Hertz Potentials in Cylindrical Coordinates

Transcription

1 Introduction 26 June 2009

2 Motivation Introduction Motivation Differential Forms and the Hodge Star Motivation 1 To extend Hertz Potentials to general curvilinear coordinates. 2 To explore the usefulness of differential forms.

3 Introduction Differential Forms and the Hodge Star Motivation Differential Forms and the Hodge Star Why use differential forms? 1 Maxwell s equations take a more elegant form. 2 Differential forms are intrinsically coordinate-independent.

4 Introduction Differential Forms and the Hodge Star Motivation Differential Forms and the Hodge Star Definition For a pseudo-riemannian orientable metrizable manifold (M, g), the Hodge Star is the unique operator : Ω k (M) Ω n k (M) such that for ω, η Ω k (M) ω, η = ω η Definition For 4-dimensional Minkowski space with positive signature, define the codifferential δ as δ = d

5 Introduction Potentials Let F be the 2-form representing the electromagnetic field in vacuum. Theorem Maxwell s equations then become df = 0 δf = 0

6 Introduction Electromagnetic Potential Potentials Lemma For any form ω, d 2 ω = 0 Since df = 0, F is called closed, and for any simply connected manifold, every closed form is exact. Thus there exists a 1-form A = A µ dx µ such that F = da. Immediately from this it follows df = d 2 A = 0. It also follows that for F = 1 2 F µνdx µ dx ν, F µν = µ A ν ν A µ

7 Hertz Potentials Introduction Potentials Choosing the Lorentz gauge is equivalent to choosing δa = 0. Since δ 2 = ± d 2, this means that A is closed and exact. There exists a 2-form Π = 1 2 Π µνdx µ dx ν such that δπ = A. Consequently, F = da = dδπ.

8 Introduction Hertz Potentials (Cont.) Potentials Definition Define the operator such that ω = dδω + δdω The condition that F = dδπ solves both Maxwell equations is satisfied when Π = dδπ + δdπ = 0. Then F = dδπ = δdπ.

9 Notation Introduction Coordinate-Specific Potentials and Fields Scalar Fields Examples Choose a particular coordinate system x 0, x 1, x 2, x 3. Definition Let h µ νρσ, h µν ρσ, and h µνρ σ such that dx µ = h µ νρσdx ν dx ρ dx σ (dx µ dx ν ) = h µν ρσdx ρ dx σ (dx µ dx ν dx ρ ) = h µνρ σdx σ. In Cartesian coordinates, these reduce to factors times ɛ µνρσ.

10 Potentials Introduction Coordinate-Specific Potentials and Fields Scalar Fields Examples In these arbitrary but fixed coordinates, the 4-potential and Hertz potential become Π = 1 2 Π µνdx µ dx ν A = A µ dx µ = 1 2 ν(π ρσ h ρσ λξ)h νλξ µdx µ.

11 Field Equations Introduction Coordinate-Specific Potentials and Fields Scalar Fields Examples With the coordinates chosen and the previous definitions, the field becomes F = 1 2 µ( λ (Π ξδ h ξδ ρσ)h λρσ ν)dx µ dx ν. The condition Π = 0 also becomes Π = 1 2 [ µ(h λρσ ν λ (h ξδ ρσπ ξδ ))+h λσ µν λ (h ρξδ σ ρ Π ξδ )]dx µ dx ν = 0.

12 Introduction Unidirectional Hertz Potentials Coordinate-Specific Potentials and Fields Scalar Fields Examples Take Π = Π 01 dx 0 dx 1 + Π 23 dx 2 dx 3. For Π = 0, the dx 0 dx 1 and dx 2 dx 3 components yield the following equations. ( 0 (h ) 1 (h ))(h 01 23Π 01 ) + h 23 01( 2 (h ) 3 (h ))Π 01 = 0 h 01 23( 0 (h ) 1 (h ))Π 23 + ( 2 (h ) 3 (h ))(h 23 01Π 23 ) = 0 To contrast, for a scalar field φ, φ = [h (h ) + h (h ) + h (h ) + h (h )]φ

13 Cartesian Coordinates Introduction Coordinate-Specific Potentials and Fields Scalar Fields Examples Let x 0 = t, x 1 = x, x 2 = y, x 3 = z, and let Π 01 = φ, Π 23 = ψ. φ = 0 ψ = 0

14 Introduction Cylindrical Coordinates Coordinate-Specific Potentials and Fields Scalar Fields Examples Let x 0 = t, x 1 = z, x 2 = ρ, x 3 = ϕ, and take Π 01 = φ, Π 23 = ψ ρ. 2 t Π 01 1 ρ ρ(ρ ρ Π 01 ) 1 ρ 2 2 ϕπ 01 2 z Π 23 = 0 2 t Π 23 ρ (ρ ρ Π 23 ρ ) 1 ρ 2 2 ϕπ 23 2 z Π 23 = 0 φ = 0 ψ = 0

15 Introduction Cylindrical Coordinates (Again) Coordinate-Specific Potentials and Fields Scalar Fields Examples Now take x 0 = t, x 1 = ρ, x 2 = ϕ, x 3 = z, and Π 01 = φ ρ, Π 23 = ψ. 2 t Π 01 ρ ( 1 ρ ρ(ρπ 01 )) 1 ρ 2 2 ϕπ 01 2 z Π 01 = 0 2 t Π 23 ρ ρ ( 1 ρ ρπ 23 ) 1 ρ 2 2 ϕπ 23 2 z Π 23 = 0 ( 2 ρ ρ)φ = 0 ( 2 ρ ρ)ψ = 0

16 Spherical Introduction Coordinate-Specific Potentials and Fields Scalar Fields Examples Let x 0 = t, x 1 = r, x 2 = θ, x 3 = ϕ with Π 01 = φ, Π 23 = ψ r 2 sin θ. 2 t Π 01 r ( 1 r 2 r (r 2 Π 01 )) 1 r 2 sin θ θ(sin θ θ Π 01 ) 1 r 2 sin θ 2 ϕ Π 01 = 0 2 t Π 23 r 2 r ( 1 r 2 r Π 23) 1 r 2 θ(sin θ θ ( Π 23 sin θ ) 1 r 2 sin θ 2 ϕ Π 23 = 0 ( 2 r 2 )φ = 0 ( 2 r 2 )ψ = 0