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
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Hertz potentials in curvilinear coordinates
July 9, 2010
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Maxwell’s Equations
Constants
Vector Equations
For simplicity, take
~ =ρ
~ ·E
∇
~ − ∂t E
~ = ~
~ B
∇×
(1)
0 = µ0 = c = 1.
(2)
Potentials
~ =0
~ ·B
∇
~ + ∇×
~ =0
~ E
∂t B
(3)
(4)
(3) and (4) imply
~ =∇
~
~ ×A
B
~ = −∇V
~
~ − ∂t A
E
(5)
(6)
Charge Conservation
~ · ~ = 0.
Also note that (1) and (2) imply ∂t ρ + ∇
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Hertz potentials in curvilinear coordinates
July 9, 2010
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Hertz Potentials
Hertz Potentials
~e
~ ·Π
V = −∇
~ = ∂t Π
~ e + ∇×
~m
~ Π
A
(7)
(8)
Lorenz Condition
~ =0
~ ·A
∂t V + ∇
Definition
Inhomogeneous Maxwell Equations
~ e) = ρ
~ · (Π
∇
~ m ) + ∂t (Π
~ m ) = ~
~
∇×(
Π
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= ∂t2 − ∇2
(9)
(10)
Hertz potentials in curvilinear coordinates
July 9, 2010
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Hertz Potentials
From here on, set ρ = 0, ~ = ~0.
Equations of Motion
~ e = ∇×
~ + ∇g
~
~ W
~ + ∂t G
Π
~ m = −∂t W
~ − ∇w
~
~ + ∇×
~ G
Π
(11)
(12)
~ terms come from (9) and (10) just as V and A
~ came from
The w and W
~
(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
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Differential Geometry
Let (x 0 , . . . , x n−1 ) be the coordinate system on an n-dimensional
manifold. Then we write vectors on that manifold as
~v = v 0 ∂x 0 + · · · + v n−1 ∂x n−1 ,
and 1-forms (or covectors) as
v = v0 dx 0 + · · · vn−1 dx n−1 .
Example
For Minwoski space, we can write the electromagnetic potential Aµ as the
vector
~ = Aµ ∂x µ = V ∂t + Ax ∂x + Ay ∂y + Az ∂z
A
(not to be confused with the 3-vector from before) or as the 1-form
A = −Vdt + Ax dx + Ay dy + Az dz.
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Hertz potentials in curvilinear coordinates
July 9, 2010
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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 dx α1 ∧ · · · ∧dx αk , define the
differential of f as
df =
n−1
X
∂fα
µ=0
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1 ···αk
∂x i
dx µ ∧dx α1 ∧ · · · ∧dx αk .
Hertz potentials in curvilinear coordinates
July 9, 2010
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Differential Geometry
Definition
Let η0···n be the volume form. For a k-form of the form
f = fα1 ···αk dx α1 ∧ · · · ∧dx αk , define the Hodge dual of f as
∗f = fα1 ···αk η α1 ···αk β1 ···βn−k dx β1 ∧ · · · ∧dx βn−k .
Definition
δ = ∗d∗
Definition
= dδ + δd = d∗d∗ + ∗d∗d
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Hertz potentials in curvilinear coordinates
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Electromagnetism Revisited
Maxwell’s Equations
Define
Lorenz Condition
δA = 0
F = −Ei dt∧dx i + Bi ∗(dt∧dx i ).
Relaxed Lorenz Condition
Then (1) – (4) become
δ(A + G ) = 0
δF = J
(13)
dF = 0
(14)
Hertz Potentials
The relaxed Lorenz condition
implies
Potentials
(14) implies
A = δΠ − G
F = dA
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(16)
(15)
Hertz potentials in curvilinear coordinates
July 9, 2010
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Electromagnetism Revisited
Since
0 = J = δF = δdA = δd(δΠ − G )
(17)
= δ(Π − dG ),
(18)
we can write
Equations of Motion
Π = dG + δW .
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Hertz potentials in curvilinear coordinates
(19)
July 9, 2010
10 / 20
Cartesian Coordinates
(x 0 , x 1 , x 2 , x 3 ) = (t, x, y , z)
Π = φdt∧dz + ψ∗(dt∧dz)
= φdt∧dz + ψdx∧dy
Equations of Motion
Given Π = 0,
φ = 0
ψ = 0
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Hertz potentials in curvilinear coordinates
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Axial Cylindrical Coordinates
(x 0 , x 1 , x 2 , x 3 ) = (t, ρ, ϕ, z)
Π = φdt∧dz + ψ∗(dt∧dz)
= φdt∧dz + ρψdρ∧dϕ
Equations of Motion
Given Π = 0,
φ = 0
ψ = 0
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Hertz potentials in curvilinear coordinates
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Spherical Coordinates
(x 0 , x 1 , x 2 , x 3 ) = (t, r , θ, ϕ)
Π = φdt∧dr + ψ∗(dt∧dr )
= φdt∧dr + ψr 2 sin θdθ∧dϕ
Definition
ˆ = + 2 ∂r = ∂t2 − ∂r2 −
r
1
∂
r 2 sin θ θ
sin θ∂θ −
1
∂2
r 2 sin2 θ ϕ
Equations of Motion?
ˆ − ∂r 2φ )dt∧dr − ∂θ 2φ dt∧dθ − ∂ϕ 2φ dt∧dϕ
Π = (φ
r
r
r
2ψ
2ψ
2ψ
ˆ − ∂r
+ (ψ
)∗(dt∧dr ) − ∂θ ∗(dt∧dθ) − ∂ϕ ∗(dt∧dϕ)
r
r
r
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Hertz potentials in curvilinear coordinates
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Spherical Coordinates
2
G = φ,
r
2
∗W = ψ,
r
2φ
2φ
2φ
dt∧dr − ∂θ dt∧dθ − ∂ϕ dt∧dϕ
r
r
r
2ψ
2ψ
2ψ
δW = −∂r
∗(dt∧dr ) − ∂θ ∗(dt∧dθ) − ∂ϕ ∗(dt∧dϕ)
r
r
r
dG = −∂r
Equations of Motion
Given Π = dG + δW ,
ˆ =0
φ
ˆ =0
ψ
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Hertz potentials in curvilinear coordinates
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Schwarzchild Coordinates
(x 0 , x 1 , x 2 , x 3 ) = (t, r , θ, ϕ)
ds 2 = (1 −
rs
2
r )dt
+
1
dr 2
(1− rrs )
+ r 2 dθ2 + r 2 sin2 θdϕ2
G = 2ζ
r φ
∗W = 2ζ
r ψ
Π = φdt∧dr + ψ∗(dt∧dr )
= φdt∧dr + ψr 2 sin θdθ∧dϕ
Definition
ζ =1−
ˆ = 1 ∂t2 − ∂r ζ∂r −
ζ
rs
r
1
∂
r 2 sin θ θ
sin θ∂θ −
1
∂2
r 2 sin2 θ ϕ
Equations of Motion
Given Π = dG + δW ,
ˆ =0
φ
ˆ =0
ψ
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Hertz potentials in curvilinear coordinates
July 9, 2010
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Radial Cylindrical Coordinates
(x 0 , x 1 , x 2 , x 3 ) = (t, ρ, ϕ, z)
Π = φdt∧dρ + ψρdφ∧dz
Equations of Motion?
Π = (φ +
+(ψ +
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φ
)dt∧dρ
ρ2
− ∂ϕ 2φ
ρ dt∧dϕ
ψ
)∗(dt∧dρ)
ρ2
− ∂ϕ 2ψ
ρ ∗(dt∧dϕ)
Hertz potentials in curvilinear coordinates
July 9, 2010
16 / 20
TE Modes in Cylindrical Coordinates
Define
ΠA = φA dt∧dz + ψA ∗(dt∧dz),
ΠR = φR dt∧dρ + ψR ∗(dt∧dρ).
We start with
A = δΠA = δΠR − G .
(20)
Bz = Bkω sin(kz)g (ρ, ϕ)e −iωt ,
hence
φA = 0, ψA =
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−Bkω
sin(kz)g (ρ, ϕ)e −iωt ,
ω2 − k 2
Hertz potentials in curvilinear coordinates
July 9, 2010
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TE Modes in Cylindrical Coordinates
From (20) we obtain
Radial Modes
iBkω
sin(kz)∂ϕ g (ρ, ϕ)e −iωt
ρω(ω 2 − k 2 )
−Bkω
ψR =
cos(kz)∂ρ g (ρ, ϕ)e −iωt
k(ω 2 − k 2 )
iBkω
Gt =
sin(kz)∂ρ ∂ϕ g (ρ, ϕ)e −iωt , Gρ = 0
ρω(ω 2 − k 2 )
Bkω
Gz =
cos(kz)∂ρ ∂ϕ g (ρ, ϕ)e −iωt , Gϕ = 0
kρ(ω 2 − k 2 )
φR =
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Hertz potentials in curvilinear coordinates
July 9, 2010
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Azimuthal Cylindrical Coordinates
Define
ΠP = φP dt∧dϕ + ψP ∗(d∧dϕ),
and again start with
δΠA = δΠP − G .
(21)
This yields
Azimuthal Modes
−iBkω
sin(kz)ρ∂ρ g (ρ, ϕ)e −iωt
ω(ω 2 − k 2 )
Bkω
ψP =
cos(kz)∂ϕ g (ρ, ϕ)e i−ωt
ρk(ω 2 − k 2 )
1
Gt = − 2 ∂ϕ φP , Gρ = 0
ρ
1
Gz = ∂ρ ψP , Gϕ = 0
ρ
φP =
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Hertz potentials in curvilinear coordinates
July 9, 2010
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Ongoing and Future Work
1
Determine the equations of motion for the radial and azimuthal
cylindrical cases.
2
Consider the polar and azimuthal spherical cases.
3
Examine the boundary conditions of all of the presented cases.
4
Consider geometries with non-trivial topology.
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Hertz potentials in curvilinear coordinates
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