scalar operator hence the Laplacian of a scalar is a scalar and the Laplacian

# Scalar operator hence the laplacian of a scalar is a

• 171

This preview shows page 125 - 129 out of 171 pages.

scalar operator ; hence the Laplacian of a scalar is a scalar and the Laplacian of a vector is a vector. 5.3.2 Cylindrical Coordinate System For the cylindrical system identified by the coordinates ( ρ, φ, z ) with a set of orthonormal basis vectors e ρ , e φ and e z we have the following definitions for the nabla based operators and operations. [72] The nabla operator is given by: = e ρ ρ + e φ 1 ρ φ + e z z (198) The Laplacian operator is given by: 2 = ρρ + 1 ρ ρ + 1 ρ 2 φφ + zz (199) The gradient of a differentiable scalar f is given by: f = e ρ ∂f ∂ρ + e φ 1 ρ ∂f ∂φ + e z ∂f ∂z (200) The divergence of a differentiable vector A is given by: ∇ · A = 1 ρ ( ρA ρ ) ∂ρ + ∂A φ ∂φ + ( ρA z ) ∂z (201) [72] It should be obvious that since ρ, φ and z are labels for specific coordinates and not variable indices, the summation convention does not apply.
5.3.3 Spherical Coordinate System 125 The curl of a differentiable vector A is given by: ∇ × A = 1 ρ e ρ ρ e φ e z ∂ρ ∂φ ∂z A ρ ρA φ A z (202) For the plane polar coordinate system, these operators and operations can be obtained by dropping the z components or terms from the cylindrical form of the above operators and operations. 5.3.3 Spherical Coordinate System For the spherical system identified by the coordinates ( r, θ, φ ) with a set of orthonormal basis vectors e r , e θ and e φ we have the following definitions for the nabla based operators and operations. [73] The nabla operator is given by: = e r r + e θ 1 r θ + e φ 1 r sin θ φ (203) The Laplacian operator is given by: 2 = rr + 2 r r + 1 r 2 θθ + cos θ r 2 sin θ θ + 1 r 2 sin 2 θ φφ (204) The gradient of a differentiable scalar f is given by: f = e r ∂f ∂r + e θ 1 r ∂f ∂θ + e φ 1 r sin θ ∂f ∂φ (205) [73] Again, the summation convention does not apply to r, θ and φ .
5.3.4 General Orthogonal Coordinate System 126 The divergence of a differentiable vector A is given by: ∇ · A = 1 r 2 sin θ sin θ ( r 2 A r ) ∂r + r (sin θA θ ) ∂θ + r ∂A φ ∂φ (206) The curl of a differentiable vector A is given by: ∇ × A = 1 r 2 sin θ e r r e θ r sin θ e φ ∂r ∂θ ∂φ A r rA θ r sin θA φ (207) 5.3.4 General Orthogonal Coordinate System For general orthogonal systems in a 3D space identified by the coordinates ( u 1 , u 2 , u 3 ) with a set of unit basis vectors u 1 , u 2 and u 3 and scale factors h 1 , h 2 and h 3 where h i = r ∂u i and r = x i e i is the position vector, we have the following definitions for the nabla based operators and operations. The nabla operator is given by: = u 1 h 1 ∂u 1 + u 2 h 2 ∂u 2 + u 3 h 3 ∂u 3 (208) The Laplacian operator is given by: 2 = 1 h 1 h 2 h 3 ∂u 1 h 2 h 3 h 1 ∂u 1 + ∂u 2 h 1 h 3 h 2 ∂u 2 + ∂u 3 h 1 h 2 h 3 ∂u 3 (209) The gradient of a differentiable scalar f is given by: f = u 1 h 1 ∂f ∂u 1 + u 2 h 2 ∂f ∂u 2 + u 3 h 3 ∂f ∂u 3 (210)
5.4 Common Identities in Vector and Tensor Notation 127 The divergence of a differentiable vector A is given by: ∇ · A = 1 h 1 h 2 h 3 ∂u 1 ( h 2 h 3 A 1 ) + ∂u 2 ( h 1 h 3 A 2 ) + ∂u 3 ( h 1 h 2 A 3 ) (211) The curl of a differentiable vector A is given by: ∇ × A = 1 h 1 h 2 h 3 h 1 u 1 h 2 u 2 h 3 u 3 ∂u 1 ∂u 2 ∂u 3 h 1 A 1 h 2 A 2 h 3 A 3 (212) 5.4 Common Identities in Vector and Tensor Notation

#### You've reached the end of your free preview.

Want to read all 171 pages?

• Summer '20
• Rajendra Paramanik
• Tensor, Coordinate system, Polar coordinate system, Coordinate systems

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern