, EVER epress the position vector in either of these two was! It should be readil apparent that the two epression above cannot represent a position vector because neither is even a directed distance! Q: Wh sure it is of course readil apparent to me but wh don t ou go ahead and eplain it to those with less insight! A: Recall that the magnitude of the position vector r has units of distance. Thus, the scalar components of the position vector must also have units of distance (e.g., meters). The coordinates,,, ρ and r do have units of distance, but coordinates θ and φ do not. Thus, the vectors θ a ˆθ and φ a ˆφ cannot be vector components of a position vector or for that matter, an other directed distance!

60 8/26/2004 The Position Vector.doc 5/7 Instead, we can use coordinate transforms to show that: r = ˆ a + ˆ a + ˆ a = ρcosφ ˆ a + ρsinφ ˆ a + ˆ a = r sinθ cosφ ˆ a + r sinθ sinφ ˆ a + r cosθ ˆ a ALWAYS use one of these three epressions of a position vector!! Note that in each of the three epressions above, we use Cartesian base vectors. The scalar components can be epressed using Cartesian, clindrical, or spherical coordinates, but we must alwas use Cartesian base vectors. Q: Wh must we alwas use Cartesian base vectors? You said that we could epress an vector using spherical or base vectors. Doesn t this also appl to position vectors? A: The reason we onl use Cartesian base vectors for constructing a position vector is that Cartesian base vectors are the onl base vectors whose directions are fied independent of position in space!

61 8/26/2004 The Position Vector.doc 6/7 To see wh this is important, let s go ahead and change the base vectors used to epress the position vector from Cartesian to spherical or clindrical. If we do this, we find: r = ˆ a + ˆ a + ˆ a = ρ ˆ a + ˆ a = r ˆ a ρ r Thus, the position vector epressed with the clindrical coordinate sstem is r = ρ ˆ a + ˆ a, while with the spherical coordinate sstem we get ρ ˆr r = r a. The problem with these two epressions is that the direction of base vectors a ˆρ and a ˆr are not constant. Instead, the themselves are vector fields their direction is a function of position! Thus, an epression such as r = 6 ˆ a r does not eplicitl define a point in space, as we do not know in what direction base vector a ˆr is pointing! The epression r = 6 ˆ a r does tell us that the coordinate r =6, but how do we determine what the values of coordinates θ or φ are? (answer: we can t! ) Compare this to the epression: r = ˆ a + 2 ˆ a 3 ˆ a

62 8/26/2004 The Position Vector.doc 7/7 Here, the point described b the position vector is clear and unambiguous. This position vector identifies the point P( =1, =2, =-3). Lesson learned: Alwas epress a position vector using Cartesian base vectors (see bo on previous page)!

63 8/26/2004 Applications of the Position Vector.doc 1/2 Applications of the Position Vector Position vectors are particularl useful when we need to determine the directed distance between two arbitrar points in space. P B (,,) R AB P A (,,) r B r A If the location of point P A is denoted b position vector r A, and the location of point P B b position vector r B, then the directed distance from point P A to point P B, is: R = r - r AB B A We can use this directed distance R AB to describe much about the relative locations of point P A and P B!

64 8/26/2004 Applications of the Position Vector.doc 2/2 For eample, the phsical distance between these two points is simpl the magnitude of this directed distance: P B (,,) d P A (,,) d = R = r r AB B A r B r A Likewise, we can specif the direction toward point P B, with respect to point P A, b find the unit vector a ˆAB : P B (,,) aˆab P A (,,) R r r AB B A ˆAB = = RAB rb ra a r B r A

65 8/26/2004 Vector Field Notation.doc 1/4 Vector Field Notation A vector field describes a vector value at ever location in space. Therefore, we can denote a vector field as A(,,), or A( ρ, φ, ), or A( r, θ, φ ), eplicitl showing that vector quantit A is a function of position, as denoted b some set of coordinates. However, as we have emphasied before, the phsical realit that vector field A epresses is independent of the coordinates we use to epress it. In other words, although the math ma look ver different, we find that: A(,,) = A( ρ, φ, ) = A( r, θ, φ ). Alternativel then, we tpicall epress a vector field as simpl: A ( r ) This smbolicall sas everthing that we need to conve; vector A is a function of position it is a vector field! Note that the vector field notation A ( r ) does not eplicitl specif a coordinate sstem for epressing A. That s up to ou to decide!

66 8/26/2004 Vector Field Notation.doc 2/4 Now, in the vector field epression A ( r ) we note that there are two vectors: A and r. It is ridiculousl important that ou understand what each of these two vectors represents! Position vector r denotes the location in space where vector A is defined. For eample, consider the vector field V ( r ), which describes the wind velocit across the state of Kansas. V( r ) 1 r 1 V( r ) r 2 2 aˆ aˆ r 3 V( r ) 3 r 5 V( r ) 5 r 4 V( r ) 4 In this map, the origin has been placed at Lawrence. The locations of Kansas towns can thus be identified using position vectors (units in miles):

67 8/26/2004 Vector Field Notation.doc 3/4 r = 400 ˆ a + 20 ˆ a the location of Goodland, KS 1 r = 90 ˆ a + 70 ˆ a the location of Marsville, KS 2 r = 30 ˆ a 5 ˆ a the location of Fort Scott,KS 3 r = 40 ˆ a 90 ˆ a the location of Fort Scott,KS 4 r = 130 ˆ a 70 ˆ a the location of Newton, KS 5 Evaluating the vector field V ( r ) at these locations provides the wind velocit at each Kansas town (units of mph). V (r = 15 ˆ a 17 ˆ a the wind velocit in Goodland, KS 1 ) V (r = 15 ˆ a 9 ˆ a the wind velocit in Marsville, KS 2 ) V (r = 11 ˆ the wind velocit in Olathe, KS 3 ) a V (r = 7 ˆ the wind velocit in Fort Scott, KS 4 ) a V (r = 9 ˆ a 4 ˆ a the wind velocit in Newton, KS 5 ) Remember, a vector field A ( r ) describes the magnitude and direction of the vector A that is located at the point defined b position vector r.

68 8/26/2004 Vector Field Notation.doc 4/4 Vector A does not etend from the origin to the point described b position vector r. Rather, the vector A describes a quantit at that point, and that point onl. The magnitude of vector A does not have units of distance! The length of the arrow that represents vector A is merel smbolic its length has no direct phsical meaning. On the other hand, the position vector r, being a directed distance, does etend from the origin to a specific point in space. The magnitude of a position vector r is distance the length of the position vector arrow has a direct phsical meaning.

69 8/18/2004 A Galler of Vector Fields.doc 1/12 A Galler of Vector Fields To help understand how a vector field relates to its mathematical representation using base vectors, carefull eamine and consider these eamples, plotted on either the - plane (i.e, the plane with all points whose coordinate =0) or the - plane (i.e, the plane with all points whose coordinate =0). Spend some time studing each of these eamples, until ou see how the math relates to the vector field plot and vice versa. Remember, vector fields epressed in terms of scalar components and base vectors are the mathematical language that we will use to describe much of electromagnetics ou must learn how to speak and interpret this language!

70 8/18/2004 A Galler of Vector Fields.doc 2/ A( r ) = ˆ a A( r ) = ˆ a -5-10

71 8/18/2004 A Galler of Vector Fields.doc 3/ A( r ) = ˆ a ˆ a A( r ) = ˆ a -5-10

72 8/18/2004 A Galler of Vector Fields.doc 4/ A( r ) = ˆ a A( r ) = ˆ a ˆ + a -5-10

73 8/18/2004 A Galler of Vector Fields.doc 5/ A( r ) = (10 + ) ˆ a A( r ) = (10 + ) ˆ a -5-10

74 8/18/2004 A Galler of Vector Fields.doc 6/ A( r ) = (10 + ) ˆ a + (10 + ) ˆ a A( r ) = (10 + ) a 3 ˆ

75 8/18/2004 A Galler of Vector Fields.doc 7/ A( r ) = (10 + ) a 3 ˆ A( r ) = (10 + ) + (10 + ) 3 3 ˆ a ˆ a -5-10

76 8/18/2004 A Galler of Vector Fields.doc 8/ A( r ) = ˆ a A( r ) = ˆ a -5-10

77 8/18/2004 A Galler of Vector Fields.doc 9/ A( r ) = ˆ a ˆ + a A( r ) = ˆ a ˆ + a -10

78 8/18/2004 A Galler of Vector Fields.doc 10/ A( r ) = ˆ a ˆ a A( r ) = ˆ a ˆ a -5-10

79 8/18/2004 A Galler of Vector Fields.doc 11/ A( r ) = ˆ a ρ = cosφ ˆ a + sinφ ˆ a A( r ) = ˆ a φ = sinφ ˆ a cosφ ˆ a -5-10

80 8/18/2004 A Galler of Vector Fields.doc 12/ A( r ) = ˆ a r = sinθ cosφ ˆ a + cosθ ˆ a A( r ) = ˆ a θ = cosθ cosφ ˆ a sinθ ˆ a -5-10