2.1. Variational derivation of the fluid equations in Eulerian form

As mentioned in the introduction, the variational principle for a compressible fluid in Eulerian variables is obtained after reduction of the general principle in Lagrangian variables. While the latter involves free variations of the Lagrangian flow $\varphi _t = \varphi (\boldsymbol{\cdot },t)$ in the group of diffeomorphisms $D(\Omega )$ , the reduced variational principle involves constrained variations of the Eulerian variables, and in particular of the Eulerian velocity field determined by the relation

(2.1) \begin{equation} {\boldsymbol{u}}(\varphi _t(\boldsymbol{x}_0),t) = \partial _t \varphi (\boldsymbol{x}_0,t). \end{equation}

Following Newcomb (Reference Newcomb1962), let us recall how these ‘Lin constraints’ are derived. One first considers free variations $\delta \varphi$ of the flow, such that $\delta \varphi (t=0) = \delta \varphi (t=T) = 0$ , and denote by $\boldsymbol{v}$ the corresponding field in Eulerian variables, i.e.

(2.2) \begin{equation} {\boldsymbol{v}}(\varphi _t(\boldsymbol{x}_0),t) := \delta \varphi (\boldsymbol{x}_0,t). \end{equation}

The key observation is that $\boldsymbol{v}$ does not represent the variation of $\boldsymbol{u}$ , because of the compositional structure of the latter. By taking the variations of (2.1), one finds indeed

(2.3) \begin{align} \delta {\boldsymbol{u}}(\varphi _t) + \delta \varphi \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}(\varphi _t) = \partial _t \delta \varphi \end{align}

(with implicit dependencies on $\boldsymbol{x}_0$ and $t$ ) while the time derivative of (2.2) yields

(2.4) \begin{align} \partial _t {\boldsymbol{v}}(\varphi _t) + \partial _t \varphi \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{v}}(\varphi _t) = \partial _t \delta \varphi . \end{align}

After right-composition with $\varphi _t^{-1}$ we find the relation

(2.5) \begin{equation} \delta {\boldsymbol{u}} = \partial _t {\boldsymbol{v}} + {\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{v}} - {\boldsymbol{v}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}, \end{equation}

which gives the precise form of the variations of the Eulerian velocity field, given that $\boldsymbol{v}$ can be chosen as an arbitrary vector field tangent to the boundary $\partial \Omega$ . For a more detailed description on this reduction and the associated Euler–Poincaré equations, we refer the reader to Holm et al. (Reference Holm, Marsden and Ratiu1998).

We are now in a position to state a first classical result which will serve as a basis for our derivations. There we denote by $X(\Omega )$ the set of (smooth) vector fields of $\Omega$ that are tangent to the boundary, and by $F(\Omega )$ the set of (smooth) functions on $\Omega$ .

Theorem 1. Consider the following Lagrangian:

(2.6) \begin{equation} l({\boldsymbol{u}}, n, s) = \int _\Omega \frac {m_s n|{\boldsymbol{u}}|^2}{2} - n e(n, s)\text{d} \boldsymbol{x}, \end{equation}

where ${\boldsymbol{u}} \in X(\Omega )$ , $n, s \in F(\Omega )$ and $e$ is the internal energy of the fluid, as a function of $n$ the density and $s$ the entropy density. Here $m_s$ is the mass of a particle constituting the fluid. Then extremisers of the action $\mathfrak{S} = \int _0^T l\text{d}t$ under constrained variations

(2.7) \begin{equation} \delta {\boldsymbol{u}} = \partial _t {\boldsymbol{v}} + [{\boldsymbol{u}},{\boldsymbol{v}}], \qquad \delta n = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n {\boldsymbol{v}}), \qquad \delta s = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s {\boldsymbol{v}}) \end{equation}

(where ${\boldsymbol{v}} = {\boldsymbol{v}}(t)$ is a curve in $X(\Omega )$ , i.e. a time-dependent vector field, vanishing at $t=0$ and $T$ , and $[{\boldsymbol{u}},{\boldsymbol{v}}] = {\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{v}} - {\boldsymbol{v}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}$ is the Lie bracket of vector fields) supplemented with the advection equations

(2.8) \begin{equation} \partial _t n + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}) = 0 , \qquad \partial _t s + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}) = 0 , \end{equation}

are solution to the following system of equations:

(2.9a) \begin{align} \partial _t n + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}) &= 0 , \end{align}

(2.9b) \begin{align} \partial _t s + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}) &= 0 , \end{align}

(2.9c) \begin{align} m_s n \partial _t {\boldsymbol{u}} + m_s n ({\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}) &+ \boldsymbol{\nabla }p = 0 , \end{align}

where $p = p(n, s)$ is the associated pressure, given by $p=n(n \partial _{n} e+ s \partial _s e)$ .

Remark 2 (Definition of the pressure). Our definition of the pressure is equivalent to the standard thermodynamical definition ${\textrm{d}}e = -p {\textrm{d}}\tau + T {\textrm{d}}\zeta$ , where $\tau = 1/n$ and $\zeta = s/n$ , or $n = 1/\tau$ , $s = \zeta /\tau$ :

(2.10) \begin{equation} p = -\frac {\partial e}{\partial \tau } = -\Big (\frac {\partial n}{\partial \tau }\frac {\partial e}{\partial n} + \frac {\partial s}{\partial \tau }\frac {\partial e}{\partial s}\Big ) = \frac {1}{\tau ^2}\frac {\partial e}{\partial n} + \frac {\zeta }{\tau ^2}\frac {\partial e}{\partial s} = n^2 \frac {\partial e}{\partial n} + ns \frac {\partial e}{\partial s}. \end{equation}

Proof. The first two equations are simply the advection equations. To find (2.9c ), we write the variational principle:

(2.11) \begin{align} 0 = \delta \mathfrak{S} &= \int _0^T \int _\Omega \delta l\text{d}t\nonumber \\ &= \int _0^T \int _\Omega \frac {\delta l}{\delta u}\delta u + \frac {\delta l}{\delta n}\delta n + \frac {\delta l}{\delta s}\delta s \,\text{d} \boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega \frac {\delta l}{\delta u}(\partial _t {\boldsymbol{v}} + [{\boldsymbol{u}},{\boldsymbol{v}}]) - \frac {\delta l}{\delta n}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{v}}) - \frac {\delta l}{\delta s}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{v}})\text{d} \boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega m_s n {\boldsymbol{u}} \boldsymbol{\cdot }(\partial _t {\boldsymbol{v}} + [{\boldsymbol{u}},{\boldsymbol{v}}]) - \left(m_s \frac {|{\boldsymbol{u}}|^2}{2} - \partial _n (n e)\right) {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{v}}) \nonumber \\ & \quad +\, \partial _s (n e){\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{v}})\text{d} \boldsymbol{x}\text{d}t. \end{align}

Using integration by parts we develop the first term as

(2.12) \begin{align} &\int _0^T \int _\Omega m_s n {\boldsymbol{u}} \boldsymbol{\cdot }(\partial _t {\boldsymbol{v}} + {\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{v}} - {\boldsymbol{v}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}})\text{d} \boldsymbol{x}\text{d}t \nonumber \\ &= - m_s \int _0^T \int _\Omega \partial _t(n {\boldsymbol{u}}) \boldsymbol{\cdot }{\boldsymbol{v}} + n ({\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}) \boldsymbol{\cdot }{\boldsymbol{v}} + {\boldsymbol{u}} {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}) \boldsymbol{\cdot }{\boldsymbol{v}} + (\boldsymbol{\nabla }{\boldsymbol{u}})^T (n {\boldsymbol{u}}) \boldsymbol{\cdot }{\boldsymbol{v}}\text{d} \boldsymbol{x}\text{d}t, \end{align}

the second term as

(2.13) \begin{align} & - \int _0^T \int _\Omega \bigg( m_s \frac {|{\boldsymbol{u}}|^2}{2}-e(n, s)- n \partial _{n} e(n, s)\bigg) {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= \int _0^T \int _\Omega \boldsymbol{\nabla }\bigg( m_s \frac {|{\boldsymbol{u}}|^2}{2} -e(n, s)-n \partial _{n} e(n, s)\bigg) \nonumber \\ &= \int _0^T \int _\Omega \big ( m_s (\boldsymbol{\nabla }{\boldsymbol{u}})^T ({\boldsymbol{u}}) - \boldsymbol{\nabla }n \partial _{n} e(n, s) - \boldsymbol{\nabla }s \partial _{s} e(n, s) - \boldsymbol{\nabla }(n \partial _{n} e(n, s)) \big ) \boldsymbol{\cdot }(n {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t \end{align}

and the third term as

(2.14) \begin{align} \int _0^T \int _\Omega n \partial _s e(n, s) {\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t = - \int _0^T \int _\Omega \boldsymbol{\nabla }(n \partial _s e(n, s)) \boldsymbol{\cdot }(s {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t . \end{align}

Using $p=n(n \partial _{n} e+ s \partial _s e)$ and (2.8) we next observe that

(2.15) \begin{align} n \big (\boldsymbol{\nabla }n \partial _{n} e(n, s) + \boldsymbol{\nabla }s \partial _{s} e(n, s) &+ \boldsymbol{\nabla }(n \partial _{n} e(n, s))\big ) + s \boldsymbol{\nabla }(n \partial _s e(n, s)) = \boldsymbol{\nabla }p , \end{align}

(2.16) \begin{align} \partial _t(n {\boldsymbol{u}}) + {\boldsymbol{u}}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}) &= {\boldsymbol{u}} \partial _t n + n \partial _t ({\boldsymbol{u}}) + {\boldsymbol{u}}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}) = n \partial _t {\boldsymbol{u}} , \end{align}

so that summing the three contributions above yields

(2.17) \begin{equation} 0 = \int _0^T \int _\Omega \big (\!- m_s n (\partial _t {\boldsymbol{u}} - {\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}) - \boldsymbol{\nabla }p \big ) \boldsymbol{\cdot }{\boldsymbol{v}}\text{d}\boldsymbol{x}\text{d}t. \end{equation}

Equation (2.9c ) follows from the fact that the latter holds for all $\boldsymbol{v}$ vanishing on the boundary.

2.2. Single charged fluid in an electromagnetic field

A variational principle for a charged fluid can next be obtained by coupling the above fluid least-action principle with an electromagnetic Lagrangian. The latter involves a potential $\phi$ and a vector potential $\boldsymbol{A}$ , and corresponds to the energy of the electric field minus that of the magnetic field. The coupling term is standard for particles in an electromagnetic field, considering here the fluid as a particle distribution. We also denote $q_s$ as the electric charge of an individual particle of species $s$ , so that $q_s n$ is the charge density.

Theorem 2. Consider the following Lagrangian:

(2.18a) \begin{align} l({\boldsymbol{u}}, n, s, {\boldsymbol{A}}, \phi ) &= l_f({\boldsymbol{u}}, n, s, {\boldsymbol{A}}, \phi ) + l_{max}({\boldsymbol{A}}, \phi ), \end{align}

(2.18b) \begin{align} l_f({\boldsymbol{u}}, n, s, {\boldsymbol{A}}, \phi ) &= \int _\Omega \frac {m_s n |{\boldsymbol{u}}|^2}{2} - n e(n, s) + q_s n {\boldsymbol{u}} \boldsymbol{\cdot }{\boldsymbol{A}} - q_s n \phi\text{d}\boldsymbol{x}, \end{align}

(2.18c) \begin{align} l_{max}({\boldsymbol{A}}, \phi ) &= \int _\Omega \epsilon _0 \frac {|\boldsymbol{\nabla }\phi + \partial _t {\boldsymbol{A}} |^2}{2}- \frac {|\boldsymbol{\nabla }\times {\boldsymbol{A}}|^2}{2 \mu _0}\text{d}\boldsymbol{x} . \end{align}

Then extremisers of the action $\mathfrak{S} = \int _0^T l\text{d}t$ under constrained variations

(2.19) \begin{equation} \delta {\boldsymbol{u}} = \partial _t {\boldsymbol{v}} + [{\boldsymbol{u}},{\boldsymbol{v}}], \qquad \delta n = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n {\boldsymbol{v}}), \qquad \delta s = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s {\boldsymbol{v}}), \end{equation}

supplemented with the advection equations

(2.20) \begin{equation} \partial _t n + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}) = 0 , \qquad \partial _t s + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}) = 0 , \end{equation}

are solution to the following system of equations:

(2.21a) \begin{align} \partial _t n &+ {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}) = 0 , \end{align}

(2.21b) \begin{align} \partial _t s &+ {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}) = 0 , \end{align}

(2.21c) \begin{align} m_s n \partial _t {\boldsymbol{u}} + m_s n ({\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}) &+ \boldsymbol{\nabla }p - q_s n (\boldsymbol{E} + {\boldsymbol{u}} \times {\boldsymbol{B}}) = 0 , \end{align}

(2.21d) \begin{align} \epsilon _0 \partial _t \boldsymbol{E} &= \frac {\boldsymbol{\nabla }\times {\boldsymbol{B}}}{\mu _0} - q_s n {\boldsymbol{u}} , \end{align}

(2.21e) \begin{align} \epsilon _0 \boldsymbol{\nabla }\boldsymbol{\cdot} & \boldsymbol{E} = q_s n, \end{align}

where $p$ is defined as in Theorem 1 , $E = -\boldsymbol{\nabla }\phi -\partial _t {\boldsymbol{A}}$ and ${\boldsymbol{B}} = \boldsymbol{\nabla }\times {\boldsymbol{A}}$ .

Proof. The first two equations follow directly from the advection equations we consider; for the last three, we write the variational principle: $0 = \delta \mathfrak{S}$ with

(2.22) \begin{align} \delta \mathfrak{S} &= \int _0^T \int _\Omega \delta l\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega \delta l_f + \delta l_{max}\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega \frac {\delta l_f}{\delta u}\delta u + \frac {\delta l_f}{\delta n}\delta n + \frac {\delta l_f}{\delta s}\delta s + \frac {\delta l_f}{\delta \phi }\delta \phi + \frac {\delta l_f}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}} + \frac {\delta l_{max}}{\delta \phi }\delta \phi + \frac {\delta l_{max}}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega \frac {\delta l_f}{\delta u}(\partial _t {\boldsymbol{v}} + [{\boldsymbol{u}},{\boldsymbol{v}}]) - \frac {\delta l_f}{\delta n}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{v}}) - \frac {\delta l_f}{\delta s}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{v}}) + \frac {\delta l_f}{\delta \phi }\delta \phi + \frac {\delta l_f}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}} \nonumber \\ & \quad +\, \frac {\delta l_{max}}{\delta \phi }\delta \phi + \frac {\delta l_{max}}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t. \end{align}

Since variations in $\boldsymbol{A}$ and $\phi$ are independent of $\boldsymbol{v}$ , we may assume $\delta {\boldsymbol{A}} =0$ and $\delta \phi = 0$ . In particular we may consider only variations implying $\boldsymbol{v}$ . Hence $\delta \mathfrak{S} = 0$ for all variations implies:

(2.23) \begin{equation} \begin{aligned} 0 &= \int _0^T \int _\Omega \frac {\delta l_f}{\delta u}(\partial _t {\boldsymbol{v}} + [{\boldsymbol{u}},{\boldsymbol{v}}]) - \frac {\delta l_f}{\delta n}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{v}}) - \frac {\delta l_f}{\delta s}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t \\ &= (I) + (II) + (III), \end{aligned} \end{equation}

where the first term is (using integration by parts as in the proof of Theorem 1)

(2.24) \begin{align} (I) &= \int _0^T \int _\Omega (m_s n {\boldsymbol{u}} + q_s n A)(\partial _t {\boldsymbol{v}} + [{\boldsymbol{u}},{\boldsymbol{v}}])\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= -\int _0^T \int _\Omega \partial _t(m_s n {\boldsymbol{u}} + q_s n {\boldsymbol{A}}) \boldsymbol{\cdot }{\boldsymbol{v}} + n ({\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }(m_s {\boldsymbol{u}} + q_s {\boldsymbol{A}})) \boldsymbol{\cdot }{\boldsymbol{v}} \nonumber \\ &\quad +\, (m_s {\boldsymbol{u}} + q_s {\boldsymbol{A}}){\boldsymbol{\nabla }{\boldsymbol{\cdot }}}( n {\boldsymbol{u}}) \boldsymbol{\cdot }{\boldsymbol{v}} + (\boldsymbol{\nabla }{\boldsymbol{u}})^T (n (m_s {\boldsymbol{u}}+ q_s {\boldsymbol{A}})) \boldsymbol{\cdot }{\boldsymbol{v}}\text{d}\boldsymbol{x}\text{d}t ,\end{align}

the second one is

(2.25) \begin{align} (II) &= - \int _0^T \int _\Omega \bigg(m_s \frac {|u|^2}{2} - \partial _n (n e) - q_s \phi + q_s {\boldsymbol{u}} \boldsymbol{\cdot }{\boldsymbol{A}}\bigg) {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= \int _0^T \int _\Omega (m_s (\boldsymbol{\nabla }{\boldsymbol{u}})^T ({\boldsymbol{u}}) - \boldsymbol{\nabla }n \partial _{n} e(n, s) - \boldsymbol{\nabla }s \partial _{s} e(n, s) - \boldsymbol{\nabla }(n \partial _{n} e(n, s)) \nonumber\\ & \quad +\, q_s \boldsymbol{\nabla }({\boldsymbol{A}} \boldsymbol{\cdot }{\boldsymbol{u}}) - q_s \boldsymbol{\nabla }\phi ) \boldsymbol{\cdot }( {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t \end{align}

and the third is

(2.26) \begin{align} (III) &= \int _0^T \int _\Omega \partial _s (n e){\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t = - \int _0^T \int _\Omega \boldsymbol{\nabla }(n \partial _s e(n, s)) \boldsymbol{\cdot }(s {\boldsymbol{v}})\text{d}\boldsymbol{x}\text{d}t. \end{align}

We next observe that

(2.27) \begin{equation} \boldsymbol{\nabla }({\boldsymbol{A}} \boldsymbol{\cdot }{\boldsymbol{u}}) - {\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{A}} - \boldsymbol{\nabla }{\boldsymbol{u}} ^T {\boldsymbol{A}} = {\boldsymbol{u}} \times \boldsymbol{\nabla }\times {\boldsymbol{A}} , \end{equation}

so that summing the above three contributions and using the latter identity, (2.15) and a version of (2.16) where $\boldsymbol{u}$ is replaced by $m_s {\boldsymbol{u}} + q_s {\boldsymbol{A}}$ , we find that

(2.28) \begin{equation} 0 = \!\int _0^T \!\!\int _\Omega \! \big (\! - n \partial _t (m_s {\boldsymbol{u}} + q_s {\boldsymbol{A}}) - m_s n {\boldsymbol{u}} \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}} \!-\! \boldsymbol{\nabla}\!p \!-\! q_s n \boldsymbol{\nabla }\phi + m_s n {\boldsymbol{u}} \times \boldsymbol{\nabla }\times {\boldsymbol{A}}\! \big ) \boldsymbol{\cdot }{\boldsymbol{v}}\text{d}\boldsymbol{x}\text{d}t. \end{equation}

Again the claimed momentum equation (2.21c ) follows from the fact that (2.28) holds for all $\boldsymbol{v}$ null at the boundaries, and by using the definitions for $\boldsymbol{E}$ and $\boldsymbol{B}$ .

Taking next variations in $\delta {\boldsymbol{A}}$ , we write

(2.29) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta l_f}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}} + \frac {\delta l_{max}}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega q_s n {\boldsymbol{u}} \boldsymbol{\cdot }\delta {\boldsymbol{A}} + \epsilon _0 (\boldsymbol{\nabla }\phi + \partial _t {\boldsymbol{A}}) \boldsymbol{\cdot }\partial _t \delta {\boldsymbol{A}} - \frac {1}{\mu _0}\boldsymbol{\nabla }\times {\boldsymbol{A}} \boldsymbol{\cdot }\boldsymbol{\nabla }\times \delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= \int _0^T \int _\Omega \left ( q_s n {\boldsymbol{u}} - \epsilon _0 \partial _t (\boldsymbol{\nabla }\phi + \partial _t {\boldsymbol{A}}) - \frac {1}{\mu _0}\boldsymbol{\nabla }\times \boldsymbol{\nabla }\times {\boldsymbol{A}} \right ) \boldsymbol{\cdot }\delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t . \end{align}

So that we obtain (2.21d ) since the above equality holds for every $\delta {\boldsymbol{A}}$ that is null at the boundaries.

Finally for the variations in $\delta \phi$ , we write

(2.30) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta l_f}{\delta \phi }\delta \phi + \frac {\delta l_{max}}{\delta \phi }\delta \phi \text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega -q_s n \delta \phi + \epsilon _0 (\boldsymbol{\nabla }\phi + \partial _t {\boldsymbol{A}}) \boldsymbol{\cdot }\boldsymbol{\nabla }\delta \phi \text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega -q_s n \delta \phi - \epsilon _0 \boldsymbol{\nabla }\boldsymbol{\cdot }(\boldsymbol{\nabla }\phi + \partial _t {\boldsymbol{A}}) \delta \phi \text{d}\boldsymbol{x}\text{d}t, \end{align}

which yields (2.21e ).

2.3. Two charged fluids in an electromagnetic field

We finish this section by presenting a variational formulation for a mixture of two charged fluids. In this first model the two fluids only interact through the electromagnetic field: it does not bring any conceptual novelty compared with a single charged fluid (in fact we only sum the two fluid Lagrangians without changing the Maxwell part). It is nevertheless useful as a basis for the Hall MHD model that will be presented next. Without loss of generality we denote the two species with subscripts $e$ and $i$ , for electrons and ions.

Theorem 3. Consider the following Lagrangian:

(2.31a) \begin{align} l({\boldsymbol{u}}_i, n_i, s_i, u_e, n_e, s_e, {\boldsymbol{A}}, \phi ) &= l_i({\boldsymbol{u}}_i, n_i, s_i, {\boldsymbol{A}}, \phi ) + l_e({\boldsymbol{u}}_e, n_e, s_e, {\boldsymbol{A}}, \phi ) + l_{max}({\boldsymbol{A}}, \phi ), \end{align}

(2.31b) \begin{align} l_i({\boldsymbol{u}}_i, n_i, s_i, {\boldsymbol{A}}, \phi ) & = \int _\Omega \frac {m_i n_i|{\boldsymbol{u}}_i|^2}{2} - n_i e_i(n_i, s_i) + q_i n_i {\boldsymbol{u}}_i \boldsymbol{\cdot }{\boldsymbol{A}} - q_i n_i \phi \text{d}\boldsymbol{x}, \end{align}

(2.31c) \begin{align} l_e({\boldsymbol{u}}_e, n_e, s_e, {\boldsymbol{A}}, \phi ) &= \int _\Omega \frac {m_e n_e|{\boldsymbol{u}}_e|^2}{2} - n_e e_e(n_e, s_e) + q_e n_e {\boldsymbol{u}}_e \boldsymbol{\cdot }{\boldsymbol{A}} - q_e n_e \phi \text{d}\boldsymbol{x}, \end{align}

(2.31d) \begin{align} l_{max}({\boldsymbol{A}}, \phi ) &= \int _\Omega \epsilon _0 \frac {|\boldsymbol{\nabla }\phi + \partial _t {\boldsymbol{A}} |^2}{2}- \frac {|\boldsymbol{\nabla }\times {\boldsymbol{A}}|^2}{2 \mu _0}\text{d}\boldsymbol{x}. \end{align}

Then extremisers of the action $\mathfrak{S} = \int _0^T l\text{d}t$ under constrained variations

(2.32) \begin{align} \delta {\boldsymbol{u}}_i = \partial _t {\boldsymbol{v}}_i + [{\boldsymbol{u}}_i,{\boldsymbol{v}}_i], \qquad \delta n_i = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n_i {\boldsymbol{v}}_i), \qquad \delta s_i = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s_i {\boldsymbol{v}}_i), \end{align}

(2.33) \begin{align} \delta {\boldsymbol{u}}_e = \partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e,{\boldsymbol{v}}_e], \qquad \delta n_e = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n_e {\boldsymbol{v}}_e), \qquad \delta s_e = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s_e {\boldsymbol{v}}_e) \end{align}

supplemented with the advection equations

(2.34a) \begin{align} \partial _t n_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_i {\boldsymbol{u}}_i) = 0 , \end{align}

(2.34b) \begin{align} \partial _t s_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s_i {\boldsymbol{u}}_i) = 0 , \end{align}

(2.34c) \begin{align} \partial _t n_e + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_e {\boldsymbol{u}}_e) = 0 , \end{align}

(2.34d) \begin{align} \partial _t s_e + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s_e {\boldsymbol{u}}_e) = 0 , \end{align}

are solution to the following system of equations:

(2.35a) \begin{align} \partial _t n_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_i {\boldsymbol{u}}_i) &= 0 , \end{align}

(2.35b) \begin{align} \partial _t s_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s_i {\boldsymbol{u}}_i) &= 0 , \end{align}

(2.35c) \begin{align} \partial _t n_e + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_e {\boldsymbol{u}}_e) &= 0 , \end{align}

(2.35d) \begin{align} \partial _t s_e + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s_e {\boldsymbol{u}}_e) &= 0 , \end{align}

(2.35e) \begin{align} m_i n_i ( \partial _t {\boldsymbol{u}}_i + ({\boldsymbol{u}}_i \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_i)) + \boldsymbol{\nabla }p_i & - q_i n_i (\boldsymbol{E} + {\boldsymbol{u}}_i \times {\boldsymbol{B}}) = 0 , \end{align}

(2.35f) \begin{align} m_e n_e ( \partial _t {\boldsymbol{u}}_e + ({\boldsymbol{u}}_e \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_e)) + \boldsymbol{\nabla }p_e &- q_e n_e (\boldsymbol{E} + {\boldsymbol{u}}_e \times {\boldsymbol{B}}) = 0 , \end{align}

(2.35g) \begin{align} \epsilon _0 \partial _t \boldsymbol{E} = \frac {\boldsymbol{\nabla }\times {\boldsymbol{B}}}{\mu _0} &- q_i n_i {\boldsymbol{u}}_i - q_e n_e {\boldsymbol{u}}_e , \end{align}

(2.35h) \begin{align} \epsilon _0 \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{E} &= q_i n_i + q_e n_e . \end{align}

3.1. From two charged fluids to Hall MHD

The equations for Hall MHD can be derived from the two-fluid model above by making two assumptions: the electrons are massless ( $m_e=0$ ) and the fluid is quasi-neutral (which is often imposed by formally letting $\epsilon _0 = 0$ ). We also assume for simplicity that the electrons are pressureless ( $e_e=0$ , so $p_e=0$ ): this hypothesis is not standard for Hall MHD but is made here to simplify the exposition. The present work could by adapted to the case of finite electron pressure, without further complications. The first and last assumptions allow one to rewrite (2.35f ) as

(3.1) \begin{equation} \boldsymbol{E} + {\boldsymbol{u}}_e \times {\boldsymbol{B}} = 0 , \end{equation}

while the second assumption transforms (2.35g ) and (2.35h ) into

(3.2) \begin{equation} \frac {\boldsymbol{\nabla }\times {\boldsymbol{B}}}{\mu _0} = q_i n_i {\boldsymbol{u}}_i + q_e n_e {\boldsymbol{u}}_e =: \boldsymbol{J} \qquad \text{ and } \qquad 0 = q_i n_i + q_e n_e , \end{equation}

which can be combined into

(3.3) \begin{equation} {\boldsymbol{u}}_e = \frac {1}{q_e n_e} \boldsymbol{J} + {\boldsymbol{u}}_i . \end{equation}

Injecting this relation in (3.1) we find Ohm’s law of Hall MHD:

(3.4) \begin{equation} \boldsymbol{E} =\bigg (\frac {1}{q_i n_i} \boldsymbol{J} - {\boldsymbol{u}}_i\bigg ) \times {\boldsymbol{B}} , \end{equation}

and substituting in (2.35e ) we recover the standard momentum equation

(3.5) \begin{equation} m_i n_i (\partial _t {\boldsymbol{u}}_i + ({\boldsymbol{u}}_i \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_i)) = \boldsymbol{J} \times {\boldsymbol{B}} - \boldsymbol{\nabla }p_i . \end{equation}

Moreover, Faraday’s equation follows from the definition of $\boldsymbol{E}$ and $\boldsymbol{B}$ :

(3.6) \begin{equation} \begin{aligned} \partial _t {\boldsymbol{B}} & = \partial _t (\boldsymbol{\nabla }\times {\boldsymbol{A}}) = \boldsymbol{\nabla }\times (\partial _t {\boldsymbol{A}}) = \boldsymbol{\nabla }\times (\partial _t {\boldsymbol{A}} + \boldsymbol{\nabla }\phi ) = - \boldsymbol{\nabla }\times \boldsymbol{E} = \boldsymbol{\nabla } \\ & \quad \times \bigg( \bigg(-\frac {\boldsymbol{J}}{q_i n_i}+ {\boldsymbol{u}}_i\bigg) \times {\boldsymbol{B}} \bigg) . \end{aligned} \end{equation}

Finally the remaining equations (advections of $n$ and $s$ ) are left unchanged, so that we indeed recover the set of Hall MHD equations (see e.g. D’Avignon et al. Reference D’Avignon, Morrison and Lingam2016).

3.2. A first variational principle for Hall MHD

By transcribing our three hypotheses in the two-fluid Lagrangian we obtain a first variational principle for the equations of Hall MHD. To do so we directly set $m_e=0$ , $e_e = 0$ and $\epsilon _0=0$ in the Lagrangian from Theorem 3 and derive the new equations satisfied by the extremisers. Note that the electron Lagrangian no longer depends on $s_e$ ; hence we can remove this term from the dynamical variables.

Theorem 4. Consider the following Lagrangian:

(3.7a) \begin{align} l({\boldsymbol{u}}_i, n_i, s_i, {\boldsymbol{u}}_e, n_e, {\boldsymbol{A}}, \phi ) &= l_i({\boldsymbol{u}}_i, n_i, s_i, {\boldsymbol{A}}, \phi ) + l_e({\boldsymbol{u}}_e, n_e, {\boldsymbol{A}}, \phi ) + l_{max}({\boldsymbol{A}}), \end{align}

(3.7b) \begin{align} l_i({\boldsymbol{u}}_i, n_i, s_i, {\boldsymbol{A}}, \phi ) &= \int _\Omega \frac {m_i n_i|{\boldsymbol{u}}_i|^2}{2} - n_i e_i(n_i, s_i) + q_i n_i {\boldsymbol{u}}_i \boldsymbol{\cdot }{\boldsymbol{A}} - q_i n_i \phi \text{d}\boldsymbol{x}, \end{align}

(3.7c) \begin{align} l_e({\boldsymbol{u}}_e, n_e, {\boldsymbol{A}}, \phi ) &= \int _\Omega q_e n_e {\boldsymbol{u}}_e \boldsymbol{\cdot }{\boldsymbol{A}} - q_e n_e \phi \text{d}\boldsymbol{x}, \end{align}

(3.7d) \begin{align} l_{max}({\boldsymbol{A}}) &= \int _\Omega - \frac {|\boldsymbol{\nabla }\times {\boldsymbol{A}}|^2}{2 \mu 0}\text{d}\boldsymbol{x}. \end{align}

Then extremisers of the action $\mathfrak{S} = \int _0^T l\text{d}t$ under constrained variations

(3.8) \begin{align} &\delta {\boldsymbol{u}}_i = \partial _t {\boldsymbol{v}}_i + [{\boldsymbol{u}}_i,{\boldsymbol{v}}_i], \qquad \delta n_i = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n_i {\boldsymbol{v}}_i), \qquad \delta s_i = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s_i {\boldsymbol{v}}_i), \end{align}

(3.9) \begin{align} &\delta {\boldsymbol{u}}_e = \partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e,{\boldsymbol{v}}_e], \qquad \delta n_e = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n_e {\boldsymbol{v}}_e), \end{align}

supplemented with the advection equations

(3.10a) \begin{align} \partial _t n_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_i {\boldsymbol{u}}_i) &= 0 , \end{align}

(3.10b) \begin{align} \partial _t s_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s_i {\boldsymbol{u}}_i) &= 0 , \end{align}

(3.10c) \begin{align} \partial _t n_e + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_e {\boldsymbol{u}}_e) &= 0 , \end{align}

are solution to the following system of equations:

(3.11a) \begin{align} \partial _t n_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_i {\boldsymbol{u}}_i) &= 0 , \end{align}

(3.11b) \begin{align} \partial _t s_i + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s_i {\boldsymbol{u}}_i) &= 0 , \end{align}

(3.11c) \begin{align} \partial _t n_e + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_e {\boldsymbol{u}}_e) &= 0 , \end{align}

(3.11d) \begin{align} m_i n_i (\partial _t {\boldsymbol{u}}_i + {\boldsymbol{u}}_i \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_i) + \boldsymbol{\nabla }p_i & - q_i n_i (\boldsymbol{E} + {\boldsymbol{u}}_i \times {\boldsymbol{B}}) = 0 , \end{align}

(3.11e) \begin{align} q_e n_e (\boldsymbol{E} + {\boldsymbol{u}}_e \times {\boldsymbol{B}}) &= 0 , \end{align}

(3.11f) \begin{align} 0 = \frac {\boldsymbol{\nabla }\times {\boldsymbol{B}}}{\mu _0} &- q_i n_i {\boldsymbol{u}}_i - q_e n_e {\boldsymbol{u}}_e , \end{align}

(3.11g) \begin{align} 0 = q_i n_i & + q_e n_e . \end{align}

Proof. Equations (3.11a )–(3.11d ) are obtained the same way as in the proof of Theorem 2. For (3.11e ) we consider variations in ${\boldsymbol{v}}_e$ , that is, in $\delta {\boldsymbol{u}}_e$ , and $\delta n_e$ :

(3.12) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta l_e}{\delta {\boldsymbol{u}}_e}\delta {\boldsymbol{u}}_e + \frac {\delta l_e}{\delta n_e}\delta n_e\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega \frac {\delta l_e}{\delta {\boldsymbol{u}}_e}(\partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e,{\boldsymbol{v}}_e]) - \frac {\delta l_e}{\delta n_e}{\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_e {\boldsymbol{v}}_e)\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega (q_e n_e {\boldsymbol{A}})(\partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e,{\boldsymbol{v}}_e]) - (q_e \phi + q_e {\boldsymbol{u}}_e \boldsymbol{\cdot }{\boldsymbol{A}}) {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_e {\boldsymbol{v}}_e)\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \!\int _0^T \!\!\int _\Omega \!-\partial _t (q_e n_e {\boldsymbol{A}})\boldsymbol{\cdot }{\boldsymbol{v}}_e \!-\! q_e n_e ({\boldsymbol{u}}_e \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{A}}) \boldsymbol{\cdot }{\boldsymbol{v}}_e -q_e {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n_e {\boldsymbol{u}}_e) {\boldsymbol{A}} \boldsymbol{\cdot }{\boldsymbol{v}}_e \!-\! q_e n_e (\boldsymbol{\nabla }{\boldsymbol{u}}_e)^T {\boldsymbol{A}} \boldsymbol{\cdot }{\boldsymbol{v}}_e \nonumber \\ & \quad +\, q_e \boldsymbol{\nabla }({\boldsymbol{u}}_e \boldsymbol{\cdot }{\boldsymbol{A}}) \boldsymbol{\cdot }(n_e {\boldsymbol{v}}_e) + q_e n_e (\boldsymbol{\nabla }\phi ) \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega q_e n_e (-\partial _t {\boldsymbol{A}} - \boldsymbol{\nabla }\phi + {\boldsymbol{u}}_e \times \boldsymbol{\nabla }\times {\boldsymbol{A}}) \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x}\text{d}t, \end{align}

using identities (2.16) and (2.27), which is (3.11e ). Variations in $\delta \phi$ now give

(3.13) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta l_e}{\delta \phi }\delta \phi + \frac {\delta l_i}{\delta \phi }\delta \phi \text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega - q_e n_e \delta \phi - q_i n_i \delta \phi \text{d}\boldsymbol{x}\text{d}t, \end{align}

which is (3.11g ), and variations in $\delta A$ give

(3.14) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta l_e}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}} + \frac {\delta l_i}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}} + \frac {\delta l_{max}}{\delta {\boldsymbol{A}}}\delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega q_e n_e {\boldsymbol{u}}_e \boldsymbol{\cdot }\delta {\boldsymbol{A}} + q_i n_i {\boldsymbol{u}}_i \boldsymbol{\cdot }\delta {\boldsymbol{A}} - \frac {1}{\mu _0} \boldsymbol{\nabla }\times {\boldsymbol{A}} \boldsymbol{\cdot }\boldsymbol{\nabla }\times \delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega q_e n_e {\boldsymbol{u}}_e \boldsymbol{\cdot }\delta {\boldsymbol{A}} + q_i n_i {\boldsymbol{u}}_i \boldsymbol{\cdot }\delta {\boldsymbol{A}} - \frac {1}{\mu _0} \boldsymbol{\nabla }\times \boldsymbol{\nabla }\times {\boldsymbol{A}} \boldsymbol{\cdot }\delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t, \end{align}

which yields (3.11f ).

3.3. Simplified Hall MHD

In this section we propose a simplified variational principle for the Hall MHD equations, which involves only five dynamical variables. To do so we observe that in the previous Lagrangian formulation, $\phi$ plays the role of a Lagrange multiplier enforcing the neutrality equation (3.11g ) and that the two densities $n_i$ , $n_e$ are linked via this equation. This allows us to simplify the Lagrangian by removing the variables $\phi$ and $n_e$ (and renaming $n = n_i$ , $s =s_i$ , $e = e_i$ ), while keeping the same equations of motion.

Theorem 5. Consider the following Lagrangian:

(3.15) \begin{equation} L({\boldsymbol{u}}_i, n, s, {\boldsymbol{u}}_e, {\boldsymbol{A}}) = \int _\Omega \frac {m_i n |{\boldsymbol{u}}_i|^2}{2} - n e(n, s) + q_i n ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }{\boldsymbol{A}} - \frac {|\boldsymbol{\nabla }\times {\boldsymbol{A}}|^2}{2 \mu _0}\text{d}\boldsymbol{x}. \end{equation}

Then extremisers of the action $\mathfrak{S} = \int _0^T L\text{d}t$ under constrained variations

(3.16) \begin{align} &\delta {\boldsymbol{u}}_i = \partial _t {\boldsymbol{v}}_i + [{\boldsymbol{u}}_i,{\boldsymbol{v}}_i], \qquad \delta n = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n {\boldsymbol{v}}_i), \qquad \delta s = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s {\boldsymbol{v}}_i), \end{align}

(3.17) \begin{align} &\delta {\boldsymbol{u}}_e = \partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e,{\boldsymbol{v}}_e], \end{align}

supplemented with the advection equations

(3.18a) \begin{align} \partial _t n + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}_i) &= 0 , \end{align}

(3.18b) \begin{align} \partial _t s + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}_i) &= 0 , \end{align}

are solutions to the following system of equations:

(3.19a) \begin{align} \partial _t n + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}_i) &= 0 , \end{align}

(3.19b) \begin{align} \partial _t s + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}_i) &= 0 , \end{align}

(3.19c) \begin{align} m_i n ( \partial _t {\boldsymbol{u}}_i + {\boldsymbol{u}}_i \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_i) + \boldsymbol{\nabla }p & - q_i n (\boldsymbol{E} + {\boldsymbol{u}}_i \times {\boldsymbol{B}}) = 0 , \end{align}

(3.19d) \begin{align} q_i n (\boldsymbol{E} + {\boldsymbol{u}}_e \times {\boldsymbol{B}}) &= 0 , \end{align}

(3.19e) \begin{align} 0 = \frac {\boldsymbol{\nabla }\times {\boldsymbol{B}}}{\mu _0} & - q_i n({\boldsymbol{u}}_i - {\boldsymbol{u}}_e) , \end{align}

where $p = p_i$ is the ion pressure (defined as in Theorem 1 ) and $\boldsymbol{E} = -\partial _t {\boldsymbol{A}} - \boldsymbol{\nabla }\phi$ is defined by setting $\phi = {\boldsymbol{A}} \boldsymbol{\cdot }{\boldsymbol{u}}_e$ .

Proof. Equations (3.19a ) and (3.19b ) are again the given advection equations. Considering next the variations in $\delta {\boldsymbol{A}}$ , we write

(3.20) \begin{align} 0 =& \int _0^T \int _\Omega q_i n ({\boldsymbol{u}}_i - {\boldsymbol{u}}_e) \boldsymbol{\cdot }\delta {\boldsymbol{A}} - \frac {1}{\mu _0} \boldsymbol{\nabla }\times {\boldsymbol{A}} \boldsymbol{\cdot }\boldsymbol{\nabla }\times \delta {\boldsymbol{A}}\text{d}\boldsymbol{x}\text{d}t, \end{align}

which directly gives (3.19e ). Since now only $\delta {\boldsymbol{u}}_e$ has variations in ${\boldsymbol{v}}_e$ , the corresponding term reads

(3.21) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta L}{\delta {\boldsymbol{u}}_e} \boldsymbol{\cdot }\delta {\boldsymbol{u}}_e\text{d}\boldsymbol{x}\text{d}t = \int _0^T \int _\Omega \frac {\delta L}{\delta {\boldsymbol{u}}_e} \boldsymbol{\cdot }(\partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e, {\boldsymbol{v}}_e])\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= \int _0^T \int _\Omega - q_i n {\boldsymbol{A}} \boldsymbol{\cdot }(\partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e, {\boldsymbol{v}}_e])\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= \int _0^T \int _\Omega (q_i \partial _t(n {\boldsymbol{A}}) + q_i n ({\boldsymbol{u}}_e \boldsymbol{\cdot }\boldsymbol{\nabla }) {\boldsymbol{A}} + q_i {\boldsymbol{A}} \boldsymbol{\nabla }\boldsymbol{\cdot }(n {\boldsymbol{u}}_e) + q_i n (\boldsymbol{\nabla }{\boldsymbol{u}}_e)^T \boldsymbol{\cdot }{\boldsymbol{A}}) \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x}\text{d}t . \end{align}

From (3.19e ) we next infer that $\boldsymbol{\nabla }\boldsymbol{\cdot }(n {\boldsymbol{u}}_i) = \boldsymbol{\nabla }\boldsymbol{\cdot }(n {\boldsymbol{u}}_e)$ , so that we can use the same identity as in (2.27) and (2.16), and write

(3.22) \begin{align} 0 &= \int _0^T \int _\Omega (q_i n \partial _t {\boldsymbol{A}} + q_i n ({\boldsymbol{u}}_e \boldsymbol{\cdot }\boldsymbol{\nabla }) {\boldsymbol{A}} + q_i n (\boldsymbol{\nabla }{\boldsymbol{u}}_e)^T \boldsymbol{\cdot }{\boldsymbol{A}}) \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= \int _0^T \int _\Omega q_i n (\partial _t {\boldsymbol{A}} + \boldsymbol{\nabla }({\boldsymbol{A}} \boldsymbol{\cdot }{\boldsymbol{u}}_e) - {\boldsymbol{u}}_e \times (\boldsymbol{\nabla }\times {\boldsymbol{A}})) \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= -\int _0^T \int _\Omega q_i n ( \boldsymbol{E} + {\boldsymbol{u}}_e \times {\boldsymbol{B}}) \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x}\text{d}t, \end{align}

which is (3.19d ), by using the definition of $\boldsymbol{E} = -\partial _t {\boldsymbol{A}} - \boldsymbol{\nabla }\phi$ with $\phi = {\boldsymbol{A}} \boldsymbol{\cdot }{\boldsymbol{u}}_e$ and ${\boldsymbol{B}} = \boldsymbol{\nabla }\times {\boldsymbol{A}}$ . Finally by taking variations in ${\boldsymbol{v}}_i$ we recognise exactly the same equation as in (2.23): therefore the same computation gives (3.19c ).

3.4. Canonical Hamiltonian and bracket formulation

Turning to the Hamiltonian point of view, we next derive a Hamiltonian formulation for the above Hall MHD equations. To do so we apply a standard Legendre transform, while the brackets are found by combining standard brackets for systems with advected parameters (Holm et al. Reference Holm, Marsden and Ratiu1998).

Theorem 6. Given the Lagrangian defined by ( 3.15 ), let the associated canonical momenta

(3.23) \begin{align} \boldsymbol{m}_i = \frac {\delta L}{\delta {\boldsymbol{u}}_i}= n (m_i {\boldsymbol{u}}_i + q_i {\boldsymbol{A}}) \quad and \quad \boldsymbol{m}_e = \frac {\delta L }{\delta {\boldsymbol{u}}_e} = -q_i n {\boldsymbol{A}}. \end{align}

Then, a curve $t \to ({\boldsymbol{u}}_i, n, s, {\boldsymbol{A}}, {\boldsymbol{u}}_e)(t)$ is a solution to the variational equations in Theorem 5 iff the corresponding curve $t \to (\boldsymbol{m}_i, \boldsymbol{m}_e, n, s)(t)$ solves the Hamiltonian system

(3.24) \begin{align} \dot {F} = \{F, H\} \end{align}

with Hamiltonian functional $H = \langle \boldsymbol{m}_i, {\boldsymbol{u}}_i \rangle + \langle \boldsymbol{m}_e, {\boldsymbol{u}}_e \rangle - L({\boldsymbol{u}}_i, n, s, {\boldsymbol{u}}_e, {\boldsymbol{A}})$ , namely

(3.25) \begin{equation} H(\boldsymbol{m}_i, \boldsymbol{m}_e, n, s) = \int _\Omega \frac {|\boldsymbol{m}_i+\boldsymbol{m}_e|^2}{2 m_i n} + n e(n, s) + \frac {1}{2 \mu _0 q_i^2}|\boldsymbol{\nabla }\times \frac {\boldsymbol{m}_e}{n}|^2\text{d}\boldsymbol{x}, \end{equation}

and bracket

(3.26a) \begin{equation} \{F,G\} = \{F,G\}_{\boldsymbol{m}_i} + \{F,G\}_{\boldsymbol{m}_e} + \{F,G\}_{n} + \{F,G\}_{s} \end{equation}

defined by

(3.26b) \begin{align} \{F,G\}_{\boldsymbol{m}_i} &= \int _\Omega \boldsymbol{m}_i \boldsymbol{\cdot }\left[\frac {\delta G}{\delta \boldsymbol{m}_i}, \frac {\delta F}{\delta \boldsymbol{m}_i}\right]\mathrm{d}\boldsymbol{x}, \end{align}

(3.26c) \begin{align} \{F,G\}_{\boldsymbol{m}_e} &= \int _\Omega \boldsymbol{m}_e \boldsymbol{\cdot }\left[\frac {\delta G}{\delta \boldsymbol{m}_e}, \frac {\delta F}{\delta \boldsymbol{m}_e}\right]\mathrm{d}\boldsymbol{x}, \end{align}

(3.26d) \begin{align} \{F,G\}_{n} = \int _\Omega \frac {\delta G}{\delta n} \boldsymbol{\nabla } & \boldsymbol{\cdot }\left(n \frac {\delta F}{\delta \boldsymbol{m}_i}\right) - \frac {\delta F}{\delta n} \boldsymbol{\nabla }\boldsymbol{\cdot }\left(n \frac {\delta G}{\delta \boldsymbol{m}_i}\right)\mathrm{d}\boldsymbol{x}, \end{align}

(3.26e) \begin{align} \{F,G\}_{s} = \int _\Omega \frac {\delta G}{\delta s} \boldsymbol{\nabla } & \boldsymbol{\cdot }\left(s \frac {\delta F}{\delta \boldsymbol{m}_i}\right) - \frac {\delta F}{\delta s} \boldsymbol{\nabla }\boldsymbol{\cdot }\left(s \frac {\delta G}{\delta \boldsymbol{m}_i}\right)\mathrm{d}\boldsymbol{x}. \end{align}

Proof. Solutions defined by Theorem 5 satisfy

(3.27) \begin{equation} \int _\Omega \frac {\text{d}}{\,\text{d}t}\frac {\delta L}{\delta {\boldsymbol{u}}_i} \boldsymbol{\cdot }{\boldsymbol{v}}_i\text{d}\boldsymbol{x} = \int _\Omega \frac {\delta L}{\delta {\boldsymbol{u}}_i} \boldsymbol{\cdot }[{\boldsymbol{u}}_i, {\boldsymbol{v}}_i] - \frac {\delta L}{\delta n} \boldsymbol{\nabla }\boldsymbol{\cdot }(n {\boldsymbol{v}}_i) - \frac {\delta L}{\delta s} \boldsymbol{\nabla }\boldsymbol{\cdot }(s {\boldsymbol{v}}_i)\text{d}\boldsymbol{x}. \end{equation}

Now, since $ {\delta L}/{\delta {\boldsymbol{u}}_i} = \boldsymbol{m}_i$ , ${\delta L}/{\delta n} = -{\delta H}/{\delta n}$ , ${\delta L}/{\delta s} = -{\delta H}/{\delta s}$ and ${\boldsymbol{u}}_i = {\delta H}/{\delta \boldsymbol{m}_i}$ , we can rewrite this as

(3.28) \begin{equation} \int _\Omega \dot {\boldsymbol{m}_i} \boldsymbol{\cdot }{\boldsymbol{v}}_i\text{d}\boldsymbol{x}= \int _\Omega \boldsymbol{m}_i \boldsymbol{\cdot }\left[\frac {\delta H}{\delta \boldsymbol{m}_i}, {\boldsymbol{v}}_i\right] + \frac {\delta H}{\delta n} \boldsymbol{\nabla }\boldsymbol{\cdot }(n {\boldsymbol{v}}_i) + \frac {\delta H}{\delta s} \boldsymbol{\nabla }\boldsymbol{\cdot }(s {\boldsymbol{v}}_i)\text{d}\boldsymbol{x}. \end{equation}

We also have that ${\boldsymbol{u}}_e= {\delta H}/{\delta \boldsymbol{m}_e}$ , so that the equation for variations in ${\boldsymbol{v}}_e$ can be rewritten as

(3.29) \begin{equation} \int _\Omega \dot {\boldsymbol{m}_e} \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x} = \int _\Omega \boldsymbol{m}_e \boldsymbol{\cdot }\left[\frac {\delta H}{\delta \boldsymbol{m}_e}, {\boldsymbol{v}}_e\right]\text{d}\boldsymbol{x}. \end{equation}

Let us next consider a functional $F = F(\boldsymbol{m}_i, \boldsymbol{m}_e, n, s)$ : we have

(3.30) \begin{align} \dot {F} &= \int _\Omega \frac {\delta F}{\delta \boldsymbol{m}_i} \dot {\boldsymbol{m}_i} + \frac {\delta F}{\delta \boldsymbol{m}_e} \dot {\boldsymbol{m}_e} + \frac {\delta F}{\delta n} \dot {n} + \frac {\delta F}{\delta s} \dot {s}\text{d}\boldsymbol{x} \nonumber \\ &= \int _\Omega \boldsymbol{m}_i \boldsymbol{\cdot }\left[\frac {\delta H}{\delta \boldsymbol{m}_i}, \frac {\delta F}{\delta \boldsymbol{m}_i}\right] + \frac {\delta H}{\delta n} \boldsymbol{\nabla }\boldsymbol{\cdot }\left(n \frac {\delta F}{\delta \boldsymbol{m}_i}\right) + \frac {\delta H}{\delta s} \boldsymbol{\nabla }\boldsymbol{\cdot }\left(s \frac {\delta F}{\delta \boldsymbol{m}_i}\right)\text{d}\boldsymbol{x} \nonumber \\ &\quad + \int _\Omega \boldsymbol{m}_e \boldsymbol{\cdot }\left[\frac {\delta H}{\delta \boldsymbol{m}_e}, \frac {\delta F}{\delta \boldsymbol{m}_e}\right] - \frac {\delta F}{\delta n} \boldsymbol{\nabla }\boldsymbol{\cdot }\left(n \frac {\delta H}{\delta \boldsymbol{m}_i}\right) - \frac {\delta F}{\delta s} \boldsymbol{\nabla }\boldsymbol{\cdot }\left(s \frac {\delta H}{\delta \boldsymbol{m}_i}\right)\text{d}\boldsymbol{x} \nonumber \\ &= \{F, H\}. \end{align}

The remaining equations are obtained by taking functionals of the form $F(\boldsymbol{m}_i) = \int _\Omega \boldsymbol{m}_i \boldsymbol{\cdot }{\boldsymbol{v}}_i\text{d}\boldsymbol{x}$ , $F(n) = \int _\Omega n v\text{d}\boldsymbol{x}$ and so forth, for given test functions ${\boldsymbol{v}}_i$ and $v$ .

Although the above Hamiltonian formulation is derived from a variational principle, it coincides with the standard ones (D’Avignon et al. Reference D’Avignon, Morrison and Lingam2016; Coquinot & Morrison Reference Coquinot and Morrison2020) when expressed in variables $({\boldsymbol{u}}, n, s, {\boldsymbol{B}})$ .

3.5. Comparison with other Hamiltonian formulations

Several Hamiltonian formulations have been proposed for the Hall MHD equations. The one derived by Holm (Reference Holm1987) involves a bracket with $(\boldsymbol{m}_i, n, s, n_e, {\boldsymbol{A}})$ as dynamical variables.

Yoshida & Hameiri (Reference Yoshida and Hameiri2013) proposed a bracket expressed in $({\boldsymbol{u}}, {\boldsymbol{B}}, \rho )$ variables with

(3.31) \begin{equation} \left \{ \begin{aligned} &{\boldsymbol{u}} = (\boldsymbol{m}_i+\boldsymbol{m}_e)/(m_i n), \\ &{\boldsymbol{B}} = \boldsymbol{\nabla }\times {\boldsymbol{A}} = - \boldsymbol{\nabla }\times \Big (\frac {\boldsymbol{m}_e}{q_i n}\Big ), \\ &\rho = m_i n, \end{aligned} \right . \end{equation}

and Coquinot & Morrison (Reference Coquinot and Morrison2020) derived a Hamiltonian formulation in $(\boldsymbol{m}, {\boldsymbol{B}}, n)$ variables:

(3.32) \begin{equation} \left \{ \begin{aligned} &\boldsymbol{m} = \boldsymbol{m}_i + \boldsymbol{m}_e \\ &{\boldsymbol{B}} = \boldsymbol{\nabla }\times {\boldsymbol{A}} = - \boldsymbol{\nabla }\times \Big (\frac {\boldsymbol{m}_e}{q_i n}\Big ). \end{aligned} \right . \end{equation}

The formulations in Yoshida & Hameiri (Reference Yoshida and Hameiri2013) and Coquinot & Morrison (Reference Coquinot and Morrison2020) have the same Hamiltonian functional, which coincides with ours. Moreover, denoting

(3.33) \begin{align} \tilde {f}(\tilde {y}) := f(y) := F(Y) \qquad \text{ and } \qquad \tilde {g}(\tilde {y}) := g(y) := G(Y) \end{align}

as the functions corresponding to the change of variables

(3.34) \begin{align} Y = (\boldsymbol{m}_i, \boldsymbol{m}_e, n, s) \to y = (\boldsymbol{m}, {\boldsymbol{B}}, n, s) \to \tilde {y} = ({\boldsymbol{u}}, {\boldsymbol{B}}, \rho , s), \end{align}

we compute

(3.35) \begin{equation} \left \{ \begin{aligned} &\frac {\delta F}{\delta \boldsymbol{m}_i} = \frac {\delta f}{\delta \boldsymbol{m}} \\ &\frac {\delta F}{\delta \boldsymbol{m}_e} = \frac {\delta f}{\delta \boldsymbol{m}} - \frac {1}{q_i n} \boldsymbol{\nabla }\times \frac {\delta f}{\delta {\boldsymbol{B}}} \\ &\frac {\delta F}{\delta n} = \frac {\delta f}{\delta n} -\frac {{\boldsymbol{A}} }{n}\boldsymbol{\cdot }\boldsymbol{\nabla }\times \frac {\delta f}{\delta {\boldsymbol{B}}} \\ &\frac {\delta F}{\delta s} = \frac {\delta f}{\delta s} \end{aligned} \right . \qquad \text{ and } \qquad \left \{ \begin{aligned} &\frac {\delta f}{\delta \boldsymbol{m}} = \frac {1}{\rho }\frac {\delta \tilde {f}}{\delta {\boldsymbol{u}}}, \\ &\frac {\delta f}{\delta n} = m_i \left(\frac {\delta \tilde {f}}{\delta \rho } - \frac {{\boldsymbol{u}}}{\rho }\frac {\delta \tilde {f}}{\delta {\boldsymbol{u}}}\right)\!. \end{aligned} \right . \end{equation}

In particular, our bracket is rewritten as

(3.36) \begin{equation} \{F,G\} =\{f,g\}^{\mathrm{CM}}_{\boldsymbol{m}} + \{f,g\}^{\mathrm{CM}}_{\boldsymbol{m},{\boldsymbol{B}}} + \{f,g\}^{\mathrm{CM}}_{{\boldsymbol{B}},\boldsymbol{m}} + \{f,g\}^{\mathrm{CM}}_{{\boldsymbol{B}}} + \{f,g\}^{\mathrm{CM}}_{n} + \{f,g\}^{\mathrm{CM}}_{s} \end{equation}

with

(3.37) \begin{equation} \left \{ \begin{aligned} &{\{f,g\}}^{\mathrm{CM}}_{\boldsymbol{m}} = - \int _\Omega \boldsymbol{m} \boldsymbol{\cdot }\left (\frac {\delta f}{\delta \boldsymbol{m}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta g}{\delta \boldsymbol{m}}\right )\text{d}\boldsymbol{x} + \int _\Omega \boldsymbol{m} \boldsymbol{\cdot }\left (\frac {\delta g}{\delta \boldsymbol{m}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta f}{\delta \boldsymbol{m}}\right )\text{d}\boldsymbol{x} \\ &{\{f,g\}}^{\mathrm{CM}}_{\boldsymbol{m}, {\boldsymbol{B}}} = \int _\Omega \bigg ({\boldsymbol{B}} \times \frac {\delta f}{\delta \boldsymbol{m}}\bigg ) \boldsymbol{\cdot }\bigg (\boldsymbol{\nabla }\times \frac {\delta g}{\delta {\boldsymbol{B}}}\bigg )\text{d}\boldsymbol{x} - \int _\Omega \bigg ({\boldsymbol{B}} \times \frac {\delta g}{\delta \boldsymbol{m}}\bigg ) \boldsymbol{\cdot }\bigg (\boldsymbol{\nabla }\times \frac {\delta f}{\delta {\boldsymbol{B}}}\bigg )\text{d}\boldsymbol{x} \\ &{\{f,g\}}^{\mathrm{CM}}_{{\boldsymbol{B}}} = \int _\Omega \frac {1}{q_i n} {\boldsymbol{B}} \boldsymbol{\cdot }\left (\bigg (\boldsymbol{\nabla }\times \frac {\delta f}{\delta {\boldsymbol{B}}}\bigg ) \times \bigg (\boldsymbol{\nabla }\times \frac {\delta g}{\delta {\boldsymbol{B}}} \bigg )\right )\text{d}\boldsymbol{x} \\ &{\{f,g\}}^{\mathrm{CM}}_{n} = - \int _\Omega n \frac {\delta f}{\delta \boldsymbol{m}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta g}{\delta n}\text{d}\boldsymbol{x} + \int _\Omega n\frac {\delta g}{\delta \boldsymbol{m}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta f}{\delta n}\text{d}\boldsymbol{x} \\ &{\{f,g\}}^{\mathrm{CM}}_{s} = - \int _\Omega s \frac {\delta f}{\delta \boldsymbol{m}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta g}{\delta s}\text{d}\boldsymbol{x} + \int _\Omega s \frac {\delta g}{\delta \boldsymbol{m}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta f}{\delta s}\text{d}\boldsymbol{x} \end{aligned} \right . \end{equation}

and as

(3.38) \begin{equation} \{F, G\} = {\{\tilde {f}, \tilde {g}\}}^{\mathrm{YH}}_{{\boldsymbol{u}}} + \{\tilde {f}, \tilde {g}\}^{\mathrm{YH}}_{{\boldsymbol{u}}, {\boldsymbol{B}}} + \{\tilde {f}, \tilde {g}\}^{\mathrm{YH}}_{{\boldsymbol{B}}} + \{\tilde {f}, \tilde {g}\}^{\mathrm{YH}}_{\rho } + \{\tilde {f}, \tilde {g}\}^{\mathrm{YH}}_{s} \end{equation}

with

(3.39) \begin{align} \left \{ \begin{aligned} \{\tilde {f},\tilde {g}\}^{\mathrm{YH}}_{{\boldsymbol{u}}} &:= \int _\Omega \frac {1}{\rho } \frac {\delta \tilde {g}}{\delta {\boldsymbol{u}}} \times \frac {\delta \tilde {f}}{\delta {\boldsymbol{u}}} \boldsymbol{\cdot }\boldsymbol{\nabla }\times {\boldsymbol{u}}\text{d}\boldsymbol{x} \\ \{\tilde {f},\tilde {g}\}^{\mathrm{YH}}_{{\boldsymbol{u}}, {\boldsymbol{B}}} &:= \{f,g\}^{\mathrm{CM}}_{\boldsymbol{m},{\boldsymbol{B}}} = \int _\Omega \frac {1}{\rho }\bigg ({\boldsymbol{B}} \times \frac {\delta \tilde {f}}{\delta {\boldsymbol{u}}} \bigg ) \boldsymbol{\cdot }\bigg (\boldsymbol{\nabla }\times \frac {\delta \tilde {g}}{\delta {\boldsymbol{B}}} \bigg )\text{d}\boldsymbol{x} \\ & \quad - \int _\Omega \frac {1}{\rho }\bigg ({\boldsymbol{B}} \times \frac {\delta \tilde {g}}{\delta {\boldsymbol{u}}} \bigg ) \boldsymbol{\cdot }\bigg (\boldsymbol{\nabla }\times \frac {\delta \tilde {f}}{\delta {\boldsymbol{B}}} \bigg )\text{d}\boldsymbol{x} \\ \{\tilde {f},\tilde {g}\}^{\mathrm{YH}}_{{\boldsymbol{B}}} &:= \{f,g\}^{\mathrm{CM}}_{{\boldsymbol{B}}} = \int _\Omega \frac {m_i}{q_i \rho } {\boldsymbol{B}} \boldsymbol{\cdot }\bigg [\bigg (\boldsymbol{\nabla }\times \frac {\delta \tilde {f}}{\delta {\boldsymbol{B}}}\bigg ) \times \bigg (\boldsymbol{\nabla }\times \frac {\delta \tilde {g}}{\delta {\boldsymbol{B}}} \bigg )\bigg ]\text{d}\boldsymbol{x} \\ \{\tilde {f},\tilde {g}\}^{\mathrm{YH}}_{\rho } &:= \int _\Omega \frac {\delta \tilde {g}}{\delta {\boldsymbol{u}}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta \tilde {f}}{\delta \rho }\text{d}\boldsymbol{x} - \int _\Omega \frac {\delta \tilde {f}}{\delta {\boldsymbol{u}}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta \tilde {g}}{\delta \rho }\text{d}\boldsymbol{x} \\ \{\tilde {f},\tilde {g}\}^{\mathrm{YH}}_{s} &:= \{f,g\}^{\mathrm{CM}}_{s} = \int _\Omega \frac {s}{\rho } \frac {\delta \tilde {g}}{\delta {\boldsymbol{u}}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta \tilde {f}}{\delta s}\text{d}\boldsymbol{x} - \int _\Omega \frac {s}{\rho } \frac {\delta \tilde {f}}{\delta {\boldsymbol{u}}} \boldsymbol{\cdot }\boldsymbol{\nabla }\frac {\delta \tilde {g}}{\delta s}\text{d}\boldsymbol{x} \, . \end{aligned} \right . \end{align}

By inspection of the brackets appearing on the right-hand sides of (3.36) and (3.38) we obtain the following result.

4.1. Lagrangian variational approach

We now use the formalism of Gay-Balmaz & Yoshimura (Reference Gay-Balmaz and Yoshimura2017a , Reference Gay-Balmaz and Yoshimurab ) to add dissipation to our variational model and recover visco-resistive Hall MHD. Specifically, this amounts to adding viscous and resistive terms in the constrained entropy density variations, which involve, respectively, a stress tensor $\sigma$ of order 2 and a scalar resistivity $\eta \ge 0$ . Here the viscous term is incorporated exactly as in these works, while for the resistive part we add a force corresponding to the interpretation of Drude’s model, which depends on the difference between both velocities. This models a resistive force that will draw the electron flow closer to the ion flow.

Theorem 8. Consider the following Lagrangian:

(4.1a) \begin{equation} l({\boldsymbol{u}}_i, n, s, {\boldsymbol{u}}_e, {\boldsymbol{A}}) = \int _\Omega \frac {m_i n |{\boldsymbol{u}}_i|^2}{2} - n e(n, s) + q_i n ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }{\boldsymbol{A}} - \frac {|\boldsymbol{\nabla }\times {\boldsymbol{A}}|^2}{2 \mu _0}\text{d}\boldsymbol{x} . \end{equation}

Then extremisers of the action $\mathfrak{S} = \int _0^T l\text{d}t$ under constrained variations

(4.2) \begin{align} \delta {\boldsymbol{u}}_i = \partial _t {\boldsymbol{v}}_i + [{\boldsymbol{u}}_i,{\boldsymbol{v}}_i], \qquad \delta n = -{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (n {\boldsymbol{v}}_i),\nonumber \\ \frac {\delta l}{\delta s}(\delta s +{\boldsymbol{\nabla }{\boldsymbol{\cdot }}} (s {\boldsymbol{v}}_i)) = -\sigma : \boldsymbol{\nabla }{\boldsymbol{v}}_i - \eta (q n)^2 ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }({\boldsymbol{v}}_i-{\boldsymbol{v}}_e), \nonumber \\ \delta {\boldsymbol{u}}_e = \partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e,{\boldsymbol{v}}_e], \end{align}

with arbitrary time-dependent vector fields ${\boldsymbol{v}}_i$ and ${\boldsymbol{v}}_e$ vanishing at $t=0$ and $T$ , supplemented with the advection equations

(4.3a) \begin{align} \partial _t n + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}_i) &= 0 , \end{align}

(4.3b) \begin{align} \frac {\delta l}{\delta s}(\partial _t s + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}_i)) &= -\sigma : \boldsymbol{\nabla }{\boldsymbol{u}}_i - \eta (q_in)^2 \lvert {\boldsymbol{u}}_i-{\boldsymbol{u}}_e\rvert ^2, \end{align}

are solution to the following system of equations:

(4.4a) \begin{align} \partial _t n + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(n {\boldsymbol{u}}_i) &= 0 , \end{align}

(4.4b) \begin{align} (\partial _t s + {\boldsymbol{\nabla }{\boldsymbol{\cdot }}}(s {\boldsymbol{u}}_i)) &= \frac {1}{T} (\sigma : \boldsymbol{\nabla }{\boldsymbol{u}}_i + \eta \lvert \boldsymbol{J}\rvert ^2) , \end{align}

(4.4c) \begin{align} m_i n ( \partial _t {\boldsymbol{u}}_i + {\boldsymbol{u}}_i \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_i) + \boldsymbol{\nabla }p & - q_in (\boldsymbol{E} + {\boldsymbol{u}}_i \times {\boldsymbol{B}}) - \boldsymbol{\nabla }\boldsymbol{\cdot }\sigma + \eta q_i n \boldsymbol{J} = 0 , \end{align}

(4.4d) \begin{align} \boldsymbol{E} + {\boldsymbol{u}}_e \times {\boldsymbol{B}} &= \eta \boldsymbol{J} , \end{align}

(4.4e) \begin{align} \frac {\boldsymbol{\nabla }\times {\boldsymbol{B}}}{\mu _0} &= \boldsymbol{J}, \end{align}

where $\boldsymbol{J} := q_i n ({\boldsymbol{u}}_i - {\boldsymbol{u}}_e)$ is the current density and $T = -{\delta l}/{\delta s} = {\delta e}/{\delta s}$ the temperature.

Proof. Equations (4.4a ) and (4.4b ) are the advection equation imposed and (4.4e ) are obtained exactly as in the proof of Theorem 5. Now looking at all the contributions in ${\boldsymbol{v}}_i$ and ${\boldsymbol{v}}_e$ ,

(4.5) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta l}{\delta {\boldsymbol{u}}_i} \delta {\boldsymbol{u}}_i + \frac {\delta l}{\delta n} \delta n + \frac {\delta l}{\delta s} \delta s + \frac {\delta l}{\delta {\boldsymbol{u}}_e} \delta {\boldsymbol{u}}_e \text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega \frac {\delta l}{\delta {\boldsymbol{u}}_i} (\partial _t {\boldsymbol{v}}_i + [{\boldsymbol{u}}_i, {\boldsymbol{v}}_i]) - \frac {\delta l}{\delta n} \boldsymbol{\nabla }\boldsymbol{\cdot }(n {\boldsymbol{v}}_i) - \frac {\delta l}{\delta s} \boldsymbol{\nabla }\boldsymbol{\cdot }(s {\boldsymbol{v}}_i) -\sigma : \boldsymbol{\nabla }{\boldsymbol{v}}_i \nonumber \\ & \quad - \eta (q_i n)^2 ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }{\boldsymbol{v}}_i + \eta (q_i n)^2 ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }{\boldsymbol{v}}_e + \frac {\delta l}{\delta {\boldsymbol{u}}_e} (\partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e, {\boldsymbol{v}}_e])\text{d}\boldsymbol{x}\text{d}t, \end{align}

considering variations only in ${\boldsymbol{v}}_i$ and using the same computation as before (as we only added one term):

(4.6) \begin{align} 0 &= \int _0^T \int _\Omega \frac {\delta l}{\delta {\boldsymbol{u}}_i} (\partial _t {\boldsymbol{v}}_i + [{\boldsymbol{u}}_i, {\boldsymbol{v}}_i]) - \frac {\delta l}{\delta n} \boldsymbol{\nabla }\boldsymbol{\cdot }(n {\boldsymbol{v}}_i) - \frac {\delta l}{\delta s} \boldsymbol{\nabla }\boldsymbol{\cdot }(s {\boldsymbol{v}}_i) -\sigma : \boldsymbol{\nabla }{\boldsymbol{v}}_i \nonumber \\ & \quad - \eta (q_i n)^2 ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }{\boldsymbol{v}}_i\text{d}\boldsymbol{x}\text{d}t \nonumber \\ &= \int _0^T \int _\Omega -\big (m_i n( \partial _t {\boldsymbol{u}}_i + {\boldsymbol{u}}_i \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_i) + \boldsymbol{\nabla }p - q_i n (\boldsymbol{E} + {\boldsymbol{u}}_i \times {\boldsymbol{B}})\big ) \boldsymbol{\cdot }{\boldsymbol{v}}_i + \boldsymbol{\nabla }\boldsymbol{\cdot }\sigma \boldsymbol{\cdot }{\boldsymbol{v}}_i \nonumber \\ & \quad - \eta (q_i n)^2 ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }{\boldsymbol{v}}_i\text{d}\boldsymbol{x}\text{d}t . \end{align}

This being true for all ${\boldsymbol{v}}_i$ , we find (4.4c ). Considering next variations in ${\boldsymbol{v}}_e$ and computing as in the proof of Theorem 5, we write

(4.7) \begin{align} 0 &= \int _0^T \int _\Omega \eta q_i n \boldsymbol{J} \boldsymbol{\cdot }{\boldsymbol{v}}_e + \frac {\delta l}{\delta {\boldsymbol{u}}_e} (\partial _t {\boldsymbol{v}}_e + [{\boldsymbol{u}}_e, {\boldsymbol{v}}_e])\text{d}\boldsymbol{x}\text{d}t\nonumber \\ &= \int _0^T \int _\Omega q_i n \big (\eta \boldsymbol{J} - (\boldsymbol{E} + {\boldsymbol{u}}_e \times {\boldsymbol{B}})\big ) \boldsymbol{\cdot }{\boldsymbol{v}}_e\text{d}\boldsymbol{x}\text{d}t, \end{align}

which gives (4.4d ).

To recover the visco-resistive Hall MHD equations from (4.4) we next proceed as follows. Summing (4.4c ) and (4.4d ), and using the definition of the current ( $\boldsymbol{J} = q_i n ({\boldsymbol{u}}_i - {\boldsymbol{u}}_e)$ ), we find

(4.8) \begin{equation} m_i n ( \partial _t {\boldsymbol{u}}_i + {\boldsymbol{u}}_i \boldsymbol{\cdot }\boldsymbol{\nabla }{\boldsymbol{u}}_i) + \boldsymbol{\nabla }p - \boldsymbol{J}\times {\boldsymbol{B}} - \boldsymbol{\nabla }\boldsymbol{\cdot }\sigma = 0 . \end{equation}

Ohm’s law is obtained by combining (4.4d ) with the definition of the current:

(4.9) \begin{equation} \boldsymbol{E} + {\boldsymbol{u}}_i \times {\boldsymbol{B}} = \frac {1}{q_i n} \boldsymbol{J} \times {\boldsymbol{B}} + \eta \boldsymbol{J} , \end{equation}

and from the definition of the fields we recover Faraday’s equation:

(4.10) \begin{equation} \partial _t {\boldsymbol{B}} = - \boldsymbol{\nabla }\times \boldsymbol{E} = -\boldsymbol{\nabla }\times \bigg (\bigg (\frac {1}{q_i n} \boldsymbol{J} - {\boldsymbol{u}}_i\bigg ) \times {\boldsymbol{B}} + \eta \boldsymbol{J} \bigg ). \end{equation}

4.2. Metriplectic bracket, Hamiltonian and entropy for visco-resistive Hall MHD

We finally turn to the metriplectic reformulation of the previous equations. The conservative bracket and Hamiltonian will remain the same as in the non-dissipative case of the previous section, while we will now consider an entropy (being simply the total entropy of the system) and a dissipative bracket, that will enable us to take into account the two dissipative mechanisms we incorporated in our equations. We use the formalism introduced in Morrison & Updike (Reference Morrison and Updike2024) where the dissipation is included using a 4-bracket that has the same symmetries as a curvature tensor. This bracket takes as argument twice the Hamiltonian on the second and fourth slot, and an entropy to be dissipated on the third position. The first slot is as in the standard Poisson bracket/Hamiltonian formulation occupied by the function whose dynamic is described. We make the assumption that the viscous tensor has the form

(4.11) \begin{equation} \sigma = \Lambda \boldsymbol{\nabla }{\boldsymbol{u}}_i , \end{equation}

where $\Lambda$ is a 4-tensor that has the symmetry $\Lambda _{abcd} = \Lambda _{cdab}$ and is non-negative in the sense that $v : \Lambda v = \sum _{abcd} v_{ab} \Lambda _{abcd} v_{cd} \geq 0$ for any 2-tensor $v$ . Here we write $(\Lambda v)_{ab} = \sum _{cd} \Lambda _{abcd} v_{cd}$ , which corresponds to seeing 4-tensors as matrices over the vector space of 2-tensors. Our assumption is made to ensure that the viscous forces are really dissipative, as it implies that the term $\sigma : \boldsymbol{\nabla }{\boldsymbol{u}}_i$ is non-negative, hence leading to entropy creation in (4.4b ).

Theorem 9. Let $H = H(\boldsymbol{m}_i, \boldsymbol{m}_e, n, s)$ and $\{\boldsymbol{\cdot }, \boldsymbol{\cdot }\}$ be the Hamiltonian and bracket from Theorem 6 . A curve $t \to ({\boldsymbol{u}}, n, s, {\boldsymbol{A}})(t)$ is a solution to the equations of Theorem 5 iff the corresponding curve $t \to (\boldsymbol{m}_i, \boldsymbol{m}_e, n, s)(t)$ solves the metriplectic system

(4.12) \begin{equation} \dot {F} = \{F, H\} + (F,H;S,H) \end{equation}

with the total entropy functional

(4.13) \begin{equation} S= \int _\Omega s\text{d}\boldsymbol{x}, \end{equation}

and a dissipative bracket defined by

(4.14a) \begin{equation} (F,M;G,N)=(F,M;G,N)_{visc}+(F,M;G,N)_{res} , \end{equation}

with

(4.14b) \begin{align} &\!(F,M;G,N)_{visc} = \int _{\Omega }\frac {1}{T} \left (\!\bigg ( \frac {\delta F}{\delta s} \boldsymbol{\nabla }\frac {\delta M}{\delta \boldsymbol{m}_i} - \frac {\delta M}{\delta s} \boldsymbol{\nabla }\frac {\delta F}{\delta \boldsymbol{m}_i}\! \bigg ) : \Lambda \bigg ( \frac {\delta G}{\delta s} \boldsymbol{\nabla }\frac {\delta N}{\delta \boldsymbol{m}_i} - \frac {\delta N}{\delta s} \boldsymbol{\nabla }\frac {\delta G}{\delta \boldsymbol{m}_i}\! \bigg )\text{d}\boldsymbol{x}\! \right )\nonumber\\& \end{align}

and

(4.14c) \begin{equation} \begin{aligned} (F,M;G,N)_{res} &= \int _{\Omega }\frac {\eta (q_i n)^2}{T} \left ( \bigg ( \frac {\delta F}{\delta s} \bigg (\frac {\delta M}{\delta \boldsymbol{m}_i}-\frac {\delta M}{\delta \boldsymbol{m}_e} \bigg ) - \frac {\delta M}{\delta s} \bigg ( \frac {\delta F}{\delta \boldsymbol{m}_i}-\frac {\delta F}{\delta \boldsymbol{m}_e} \bigg ) \bigg ) \right . \\ &\left . \quad \boldsymbol{\cdot }\bigg ( \frac {\delta G}{\delta s} \bigg ( \frac {\delta N}{\delta \boldsymbol{m}_i}-\frac {\delta N}{\delta \boldsymbol{m}_e} \bigg ) - \frac {\delta N}{\delta s} \bigg ( \frac {\delta G}{\delta \boldsymbol{m}_i}-\frac {\delta G}{\delta \boldsymbol{m}_e} \bigg ) \bigg ) \right )\text{d}\boldsymbol{x}, \end{aligned} \end{equation}

where the temperature is defined as $ T = \delta H/\delta s.$

Remark 6. Using ${\delta H}/{\delta s} = T$ , ${\delta H}/{\delta \boldsymbol{m}_i} = {\boldsymbol{u}}_i$ and ${\delta H}/{\delta \boldsymbol{m}_e} = {\boldsymbol{u}}_e$ , we find that the symmetric 2-bracket (Morrison & Updike Reference Morrison and Updike2024) corresponding to ( 4.14 ) reads

(4.15a) \begin{equation} (F,G)_H = (F,G)_{H, visc} + (F,G)_{H, res} \end{equation}

with

(4.15b) \begin{equation} \begin{aligned} (F,G)_{H, visc} & = (F,H;G,H)_{visc} \\ &= \int _\varOmega \frac {1}{T} \left ( \Bigg ( \frac {\delta F}{\delta s} \boldsymbol{\nabla }{\boldsymbol{u}}_i - T \boldsymbol{\nabla }\frac {\delta F}{\delta \boldsymbol{m}_i} \Bigg ) : \varLambda \Bigg ( \frac {\delta G}{\delta s} \boldsymbol{\nabla }{\boldsymbol{u}}_i - T \boldsymbol{\nabla }\frac {\delta G}{\delta \boldsymbol{m}_i} \Bigg ) \right )\text{d}\boldsymbol{x} \end{aligned} \end{equation}

and

(4.15c) \begin{equation} \begin{aligned} (F,G)_{H, res} &= (F,H;G,H)_{res} = \int _{\varOmega }\frac {\eta (q_i n)^2}{T} \left ( \Bigg ( \frac {\delta F}{\delta s} ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) - T \Bigg ( \frac {\delta F}{\delta \boldsymbol{m}_i}-\frac {\delta F}{\delta \boldsymbol{m}_e} \Bigg ) \Bigg ) \right . \\ &\left . \quad \boldsymbol{\cdot }\Bigg ( \frac {\delta G}{\delta s} ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) - T \Bigg ( \frac {\delta G}{\delta \boldsymbol{m}_i}-\frac {\delta G}{\delta \boldsymbol{m}_e} \Bigg ) \Bigg ) \right )\text{d}\boldsymbol{x} . \end{aligned} \end{equation}

We may now prove the theorem.

Proof. Using $ {\delta S}/{\delta \boldsymbol{m}_i} = 0$ and ${\delta S}/{\delta s}=1$ , we derive from (4.15b ) and (4.15c ) that

(4.16) \begin{align} \begin{aligned} (F,H;S,H)_{visc} &= \int _\Omega \frac {1}{T} \left ( \bigg ( \frac {\delta F}{\delta s} \boldsymbol{\nabla }{\boldsymbol{u}}_i - T \boldsymbol{\nabla }\frac {\delta F}{\delta \boldsymbol{m}_i} \bigg ) : \Lambda \boldsymbol{\nabla }{\boldsymbol{u}}_i \right)\text{d}\boldsymbol{x} \\ &= \int _{\Omega } \frac {1}{T} \boldsymbol{\nabla }{\boldsymbol{u}}_i : \sigma \frac {\delta F}{\delta s} + \boldsymbol{\nabla }\boldsymbol{\cdot }\sigma \boldsymbol{\cdot }\frac {\delta F}{\delta \boldsymbol{m}_i}\text{d}\boldsymbol{x}, \end{aligned} \end{align}

where we recall that $\sigma$ is given by (4.11), and

(4.17) \begin{align} (F,H;S,H)_{res} = \int _\Omega \eta (q_i n)^2 \left (\frac {\lvert {\boldsymbol{u}}_i-{\boldsymbol{u}}_e\rvert ^2}{T} \frac {\delta F}{\delta s} - ({\boldsymbol{u}}_i-{\boldsymbol{u}}_e) \boldsymbol{\cdot }\bigg (\frac {\delta F}{\delta \boldsymbol{m}_i}-\frac {\delta F}{\delta \boldsymbol{m}_e}\bigg ). \right )\text{d}\boldsymbol{x}. \end{align}

Doing the same computations as in the proof of Theorem 6 and using the definitions, one then finds

(4.18) \begin{equation} \dot {F} = \int _\Omega \frac {\delta F}{\delta \boldsymbol{m}_i} \dot {\boldsymbol{m}_i} + \frac {\delta F}{\delta \boldsymbol{m}_e} \dot {\boldsymbol{m}_e} + \frac {\delta F}{\delta n} \dot {n} + \frac {\delta F}{\delta s} \dot{s}\text{d}\boldsymbol{x} = \{F,H\} + (F,H;S,H). \end{equation}

Substituting $S$ in the equation above we also recover the total entropy variation (that could already be obtained from (4.4b )):

(4.19) \begin{equation} \frac {{\textrm{d}}S}{\,\text{d}t}= \int _\Omega \frac {\boldsymbol{\nabla }{\boldsymbol{u}}_i : \Lambda \boldsymbol{\nabla }{\boldsymbol{u}}_i}{T} + \frac {\eta (q_i n)^2 \lvert {\boldsymbol{u}}_i-{\boldsymbol{u}}_e\rvert ^2}{T}\text{d}\boldsymbol{x}, \end{equation}

which is positive due to our assumption on $\Lambda$ and the fact that $\eta \ge 0$ .

4.3. Relation with the metric bracket of Coquinot and Morrison

We reformulate our 2-bracket (4.15) in $y = (\boldsymbol{m}, {\boldsymbol{B}}, n, s)$ variables. Writing again $f(y) = F(Y)$ , we recall that (3.35) gives $ {\delta F}/{\delta \boldsymbol{m}_i} = {\delta f}/{\delta \boldsymbol{m}}$ , $ {\delta F}/{\delta s} = {\delta f}/{\delta s}$ and ${\delta F}/{\delta \boldsymbol{m}_i}-{\delta F}/{\delta \boldsymbol{m}_e} = ( {1}/{q_i n}) \boldsymbol{\nabla }\times {\delta f}/{\delta {\boldsymbol{B}}}$ . Together with ${\boldsymbol{u}}_i - {\boldsymbol{u}}_e = ({1}/{q_i n}) \boldsymbol{J}$ , this yields

(4.20a) \begin{equation} (F,G)_H = (f,g)^{CM} = (f,g)^{CM}_{visc} + (f,g)^{CM}_{res} \end{equation}

with

(4.20b) \begin{equation} (f,g)^{CM}_{visc} = \int _\Omega T \bigg ( \frac {1}{T} \frac {\delta f}{\delta s} \boldsymbol{\nabla }{\boldsymbol{u}}_i - \boldsymbol{\nabla }\frac {\delta f}{\delta \boldsymbol{m}} \bigg ) : \Lambda \bigg ( \frac {1}{T} \frac {\delta g}{\delta s} \boldsymbol{\nabla }{\boldsymbol{u}}_i - \boldsymbol{\nabla }\frac {\delta g}{\delta \boldsymbol{m}} \bigg )\text{d}\boldsymbol{x} \end{equation}

and

(4.20c) \begin{equation} \begin{aligned} (f,g)^{CM}_{res} \int _{\Omega } T & \bigg ( \frac {1}{T}\frac {\delta f}{\delta s} \boldsymbol{J} - \boldsymbol{\nabla }\times \frac {\delta f}{\delta {\boldsymbol{B}}} \bigg ) \boldsymbol{\cdot }\eta \bigg ( \frac {1}{T} \frac {\delta g}{\delta s} \boldsymbol{J} - \boldsymbol{\nabla }\times \frac {\delta g}{\delta {\boldsymbol{B}}} \bigg )\text{d}\boldsymbol{x}, \end{aligned} \end{equation}

which corresponds to the metric bracket (4.23) of Coquinot & Morrison (Reference Coquinot and Morrison2020), with

(4.21) \begin{align} \left \{ \begin{aligned} &\Lambda _{vv} = \Lambda \\ &\Lambda _{vj} = \Lambda _{jv} = \Lambda _{jj} = 0 \\ \end{aligned} \right . \qquad \text{ and } \qquad \left \{ \begin{aligned} &\eta _{jj} = \eta \\ &\eta _{TT} = \eta _{Tj} = \eta _{jT} = 0 . \end{aligned} \right . \end{align}