Laplacian

Potential solutions to the static Maxwell’s equation using geometric algebra

March 20, 2018 math and physics play , , , , , , , , , , , , , , , , ,

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When neither the electromagnetic field strength \( F = \BE + I \eta \BH \), nor current \( J = \eta (c \rho – \BJ) + I(c\rho_m – \BM) \) is a function of time, then the geometric algebra form of Maxwell’s equations is the first order multivector (gradient) equation
\begin{equation}\label{eqn:staticPotentials:20}
\spacegrad F = J.
\end{equation}

While direct solutions to this equations are possible with the multivector Green’s function for the gradient
\begin{equation}\label{eqn:staticPotentials:40}
G(\Bx, \Bx’) = \inv{4\pi} \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3 },
\end{equation}
the aim in this post is to explore second order (potential) solutions in a geometric algebra context. Can we assume that it is possible to find a multivector potential \( A \) for which
\begin{equation}\label{eqn:staticPotentials:60}
F = \spacegrad A,
\end{equation}
is a solution to the Maxwell statics equation? If such a solution exists, then Maxwell’s equation is simply
\begin{equation}\label{eqn:staticPotentials:80}
\spacegrad^2 A = J,
\end{equation}
which can be easily solved using the scalar Green’s function for the Laplacian
\begin{equation}\label{eqn:staticPotentials:240}
G(\Bx, \Bx’) = -\inv{\Norm{\Bx – \Bx’} },
\end{equation}
a beastie that may be easier to convolve than the vector valued Green’s function for the gradient.

It is immediately clear that some restrictions must be imposed on the multivector potential \(A\). In particular, since the field \( F \) has only vector and bivector grades, this gradient must have no scalar, nor pseudoscalar grades. That is
\begin{equation}\label{eqn:staticPotentials:100}
\gpgrade{\spacegrad A}{0,3} = 0.
\end{equation}
This constraint on the potential can be avoided if a grade selection operation is built directly into the assumed potential solution, requiring that the field is given by
\begin{equation}\label{eqn:staticPotentials:120}
F = \gpgrade{\spacegrad A}{1,2}.
\end{equation}
However, after imposing such a constraint, Maxwell’s equation has a much less friendly form
\begin{equation}\label{eqn:staticPotentials:140}
\spacegrad^2 A – \spacegrad \gpgrade{\spacegrad A}{0,3} = J.
\end{equation}
Luckily, it is possible to introduce a transformation of potentials, called a gauge transformation, that eliminates the ugly grade selection term, and allows the potential equation to be expressed as a plain old Laplacian. We do so by assuming first that it is possible to find a solution of the Laplacian equation that has the desired grade restrictions. That is
\begin{equation}\label{eqn:staticPotentials:160}
\begin{aligned}
\spacegrad^2 A’ &= J \\
\gpgrade{\spacegrad A’}{0,3} &= 0,
\end{aligned}
\end{equation}
for which \( F = \spacegrad A’ \) is a grade 1,2 solution to \( \spacegrad F = J \). Suppose that \( A \) is any formal solution, free of any grade restrictions, to \( \spacegrad^2 A = J \), and \( F = \gpgrade{\spacegrad A}{1,2} \). Can we find a function \( \tilde{A} \) for which \( A = A’ + \tilde{A} \)?

Maxwell’s equation in terms of \( A \) is
\begin{equation}\label{eqn:staticPotentials:180}
\begin{aligned}
J
&= \spacegrad \gpgrade{\spacegrad A}{1,2} \\
&= \spacegrad^2 A
– \spacegrad \gpgrade{\spacegrad A}{0,3} \\
&= \spacegrad^2 (A’ + \tilde{A})
– \spacegrad \gpgrade{\spacegrad A}{0,3}
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:staticPotentials:200}
\spacegrad^2 \tilde{A} = \spacegrad \gpgrade{\spacegrad A}{0,3}.
\end{equation}
This non-homogeneous Laplacian equation that can be solved as is for \( \tilde{A} \) using the Green’s function for the Laplacian. Alternatively, we may also solve the equivalent first order system using the Green’s function for the gradient.
\begin{equation}\label{eqn:staticPotentials:220}
\spacegrad \tilde{A} = \gpgrade{\spacegrad A}{0,3}.
\end{equation}
Clearly \( \tilde{A} \) is not unique, as we can add any function \( \psi \) satisfying the homogeneous Laplacian equation \( \spacegrad^2 \psi = 0 \).

In summary, if \( A \) is any multivector solution to \( \spacegrad^2 A = J \), that is
\begin{equation}\label{eqn:staticPotentials:260}
A(\Bx)
= \int dV’ G(\Bx, \Bx’) J(\Bx’)
= -\int dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} },
\end{equation}
then \( F = \spacegrad A’ \) is a solution to Maxwell’s equation, where \( A’ = A – \tilde{A} \), and \( \tilde{A} \) is a solution to the non-homogeneous Laplacian equation or the non-homogeneous gradient equation above.

Integral form of the gauge transformation.

Additional insight is possible by considering the gauge transformation in integral form. Suppose that
\begin{equation}\label{eqn:staticPotentials:280}
A(\Bx) = -\int_V dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \tilde{A}(\Bx),
\end{equation}
is a solution of \( \spacegrad^2 A = J \), where \( \tilde{A} \) is a multivector solution to the homogeneous Laplacian equation \( \spacegrad^2 \tilde{A} = 0 \). Let’s look at the constraints on \( \tilde{A} \) that must be imposed for \( F = \spacegrad A \) to be a valid (i.e. grade 1,2) solution of Maxwell’s equation.
\begin{equation}\label{eqn:staticPotentials:300}
\begin{aligned}
F
&= \spacegrad A \\
&=
-\int_V dV’ \lr{ \spacegrad \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_V dV’ \lr{ \spacegrad’ \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_V dV’ \spacegrad’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V dV’ \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
– \spacegrad \tilde{A}(\Bx) \\
&=
\int_{\partial V} dA’ \ncap’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
– \spacegrad \tilde{A}(\Bx).
\end{aligned}
\end{equation}
Where \( \ncap’ = (\Bx’ – \Bx)/\Norm{\Bx’ – \Bx} \), and the fundamental theorem of geometric calculus has been used to transform the gradient volume integral into an integral over the bounding surface. Operating on Maxwell’s equation with the gradient gives \( \spacegrad^2 F = \spacegrad J \), which has only grades 1,2 on the left hand side, meaning that \( J \) is constrained in a way that requires \( \spacegrad J \) to have only grades 1,2. This means that \( F \) has grades 1,2 if
\begin{equation}\label{eqn:staticPotentials:320}
\spacegrad \tilde{A}(\Bx)
= \int_{\partial V} dA’ \frac{ \gpgrade{\ncap’ J(\Bx’)}{0,3} }{\Norm{\Bx – \Bx’} }.
\end{equation}
The product \( \ncap J \) expands to
\begin{equation}\label{eqn:staticPotentials:340}
\begin{aligned}
\ncap J
&=
\gpgradezero{\ncap J_1} + \gpgradethree{\ncap J_2} \\
&=
\ncap \cdot (-\eta \BJ) + \gpgradethree{\ncap (-I \BM)} \\
&=- \eta \ncap \cdot \BJ -I \ncap \cdot \BM,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:staticPotentials:360}
\spacegrad \tilde{A}(\Bx)
=
-\int_{\partial V} dA’ \frac{ \eta \ncap’ \cdot \BJ(\Bx’) + I \ncap’ \cdot \BM(\Bx’)}{\Norm{\Bx – \Bx’} }.
\end{equation}
Observe that if there is no flux of current density \( \BJ \) and (fictitious) magnetic current density \( \BM \) through the surface, then \( F = \spacegrad A \) is a solution to Maxwell’s equation without any gauge transformation. Alternatively \( F = \spacegrad A \) is also a solution if \( \lim_{\Bx’ \rightarrow \infty} \BJ(\Bx’)/\Norm{\Bx – \Bx’} = \lim_{\Bx’ \rightarrow \infty} \BM(\Bx’)/\Norm{\Bx – \Bx’} = 0 \) and the bounding volume is taken to infinity.

References

A comparison of Geometric Algebra electrodynamic potential methods

January 7, 2017 math and physics play , , , , , , , , , , , , , , , , , , ,

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Motivation

Geometric algebra (GA) allows for a compact description of Maxwell’s equations in either an explicit 3D representation or a STA (SpaceTime Algebra [2]) representation. The 3D GA and STA representations Maxwell’s equation both the form

\begin{equation}\label{eqn:potentialMethods:1280}
L \boldsymbol{\mathcal{F}} = J,
\end{equation}

where \( J \) represents the sources, \( L \) is a multivector gradient operator that includes partial derivative operator components for each of the space and time coordinates, and

\begin{equation}\label{eqn:potentialMethods:1020}
\boldsymbol{\mathcal{F}} = \boldsymbol{\mathcal{E}} + \eta I \boldsymbol{\mathcal{H}},
\end{equation}

is an electromagnetic field multivector, \( I = \Be_1 \Be_2 \Be_3 \) is the \R{3} pseudoscalar, and \( \eta = \sqrt{\mu/\epsilon} \) is the impedance of the media.

When Maxwell’s equations are extended to include magnetic sources in addition to conventional electric sources (as used in antenna-theory [1] and microwave engineering [3]), they take the form

\begin{equation}\label{eqn:chapter3Notes:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \PD{t}{\boldsymbol{\mathcal{B}}}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = q_{\textrm{e}}
\end{equation}
\begin{equation}\label{eqn:chapter3Notes:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = q_{\textrm{m}}.
\end{equation}

The corresponding GA Maxwell equations in their respective 3D and STA forms are

\begin{equation}\label{eqn:potentialMethods:300}
\lr{ \spacegrad + \inv{v} \PD{t}{} } \boldsymbol{\mathcal{F}}
=
\eta
\lr{ v q_{\textrm{e}} – \boldsymbol{\mathcal{J}} }
+ I \lr{ v q_{\textrm{m}} – \boldsymbol{\mathcal{M}} }
\end{equation}
\begin{equation}\label{eqn:potentialMethods:320}
\grad \boldsymbol{\mathcal{F}} = \eta J – I M,
\end{equation}

where the wave group velocity in the medium is \( v = 1/\sqrt{\epsilon\mu} \), and the medium is isotropic with
\( \boldsymbol{\mathcal{B}} = \mu \boldsymbol{\mathcal{H}} \), and \( \boldsymbol{\mathcal{D}} = \epsilon \boldsymbol{\mathcal{E}} \). In the STA representation, \( \grad, J, M \) are all four-vectors, the specific meanings of which will be spelled out below.

How to determine the potential equations and the field representation using the conventional distinct Maxwell’s \ref{eqn:chapter3Notes:20}, … is well known. The basic procedure is to consider the electric and magnetic sources in turn, and observe that in each case one of the electric or magnetic fields must have a curl representation. The STA approach is similar, except that it can be observed that the field must have a four-curl representation for each type of source. In the explicit 3D GA formalism
\ref{eqn:potentialMethods:300} how to formulate a natural potential representation is not as obvious. There is no longer an reason to set any component of the field equal to a curl, and the representation of the four curl from the STA approach is awkward. Additionally, it is not obvious what form gauge invariance takes in the 3D GA representation.

Ideas explored in these notes

  • GA representation of Maxwell’s equations including magnetic sources.
  • STA GA formalism for Maxwell’s equations including magnetic sources.
  • Explicit form of the GA potential representation including both electric and magnetic sources.
  • Demonstration of exactly how the 3D and STA potentials are related.
  • Explore the structure of gauge transformations when magnetic sources are included.
  • Explore the structure of gauge transformations in the 3D GA formalism.
  • Specify the form of the Lorentz gauge in the 3D GA formalism.

Traditional vector algebra

No magnetic sources

When magnetic sources are omitted, it follows from \ref{eqn:chapter3Notes:80} that there is some \( \boldsymbol{\mathcal{A}}^{\mathrm{e}} \) for which

\begin{equation}\label{eqn:potentialMethods:20}
\boxed{
\boldsymbol{\mathcal{B}} = \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{e}},
}
\end{equation}

Substitution into Faraday’s law \ref{eqn:chapter3Notes:20} gives

\begin{equation}\label{eqn:potentialMethods:40}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \PD{t}{}\lr{ \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{e}} },
\end{equation}

or
\begin{equation}\label{eqn:potentialMethods:60}
\spacegrad \cross \lr{ \boldsymbol{\mathcal{E}} + \PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} } } = 0.
\end{equation}

A gradient representation of this curled quantity, say \( -\spacegrad \phi \), will provide the required zero

\begin{equation}\label{eqn:potentialMethods:80}
\boxed{
\boldsymbol{\mathcal{E}} = -\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }.
}
\end{equation}

The final two Maxwell equations yield

\begin{equation}\label{eqn:potentialMethods:100}
\begin{aligned}
-\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \spacegrad \lr{ \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} } &= \mu \lr{ \boldsymbol{\mathcal{J}} + \epsilon \PD{t}{} \lr{ -\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} } } } \\
\spacegrad \cdot \lr{ -\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} } } &= q_e/\epsilon,
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:potentialMethods:120}
\boxed{
\begin{aligned}
\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{e}} – \inv{v^2} \PDSq{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
– \spacegrad \lr{
\inv{v^2} \PD{t}{\phi}
+\spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}}
}
&= -\mu \boldsymbol{\mathcal{J}} \\
\spacegrad^2 \phi + \PD{t}{} \lr{ \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} } &= -q_e/\epsilon.
\end{aligned}
}
\end{equation}

Note that the Lorentz condition \( \PDi{t}{(\phi/v^2)} + \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} = 0 \) can be imposed to decouple these, leaving non-homogeneous wave equations for the vector and scalar potentials respectively.

No electric sources

Without electric sources, a curl representation of the electric field can be assumed, satisfying Gauss’s law

\begin{equation}\label{eqn:potentialMethods:140}
\boxed{
\boldsymbol{\mathcal{D}} = – \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}}.
}
\end{equation}

Substitution into the Maxwell-Faraday law gives
\begin{equation}\label{eqn:potentialMethods:160}
\spacegrad \cross \lr{ \boldsymbol{\mathcal{H}} + \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}} } = 0.
\end{equation}

This is satisfied with any gradient, say, \( -\spacegrad \phi_m \), providing a potential representation for the magnetic field

\begin{equation}\label{eqn:potentialMethods:180}
\boxed{
\boldsymbol{\mathcal{H}} = -\spacegrad \phi_m – \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}}.
}
\end{equation}

The remaining Maxwell equations provide the required constraints on the potentials

\begin{equation}\label{eqn:potentialMethods:220}
-\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{m}} + \spacegrad \lr{ \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} } = -\epsilon
\lr{
-\boldsymbol{\mathcal{M}} – \mu \PD{t}{}
\lr{
-\spacegrad \phi_m – \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}}
}
}
\end{equation}
\begin{equation}\label{eqn:potentialMethods:240}
\spacegrad \cdot
\lr{
-\spacegrad \phi_m – \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}}
}
= \inv{\mu} q_m,
\end{equation}

or
\begin{equation}\label{eqn:potentialMethods:260}
\boxed{
\begin{aligned}
\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{m}} – \inv{v^2} \PDSq{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}} – \spacegrad \lr{ \inv{v^2} \PD{t}{\phi_m} + \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} } &= -\epsilon \boldsymbol{\mathcal{M}} \\
\spacegrad^2 \phi_m + \PD{t}{}\lr{ \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} } &= -\inv{\mu} q_m.
\end{aligned}
}
\end{equation}

The general solution to Maxwell’s equations is therefore
\begin{equation}\label{eqn:potentialMethods:280}
\begin{aligned}
\boldsymbol{\mathcal{E}} &=
-\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
– \inv{\epsilon} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
\boldsymbol{\mathcal{H}} &=
\inv{\mu} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{e}}
-\spacegrad \phi_m – \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}},
\end{aligned}
\end{equation}

subject to the constraints \ref{eqn:potentialMethods:120} and \ref{eqn:potentialMethods:260}.

Potential operator structure

Knowing that there is a simple underlying structure to the potential representation of the electromagnetic field in the STA formalism inspires the question of whether that structure can be found directly using the scalar and vector potentials determined above.

Specifically, what is the multivector representation \ref{eqn:potentialMethods:1020} of the electromagnetic field in terms of all the individual potential variables, and can an underlying structure for that field representation be found? The composite field is

\begin{equation}\label{eqn:potentialMethods:280b}
\boldsymbol{\mathcal{F}}
=
-\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
– \inv{\epsilon} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
+ I \eta
\lr{
\inv{\mu} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{e}}
-\spacegrad \phi_m – \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}}
}.
\end{equation}

Can this be factored into into multivector operator and multivector potentials? Expanding the cross products provides some direction

\begin{equation}\label{eqn:potentialMethods:1040}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
– \PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
– \eta \PD{t}{I \boldsymbol{\mathcal{A}}^{\mathrm{m}}}
– \spacegrad \lr{ \phi – \eta I \phi_m } \\
&\quad + \frac{\eta}{2 \mu} \lr{ \rspacegrad \boldsymbol{\mathcal{A}}^{\mathrm{e}} – \boldsymbol{\mathcal{A}}^{\mathrm{e}} \lspacegrad }
+ \frac{1}{2 \epsilon} \lr{ \rspacegrad I \boldsymbol{\mathcal{A}}^{\mathrm{m}} – I \boldsymbol{\mathcal{A}}^{\mathrm{m}} \lspacegrad }.
\end{aligned}
\end{equation}

Observe that the
gradient and the time partials can be grouped together

\begin{equation}\label{eqn:potentialMethods:1060}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
– \PD{t}{ } \lr{\boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \boldsymbol{\mathcal{A}}^{\mathrm{m}}}
– \spacegrad \lr{ \phi + \eta I \phi_m }
+ \frac{v}{2} \lr{ \rspacegrad (\boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \boldsymbol{\mathcal{A}}^{\mathrm{m}}) – (\boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \boldsymbol{\mathcal{A}}^{\mathrm{m}}) \lspacegrad } \\
&=
\inv{2} \lr{
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} } \lr{ v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}} }

\lr{ v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}} \lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
} \\
&+\quad \inv{2} \lr{
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} } \lr{ -\phi – \eta I \phi_m }
– \lr{ \phi + \eta I \phi_m } \lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
}
,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:potentialMethods:1080}
\boxed{
\boldsymbol{\mathcal{F}}
=
\inv{2} \Biglr{
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} }
\lr{
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
}

\lr{
\phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
+ \eta I \phi_m
}
\lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
}
.
}
\end{equation}

There’s a conjugate structure to the potential on each side of the curl operation where we see a sign change for the scalar and pseudoscalar elements only. The reason for this becomes more clear in the STA formalism.

Potentials in the STA formalism.

Maxwell’s equation in its explicit 3D form \ref{eqn:potentialMethods:300} can be
converted to STA form, by introducing a four-vector basis \( \setlr{ \gamma_\mu } \), where the spatial basis
\( \setlr{ \Be_k = \gamma_k \gamma_0 } \)
is expressed in terms of the Dirac basis \( \setlr{ \gamma_\mu } \).
By multiplying from the left with \( \gamma_0 \) a STA form of Maxwell’s equation
\ref{eqn:potentialMethods:320}
is obtained,
where
\begin{equation}\label{eqn:potentialMethods:340}
\begin{aligned}
J &= \gamma^\mu J_\mu = ( v q_e, \boldsymbol{\mathcal{J}} ) \\
M &= \gamma^\mu M_\mu = ( v q_m, \boldsymbol{\mathcal{M}} ) \\
\grad &= \gamma^\mu \partial_\mu = ( (1/v) \partial_t, \spacegrad ) \\
I &= \gamma_0 \gamma_1 \gamma_2 \gamma_3,
\end{aligned}
\end{equation}

Here the metric choice is \( \gamma_0^2 = 1 = -\gamma_k^2 \). Note that in this representation the electromagnetic field \( \boldsymbol{\mathcal{F}} = \boldsymbol{\mathcal{E}} + \eta I \boldsymbol{\mathcal{H}} \) is a bivector, not a multivector as it is explicit (frame dependent) 3D representation of \ref{eqn:potentialMethods:300}.

A potential representation can be obtained as before by considering electric and magnetic sources in sequence and using superposition to assemble a complete potential.

No magnetic sources

Without magnetic sources, Maxwell’s equation splits into vector and trivector terms of the form

\begin{equation}\label{eqn:potentialMethods:380}
\grad \cdot \boldsymbol{\mathcal{F}} = \eta J
\end{equation}
\begin{equation}\label{eqn:potentialMethods:400}
\grad \wedge \boldsymbol{\mathcal{F}} = 0.
\end{equation}

A four-vector curl representation of the field will satisfy \ref{eqn:potentialMethods:400} allowing an immediate potential solution

\begin{equation}\label{eqn:potentialMethods:560}
\boxed{
\begin{aligned}
&\boldsymbol{\mathcal{F}} = \grad \wedge {A^{\mathrm{e}}} \\
&\grad^2 {A^{\mathrm{e}}} – \grad \lr{ \grad \cdot {A^{\mathrm{e}}} } = \eta J.
\end{aligned}
}
\end{equation}

This can be put into correspondence with \ref{eqn:potentialMethods:120} by noting that

\begin{equation}\label{eqn:potentialMethods:460}
\begin{aligned}
\grad^2 &= (\gamma^\mu \partial_\mu) \cdot (\gamma^\nu \partial_\nu) = \inv{v^2} \partial_{tt} – \spacegrad^2 \\
\gamma_0 {A^{\mathrm{e}}} &= \gamma_0 \gamma^\mu {A^{\mathrm{e}}}_\mu = {A^{\mathrm{e}}}_0 + \Be_k {A^{\mathrm{e}}}_k = {A^{\mathrm{e}}}_0 + \BA^{\mathrm{e}} \\
\gamma_0 \grad &= \gamma_0 \gamma^\mu \partial_\mu = \inv{v} \partial_t + \spacegrad \\
\grad \cdot {A^{\mathrm{e}}} &= \partial_\mu {A^{\mathrm{e}}}^\mu = \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}},
\end{aligned}
\end{equation}

so multiplying from the left with \( \gamma_0 \) gives

\begin{equation}\label{eqn:potentialMethods:480}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \lr{ {A^{\mathrm{e}}}_0 + \BA^{\mathrm{e}} } – \lr{ \inv{v} \partial_t + \spacegrad }\lr{ \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}} } = \eta( v q_e – \boldsymbol{\mathcal{J}} ),
\end{equation}

or

\begin{equation}\label{eqn:potentialMethods:520}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \BA^{\mathrm{e}} – \spacegrad \lr{ \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}} } = -\eta \boldsymbol{\mathcal{J}}
\end{equation}
\begin{equation}\label{eqn:potentialMethods:540}
\spacegrad^2 {A^{\mathrm{e}}}_0 – \inv{v} \partial_t \lr{ \spacegrad \cdot \BA^{\mathrm{e}} } = -q_e/\epsilon.
\end{equation}

So \( {A^{\mathrm{e}}}_0 = \phi \) and \( -\ifrac{\BA^{\mathrm{e}}}{v} = \boldsymbol{\mathcal{A}}^{\mathrm{e}} \), or

\begin{equation}\label{eqn:potentialMethods:600}
\boxed{
{A^{\mathrm{e}}} = \gamma_0\lr{ \phi – v \boldsymbol{\mathcal{A}}^{\mathrm{e}} }.
}
\end{equation}

No electric sources

Without electric sources, Maxwell’s equation now splits into

\begin{equation}\label{eqn:potentialMethods:640}
\grad \cdot \boldsymbol{\mathcal{F}} = 0
\end{equation}
\begin{equation}\label{eqn:potentialMethods:660}
\grad \wedge \boldsymbol{\mathcal{F}} = -I M.
\end{equation}

Here the dual of an STA curl yields a solution

\begin{equation}\label{eqn:potentialMethods:680}
\boxed{
\boldsymbol{\mathcal{F}} = I ( \grad \wedge {A^{\mathrm{m}}} ).
}
\end{equation}

Substituting this gives

\begin{equation}\label{eqn:potentialMethods:720}
\begin{aligned}
0
&=
\grad \cdot (I ( \grad \wedge {A^{\mathrm{m}}} ) ) \\
&=
\gpgradeone{ \grad I ( \grad \wedge {A^{\mathrm{m}}} ) } \\
&=
-I \grad \wedge ( \grad \wedge {A^{\mathrm{m}}} ).
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:potentialMethods:740}
\begin{aligned}
-I M
&=
\grad \wedge (I ( \grad \wedge {A^{\mathrm{m}}} ) ) \\
&=
\gpgradethree{ \grad I ( \grad \wedge {A^{\mathrm{m}}} ) } \\
&=
-I \grad \cdot ( \grad \wedge {A^{\mathrm{m}}} ).
\end{aligned}
\end{equation}

The \( \grad \cdot \boldsymbol{\mathcal{F}} \) relation \ref{eqn:potentialMethods:720} is identically zero as desired, leaving

\begin{equation}\label{eqn:potentialMethods:760}
\boxed{
\grad^2 {A^{\mathrm{m}}} – \grad \lr{ \grad \cdot {A^{\mathrm{m}}} }
=
M.
}
\end{equation}

So the general solution with both electric and magnetic sources is

\begin{equation}\label{eqn:potentialMethods:800}
\boxed{
\boldsymbol{\mathcal{F}} = \grad \wedge {A^{\mathrm{e}}} + I (\grad \wedge {A^{\mathrm{m}}}),
}
\end{equation}

subject to the constraints of \ref{eqn:potentialMethods:560} and \ref{eqn:potentialMethods:760}. As before the four-potential \( {A^{\mathrm{m}}} \) can be put into correspondence with the conventional scalar and vector potentials by left multiplying with \( \gamma_0 \), which gives

\begin{equation}\label{eqn:potentialMethods:820}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \lr{ {A^{\mathrm{m}}}_0 + \BA^{\mathrm{m}} } – \lr{ \inv{v} \partial_t + \spacegrad }\lr{ \inv{v} \partial_t {A^{\mathrm{m}}}_0 – \spacegrad \cdot \BA^{\mathrm{m}} } = v q_m – \boldsymbol{\mathcal{M}},
\end{equation}

or
\begin{equation}\label{eqn:potentialMethods:860}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \BA^{\mathrm{m}} – \spacegrad \lr{ \inv{v} \partial_t {A^{\mathrm{m}}}_0 – \spacegrad \cdot \BA^{\mathrm{m}} } = – \boldsymbol{\mathcal{M}}
\end{equation}
\begin{equation}\label{eqn:potentialMethods:880}
\spacegrad^2 {A^{\mathrm{m}}}_0 – \inv{v} \partial_t \spacegrad \cdot \BA^{\mathrm{m}} = -v q_m.
\end{equation}

Comparing with \ref{eqn:potentialMethods:260} shows that \( {A^{\mathrm{m}}}_0/v = \mu \phi_m \) and \( -\ifrac{\BA^{\mathrm{m}}}{v^2} = \mu \boldsymbol{\mathcal{A}}^{\mathrm{m}} \), or

\begin{equation}\label{eqn:potentialMethods:900}
\boxed{
{A^{\mathrm{m}}} = \gamma_0 \eta \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} }.
}
\end{equation}

Potential operator structure

Observe that there is an underlying uniform structure of the differential operator that acts on the potential to produce the electromagnetic field. Expressed as a linear operator of the
gradient and the potentials, that is

\( \boldsymbol{\mathcal{F}} = L(\lrgrad, {A^{\mathrm{e}}}, {A^{\mathrm{m}}}) \)

\begin{equation}\label{eqn:potentialMethods:980}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
L(\grad, {A^{\mathrm{e}}}, {A^{\mathrm{m}}}) \\
&= \grad \wedge {A^{\mathrm{e}}} + I (\grad \wedge {A^{\mathrm{m}}}) \\
&=
\inv{2} \lr{ \rgrad {A^{\mathrm{e}}} – {A^{\mathrm{e}}} \lgrad }
+ \frac{I}{2} \lr{ \rgrad {A^{\mathrm{m}}} – {A^{\mathrm{m}}} \lgrad } \\
&=
\inv{2} \lr{ \rgrad {A^{\mathrm{e}}} – {A^{\mathrm{e}}} \lgrad }
+ \frac{1}{2} \lr{ -\rgrad I {A^{\mathrm{m}}} – I {A^{\mathrm{m}}} \lgrad } \\
&=
\inv{2} \lr{ \rgrad ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}}) – ({A^{\mathrm{e}}} + I {A^{\mathrm{m}}}) \lgrad }
,
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:potentialMethods:1000}
\boxed{
\boldsymbol{\mathcal{F}}
=
\inv{2} \lr{ \rgrad ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}}) – ({A^{\mathrm{e}}} – I {A^{\mathrm{m}}})^\dagger \lgrad }
.
}
\end{equation}

Observe that \ref{eqn:potentialMethods:1000} can be
put into correspondence with \ref{eqn:potentialMethods:1080} using a factoring of unity \( 1 = \gamma_0 \gamma_0 \)

\begin{equation}\label{eqn:potentialMethods:1100}
\boldsymbol{\mathcal{F}}
=
\inv{2} \lr{ (-\rgrad \gamma_0) (-\gamma_0 ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}})) – (({A^{\mathrm{e}}} + I {A^{\mathrm{m}}}) \gamma_0)(\gamma_0 \lgrad) },
\end{equation}

where

\begin{equation}\label{eqn:potentialMethods:1140}
\begin{aligned}
-\grad \gamma_0
&=
-(\gamma^0 \partial_0 + \gamma^k \partial_k) \gamma_0 \\
&=
-\partial_0 – \gamma^k \gamma_0 \partial_k \\
&=
\spacegrad
-\inv{v} \partial_t
,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:potentialMethods:1160}
\begin{aligned}
\gamma_0 \grad
&=
\gamma_0 (\gamma^0 \partial_0 + \gamma^k \partial_k) \\
&=
\partial_0 – \gamma^k \gamma_0 \partial_k \\
&=
\spacegrad
+ \inv{v} \partial_t
,
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:potentialMethods:1200}
\begin{aligned}
-\gamma_0 ( {A^{\mathrm{e}}} – I {A^{\mathrm{m}}} )
&=
-\gamma_0 \gamma_0 \lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \\
&=
-\lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \phi_m – \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}} } \\
&=
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:potentialMethods:1220}
\begin{aligned}
( {A^{\mathrm{e}}} + I {A^{\mathrm{m}}} )\gamma_0
&=
\lr{ \gamma_0 \lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} } + I \gamma_0 \eta \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \gamma_0 \\
&=
\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \phi_m + I \eta v \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
&=
\phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}
+ \eta I \phi_m
,
\end{aligned}
\end{equation}

This recovers \ref{eqn:potentialMethods:1080} as desired.

Potentials in the 3D Euclidean formalism

In the conventional scalar plus vector differential representation of Maxwell’s equations \ref{eqn:chapter3Notes:20}…, given electric(magnetic) sources the structure of the electric(magnetic) potential follows from first setting the magnetic(electric) field equal to the curl of a vector potential. The procedure for the STA GA form of Maxwell’s equation was similar, where it was immediately evident that the field could be set to the four-curl of a four-vector potential (or the dual of such a curl for magnetic sources).

In the 3D GA representation, there is no immediate rationale for introducing a curl or the equivalent to a four-curl representation of the field. Reconciliation of this is possible by recognizing that the fact that the field (or a component of it) may be represented by a curl is not actually fundamental. Instead, observe that the two sided gradient action on a potential to generate the electromagnetic field in the STA representation of \ref{eqn:potentialMethods:1000} serves to select the grade two component product of the gradient and the multivector potential \( {A^{\mathrm{e}}} – I {A^{\mathrm{m}}} \), and that this can in fact be written as
a single sided gradient operation on a potential, provided the multivector product is filtered with a four-bivector grade selection operation

\begin{equation}\label{eqn:potentialMethods:1240}
\boxed{
\boldsymbol{\mathcal{F}} = \gpgradetwo{ \grad \lr{ {A^{\mathrm{e}}} – I {A^{\mathrm{m}}} } }.
}
\end{equation}

Similarly, it can be observed that the
specific function of the conjugate structure in the two sided potential representation of
\ref{eqn:potentialMethods:1080}
is to discard all the scalar and pseudoscalar grades in the multivector product. This means that a single sided potential can also be used, provided it is wrapped in a grade selection operation

\begin{equation}\label{eqn:potentialMethods:1260}
\boxed{
\boldsymbol{\mathcal{F}} =
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} }
\lr{
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
} }{1,2}.
}
\end{equation}

It is this grade selection operation that is really the fundamental defining action in the potential of the STA and conventional 3D representations of Maxwell’s equations. So, given Maxwell’s equation in the 3D GA representation, defining a potential representation for the field is really just a demand that the field have the structure

\begin{equation}\label{eqn:potentialMethods:1320}
\boldsymbol{\mathcal{F}} = \gpgrade{ (\alpha \spacegrad + \beta \partial_t)( A_0 + A_1 + I( A_0′ + A_1′ ) }{1,2}.
\end{equation}

This is a mandate that the electromagnetic field is the grades 1 and 2 components of the vector product of space and time derivative operators on a multivector field \( A = \sum_{k=0}^3 A_k = A_0 + A_1 + I( A_0′ + A_1′ ) \) that can potentially have any grade components. There are more degrees of freedom in this specification than required, since the multivector can absorb one of the \( \alpha \) or \( \beta \) coefficients, so without loss of generality, one of these (say \( \alpha\)) can be set to 1.

Expanding \ref{eqn:potentialMethods:1320} gives

\begin{equation}\label{eqn:potentialMethods:1340}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
\spacegrad A_0
+ \beta \partial_t A_1
– \spacegrad \cross A_1′
+ I (\spacegrad \cross A_1
+ \beta \partial_t A_1′
+ \spacegrad A_0′) \\
&=
\boldsymbol{\mathcal{E}} + I \eta \boldsymbol{\mathcal{H}}.
\end{aligned}
\end{equation}

This naturally has all the right mixes of curls, gradients and time derivatives, all following as direct consequences of applying a grade selection operation to the action of a “spacetime gradient” on a general multivector potential.

The conclusion is that the potential representation of the field is

\begin{equation}\label{eqn:potentialMethods:1360}
\boldsymbol{\mathcal{F}} =
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } A }{1,2},
\end{equation}

where \( A \) is a multivector potentially containing all grades, where grades 0,1 are required for electric sources, and grades 2,3 are required for magnetic sources. When it is desirable to refer back to the conventional scalar and vector potentials this multivector potential can be written as \( A = -\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ -\phi_m + v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } \).

Gauge transformations

Recall that for electric sources the magnetic field is of the form

\begin{equation}\label{eqn:potentialMethods:1380}
\boldsymbol{\mathcal{B}} = \spacegrad \cross \boldsymbol{\mathcal{A}},
\end{equation}

so adding the gradient of any scalar field to the potential \( \boldsymbol{\mathcal{A}}’ = \boldsymbol{\mathcal{A}} + \spacegrad \psi \)
does not change the magnetic field

\begin{equation}\label{eqn:potentialMethods:1400}
\begin{aligned}
\boldsymbol{\mathcal{B}}’
&= \spacegrad \cross \lr{ \boldsymbol{\mathcal{A}} + \spacegrad \psi } \\
&= \spacegrad \cross \boldsymbol{\mathcal{A}} \\
&= \boldsymbol{\mathcal{B}}.
\end{aligned}
\end{equation}

The electric field with this changed potential is

\begin{equation}\label{eqn:potentialMethods:1420}
\begin{aligned}
\boldsymbol{\mathcal{E}}’
&= -\spacegrad \phi – \partial_t \lr{ \BA + \spacegrad \psi} \\
&= -\spacegrad \lr{ \phi + \partial_t \psi } – \partial_t \BA,
\end{aligned}
\end{equation}

so if
\begin{equation}\label{eqn:potentialMethods:1440}
\phi = \phi’ – \partial_t \psi,
\end{equation}

the electric field will also be unaltered by this transformation.

In the STA representation, the field can similarly be altered by adding any (four)gradient to the potential. For example with only electric sources

\begin{equation}\label{eqn:potentialMethods:1460}
\boldsymbol{\mathcal{F}} = \grad \wedge (A + \grad \psi) = \grad \wedge A
\end{equation}

and for electric or magnetic sources

\begin{equation}\label{eqn:potentialMethods:1480}
\boldsymbol{\mathcal{F}} = \gpgradetwo{ \grad (A + \grad \psi) } = \gpgradetwo{ \grad A }.
\end{equation}

In the 3D GA representation, where the field is given by \ref{eqn:potentialMethods:1360}, there is no field that is being curled to add a gradient to. However, if the scalar and vector potentials transform as

\begin{equation}\label{eqn:potentialMethods:1500}
\begin{aligned}
\boldsymbol{\mathcal{A}} &\rightarrow \boldsymbol{\mathcal{A}} + \spacegrad \psi \\
\phi &\rightarrow \phi – \partial_t \psi,
\end{aligned}
\end{equation}

then the multivector potential transforms as
\begin{equation}\label{eqn:potentialMethods:1520}
-\phi + v \boldsymbol{\mathcal{A}}
\rightarrow -\phi + v \boldsymbol{\mathcal{A}} + \partial_t \psi + v \spacegrad \psi,
\end{equation}

so the electromagnetic field is unchanged when the multivector potential is transformed as

\begin{equation}\label{eqn:potentialMethods:1540}
A \rightarrow A + \lr{ \spacegrad + \inv{v} \partial_t } \psi,
\end{equation}

where \( \psi \) is any field that has scalar or pseudoscalar grades. Viewed in terms of grade selection, this makes perfect sense, since the transformed field is

\begin{equation}\label{eqn:potentialMethods:1560}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&\rightarrow
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } \lr{ A + \lr{ \spacegrad + \inv{v} \partial_t } \psi } }{1,2} \\
&=
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } A + \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi }{1,2} \\
&=
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } A }{1,2}.
\end{aligned}
\end{equation}

The \( \psi \) contribution to the grade selection operator is killed because it has scalar or pseudoscalar grades.

Lorenz gauge

Maxwell’s equations are completely decoupled if the potential can be found such that

\begin{equation}\label{eqn:potentialMethods:1580}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } A }{1,2} \\
&=
\lr{ \spacegrad – \inv{v} \PD{t}{} } A.
\end{aligned}
\end{equation}

When this is the case, Maxwell’s equations are reduced to four non-homogeneous potential wave equations

\begin{equation}\label{eqn:potentialMethods:1620}
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } A = J,
\end{equation}

that is

\begin{equation}\label{eqn:potentialMethods:1600}
\begin{aligned}
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \phi &= – \inv{\epsilon} q_e \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \boldsymbol{\mathcal{A}}^{\mathrm{e}} &= – \mu \boldsymbol{\mathcal{J}} \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \phi_m &= – \frac{I}{\mu} q_m \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \boldsymbol{\mathcal{A}}^{\mathrm{m}} &= – I \epsilon \boldsymbol{\mathcal{M}}.
\end{aligned}
\end{equation}

There should be no a-priori assumption that such a field representation has no scalar, nor no pseudoscalar components. That explicit expansion in grades is

\begin{equation}\label{eqn:potentialMethods:1640}
\begin{aligned}
\lr{ \spacegrad – \inv{v} \PD{t}{} } A
&=
\lr{ \spacegrad – \inv{v} \PD{t}{} } \lr{ -\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ -\phi_m + v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \\
&=
\inv{v} \partial_t \phi
+ v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
&-\spacegrad \phi
+ I \eta v \spacegrad \wedge \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \partial_t \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
&+ v \spacegrad \wedge \boldsymbol{\mathcal{A}}^{\mathrm{e}}
– \eta I \spacegrad \phi_m
– I \eta \partial_t \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
&+ \eta I \inv{v} \partial_t \phi_m
+ I \eta v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}},
\end{aligned}
\end{equation}

so if this potential representation has only vector and bivector grades, it must be true that

\begin{equation}\label{eqn:potentialMethods:1660}
\begin{aligned}
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} &= 0 \\
\inv{v} \partial_t \phi_m + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} &= 0.
\end{aligned}
\end{equation}

The first is the well known Lorenz gauge condition, whereas the second is the dual of that condition for magnetic sources.

Should one of these conditions, say the Lorenz condition for the electric source potentials, be non-zero, then it is possible to make a potential transformation for which this condition is zero

\begin{equation}\label{eqn:potentialMethods:1680}
\begin{aligned}
0
&\ne
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
&=
\inv{v} \partial_t (\phi’ – \partial_t \psi) + v \spacegrad \cdot (\boldsymbol{\mathcal{A}}’ + \spacegrad \psi) \\
&=
\inv{v} \partial_t \phi’ + v \spacegrad \boldsymbol{\mathcal{A}}’
+ v \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi,
\end{aligned}
\end{equation}

so if \( \inv{v} \partial_t \phi’ + v \spacegrad \boldsymbol{\mathcal{A}}’ \) is zero, \( \psi \) must be found such that
\begin{equation}\label{eqn:potentialMethods:1700}
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}}
= v \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi.
\end{equation}

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.

Vector wave equation in spherical coordinates

November 10, 2016 math and physics play , , ,

[Click here for a PDF of this post with nicer formatting]

For a vector \( \BA \) in spherical coordinates, let’s compute the Laplacian

\begin{equation}\label{eqn:vectorWaveEquationSpherical:20}
\spacegrad^2 \BA,
\end{equation}

to see the form of the wave equation. The spherical vector representation has a curvilinear basis
\begin{equation}\label{eqn:vectorWaveEquationSpherical:40}
\BA = \rcap A_r + \thetacap A_\theta + \phicap A_\phi,
\end{equation}

and the spherical Laplacian has been found to have the representation

\begin{equation}\label{eqn:vectorWaveEquationSpherical:60}
\spacegrad^2 \psi
=
\inv{r^2} \PD{r}{} \lr{ r^2 \PD{r}{ \psi} }
+ \frac{1}{r^2 \sin\theta} \PD{\theta}{} \lr{ \sin\theta \PD{\theta}{ \psi } }
+ \frac{1}{r^2 \sin^2\theta} \PDSq{\phi}{ \psi}.
\end{equation}

Evaluating the Laplacian will require the following curvilinear basis derivatives

\begin{equation}\label{eqn:vectorWaveEquationSpherical:80}
\begin{aligned}
\partial_\theta \rcap &= \thetacap \\
\partial_\theta \thetacap &= -\rcap \\
\partial_\theta \phicap &= 0 \\
\partial_\phi \rcap &= S_\theta \phicap \\
\partial_\phi \thetacap &= C_\theta \phicap \\
\partial_\phi \phicap &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}
\end{equation}

We’ll need to evaluate a number of derivatives. Starting with the \( \rcap \) components

\begin{equation}\label{eqn:vectorWaveEquationSpherical:120}
\partial_r \lr{ r^2 \partial_r \lr{ \rcap \psi} }
=
\rcap \partial_r \lr{ r^2 \partial_r \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:140}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \rcap \psi } }
&=
\partial_\theta \lr{ S_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) } \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \psi + \rcap \partial_\theta \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta \partial_\theta ((\partial_\theta \thetacap) \psi + (\partial_\theta \rcap) \partial_\theta \psi )
+ S_\theta \partial_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
C_\theta (\thetacap \psi + \rcap \partial_\theta \psi )
+ S_\theta ((-\rcap) \psi + (\thetacap) \partial_\theta \psi )
+ S_\theta (\thetacap \partial_\theta \psi + \rcap \partial_{\theta \theta} \psi ) \\
&=
\rcap \lr{
C_\theta \partial_\theta \psi
– S_\theta \psi
+ S_\theta \partial_{\theta \theta} \psi
}
+\thetacap \lr{
C_\theta \psi
+ 2 S_\theta \partial_\theta \psi
}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:160}
\begin{aligned}
\partial_{\phi \phi} \lr{ \rcap \psi}
&=
\partial_\phi \lr{ (\partial_\phi \rcap) \psi + \rcap \partial_\phi \psi } \\
&=
\partial_\phi \lr{ (S_\theta \phicap) \psi + \rcap \partial_\phi \psi } \\
&=
S_\theta \partial_\phi (\phicap \psi)
+ \partial_\phi \lr{ \rcap \partial_\phi \psi } \\
&=
S_\theta (\partial_\phi \phicap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (\partial_\phi \rcap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
S_\theta (-S_\theta \rcap – C_\theta \thetacap) \psi
+ S_\theta \phicap \partial_\phi \psi
+ (S_\theta \phicap) \partial_\phi \psi
+ \rcap \partial_{\phi\phi} \psi \\
&=
\rcap \lr{
– S_\theta^2 \psi
+ \partial_{\phi\phi} \psi
}
+
\thetacap \lr{
– S_\theta C_\theta \psi
}
+
\phicap \lr{
2 S_\theta \phicap \partial_\phi \psi
}
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:vectorWaveEquationSpherical:180}
\begin{aligned}
\spacegrad^2 (\rcap A_r)
&=
\rcap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_r }
+
\inv{r^2 S_\theta}
\lr{
C_\theta \partial_\theta A_r
– S_\theta A_r
+ S_\theta \partial_{\theta \theta} A_r
}
+ \inv{r^2 S_\theta^2}
\lr{
– S_\theta^2 A_r
+ \partial_{\phi\phi} A_r
}
} \\
&\quad +
\thetacap
\lr{
\inv{r^2 S_\theta}
\lr{
C_\theta A_r
+ 2 S_\theta \partial_\theta A_r
}

\inv{r^2 S_\theta}
S_\theta C_\theta A_r
} \\
&\quad +
\phicap
\lr{
\inv{r^2 S_\theta^2}
2 S_\theta \partial_\phi A_r
} \\
&=
\rcap \lr{
\spacegrad^2 A_r
-\frac{2}{r^2 } A_r
}
+
\frac{\thetacap}{r^2}
\lr{
\frac{C_\theta}{S_\theta} A_r
+ 2 \partial_\theta A_r
– C_\theta A_r
}
+
\phicap
\frac{2}{r^2 S_\theta} \partial_\phi A_r.
\end{aligned}
\end{equation}

Next, let’s compute the derivatives of the \( \thetacap \) projection.

\begin{equation}\label{eqn:vectorWaveEquationSpherical:220}
\partial_r \lr{ r^2 \partial_r \lr{ \thetacap \psi} }
=
\thetacap \partial_r \lr{ r^2 \partial_r \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:240}
\begin{aligned}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \thetacap \psi } }
&=
\partial_\theta \lr{ S_\theta
\lr{
(\partial_\theta \thetacap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
\partial_\theta
\lr{ S_\theta
\lr{
(-\rcap ) \psi
+\thetacap \partial_\theta \psi
}
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\partial_\theta \rcap) \psi
-\rcap \partial_\theta \psi
+(\partial_\theta \thetacap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
C_\theta \lr{
-\rcap \psi
+\thetacap \partial_\theta \psi
}
+
S_\theta
\lr{
-(\thetacap) \psi
-\rcap \partial_\theta \psi
+(-\rcap) \partial_\theta \psi
+\thetacap \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+C_\theta \partial_\theta \psi
-S_\theta \psi
+S_\theta \partial_{\theta \theta} \psi
} \\
&=
\rcap \lr{
-C_\theta \psi
-2 S_\theta \partial_\theta \psi
}
+
\thetacap \lr{
+\partial_\theta (S_\theta \partial_\theta \psi)
-S_\theta \psi
}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:260}
\begin{aligned}
\partial_{\phi \phi} \lr{ \thetacap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \thetacap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(C_\theta \phicap) \psi
+\thetacap \partial_\phi \psi
} \\
&=
C_\theta \partial_{\phi} (\phicap \psi)
+
\partial_{\phi} ( \thetacap \partial_\phi \psi ) \\
&=
C_\theta (\partial_\phi \phicap) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (\partial_\phi \thetacap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
C_\theta (-\rcap S_\theta – \thetacap C_\theta) \psi
+C_\theta \phicap \partial_{\phi} \psi
+ (C_\theta \phicap) \partial_\phi \psi
+\thetacap \partial_{\phi\phi} \psi \\
&=
-\rcap C_\theta S_\theta \psi
+\thetacap \lr{
-C_\theta C_\theta \psi
+\partial_{\phi\phi} \psi
}
+2 \phicap C_\theta \partial_\phi \psi,
\end{aligned}
\end{equation}

which gives
\begin{equation}\label{eqn:vectorWaveEquationSpherical:360}
\begin{aligned}
\spacegrad^2 (\thetacap A_\theta)
&=
\rcap
\lr{
\inv{r^2 S_\theta}
\lr{
-C_\theta A_\theta
-2 S_\theta \partial_\theta A_\theta
}

\inv{r^2 S_\theta^2}
C_\theta S_\theta A_\theta
} \\
&\quad +
\thetacap \lr{
\inv{r^2} \partial_r \lr{ r^2 \partial_r A_\theta }
+
\inv{r^2 S_\theta}
\lr{
+\partial_\theta (S_\theta \partial_\theta A_\theta)
-S_\theta A_\theta
}
+\inv{r^2 S_\theta^2}
\lr{
-C_\theta C_\theta A_\theta
+\partial_{\phi\phi} A_\theta
}
} \\
&\quad +
\phicap \lr{
\inv{r^2 S_\theta^2}
2 C_\theta \partial_\phi A_\theta
} \\
&=
-2 \rcap
\inv{r^2 S_\theta}
\partial_\theta (S_\theta A_\theta)
+
\thetacap \lr{
\spacegrad^2 A_\theta
-\inv{r^2}
A_\theta
-\inv{r^2 S_\theta^2} C_\theta^2 A_\theta
}
+
2 \phicap \lr{
\inv{r^2 S_\theta^2}
C_\theta \partial_\phi A_\theta
}.
\end{aligned}
\end{equation}

Finally, we can compute the derivatives of the \( \phicap \) projection.

\begin{equation}\label{eqn:vectorWaveEquationSpherical:300}
\partial_r \lr{ r^2 \partial_r \lr{ \phicap \psi} }
=
\phicap \partial_r \lr{ r^2 \partial_r \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:320}
\partial_\theta \lr{ S_\theta \partial_\theta \lr{ \phicap \psi } }
=
\phicap \partial_\theta \lr{ S_\theta \partial_\theta \psi }
\end{equation}

\begin{equation}\label{eqn:vectorWaveEquationSpherical:340}
\begin{aligned}
\partial_{\phi \phi} \lr{ \phicap \psi}
&=
\partial_{\phi} \lr{
(\partial_\phi \phicap) \psi
+\phicap \partial_\phi \psi
} \\
&=
\partial_{\phi} \lr{
(-\rcap S_\theta – \thetacap C_\theta) \psi
+\phicap \partial_\phi \psi
} \\
&=
-((\partial_\phi \rcap) S_\theta + (\partial_\phi \thetacap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(\partial_\phi \phicap \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
-((S_\theta \phicap) S_\theta + (C_\theta \phicap) C_\theta) \psi
-(\rcap S_\theta + \thetacap C_\theta) \partial_\phi \psi
+(-\rcap S_\theta – \thetacap C_\theta) \partial_\phi \psi
+\phicap \partial_{\phi \phi} \psi \\
&=
– 2 \rcap S_\theta \partial_\phi \psi
– 2 \thetacap C_\theta \partial_\phi \psi
+ \phicap \lr{
\partial_{\phi \phi} \psi
-\psi
},
\end{aligned}
\end{equation}

which gives
\begin{equation}\label{eqn:vectorWaveEquationSpherical:380}
\begin{aligned}
\spacegrad^2 \lr{ \phicap A_\phi }
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
&\quad +
\phicap \lr{
\inv{r^2}
\partial_r \lr{ r^2 \partial_r A_\phi }
+
\inv{r^2 S_\theta}
\partial_\theta \lr{ S_\theta \partial_\theta A_\phi }
+
\inv{r^2 S_\theta^2}
\lr{
\partial_{\phi \phi} A_\phi -A_\phi
}
} \\
&=
-2 \rcap \inv{r^2 S_\theta} \partial_\phi A_\phi
-2 \thetacap \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi
+
\phicap \lr{
\spacegrad^2 A_\phi – \inv{r^2} A_\phi
}.
\end{aligned}
\end{equation}

The vector Laplacian resolves into three augmented scalar wave equations, all highly coupled

\begin{equation}\label{eqn:vectorWaveEquationSpherical:420}
\boxed{
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
\spacegrad^2 A_r
-\frac{2}{r^2 } A_r
– \frac{2}{r^2 S_\theta} \partial_\theta (S_\theta A_\theta)
– \frac{2}{r^2 S_\theta} \partial_\phi A_\phi \\
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{1}{r^2} \frac{C_\theta}{S_\theta} A_r
+ \frac{2}{r^2} \partial_\theta A_r
– \frac{1}{r^2} C_\theta A_r
+ \spacegrad^2 A_\theta
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-2 \inv{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
\frac{2}{r^2 S_\theta} \partial_\phi A_r
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi.
\end{aligned}
}
\end{equation}

I’d guess one way to decouple these equations would be to impose a constraint that allows all the non-wave equation terms in one of the component equations to be killed, and then substitute that constraint into the remaining equations. Let’s try one such constraint

\begin{equation}\label{eqn:vectorWaveEquationSpherical:480}
A_r
=
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi.
\end{equation}

This gives

\begin{equation}\label{eqn:vectorWaveEquationSpherical:520}
\begin{aligned}
\rcap \cdot \lr{ \spacegrad^2 \BA }
&=
\spacegrad^2 A_r \\
\thetacap \cdot \lr{ \spacegrad^2 \BA }
&=
\lr{
\frac{1}{r^2} \frac{C_\theta}{S_\theta}
+ \frac{2}{r^2} \partial_\theta
– \frac{1}{r^2} C_\theta
}
\lr{
– \inv{S_\theta} \partial_\theta (S_\theta A_\theta)
– \inv{S_\theta} \partial_\phi A_\phi
} \\
&\quad+ \spacegrad^2 A_\theta
– \inv{r^2} A_\theta
– \inv{r^2 S_\theta^2} C_\theta^2 A_\theta
-\frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\phi \\
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
– \frac{2}{r^2 S_\theta} \partial_\phi
\lr{
\inv{S_\theta} \partial_\theta (S_\theta A_\theta)
+ \inv{S_\theta} \partial_\phi A_\phi
}
+ \frac{2}{r^2 S_\theta^2} C_\theta \partial_\phi A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi \\
&=
-\frac{2}{r^2 S_\theta} \partial_\theta A_\theta
-\frac{2}{r^2 S_\theta^2} \partial_{\phi\phi} A_\theta
+ \spacegrad^2 A_\phi – \inv{r^2} A_\phi
\end{aligned}
\end{equation}

It looks like some additional cancellations may be had in the \( \thetacap \) projection of this constrained vector Laplacian. I’m not inclined to try to take this reduction any further without a thorough check of all the algebra (using Mathematica to do so would make sense).

I also guessing that such a solution might be how the \( \textrm{TE}^r \) and \( \textrm{TM}^r \) modes were defined, but that doesn’t appear to be the case according to [1]. There the wave equation is formulated in terms of the vector potentials (picking one to be zero and the other to be radial only). The solution obtained from such a potential wave equation then directly defines the \( \textrm{TE}^r \) and \( \textrm{TM}^r \) modes. It would be interesting to see how the modes derived in that analysis transform with application of the vector Laplacian derived above.

References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

Spherical gradient, divergence, curl and Laplacian

November 9, 2016 math and physics play , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

Unit vectors

Two of the spherical unit vectors we can immediately write by inspection.

\begin{equation}\label{eqn:sphericalLaplacian:20}
\begin{aligned}
\rcap &= \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta \\
\phicap &= -\Be_1 \sin\theta + \Be_2 \cos\phi
\end{aligned}
\end{equation}

We can compute \( \thetacap \) by utilizing the right hand triplet property

\begin{equation}\label{eqn:sphericalLaplacian:40}
\begin{aligned}
\thetacap
&=
\phicap \cross \rcap \\
&=
\begin{vmatrix}
\Be_1 & \Be_2 & \Be_3 \\
-S_\phi & C_\phi & 0 \\
S_\theta C_\phi & S_\theta S_\phi & C_\theta \\
\end{vmatrix} \\
&=
\Be_1 \lr{ C_\theta C_\phi }
+\Be_2 \lr{ C_\theta S_\phi }
+\Be_3 \lr{ -S_\theta \lr{ S_\phi^2 + C_\phi^2 } } \\
&=
\Be_1 \cos\theta \cos\phi
+\Be_2 \cos\theta \sin\phi
-\Be_3 \sin\theta.
\end{aligned}
\end{equation}

Here I’ve used \( C_\theta = \cos\theta, S_\phi = \sin\phi, \cdots \) as a convenient shorthand. Observe that with \( i = \Be_1 \Be_2 \), these unit vectors admit a small factorization that makes further manipulation easier

\begin{equation}\label{eqn:sphericalLaplacian:80}
\boxed{
\begin{aligned}
\rcap &= \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta \\
\thetacap &= \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 \\
\phicap &= \Be_2 e^{i\phi}
\end{aligned}
}
\end{equation}

It should also be the case that \( \rcap \thetacap \phicap = I \), where \( I = \Be_1 \Be_2 \Be_3 = \Be_{123}\) is the \R{3} pseudoscalar, which is straightforward to check

\begin{equation}\label{eqn:sphericalLaplacian:60}
\begin{aligned}
\rcap \thetacap \phicap
&=
\lr{ \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta }
\lr{ \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 }
\Be_2 e^{i\phi} \\
&=
\lr{ \sin\theta \cos\theta – \cos\theta \sin\theta + \Be_{31} e^{i\phi} \lr{ \cos^2\theta + \sin^2\theta } }
\Be_2 e^{i\phi} \\
&=
\Be_{31} \Be_2 e^{-i\phi} e^{i\phi} \\
&=
\Be_{123}.
\end{aligned}
\end{equation}

This property could also have been used to compute \(\thetacap\).

Gradient

To compute the gradient, note that the coordinate vectors for the spherical parameterization are
\begin{equation}\label{eqn:sphericalLaplacian:120}
\begin{aligned}
\Bx_r
&= \PD{r}{\Br} \\
&= \PD{r}{\lr{r \rcap}} \\
&= \rcap + r \PD{r}{\rcap} \\
&= \rcap,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:140}
\begin{aligned}
\Bx_\theta
&= \PD{\theta}{\lr{r \rcap} } \\
&= r \PD{\theta}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r \PD{\theta}{} \lr{ C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3 } \\
&= r \thetacap,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:160}
\begin{aligned}
\Bx_\phi
&= \PD{\phi}{\lr{r \rcap} } \\
&= r \PD{\phi}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r S_\theta \Be_2 e^{i\phi} \\
&= r \sin\theta \phicap.
\end{aligned}
\end{equation}

Since these are all normal, the dual vectors defined by \( \Bx^j \cdot \Bx_k = \delta^j_k \), can be obtained by inspection
\begin{equation}\label{eqn:sphericalLaplacian:180}
\begin{aligned}
\Bx^r &= \rcap \\
\Bx^\theta &= \inv{r} \thetacap \\
\Bx^\phi &= \inv{r \sin\theta} \phicap.
\end{aligned}
\end{equation}

The gradient follows immediately
\begin{equation}\label{eqn:sphericalLaplacian:200}
\spacegrad =
\Bx^r \PD{r}{} +
\Bx^\theta \PD{\theta}{} +
\Bx^\phi \PD{\phicap}{},
\end{equation}

or
\begin{equation}\label{eqn:sphericalLaplacian:240}
\boxed{
\spacegrad
=
\rcap \PD{r}{} +
\frac{\thetacap}{r} \PD{\theta}{} +
\frac{\phicap}{r\sin\theta} \PD{\phicap}{}.
}
\end{equation}

More information on this general dual-vector technique of computing the gradient in curvilinear coordinate systems can be found in
[2].

Partials

To compute the divergence, curl and Laplacian, we’ll need the partials of each of the unit vectors \( \PDi{\theta}{\rcap}, \PDi{\phi}{\rcap}, \PDi{\theta}{\thetacap}, \PDi{\phi}{\thetacap}, \PDi{\phi}{\phicap} \).

The \( \thetacap \) partials are

\begin{equation}\label{eqn:sphericalLaplacian:260}
\begin{aligned}
\PD{\theta}{\thetacap}
&=
\PD{\theta}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
-S_\theta \Be_1 e^{i\phi} – C_\theta \Be_3 \\
&=
-\rcap,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:280}
\begin{aligned}
\PD{\phi}{\thetacap}
&=
\PD{\phi}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
C_\theta \Be_2 e^{i\phi} \\
&=
C_\theta \phicap.
\end{aligned}
\end{equation}

The \( \phicap \) partials are

\begin{equation}\label{eqn:sphericalLaplacian:300}
\begin{aligned}
\PD{\theta}{\phicap}
&=
\PD{\theta}{} \Be_2 e^{i\phi} \\
&=
0.
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:320}
\begin{aligned}
\PD{\phi}{\phicap}
&=
\PD{\phi}{} \Be_2 e^{i \phi} \\
&=
-\Be_1 e^{i \phi} \\
&=
-\rcap \gpgradezero{ \rcap \Be_1 e^{i \phi} }
– \thetacap \gpgradezero{ \thetacap \Be_1 e^{i \phi} }
– \phicap \gpgradezero{ \phicap \Be_1 e^{i \phi} } \\
&=
-\rcap \gpgradezero{ \lr{
\Be_1 e^{i\phi} S_\theta + \Be_3 C_\theta
} \Be_1 e^{i \phi} }
– \thetacap \gpgradezero{ \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \Be_1 e^{i \phi} } \\
&=
-\rcap \gpgradezero{ e^{-i\phi} S_\theta e^{i \phi} }
– \thetacap \gpgradezero{ C_\theta e^{-i\phi} e^{i \phi} } \\
&=
-\rcap S_\theta
– \thetacap C_\theta.
\end{aligned}
\end{equation}

The \( \rcap \) partials are were computed as a side effect of evaluating \( \Bx_\theta \), and \( \Bx_\phi \), and are

\begin{equation}\label{eqn:sphericalLaplacian:340}
\PD{\theta}{\rcap}
=
\thetacap,
\end{equation}
\begin{equation}\label{eqn:sphericalLaplacian:360}
\PD{\phi}{\rcap}
=
S_\theta \phicap.
\end{equation}

In summary
\begin{equation}\label{eqn:sphericalLaplacian:380}
\boxed{
\begin{aligned}
\partial_{\theta}{\rcap} &= \thetacap \\
\partial_{\phi}{\rcap} &= S_\theta \phicap \\
\partial_{\theta}{\thetacap} &= -\rcap \\
\partial_{\phi}{\thetacap} &= C_\theta \phicap \\
\partial_{\theta}{\phicap} &= 0 \\
\partial_{\phi}{\phicap} &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}
}
\end{equation}

Divergence and curl.

The divergence and curl can be computed from the vector product of the spherical coordinate gradient and the spherical representation of a vector. That is

\begin{equation}\label{eqn:sphericalLaplacian:400}
\spacegrad \BA
= \spacegrad \cdot \BA + \spacegrad \wedge \BA
= \spacegrad \cdot \BA + I \spacegrad \cross \BA.
\end{equation}

That gradient vector product is

\begin{equation}\label{eqn:sphericalLaplacian:420}
\begin{aligned}
\spacegrad \BA
&=
\lr{
\rcap \partial_{r}
+ \frac{\thetacap}{r} \partial_{\theta}
+ \frac{\phicap}{rS_\theta} \partial_{\phi}
}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\rcap \partial_{r}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\thetacap}{r} \partial_{\theta}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\phicap}{rS_\theta} \partial_{\phicap}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\partial_\theta \rcap) A_r + \thetacap (\partial_\theta \thetacap) A_\theta + \thetacap (\partial_\theta \phicap) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{rS_\theta}
\lr{
\phicap (\partial_\phi \rcap) A_r + \phicap (\partial_\phi \thetacap) A_\theta + \phicap (\partial_\phi \phicap) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\thetacap) A_r + \thetacap (-\rcap) A_\theta + \thetacap (0) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{r S_\theta}
\lr{
\phicap (S_\theta \phicap) A_r + \phicap (C_\theta \phicap) A_\theta – \phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
}.
\end{aligned}
\end{equation}

The scalar component of this is the divergence
\begin{equation}\label{eqn:sphericalLaplacian:440}
\begin{aligned}
\spacegrad \cdot \BA
&=
\partial_r A_r
+ \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
\lr{ S_\theta A_r + C_\theta A_\theta + \partial_\phi A_\phi
} \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi,
\end{aligned}
\end{equation}

which can be factored as
\begin{equation}\label{eqn:sphericalLaplacian:460}
\boxed{
\spacegrad \cdot \BA
=
\inv{r^2} \partial_r (r^2 A_r)
+ \inv{r S_\theta} \partial_\theta (S_\theta A_\theta)
+ \frac{1}{r S_\theta} \partial_\phi A_\phi.
}
\end{equation}

The bivector grade of \( \spacegrad \BA \) is the bivector curl
\begin{equation}\label{eqn:sphericalLaplacian:480}
\begin{aligned}
\spacegrad \wedge \BA
&=
\lr{
\rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi
} \\
&\quad + \frac{1}{r}
\lr{
\thetacap (-\rcap) A_\theta
+\thetacap \rcap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
&\quad +
\frac{1}{r S_\theta}
\lr{
-\phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta
} \\
&=
\lr{
\rcap \thetacap \partial_r A_\theta – \phicap \rcap \partial_r A_\phi
} \\
&\quad + \frac{1}{r}
\lr{
\rcap \thetacap A_\theta
-\rcap \thetacap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
&\quad +
\frac{1}{r S_\theta}
\lr{
-\phicap \rcap S_\theta A_\phi + \thetacap \phicap C_\theta A_\phi
+\phicap \rcap \partial_\phi A_r – \thetacap \phicap \partial_\phi A_\theta
} \\
&=
\thetacap \phicap \lr{
\inv{r S_\theta} C_\theta A_\phi
+\frac{1}{r} \partial_\theta A_\phi
-\frac{1}{r S_\theta} \partial_\phi A_\theta
} \\
&\quad +\phicap \rcap \lr{
-\partial_r A_\phi
+
\frac{1}{r S_\theta}
\lr{
-S_\theta A_\phi
+ \partial_\phi A_r
}
} \\
&\quad +\rcap \thetacap \lr{
\partial_r A_\theta
+ \frac{1}{r} A_\theta
– \inv{r} \partial_\theta A_r
} \\
&=
I
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ I \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ I \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}
\end{aligned}
\end{equation}

This gives
\begin{equation}\label{eqn:sphericalLaplacian:500}
\boxed{
\spacegrad \cross \BA
=
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}.
}
\end{equation}

This and the divergence result above both check against the back cover of [1].

Laplacian

Using the divergence and curl it’s possible to compute the Laplacian from those, but we saw in cylindrical coordinates that it was much harder to do it that way than to do it directly.

\begin{equation}\label{eqn:sphericalLaplacian:540}
\begin{aligned}
\spacegrad^2 \psi
&=
\lr{
\rcap \partial_{r} +
\frac{\thetacap}{r} \partial_{\theta} +
\frac{\phicap}{r S_\theta} \partial_{\phi}
}
\lr{
\rcap \partial_{r} \psi
+ \frac{\thetacap}{r} \partial_{\theta} \psi
+ \frac{\phicap}{r S_\theta} \partial_{\phi} \psi
} \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap}{r} \partial_{\theta} \lr{ \rcap \partial_{r} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \frac{\phicap}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap}{r S_\theta} \partial_{\phi} \lr{ \rcap \partial_{r} \psi }
+ \frac{\phicap}{r^2 S_\theta} \partial_{\phi} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\phicap}{r^2 S_\theta^2} \partial_{\phi} \lr{ \phicap \partial_{\phi} \psi } \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\partial_\theta \rcap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \thetacap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \phicap) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (\partial_\phi \rcap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (\partial_\phi \thetacap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (\partial_\phi \phicap) \partial_{\phi} \psi \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\thetacap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (-\rcap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (0) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (S_\theta \phicap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (C_\theta \phicap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (-\rcap S_\theta – \thetacap C_\theta) \partial_{\phi} \psi
\end{aligned}
\end{equation}

All the bivector factors are expected to cancel out, but this should be checked. Those with an \( \rcap \thetacap \) factor are

\begin{equation}\label{eqn:sphericalLaplacian:560}
\partial_r \lr{ \inv{r} \partial_\theta \psi}
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
=
-\inv{r^2} \partial_\theta \psi
+\inv{r} \partial_{r \theta} \psi
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
= 0,
\end{equation}

and those with a \( \thetacap \phicap \) factor are
\begin{equation}\label{eqn:sphericalLaplacian:580}
\frac{1}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi }
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
=
– \frac{1}{r^2} \frac{C_\theta}{S_\theta^2} \partial_{\phi} \psi
+ \frac{1}{r^2 S_\theta} \partial_{\theta \phi} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
= 0,
\end{equation}

and those with a \( \phicap \rcap \) factor are
\begin{equation}\label{eqn:sphericalLaplacian:600}
– \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi }
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta^2} S_\theta \partial_{\phi} \psi
=
\inv{S_\theta} \frac{1}{r^2} \partial_\phi \psi
– \inv{r S_\theta} \partial_{r \phi} \psi
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi} \psi
= 0.
\end{equation}

This leaves
\begin{equation}\label{eqn:sphericalLaplacian:620}
\spacegrad^2 \psi
=
\partial_{rr} \psi
+ \frac{2}{r} \partial_{r} \psi
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{1}{r^2 S_\theta} C_\theta \partial_{\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi.
\end{equation}

This factors nicely as

\begin{equation}\label{eqn:sphericalLaplacian:640}
\boxed{
\spacegrad^2 \psi
=
\inv{r^2} \PD{r}{} \lr{ r^2 \PD{r}{ \psi} }
+ \frac{1}{r^2 \sin\theta} \PD{\theta}{} \lr{ \sin\theta \PD{\theta}{ \psi } }
+ \frac{1}{r^2 \sin\theta^2} \PDSq{\phi}{ \psi}
,
}
\end{equation}

which checks against the back cover of Jackson. Here it has been demonstrated explicitly that this operator expression is valid for multivector fields \( \psi \) as well as scalar fields \( \psi \).

References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

Gradient, divergence, curl and Laplacian in cylindrical coordinates

November 6, 2016 math and physics play , , , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

In class it was suggested that the identity

\begin{equation}\label{eqn:laplacianCylindrical:20}
\spacegrad^2 \BA =
\spacegrad \lr{ \spacegrad \cdot \BA }
-\spacegrad \cross \lr{ \spacegrad \cross \BA },
\end{equation}

can be used to compute the Laplacian in non-rectangular coordinates. Is that the easiest way to do this?

How about just sequential applications of the gradient on the vector? Let’s start with the vector product of the gradient and the vector. First recall that the cylindrical representation of the gradient is

\begin{equation}\label{eqn:laplacianCylindrical:80}
\spacegrad = \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z,
\end{equation}

where
\begin{equation}\label{eqn:laplacianCylindrical:100}
\begin{aligned}
\rhocap &= \Be_1 e^{\Be_1 \Be_2 \phi} \\
\phicap &= \Be_2 e^{\Be_1 \Be_2 \phi} \\
\end{aligned}
\end{equation}

Taking \( \phi \) derivatives of \ref{eqn:laplacianCylindrical:100}, we have

\begin{equation}\label{eqn:laplacianCylindrical:120}
\begin{aligned}
\partial_\phi \rhocap &= \Be_1 \Be_1 \Be_2 e^{\Be_1 \Be_2 \phi} = \Be_2 e^{\Be_1 \Be_2 \phi} = \phicap \\
\partial_\phi \phicap &= \Be_2 \Be_1 \Be_2 e^{\Be_1 \Be_2 \phi} = -\Be_1 e^{\Be_1 \Be_2 \phi} = -\rhocap.
\end{aligned}
\end{equation}

The gradient of a vector \( \BA = \rhocap A_\rho + \phicap A_\phi + \zcap A_z \) is

\begin{equation}\label{eqn:laplacianCylindrical:60}
\begin{aligned}
\spacegrad \BA
&=
\lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z }
\lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&=
\quad \rhocap \partial_\rho \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&\quad + \frac{\phicap}{\rho} \partial_\phi \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&\quad + \zcap \partial_z \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z } \\
&=
\quad \rhocap \lr{ \rhocap \partial_\rho A_\rho + \phicap \partial_\rho A_\phi + \zcap \partial_\rho A_z } \\
&\quad + \frac{\phicap}{\rho} \lr{ \partial_\phi(\rhocap A_\rho) + \partial_\phi(\phicap A_\phi) + \zcap \partial_\phi A_z } \\
&\quad + \zcap \lr{ \rhocap \partial_z A_\rho + \phicap \partial_z A_\phi + \zcap \partial_z A_z } \\
&=
\quad \partial_\rho A_\rho + \rhocap \phicap \partial_\rho A_\phi + \rhocap \zcap \partial_\rho A_z \\
&\quad +\frac{1}{\rho} \lr{ A_\rho + \phicap \rhocap \partial_\phi A_\rho – \phicap \rhocap A_\phi + \partial_\phi A_\phi + \phicap \zcap \partial_\phi A_z } \\
&\quad + \zcap \rhocap \partial_z A_\rho + \zcap \phicap \partial_z A_\phi + \partial_z A_z \\
&=
\quad \partial_\rho A_\rho + \frac{1}{\rho} \lr{ A_\rho + \partial_\phi A_\phi } + \partial_z A_z \\
&\quad +
\zcap \rhocap \lr{
\partial_z A_\rho
-\partial_\rho A_z
} \\
&\quad +
\phicap \zcap \lr{
\inv{\rho} \partial_\phi A_z
– \partial_z A_\phi
} \\
&\quad +
\rhocap \phicap \lr{
\partial_\rho A_\phi
– \inv{\rho} \lr{ \partial_\phi A_\rho – A_\phi }
},
\end{aligned}
\end{equation}

As expected, we see that the gradient splits nicely into a dot and curl

\begin{equation}\label{eqn:laplacianCylindrical:160}
\begin{aligned}
\spacegrad \BA
&= \spacegrad \cdot \BA + \spacegrad \wedge \BA \\
&= \spacegrad \cdot \BA + \rhocap \phicap \zcap (\spacegrad \cross \BA ),
\end{aligned}
\end{equation}

where the cylindrical representation of the divergence is seen to be

\begin{equation}\label{eqn:laplacianCylindrical:140}
\spacegrad \cdot \BA
=
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z,
\end{equation}

and the cylindrical representation of the curl is

\begin{equation}\label{eqn:laplacianCylindrical:180}
\spacegrad \cross \BA
=
\rhocap
\lr{
\inv{\rho} \partial_\phi A_z
– \partial_z A_\phi
}
+
\phicap
\lr{
\partial_z A_\rho
-\partial_\rho A_z
}
+
\inv{\rho} \zcap \lr{
\partial_\rho ( \rho A_\phi )
– \partial_\phi A_\rho
}.
\end{equation}

Should we want to, it is now possible to evaluate the Laplacian of \( \BA \) using
\ref{eqn:laplacianCylindrical:20}
, which will have the following components

\begin{equation}\label{eqn:laplacianCylindrical:220}
\begin{aligned}
\rhocap \cdot \lr{ \spacegrad^2 \BA }
&=
\partial_\rho
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}

\lr{
\inv{\rho} \partial_\phi \lr{
\inv{\rho} \lr{
\partial_\rho ( \rho A_\phi ) – \partial_\phi A_\rho
}
}
– \partial_z \lr{
\partial_z A_\rho -\partial_\rho A_z
}
} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \partial_\rho \lr{ \frac{1}{\rho} \partial_\phi A_\phi}
+ \partial_{\rho z} A_z
– \inv{\rho^2}\partial_{\phi \rho} ( \rho A_\phi )
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \partial_{z\rho} A_z \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{1}{\rho^2} \partial_\phi A_\phi
+ \frac{1}{\rho} \partial_{\rho\phi} A_\phi
– \inv{\rho^2}\partial_{\phi} A_\phi
– \inv{\rho}\partial_{\phi\rho} A_\phi \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho (\rho A_\rho)}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{2}{\rho^2} \partial_\phi A_\phi \\
&=
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho}
+ \inv{\rho^2}\partial_{\phi\phi} A_\rho
+ \partial_{zz} A_\rho
– \frac{A_\rho}{\rho^2}
– \frac{2}{\rho^2} \partial_\phi A_\phi,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:laplacianCylindrical:240}
\begin{aligned}
\phicap \cdot \lr{ \spacegrad^2 \BA }
&=
\inv{\rho} \partial_\phi
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}

\lr{
\lr{
\partial_z \lr{
\inv{\rho} \partial_\phi A_z – \partial_z A_\phi
}
-\partial_\rho \lr{
\inv{\rho} \lr{ \partial_\rho ( \rho A_\phi ) – \partial_\phi A_\rho}
}
}
} \\
&=
\inv{\rho^2} \partial_{\phi\rho} (\rho A_\rho)
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \inv{\rho}\partial_{\phi z} A_z
– \inv{\rho} \partial_{z\phi} A_z
+ \partial_{z z} A_\phi
+\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
– \partial_\rho \lr{ \inv{\rho} \partial_\phi A_\rho} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \inv{\rho^2} \partial_{\phi\rho} (\rho A_\rho)
+ \inv{\rho}\partial_{\phi z} A_z
– \inv{\rho} \partial_{z\phi} A_z
– \partial_\rho \lr{ \inv{\rho} \partial_\phi A_\rho} \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \inv{\rho^2} \partial_{\phi} A_\rho
+ \inv{\rho} \partial_{\phi\rho} A_\rho
+ \inv{\rho^2} \partial_\phi A_\rho
– \inv{\rho} \partial_{\rho\phi} A_\rho \\
&=
\partial_\rho \lr{ \inv{\rho} \partial_\rho ( \rho A_\phi ) }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \frac{2}{\rho^2} \partial_{\phi} A_\rho \\
&=
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \frac{1}{\rho^2} \partial_{\phi\phi} A_\phi
+ \partial_{z z} A_\phi
+ \frac{2}{\rho^2} \partial_{\phi} A_\rho
– \frac{A_\phi}{\rho^2},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:laplacianCylindrical:260}
\begin{aligned}
\zcap \cdot \lr{ \spacegrad^2 \BA }
&=
\partial_z
\lr{
\inv{\rho} \partial_\rho (\rho A_\rho) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z
}

\inv{\rho} \lr{
\partial_\rho \lr{ \rho \lr{
\partial_z A_\rho -\partial_\rho A_z
}
}
– \partial_\phi \lr{
\inv{\rho} \partial_\phi A_z – \partial_z A_\phi
}
} \\
&=
\inv{\rho} \partial_{z\rho} (\rho A_\rho)
+ \frac{1}{\rho} \partial_{z\phi} A_\phi
+ \partial_{zz} A_z
– \inv{\rho}\partial_\rho \lr{ \rho \partial_z A_\rho }
+ \inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
– \inv{\rho} \partial_{\phi z} A_\phi \\
&=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
+ \partial_{zz} A_z
+ \inv{\rho} \partial_{z} A_\rho
+\partial_{z\rho} A_\rho
+ \frac{1}{\rho} \partial_{z\phi} A_\phi
– \inv{\rho}\partial_z A_\rho
– \partial_{\rho z} A_\rho
– \inv{\rho} \partial_{\phi z} A_\phi \\
&=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho A_z }
+ \inv{\rho^2} \partial_{\phi\phi} A_z
+ \partial_{zz} A_z
\end{aligned}
\end{equation}

Evaluating these was a fairly tedious and mechanical job, and would have been better suited to a computer algebra system than by hand as done here.

Explicit cylindrical Laplacian

Let’s try this a different way. The most obvious potential strategy is to just apply the Laplacian to the vector itself, but we need to include the unit vectors in such an operation

\begin{equation}\label{eqn:laplacianCylindrical:280}
\spacegrad^2 \BA =
\spacegrad^2 \lr{ \rhocap A_\rho + \phicap A_\phi + \zcap A_z }.
\end{equation}

First we need to know the explicit form of the cylindrical Laplacian. From the painful expansion, we can guess that it is

\begin{equation}\label{eqn:laplacianCylindrical:300}
\spacegrad^2 \psi
=
\inv{\rho}\partial_\rho \lr{ \rho \partial_\rho \psi }
+ \inv{\rho^2} \partial_{\phi\phi} \psi
+ \partial_{zz} \psi.
\end{equation}

Let’s check that explicitly. Here I use the vector product where \( \rhocap^2 = \phicap^2 = \zcap^2 = 1 \), and these vectors anticommute when different

\begin{equation}\label{eqn:laplacianCylindrical:320}
\begin{aligned}
\spacegrad^2 \psi
&=
\lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi + \zcap \partial_z }
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi } \\
&=
\rhocap \partial_\rho
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi }
+ \frac{\phicap}{\rho} \partial_\phi
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi }
+ \zcap \partial_z
\lr{ \rhocap \partial_\rho \psi + \frac{\phicap}{\rho} \partial_\phi \psi + \zcap \partial_z \psi } \\
&=
\partial_{\rho\rho} \psi
+ \rhocap \phicap \partial_\rho \lr{ \frac{1}{\rho} \partial_\phi \psi}
+ \rhocap \zcap \partial_{\rho z} \psi
+ \frac{\phicap}{\rho} \partial_\phi \lr{ \rhocap \partial_\rho \psi }
+ \frac{\phicap}{\rho} \partial_\phi \lr{ \frac{\phicap}{\rho} \partial_\phi \psi }
+ \frac{\phicap \zcap }{\rho} \partial_{\phi z} \psi
+ \zcap \rhocap \partial_{z\rho} \psi
+ \frac{\zcap \phicap}{\rho} \partial_{z\phi} \psi
+ \partial_{zz} \psi \\
&=
\partial_{\rho\rho} \psi
+ \inv{\rho} \partial_\rho \psi
+ \frac{1}{\rho^2} \partial_{\phi \phi} \psi
+ \partial_{zz} \psi
+ \rhocap \phicap
\lr{
-\frac{1}{\rho^2} \partial_\phi \psi
+\frac{1}{\rho} \partial_{\rho \phi} \psi
-\inv{\rho} \partial_{\phi \rho} \psi
+ \frac{1}{\rho^2} \partial_\phi \psi
}
+ \zcap \rhocap \lr{
-\partial_{\rho z} \psi
+ \partial_{z\rho} \psi
}
+ \phicap \zcap \lr{
\inv{\rho} \partial_{\phi z} \psi
– \inv{\rho} \partial_{z\phi} \psi
} \\
&=
\partial_{\rho\rho} \psi
+ \inv{\rho} \partial_\rho \psi
+ \frac{1}{\rho^2} \partial_{\phi \phi} \psi
+ \partial_{zz} \psi,
\end{aligned}
\end{equation}

so the Laplacian operator is

\begin{equation}\label{eqn:laplacianCylindrical:340}
\boxed{
\spacegrad^2
=
\inv{\rho} \PD{\rho}{} \lr{ \rho \PD{\rho}{} }
+ \frac{1}{\rho^2} \PDSq{\phi}{}
+ \PDSq{z}{}.
}
\end{equation}

All the bivector grades of the Laplacian operator are seen to explicitly cancel, regardless of the grade of \( \psi \), just as if we had expanded the scalar Laplacian as a dot product
\( \spacegrad^2 \psi = \spacegrad \cdot \lr{ \spacegrad \psi} \).
Unlike such a scalar expansion, this derivation is seen to be valid for any grade \( \psi \). We know now that we can trust this result when \( \psi \) is a scalar, a vector, a bivector, a trivector, or even a multivector.

Vector Laplacian

Now that we trust that the typical scalar form of the Laplacian applies equally well to multivectors as it does to scalars, that cylindrical coordinate operator can now be applied to a
vector. Consider the projections onto each of the directions in turn

\begin{equation}\label{eqn:laplacianCylindrical:360}
\spacegrad^2 \lr{ \rhocap A_\rho }
=
\rhocap \inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho }
+ \frac{1}{\rho^2} \partial_{\phi\phi} \lr{\rhocap A_\rho}
+ \rhocap \partial_{zz} A_\rho
\end{equation}

\begin{equation}\label{eqn:laplacianCylindrical:380}
\begin{aligned}
\partial_{\phi\phi} \lr{\rhocap A_\rho}
&=
\partial_\phi \lr{ \phicap A_\rho + \rhocap \partial_\phi A_\rho } \\
&=
-\rhocap A_\rho
+\phicap \partial_\phi A_\rho
+ \phicap \partial_\phi A_\rho
+ \rhocap \partial_{\phi\phi} A_\rho \\
&=
\rhocap \lr{ \partial_{\phi\phi} A_\rho -A_\rho }
+ 2 \phicap \partial_\phi A_\rho
\end{aligned}
\end{equation}

so this component of the vector Laplacian is

\begin{equation}\label{eqn:laplacianCylindrical:400}
\begin{aligned}
\spacegrad^2 \lr{ \rhocap A_\rho }
&=
\rhocap
\lr{
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\rho }
+ \inv{\rho^2} \partial_{\phi\phi} A_\rho
– \inv{\rho^2} A_\rho
+ \partial_{zz} A_\rho
}
+
\phicap
\lr{
2 \inv{\rho^2} \partial_\phi A_\rho
} \\
&=
\rhocap \lr{
\spacegrad^2 A_\rho
– \inv{\rho^2} A_\rho
}
+
\phicap
\frac{2}{\rho^2} \partial_\phi A_\rho
.
\end{aligned}
\end{equation}

The Laplacian for the projection of the vector onto the \( \phicap \) direction is

\begin{equation}\label{eqn:laplacianCylindrical:420}
\spacegrad^2 \lr{ \phicap A_\phi }
=
\phicap \inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \frac{1}{\rho^2} \partial_{\phi\phi} \lr{\phicap A_\phi}
+ \phicap \partial_{zz} A_\phi,
\end{equation}

Again, since the unit vectors are \( \phi \) dependent, the \( \phi \) derivatives have to be treated carefully

\begin{equation}\label{eqn:laplacianCylindrical:440}
\begin{aligned}
\partial_{\phi\phi} \lr{\phicap A_\phi}
&=
\partial_{\phi} \lr{-\rhocap A_\phi + \phicap \partial_\phi A_\phi} \\
&=
-\phicap A_\phi
-\rhocap \partial_\phi A_\phi
– \rhocap \partial_\phi A_\phi
+ \phicap \partial_{\phi \phi} A_\phi \\
&=
– 2 \rhocap \partial_\phi A_\phi
+
\phicap
\lr{
\partial_{\phi \phi} A_\phi
– A_\phi
},
\end{aligned}
\end{equation}

so the Laplacian of this projection is
\begin{equation}\label{eqn:laplacianCylindrical:460}
\begin{aligned}
\spacegrad^2 \lr{ \phicap A_\phi }
&=
\phicap
\lr{
\inv{\rho} \partial_\rho \lr{ \rho \partial_\rho A_\phi }
+ \phicap \partial_{zz} A_\phi,
\inv{\rho^2} \partial_{\phi \phi} A_\phi
– \frac{A_\phi }{\rho^2}
}
– \rhocap \frac{2}{\rho^2} \partial_\phi A_\phi \\
&=
\phicap \lr{
\spacegrad^2 A_\phi
– \frac{A_\phi}{\rho^2}
}
– \rhocap \frac{2}{\rho^2} \partial_\phi A_\phi.
\end{aligned}
\end{equation}

Since \( \zcap \) is fixed we have

\begin{equation}\label{eqn:laplacianCylindrical:480}
\spacegrad^2 \zcap A_z
=
\zcap \spacegrad^2 A_z.
\end{equation}

Putting all the pieces together we have
\begin{equation}\label{eqn:laplacianCylindrical:500}
\boxed{
\spacegrad^2 \BA
=
\rhocap \lr{
\spacegrad^2 A_\rho
– \inv{\rho^2} A_\rho
– \frac{2}{\rho^2} \partial_\phi A_\phi
}
+\phicap \lr{
\spacegrad^2 A_\phi
– \frac{A_\phi}{\rho^2}
+ \frac{2}{\rho^2} \partial_\phi A_\rho
}
+
\zcap \spacegrad^2 A_z.
}
\end{equation}

This matches the results of \ref{eqn:laplacianCylindrical:220}, …, from the painful expansion of
\( \spacegrad \lr{ \spacegrad \cdot \BA } – \spacegrad \cross \lr{ \spacegrad \cross \BA } \).