Hertz potentials in curvilinear coordinates - Texas A&M ...

Hertz potentials in curvilinear coordinates

Jeff Bouas

Texas A&M University

July 9, 2010

Quantum Vacuum Workshop

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 1 / 20

Purpose and Outline

Purpose
To describe any arbitrary electromagnetic field in a bounded geometry
in terms of two scalar fields, and
To define these fields such that the boundary conditions consist of at
most first-derivatives of the fields.

Outline
1 Review of Electromagnetism and Hertz Potentials in Vector Formalism
2 Overview of Differential Form Formalism
3 Formulation of Electromagnetism in Differential Form Formalism
4 “Scalar” Hertz Potential Examples

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 2 / 20

Maxwell’s Equations

Vector Equations Constants (5)
(6)
∇ · E = ρ (1) For simplicity, take
∇×B − ∂tE =  (2)
0 = µ0 = c = 1.
∇ · B = 0 (3)
∂tB + ∇×E = 0 (4) Potentials

(3) and (4) imply

B =∇×A
E = −∇V − ∂t A

Charge Conservation
Also note that (1) and (2) imply ∂t ρ + ∇ ·  = 0.

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 3 / 20

Hertz Potentials

Hertz Potentials

V = −∇ · Πe (7)
A = ∂t Πe + ∇×Πm
(8) Lorenz Condition

∂t V + ∇ · A = 0

Inhomogeneous Maxwell Equations Definition

∇ · ( Πe) = ρ = ∂t2 − ∇2
∇×( Πm) + ∂t ( Πm) = 
(9)
(10)

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 4 / 20

Hertz Potentials

From here on, set ρ = 0,  = 0. (11)
(12)
Equations of Motion

Πe = ∇×W + ∇g + ∂t G
Πm = −∂t W − ∇w + ∇×G

The w and W terms come from (9) and (10) just as V and A came from
(3) and (4). The g and G terms come from relaxing the Lorenz condition.

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 5 / 20

Differential Geometry

Let (x0, . . . , xn−1) be the coordinate system on an n-dimensional
manifold. Then we write vectors on that manifold as

v = v 0∂x0 + · · · + v n−1∂xn−1 ,

and 1-forms (or covectors) as

v = v0dx 0 + · · · vn−1dx n−1.

Example

For Minwoski space, we can write the electromagnetic potential Aµ as the

vector

A = Aµ∂xµ = V ∂t + Ax ∂x + Ay ∂y + Az ∂z

(not to be confused with the 3-vector from before) or as the 1-form

A = −Vdt + Ax dx + Ay dy + Az dz.

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 6 / 20

Differential Geometry

Definition

The wedge product of two forms, written f ∧g , is the antisymmetrized
tensor product.

Example

dx∧dy = dx⊗dy − dy ⊗dx

dx∧dy ∧dz = dx⊗dy ⊗dz − dx⊗dz⊗dy + dz⊗dx⊗dy + . . .

Definition

For a k-form of the form f = fα1···αk dxα1∧ · · · ∧dxαk , define the
differential of f as

n−1 ∂ fα1 ···αk
∂xi
df = dx µ∧dx α1 ∧ · · · ∧dx αk .

µ=0

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 7 / 20

Differential Geometry

Definition

Let η0···n be the volume form. For a k-form of the form
f = fα1···αk dxα1∧ · · · ∧dxαk , define the Hodge dual of f as

∗f = fα1···αk ηα1···αk β1···βn−k dx β1 ∧ · · · ∧dx βn−k .

Definition

δ = ∗d∗

Definition

= dδ + δd = d∗d∗ + ∗d∗d

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 8 / 20

Electromagnetism Revisited

Maxwell’s Equations Lorenz Condition

Define δA = 0

F = −Ei dt∧dxi + Bi ∗(dt∧dxi ). Relaxed Lorenz Condition

Then (1) – (4) become δ(A + G ) = 0

δF = J (13) Hertz Potentials
dF = 0 (14)
The relaxed Lorenz condition
Potentials implies

(14) implies A = δΠ − G (16)

F = dA (15)

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 9 / 20

Electromagnetism Revisited (17)
(18)
Since
0 = J = δF = δdA = δd(δΠ − G ) (19)
= δ( Π − dG ),

we can write

Equations of Motion

Π = dG + δW .

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 10 / 20

Cartesian Coordinates

(x0, x1, x2, x3) = (t, x, y , z)

Π = φdt∧dz + ψ∗(dt∧dz)
= φdt∧dz + ψdx∧dy

Equations of Motion

Given Π = 0,

φ=0
ψ=0

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 11 / 20

Axial Cylindrical Coordinates

(x0, x1, x2, x3) = (t, ρ, ϕ, z)

Π = φdt∧dz + ψ∗(dt∧dz)
= φdt∧dz + ρψdρ∧dϕ

Equations of Motion

Given Π = 0,

φ=0
ψ=0

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 12 / 20

Spherical Coordinates

(x0, x1, x2, x3) = (t, r , θ, ϕ)

Π = φdt∧dr + ψ∗(dt∧dr )
= φdt∧dr + ψr 2 sin θdθ∧dϕ

Definition + 2 ∂r = ∂t2 − ∂r2 − r 2 1 θ ∂θ sin θ∂θ − r 2 1 θ ∂ϕ2
r sin sin2
ˆ=

Equations of Motion?

Π = ( ˆ φ − ∂r 2φ )dt∧dr − ∂θ 2φ dt∧dθ − ∂ϕ 2φ dt∧dϕ
r r r

+ ( ˆ ψ − ∂r 2ψ )∗(dt ∧dr ) − ∂θ 2ψ ∗(dt ∧d θ) − ∂ϕ 2ψ ∗(dt ∧d ϕ)
r r r

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 13 / 20

Spherical Coordinates

2 dG = −∂r 2φ dt∧dr − ∂θ 2φ dt ∧d θ − ∂ϕ 2φ dt ∧d ϕ
G = φ, r r r

r

∗W = 2 δW = −∂r 2ψ ∗(dt ∧dr ) − ∂θ 2ψ ∗(dt ∧d θ) − ∂ϕ 2ψ ∗(dt ∧d ϕ)
ψ, r r r
r

Equations of Motion

Given Π = dG + δW ,

ˆφ = 0
ˆψ = 0

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 14 / 20

Schwarzchild Coordinates

(x0, x1, x2, x3) = (t, r , θ, ϕ)

ds 2 = (1 − rs )dt 2 + 1 rs ) dr 2 + r 2d θ2 + r2 sin2 θd ϕ2
r r
(1−

Π = φdt∧dr + ψ∗(dt∧dr ) G = 2ζ φ
= φdt∧dr + ψr 2 sin θdθ∧dϕ r
2ζ
∗W = r ψ

Definition

ζ = 1 − rs
r

ˆ = 1 ∂t2 − ∂r ζ∂r − r 2 1 θ ∂θ sin θ∂θ − r 2 1 θ ∂ϕ2
ζ sin sin2

Equations of Motion

Given Π = dG + δW ,

Jeff Bouas (Texas A&M University) ˆφ = 0 July 9, 2010 15 / 20
ˆψ = 0

Hertz potentials in curvilinear coordinates

Radial Cylindrical Coordinates

(x0, x1, x2, x3) = (t, ρ, ϕ, z)

Π = φdt∧dρ + ψρdφ∧dz

Equations of Motion?

Π=( φ + φ )dt ∧d ρ − ∂ϕ 2φ dt ∧d ϕ
ρ2 ρ

+( ψ + ψ )∗(dt ∧d ρ) − ∂ϕ 2ψ ∗(dt ∧d ϕ)
ρ2 ρ

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 16 / 20

TE Modes in Cylindrical Coordinates

Define ΠA = φAdt∧dz + ψA∗(dt∧dz),
We start with ΠR = φR dt∧dρ + ψR ∗(dt∧dρ).

A = δΠA = δΠR − G . (20)

Bz = Bkω sin(kz)g (ρ, ϕ)e−iωt ,

hence −Bkω
ω2 − k2
φA = 0, ψA = sin(kz)g (ρ, ϕ)e−iωt ,

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 17 / 20

TE Modes in Cylindrical Coordinates

From (20) we obtain

Radial Modes

φR = iBkω k2) sin(kz)∂ϕg (ρ, ϕ)e −i ωt
ρω(ω2 −

ψR = −Bkω cos(kz)∂ρg (ρ, ϕ)e−iωt
k(ω2 − k2)

Gt = iBkω k2) sin(kz)∂ρ∂ϕg (ρ, ϕ)e−iωt , Gρ = 0
ρω(ω2 −

Gz = Bk ω k2) cos(kz)∂ρ∂ϕg (ρ, ϕ)e−iωt , Gϕ = 0
kρ(ω2 −

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 18 / 20

Azimuthal Cylindrical Coordinates

Define

ΠP = φP dt∧dϕ + ψP ∗(d∧dϕ),

and again start with

δΠA = δΠP − G . (21)

This yields

Azimuthal Modes

φP = −iBkω sin(kz)ρ∂ρg (ρ, ϕ)e −i ωt
ω(ω2 − k2)

ψP = Bk ω k2) cos(kz)∂ϕg (ρ, ϕ)e i −ωt
ρk(ω2 −

1
Gt = − ρ2 ∂ϕφP , Gρ = 0

1
Gz = ρ ∂ρψP , Gϕ = 0

Jeff Bouas (Texas A&M University) Hertz potentials in curvilinear coordinates July 9, 2010 19 / 20