A Hamiltonian model for the macroscopic Maxwell equations using exterior calculus

Maxwell’s equations may be written in two equivalent formats:

$$\begin{aligned} \nabla\cdot\bm{D}&=4\pi\rho_{f},\\ \nabla\cdot\bm{B}&=0,\\ -c\nabla\times\bm{E}&=\frac{\partial\bm{B}}{\partial t},\\ c\nabla\times\bm{H}&=\frac{\partial\bm{D}}{\partial t}+4\pi\bm{J}_{f},\end{aligned}\qquad\text{or}\qquad\begin{aligned} \nabla\cdot\bm{E}&=4\pi\rho\\ \nabla\cdot\bm{B}&=0\\ -c\nabla\times\bm{E}&=\frac{\partial\bm{B}}{\partial t}\\ c\nabla\times\bm{B}&=\frac{\partial\bm{E}}{\partial t}+4\pi\bm{J}.\end{aligned}$$

(1)

The left column is frequently called the macroscopic Maxwell equations, and the right column are just the standard Maxwell equations. One relates the displacement field, $\bm{D}$, and the magnetic field intensity, $\bm{H}$, with the electric field, $\bm{E}$, and magnetic field, $\bm{B}$, by

$$\bm{D}=\bm{E}+4\pi\bm{P}\quad\text{and}\quad\bm{H}=\bm{B}-4\pi\bm{M}$$

(2)

where $\bm{P}$ is the polarization and $\bm{M}$ is the magnetization. Moreover, $\rho_{f}$ and $\bm{J}_{f}$, the free charge and free current densities respectively, are related to $\rho$ and $\bm{J}$ by

$$\rho=\rho_{b}+\rho_{f}\quad\text{and}\quad\bm{J}=\bm{J}_{b}+\bm{J}_{f}$$

(3)

where

$$\rho_{b}=-\nabla\cdot\bm{P}\quad\text{and}\quad\bm{J}_{b}=\bm{J}_{p}+\bm{J}_{m}=\frac{\partial\bm{P}}{\partial t}+c\nabla\times\bm{M}.$$

(4)

It is straightforward to show that the two forms of Maxwell’s equations are equivalent using these definitions. The utility of the macroscopic Maxwell equations is that the terms which couple Maxwell’s equations to a matter model, $\rho_{f}$ and $\bm{J}_{f}$, are prescribed by freely propagating charged particles rather than the, possibly complicated, bound charges and currents in the medium. These microscopic features bound to the medium are instead encoded in constitutive relations for $\bm{P}$ and $\bm{M}$.

This paper is concerned with combining two distinct modeling frameworks to study this system. First, we translate this system into the language of split exterior calculus [6]. Second, we translate the Hamiltonian structure of this model, discovered in [14], into the language of split exterior calculus. We shall see that this approach helps to elucidate the rich structure of these equations.

The modeling framework employed in this paper utilizing the double de Rham complex is inspired by the split exterior calculus framework of [6]. Discussion of the differential geometric framework is kept to a minimum here, and the reader is referred to the aforementioned split exterior calculus paper and to refereces such as [7] and [12]. All notation used herein will either be standard as found in the previous references, or shall use notation defined below.

Let $(\Omega,g)$ be a Riemannian manifold of dimension $n$. Throughout this paper, we assume either periodic, or homogeneous boundary conditions. Let $\{(\Lambda^{k},\mathsf{d}_{k})\}_{k=0}^{n}$ be the vector spaces of differential forms on $\Omega$. We may define a second complex, $\{(\tilde{\Lambda}^{k},\tilde{\mathsf{d}}_{k})\}_{k=0}^{n}$, called the complex of twisted differential forms. This dual complex differs from the first in that twisted forms change sign under orientation changing transformations. The two complexes are related to each other through the Hodge star operator. Diagrammatically, this is given by

(5)

While the primal de Rham complex of straight differential forms, $\{\Lambda^{k}\}_{k=0}^{n}$, is a familiar object in differential geometry, the dual complex of twisted differential forms, $\{\tilde{\Lambda}^{k}\}_{k=0}^{n}$ would benefit from further discussion. Let $\{\partial_{i}\}_{i=1}^{n}$ be a coordinate system on the $T\Omega$, the tangent space of $\Omega$. We define an operator $o(\bm{\partial})$, which encodes the orientation of the manifold, such that it transforms as

$$o(T\bm{\partial})=o(T\partial_{1},\ldots,T\partial_{n})=\text{sign}(\det(T))o(\partial_{1},\ldots,\partial_{n})=\text{sign}(\det(T))o(\bm{\partial}).$$

(6)

We may consider twisted differential forms to be straight forms which have been multiplied by this operator. Because the orientation rarely plays a role in the algebraic manipulation of differential forms, for conciseness, we suppress any further use of this operator to the background.

We can see that the sign of a twisted form depends on the parity of the coordinate system of the ambient space. Therefore, integration over a set $U\subset\Omega$ of a twisted $n$-form will return a value which is independent of the orientation of space. Hence, densities are modeled as twisted forms. As an example, the volume form, that $n$-form such that $\bm{\mathrm{{vol}}}(\partial_{1},\ldots,\partial_{n})=1$ for any orthonormal frame $\{\partial_{i}\}_{i=1}^{n}$, is a twisted form. We shall typically denote a straight $k$-form as $\omega^{k}$ and a twisted $k$-form as $\tilde{\omega}^{k}$.

An important feature of the split exterior calculus is two distinct notions of duality on the double de Rham complex. First, we have the standard $L^{2}$ inner product, $(\cdot,\cdot):V^{k}\times V^{k}\to\mathbb{R}$, which is defined

$$(\omega^{k},\eta^{k})=\int_{\Omega}g_{x}(\omega^{k},\eta^{k})\bm{\mathrm{{vol}}}^{n}$$

(7)

where $g_{x}$ is the pointwise inner product on $k$-forms induced by the Riemannian metric. The second notion of duality is Poincaré duality, $\langle\cdot,\cdot\rangle:V^{k}\times\tilde{V}^{n-k}\to\mathbb{R}$, which is defined

$$\left\langle\omega^{k},\tilde{\eta}^{n-k}\right\rangle=\int_{\Omega}\omega^{k}\wedge\tilde{\eta}^{n-k}.$$

(8)

The $L^{2}$ inner product, because of its dependence on the Riemannian metric and volume form, is a metric dependent quantity. On the other hand, the Poincaré duality pairing, built from the wedge product structure alone, is purely topological. Moreover, as both duality pairings are expressed as an integral of a twisted $n$-form, they are independent of the orientation of the coordinate system.

Finally, the Hodge star operator $\star:V^{k}\to\tilde{V}^{n-k}$ is defined such that

$$\left(\omega^{k},\eta^{k}\right)=\left\langle\omega^{k},\star\eta^{k}\right\rangle.$$

(9)

Note that the Hodge star is not a single operator, but rather a family of operators. A more precise notation might be $\star_{n-k,k}:V^{k}\to\tilde{V}^{n-k}$, however we opt for the more concise notation where confusion is unlikely. These notions of duality are essential for discussing the Hamiltonian structure of the macroscopic Maxwell equations.

We denote the space of vector fields on $\Omega$ by $\mathfrak{X}$. Vector fields and $1$-forms are isomorphic to each other. The index lowering operator or flat operator $(\cdot)^{\flat}:\mathfrak{X}\to\Lambda^{1}$ is defined by

$$v^{1}=g(\cdot,V)=g_{ij}V^{j}\mathsf{d}x^{i}:=V^{\flat}.$$

(10)

The inverse operation is called the index raising operator or sharp operator $(\cdot)^{\sharp}:\Lambda^{1}\to\mathfrak{X}$:

$$V=g^{*}(\cdot,v^{1})=g^{ij}v^{1}_{j}\frac{\partial}{\partial x^{i}}:=(v^{1})^{\sharp}$$

(11)

where $g^{*}$ is the dual metric on the covectors. We may likewise define an isomorphism between vector fields and twisted $(n-1)$-forms. We define $\textbf{i}_{(\cdot)}\bm{\mathrm{{vol}}}^{n}:\mathfrak{X}\to\tilde{\Lambda}^{n-1}$ by

$$\tilde{v}^{n-1}=\textbf{i}_{V}\bm{\mathrm{{vol}}}^{n}=\sum_{i}V^{i}\sqrt{\det(g)}\mathsf{d}x^{1}\wedge\ldots\wedge\widehat{\mathsf{d}x^{i}}\wedge\ldots\wedge\mathsf{d}x^{n}$$

(12)

where the hat symbol means omission of “$\mathsf{d}x^{i}$” from the wedge product and $\textbf{i}_{V}\alpha$ is the interior product. It is possible to show that $\textbf{i}_{V}\bm{\mathrm{{vol}}}^{n}=\star V^{\flat}$. Hence, the inverse operation is given by

$$V=\left(\star\tilde{v}^{n-1}\right)^{\sharp}.$$

(13)

Many physically significant quantities are orientation dependent. The magnetic field and vorticity of a fluid are notable examples of so called pseudovectors. If $V$ is a pseudovector, then one can see that $\tilde{v}^{1}=V^{\flat}$ is twisted while $v^{n-1}=\textbf{i}_{V}\bm{\mathrm{{vol}}}^{n}$ is straight. Hence, pseudovectors are naturally identified with straight $(n-1)$-forms.

It is helpful to establish the following adjoint relationships between the isomorphisms we just defined. We define the $L^{2}$ inner product of vector fields to be

$$(W,V)=\int_{M}W\cdot V\bm{\mathrm{{vol}}}^{n}$$

where $W\cdot V=W^{i}g_{ij}V^{j}$.

We now turn our attention to vector calculus in $\mathbb{R}^{3}$. Exterior differentiation is inherently metric independent whereas it is well known that vector calculus operators such as the gradient, divergence, and curl depend on one’s coordinate system – often in a very complicated manner. By defining a differential form related to a vectorial quantity, one may obtain simpler expressions for their derivatives which disentangle the metric dependent and independent portions of the operation. The following definitions allow us to reconstruct the standard vector calculus operations from the exterior derivative.

If $f:\Omega\to\mathbb{R}$ be a scalar field on a Riemannian manifold, its gradient and exterior derivative are related to each other via

$$\mathsf{d}f=(\nabla f)^{\flat}\iff\nabla f=(\mathsf{d}f)^{\sharp}.$$

(15)

Let $V$ be a vector field and $v^{1}=V^{\flat}$. Then the curl of $V$ is defined by

$$\mathsf{d}v^{1}=\textbf{i}_{\nabla\times V}\bm{\mathrm{{vol}}}^{3}=\star(\nabla\times V)^{\flat}\iff\nabla\times V=(\star(\mathsf{d}v^{1}))^{\sharp}.$$

(16)

Letting $\tilde{v}^{2}=\textbf{i}_{V}\bm{\mathrm{{vol}}}^{3}$, then the divergence is defined to be

$$\mathsf{d}\tilde{v}^{2}=\mathsf{d}\textbf{i}_{V}\bm{\mathrm{{vol}}}^{3}=\pounds_{V}\bm{\mathrm{{vol}}}^{3}=(\nabla\cdot V)\bm{\mathrm{{vol}}}^{3}=\star(\nabla\cdot V)\iff\nabla\cdot V=\star\mathsf{d}\tilde{v}^{2}$$

(17)

where $\pounds_{V}$ is the Lie derivative. The divergence of a vector field physically corresponds with the the change in the volume element as it is dragged by that vector field. It is relatively straightforward to show that these formulas obtain the correct expressions for the gradient, divergence, and curl in Euclidean space and standard curvilinear coordinate systems (e.g. cylindrical and spherical coordinates).

We next establish a relationship between the wedge product of differential forms and the typical vector and scalar products in $\mathbb{R}^{3}$.

We now have the machinery to directly translate the standard vector calculus formulation of Maxwell’s equations into the language of split exterior calculus. Let

$$\bm{e}^{1}=\bm{E}^{\flat},\quad\bm{b}^{2}=\textbf{i}_{\bm{B}}\bm{\mathrm{{vol}}}^{3},\quad\tilde{\bm{d}}^{2}=\textbf{i}_{\bm{D}}\bm{\mathrm{{vol}}}^{3},\quad\tilde{\bm{h}}^{1}=\bm{H}^{\flat},\quad\tilde{\rho}=\star\rho,\quad\text{and}\quad\tilde{\bm{j}}=\textbf{i}_{\bm{J}}\bm{\mathrm{{vol}}}^{3}.$$

(19)

If we transform the equations by applying the appropriate isomorphisms as follows:

$$\begin{split}\star\left(c\nabla\times\bm{E}-\frac{\partial\bm{B}}{\partial t}\right)^{\flat}&=0\\ \star(\nabla\cdot\bm{B})&=0\end{split}\hskip 40.00006pt\begin{split}\star\left(c\nabla\times\bm{H}-\frac{\partial\bm{D}}{\partial t}-4\pi\bm{J}_{f}\right)^{\flat}&=0\\ \star(\nabla\cdot\bm{D}-4\pi\rho_{f})&=0.\end{split}$$

Then, we may write the equations in either of two different formats:

$$\begin{aligned} \mathsf{d}\tilde{\bm{d}}^{2}&=4\pi\tilde{\rho}^{3}_{f},\\ \mathsf{d}\bm{b}^{2}&=0,\\ -c\mathsf{d}\bm{e}^{1}&=\frac{\partial\bm{b}^{2}}{\partial t},\\ c\mathsf{d}\tilde{\bm{h}}^{1}&=\frac{\partial\tilde{\bm{d}}^{2}}{\partial t}+4\pi\tilde{\bm{j}}^{2}_{f},\end{aligned}\qquad\text{or}\qquad\begin{aligned} \mathsf{d}\star\bm{e}^{1}&=4\pi\tilde{\rho}^{3}\\ \mathsf{d}\bm{b}^{2}&=0\\ -c\mathsf{d}\bm{e}^{1}&=\frac{\partial\bm{b}^{2}}{\partial t}\\ c\mathsf{d}\star\bm{b}^{2}&=\frac{\partial\star\bm{e}^{1}}{\partial t}+4\pi\tilde{\bm{j}}^{2}.\end{aligned}$$

(20)

The displacement field twisted $2$-form and magnetic intensity field twisted $1$-form, $\tilde{\bm{d}}^{2}$ and $\tilde{\bm{h}}^{1}$ respectively, are related to the electric field $1$-form, $\bm{e}^{1}$, and magnetic field $2$-form, $\bm{b}^{2}$, by

$$\tilde{\bm{d}}^{2}=\tilde{\star}_{1}\bm{e}^{1}+4\pi\tilde{\bm{p}}^{2}\quad\text{and}\quad\tilde{\bm{h}}^{1}=\tilde{\star}_{2}\bm{b}^{2}-4\pi\tilde{\bm{m}}^{1}$$

(21)

where $\tilde{\bm{p}}^{2}=\textbf{i}_{\bm{P}}\bm{\mathrm{{vol}}}^{3}$ is the polarization twisted $2$-form, and $\tilde{\bm{m}}^{1}=\bm{M}^{\flat}$ is the magnetization twisted $1$-form. Moreover, the free charge twisted $3$-form and free current twisted $2$-form, $\tilde{\rho}^{3}_{f}$ and $\tilde{\bm{j}}^{2}_{f}$ respectively, are related to $\tilde{\rho}^{3}$ and $\tilde{\bm{j}}^{2}$ by

$$\tilde{\rho}^{3}=\tilde{\rho}^{3}_{f}+\tilde{\rho}^{3}_{b}\quad\text{and}\quad\tilde{\bm{j}}^{2}=\tilde{\bm{j}}^{2}_{f}+\tilde{\bm{j}}^{2}_{b}$$

where

$$\tilde{\rho}^{3}_{b}=-\mathsf{d}\tilde{\bm{p}}^{2}\quad\text{and}\quad\tilde{\bm{j}}^{2}_{b}=\tilde{\bm{j}}^{2}_{p}+\tilde{\bm{j}}^{2}_{m}=\frac{\partial\tilde{\bm{p}}^{2}}{\partial t}+c\mathsf{d}\tilde{\bm{m}}^{1}.$$

Straightforward algebraic manipulation shows that the two columns of Maxwell’s equations are equivalent. The left column is the split exterior calculus formulation of the macroscopic Maxwell equations. The remainder of this paper is dedicated to translating the Hamiltonian structure of the above system, discovered in [14], into the language of split exterior calculus.

Let $K:\Lambda^{k}\to\mathbb{R}$. We may define a Fréchet derivative of this functional in the usual manner

$$\left|K[\omega+\eta]-K[\omega]-DK[\omega]\eta\right|=O(\|\eta\|).$$

(22)

We shall not worry about functional analytic rigor here, but only the formal correctness of our expressions. Note that $DK[\omega]\in(\Lambda^{k})^{*}$, the dual space to the $k$-forms. In order to identify the functional derivative as an element of one of the vector spaces in the double de Rham complex, we must utilize the previously defined duality pairings. Using the $L^{2}$ and Poincaré duality structures on the double de Rham complex, we may define two varieties of functional derivatives:

$$DK[\omega]\eta=\left(\eta,\frac{\delta K}{\delta\omega}\right)=\left\langle\eta,\frac{\tilde{\delta}K}{\delta\omega}\right\rangle.$$

(23)

One can see that

$$\frac{\delta K}{\delta\omega}\in\Lambda^{k},\quad\frac{\tilde{\delta}K}{\delta\omega}\in\tilde{\Lambda}^{n-k},\quad\text{and}\quad\frac{\tilde{\delta}K}{\delta\omega}=\star\frac{\delta K}{\delta\omega}.$$

Notice, $\tilde{\delta}K/\delta\omega$ is a twisted form while $\delta K/\delta\omega$ is straight. For this reason, we call $\tilde{\delta}K/\delta\omega$ a twisted functional derivative [6].

The same procedure may be done for a functional which takes a twisted form as an argument: $K:\tilde{\Lambda}^{k}\to\mathbb{R}$. In this case, the twisted functional derivative yields a straight form and the straight functional derivative yields a twisted form:

$$\frac{\delta K}{\delta\tilde{\omega}}\in\tilde{\Lambda}^{k},\quad\frac{\tilde{\delta}K}{\delta\tilde{\omega}}\in\Lambda^{n-k},\quad\text{and}\quad\frac{\tilde{\delta}K}{\delta\tilde{\omega}}=\star\frac{\delta K}{\delta\tilde{\omega}}.$$

Finally, if we let $\tilde{\omega}=\tilde{\star}\omega\in\tilde{\Lambda}^{n-k}$, then the chain rule implies that $\tilde{\delta}K/\delta\tilde{\omega}=\delta K/\delta\omega$.

The following proposition will be needed to establish the existence of Casimir invariants of the macroscopic Maxwell equations.

Because of the isomorphisms relating vectors with $1$-forms and $(n-1)$-forms, we can easily establish a relationship between functional derivatives with respect to vector fields and functional derivatives with respect to differential forms.

Consider a map $\phi:\Lambda^{k}\to\Lambda^{l}$. The Fréchet derivative of $\phi$ is defined by

$$\left\|\phi[\omega+\eta]-\phi[\omega]-D\phi[\omega]\eta\right\|=O\left(\|\eta\|\right).$$

(26)

We identify this via either of the two duality pairings:

$$D\phi[\omega](\eta,\xi)=\left(\eta,\frac{\delta\phi}{\delta\omega}\xi\right)=\left\langle\eta,\frac{\tilde{\delta}\phi}{\delta\omega}\xi\right\rangle.$$

This is the general pattern for the higher order derivatives considered in this section.

By taking the Fréchet derivative of $DK[\omega]\in(\Lambda^{k})^{*}$, one obtains the second variation:

$$\left|DK[\omega+\xi]\eta-DK[\omega]\eta-D^{2}K[\omega](\eta,\xi)\right|=O\left(\|\eta\|\|\xi\|\right).$$

where $D^{2}K[\omega](\cdot,\cdot)$ is a symmetric bilinear form. Using our two notions of duality, we may write

$$D^{2}K[\omega](\eta,\xi)=\left(\eta,\frac{\delta^{2}K}{\delta\omega\delta\omega}\xi\right)=\left\langle\eta,\frac{\tilde{\delta}^{2}K}{\delta\omega\delta\omega}\xi\right\rangle.$$

(27)

From this definition, it is clear that

$$\frac{\delta^{2}K}{\delta\omega\delta\omega}:\Lambda^{k}\to\Lambda^{k}\quad\text{and}\quad\frac{\tilde{\delta}^{2}K}{\delta\omega\delta\omega}:\Lambda^{k}\to\tilde{\Lambda}^{n-k}.$$

We can relate these two versions of the second variation via

$$\frac{\tilde{\delta}^{2}K}{\delta\omega\delta\omega}=\star\frac{\delta^{2}K}{\delta\omega\delta\omega}.$$

As previously noted, the second variation is a symmetric operator:

$$D^{2}K[\omega](\eta,\xi)=D^{2}K[\omega](\xi,\eta).$$

Hence, it follows that

$$\left(\eta,\frac{\delta^{2}K}{\delta\omega\delta\omega}\xi\right)=\left(\frac{\delta^{2}K}{\delta\omega\delta\omega}\eta,\xi\right)\\ \quad\text{and}\quad\left\langle\eta,\frac{\tilde{\delta}^{2}K}{\delta\omega\delta\omega}\xi\right\rangle=\left\langle\xi,\frac{\tilde{\delta}^{2}K}{\delta\omega\delta\omega}\eta\right\rangle.$$

If $K:\Lambda^{k}\times\Lambda^{l}\to\mathbb{R}$, then we may introduce the notion of partial functional derivatives:

$$\left|K[\omega+\nu,\eta]-K[\omega,\eta]-D_{1}K[\omega,\eta]\nu\right|=O\left(\|\nu\|\right)$$

and

$$\left|K[\omega,\eta+\xi]-K[\omega,\eta]-D_{2}K[\omega,\eta]\xi\right|=O\left(\|\xi\|\right).$$

Notice, we indicate the partial functional derivatives with subscripts to indicate with respect to which argument we are varying. From here, we may define the total functional derivative:

$$DK[\omega,\eta](\nu,\xi)=D_{1}K[\omega,\eta]\nu+D_{2}K[\omega,\eta]\xi,$$

(28)

and we may identify these using either of the two duality pairings:

$$DK[\omega,\eta](\nu,\xi)=\left(\nu,\frac{\delta K}{\delta\omega}\right)+\left(\xi,\frac{\delta K}{\delta\eta}\right)=\left\langle\nu,\frac{\tilde{\delta}K}{\delta\omega}\right\rangle+\left\langle\xi,\frac{\tilde{\delta}K}{\delta\eta}\right\rangle.$$

Hence, the total functional derivative is just the standard Fréchet derivative on a tensor product space.

Finally, we have mixed second partial derivatives. For example,

$$\left|D_{1}K[\omega,\eta+\xi]\nu-D_{1}K[\omega,\eta]\nu-D^{2}_{12}K[\omega,\eta](\nu,\xi)\right|=O\left(\|\nu\|\|\xi\|\right).$$

As before, there are two natural identifications of the second functional derivative:

$$D^{2}_{12}K(\omega,\eta)(\nu,\xi)=\left(\nu,\frac{\delta^{2}K}{\delta\eta\delta\omega}\xi\right)=\left\langle\nu,\frac{\tilde{\delta}^{2}K}{\delta\eta\delta\omega}\xi\right\rangle.$$

The mixed second partial functional derivative is symmetric with respect to interchange of order of differentiation:

$$D^{2}_{12}K=D^{2}_{21}K.$$

Suppose that $\phi:\mathfrak{X}\to\mathfrak{X}$ and let $\hat{\phi}:\Lambda^{1}\to\tilde{\Lambda}^{n-1}$ such that $\phi[\bm{V}]=(\star\hat{\phi}[\bm{V}^{\flat}])^{\sharp}$. Then letting $w^{1}=\bm{W}^{\flat}$, $v^{1}=\bm{V}^{\flat}$, and $\xi^{1}=\bm{X}^{\flat}$, we find

$$D\hat{\phi}[v^{1}](w^{1},\xi^{1})=\left\langle w^{1},\frac{\tilde{\delta}\hat{\phi}}{\delta v^{1}}\xi^{1}\right\rangle=\int_{\Omega}w^{1}\wedge\frac{\tilde{\delta}\hat{\phi}}{\delta v^{1}}\xi^{1}=\int_{\Omega}\bm{W}\cdot\left(\star\frac{\tilde{\delta}\hat{\phi}}{\delta v^{1}}\xi^{1}\right)^{\sharp}\bm{\mathrm{{vol}}}^{n}.$$

Therefore,

$$D\phi[\bm{V}](\bm{W},\bm{X})=\left(\bm{W},\frac{\delta\phi}{\delta\bm{V}}\bm{X}\right)\implies\frac{\delta\phi}{\delta\bm{V}}\bm{X}=\left(\star\frac{\tilde{\delta}\hat{\phi}}{\delta v^{1}}\xi^{1}\right)^{\sharp}.$$

(29)

Similar results hold if we identify the domain or codomain of $\hat{\phi}$ by any of the other isomorphisms between vector fields and differential forms, but we omit these results for brevity with the hope that they are obvious.

First, we review the Hamiltonian structure as presented in [14]. Consider a general matter model prescribed by

$$K[\bm{E},\bm{B}]=\int_{\Omega}\mathcal{K}(\bm{x},\bm{E},\bm{B},\nabla\bm{E},\nabla\bm{B},...)\bm{\mathrm{{vol}}}^{3}$$

(30)

so that the polarization and magnetization are defined by

$$\bm{P}(\bm{x},t)=-\frac{\delta K}{\delta\bm{E}}\quad\text{and}\quad\bm{M}(\bm{x},t)=-\frac{\delta K}{\delta\bm{B}}.$$

(31)

The Hamiltonian of the macroscopic Maxwell equations system is given by

$$H[\bm{E},\bm{B}]=K-\int_{\Omega}\bm{E}\cdot\frac{\delta K}{\delta\bm{E}}\bm{\mathrm{{vol}}}^{3}+\frac{1}{8\pi}\int_{M}(\bm{E}\cdot\bm{E}+\bm{B}\cdot\bm{B})\bm{\mathrm{{vol}}}^{3}.$$

(32)

The Poisson bracket is defined to be

$$\{F,G\}=4\pi c\int_{\Omega}\left[\frac{\delta F}{\delta\bm{D}}\cdot\nabla\times\frac{\delta G}{\delta\bm{B}}-\frac{\delta G}{\delta\bm{D}}\cdot\nabla\times\frac{\delta F}{\delta\bm{B}}\right]\bm{\mathrm{{vol}}}^{3}.$$

(33)

This bracket has Casimir invariants $\mathcal{C}_{\bm{D}}=\mathcal{F}[\nabla\cdot\bm{D}]$ and $\mathcal{C}_{\bm{B}}=\mathcal{F}[\nabla\cdot\bm{B}]$ where $\mathcal{F}$ is an arbitrary smooth functional. Hence, Gauss’s law and the absence of magnetic monopoles hold as long as they are enforced in the initial conditions.

In order to obtain the equations of motion, we need to take derivatives of the Hamiltonian with respect to $(\bm{D},\bm{B})$. As the details of this procedure are omitted in [14], it is useful to show the full calculation here since we will perform the calculation again in the language of exterior calculus in the following section.