Cylindrical and Spherical Coordinates The Cartesian coordinate - PowerPoint PPT Presentation

Cylindrical and Spherical Coordinates The Cartesian coordinate system is by far the simplest, the most universal and the most important. There are some situations for which the Cartesian coordinate sys- tem is not entirely ideal. These typically involve scalar or vector fields which exhibit some kind of inherent symmetry. The cylindrical and spherical coordinate systems are designed for just this purpose. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 1/27

Polar coordinates The point A is represented by ( r, θ ), which has a very different interpretation from the Cartesian pair ( x, y ). While −∞ < x < ∞ and −∞ < y < ∞ , the polar coordinates obey 0 ≤ r < ∞ , 0 ≤ θ < 2 π. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 2/27

Polar coordinates (cont.) The Cartesian and polar coordinates are related as x = r cos θ, y = r sin θ. Thus, in terms of the Cartesian basis, any point in R 2 can be represented as v ( r, θ ) = r cos θ i + r sin θ j . Conversely, � x 2 + y 2 r = and  arctan( y/x ) , x > 0 , y > 0 ,   θ = π + arctan( y/x ) , x < 0 , y ∈ R ,   2 π + arctan( y/x ) , x > 0 , y < 0 . Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 3/27

Standard vectors Any v ∈ R 2 can be represented in terms of the standard Carte- sian unit vectors i and j as v = x i + y j . What are the “standard” vectors which are better suited to polar coordinates? To answer this question, for A with coordinates ( r 0 , θ 0 ), we define the following two paths: γ 1 ( r ) := v ( r, θ 0 ) = r cos θ 0 i + r sin θ 0 j and γ 2 ( θ ) := v ( r 0 , θ ) = r 0 cos θ i + r 0 sin θ j . Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 4/27

Standard vectors (cont.) The tangent vectors at ( r 0 , θ 0 ) are given by 1 ( r 0 ) = dγ 1 ( r ) � γ (1) � � dr r = r 0 � 2 ( θ 0 ) = dγ 2 ( θ ) γ (1) � � dθ θ = θ 0 Thus, after normalization, we obtain the two standard vectors γ (1) 1 ( r 0 ) e r ( r 0 , θ 0 ) := = cos θ 0 i + sin θ 0 j � γ (1) 1 ( r 0 ) � γ (1) 2 ( θ 0 ) e θ ( r 0 , θ 0 ) := = − sin θ 0 i + cos θ 0 j � γ (1) 2 ( θ 0 ) � The pair of orthogonal basis vectors { e r ( r 0 , θ 0 ) , e θ ( r 0 , θ 0 ) } is cal- led the coordinate frame at the point A . Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 5/27

Coordinate frame Notice that { e r ( r, θ ) , e θ ( r, θ ) } change direction as the point ( r, θ ) moves, so that we have a moving coordinate frame . This is in direct contrast to the Cartesian unit basis vectors { i , j } which are constant throughout R 2 . Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 6/27

Riemannian scale functions How does length changes when we change the polar coordinates? When “moving” from A to B , the Euclidean distance is defined by ( ds ) 2 = ( dx ) 2 + ( dy ) 2 . At the same time, in the polar co- ordinates, ds is equal to ( ds ) 2 = ( dr ) 2 + r 2 ( dθ ) 2 = = [ h r ( r, θ ) dr ] 2 + [ h θ ( r, θ ) dθ ] 2 , where h r ( r, θ ) := 1 , h θ ( r, θ ) = r are the Riemann scale functions . Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 7/27

Cylindrical coordinates Let ( r, θ ) be the polar coordinates of the point B . Then the triplet of real numbers ( r, θ, z ) (with z ∈ R ) completely specifies the point A . Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 8/27

Cylindrical coordinates (cont.) If A is given by the cylindrical coordinates ( r, θ, z ) then, in terms of the Cartesian basis { i, j, k } , it must be given by the vector v ( r, θ, z ) = r cos( θ ) i + r sin( θ ) j + z k . To find the standard coordinate frame, we define three curves re- presented by the following paths: γ 1 ( r ) := v ( r, θ 0 , z 0) = r cos( θ 0 ) i + r sin( θ 0 ) j + z 0 k , ∀ r ∈ ( −∞ , 0] , γ 2 ( θ ) := v ( r 0 , θ, z 0) = r 0 cos( θ ) i + r 0 sin( θ ) j + z 0 k , ∀ θ ∈ [0 , 2 π ) , γ 3 ( z ) := v ( r 0 , θ 0 , z ) = r 0 cos( θ 0 ) i + r 0 sin( θ 0 ) j + z k , ∀ z ∈ R . Next, we compute the tangent vectors γ (1) 1 ( r 0 ), γ (1) 2 ( θ 0 ), and γ (1) 3 ( z 0 ), followed by their normalization to obtain e r ( r 0 , θ 0 , z 0 ), e θ ( r 0 , θ 0 , z 0 ), and e z ( r 0 , θ 0 , z 0 ), respectively. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 9/27

Cylindrical coordinates (cont.) The cylindrical coordinate frame is defined by e r ( r, θ, z ) = cos θ i + sin θ j + 0 k , e θ ( r, θ, z ) = − sin θ i + cos θ j + 0 k , e z ( r, θ, z ) = 0 i + 0 j + 1 k . One can see that the unit vectors are orthogonal with respect to each other. Also, the triplet of vectors constitutes a moving coordinate frame. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 10/27

Cylindrical coordinates (cont.) Due to a small “permutation” the length ds can be defined as follows. ( ds ) 2 = (length AB) 2 = (length AD) 2 + +(length AC) 2 + (length AE) 2 = = ( dr ) 2 + r 2 ( dθ ) 2 + ( dz ) 2 . Alternatively, we can write ( ds ) 2 = [ h r ( r, θ, z ) dr ] 2 + [ h θ ( r, θ, z ) dθ ] 2 + [ h z ( r, θ, z ) dz ] 2 , where h r ( r, θ, z ) := 1 , h θ ( r, θ, z ) := r, h z ( r, θ, z ) := 1 are the Riemannian scale functions. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 11/27

Spherical coordinates The radial length of the vector from the origin O to point A is � x 2 + y 2 + z 2 . ρ = Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 12/27

Spherical coordinates (cont.) The Cartesian coordinates ( x, y, z ) are given in terms of the spherical coordinates ( ρ, θ, φ ) by x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, with ρ ∈ [0 , ∞ ), θ ∈ [0 , 2 π ), and φ ∈ [0 , π ]. Then, in terms of the Cartesian basis { i , j , k } , any point A with coordinates ( ρ, θ, φ ) can be expressed as v ( ρ, θ, φ ) = ρ sin φ cos θ i + ρ sin φ sin θ j + ρ cos φ k . This shows how the Cartesian representation of a point changes when we change its spherical coordinates. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 13/27

Spherical coordinates (cont.) To define a moving (orthonormal) frame, we start with γ 1 ( ρ ) = v ( ρ, θ 0 , φ 0 ) = ρ sin φ 0 cos θ 0 i + ρ sin φ 0 sin θ 0 j + ρ cos φ 0 k , γ 2 ( θ ) = v ( ρ 0 , θ, φ 0 ) = ρ 0 sin φ 0 cos θ i + ρ 0 sin φ 0 sin θ j + ρ 0 cos φ 0 k , γ 3 ( φ ) = v ( ρ, θ 0 , φ ) = ρ 0 sin φ cos θ 0 i + ρ 0 sin φ sin θ 0 j + ρ 0 cos φ k . Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 14/27

Spherical coordinates (cont.) The corresponding tangent vectors at ( ρ 0 , θ 0 , φ 0 ) are given by 1 ( ρ 0 ) = dγ 1 ( ρ ) � 2 ( θ 0 ) = dγ 2 ( θ ) � 3 ( φ 0 ) = dγ 3 ( φ ) � γ (1) ρ 0 , γ (1) θ 0 , γ (1) � � � φ 0 . � � � dρ dθ dφ Normalizing the tangent vectors results in the spherical coordi- nate frame { e ρ , e θ , e φ } which are e ρ ( ρ, θ, φ ) = sin φ cos θ i + sin φ sin θ j + cos φ k , e θ ( ρ, θ, φ ) = − sin θ i + cos θ j + 0 k , e φ ( ρ, θ, φ ) = cos φ cos θ i + cos φ sin θ j − sin φ k , under the replacement of the generic (i.e., “naughted”) spherical coordinates ( ρ 0 , θ 0 , φ 0 ) with ( ρ, θ, φ ) (to simplify the notations). One can check that the unit vectors indeed orthogonal w.r.t. each other. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 15/27

Spherical coordinates (cont.) Suppose a point A is given by v ( ρ, θ, φ ), while a close-by point B is given by ( ρ + dρ, θ + dθ, φ + dφ ). Let’s find an expression of ds in terms of ( ρ, θ, φ ). Formally we have ( ds ) 2 = (length AB) 2 = = � v ( ρ + dρ, θ + dθ, φ + dφ ) − v ( ρ, θ, φ ) � 2 Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 16/27

Spherical coordinates (cont.) First we note that v ( ρ + dρ, θ + dθ, φ + dφ ) − v ( ρ, θ, φ ) = = ∂ v ∂ρ ( ρ, θ, φ ) dρ + ∂ v ∂θ ( ρ, θ, φ ) dθ + ∂ v ∂φ ( ρ, θ, φ ) dφ = = ( dρ ) e ρ ( ρ, θ, φ ) + ( ρ sin φ dθ ) e θ ( ρ, θ, φ ) + ( ρ dφ ) e φ ( ρ, θ, φ ) . Thus, we have ( ds ) 2 = ( dρ ) 2 + ( ρ sin φ ) 2 ( dθ ) 2 + ( ρ ) 2 ( dφ ) 2 , or, alternatively (in the Riemannian form), ( ds ) 2 = ( h ρ ( ρ, θ, φ ) dρ ) 2 + ( h θ ( ρ, θ, φ ) dθ ) 2 + ( h φ ( ρ, θ, φ ) dφ ) 2 , with the Riemannian scaling functions defined as h ρ ( ρ, θ, φ ) := 1 , h θ ( ρ, θ, φ ) := ρ sin φ, h φ ( ρ, θ, φ ) := ρ. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 17/27

Differential operators Let F : R 3 → R 3 be a C 1 -vector field. The divergence of F is defined as div F ( x, y, z ) = ∂F 1 ∂x ( x, y, z ) + ∂F 2 ∂y ( x, y, z ) + ∂F 3 ∂z ( x, y, z ) , for all ( x, y, z ) in R 3 . Alternatively, we can write div F = ∇ · F , where ∇ := ( ∂/∂x ) i + ( ∂/∂y ) j + ( ∂/∂z ) k . The gradient operator ∇ is of fundamental importance, since all the other operators (viz., divergence, curl, and Laplacian) are de- fined by it. Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 18/27