divergence

Vector gradients in dyadic notation and geometric algebra: Part II.

March 6, 2022 math and physics play , , , , ,

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Symmetrization and antisymmetrization of the vector differential in GA.

There was an error in yesterday’s post. This decomposition was correct:
\begin{equation}\label{eqn:dyadicVsGa:460}
d\Bv
= (d\Bx \cdot \spacegrad) \Bv
= d\Bx (\spacegrad \cdot \Bv)
+
\spacegrad \cdot \lr{ d\Bx \wedge \Bv }.
\end{equation}
However, identifying these terms with the symmetric and antisymmetric splits of \( \spacegrad \otimes \Bv \) was wrong.
Brian pointed out that a purely incompressible flow is one for which \(\spacegrad \cdot \Bv = 0\), yet, in general, an incompressible flow can have a non-zero deformation tensor.

Also, given the nature of the matrix expansion of the antisymmetric tensor, we should have had a curl term in the mix and we do not. The conclusion must be that \ref{eqn:dyadicVsGa:460} is a split into divergence and non-divergence terms, but we really wanted a split into curl and non-curl terms.

Symmetrization and antisymmetrization of the vector differential in GA: Take II.

Identification of \( \ifrac{1}{2} \lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger } \) with the divergence was incorrect.

Let’s explicitly expand out our symmetric tensor component fully to see what it really yields, without guessing.
\begin{equation}\label{eqn:dyadicVsGa:480}
\begin{aligned}
d\Bx \cdot
\inv{2}
\lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger }
&=
d\Bx \cdot
\inv{2}
\lr{
\begin{bmatrix}
\partial_i v_j
\end{bmatrix}
+
\begin{bmatrix}
\partial_j v_i
\end{bmatrix}
} \\
&=
dx_i
\inv{2}
\begin{bmatrix}
\partial_i v_j +
\partial_j v_i
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix}.
\end{aligned}
\end{equation}
The symmetric matrix that represents this direct product tensor is
\begin{equation}\label{eqn:dyadicVsGa:500}
\inv{2}
\begin{bmatrix}
\partial_i v_j +
\partial_j v_i
\end{bmatrix}
=
\inv{2}
\begin{bmatrix}
2 \partial_1 v_1 & \partial_1 v_2 + \partial_2 v_1 & \partial_1 v_3 + \partial_3 v_1 \\
\partial_2 v_1 + \partial_1 v_2 & 2 \partial_2 v_2 & \partial_2 v_3 + \partial_3 v_2 \\
\partial_3 v_1 + \partial_1 v_3 & \partial_3 v_2 + \partial_2 v_3 & \partial_3 v_1 + \partial_1 v_3 \\
\end{bmatrix}
.
\end{equation}
This certainly isn’t isomorphic to the divergence. Instead, the trace of this matrix is the portion that is isomorphic to the divergence. The rest is something else. Let’s put the tensors into vector form to understand what they really represent.

For the symmetric part we have
\begin{equation}\label{eqn:dyadicVsGa:520}
\begin{aligned}
d\Bx \cdot
\inv{2}
\lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger }
&=
dx_i
\inv{2}
\begin{bmatrix}
\partial_i v_j +
\partial_j v_i
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix} \\
&=
\inv{2} \lr{
\lr{ d\Bx \cdot \spacegrad } \Bv + \spacegrad \lr{ d\Bx \cdot \Bv }
},
\end{aligned}
\end{equation}
and, similarily, for the antisymmetric tensor component, we have
\begin{equation}\label{eqn:dyadicVsGa:540}
\begin{aligned}
d\Bx \cdot
\inv{2}
\lr{ \spacegrad \otimes \Bv – \lr{ \spacegrad \otimes \Bv }^\dagger }
&=
dx_i
\inv{2}
\begin{bmatrix}
\partial_i v_j –
\partial_j v_i
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix} \\
&=
\inv{2} \lr{
\lr{ d\Bx \cdot \spacegrad } \Bv – \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
d\Bx \cdot \lr{ \spacegrad \wedge \Bv}.
\end{aligned}
\end{equation}
We find an isomorphism of the antisymmetric term with the curl, but the symmetric term has a divergence component, plus more.

If we want to we can split the symmetric component into it’s divergence and non-divergence terms, we get
\begin{equation}\label{eqn:dyadicVsGa:560}
\begin{aligned}
d\Bx \cdot \Bd
&=
\inv{2}
\lr{
\lr{ d\Bx \cdot \spacegrad } \Bv + \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
\lr{
d\Bx \lr{ \spacegrad \cdot \Bv } + \spacegrad \cdot \lr{ d\Bx \wedge \Bv } + \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
\lr{
d\Bx \lr{ \spacegrad \cdot \Bv } + \gpgradeone{ \spacegrad \lr{ d\Bx \wedge \Bv } + \spacegrad \lr{ d\Bx \cdot \Bv } }
} \\
&=
\inv{2}
\lr{
d\Bx \lr{ \spacegrad \cdot \Bv } + \gpgradeone{ \spacegrad d\Bx\, \Bv }
},
\end{aligned}
\end{equation}
so for incompressible flow, the GA representation is a single grade one selection
\begin{equation}\label{eqn:dyadicVsGa:600}
d\Bx \cdot \Bd = \inv{2} \gpgradeone{ \spacegrad d\Bx\, \Bv }.
\end{equation}
It is a little unfortunate that we cannot factor out the \( d\Bx \) term. We can do that for the
GA representation of the antisymmetric tensor contribution, which is just
\begin{equation}\label{eqn:dyadicVsGa:580}
\BOmega
=
\inv{2} \spacegrad \wedge \Bv.
\end{equation}

Let’s see what the antisymmetric tensor equivalent looks like in the incompressible case, by subtracting a divergence term
\begin{equation}\label{eqn:dyadicVsGa:680}
\begin{aligned}
d\Bx \cdot \lr{ \spacegrad \wedge \Bv } – d\Bx \lr{ \spacegrad \cdot \Bv }
&=
\gpgradeone{ d\Bx \lr{ \spacegrad \wedge \Bv } – d\Bx \lr{ \spacegrad \cdot \Bv } } \\
&=
\gpgradeone{ -d\Bx \lr{ \Bv \wedge \spacegrad } – d\Bx \lr{ \Bv \cdot \spacegrad } } \\
&=
-\gpgradeone{ d\Bx \Bv \spacegrad },
\end{aligned}
\end{equation}
so we have
\begin{equation}\label{eqn:dyadicVsGa:700}
d\Bx \cdot \lr{ \spacegrad \wedge \Bv } = d\Bx \lr{ \spacegrad \cdot \Bv } – \gpgradeone{ d\Bx\, \Bv \spacegrad }.
\end{equation}
Both the symmetric and antisymmetric tensors have compressible components.

Summary.

We found that it was possible to split the vector differential into a divergence and incompressible components, as follows
\begin{equation}\label{eqn:dyadicVsGa:620}
\begin{aligned}
d\Bv
&= \lr{ d\Bx \cdot \spacegrad } \Bv \\
&= d\Bx (\spacegrad \cdot \Bv)
+
\spacegrad \cdot \lr{ d\Bx \wedge \Bv }.
\end{aligned}
\end{equation}

With
\begin{equation}\label{eqn:dyadicVsGa:720}
\begin{aligned}
d\Bv
&= d\Bx \cdot
\lr{
\inv{2} \lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger }
+
\inv{2} \lr{ \spacegrad \otimes \Bv – \lr{ \spacegrad \otimes \Bv }^\dagger }
} \\
&= d\Bx \cdot \lr{ \Bd + \BOmega },
\end{aligned}
\end{equation}
we found the following correspondences between the symmetric and antisymmetric tensor product components
\begin{equation}\label{eqn:dyadicVsGa:640}
\begin{aligned}
d\Bx \cdot \Bd &=
\inv{2} \lr{
\lr{ d\Bx \cdot \spacegrad } \Bv + \spacegrad \lr{ d\Bx \cdot \Bv }
} \\
&=
\inv{2}
\lr{
d\Bx \lr{ \spacegrad \cdot \Bv } + \gpgradeone{ \spacegrad d\Bx\, \Bv }
}
\end{aligned},
\end{equation}
and
\begin{equation}\label{eqn:dyadicVsGa:660}
\begin{aligned}
d\Bx \cdot \BOmega
&=
\inv{2} d\Bx \cdot \lr{ \spacegrad \wedge \Bv } \\
&=
\inv{2} \lr{
d\Bx \lr{ \spacegrad \cdot \Bv } – \gpgradeone{ d\Bx\, \Bv \spacegrad }
}.
\end{aligned}
\end{equation}

In the incompressible case where \( \spacegrad \cdot \Bv = 0 \), we have
\begin{equation}\label{eqn:dyadicVsGa:740}
\begin{aligned}
d\Bx \cdot \Bd &= \inv{2} \gpgradeone{ \spacegrad d\Bx\, \Bv } \\
d\Bx \cdot \BOmega &= -\inv{2} \gpgradeone{ d\Bx\, \Bv \spacegrad },
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:dyadicVsGa:760}
\begin{aligned}
d\Bv
&= d\Bx \cdot \lr{ \Bd + \BOmega } \\
&= \inv{2} \gpgradeone{ \spacegrad d\Bx\, \Bv – d\Bx\, \Bv \spacegrad } \\
&= \spacegrad \cdot \lr{ d\Bx \wedge \Bv }.
\end{aligned}
\end{equation}

Vector gradients in dyadic notation and geometric algebra.

March 5, 2022 math and physics play , , , ,

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

This is an exploration of the dyadic representation of the gradient acting on a vector in \(\mathbb{R}^3\), where we determine a tensor product formulation of a vector differential. Such a tensor product formulation can be split into symmetric and antisymmetric components. The geometric algebra (GA) equivalents of such a split are determined.

There is an error in part of the analysis below, which is addressed in a followup post made the next day.

GA gradient of a vector.

In GA we are free to express the product of the gradient and a vector field by adjacency. In coordinates (summation over repeated indexes assumed), such a product has the form
\begin{equation}\label{eqn:dyadicVsGa:20}
\spacegrad \Bv
= \lr{ \Be_i \partial_i } \lr{ v_j \Be_j }
= \lr{ \partial_i v_j } \Be_i \Be_j.
\end{equation}
In this sum, any terms with \( i = j \) are scalars since \( \Be_i^2 = 1 \), and the remaining terms are bivectors. This can be written compactly as
\begin{equation}\label{eqn:dyadicVsGa:40}
\spacegrad \Bv = \spacegrad \cdot \Bv + \spacegrad \wedge \Bv,
\end{equation}
or for \(\mathbb{R}^3\)
\begin{equation}\label{eqn:dyadicVsGa:60}
\spacegrad \Bv = \spacegrad \cdot \Bv + I \lr{ \spacegrad \cross \Bv},
\end{equation}
either of which breaks the gradient into into divergence and curl components. In \ref{eqn:dyadicVsGa:40} this vector gradient is expressed using the bivector valued curl operator \( (\spacegrad \wedge \Bv) \), whereas \ref{eqn:dyadicVsGa:60} is expressed using the vector valued dual form of the curl \( (\spacegrad \cross \Bv) \) from convential vector algebra.

It is worth noting that order matters in the GA coordinate expansion of \ref{eqn:dyadicVsGa:20}. It is not correct to write
\begin{equation}\label{eqn:dyadicVsGa:80}
\spacegrad \Bv
= \lr{ \partial_i v_j } \Be_j \Be_i,
\end{equation}
which is only true when the curl, \( \spacegrad \wedge \Bv = 0 \), is zero.

Dyadic representation.

Given a vector field \( \Bv = \Bv(\Bx) \), the differential of that field can be computed by chain rule
\begin{equation}\label{eqn:dyadicVsGa:100}
d\Bv = \PD{x_i}{\Bv} dx_i = \lr{ d\Bx \cdot \spacegrad} \Bv,
\end{equation}
where \( d\Bx = \Be_i dx_i \). This is a representation invariant form of the differential, where we have a scalar operator \( d\Bx \cdot \spacegrad \) acting on the vector field \( \Bv \). The matrix representation of this differential can be written as
\begin{equation}\label{eqn:dyadicVsGa:120}
d\Bv = \lr{
{\begin{bmatrix}
d\Bx
\end{bmatrix}}^\dagger
\begin{bmatrix}
\spacegrad
\end{bmatrix}
}
\begin{bmatrix}
\Bv
\end{bmatrix}
,
\end{equation}
where we are using the dagger to designate transposition, and each of the terms on the right are the coordinate matrixes of the vectors with respect to the standard basis
\begin{equation}\label{eqn:dyadicVsGa:140}
\begin{bmatrix}
d\Bx
\end{bmatrix}
=
\begin{bmatrix}
dx_1 \\
dx_2 \\
dx_3
\end{bmatrix},\quad
\begin{bmatrix}
\Bv
\end{bmatrix}
=
\begin{bmatrix}
v_1 \\
v_2 \\
v_3
\end{bmatrix},\quad
\begin{bmatrix}
\spacegrad
\end{bmatrix}
=
\begin{bmatrix}
\partial_1 \\
\partial_2 \\
\partial_3
\end{bmatrix}.
\end{equation}

In \ref{eqn:dyadicVsGa:120} the parens are very important, as the expression is meaningless without them. With the parens we have a \((1 \times 3)(3 \times 1)\) matrix (i.e. a scalar) multiplied with a \(3\times 1\) matrix. That becomes ill-formed if we drop the parens since we are left with an incompatible product of a \((3\times1)(3\times1)\) matrix on the right. The dyadic notation, which introducing a tensor product into the mix, is a mechanism to make sense of the possibility of such a product. Can we make sense of an expression like \( \spacegrad \Bv \) without the geometric product in our toolbox?

Stepping towards that question, let’s examine the coordinate expansion of our vector differential \ref{eqn:dyadicVsGa:100}, which is
\begin{equation}\label{eqn:dyadicVsGa:160}
d\Bv = dx_i \lr{ \partial_i v_j } \Be_j.
\end{equation}
If we allow a matrix of vectors, this has a block matrix form
\begin{equation}\label{eqn:dyadicVsGa:180}
d\Bv =
{\begin{bmatrix}
d\Bx
\end{bmatrix}}^\dagger
\begin{bmatrix}
\spacegrad \otimes \Bv
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix}
.
\end{equation}
Here we introduce the tensor product
\begin{equation}\label{eqn:dyadicVsGa:200}
\spacegrad \otimes \Bv
= \partial_i v_j \, \Be_i \otimes \Be_j,
\end{equation}
and designate the matrix of coordinates \( \partial_i v_j \), a second order tensor, by \(
\begin{bmatrix}
\spacegrad \otimes \Bv
\end{bmatrix}
\).

We have succeeded in factoring out a vector gradient. We can introduce dot product between vectors and a direct product of vectors, by observing that \ref{eqn:dyadicVsGa:180} has the structure of a quadradic form, and define
\begin{equation}\label{eqn:dyadicVsGa:220}
\Bx \cdot (\Ba \otimes \Bb) \equiv
{\begin{bmatrix}
\Bx
\end{bmatrix}}^\dagger
\begin{bmatrix}
\Ba \otimes \Bb
\end{bmatrix}
\begin{bmatrix}
\Be_1 \\
\Be_2 \\
\Be_3
\end{bmatrix},
\end{equation}
so that \ref{eqn:dyadicVsGa:180} takes the form
\begin{equation}\label{eqn:dyadicVsGa:240}
d\Bv = d\Bx \cdot \lr{ \spacegrad \otimes \Bv }.
\end{equation}
Such a dot product gives operational meaning to the gradient-vector tensor product.

Symmetrization and antisymmetrization of the vector differential in GA.

Using the dyadic notation, it’s possible to split a vector derivative into symmetric and antisymmetric components with respect to the gradient-vector direct product
\begin{equation}\label{eqn:dyadicVsGa:260}
d\Bv
= d\Bx \cdot
\lr{
\inv{2} \lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger }
+
\inv{2} \lr{ \spacegrad \otimes \Bv – \lr{ \spacegrad \otimes \Bv }^\dagger }
},
\end{equation}
or \( d\Bv = d\Bx \cdot \lr{ \Bd + \BOmega } \), where \( \Bd \) is a symmetric tensor, and \( \BOmega \) is a traceless antisymmetric tensor.

A question of potential interest is “what GA equvivalent of this expression?”. There are two identities that are helpful for extracting this equivalence, the first of which is the k-blade vector product identities. Given a k-blade \( B_k \) (i.e.: a product of \( k \) orthogonal vectors, or the wedge of \( k \) vectors), and a vector \( \Ba \), the dot product of the two is
\begin{equation}\label{eqn:dyadicVsGa:280}
B_k \cdot \Ba = \inv{2} \lr{ B_k \Ba + (-1)^{k+1} \Ba B_k }
\end{equation}
Specifically, given two vectors \( \Ba, \Bb \), the vector dot product can be written as a symmetric sum
\begin{equation}\label{eqn:dyadicVsGa:300}
\Ba \cdot \Bb = \inv{2} \lr{ \Ba \Bb + \Bb \Ba } = \Bb \cdot \Ba,
\end{equation}
and given a bivector \( B \) and a vector \( \Ba \), the bivector-vector dot product can be written as an antisymmetric sum
\begin{equation}\label{eqn:dyadicVsGa:320}
B \cdot \Ba = \inv{2} \lr{ B \Ba – \Ba B } = – \Ba \cdot B.
\end{equation}

We may apply these to expressions where one of the vector terms is the gradient, but must allow for the gradient to act bidirectionally. That is, given multivectors \( M, N \)
\begin{equation}\label{eqn:dyadicVsGa:340}
M \spacegrad N
=
\partial_i (M \Be_i N)
=
(\partial_i M) \Be_i N + M \Be_i (\partial_i N),
\end{equation}
where parens have been used to indicate the scope of applicibility of the partials. In particular, this means that we may write the divergence as a GA symmetric sum
\begin{equation}\label{eqn:dyadicVsGa:360}
\spacegrad \cdot \Bv = \inv{2} \lr{
\spacegrad \Bv + \Bv \spacegrad },
\end{equation}
which clearly corresponds to the symmetric term \( \Bd = (1/2) \lr{ \spacegrad \otimes \Bv + \lr{ \spacegrad \otimes \Bv }^\dagger } \) from \ref{eqn:dyadicVsGa:260}.

Let’s assume that we can write our vector differential in terms of a divergence term isomorphic to the symmetric sum in \ref{eqn:dyadicVsGa:260}, and a “something else”, \(\BX\). That is
\begin{equation}\label{eqn:dyadicVsGa:380}
\begin{aligned}
d\Bv
&= \lr{ d\Bx \cdot \spacegrad } \Bv \\
&= d\Bx (\spacegrad \cdot \Bv) + \BX,
\end{aligned}
\end{equation}
where
\begin{equation}\label{eqn:dyadicVsGa:400}
\BX = \lr{ d\Bx \cdot \spacegrad } \Bv – d\Bx (\spacegrad \cdot \Bv),
\end{equation}
is a vector expression to be reduced to something simpler. That reduction is possible using the distribution identity
\begin{equation}\label{eqn:dyadicVsGa:420}
\Bc \cdot (\Ba \wedge \Bb)
=
(\Bc \cdot \Ba) \Bb
– (\Bc \cdot \Bb) \Ba,
\end{equation}
so we find
\begin{equation}\label{eqn:dyadicVsGa:440}
\BX = \spacegrad \cdot \lr{ d\Bx \wedge \Bv }.
\end{equation}

We find the following GA split of the vector differential into symmetric and antisymmetric terms
\begin{equation}\label{eqn:dyadicVsGa:460}
\boxed{
d\Bv
= (d\Bx \cdot \spacegrad) \Bv
= d\Bx (\spacegrad \cdot \Bv)
+
\spacegrad \cdot \lr{ d\Bx \wedge \Bv }.
}
\end{equation}
Such a split avoids the indeterminant nature of the tensor product, which we only give meaning by introducing the quadratic form based dot product given by \ref{eqn:dyadicVsGa:220}.

Transverse gauge

November 16, 2016 math and physics play , , , , , , , , , , , , , , , , ,

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

Jackson [1] has an interesting presentation of the transverse gauge. I’d like to walk through the details of this, but first want to translate the preliminaries to SI units (if I had the 3rd edition I’d not have to do this translation step).

Gauge freedom

The starting point is noting that \( \spacegrad \cdot \BB = 0 \) the magnetic field can be expressed as a curl

\begin{equation}\label{eqn:transverseGauge:20}
\BB = \spacegrad \cross \BA.
\end{equation}

Faraday’s law now takes the form
\begin{equation}\label{eqn:transverseGauge:40}
\begin{aligned}
0
&= \spacegrad \cross \BE + \PD{t}{\BB} \\
&= \spacegrad \cross \BE + \PD{t}{} \lr{ \spacegrad \cross \BA } \\
&= \spacegrad \cross \lr{ \BE + \PD{t}{\BA} }.
\end{aligned}
\end{equation}

Because this curl is zero, the interior sum can be expressed as a gradient

\begin{equation}\label{eqn:transverseGauge:60}
\BE + \PD{t}{\BA} \equiv -\spacegrad \Phi.
\end{equation}

This can now be substituted into the remaining two Maxwell’s equations.

\begin{equation}\label{eqn:transverseGauge:80}
\begin{aligned}
\spacegrad \cdot \BD &= \rho_v \\
\spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\
\end{aligned}
\end{equation}

For Gauss’s law, in simple media, we have

\begin{equation}\label{eqn:transverseGauge:140}
\begin{aligned}
\rho_v
&=
\epsilon \spacegrad \cdot \BE \\
&=
\epsilon \spacegrad \cdot \lr{ -\spacegrad \Phi – \PD{t}{\BA} }
\end{aligned}
\end{equation}

For simple media again, the Ampere-Maxwell equation is

\begin{equation}\label{eqn:transverseGauge:100}
\inv{\mu} \spacegrad \cross \lr{ \spacegrad \cross \BA } = \BJ + \epsilon \PD{t}{} \lr{ -\spacegrad \Phi – \PD{t}{\BA} }.
\end{equation}

Expanding \( \spacegrad \cross \lr{ \spacegrad \cross \BA } = -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } \) gives
\begin{equation}\label{eqn:transverseGauge:120}
-\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } + \epsilon \mu \PDSq{t}{\BA} = \mu \BJ – \epsilon \mu \spacegrad \PD{t}{\Phi}.
\end{equation}

Maxwell’s equations are now reduced to
\begin{equation}\label{eqn:transverseGauge:180}
\boxed{
\begin{aligned}
\spacegrad^2 \BA – \spacegrad \lr{ \spacegrad \cdot \BA + \epsilon \mu \PD{t}{\Phi}} – \epsilon \mu \PDSq{t}{\BA} &= -\mu \BJ \\
\spacegrad^2 \Phi + \PD{t}{\spacegrad \cdot \BA} &= -\frac{\rho_v }{\epsilon}.
\end{aligned}
}
\end{equation}

There are two obvious constraints that we can impose
\begin{equation}\label{eqn:transverseGauge:200}
\spacegrad \cdot \BA – \epsilon \mu \PD{t}{\Phi} = 0,
\end{equation}

or
\begin{equation}\label{eqn:transverseGauge:220}
\spacegrad \cdot \BA = 0.
\end{equation}

The first constraint is the Lorentz gauge, which I’ve played with previously. It happens to be really nice in a relativistic context since, in vacuum with a four-vector potential \( A = (\Phi/c, \BA) \), that is a requirement that the four-divergence of the four-potential vanishes (\( \partial_\mu A^\mu = 0 \)).

Transverse gauge

Jackson identifies the latter constraint as the transverse gauge, which I’m less familiar with. With this gauge selection, we have

\begin{equation}\label{eqn:transverseGauge:260}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \epsilon\mu \spacegrad \PD{t}{\Phi}
\end{equation}
\begin{equation}\label{eqn:transverseGauge:280}
\spacegrad^2 \Phi = -\frac{\rho_v }{\epsilon}.
\end{equation}

What’s not obvious is the fact that the irrotational (zero curl) contribution due to \(\Phi\) in \ref{eqn:transverseGauge:260} cancels the corresponding irrotational term from the current. Jackson uses a transverse and longitudinal decomposition of the current, related to the Helmholtz theorem to allude to this.

That decomposition follows from expanding \( \spacegrad^2 J/R \) in two ways using the delta function \( -4 \pi \delta(\Bx – \Bx’) = \spacegrad^2 1/R \) representation, as well as directly

\begin{equation}\label{eqn:transverseGauge:300}
\begin{aligned}
– 4 \pi \BJ(\Bx)
&=
\int \spacegrad^2 \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\spacegrad
\int \spacegrad \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+
\spacegrad \cdot
\int \spacegrad \wedge \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
-\spacegrad
\int \BJ(\Bx’) \cdot \spacegrad’ \inv{\Abs{\Bx – \Bx’}} d^3 x’
+
\spacegrad \cdot \lr{ \spacegrad \wedge
\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
} \\
&=
-\spacegrad
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+\spacegrad
\int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

\spacegrad \cross \lr{
\spacegrad \cross
\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
}
\end{aligned}
\end{equation}

The first term can be converted to a surface integral

\begin{equation}\label{eqn:transverseGauge:320}
-\spacegrad
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
=
-\spacegrad
\int d\BA’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}},
\end{equation}

so provided the currents are either localized or \( \Abs{\BJ}/R \rightarrow 0 \) on an infinite sphere, we can make the identification

\begin{equation}\label{eqn:transverseGauge:340}
\BJ(\Bx)
=
-\spacegrad \inv{4 \pi} \int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+
\spacegrad \cross \spacegrad \cross \inv{4 \pi} \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
\equiv
\BJ_l +
\BJ_t,
\end{equation}

where \( \spacegrad \cross \BJ_l = 0 \) (irrotational, or longitudinal), whereas \( \spacegrad \cdot \BJ_t = 0 \) (solenoidal or transverse). The irrotational property is clear from inspection, and the transverse property can be verified readily

\begin{equation}\label{eqn:transverseGauge:360}
\begin{aligned}
\spacegrad \cdot \lr{ \spacegrad \cross \lr{ \spacegrad \cross \BX } }
&=
-\spacegrad \cdot \lr{ \spacegrad \cdot \lr{ \spacegrad \wedge \BX } } \\
&=
-\spacegrad \cdot \lr{ \spacegrad^2 \BX – \spacegrad \lr{ \spacegrad \cdot \BX } } \\
&=
-\spacegrad \cdot \lr{\spacegrad^2 \BX} + \spacegrad^2 \lr{ \spacegrad \cdot \BX } \\
&= 0.
\end{aligned}
\end{equation}

Since

\begin{equation}\label{eqn:transverseGauge:380}
\Phi(\Bx, t)
=
\inv{4 \pi \epsilon} \int \frac{\rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’,
\end{equation}

we have

\begin{equation}\label{eqn:transverseGauge:400}
\begin{aligned}
\spacegrad \PD{t}{\Phi}
&=
\inv{4 \pi \epsilon} \spacegrad \int \frac{\partial_t \rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\inv{4 \pi \epsilon} \spacegrad \int \frac{-\spacegrad’ \cdot \BJ}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\frac{\BJ_l}{\epsilon}.
\end{aligned}
\end{equation}

This means that the Ampere-Maxwell equation takes the form

\begin{equation}\label{eqn:transverseGauge:420}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA}
= -\mu \BJ + \mu \BJ_l
= -\mu \BJ_t.
\end{equation}

This justifies the transverse in the label transverse gauge.

References

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

Spherical gradient, divergence, curl and Laplacian

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

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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.

Frequency domain time averaged Poynting theorem

November 8, 2016 math and physics play , , ,

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

The time domain Poynting relationship was found to be

\begin{equation}\label{eqn:poyntingTimeHarmonic:20}
0
=
\spacegrad \cdot \lr{ \BE \cross \BH }
+ \frac{\epsilon}{2} \BE \cdot \PD{t}{\BE}
+ \frac{\mu}{2} \BH \cdot \PD{t}{\BH}
+ \BH \cdot \BM_i
+ \BE \cdot \BJ_i
+ \sigma \BE \cdot \BE.
\end{equation}

Let’s derive the equivalent relationship for the time averaged portion of the time-harmonic Poynting vector. The time domain representation of the Poynting vector in terms of the time-harmonic (phasor) vectors is

\begin{equation}\label{eqn:poyntingTimeHarmonic:40}
\begin{aligned}
\boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}
&= \inv{4}
\lr{
\BE e^{j\omega t}
+ \BE^\conj e^{-j\omega t}
}
\cross
\lr{
\BH e^{j\omega t}
+ \BH^\conj e^{-j\omega t}
} \\
&=
\inv{2} \textrm{Re} \lr{ \BE \cross \BH^\conj + \BE \cross \BH e^{2 j \omega t} },
\end{aligned}
\end{equation}

so if we are looking for the relationships that effect only the time averaged Poynting vector, over integral multiples of the period, we are interested in evaluating the divergence of

\begin{equation}\label{eqn:poyntingTimeHarmonic:60}
\inv{2} \BE \cross \BH^\conj.
\end{equation}

The time-harmonic Maxwell’s equations are
\begin{equation}\label{eqn:poyntingTimeHarmonic:80}
\begin{aligned}
\spacegrad \cross \BE &= – j \omega \mu \BH – \BM_i \\
\spacegrad \cross \BH &= j \omega \epsilon \BE + \BJ_i + \sigma \BE \\
\end{aligned}
\end{equation}

The latter after conjugation is

\begin{equation}\label{eqn:poyntingTimeHarmonic:100}
\spacegrad \cross \BH^\conj = -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj.
\end{equation}

For the divergence we have

\begin{equation}\label{eqn:poyntingTimeHarmonic:120}
\begin{aligned}
\spacegrad \cdot \lr{ \BE \cross \BH^\conj }
&=
\BH^\conj \cdot \lr{ \spacegrad \cdot \BE }
-\BE \cdot \lr{ \spacegrad \cdot \BH^\conj } \\
&=
\BH^\conj \cdot \lr{ – j \omega \mu \BH – \BM_i }
– \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:poyntingTimeHarmonic:140}
0
=
\spacegrad \cdot \lr{ \BE \cross \BH^\conj }
+
\BH^\conj \cdot \lr{ j \omega \mu \BH + \BM_i }
+ \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },
\end{equation}

so
\begin{equation}\label{eqn:poyntingTimeHarmonic:160}
\boxed{
0
=
\spacegrad \cdot \inv{2} \lr{ \BE \cross \BH^\conj }
+ \inv{2} \lr{ \BH^\conj \cdot \BM_i
+ \BE \cdot \BJ_i^\conj }
+ \inv{2} j \omega \lr{ \mu \Abs{\BH}^2 – \epsilon^\conj \Abs{\BE}^2 }
+ \inv{2} \sigma^\conj \Abs{\BE}^2.
}
\end{equation}