divergence

Geometric calculus relationships to differential forms, and vector calculus identities.

November 12, 2023 math and physics play , , , , , ,

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Motivation.

I was asked about the geometric algebra equivalents of some of the vector calculus identities from [1]. I’ll call the specific page of those calculus notes “the article”. The article includes identities like
\begin{equation}\label{eqn:formAndCurl:20}
\begin{aligned}
\spacegrad (f g) &= f \spacegrad g + g \spacegrad f \\
\spacegrad \cross (f \BF) &= f \spacegrad \cross \BF + (\spacegrad f) \cross \BF \\
\spacegrad \cdot (f \BF) &= f \spacegrad \cdot \BF + (\spacegrad f) \cdot \BF \\
\spacegrad \cdot \lr{ \BF \cross \BG } &= \BG \cdot \lr{ \spacegrad \cross \BF } – \BF \cdot \lr{ \spacegrad \cross \BG },
\end{aligned}
\end{equation}
but the point of these particular lecture notes is the interface between traditional Gibbs vector calculus and differential forms. That’s a much bigger topic, and perhaps not what I was actually being asked about. It is, however, an interesting topic, so let’s explore it.

Comparisons.

The article identifies the cross product representation of the curl \( \spacegrad \cross \BF \) as the equivalent to the exterior derivative of a one form (which has been mapped to a vector function.) In geometric algebra, this isn’t the identification we would use. Instead we should identify the “bivector curl” \( \spacegrad \wedge \BF \) as the logical equivalent of the exterior derivative of that one form, and in general identify \( \spacegrad \wedge A_k \) as the exterior derivative of a k-form (k-blade). In my notes to follow, the wedge of the gradient with a function, will be called the curl of that function, even if we are operating in \(\mathbb{R}^3\) where the cross product is defined.

The starting place of the article was to define a one form and it’s exterior derivative was essentially as follows

Definition 1.1: The exterior derivative of a one form.

Let \( f : \mathbb{R}^N \rightarrow \mathbb{R} \) be a zero form. It’s exterior derivative is
\begin{equation*}
df = \sum_i dx_i \PD{x_i}{f}.
\end{equation*}

I’ve stated that the GA equivalent of the exterior derivative was a curl \( \spacegrad \wedge A \), and this doesn’t look anything curl like, so right away, we have some trouble to deal with. To resolve that trouble, let’s step back to the gradient, which we haven’t defined yet. In the article, the gradient (of a scalar function) was defined as a coordinate triplet
\begin{equation}\label{eqn:formAndCurl:60}
\spacegrad \Bf = \lr{ \PD{x}{f}, \PD{y}{f}, \PD{z}{f} }.
\end{equation}
In GA we don’t like representations where the basis vectors are implicit, so we’d prefer to define

Definition 1.2: The gradient of a function.

We define the gradient of multivector \( f(x_1, x_2, \cdots, x_N) \), and denote it by \( \spacegrad f \)
\begin{equation*}
\spacegrad f = \sum_{i=1}^N \Be_i \PD{x_i}{f},
\end{equation*}
where \( \setlr{ \Be_1, \cdots \Be_N } \) is an orthonormal basis for \(\mathbb{R}^N\).

Unlike the article, we do not restrict \( f \) to be a scalar function, since we do not have a problem with a vector valued operator directly multiplying a vector or any product of vectors. Instead \( f \) can be a multivector function, with scalar, vector, bivector, trivector, … components, and we define the gradient the same way.

In order to define the curl of a k-blade, we need a reminder of how we define the wedge of a vector with a k-blade. Recall that this is how we generally define the wedge between two blades.

Definition 1.3:

Let \( A_r \) be a r-blade, and \( B_s \) a s-blade. The wedge of \( A_r \) with \( B_s \) is
\begin{equation}\label{eqn:formAndCurl:120}
A_r \wedge B_s = \gpgrade{A_r B_s}{r+s}.
\end{equation}

In particular, if \( \Ba \) is a vector, then the wedge with an s-blade \( B_s \) is
\begin{equation}\label{eqn:formAndCurl:140}
\Ba \wedge B_s = \gpgrade{\Ba B_s}{s+1},
\end{equation}
which is just the \( s+1 \) grade selection of their product. Furthermore, if \( f \) is a scalar, then
\begin{equation}\label{eqn:formAndCurl:160}
\Ba \wedge f = \gpgrade{\Ba f}{1} = \Ba f.
\end{equation}
We can now state the curl of a k-blade

Definition 1.4: Curl of a k-blade.

Let \( A_k \) be a k-blade. We define the curl of a k-blade as the wedge product of the gradient with that k-blade, designated
\begin{equation*}
\spacegrad \wedge A_k.
\end{equation*}

Observe, given our generalized wedge product definition above, that the curl of a scalar function \( f \), is in fact just the gradient of that function
\begin{equation}\label{eqn:formAndCurl:200}
\spacegrad \wedge f = \spacegrad f = \sum_i \Be_i \PD{x_i}{f}.
\end{equation}
This has exactly the structure of the exterior derivative of a one form, as stated in “Definition: The exterior derivative of a one form”, but we have replaced \( dx_i \) with a basis vector \( \Be_i \).

Definition 1.5: Exterior derivative of a one-form.

Let \( \omega = f_i dx_i \) be a one-form. The exterior derivative of \( d \omega \) is
\begin{equation*}
d\omega = \sum_i d( f_i ) \wedge dx_i.
\end{equation*}

Lemma 1.1: Exterior derivative of a one-form.

Let \( \omega = f_i dx_i \) be a one-form. The exterior derivative \( d \omega \) can be expanding into a Jacobian form
\begin{equation*}
d\omega
=
\sum_{i < j} \lr{
\PD{x_i}{f_j}
–
\PD{x_j}{f_i}
} dx_i \wedge dx_j.
\end{equation*}

Start proof:

\begin{equation}\label{eqn:formAndCurl:220}
\begin{aligned}
d\omega
&= \sum_j d( f_j dx_j ) \\
&= \sum_j d( f_j ) \wedge dx_j \\
&= \sum_j \lr{ \sum_i dx_i \PD{x_i}{f_j} } \wedge dx_j \\
&= \sum_{ji} \PD{x_i}{f_j} dx_i \wedge dx_j \\
&= \sum_{j \ne i} \PD{x_i}{f_j} dx_i \wedge dx_j \\
&=
\sum_{i < j} \PD{x_i}{f_j} dx_i \wedge dx_j
+
\sum_{j < i} \PD{x_i}{f_j} dx_i \wedge dx_j \\
&=
\sum_{i < j} \PD{x_i}{f_j} dx_i \wedge dx_j
+
\sum_{i < j} \PD{x_j}{f_i} dx_j \wedge dx_i \\
&=
\sum_{i < j} \lr{
\PD{x_i}{f_j}
–
\PD{x_j}{f_i}
} dx_i \wedge dx_j.
\end{aligned}
\end{equation}

End proof.

Lemma 1.2: Curl of a vector.

Let \( \Bf = \sum_i \Be_i f_i \in \mathbb{R}^N \) be a vector. The curl of \( \Bf \) has a Jacobian structure
\begin{equation*}
\spacegrad \wedge \Bf
=
\sum_{i < j}
\lr{ \PD{x_i}{f_j} – \PD{x_j}{f_i} }
\lr{ \Be_i \wedge \Be_j }
.
\end{equation*}

Start proof:

The antisymmetry of the wedges of differentials in the exterior derivative and the curl clearly has a one to one correspondence. Let’s show this explicitly by expansion
\begin{equation}\label{eqn:formAndCurl:240}
\begin{aligned}
\spacegrad \wedge \Bf
&=
\sum_{ij} \lr{ \Be_i \PD{x_i}{} } \wedge \lr{ \Be_j f_j } \\
&=
\sum_{ij} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j} \\
&=
\sum_{i \ne j} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j} \\
&=
\sum_{i < j} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j}
+
\sum_{j < i} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j} \\
&=
\sum_{i < j} \lr{ \Be_i \wedge \Be_j } \PD{x_i}{f_j}
+
\sum_{i < j} \lr{ \Be_j \wedge \Be_i } \PD{x_j}{f_i} \\
&=
\sum_{i < j} \lr{ \Be_i \wedge \Be_j } \lr{ \PD{x_i}{f_j} – \PD{x_j}{f_i} }.
\end{aligned}
\end{equation}

End proof.

If we are translating from differential forms, again, we see that we simply replace any differentials \( dx_i \) with the basis vectors \( \Be_i \) (at least for the zero-form and one-form cases, which is all that we have looked at here.)

Note that in differential forms, we often assume that there is an implicit wedge product between any different one form elements, writing
\begin{equation}\label{eqn:formAndCurl:260}
dx_1 \wedge dx_2 = dx_1 dx_2.
\end{equation}
This works out fine when we map differentials to basis vectors, since
\begin{equation}\label{eqn:formAndCurl:280}
\Be_1 \Be_2 =
\Be_1 \cdot \Be_2
+
\Be_1 \wedge \Be_2
=
\Be_1 \wedge \Be_2.
\end{equation}
However, we have to be more careful in GA when using indexed expressions, since
\begin{equation}\label{eqn:formAndCurl:300}
\Be_i \Be_j = \Be_i \cdot \Be_j + \Be_i \wedge \Be_j.
\end{equation}
The dot product portion of the RHS is only zero if \( i \ne j \).

Now let’s look at the equivalence between the exterior derivative of a two-form with the curl.

Definition 1.6: Exterior derivative of a two-form.

Let \( \eta = \sum_{ij} f_{ij} dx_i \wedge dx_j \) be a two-form. The exterior derivative of \( \eta \) is
\begin{equation*}
d\eta =
\sum_{ij} d( f_{ij} ) \wedge dx_i \wedge dx_j.
\end{equation*}

Lemma 1.3: Exterior derivative of a two-form.

Let \( \eta = \sum_{ij} f_{ij} dx_i \wedge dx_j \) be a two-form. The exterior derivative of \( \eta \) can be expanded as
\begin{equation*}
d \eta
=
\sum_{i,j,k} \PD{x_k}{f_{ij}} dx_i \wedge dx_j \wedge dx_k.
\end{equation*}

Start proof:

The exterior derivative of \( \eta \) is
\begin{equation}\label{eqn:formAndCurl:340}
\begin{aligned}
d \eta
&=
\sum_{i,j} d( f_{ij} dx_i \wedge dx_j ) \\
&=
\sum_{i,j,k} \lr{ \PD{x_k}{f_{ij}} dx_k } \wedge dx_i \wedge dx_j \\
&=
\sum_{i,j,k} \PD{x_k}{f_{ij}} dx_i \wedge dx_j \wedge dx_k.
\end{aligned}
\end{equation}

End proof.

Let’s compare that to the curl of a bivector valued function.

Lemma 1.4: Curl of a 2-blade.

Let \( B = \sum_{i \ne j} f_{ij} \Be_i \wedge \Be_j \) be a 2-blade. The curl of \( B \) is
\begin{equation*}
\spacegrad \wedge B
=
\sum_{i,j,k} \PD{x_k}{f_{ij}} \Be_i \wedge \Be_j \wedge \Be_k.
\end{equation*}

Start proof:

\begin{equation}\label{eqn:formAndCurl:380}
\begin{aligned}
\spacegrad \wedge B
&=
\lr{ \sum_k \Be_k \PD{x_k}{} } \wedge \lr{ \sum_{i \ne j} f_{ij} \Be_i \wedge \Be_j } \\
&=
\sum_{k, i \ne j} \PD{x_k}{f_{ij}} \Be_k \wedge \Be_i \wedge \Be_j \\
&=
\sum_{i,j,k} \PD{x_k}{f_{ij}} \Be_i \wedge \Be_j \wedge \Be_k.
\end{aligned}
\end{equation}

End proof.

Again, we see an exact correspondence with the exterior derivative \( d \eta \) of a two-form, and the curl \( \spacegrad \wedge B \), of a 2-blade.

The article established a coorespondence between the exterior derivative of a two form over \(\mathbb{R}^3\) to the divergence. The way we would express this in GA (also for \R{3}) is to write
\begin{equation}\label{eqn:formAndCurl:400}
B = I \Bb,
\end{equation}
where \( I = \Be_1 \Be_2 \Be_3 \) is the \(\mathbb{R}^3\) pseudoscalar (a “unit” trivector.) Forming the curl of \( B \) we have
\begin{equation}\label{eqn:formAndCurl:420}
\begin{aligned}
\spacegrad \wedge B
&= \gpgradethree{ \spacegrad B } \\
&= \gpgradethree{ \spacegrad I \Bb } \\
&= \gpgradethree{ I (\spacegrad \Bb) } \\
&= \gpgradethree{ I (\spacegrad \cdot \Bb + \spacegrad \wedge \Bb) } \\
&= I (\spacegrad \cdot \Bb).
\end{aligned}
\end{equation}
The equivalence relationships that we have developed must then imply that the differential forms representation of this relationship is
\begin{equation}\label{eqn:formAndCurl:440}
d B = dx_1 \wedge dx_2 \wedge dx_3 (\spacegrad \cdot \Bb)
= dx \wedge dy \wedge dz \lr{ \PD{x}{b_1} + \PD{y}{b_2} + \PD{z}{b_3} },
\end{equation}
as defined in the article.

Here is the GA equivalent of Lemma 4.4.10 from the article

Lemma 1.5: Repeated curl identities.

Let \( A \) be a smooth k-blade, then
\begin{equation*}
\spacegrad \wedge \lr{ \spacegrad \wedge A } = 0.
\end{equation*}
For \(\mathbb{R}^3\), this result, for a scalar function \( f \), and a vector function \( \Bf \), in terms of the cross product, as
\begin{equation}\label{eqn:formAndCurl:560}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad f } &= 0 \\
\spacegrad \cdot \lr{ \spacegrad \cross \Bf } &= 0.
\end{aligned}
\end{equation}

It shouldn’t be surprising that this is the equivalent of \( d^2 A = 0 \) from differential forms. Let’s prove this, first considering the 0-blade case

Start proof:

\begin{equation}\label{eqn:formAndCurl:480}
\begin{aligned}
\spacegrad \wedge \lr{ \spacegrad \wedge A }
&=
\spacegrad \wedge \lr{ \spacegrad A } \\
&=
\sum_{ij} \Be_i \wedge \Be_j \frac{\partial^2 A}{\partial x_i \partial x_j} \\
&= 0.
\end{aligned}
\end{equation}
The smooth criteria of for the function \( A \) is assumed to imply that we have equality of mixed partials, and since this is a sum of an antisymmetric term with respect to indexes \( i, j \) (the wedge) and a symmetric term in indexes \( i, j \) (the partials), we have zero overall.

Now consider a k-blade \( A, k > 0 \). Expanding the gradients, we have
\begin{equation}\label{eqn:formAndCurl:500}
\spacegrad \wedge \lr{ \spacegrad \wedge A }
=
\sum_{ij} \Be_i \wedge \Be_j \wedge \frac{\partial^2 A}{\partial x_i \partial x_j}.
\end{equation}
It may be obvious that this is zero for the same reasons as above (sum of product of symmetric and antisymmetric entities). We can, however, make it more obvious, at the cost of some hellish indexing, by expressing \( A \) in coordinate form. Let
\begin{equation}\label{eqn:formAndCurl:520}
A = \sum_{i_1, i_2, \cdots, i_k}
A_{i_1, i_2, \cdots, i_k} \Be_{i_1} \wedge \Be_{i_2} \wedge \cdots \wedge \Be_{i_k},
\end{equation}
then
\begin{equation}\label{eqn:formAndCurl:540}
\begin{aligned}
\spacegrad \wedge \lr{ \spacegrad \wedge A }
&=
\sum_{i,j,i_1, i_2, \cdots, i_k} \Be_i \wedge \Be_j \wedge \Be_{i_1} \wedge \Be_{i_2} \wedge \cdots \wedge \Be_{i_k}
\frac{\partial^2 }{\partial x_i \partial x_j} A_{i_1, i_2, \cdots, i_k} \\
&= 0.
\end{aligned}
\end{equation}
Now we clearly have a sum of an antisymmetric term (the wedges), and a symmetric term (assuming smooth \( A \) means that we have equality of mixed partials), so the sum is zero.

Finally, for the \(\mathbb{R}^3\) identities, we have
\begin{equation}\label{eqn:formAndCurl:580}
\begin{aligned}
\spacegrad \cross \lr{ \spacegrad f}
&=
-I \lr{ \spacegrad \wedge \lr{ \spacegrad f } }
&=
0,
\end{aligned}
\end{equation}
since \( \spacegrad \wedge \lr{ \spacegrad f } = 0 \). For a vector \( \Bf \), we have
\begin{equation}\label{eqn:formAndCurl:600}
\begin{aligned}
\spacegrad \cdot \lr{ \spacegrad \cross \Bf}
&=
\gpgradezero{
\spacegrad \lr{ \spacegrad \cross \Bf}
} \\
&=
\gpgradezero{
\spacegrad (-I) \lr{ \spacegrad \wedge \Bf}
} \\
&=
-\gpgradezero{
I \spacegrad \lr{ \spacegrad \wedge \Bf}
} \\
&=
–
I \spacegrad \wedge \lr{ \spacegrad \wedge \Bf} \\
&= 0,
\end{aligned}
\end{equation}
again, because \( \spacegrad \wedge \lr{ \spacegrad \wedge \Bf} = 0 \).

End proof.

Identities.

We have a number of chain rule identities in the article. Here is the GA equivalent of that, and its corollaries

Lemma 1.6: Chain rule identities.

Let \( f \) be a scalar function and \( A \) be a k-blade, then
\begin{equation*}
\spacegrad \lr{ f A } = \lr{ \spacegrad f } A + f \lr{ \spacegrad A }.
\end{equation*}
For \( A \) with grade \( k > 0 \), the grade \( k-1 \) and \( k+1 \) components of this product are
\begin{equation*}
\begin{aligned}
\spacegrad \cdot \lr{ f A } &= \lr{ \spacegrad f } \cdot A + f \lr{ \spacegrad \cdot A } \\
\spacegrad \wedge \lr{ f A } &= \lr{ \spacegrad f } \wedge A + f \lr{ \spacegrad \wedge A }.
\end{aligned}
\end{equation*}
For \(\mathbb{R}^3\), the wedge product relation above can be written in dual form as
\begin{equation*}
\spacegrad \cross \lr{ f A } = \lr{ \spacegrad f } \cross A + f \lr{ \spacegrad \cross A }.
\end{equation*}

Proving this is left to the reader.

We have some chain rule identities left in the article to verify and to find GA equivalents of. Before doing so, we need a couple miscellaneous identities relating triple cross products to wedge-dots.

Lemma 1.7: Triple cross products.

Let \( \Ba, \Bb, \Bc \) be vectors in \(\mathbb{R}^3\). Then
\begin{equation*}
\begin{aligned}
\Ba \cross \lr{ \Bb \cross \Bc } &= – \Ba \cdot \lr{ \Bb \wedge \Bc } \\
\lr{ \Ba \cross \Bb } \cross \Bc &= – \lr{ \Ba \wedge \Bb } \cdot \Bc.
\end{aligned}
\end{equation*}

Start proof:

\begin{equation}\label{eqn:formAndCurl:720}
\begin{aligned}
\Ba \cross \lr{ \Bb \cross \Bc }
&=
\gpgradeone{ -I \lr{ \Ba \wedge \lr{ \Bb \cross \Bc } } } \\
&=
\gpgradeone{ -I \lr{ \Ba \lr{ \Bb \cross \Bc } } } \\
&=
\gpgradeone{ (-I)^2 \lr{ \Ba \lr{ \Bb \wedge \Bc } } } \\
&=
-\Ba \cdot \lr{ \Bb \wedge \Bc },
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:formAndCurl:740}
\begin{aligned}
\lr{ \Ba \cross \Bb } \cross \Bc
&=
\gpgradeone{ -I \lr{ \Ba \cross \Bb } \wedge \Bc } \\
&=
\gpgradeone{ -I \lr{ \Ba \cross \Bb } \Bc } \\
&=
\gpgradeone{ (-I)^2 \lr{ \Ba \wedge \Bb } \Bc } \\
&=
– \lr{ \Ba \wedge \Bb } \cdot \Bc.
\end{aligned}
\end{equation}

End proof.

Next up is another chain rule identity

Lemma 1.8: Gradient of dot product.

If \( \Ba, \Bb \) are vectors, then
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb } =
\lr{ \Ba \cdot \spacegrad } \Bb
+
\lr{ \Bb \cdot \spacegrad } \Ba
+
\lr{ \spacegrad \wedge \Bb }
\cdot
\Ba
+
\lr{ \spacegrad \wedge \Ba }
\cdot
\Bb
\end{equation*}
For \(\mathbb{R}^3\), this can be written as
\begin{equation*}
\spacegrad \lr{ \Ba \cdot \Bb }
=
\lr{ \Ba \cdot \spacegrad } \Bb
+
\lr{ \Bb \cdot \spacegrad } \Ba
+
\Ba \cross \lr{ \spacegrad \cross \Bb }
+
\Bb \cross \lr{ \spacegrad \cross \Ba }
\end{equation*}

Start proof:

We will use \( \rspacegrad \) to indicate that the gradient operates on everything to the right, \( \lrspacegrad \) to indicate that the gradient operates bidirectionally, and \( \spacegrad’ A B’ \) to indicate that the gradient’s scope is limited to the ticked entity (just on \( B \) in this case.)
\begin{equation}\label{eqn:formAndCurl:760}
\begin{aligned}
\rspacegrad \lr{ \Ba \cdot \Bb }
&=
\gpgradeone{
\rspacegrad \lr{ \Ba \Bb – \Ba \wedge \Bb }
} \\
&=
\gpgradeone{
\spacegrad’ \Ba’ \Bb
+
\spacegrad’ \Ba \Bb’
}
– \rspacegrad \cdot \lr{ \Ba \wedge \Bb }
\\
&=
\lr{ \spacegrad \cdot \Ba} \Bb
+
\lr{ \spacegrad \wedge \Ba} \cdot \Bb
+
\gpgradeone{
– \Ba \spacegrad \Bb + 2 \lr{ \Ba \cdot \spacegrad } \Bb
}
– \spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
– \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ }
\\
&=
\lr{ \spacegrad \cdot \Ba} \Bb
+
\lr{ \spacegrad \wedge \Ba} \cdot \Bb
–
\Ba \lr{ \spacegrad \cdot \Bb }
–
\Ba \cdot \lr{ \spacegrad \wedge \Bb }
+ 2 \lr{ \Ba \cdot \spacegrad } \Bb
– \spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
– \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ }.
\end{aligned}
\end{equation}
We are running out of room, and have not had any cancellation yet, so let’s expand those last two terms separately
\begin{equation}\label{eqn:formAndCurl:780}
\begin{aligned}
– \spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
– \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ }
&=
– \lr{ \spacegrad’ \cdot \Ba’ } \Bb
+ \lr{ \spacegrad’ \cdot \Bb } \Ba’
– \lr{ \spacegrad’ \cdot \Ba } \Bb’
+ \lr{ \spacegrad’ \cdot \Bb’ } \Ba
\\
&=
– \lr{ \spacegrad \cdot \Ba } \Bb
+ \lr{ \Bb \cdot \spacegrad } \Ba
– \lr{ \Ba \cdot \spacegrad } \Bb
+ \lr{ \spacegrad \cdot \Bb } \Ba.
\end{aligned}
\end{equation}
Now we can cancel some terms, leaving
\begin{equation}\label{eqn:formAndCurl:800}
\begin{aligned}
\rspacegrad \lr{ \Ba \cdot \Bb }
&=
\lr{ \spacegrad \wedge \Ba} \cdot \Bb
–
\Ba \cdot \lr{ \spacegrad \wedge \Bb }
+ \lr{ \Ba \cdot \spacegrad } \Bb
+ \lr{ \Bb \cdot \spacegrad } \Ba.
\end{aligned}
\end{equation}
After adjustment of the order and sign of the second term, we see that this is the result we wanted. To show the \(\mathbb{R}^3\) formulation, we have only apply “Lemma: Triple cross products”.

End proof.

Lemma 1.9: Divergence of a bivector.

Let \( \Ba, \Bb \in \mathbb{R}^N \) be vectors. The divergence of their wedge can be written
\begin{equation*}
\spacegrad \cdot \lr{ \Ba \wedge \Bb }
=
\Bb \lr{ \spacegrad \cdot \Ba }
– \Ba \lr{ \spacegrad \cdot \Bb }
– \lr{ \Bb \cdot \spacegrad } \Ba
+ \lr{ \Ba \cdot \spacegrad } \Bb.
\end{equation*}
For \(\mathbb{R}^3\), this can also be written in triple cross product form
\begin{equation*}
\spacegrad \cdot \lr{ \Ba \wedge \Bb }
=
-\spacegrad \cross \lr{ \Ba \cross \Bb }.
\end{equation*}

Start proof:

\begin{equation}\label{eqn:formAndCurl:860}
\begin{aligned}
\rspacegrad \cdot \lr{ \Ba \wedge \Bb }
&=
\spacegrad’ \cdot \lr{ \Ba’ \wedge \Bb }
+ \spacegrad’ \cdot \lr{ \Ba \wedge \Bb’ } \\
&=
\lr{ \spacegrad’ \cdot \Ba’ } \Bb
– \lr{ \spacegrad’ \cdot \Bb } \Ba’
+ \lr{ \spacegrad’ \cdot \Ba } \Bb’
– \lr{ \spacegrad’ \cdot \Bb’ } \Ba
\\
&=
\lr{ \spacegrad \cdot \Ba } \Bb
– \lr{ \Bb \cdot \spacegrad } \Ba
+ \lr{ \Ba \cdot \spacegrad } \Bb
– \lr{ \spacegrad \cdot \Bb } \Ba.
\end{aligned}
\end{equation}
For the \(\mathbb{R}^3\) part of the story, we have
\begin{equation}\label{eqn:formAndCurl:870}
\begin{aligned}
\spacegrad \cross \lr{ \Ba \cross \Bb }
&=
\gpgradeone{
-I \lr{ \spacegrad \wedge \lr{ \Ba \cross \Bb } }
} \\
&=
\gpgradeone{
-I \spacegrad \lr{ \Ba \cross \Bb }
} \\
&=
\gpgradeone{
(-I)^2 \spacegrad \lr{ \Ba \wedge \Bb }
} \\
&=
–
\spacegrad \cdot \lr{ \Ba \wedge \Bb }
\end{aligned}
\end{equation}

End proof.

We have just one identity left in the article to find the GA equivalent of, but will split that into two logical pieces.

Lemma 1.10: Dual of triple wedge.

If \( \Ba, \Bb, \Bc \in \mathbb{R}^3 \) are vectors, then
\begin{equation*}
\Ba \cdot \lr{ \Bb \cross \Bc } = -I \lr{ \Ba \wedge \Bb \wedge \Bc }.
\end{equation*}

Start proof:

\begin{equation}\label{eqn:formAndCurl:680}
\begin{aligned}
\Ba \cdot \lr{ \Bb \cross \Bc }
&=
\gpgradezero{
\Ba \lr{ \Bb \cross \Bc }
} \\
&=
\gpgradezero{
\Ba (-I) \lr{ \Bb \wedge \Bc }
} \\
&=
\gpgradezero{
-I \lr{
\Ba \cdot \lr{ \Bb \wedge \Bc }
+
\Ba \wedge \lr{ \Bb \wedge \Bc }
}
} \\
&=
\gpgradezero{
-I \lr{ \Ba \wedge \lr{ \Bb \wedge \Bc } }
} \\
&=
-I \lr{ \Ba \wedge \lr{ \Bb \wedge \Bc } }.
\end{aligned}
\end{equation}

End proof.

Lemma 1.11: Curl of a wedge of gradients (divergence of a gradient cross products.)

Let \( f, g, h \) be smooth functions with smooth derivatives. Then
\begin{equation*}
\spacegrad \wedge \lr{ f \lr{ \spacegrad g \wedge \spacegrad h } }
=
\spacegrad f
\wedge
\spacegrad g
\wedge
\spacegrad h.
\end{equation*}
For \(\mathbb{R}^3\) this can be written as
\begin{equation*}
\spacegrad \cdot \lr{ f \lr{ \spacegrad g \cross \spacegrad h } }
=
\spacegrad f
\cdot
\lr{
\spacegrad g
\cross
\spacegrad h
}.
\end{equation*}

Start proof:

The GA identity follows by chain rule and application of “Lemma: Repeated curl identities”.
\begin{equation}\label{eqn:formAndCurl:910}
\begin{aligned}
\spacegrad \wedge \lr{ f \lr{ \spacegrad g \wedge \spacegrad h } }
&=
\spacegrad f \wedge \lr{ \spacegrad g \wedge \spacegrad h }
+
f
\spacegrad \wedge \lr{ \spacegrad g \wedge \spacegrad h } \\
&=
\spacegrad f \wedge \spacegrad g \wedge \spacegrad h.
\end{aligned}
\end{equation}
The \(\mathbb{R}^3\) part of the lemma follows from “Lemma: Dual of triple wedge”, applied twice
\begin{equation}\label{eqn:formAndCurl:970}
\begin{aligned}
\spacegrad \cdot \lr{ (f \spacegrad g) \cross \spacegrad h }
&=
-I \lr{ \spacegrad \wedge (f \spacegrad g) \wedge \spacegrad h } \\
&=
-I \lr{ \spacegrad f \wedge \spacegrad g \wedge \spacegrad h } \\
&=
\spacegrad f \cdot \lr{ \spacegrad g \cross \spacegrad h}.
\end{aligned}
\end{equation}

End proof.

Lemma 1.12: Curl of a bivector.

Let \( \Ba, \Bb \) be vectors. The curl of their wedge is
\begin{equation*}
\spacegrad \wedge \lr{ \Ba \wedge \Bb } = \Bb \wedge \lr{ \spacegrad \wedge \Ba } – \Ba \wedge \lr{ \spacegrad \wedge \Bb }
\end{equation*}
For \(\mathbb{R}^3\), this can be expressed as the divergence of a cross product
\begin{equation*}
\spacegrad \cdot \lr{ \Ba \cross \Bb } = \Bb \cdot \lr{ \spacegrad \cross \Ba } – \Ba \cdot \lr{ \spacegrad \cross \Bb }
\end{equation*}

Start proof:

The GA case is a trivial chain rule application
\begin{equation}\label{eqn:formAndCurl:950}
\begin{aligned}
\rspacegrad \wedge \lr{ \Ba \wedge \Bb }
&=
\lr{ \spacegrad’ \wedge \Ba’} \wedge \Bb
+
\lr{ \spacegrad’ \wedge \Ba } \wedge \Bb’ \\
&= \Bb \wedge \lr{ \spacegrad \wedge \Ba } – \Ba \wedge \lr{ \spacegrad \wedge \Bb }.
\end{aligned}
\end{equation}
The \(\mathbb{R}^3\) case, is less obvious by inspection, but follows from “Lemma: Dual of triple wedge”.

End proof.

Summary.

We found that we have an isomorphism between the exterior derivative of differential forms and the curl operation of geometric algebra, as follows
\begin{equation}\label{eqn:formAndCurl:990}
\begin{aligned}
df &\rightleftharpoons \spacegrad \wedge f \\
dx_i &\rightleftharpoons \Be_i.
\end{aligned}
\end{equation}
We didn’t look at how the Hodge dual translates to GA duality (pseudoscalar multiplication.) The divergence relationship between the exterior derivative of an \(\mathbb{R}^3\) two-form really requires that formalism, and has only been examined in a cursory fashion.

We also translated a number of vector and gradient identities from conventional vector algebra (i.e.: using cross products) and wedge product equivalents of the same. The GA identities are often simpler, and in some cases, provide nice mechanisms to derive the conventional identities that would be more cumbersome to determine without the GA toolbox.

References

[1] Vincent Bouchard. Math 215: Calculus iv: 4.4 the exterior derivative and vector calculus, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_exterior_vector.html. [Online; accessed 11-November-2023].

Gauge freedom and four-potentials in the STA form of Maxwell’s equation.

March 27, 2022 math and physics play , , , , , , , , , , , , , , , , , , , , ,

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

Motivation.

In a recent video on the tensor structure of Maxwell’s equation, I made a little side trip down the road of potential solutions and gauge transformations. I thought that was worth writing up in text form.

The initial point of that side trip was just to point out that the Faraday tensor can be expressed in terms of four potential coordinates
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:20}
F_{\mu\nu} = \partial_\mu A_\nu – \partial_\nu A_\mu,
\end{equation}
but before I got there I tried to motivate this. In this post, I’ll outline the same ideas.

STA representation of Maxwell’s equation.

We’d gone through the work to show that Maxwell’s equation has the STA form
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:40}
\grad F = J.
\end{equation}
This is a deceptively compact representation, as it requires all of the following definitions
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:60}
\grad = \gamma^\mu \partial_\mu = \gamma_\mu \partial^\mu,
\end{equation}
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:80}
\partial_\mu = \PD{x^\mu}{},
\end{equation}
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:100}
\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu,
\end{equation}
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:160}
\gamma_\mu \cdot \gamma_\nu = g_{\mu\nu},
\end{equation}
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:120}
\begin{aligned}
F
&= \BE + I c \BB \\
&= -E^k \gamma^k \gamma^0 – \inv{2} c B^r \gamma^s \gamma^t \epsilon^{r s t} \\
&= \inv{2} \gamma^{\mu} \wedge \gamma^{\nu} F_{\mu\nu},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:140}
\begin{aligned}
J &= \gamma_\mu J^\mu \\
J^\mu &= \frac{\rho}{\epsilon} \gamma_0 + \eta (\BJ \cdot \Be_k).
\end{aligned}
\end{equation}

Four-potentials in the STA representation.

In order to find the tensor form of Maxwell’s equation (starting from the STA representation), we first split the equation into two, since
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:180}
\grad F = \grad \cdot F + \grad \wedge F = J.
\end{equation}
The dot product is a four-vector, the wedge term is a trivector, and the current is a four-vector, so we have one grade-1 equation and one grade-3 equation
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:200}
\begin{aligned}
\grad \cdot F &= J \\
\grad \wedge F &= 0.
\end{aligned}
\end{equation}
The potential comes into the mix, since the curl equation above means that \( F \) necessarily can be written as the curl of some four-vector
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:220}
F = \grad \wedge A.
\end{equation}
One justification of this is that \( a \wedge (a \wedge b) = 0 \), for any vectors \( a, b \). Expanding such a double-curl out in coordinates is also worthwhile
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:240}
\begin{aligned}
\grad \wedge \lr{ \grad \wedge A }
&=
\lr{ \gamma_\mu \partial^\mu }
\wedge
\lr{ \gamma_\nu \partial^\nu }
\wedge
A \\
&=
\gamma^\mu \wedge \gamma^\nu \wedge \lr{ \partial_\mu \partial_\nu A }.
\end{aligned}
\end{equation}
Provided we have equality of mixed partials, this is a product of an antisymmetric factor and a symmetric factor, so the full sum is zero.

Things get interesting if one imposes a \( \grad \cdot A = \partial_\mu A^\mu = 0 \) constraint on the potential. If we do so, then
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:260}
\grad F = \grad^2 A = J.
\end{equation}
Observe that \( \grad^2 \) is the wave equation operator (often written as a square-box symbol.) That is
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:280}
\begin{aligned}
\grad^2
&= \partial^\mu \partial_\mu \\
&= \partial_0 \partial_0
– \partial_1 \partial_1
– \partial_2 \partial_2
– \partial_3 \partial_3 \\
&= \inv{c^2} \PDSq{t}{} – \spacegrad^2.
\end{aligned}
\end{equation}
This is also an operator for which the Green’s function is well known ([1]), which means that we can immediately write the solutions
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:300}
A(x) = \int G(x,x’) J(x’) d^4 x’.
\end{equation}
However, we have no a-priori guarantee that such a solution has zero divergence. We can fix that by making a gauge transformation of the form
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:320}
A \rightarrow A – \grad \chi.
\end{equation}
Observe that such a transformation does not change the electromagnetic field
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:340}
F = \grad \wedge A \rightarrow \grad \wedge \lr{ A – \grad \chi },
\end{equation}
since
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:360}
\grad \wedge \grad \chi = 0,
\end{equation}
(also by equality of mixed partials.) Suppose that \( \tilde{A} \) is a solution of \( \grad^2 \tilde{A} = J \), and \( \tilde{A} = A + \grad \chi \), where \( A \) is a zero divergence field to be determined, then
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:380}
\grad \cdot \tilde{A}
=
\grad \cdot A + \grad^2 \chi,
\end{equation}
or
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:400}
\grad^2 \chi = \grad \cdot \tilde{A}.
\end{equation}
So if \( \tilde{A} \) does not have zero divergence, we can find a \( \chi \)
\begin{equation}\label{eqn:gaugeFreedomAndPotentialsMaxwell:420}
\chi(x) = \int G(x,x’) \grad’ \cdot \tilde{A}(x’) d^4 x’,
\end{equation}
so that \( A = \tilde{A} – \grad \chi \) does have zero divergence.

References

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

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

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

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

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 , , , , , , , , , , , , , , , , ,

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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.